All right. Now, we were discussing about the
velocities in a given cross section in the
deformed state. So, I will draw the section
of the aerofoil. I will write first. Please
note this is your Y 2 axis, this is your Z
2 axis, X 2 is going in. And we had the velocity
components, U T later I will write what it
is, that is the velocity along Y 2 direction.
But since it is coming towards it, minus sign
is there that is why I put the U T this way.
And then there is an U P, which is the perpendicular
that is 
along the Z 2 direction which is again a negative
quantity minus sign, that minus sign I have
taken here. Then there is a U R, there is
a radial which you may I will put a circle
here, it is U R. Velocity going into the port,
all are air velocity.
Now, your U T we wrote last time, we make
an assumption omega R. Please note beta is
small. So, cosine beta is 1, sin beta is beta
and lambda is also small. Based on that, we
will write the assumption. This is omega r
plus mu omega capital R. We will write it
as… So, that I take out this omega R outside
and write it as R bar plus mu. So, this is
basically this symbolically because non-dimensional
U T, because later I will use U bar T everywhere.
Similarly, you go to U P. U P is actually
minus, that is why I have taken that. U P
becomes lambda omega R plus… and which you
write it as again omega R lambda plus… There
is a mu, beta 
which you call it as omega R U P bar. So,
that it becomes easy for later representation,
U P bar U T bar. Hut he saying is that this
derivative, since you brought it that is good.
It is a time derivative we take, but here
I non-dimensionalize this. What you said is
correct.
So, there is a omega divided. And if I write
it in this fashion what will happen is, even
though this is that, if you want I can write
this as a star just to denote this is a derivative
with respect to omega t, then what will I
do? So, you can write it like this, d over
d t is omega.
Now, this becomes non-dimensional time which
is our symbol. Is it clear? Therefore, in
all our later things the derivatives are with
respect to non-dimensional time. But I do
not put star, I use the same symbol as dot.
So, take it that the time derivative is with
respect to this omega t because I want to
take out omega outside. Because in our operation
omega is a constant, all these derivations.
Is it clear? And then you will have the U
R term. Basically we neglect one particular
term, which is we know that it is mu omega
R cosine psi. But I neglect the beta small,
so this becomes lambda beta. So, product of
lambda and beta, I am neglecting the term
please understand. So, this essentially becomes
mu and your omega R is there. Now, you got
all the expressions for your velocities. You
have to go back and write your lift and drag
and if you have aerodynamic movement you have
to use the moment. But for our calculation
we assume that the aerodynamic moment is 0.
So, we neglect the term.
Now, this is our and this is the theta, which
is the pitch input given by the pilot. Now,
you have U T, U P. So, there is an induced
because this is your, resultant velocity resultant
velocity. Resultant velocity is 
square root of… And this angle you call
it phi, which is tan phi is…
If you want to compute numerically because
please understand these are the expressions.
Every section this is changing. And all these
quantities are, we assume lambda is the constant
that is the most important thing. If you want
to put the lambda also varies along the radial,
along the azimuth, then you should have a
model for that. That is your model. But right
now we assume lambda is constant over the
entire disk. You need to calculate this that
means, with psi, every time instant things
are changing. So, what we do I,s we further
make assumptions here. The further assumption
is, I say my U P which is this term. I assign
some order, lambda is small and the flap is
also small and this is a flap angle is small.
Therefore, you see all these three terms they
are of the same order, because mu is forward
speed that is v cosine alpha by omega R. And
the that is a little higher term and that
is sitting here.
So, I always say my U tangent is much larger
than U P, much larger I am just make this
statement. It is true as you go R far away,
but when you come near the root that assumption
may not be valid, but still I make that assumption.
That means, I am going to write, that is all.
And this is phi is… Because U P is small,
therefore U P over U T is small and tan phi
I call it as phi. Now, my angle of attack,
alpha effective, that becomes theta minus
phi, which is theta minus U P over U T, very
simple. Now, I have this alpha, now I can
get the lift expression, drag expression.
