I thought just to give you a recap of what
we did in the last class. I will just write
them all the quantities. And then F Z 1 is
L… These are our basic quantities and we
know tan phi is U P over U T. And we had the
expressions for lift, drag everything. Now,
you can add radial drag to this term. That
means, that is because of the flow along the
span of the plate, but that is purely an approximation.
That approximation is given by… I will just
give you, but we will neglect that term in
this rest of the formulation. I will write
it D radial. This is drag due to like a skewed
flow, flow past the aerofoil, actually along
the blade. This is written just approximately,
it is taken as U radial over total resultant
flow into drag on the blade. Drag on the blade,
you know that D is given by half rho U square
C C d. And this is just some fraction of the
component, that is all. It is a very crude
approximation which is done. Now, if you take
that U, you can substitute. This is the distributed
radial force, very crude approximation. But
we will be neglecting it. I just want to say
that how it is represented and then of course,
we neglect it. Now, we throw this off. And
we have to substitute for lift, drag expressions.
And then we know the forces, transfer them
to hub loads, which you already did last class.
So, our hub loads are… But only you have
to integrate. But before that I am just giving
the F X is F X 1 cosine psi minus F Y 1 sin
psi. And F Y is sin psi and then plus… And
your F Z will be F Z 1. Now, we have written
the expressions in different forms. The reason
I am writing this again is just for clarity.
Because we wrote L half rho U square C C l.
And d we wrote half rho U square C C d naught.
And C l in one form, we wrote as C l is C
l alpha into theta minus phi, which is U P
over U T. And then, U we approximated U T.
Because why I am writing everything is, when
I display the expressions, in some of them
we make the approximation. Only in one, that
is a little interesting derivation. That is
particularly in getting the torque.
Now, these are distributed forces. You have
to integrate over the blade and then sum it
up over all the blades at a particular time.
Then you are doing the mean value, but this
is only forces. These forces are acting on
the typical section which we have converted
into hub loads. But when you transfer them
to the hub, they will also give hub moment.
So, you have to take the moment also into
account. But when we do the moment expression,
I will write moment in hub coordinate directly.
When I write that means, we had our blade.
This is our X 1, Z 1 and this the blade, this
is X 2, this is Z 2 and this angle was beta.
This force is acting at a location which is
r. Now, when I want to take the moment, you
can write this is moment is r cross F. Now,
the r you can write r cosine beta e X 1 plus
r sin beta e X e Z 1 cross F X 1 Y 1 Z 1.
They are here X 1, Y 1, Z 1. You will have
F X 1 e X 1 plus F Y 1 e Y 1 plus F Z 1 e
Z 1. This is my moment expression.
Now, you can make approximation that beta
is small, so cos beta is one, sin beta is
beta. Then make the cross product and I will
write the expression for the moment here in
vectorial form. And then later I will… This
is e X 1. Because you just have to take the
cross product, because y is zero. So, Y Z
is zero and this is r beta F Y 1 plus r beta
F X 1 minus r F Z 1. This is along e Y 1 direction.
Please understand. These are all in X 1 Y
1 direction, still they are rotating, it is
not actually the hub loads. It is still at
the root of the blade. Then I have to convert.
And then plus you will have r f Y 1 e Z 1.
These are my root moments. May be I will put
a line here, this is root moment.
Now, for simplicity because later I will bring
out. Instead of writing in these two quantities,
I will replace them by another term. That
we will come to that later. Because when we
model the blade, how we model the blade? That
will come. That is only a moment along X Y
directions.
Let us first take only e Z 1. e Z 1 because
this is the drag force and this is my position.
So, that is going to give a torque. And you
know that e Z 1 is same as e Z. So, my torque
acting on the shaft is basically r F Y 1.
And you can take this quantity, calculate
torque at the hub due to all the blades. Because
all the blades will sum it up. There is no
cosines and sins because all of them are in
the Z direction.
Now, using this I will represent all the quantities
because there is only algebra now. I do not
want to again spend time on algebra. So, if
you have any questions you can ask me because
if this is clear, we got the sectional load
and transfer to the hub and then we also got
the hub moment. Now, we are going to get all
the hub loads, but non-dimensionalise rho
pi R square omega R square for forces, moment
is rho pi R square omega R square into R.
This is the moment, non-dimensional.
