When you start describing the motion of the
rotor blade in forward flight, you can refer
to several planes, in the sense reference
frame. See, this is the rotor shaft axis and
this is actually swash plate and there is
a tip path plane. And this is perpendicular
to the flight path. We can have a flight path
and then whatever is the oncoming velocity
you can define a coordinate system with respective
to that.
Now, you can have several coordinate systems.
But for ease of formulation, you have to decide
what you are choosing. And if you decide to
choose a particular coordinate system, then
you have to stick to that throughout the development
of forward flight equations. Because the blade
is flapping lead-lag torsion about the hub
and the blade also is given a pitch angle
which is varying as it goes around the azimuth.
And more complicated situation, which the
fuselage itself is moving up and down, three
translation, three rotation. So, this is the
actual full dynamics. Now, because the roll
or pitch and yaw motion plus up and down,
this can be any large angle also. So, you
have to describe the motion of the blade,
get the loads on that, transfer to the hub
and then how the fuselage will move. Unless,
you are very systematic in the development,
it becomes quite complicated.
So, what normally done. Normally in the sense,
it depends on the person who uses that. But
it is preferable to choose the hub plane.
That means, hub plane is, I am choosing the
shaft axis. So, the shaft is there, hub is
there. So, I decide that my coordinate system
is fixed. And with respect to that coordinate
system only I will start describing all the
motions. So, coordinate system, we will say,
if this is the hub, this is my z axis and
I may have my x and my y.
And the blade can be inclined, flapped, it
can do anything it wants. But my motion is
describe in this plane. This is my reference
plane, which is same as what I do in the fuselage,
because this is attached to the fuselage,
so once you decide this frame, you can say
this is parallel to the c g frame also if
you want, but it can be little different also.
But when you are calculating the aerodynamic
load, what we have learnt in aerodynamics
is, you have an aerofoil, there is a oncoming
flow and there is a angle of attack may be,
and you get the lift and drag. But in the
present case, you will have of course, you
have a normal flow because this is not the
only flow, you are going to have a flow like
this, which is due to the inflow plus the
motion of the blade, because the blade is
flapping up and down. That mean there is a
relative velocity due to flap motion. And
then this is, all these are in the plane of
the aerofoil please understand. That is, if
you cut a section of the blade, that section
you are describing all these aerodynamic loads.
But my blade, when it rotates, so I will write
some. This is the hub, this is my capital
X and here z is coming, that is the hub plane
which I have drawn.
But if I look at, view it from the top my
blade may be going like this. I am drawing
only the projection of the blade which is
in this. But, actually I will call this as
may be x 1, y 1. x 1, y 1, z 1 is actually
a rotating coordinate system. It rotates with
the blade. But the blade can flap. Now, if
I look at from y 1, assuming I am viewing
from y 1, I will have, this is my x 1, this
is my z 1 and my blade can be like this. This
is my flap beta. So, I call this as x 2, z
2 and y is into the. Is it clear?
So, you see I am defining several coordinate
systems. Because the idea is, if you are very
clear about the, this is the most I would
call it the simplest problem, hover is the
simplest of all the cases. And when you go
to the forward flight, taking only flap motion
is the simplest. But, if you want to really
understand the full, if you understand this
then you will learn more and more as you go
along, if you are interested in this field,
otherwise it stops right here. Because if
you have all the other motions, here we are
not, in this what we developed now, we assume
the blade is only flapping. That means it
is coming out of the plane of hub. I define
that flap as the angle beta, x 2, y 2. Now,
you see which is my, if I look at this is
my aerofoil, cross section of the rotor blade,
my x 2 is coming out here. Because that is
the cross section of the blade. And I know
aerodynamic theory is for the section. So,
I will simply apply. When I want to get the
aerodynamic loads, I use this with what is
the angle I am having with respect to, this
is the u tangential, this is u perpendicular.
And then of course, I get the induced angle
of attack. So, I will subtract. Finally, I
will get a lift and a drag in this frame in
the z 2 x 2 frame, the lift.