Lift is perpendicular to the resultant flow,
there I will take this. So, my lift is, this
is my lift, and the drag is, this is my drag.
Actually perpendicular means, well it looks
like that, but this is perpendicular to not
to the chord, but the resultant velocity.
So, this angle is phi, this angle is phi.
Now, it should not look like. So, this is
lift and drag.
So, your lift expression is half, I am writing
approximately. Because I already make this
assumption u square, I may take this, but
I am going to put it half rho. Even though
exactly it is U square, sorry U square chord
C l, it is lift per unit square. But I am
writing this as… which I can simplify as.
Let me erase this part. Half rho U T square
C C l alpha, that is lift curve slope which
we call it a, into theta minus phi. This is
my lift expression now. And drag is half rho
u square C C d 0, drag coefficient. But hear
again I make that assumption that U is U T.
So, this also becomes half rho U T square
C C d 0. Now, I have lift expression, drag
expression. I need to get the forces now.
Forces I get it along f z 2. Now, I erase
this part. So, I will use only U P U T expression
fully.
So, my F z 0 is l cosine phi d sin phi and
F y 2 is minus of, because Y 2 is this way,
so that will be minus l sin phi plus… These
are forces in Z 2 and Y 2. I have to convert
them into 
X 1 Y 1 Z 1. Finally, it transfer them back
to X Y Z, which are basically the hub coordinates.
Because if you remember last class, we said
this 
is hub and this is the projection, this we
said this is Y 1 and this is my… And then
of course, these two, x 2, y 2. So, we had
x 1m this is z 1 and we put this is our blade
really, this is x 2 and this is z 2, this
is beta.
You have this transformation. So, what we
have to do is, you get the force in x 2 y
2 z 2 coordinate, transfer the forces back
to x 1 y 1, finally back to x and y. That
means, you are doing blade sectional load,
this is sectional load per unit length. If
you want to get that total load acting at
the root of the blade, you will integrate
this along the span of the blade, then you
will say this is my root load on the blade.
You can give it in x 1 y 2 direction, that
is root load in the undeformed rotating coordinate
system.
But if you want hub load because x 1 y 1 you
keeps on rotating. So, you have to get the
hub load means hub load along fixed direction
x and y and z. But if you want to all our
expressions. So, in the design when you go,
you need to know what is the blade load, then
you also need to know what is the hub load.
Because blade load you require further the
design of the blade, but when you translate
into hub, this is not the only blade you will
have four or five blades, every blade you
have to add. So, when you go to the hub load,
you have to put a summation of all the blades.
So, each blade contribution you will take,
transform all of them along x and y, then
you will say this is my hub load. Otherwise,
each blade will have its own load. Now, we
make certain assumptions, we are going to
do one by one. And then finally, we will write
the hub load expression.
Now, let us take the, we make now approximations.
This is what we are doing now, lot of approximations
which are done here so that you can get something
like a closed form solution. Otherwise, if
you do not make any assumption, that is all
it will be left like this. Then I will say
you will take it from some aerodynamic section
where it starts, from some root of z to tip
of the blade and you get the loads, transform
it, put it. Otherwise, you cant get a nice
little expression. This is the reality. If
you want to really do practical blades they
stop right here and you do not even have to
make. If you are doing numerically, you do
not make even this.
But please remember I am making here. c l
is c l alpha this fine. But how c l alpha
varies ?c l alpha can be a function of mark
number. So, as you go along the span, mark
number is changing. That means, your c l alpha
will change and c d that will also change.
So, you need to take actual aerofoil characteristic.
How it varies with mark number? How it varies
with angle of attack for the given aerofoil?
This is what is done because this they call
it polar, that is all, that is the aerodynamic
polar. But it is a static data that means,
you put an aerofoil, whatever aerofoil which
you have chosen for your rotor blade cross
section, put it in a wind tunnel, get that
data. And in the earlier days they were using
not 0 0 1 2 etcetera, symmetric aerofoil.