Now, I have to write thrust is F Z. And when
I am using the hub, I always call this load
as thrust. And F Y, this is my side force.
And this I call it H, which is the longitudinal
force. This is the symbol convention we use
H, Y, T; that means, H is F X which is this.
But please remember this H is towards the
tail acting on the hub. Because you remember
the coordinate system, X 1. X is you know
that this is a rotor disk, so your X is this
and this is Y and Z is up and here is your
tail. H is the longitudinal force, this is
lateral force, this is the thrust.
But please remember all of them are acting
at the hub in the hub coordinate system, hub
fixed coordinate system. T is perpendicular
to the hub, H and Y are in the plane of the
hub. If you tilt the hub, it is tilt T is
in this direction. That is why you have to
follow very systematically the coordinate
system and finally, you define these are my
forces. Moment when you go I will write only
the pitching and rolling moment, which are
actually M X will be the roll moment, M Y
will be the pitch moment and M Z is basically
yaw moment or you call it the torque, due
to the only the main rotor system. That is
all.
So, you have correspondingly M X. May be I
will write here, M X, M Y and M Z. Sometimes
this is written as Q capital Q also, this
is also used. So, this is the roll moment,
pitch moment, torque acting at the hub center.
But please remember this is one section. When
I want get these loads, you have to integrate
along the blade, you have to add all the blades
and then take the mean value. So, the entire
thing is done before you write these six expressions.
Of the six, right now I will show only four.
These two I will keep it pending. Because
one way is you can do this way, another way
is there is some easier way. These all elements.
When you get this, you have to integrate everything,
integrate along the blade, add all the blades
and finally, write that as the expression.
That is why I used the word correspondingly
means this side I should not put a equality.
This is basically, you can take it like this.
They lead to this quantities.
Now, imagine I have to first get these expressions,
which I get it from here. And then I will
substitute for L and D, then I put them here,
then I will do integral over the span of one
blade, then put summation. Is it right? So,
all these steps I just mentioned in words
because this you have to do by algebra. This
is where the time, number one. Number two,
people may make mistakes. Usually because
this is not very long, but still it is reasonably
long to make mistakes. So, what I will show
you is, I do not want to go through these
derivation, I am only giving words. These
days you can do in computation, no problem.
Just take L and D, substitute that every section,
then put it here, integrate, then use an integration
scheme and then get it. But what is done is
to get a closed form expression. Please understand,
closed form, that is only number. Here I will
provide a closed form expression and that
expression you are actually doing simplifications.
The simplification is I neglect in writing
F Z, I throw away the drag term. And I will
take only l. cosine phi phi is small, I will
say F Z is l. You follow? These are the assumptions
I make in trying to get a closed form expression.
Now, I will show those closed form expressions.
So, I have non-dimensionalized the quantities.
C T is thrust divided by rho pi R square omega
R square. You can see here, N is the number
of blades. I am assuming that all the blades
are doing the same motion etcetera. So, thrust
is a function of azimuth, that is what I told
you earlier. Similarly, when you go to lateral,
that is the in-plane force, F Y 1 cosine psi
F X 1 sin psi. This is basically this quantity,
F Y side force, lateral force. And again you
will have all the blades.
And you are substituting, I have actually
gotten this expression after substituting
for F Y 1 F X 1, collect the terms. And I
actually split the terms into two parts, one
due to C d, C d naught actually, here I used
the symbol C d. One due to only profile drag.
And other terms we call them as induced drag
term, which is independent of C d.
So, you collect C d term and the remaining
term basically comes from the lift, that is
the induced drag term. Same way I am representing
my longitudinal force here again F X 1 cosine
psi minus F Y 1 sin psi. I again substitute
all the expressions, get it, derive. This
also I split into C d naught and then remaining
terms.
Now, you see here there is a theta, beta,
all the beta dot. And you have to substitute
for… I told you earlier assume that theta
is theta naught 1 c cosine psi plus 1 sin
psi. And beta is beta naught plus 1 c cosine
psi plus beta 1 s sine psi. So, these are…
I am not substituted yet. I will be doing
later.
Just to give you a glimpse of how they look.
And here the torque. I put a minus sign here
because this is basically just to make it
plus, that is nothing else. Because if the
rotor blade is rotating in the counterclockwise
direction, the drag force is actually you
know that the drag force is minus Y 1 direction.