And then I again transfer back to, if I want
to get the hub loads, I have to transfer back
the aerodynamic load which is getting in this
axis system, I should transfer back to this
system. So, we have a transformation of coordinates
which we have to apply. That is systematic,
these are all orthogonal system. So, the transformation
you just use. So, you define motion in one
coordinate system and then you describe a
position vector, velocity, then you get the
aerodynamic load and again you can transfer
back. So, if you are very systematic, then
you will find everything follows a very clear
cut procedure.
But most of the books, as a matter no books,
they do not give any of these things. They
will finally write this is the velocity component.
But in writing that component they make lot
of assumptions. So, what we will do is, we
will use this. I will go through math part
here up to some point, describe how you really
get the lift, drag, moment everything and
how they are converted and then what approximation
we make. And at the end of it you will get
loads here, like your thrust. And you will
have a longitudinal force, you will have a
lateral force, you will have moment also,
yaw moment, pitch moment, role moment everything
will come in forward flight. And that is where
the key development in forward flight is,
first part. And this is like assuming my helicopter
is in a steady level flight. Steady, that
means it is flying with the constant speed.
Now, before you go to describe a little bit,
I wanted to show you a just a brief very brief
introduction to modeling the blade.
Because if you remember right at the beginning
of the course we mentioned the blade is attached
to the hub and then a if it is articulated,
you have a flap hinge and you have a lead-lag
hinge and you have a pitch bearing. And in
the case of a hingeless, you do not have any
hinge, but you have a pitch bearing, because
you need to change the pitch angle of the
blade cyclically, as it goes around.
Now, this is a very interesting thing. In
this diagram, this is a rotor shaft, blade
is attached to the hub. It is very schematic,
I am just showing schematic. I put the flap
hinge near the root, after that lead-lag hinge,
after that the feathering. Feathering is the
pitch bearing, Please understand. They are
kept in a different particular sequence moving
from the root to the blade. And feathering
axis is the axis about which you change the
angle of attack of the blade, basically the
pitch angle of the blade, by moving this control.
Because if you move it up and down, you change
the pitch angle.
This is a symmetric. And now is this really
important that they have to be in this sequence?
No. It need not be because if you take hinge
less blade, there will no hinge, hinge is
somewhere in the design of the blade. When
you make the blade you make it such that this
portion acts like a hinge. But you will change
the pitch angle here by putting some bracket
and taking it you rotate it here. That means,
your pitch bearing is inboard of flap and
lag hinges. In this case, it is inboard of
flap and lag hinges. Please understand, inboard,
outboard. What is the big deal you can ask.
If it is an outboard, suppose flap hinge is
this, it is flapping like this. If my pitch
is outboard, I will change the pitch angle
to any value, but the flap hinge will not
rotate. It will be still in the same. So,
the blade will flap like this. On the other
hand, if my pitch is here, flap is outboard.
If I rotate, the flap will be going, then
the blade will flap like this. You follow?
Now, these are very critical. So, you will
learn about these things slowly later. If
you have outboard, if you have inboard, what
happens? Because that introduces something
called a structural coupling. Because flap
is what? Flap is something out of plane motion
of the blade. This is the hub plane. But the
blade will instead of going like this, it
will go like this. That means, you have both
up motion, lag motion.
So, you have essentially coupled flap lag
motion. You have a coupling. But is beneficial
or is it determental? Unfortunately, it is
beneficial in some case, it is detrimental
in some other case. So, these studies has
been performed earlier. And then that is why
the choice of the rotor for a particular helicopter,
the industry, they go by their experience
and what they have learnt in designing one
rotor system, they will stick to the same
thing. Only thing is when they went from articulated,
because articulated rotor was there in all
most of the articulated rotors. And I do not
know you may correct the pitch barring is
outboard. In,is it outboard or inboard? Outboard.
Inboard? Outboard. Whereas in the, it goes
inboard. Inboard it introduces coupling. So,
you change it from articulated to a hinge
blade. It is not that I take it and I put
it, then let me try what happens. Because
if you try what happens, the blade may break
because of the aero-elastic coupling. All
are very small effects only, please understand.