But nowadays nobody uses that. Each company
has its own aerofoil. They have some name
they give and the rotor blade aerofoil they
design and they use that and they will know
the aerodynamic data of that aerofoil, please
understand, it is a static data only.
Now, if the angle of attack exceeds the stall,
so blade will stall. So, you have to take
the stalled value. That is why I am saying
actual blade calculation to highly idealized
situation, what we are doing? We throw away
c l alpha is constant, c d naught is going
to be constant, I make U P as small comparison
to U T, then I do all these assumptions.
Now, I get this. Here also I make further
assumptions. The further is, since phi is
small, I will write this as l. I am making
and throwing this out because d phi, d is
anyway small, phi is also small, so I neglecting.
When I go here this will be minus l phi plus
d. Now, you see l phi. This is d is drag which
is due to the profile drag, c d naught this
term comes from lift, that is why you call
this as the induced drag. So, you have in
the, you have induced drag and you have profile
drag, two expressions. This is per section,
so keep doing and integrate the whole thing.
But integrate, before integration you have
to transfer this to f y 1 and z 1 and now
that also made lot of assumptions again. Now,
we will come to one by one. What is f? Can
I erase this part? This is not necessary now.
Because you know the transformation which
I wrote last time.
That is e x, I am actually I make the transformation.
This is what I wrote last time. This is what
was written last time. And another expression
was e x 2 y 2 z 2, you wrote cosine beta 0…
e x 1 e y 1 e z 1.
Now, I have the forces in this direction,
z 2 y 2, they are here. z 2 y 2 x 2, actually
x 2 will you have a force? Well you can have
a force, that is the drag because it is along
the span. And how do you get that expression?
That I will tell you later. Right now we say
there is no x 2. So, you can get e x 2 because
this is all orthogonal transformation. So,
you will have just cosine beta 0, just transpose,
e x 2 e y 2 e z 2.
Now, I have f z 2 f y 2. So, f y 1 and e x
1 that means, I will have f x 1 is minus sin
beta force along f z 2. So, minus beta, in
the sense, sin beta I am writing it as, sorry
let me put it clearly, minus sin beta f z
2. I am writing it as minus beta f z 2, which
is minus beta l from there. And I know the
lift expression. And then f y 1 is nothing,
but f y 2. There is no change because that
is only one. And your f z 2, f z 1 is beta
into e x 2, there is no force. So, cosine
beta f z 2, that is just f z 2 which is just
simply l. That means, I making beta is very
small. So, I get this. Now, you go back with
these three. Because you know that lift is
nothing, but f z 2. That is all. You go back
and get the transformation in the, let me
erase this part this is not now required.
Along x y z, this is hub fixed. So, you will
again make the transformation cosine psi minus
sin psi 0. Here you will have what? x 1 y
1 z 1. Now, you know f x 1, f y 1, f z 1.
Put them here, you get the f x y z. So, you
will write f x cosine psi f x 1. So, minus
beta l cosine psi. And then sin psi y 1, so
minus sin psi f y 1. f y 1 is nothing, but
f y 2. Which if you write everything in terms
of minus, what is this? This is sin psi, f
y 1 is f y 2 and minus and minus will become
plus, this will become l phi plus d.
You got it? Now, then f y sin psi e x 1 f
x 1 that is minus beta l sin psi. And then
cosine psi f y 1. So, you will have plus cosine
psi f y 1. And f y 1 you know minus of that,
so this will become minus beta l sin psi minus
cosine psi l phi plus d. And then f z, that
is nothing, but f z is f z 1. f z 1 is f z
2, which is l. So, you will write your f z
is f z 1, which is f z 2, which is basically
l. So, you see this is my full expression
of transformed load along the hub fixed non-rotating
coordinate system per section. Basically,
all these things you have to now get. Because
this is the hub load, so we need to integrate
over the span of the blade and then you must
also add the value from each blade. Because
when you get the hub load, every blade will
give some load. That blade will depends on
where its location is. Now, you see my hub
loads even if you look at the simple expression,
these loads are functions of psi, that means
with position. Psi is omega t, that means
my hub load is the function of time. It is
not a constant, you understand. So, as the
blade goes round my load is not fixed, it
is varying with time. So, how do we really
proceed in the entire formulation?