So, minus Y 1 direction in this. This is the
blade, so this is my Y 1. So, minus Y 1 is
this; that means, my drag force is like this.
If I take a cross product, my torque is clockwise.
But to get a positive, just to get a positive
number, that is why I will put a minus sign.
So, that clockwise moment is negative. But
put one more minus sign, I will get everything
positive. That is how I get the torque. Now,
you see these are my expressions. But once
I get torque coefficient, I know torque coefficient
is power coefficient. So, C P equals C Q.
So, directly for… But I have not taken the
tail rotor. Please understand. This is a main
rotor, power main rotor torque.
Now, since these are all functions of time,
I have integrated, substituted that theta
naught 1 c 1 s. Because please remember all
these loads. I call it loads, they vary every
instant. Is not that they are fixed. Your
thrust force is varying, side force is varying,
longitudinal torque moment. All these quantities
every instant they are changing. That means,
what is my thrust value? You have to take
only mean value. What is my longitudinal force?
Mean value of the longitudinal force. So,
that is why when we get these loads, we take
average values.
So, the average, how it is done is, you have
summed up over all the blades, but then you
integrate. In one revolution, how the loads
change? Take the mean value and the mean value
is same for all the blades. So, just multiply
by number of blades. So, that is what is done
in getting the loads.
So, please understand there are several steps
before you arrive at these expressions. So,
it is not very simplistic like what we have
done in hover, it is much more involved. But
you made assumptions that you know these quantities.
Even though you do not know the number, you
say I know that, I assume that. Similarly,
beta naught 1 c 1 s, you have assumed, well
this is how my flap is going to be. But please
note I am neglected all the higher harmonics.
If you include, that they will also come and
sit here, but then these expressions will
be very long. Numerically you can do that.
That is why this becomes algebraically cumbersome,
really cumbersome. And just to give an idea
of what is the minimum one can get an equation
which you can have a look at it, that is this.
With all these assumptions this is made. And
you make small angle assumption, phi also
you make small and you get C T and twist is
added. You can add a twist also, plus theta
twist R over R. So, this is my thrust, mean
value of the thrust. Please understand. And
then this is my in-plane force coefficient
in the longitudinal direction C H. If you
look at here, sigma a is lift for slope. Sigma
contains number of the blades, that is lag
number. You will see lambda, mu, theta naught,
beta 1 c and then you will have beta 1 s and
1 c square, beta naught square so many other
terms. And then there is one term with the
C d naught, which is the profile drag term.
So, one term with profile drag, rest of the
other due to induced drag. That is all. So,
you will have all these terms. And there all
small values, but finally, you have to use
them. And now similarly I have written the
in-plane this long and you can write the torque.
Now, this is just for, you cannot make out
anything from here I am telling. You here
only I am telling you what was done. If you
look at any book, if they have written all
these things, that means where did they come
from, that is all. Because from here you just
make out anything. Here it is all some mu
beta naught, simply whether plus sign is there,
minus sign is there. If there is an error,
sorry you cannot make out. This is just for
you to tell the procedure and this is how
they look like, just to give you a glimpse
of how these loads will look like, expressions
for the loads, closed form. And here you have
integrated full, zero to tip. All those assumptions
you have made. Now, the question is how does
the pilot balance? Because if you want to
hover or forward fly, all the forces must
be balanced and the moment about the center
of mass is zero. Force also zero, moment is
zero.
The tail rotor, not only it gives a torque,
it also gives a side force. So, the side force
has to be balanced. Now, what is the attitude
of the helicopter? Please understand. Because
these are all hub loads. You have to transfer
the hub load to the C g. And C g you have
C g coordinate system. And transfer the tail
load to the C g. Since you have asked, that
is what next in the another two, three lectures
I will bring in those loads. Then I will say
this is how your full expression will be.
And that is the assignment which I will give,
you people have to write a code, get it and
I know you may make. It is an iterative procedure,
it is not just a straight forward, that is
why unless you solve the problem yourself
you will not be able to appreciate the complexity
involved in the helicopter dynamics, aerodynamics
or whatever it may be.
I am saying this is the simplest problem,
this is not the complicated problem because
there are more complicated problems. But simplest
in terms of, we have reduced everything, made
lot of assumptions and then finally, say take
this and then try to do some simple trim calculation.
So, this is only the load due to main rotor.
That is all. Now, if you want tail rotor,
you can use the same expression. Please understand.