That is why I am just giving you a glimpse
of the physics involved in this. But later
we will time permits, we will introduce a
little bit.
But in the beginning, when we are starting,
what you do for this? And then the question
is, where is the hinge? Is the hinge right
here or here? Here it is moved from the hub.
So, as a first approximation, please understand,
now this is where I say first approximation
I do not bother about any of these things.
I will simply say all of them are coincident
and sitting there right at the center, at
the center of the hub. So, I consider only
flapping motion, somehow I can change the
pitch angle, that is all. I do not consider
my flap axis is tilting, there is no lead
lag motion, these are assumptions I am making
right at the, that is why I said reality when
you see, when you try to model when you see
some of the literature or books or anything,
there is a tremendous gap between what is
their what is modern, because you make assumptions,
lots of assumptions. So, the first assumption
which we make in modeling the rotor blade
is it is centrally hinged and I am changing
the pitch angle. This is the flap, centrally
hinged it is. Later we will study blade modeling,
after we finish the forward flight which will
take a few lectures, then we will go to the
blade modeling. So, right now for our development
we make this approximation that. And here
you see I make coincident hinges, flap and
lag same. Actually there is a interesting
thing, they made a blade because most of the
analysis we do coincident that we do not distinguish
between this end and that end, lead lag flap
hinge they are all the same place. But technically
they can be at different place. So, what was
done is, they made a blade, designed a hub
such that the flap and lag hinge exactly at
the same location and then tested it. And
then they introduced different things to see
the effect of all this on the blade dynamics.
Now, having known made the assumption, we
now look at it. This is the feathering. Please
understand. What is the axis? Again I go here.
When the blade flaps, this axis going up that
means, I change my pitch angle along this
axis. I do not change the pitch angle about
this axis. Because if I change the pitch angle
here, what will happen? The blade will go
in a conical path. Even though I may rotate
here, the blade will do like this. And these
are all are very small things, but they have
tremendous aero-elastic effects. You have
to model them properly. So, the rotation about
what axis you are giving. We make very simple
assumption now, we rotate the pitch axis only
along the blade axis, not about some other
axis and this is what I put phi. And phi is
of course, elastic torsion plus in this case
we take it there is no twist of the blade
happening by elastic twist.
Normally those things happen that is why I
put flap, lead lag elastic torsion. All of
them are happening on the blade. But I throw
everything. I say only flap is happening,
there is no phi, but theta is the pilot input.
And that is there he gives. And the blade
experiences that variation as it goes round
and round. So, these are the basic things
which you have to know. And now I will go
to one more new graph, I have written here
the same diagram, little bit more. So, you
see this is my hub plane x y z and the rotation
vector is along the shaft that is, the hub
plane. And the blade centrally hinged and
it is flapping beta and the pitch angle is
theta. And pitch angle is given by this expression,
you know that this is the pilot input, collective
two cyclic 1 c 1 s.
Now, my hub system is what I have, what are
the forces which I have to calculate. Later
I will derive then you will know. This is
again my hub access x y z, even though I put
a circle the blade is not in this. It is a
hub plane. I have a h force which is along
the x axis. Please note the convention is
helicopter is flying like this, x is behind
towards the tail in the hub. And y and this
the z. And I defined my mu v cosign alpha
over omega r. You know that velocity, because
earlier we said only some rotor disk, now
I am defining mu in the hub plane. Alpha is
the tilt of the helicopter. Because helicopter
normally when it flies forward, I have exaggerated.
Leave that part. This is my, I have exaggerated.
This angle that is, the latitudinal axis to
the earth, you can take it or the oncoming
flow. You define the oncoming flow as alpha.