What we make in this course is, let me look
at what is the mean value of the load. Mean
value of my hub load, that is what. Mean value
means I take average. One revolution the blade
will go, you will have some load. It will
be time varying, then I take the mean value,
that is all. What about the time variation?
I throw it out. I do not even write that because
the time variation quantity essentially represents
the vibrating part. So, the helicopter, the
moment it starts moving forward, you will
have a mean value and you will have a vibrations
value. And the vibratory part I neglect. That
is why helicopters have to vibrate. If it
is not vibrating, it is not a helicopter,
in the sense it is very crazy vibration, that
is where other problems starts, how do I reduce
vibration etcetera. So, that is a different
part which will not be part of this course
at all. Vibration is a, please note, a major
problem in helicopters that comes purely because
of this time variation in the load. But then
you are flying. How you are flying is you
use the mean value of the load. So, you say
what is my mean f z, mean lift. If that value
is equal to the weight, I support. The mean
value is equal to the weight, it is flying,
but then it will be shaking and flying. And
same thing happens f x, fy everywhere. But
this is only sectional load, please understand.
You have to get the momentum because aerofoil
theory gives you only lift, drag and pitching
moment. Pitching moment I said take it as
zero. That is for a section, but when I take
the load at a particular section, I am transferring
this load to the hub.
So, that will give you hub moments also. Because
in the route, bending will come. So, essentially
you have a lift force, you have a drag force.
The lift into this that will give you flap
bending moment for the blade. Each blade will
give its own flap bending moment. You have
to sum up all and get the hub moment. We will
come to that moment part a little later. And
then the drag force will give you lead lag,
which is the torque. So, what we will do is,
we will write the torque expression, later
we will write the bending moment expression.
So, you see at the hub if I integrate all
of them, I get the force and r cross f. Because
moment is…, vector is… and r is 
you take it as whatever location. So, into
f, f you put all these f and take the cross
product, r is r into e x 2, because r e x
2 is what you have. And if you want to convert
e x 1, you can transfer it and then take that
x 1 y 1 cross product and then transfer it
to hub loads. So, you will have at the hub
three forces, three moments. And these expressions
you need to get. They are pretty messy long
expressions. Finally, I will give only very
simplistic stuff with all assumptions made,
all integration done everything.
Now, let us go back to writing first expression
which is lift. Because lift is my thrust.
I am going to write that expression first.
So, one by one we will write the expressions.
So, I will write that and then we will non-dimensionalize
all those loads. Let me erase this part completely,
I think this is not required. So, I erase
this. Your lift f z 1 which is nothing, but
f z 2, which is lift..
So, I will take it as this is half rho. We
made an approximation that is u t square,
please understand. I am putting like this.
chord c l alpha, c l alpha you may use or
c l alpha you replace by the symbol a. So,
this is what we use a sometimes. And then
multiplied by theta minus. Now, take the U
T square inside, you will get half rho c a,
I am putting this c l alpha as symbol a, theta
U T square minus U P U T. And you know U T
you have non-dimensionalized with the omega
r outside. So, take out that omega r, you
will write half rho omega capital R square
c a, it will be theta U T bar square minus
U P bar U T bar. I am taking omega r outside
because finally, I will non-dimensionalize
the whole stuff. Let me erase this part now.
Now, I just replace the expressions here.
This becomes, please note, I am taking half
rho c a omega capital r whole square, substitute
for U T bar.