Except tail rotor does not have 1 c and 1
s, only collective. A tail rotor can have
twist also. That is all. But only thing is
that tail rotor will be vertical and it is
flying like this. Main rotor is horizontal,
it is flying like this. That is all. So, you
can use the same expressions for tail rotor
also. But if you want, usually what we do
is you neglect all the other loads, tail rotor
thrust only you take. But please understand,
tail rotor will rotate at a different rpm,
radius is different, everything is different.
So, you have to use those quantities in getting
the tail rotor force, that is they are non-dimensionalized
with respect to tail rotor quantities. If
you want to take all the forces, no problem
take everything, then it becomes a little
bit more complex.
That is why you say let me throw away rest
of the terms, only take tail rotor thrust
because that is good enough, which is reasonably
a good approximation. You know that I am just
using, you know torque coefficient C Q. C
Q is C P power coefficient, both are same.
One is you directly use the earlier expression.
Because this is the torque, you can substitute
F Y, I have substituted here and then used.
One way is you can calculate everything by
this approach, which is fine. And that expression
I will give you. Another way is you have torque,
main rotor torque I say convert to power,
because main rotor power. Now, I split that
term. Basically what I will be doing is, this
F Y 1 because this is the torque r F Y 1.
And this F Y 1, I have here. What I will do
is this is L sin phi D cos phi. And I will
replace this L in terms of F Z 1. Because
you know L from this expression. I will put
F Z here. I will write everything in terms
of F Z. And the drag. That is the lift and
the drag term. So, this jugglery is done.
But when I do that, I do not make any assumptions,
I will just take the entire expression as
it is. It is not that I throw away the drag
term and then I will take F Z is only lift,
then I cannot get this type of form. And then
I will add sum because the forward flight
you have the longitudinal force. So, the longitudinal
force, I will take and force into velocity.
I will take velocity is forward flight, that
is again another term. So, that is what I
will do. mu into C H is again a power term.
And I will subtract the same mu into C H.
This I will do because if time permits I can
start today, but it will not be done within
one short lecture. It takes time because lot
of back and forth substitution. And then you
will find a very beautiful expression. Some
other terms will drop out, some will combine
to give you essentially because then there
is you have to take the flight condition.
You make lot of approximation there also.
That is why I said to come to this form is
a little tricky, really tricky, you cannot.
If I give you derivative, you will not be
able to derive it, I am telling you, unless
you understand all this.
The power is split into four sub components
of the main rotor only. One of them call it
C Pi which is induced power. Induced power
is basically you want to lift the helicopter.
It supports the weight of the helicopter.
Then you have climb, that is the climb power.
Then you call it parasite drag because you
are dragging the helicopter. So, the helicopter
will have its own drag force. And that drag
force, a power is required. And that is what
is called the parasite power. And the last
expression is profile. This is purely C d
naught. Please understand. So, C d naught
is profile drag of the blade, sectional drag,
but this is drag of the fuselage which you
are taking.
So, you have split the power into four sub
components. And each sub component, you have
a very nice expression. That expression is
what is very interesting. That is given here.
C P of course, I put an approximation symbol.
Because if you remember, we had seen the inflow
curve in forward flight.
I will just briefly draw. I will put it lambda
i and mu. It was coming down. It is around
say somewhere around 0.1 mu, it starts deviating.
Look back your… That is you can take high
speed condition for mu greater than 0.1. Assume
the high speed condition and you can write
your lambda i as C T by 2 mu.
If you go back your notes, you will find that
lambda i at high speed. You can write it as
C T by 2 mu and there is factor you add for
non-uniformity. Then your induced power is
some non-dimensional correction factor, C
T square over 2 mu. But you cannot use this
when mu is zero. Please understand. This is
for mu at least more than 0.1. Then you will
have profile drag, sigma C d over 8 1 plus,
I have put 4.6 mu square, that is again some
reverse flow etcetera, if you include 4.6.
Otherwise it will become 3, when you derive
it will become 3. 1 plus 3 mu square, that
is sigma C d. C d is the drag or C d naught
you can take it. And then there is another
term 1 over 2 mu q f over A. A is the area
of the rotor disk, f is fuselage frontal area.
So, that is normally obtained from wind tunnel,
frontal area given in terms of drag. And this
is mu cube. And then of course, you have a
climb, lambda C C T, climb.