You are now, your hub plane is what? This
is your hub plane, this is your v because
helicopter is flying forward, what do you
do? In the hub plane, you will have v cosine
alpha and normal to the hub plane you will
have a v sign alpha. And of course, you will
have an induced velocity also. Now, you know
induced velocity is defined normal to the
hub plane in the formulation. I am not really
looking at tip path plane nothing, please
understand. Then you can define the tip path
plane with respect to the hub plane. It is
very easy because tip path plane whatever
is the 1 c beta 1 c, that is the tip path
plane.
So, you see v cos alpha, v sin alpha and mu.
These are my definitions of advanced ratio
and inflow, total inflow you may call it.
Because I divided by omega r tips phi, but
only thing is I do not know what is alpha.
When I am hovering you may say alpha is 0.
Because I am having straight weight is below,
but please understand suppose weight is not
right on shaft, weight can be anywhere here
there. So, that depends on even in hover you
can tilt it. But anyway mu is 0, therefore
does not matter in hover case. So, in forward
flight you need to know alpha that means,
what is the pitch attitude of the helicopter
which you will not know beforehand up.
So, in all over formulation what we are going
to do next. We will simply say, hey I know
mu I know lambda. But please understand this
lambda is v sin alpha plus mu. Now, mu is
the induced flow. It can be any model. You
can assume uniform flow, you can assume grease
model, I can consume square model, you can
have various inflow models. Because my mu
is changing, it can change with the location.
But for simplicity of this I am assuming it
is constant. This is constant uniform which
I will get it from basically the momentum
theory in forward flight which I mention in
the last class or before. Now, you know how
do you get that inflow. Assuming that you
know this, because I will tell you just for,
see the inflow expression was this, either
this or this. Because you multiply by omega
r you get the mu, but this requires again
alpha. So, you see to get the inflow you need
to know the alpha. Only when you define the
alpha you know mu.
So, it is like one is related to other. So,
there will be an iteration in forward flight
if you want to solve the flight condition,
that is the basically the trim or equilibrium.
To do that you need to start with only symbols
and using that we will derive all the expressions
for this. We will get the expressions for
the six t, h, y and then moments torque, roll
moment, pitch moment. All of the them we will
obtain in forward flight and this is where
the lot of complexities there.
Now, of course, this I mention to you last
time. How will I describe my flap motion?
I am assuming my flap is periodic that means,
because it is going round and round my velocity
is varying. So, I make the assumption first
that this is going to be my flap motion, but
I knock out the higher harmonic terms, I neglect
them. And I try to give physical meaning to
these three which are nothing, but this is
the conning because collective all the blades
go up simultaneously, that is we call it conning.
Then if they tilt longitudinally, that you
call it 1 c and this is lateral tilt, that
is 1 s. Of course, the other things all are
rest of the perturbations you may call it
in the tip path plane.
Now, you see I have essentially kind of define
what type of motion, what my axis system is
and what my mu and lambda. Using this, how
will get my aerodynamic load first? But please
understand I am assuming mu, lambda, beta.
You may ask how will I get beta naught 1 c
1 s, how will I get it? Because that nobody
going to supply it to you. I am only saying
it is periodic and you can write it any periodic
signal, you can write it as a Fourier series,
this is essentially a Fourier series. And
your pitch input is given here. So, this is
the pitch input or you can call it pilot input
which goes to the main rotor. We are talking
about only main rotor please understand now.
And then these are the velocity components
and this is the blade motion, flap motion.
That is all. Using this how do we get the
aerodynamic load at any cross section. Is
it clear?
But this is most primitive I would call it.
Because in the helicopter this is simplest
one you can have. Because this part is, I
will do it on the board for clarity. Now,
the symbol now the symbol I use, T for thrust,
T is the thrust, H is the longitudinal force
and Y is the side force, side force to the
helicopter. This is the symbol please understand
I am using the same symbol we all know, because
this is the standard symbol which people use.
T H for longitudinal force, Y for side force,
all are uppercase. And then you can have moments.
Please understand you can have a moment m
x, m y and m z. m z is basically your torque.
And if you know the rotor rpm torque, you
can get the power. That is all.