U T bar is r bar plus theta lambda plus r
bar theta dot. Please understand this dot
is with respect to non-dimensional time derivative,
I do not put a star, plus beta mu psi multiplied
by… So, this is my sectional lift. Please
understand sectional lift. If I want my total
thrust. Thrust is number of blades I have
to take. I assume now each blade I must add,
it is essentially a summation, it is not a
integration. Because at a particular time,
one blade will be in azimuth, another blade
will be in another azimuth. It will be here,
another one can be there, third one can be
here, and if you have four bladed, it may
be like this.
At every given instant, these psi will get
replace by psi k. And I must put summation
k running from 1 to number of blades. That
is what I have to do. 1 to n, where I must
replace here all the psi by every blade. And
this blade what? I must integrate from 0 0
or some root offset to b r. It will be d r,
you understand, is it clear? But this is non-dimensionalize.
So, I will actually take a r bar, this will
become b, this will become e and 1 r will
come out. That r I will take it here. Is it
okay? And then this is a summation at every
instant. Now, please I am just slowly look,
just with the simple expression. You still
do not know what is theta. Because theta is
theta naught plus theta 1 c cosine psi plus
theta 1 s sin psi. That is the pilot input.
What about beta? You do not know beta. So,
you are assuming now this is the you, assume
I know all these quantities. That means, I
am essentially saying theta is cosine psi
plus theta 1 s sin psi. And beta is I am writing
like this cosine psi plus beta 1 s sin psi
plus I neglect all the other terms. I am throwing
away everything, I assume only this motion.
Even though technically it is not correct,
you have to take all the terms, but up to
what harmonics you will go? These are all
real practical problems. Let us take up to
one harmonic. Now, you see this I have to
put it here, here, here, here. And then I
will have a long expression, integrate from
this to this, then sum it up. Instead up doing
now, I make one more assumption that all the
blades as it goes around the azimuth, everything
does the same thing, same motion. That means
I assume all blades are identical. And the
response is also identical. Identical means
you take one blade, it goes around the azimuth,
that means it you will some flap response.
That response is same for all the blades.
And this is what is in the practical thing
they call it, all the blade first identical
in terms of mass distribution both, please
understand. Mass of the blade you can keep,
then mass distribution how the mass per unit
length is distributed. So, when you really
manufacture, all the blades are not identically.
There will always be some difference in weight.
Aerodynamic shape has to be identical. That
we take it mass, but what they do it they
measure the blade weight. If the blade mass
is not same, they have to add weight. So,
they will have boxes where they will put some
extra weight, some 200 grams something like
that. Make the blade mass is same as well
as first movement of the blade mass with respect
to the root, that is all they can do. Because
you can have, mass is what? Mass of the blade
integral m d r, some 0 to r you can take it.
First movement I call it m x c g, this is
1 integral m r d r, 0 to capital R, then I
blade, 0 to capital R m r square d r. All
of them must be same, then only you say they
are identical. Otherwise, there will be some
difference, but in reality there will be difference.
Now, these differences will appear when you
really flight, that mean no to blade is same.
But they have some criteria how much I allow.
That is why I am just deviating now, there
is something called a tracking balancing.
Balancing is balance the mass of all the blades
because otherwise, if one blade is less mass
or different then the other one, then you
know that from basic vibration you will have
a shaking because the mass is c g f the blade
system is not right at the center. So, what
it will do? It will start shaking. You will
have vibration. So, this is balancing the
blade. Tracking is, as the blade goes around
you keep every blade some pitch angle, every
blade should when at a particular azimuth
angle, they should come to the same response.