So, you see this forward flight power required…
Tail rotor power is separate which you have
to add, that is all. So, this is a very neat
expression for power. And this curve if you
plot, because you know that this is a thrust
square, anyway C T is fixed. So, you see as
you increase your forward speed, this term
will actually come down.
Now, you see if you look at the power graph.
This is the rotor power, this is forward speed.
Induced power will go like this. This is induced
because induced will keep on decreasing with
forward speed, because with mu. I am taking
starting from hover even though it is not
valid. That expression is not valid at hover.
I am taking induced power decreases. But if
you look at the profile power, this is mu
square. But please understand mu all of them
are in the range of maximum 0.5, do not take
it 1, 2, 3, 4, 5. Please understand. Here
all mu is in a very narrow range, 0.5 is very
large, actually 0.35, 0.4, that is it. That
is why do not try to put 1, 2, 3, 4, 5 like
that, you will take it 0.1, 0.2, 0.3, 0.4,
that is it.
Then your C d curve may go like this, may
be profile drag may go, because it is increasing
with mu slightly, but the square. But if you
look at this term, parasite drag, that is
mu cube. This will start like a cubic. So,
this is induced. May be I will write it here
separately. This is induced, this is profile,
this is parasite. Climb, you do not bother
about that. Now, if you add all of them. I
think I should take it like this. If you add
all of them, the curve will look like this.
It will go something like this. Maybe I should
put that dip a little down. Because you have
to add everything. This is the rotor power
in forward flight.
So, this is the parasite profile induced.
Now, you add all of them, they will go like
this. So, this is my power curve. Let us take,
we will just describe a few of the power curve.
That means, you see you fly at low power somewhere
here. Power required to fly the helicopter
is minimum at some forward speed, but hover
it requires more power. And as you go to forward
speed, the power actually decreases that is
because this power decreases, induced power
decreases drastically. But the increase of
profile power is not too much. This will start
picking up only after some speed, the parasite
power.
So, you have a minimum power at some particular
fly. Usually it is in the range of, this may
be somewhere around 0.15 something like that,
around 0.15, 0.1 to 0.2. It will be around
that. Sometimes 0.11, 0.12, it is in that
zone, minimum power forward flight. And then
for climb. Please understand. What is the
power available on the helicopter? That you
may draw some line. This is P available at
any time assuming it is sea level. But as
you go high altitude, power available will
come down. Power required because of induced
flow is more, power required will go up. Here
you will have this much excess power for climb,
if you go here it is very less. But if you
go to high speed, the power required climb
is less.
Now, you see when you want to do. Let us go
back to the autorotation. You know in autorotation,
you are descending. And basically autorotation,
the power required by the rotor is supplied
by the air in descending. That means the descend
kinetic energy is converted into rotor rotation
kinetic energy. If the required power for
the helicopter is less, then your descent
velocity also will be less because you do
not have to descend much faster.
So, if you auto rotate. You do not auto rotate
in hover. You can do, but the descend velocity
will be much larger. Whereas, if you auto
rotate at the minimum power, the descend velocity
will also will be there. Of course, he is
moving forward. Please understand. He will
come forward, but he will auto rotate, but
his descend velocity will be less. But when
he comes near the ground, usually what he
will do is he will increase the collective
so that he goes up. It is like a breaking.
He will try to go up and then he will come.
So, autorotation descend will be minimum when
the rotor power is required is minimum.
Rotor power required is minimum, somewhere
around 0.1. It is around 0.15 to somewhere
in that zone, mu advance ratio. But you see
this power curve, this raises very sharply.
You just cannot fly beyond that speed because
the power required will be tremendous. You
will not be able to do it.
So, you will find this curve will shoot up
like this. The power required will be much
more, that is number one. And your vibratory
loads also will increase very high. Sometimes,
usually you want the helicopter to fly without
any problem. You have designed for some forward
speed. You have the power, you put everything.
But when you are trying to fly forward speed,
your vibration starts increasing. Even though
I may have enough power in my engine to take
you to a higher forward speed, but my vibration
is so high I cannot even go to that speed.
So, the restriction comes from not because
of lack of power in the engine, the restrictions
comes from vibration in the helicopter.