Now, you have these defined. We have to go
and get expressions for thrust H y, m x, m
y torque, these six quantities. That means,
I know what are the loads that act on the
hub. Knowing this, but still in assuming this,
there are unknowns that is why I am again
emphasizing the unknown is alpha, I do not
know. And here I do not know beta naught ,beta
1 c, beta 1 s I am neglecting all the higher
harmonics again, please understand I am neglecting
all of them. Because if I do not neglect it
is your expressions everything will become
massive, very big ones. So, we make assumptions.
Now, let us start with the simple formulation,
the transformation. So, you understood this.
We will come back to this figure later also.
Now, how do we get the, because we will write
the transformation matrix. The transformation
matrix because e… So, this is hub 
fixed non-rotating. This is hub fixed rotating
because this is the transformation. Cosine
psi, this is x minus sine psi. This is the
orthogonal transformation. All of them orthogonal.
If you want from this to this, you simply
transpose of that. Because this is an orthogonal,
so inverse is transpose. This is a simple
transformation. Then you have to have another
transformation between this system and this
system, because this is the blade fixed coordinate
system rotating. So, I will write, now here
y 2… e x 1 e y 1 e z 1. This is blade fixed
rotating. And I am assuming my blade is rigid.
It is the rigid blade, it can only do like
this. It is not a elastic blade. So, please
understand. It is a rigid blade assumption
only flap motion. Now, you have this transformation.
You need to know for getting the aerodynamic
loads on the cross section of the blade, you
have to know the velocity components. The
velocity you have from two sources. One is
due to the motion of the blade, another one
due to the motion on the vehicle.
So, first you say motion of the blade that
means, the velocity at any particular cross
section. Because we say it is a rigid blade,
cross section is fixed, so you draw a line.
That line is you say your pitch axis line
or you may call it elastic axis line. Because
that is the pitch axis about which you are
actually changing your blade pitch angle.
I want velocities obtained at that point.
Because you may ask, if this is where, velocity
at which point I will get it, you may ask.
So, this is the axis of pitch and in the design
of the rotor blades, this is a usually for
subsonic 25 percent chord is the aerodynamic
center. Then you have learnt about shear center
in structures and you have learnt mass center
for every section. In the design of rotor
blades, what you try to do is you try to bring
every center here, 25 percent. Every center,
aerodynamic center is shaped which is a subsonic.
You can take it as 25 percent, but mass center,
that is why they put a leading edge mass to
shift the c g, onces they design. Leading
edge mass they put such that the c g also
comes center of mass at 25 percent chord.
Elastic axis of the structure is also at 25
percent and pitch axis is also at 25 percent.
Please understand everything they try to do.
It is a very complex design, it is not that
easy, it goes through several iterations.
And then finally, they will try to fix it,
put it there. The idea is you do not want
introduce too many couplings because if the
mass center is somewhere there, if it start
doing, then it will have its own inertia,
there are problems. So, let us say this is
our reference line 25 percent chord and this
is what my pitch axis is also. Usually the
pitch axis for the rotor blade is kept at
25 percent chord. Now, I define the velocity
components, u tangent and u propagate. That
is all. Because you realize you can find out
velocity at any point. This point will have
different velocity because if your pitch is
changing about this axis, this will have different
velocity components.
So, which point you will really take the velocity?
So, these are all very basic questions. Now,
you start thinking what do I do. How the aerodynamic
loads are obtained? Now, you say let me go
for CFD or wind tunnel. Wind tunnel, what
they do is they put that they will rotate
about 25 percent chord pitch, they will do
that and then get the loads.
So, these are some of the. When you really
start doing only you will start realizing
hey there are so many things which we have
to know before we really start analyzing or
understanding, otherwise books will simply
take it. Take oncoming flow because it is
not a fixed airfoil. Please understand because
this is a moving, moving one wing it is doing
something and your pitch angle is changing
continuously that means, it is no longer a
static problem, it is a dynamic problem. So,
we do not have assume, we do not have any
dynamic data, we only know static. If you
give me what is the angle of attack, I will
give you what the lift is.