But what will happen is, if you look at that
if all the fans are rotating. If one blade
is going up, another one is going down at
the same location that means, they are not
doing the same response, but this is called
tracking. So, how do they do is, they adjust
some pitch angle of the blade, individual
blade. Some blade goes up a little higher,
then they will say reduce the pitch of that
particular blade or they have a trailing edge
tab. Usually most of the blades if you see,
they will have something like this. There
will be a fixed the tab, small tab will be
there. They adjust the angle. Fixed, it is
set. It is really reverted and they just adjust
it a little bit, so that all the blades almost
track. And they give some allowance, this
much is there. Some engineers, I will tell
you, their specialization is only this keep
tracking balancing. Because if the track and
balance is not proper, when they fly it will
have lot of vibration. So, you balance it,
then again fly. Of course, you will always
have vibration, even if there all balanced
you will have vibration. If they are not balance,
you will have more vibration. This is a real
practical problems.
Now, for us easy everything is identical.
So, all blades perform the same response.
Identical blades, identical motion. Now, what
I will do is, the movement I say everything
is identical. This is per at a particular
instant this is the load, I am going to get
mean values of the loads. Mean values… some
hinge offset b r. Mean value is 0 to 2 pi
1 by 2 pi n. So, this is the mean value of
one blade. If I have n number of blades, I
will simply multiplied by n. You got it? So,
what I am doing is, I am assuming all blades
to same motion therefore, I am just multiplying
by one blade effect, multiplied by n. So,
this is what I finally get. And this I have
to do for all the loads. Please understand,
I am writing using only thrust. Because I
do not want to write everything again and
again. If I explain for one, the same thing
is valid for f x, same thing is valid for
f y, same is valid for movement and torque
etcetera everything. Now, this is what I will
do. Now, in the non-dimentionalization. So,
now let me erase all of that, I knock out
everything because I hope you have all these
expressions, that is all. This thrust is a
function of time, agreed or not? Because this
is what is coming here. This entire expression
comes here, agreed? And it is the function
of psi. Psi is omega is t. k, I have to put
wherever psi is there, I must put a psi k.
Replace psi by psi k, where psi k is 2 pi
over n k minus 1, n is the number of blades.
So, at a particular instant, each blade will
have different lift value depending on where
it is. Even though their r is fixed, assuming.
Because this is a location of the position
from the root. You take the same r bar in
all the blades. At a particular instant, they
will have different lift values. What you
are doing is, you were integrating for one
blade fully. After integration you are summing
it up, each blade value you integrate separately,
then add at that time, that is a particular
time. You got it? Then again next instant
again you do it. That means, your this t is
basically a function of time. Time is non-dimensional
it psi, agreed? Now, I say I am going to have
oscillatory force, I am going to make this
is d mean, that is 1 over 2 pi, 0 to 2 pi.
Is it clear? That is for one blade I have
done. Now, I make assumtions that all the
blades are same, identical. Therefore, I simply
multiply. I do not go and then add every time
each blade and then calculate it, I wont do.
In this case, I will simply calculate for
one blade, take the mean value multiply by
n. It is not that I will go and do for. Actually
in my code what we have developed here, aeroelastic
code, we do each blade separately. Then you
do, but it is technically same, it is basically
same. You take this t. t mean what will you
do? You integrate 1 over 2 pi to 2 pi, that
is basically the same thing, that is what
I am saying. But if you really want to capture
the vibratory load also, then you have to
do this. You will have a t, every t. That
is why the blade you keep moving it, every
4 degrees or 5 degrees like that and then
keep integrating. When one blade moves, other
blade also moves and other blade aerodynamics
is different.
When it comes to on the retreating side, this
blade may be stalling some sections. That
means, you have to take that stalled value
of that lift into account. So, when we write
the big aeroelastic code, which we have developed
of course, the h a l is interested done hence
now its a collaboratively we are adding it.
It is a very comprehensive aeroelastic analysis.
There we do every section of all the blades
and we do not assume all blades are same.
Even though for computation we do. Otherwise,
even if you want change the blade properties
the little bit, you can do it. The flexibility
is there in the program, but industrial code
they assume all blades are same, nobody does
that. And after that it is all minor adjustments
they do. Because see practical problems sometimes
may not be easy to translate into theoretical
modeling. Because sometimes it is very complicated
realistic problems are. So, we make lot assumption.