So, these are real design challenges. The
best design is one where your flight envelope
or the max speed is restricted because of
my power is not that, so I am not able to
fly. You follow? On the other hand, it is
not like that, you may have power, engine
can give. But you say I cannot go my forward
speed is restricted because I have lot of
vibrations. You follow what I am talking?
And that is happens in almost majority of
the helicopters. Because you know that we
are all talking about mean value, they have
never mentioned anything about the vibratory
loads, because we took the average. But the
vibratory loads are the ones which contribute.
So, helicopter essentially is a vibrating
machine. And unless you go sit there and see
for yourself what that vibration is, you know
you will not believe it. It is phenomenal.
You have to experience it, if you get a chance.
But not the normal flight. If you fly somewhere
here, you will say beautiful nice flight you
had. But you go here, then only you will start
seeing the vibrations, not here. Of course,
here you may say a little bit, but that is
a different thing. But here it is most of
the commercial flights, low power required
you can fly nicely etcetera. But when you
want to go here, that is why in the military
helicopter they have to sometimes go on a
high speed. That time the vibration will be
phenomenal. So, you will be restrict. Now,
with this curve I will give you one another
interesting result which can be…
So, we take this power versus curve. What
is the speed 
for max range? If you want to maximize the
range, what speed should I fly? That means
because the fuel consumption is related to
power. So, how do we decide? Because if you
fly high speed you may say well I will go
fast, but you may burn lot of fuel. So, you
will not have left with fuel to go. So, what
is the speed for max range? So, what you do
is, simple, you take the function f. Because
power is a function of what? Power is a function
of this curve. What is flight speed for max
range?
So, you take a function. Let us say that function
F is V by P. I want to basically maxima minima.
If I maximize this, that means velocity is
maximum, power is. I am minimizing the power,
I am increasing my velocity. So, what you
do is, you take this d F by d velocity. This
will be V d U minus U by can put it, but this
is 0. That means d P by, numerator d P over
d V is… d P by d V is slope of this curve.
Slope of the curve is basically P over V.
So, you draw a line, this point. If you fly
at this speed. Because this is what? This
is P over V, P over V slope of that. Is it
clear?
Now, you do not fly at minimum speed, you
actually fly at a higher speed. So, if you
want endurance, that is long time you want
to float around, then you use minimum power.
But if you want to increase the range, you
do this. Then there is one more problem in
this which I will come to that later. Is it
clear?
Because this much, this is for forward flight,
power required range. And another one is a
climb also, there is something related to
climb. That I will ask you later. Now, there
are few directions the course will proceed.
Like what we derived in the hover; thrust,
power. Here of course, it is not just thrust
alone, we had thrust, side force, forward
force, power further moments are there, which
we will get them. But in between we have used
beta naught, beta 1 c, beta 1 s. That means,
the flap motion of the blade. How are you
going to get it? Because then you need to
know the blade dynamics. How my rotor blade
is going to respond? Because that is very
essential, which means you study only the
dynamics of the blade in forward flight, you
do not bother about anything else. How my
blade will vibrate for a given pitch input?
Given pitch input is pilot gives, theta naught,
theta 1c theta 1 s.
You assume that pilot is giving this much
value and then you have to write the equation
of motion for the blade. Now, for the simple
case we will take only the flapping motion.
Because we have you included only the flap
dynamics. Because we did not include lead
lag or torsion. So, how do you get the flap
dynamics of the blade? So, you have to proceed
in the blade dynamics separately. Please understand,
this is the one direction. So, that is required
for even getting the loads. Because if you
want to get the load, because you I have put
just beta naught 1 c 1 s. Who is going to
give me those values? Similarly, theta naught,
theta 1 c, theta 1 s and then lambda. All
these quantities are only symbols now. How
do we get those values? This is one aspect.
And when you restrict the blade motion, because
we said that the blade is free to do flap,
lead lag, torsion everything. It is a long
beam it has vibration, that means you are
analyzing only the blade dynamics.
So, for simplicity in at least getting the
hub loads, we need to have a highly simplified
model. Because I am not going to analyze the
blade real complicated, because please understand
complicated analysis is one problem. So, that
will take you in a different direction completely.
You will not be then bothered about any of
these things. I am only analyzing the blade.
But here I want loads, but at least give me
a simple model. So, what we will do is, we
will make a highly simplified blade model
to describe certain quantities. Later we will
take, but at the slight complications in the
blade model itself. That will come later,
because if you want to use this you need to
have those things.