So, every instant I am assuming this is the
angle of attack. So, this is what I call it
the quasi static approximation. Because even
though my pitch angle is continuously changing
that means, I am oscillating my blade. It
is a dynamic situation, but do not consider
that. So, this is another first, another assumption
we make in getting aerodynamic loads. Purely
from what you have learned aerodynamics, given
the airfoil what is the angle of attack, take
it and then put it. When you want to complicate
a little more, then you say let me introduce
unsteady aerodynamics, but we will not do
that.
So, I am going to take this. So, for me when
I am trying to get the velocity expression,
I will just take that 25 percent which is
the reference line. So, I am going to say
what is my position vector. This is my origin.
Please note origin, all the origins are fixed,
same. Position vector, I take a point p on
the reference. I will say r is the position
vector of, this is r, any point. Because that
is in the deformed axis, I have to get the
velocity. If you want to get the velocity,
one way is you convert x 2 to, using this
transformation relation, to x 1 and then again
differentiate. Or another way is, I am not
going to do I can directly go ahead and differentiate.
You can do many ways. That is why I said basic
dynamics is very important here.
Because when I differentiate velocity of that
point, I will write velocity of point p, r
dot, but r is a constant. So, that means,
e x 2 is a varying. What is the angular velocity
of this coordinate system x 2 with respect
to fixed frame fixed, this term is capital
X capital Y capital Z. This is what you need
to be careful, is it clear? Because basic
dynamics is very important. Another way, slightly
easier way is, you know that this is my omega.
x 1 is rotating with the omega, which is the
rotor angular velocity with respect to capital
X capital Y capital Z. So, one axis rotation.
Whereas, when you go to x 2, it is not one
axis rotation because you are also rotating
about this. Because the flap y 1, flapping
is also a rotation.
So, when you want to define rotation of x
2 y 2 z 2 with respect to fixed frame you
have to include capital omega, you have to
include flap. So, that will be like this.
If you define omega here. This will be omega
e z because that is the z axis, counter clockwise
rotation positive. But my flap is about what
axis? Flap is clockwise about y 1. y 1 y 2
both are same. So, I will have minus y 1 or
y 2, both are pretty much same. You can put
it actually y 2 is better because you have
to convert y 2 only. Because y 1 y 2 both
are same
Now, you see this is along z direction, e
z. This is along y 2 direction, this is defined
along x 2. We have to get the velocity that
means, covert all these vectors in x 2 y 2
z 2, then only you can get the velocity. You
know that r dot because the position vector
is r is fixed. So, that dot is zero. So, you
will 
get omega cross… This is what the basic
dynamics, omega is the angular velocity of
the coordinate system. And this is the position
vector.
Now, you convert this omega. You know e z
that means, e z, basically e z 1. Then you
put it e z 1 here and then get it in this
coordinate system. So, you will have. Can
I erase this part? y 1 y 2. See, x 1 y 1 z
1 flapping is coming out of plane, this is
flapping. This I call it beta. But onces it
comes up, that is x 2. That is why x 2 and
z 2. This is z 1. z 1 and z are same. y 1
and y 2 are same. And now this is beta is
the angle, time rate of change is the angular
velocity, beta dot. Now, I have to convert
this into in the coordinate system of, what
I will write it as you put it omega, I am
writing directly. Because please note I have
put omega here. Omega is omega, that omega
comes here. Omega sine beta x 2, omega cos
beta z 2. Now, I know r p, I know omega. So,
I can calculate my velocity. Velocity of the
point p is… This is what and here it is.
Now, what is v p? Velocity of the point p
is x 2 is 0, y 2 omega r cos beta into e y
2. Now, this is the velocity of the body.
Now, what is the velocity of air? Minus of
this. But you have to add velocity of the
helicopters also. See, helicopter is now going
forward. So, I will have these are the two
components which are due to the vehicle. So,
I will write the vehicle velocity.
So, I erase this part now. Now, these are
all not necessary. Due to blade. What we are
writing now is, it is due to vehicle and inflow.