Because you need to get some values, if you
want to fly. See, if you really see the history
of, you all heard a history helicopter development,
all these things came much later. All the
mathematical modeling as all much later, after
the helicopter started flying.
But now you try to understand the problems.
What is the reason for this problem, how do
you fix it, how do you improve it. But improvements
have happened substantially. Is this clear?
But vibratory load you cannot get it, mean
value fine. And that is what is done in practical
things. So, if the mean values you capture
correctly, you can go ahead and design. Vibratory
values always problem because the real aerodynamics
you do not know, you understand because the
flow is so complicated. Your inflow you have
to get properly because if your inflow is
wrong, your angle of attack is wrong. If the
angle of attack is wrong, everything is wrong.
You understand? And it is a time varying.
It is an oscillating aerofoil. Unsteady aerodynamics
you have to use, but usually industrial use
just steady values. But we have a model for
unsteady aerodynamics. So, we have incorporated
a stall model. But again that is empirical.
We try to fit a curve for the dynamic stall
and then take that. It is a differential equation.
So, I get my c l, c d, c m from a differential
equation. If my blade is oscillating in a
particular fashion, at a particular time what
will be the value of my c l? What will be
my drag? What will be my moment at a instant?
So, usually the lift, drag, moment are not
just functions of angle of attack. They also
functions of time variation in angle of attack.
You understand? Time variation in angle of
attack also will come and second derivative
also may be there. So, these are more, I would
say, advanced modeling approaches. But even
then that is not precise, but you make assumption
which is a better model than assuming everything
is constant. But in the course for the basic
level, you make all these assumptions. And
then I will write the expression for the mean.
I will not use the t mean every time, please
understand. Because you take it as it is the
mean value, it is already integrated over
azimuth, 1 over 2 pi etcetera. So, let us
write the expression for the thrust.
So, your thrust I take it and I will non-dimensionalize
the thrust expression also. So, I will get
the mean value of the thrust, which is c of
t is thrust divided by rho. This is number
of blades, putting that, there will be a 0
to 2 pi and one over sorry there is the n,
1 over 2 pi is also there. Then I write this
expression. You may say e, I will put a e
bar and b. Take out r outside and I will have
this half rho c a omega capital R whole square.
I am just simple because for simplicity I
am writing like this… into… But now I
make further assumption. I will not take bar
and b. 0 to 1 it is much easier because then
you can get a neat closed form expression.
Now, I non-dimensionalize. So, I have to divide
by, sorry I forgot to rho pi r square. So,
you look at the term omega r, this will go
up, rho goes off, n c r pi r square that is
sigma. So, you will have 1 over 2 pi, I will
have n r c. So, I will put a half this is
sigma, this is a. Times your U T bar square
theta U P bar U T bar d r bar d psi. This
is my c t expression. This integral I have
to do. Oh I am sorry 0 to 1 integral. Yeah
take it 0 to 1, yeah I put it 0 to 1. Otherwise,
you put it e to the tip, whatever you may
correct at, tip correction factor 0.97, this
root of set. So, this is my expression. Here
I made root is also, I do not have any correction
I am making it 0, tip I go to integrate till
1. Is it clear? Now, I write this U T, I have
to substitute r bar plus mu sin psi whole
square. U P bar U T bar I am write these two
expression and then I have to theta.
I write that, theta naught plus theta 1 c
cosine psi plus theta 1 s sin psi. And similarly
I will write beta beta naught plus beta 1
c cosine psi plus beta 1 s sin psi. I write
all these expressions. When I take a derivative,
beta naught is constant that goes off. Beta
1 c, please understand in this expression
beta naught 1 c 1 s they are all constants.
Because it is steady state value. You are
writing the flapping motion of the blade like
a harmonic series, Fourier series. And this
is what it is. These are constants. Later
I will introduce something these are not constants
also, they are time varying. That will come
when you have to do stability, then they are
not constants, they also can varying. Right
now take it as, now you substitute these expressions
here. Theta you put it here, beta naught you
put it here and here, beta. And then integrate
the whole thing. And you write the expression.
