So, I will keep this we saw the ideal precone,
that was what we were discussineg in last
class. That is essentially, the root moment
is K beta beta naught minus beta p this is
the ideal precone. If the precone value is
this is ideal precone then, we have already
kept it at that then there one be any root
moment, because root moment will be zero value.
That is the idea of a natural design, what
they do is they will keep a little bit of
precone angle. So, that you really in the
design, structural design it is purely from
a structural design point of view, you reduce
the moment that comes at that blade root.
Because these are all design features plus
they also have certain influence on the stability,
because if you remember last clas,s because
I am just going back.
We wrote K beta into we substituted for beta
naught and we got this equation 1 by that
this, we wrote last time beta ideal minus
beta p. And you can replace that K beta directly,
because that is again written in terms of
been, we know omega bar r flap square is 1
plus omega bar non rotating flap square which
is I b omega square. So, what you were doing
is you now subtract from here this will be
I b omega square into omega bar R F minus
1. So, you substitute this here. Now, everything
will be in terms of over sorry minus 1 and
I b omega square beta ideal minus beta p.
So, you see automatically, whenever the flap
frequency rotating flap frequency is equal
to 1 root moment is 0, you would not get any
hub moment. Whereas, the moment you have a
rotating flap frequency, greater than one
then you get a hub moment, but beta ideal
is not equal to beta p. Now, you may have
any frequency, but if ideal is equal to beta
p, then you will not get any root moment,
but it is not that you do not want to get
any root moment, see it is a very conflicting
requirements. If you want to maneuver you
need to have root moment, because you want
to generate moments, other you will generate.
Later we will show you will generate only
the moment on the vehicle only by thrust vector
that will have some value.
But if you want to have highly agile, you
need to generate more moment, then you need
to have hub moments from the rotor blades.
But at the same time, you do not want to blade
to have lot of root moment from a structural
design point of view, on one hand you want
highly agile, but on the other hand you do
not want the structure to be very big. So,
this is a trade off finally, you that is why
the design of rotating flap frequency to have
a particular number is a very very interesting.
And very important in the helicopter design,
what flap frequency you will keep for your
helicopter.
Next we will look at the, if you have the
rotating flap frequency other than one, what
will happen to the phase difference between
control input and the output. Because earlier
when the centrally hinged blade, we said that
the phase is 90 degree. Now, what will happen
is if this frequency is more than one, that
phase difference will reduce it will not be
90 it will be less than 90.
That means, we will see what that values,
we will derive only for the hover case, but
you will find that, if you restrict your case
to hover. The relationship between this is
I am writing it from the nodes which I wrote
last time. And then R F square minus 1 divided
by gamma over 8 minus beta 1 c sorry beta
1 s equals theta 1 s. This is the relation
which you reduced from last time, I wrote
the long expression from there if you put
mu 0, you will get this expression particularly
in hover case.
We are looking at hover first, you can solve
for beta 1 c and beta 1 s from these 2 equations,
because your control input is this and this.
Now, if you write that I will just briefly
see from these two equations you see that
there is a coupling, coupling between longitudinal
and lateral. In the sense, if I give theta
1 c I am having these two, if omega R F rotating
flap frequency is other than one. If it is
one you got this directly, because this we
got earlier, if it is not you are introducing
coupling between longitudinal and lateral
motion of the rotor disk.
And if I you can solve this, I am just writing
this result that is theta 1 c becomes minus
theta 1 s plus minus 1 over gamma over 8 theta
1 c divided by 1 plus omega bar R F square
minus 1 over gamma over 8. And similarly,
beta 1 s equals theta 1 c plus omega bar R
F square minus 1 over the denominator is same
minus 1 over gamma over 8. Now, this is my
the moment in hover itself, I am changing
my flap frequency other than one. So, I am
coupling now whereas, when the flap frequency
is one, you know that theta 1 c gives you
beta 1 s theta 1 s will give me beta 1 c that
is all.
But most of the rotors have flap frequency
other than one, that is actually, more than
one. Now, usineg these relations you can find
out, what is the phase difference between
input output. And that I leave it to you as
an exercise, because if my input is theta
equal to theta bar cosinee psi plus psi naught
and my output I call it beta beta bar cosinee
psi plus psi naught minus delta psi. So, this
is my input, because you know theta naught
is different we are looked the cyclic inputs
affect my rotor tilt, but how much it affects
it is like this, if you pilot gives an input
in the longitudinal direction, thinking that
he is tilting the rotor disk like this, he
wants to go forward what he will do is he
will also start rolling, not only into pitch
you will start rolling. You will start moving
left or right depending on the coupling.
So, that means, the two axis are not independent,
they are coupled which is difficult to fly.
When you really want to go in forward direction,
you will start going in some other direction,
if you want to give pitch you will roll. But
you will also pitch it is not that, you do
not pitch you will roll. And particularly,
in when we I hope whether, there will be some
time for analyzing the flight dynamics. The
inertia of the vehicle is different, in pitch
direction and roll direction pitch is it larger
roll it is less. Now, if the inertia is less
what will happen is if pilot gives an input
for pitching down, actually he will roll much
more than what he intends to do in pitch itself.