This is velocity that is all. Now, this is
due to vehicle velocity or you may hub velocity
actually, the origin. Because velocity of
the origin you have to take, because the hub
is moving and the blade is rotating. So, you
have to add the velocity of the hub also into
that. So, that is what we are getting here.
But here I am directly writing instead of
hub velocity, I am putting it in terms of
air. This is body motion, this is air. But
if I want total, then I will subtract this
from here.
Now, this and then your lambda. Lambda is
in the z direction downward. So, I will put
minus lambda omega r e z. Now, you see these
two are in x and z. So, you substitute that,
get it in this, again put it, get it in this,
in x 2 y 2. So, you go through that full transformation.
So, you will get, I will write the final velocity
of relative air at airfoil at p or you may
call it point p. This is nothing, but what?
You take this velocity and you subtract this.
So, that is this expression minus v p. This
is what your final velocity expression is.
But you can write this whole thing. Now, I
will put it here. What is v air? Mu omega
r bar in the x direction that will have a
cosine psi and then it will have a minus sin
psi. Then again you have to substitute here
and then get it. So, that whole expression
will become mu 
omega r cos beta. So, I am carrying on this
here. Plus, not plus actually it will be a
minus sign. Minus mu omega r cosine psi sin
beta plus r beta dot plus lambda omega r cos
beta e hat z 2. May be I will write it again.
So, you see here mu omega r cosine psi minus
lambda omega r sine beta e x 2. Then minus
omega r cosine beta plus mu omega r sin psi,
this is along y 2. And this is a minus sign,
mu omega r cosine psi sin beta plus r beta
dot plus lambda omega r cos beta, this is
along e z 2. Is it clear? You have now three
components of velocity. You go ahead and you
substitute and get it. You take that as a
exercise. Now, I will write that and just
briefly mention then we can, onces I write
the velocity expressions. I am going to make
approximation now. And that is what is the
key because you make approximations. What
kind of approximation I am going to make?
Beta that is the flap angle is small, the
first approximation.
So, when I make beta small, sine beta is beta,
cosine beta is 1. Then I also make another
assumption. So, I will put v relative air.
The first expression will be mu omega r cosine
psi. I am going to like this, minus lambda
beta omega r. Then this is minus omega r,
cos beta is 1. Sorry this is not uppercase
r, I am sorry, this is a lowercase r, I am
sorry about that. Please change that. That
is not r, because this is a lowercase, please
that location.
Omega r cosine beta is 1 and then plus mu
omega r sin psi e y 2. And then minus 
lambda omega r because I am writing this expression
first. Then r beta dot, then this expression
plus mu omega r cosine psi into beta. Please
understand that is a beta mu omega r cosine
psi. Now, I will briefly show the diagram.
And here again I make one more assumption.
Lambda is also small. So, I say beta is small,
lambda is also small. So, the product of,
I am neglecting.
Now, you see I have written the expressions.
Please understand this is along x 2. Along
x 2 means along the span of the blade it is
going, that is this. You call it radial velocity.
But y 2 is, it is the velocity which is coming
towards the leading edge, minus sign is there
that means, it is coming towards. Because
y 2 is forward, minus means. So, your airfoil
if you take, this is my u t tangential, which
is omega r plus mu omega capital R sin psi.
This is that velocity and the cross section.
And this is z 2 with a minus sign that means,
it is coming down, so I will put it this is
u p. Because the sign is already taken because
z 2 is up. So, u p is lambda omega r plus
r beta dot plus beta mu omega r cosine psi.
Now, you have u t, u p, that means, and u
r, which is the radial velocity. Because please
understand the radial velocity is going into
the board. Because you now knew that this
is my y 2 axis, z 2 axis, x 2 axis. So, this
is y 2, x 2 is going into the board, this
is z. You have velocity component and you
say my theta because as usual I am defining
my theta which is rotation of the blade pitch
angle only about its elastic axis, which will
happen about 25 percent chord. So, I define
this is my theta. Now, you got it. This is
what I have shown here. We will continue next
class.