Now, I will give, this is I will go back there
and show you what is the thrust expression.
That is all. You do not have do all these
things. What I am saying is you will be given
the sheet how the thrust is obtained.
We basically non-Dimensionalize the quantities.
So, we non-Dimensionalize thrust, in-plane
force that is why I s c sub h, c sub y, roll
moment, pitch moment, torque all of them are
non-Dimensionalized. And then I give the final
expression which is. Here I have just given
only thrust coefficient. What I have written
there is this expression.
And c t is a function of azimuth. That is
what you asked. I did not integrate 1 over
2 pi, 0 to 2 pi, that I have not done. But
I have taken all the blades also. So, it is
like very, that is why sigma a over 2 some
function of psi. That is where all the 0 to
1 everything is sitting there.
Now, like this you have to do for in-plane
force, you follow? Side force, longitudinal
force, torque everything. Because this you
do not have to copy, because this is something
which is. I will give you the final expression.
Only thrust expression, do not bother about
the rest of the things, only thrust you look
at the thrust part.
Averaging over this is what I have done. And
then my c t, mean thrust coefficient is this
expression. And I have included theta twist
also. So, I put this plus theta twist r over
r. I have added that. Now, this is my c t
expression, After doing all, it comes out
very simple. But you will find there is no
flap term sitting here. Flap term will not
be there, but you will get all the flap terms
will come in the side force and other things.
That will worry later. I want to just look
at it only this part. If you set mu, that
is the forward speed. So, please understand
my c t, thrust coefficient is a function of
forward speed, which is advanced ratio mu,
mu square. And it is a function of theta 1
s. Theta 1 s is we call it that is why cyclic
pitch, whether you call longitudinal cyclic,
pitch or lateral cyclic pitch because always
that is a confusion will arise.
Theta 1 you say it is the longitudinal cyclic
pitch, but the problem is it will be given
at 90 degrees. Later you will understand the
relation between this and this. Because you
need to know how flap. Now, if theta 1 s is
positive, that means what? Pitch angle will
increase at 90 degree and it will degrees
at 270. So, usually you would want on the
advancing side, velocity is more, pitch angle
is less. Retracing side, velocity is less,
pitch angle is more. So, your theta 1 s is
always negative. Please understand. When you
do actual calculations you will find theta
1 s is always negative quantity.
Now, you see the moment pilot flies forward,
if you gives a theta 1 s. Because pilot only
will give theta 1 s because he will move the
stick forward. When he moves the stick forward,
pitch angle when the blade comes at 90 degrees,
it will decrease. So, what will happen? When
the blade pitch angle theta 1 s is mu is starting
forward speed at that quantity becomes negative.
That means, your thrust will decrease. So,
always pilot will find, if you are starting
or you give a cyclic input forward, he will
always go down also, he will go down. So,
they will always increase the collective.
So, that collective you increase it, so that
maintain that. So, if you give one input something
the vehicle will. So, this is how. These are
all very small terms they look like, but then
they contribute substantially in understanding
how the vehicle will behave in flight. When
simplistically what people will explain is,
it is like this. The very simple if you want
to explain you will say this is my rotor disk,
thrust is like this. If I tilt it forward,
then what? Because of the tilt in angle the
vertical component gets reduced. Therefore,
you have to increase the vertical.
So, whenever you want to tilt the disk, there
is a reduction in the thrust. It is a simple
explanation you can give. Because only when
you tilt forward, you will go forward. So,
if this is the thrust you want to tilt it,
so your vertical component decreases. But
actually mathematically if you want so, this
is what is really happening, that is the term.
And then of course, lambda I have used constant
inflow. Now, you imagine your inflow can be
a function of azimuth. Now, imagine you put
all of them, these expressions will become
more and more miss here.