That is because of the less inertia in the
roll direction. Now, this coupling is very
difficult for the pilot to fly. So, what do
they do, because one input gives me two outputs
basically, one intended another one unintended,
that means, I have to go either compensate
for that, this is where the flying requires
skill. Now, you can have automatic flight
control system, etcetera etcetera many things,
but that will aid the pilot to in assist him
slightly. This is one of the key, outside
the difference between an aircraft and the
helicopter, one of the key in flying the machine.
Now, you see if the input is I can write theta
1 c is what theta 1 s.
Similarly, beta this is the phase, usineg
this equations you solve I will just write
the answer, because that is a algebra. So,
which you can try yourself, that is the input
theta bar is actually the magnitude of input,
that is under root and beta bar is beta 1
c square plus beta 1 s square half it is basically,
the amplitude of your input. Because you can
see, if you expand this theta bar cosinee
psi cosinee psi naught cosinee psi is time
varying that theta bar cosinee psi naught
is basically theta 1 c that psi naught is
some number.
Now, the phase is actually tan delta psi is
gamma over 8 minus 1 and this can also be
written as delta psi is 90 degrees minus theta
which is 90 minus tan inverse minus 1 over
gamma. Because you just put 90 minus that
way that ratio gets changed, because you can
write this itself as tan 90 minus theta, just
then you know this will be 1 over tan theta
right; that means, tan theta is omega bar
R F square minus 1 over gamma 8. So, that
is how you just because normally, you write
the face is 90 degree when this is one. Now,
you see two parameters are controlling, one
is the flap frequency another one is the lap
number, lap number you know that it is a ratio
of aerodynamics over inertia of the blade.
So, both of them influence the coupling now
just for some number, if omega bar R F is
1.15 and gamma is 8 then delta psi will be
72 degrees. And our beta bar by theta bar
that is is the magnitude of the input to the
tilt. This actually, is 1 plus omega bar R
F square minus 1 over gamma over 8 whole square
power minus half. Usually, you will find this
reduction is not a lot, that is the magnitude
of the flap and the input is there is slight
change of 5 percent, because if you take it
as 1 minus 1 0. So, beta bar is theta bar.
But if it is 1.15 you take a square you subtract
1 gamma 8, you take again square and then
bring it it will be about 5 percent 6 percent
change whereas, here this change is substantial.
So, what happens, if pilot gives a longitudinal
he will roll. Now, what will they do, because
the pilot how this is done is there is some
control rigging, they call it control rigging
that means, pilot will still move stick forward.
But the input that goes to the blade is adjusted.
Such that, if you have the rotor disk, if
this is x sorry x this is psi, this is a 0
degree 90 degree 180 and 270. And you are
flying forward velocity is coming just I am
saying, because forward flight also will change,
we are doing just hover case. If we here theta
1 s what it will do, it will go and tilt here
this will go like this. So, what normally
done is when a pilot gives, he will give a
input somewhere here. So, that this rotor
disk tilts so he will go forward. So, try
to reduce the coupling, that rigging is done
in the mechanical arrangement of your control
rods. Now, you imagine that design, the layout
depends on what is your flap frequency.
So, it is everything is linked, even your
design, layout many of this things link to
flap, otherwise you are increasineg the load
of the pilot, because he still has to do lot
of control. Now, this phase angle is it constant
that 72 degrees is it whatever, we got for
this is it fixed, no it may vary with forward
speed.But control rigging is done for one,
that variation is not for very large, that
is why it is done at one value. And then just
leave it after that pilot has to fly, if there
is a coupling in some other forward speed
then he has to adjust. So, the flying he goes
through a training. So, the key aspect is
coupling between longitudinal and lateral.
So, when now you imagine, if you want to do
you have done in aircraft and if you have
gone through the first level course, longitudinal
dynamics separately. You do lateral dynamics
you do separately, but in the case of helicopter,
because most of the people who are trained
is aircraft first. Even here, let me analyze
their longitudinal dynamics of the helicopters,
even though it is not right. You have to analyze
the full air full helicopter. So, industry
for ease of understanding, you can do a little
boy of the decoupling, but otherwise it is
not possible to decouple, the full longitudinal
lateral degrees of freedom. So, you have to
solve the full 6 degrees of freedom all coupled
in the flight dynamics.
Now, control becomes more complex, because
usually when you decouple all this things,
you try to understand each subset. And then
make some design modifications, such that
the vehicle is good. Here, if you do something
in one it is going to affect another thing.
So, that is the coupling plays a dominant
role in the helicopters, how do you reduce
it, but of you reduce it, become you say that
why you want to have make it this is one that
is wonderful, but then the problem is. Now,
what we are going to discuss, what happens
to the hub moment, because till now, we have
studied thrust side force, longitudinal force
and the torque, I said we keep the hub moments
for a later thing. Now, let us look at the
hub moment, which arises due to flapping aerodynamics
flapping motion, I will call it flapping motion,
because you have to take the inertia also.
Now, how do you get the hub moment.
So, this is the we have our model is centrally
hinged, blade with a spring this is beta,
one way is you calculate like we did the inertia
force, aerodynamic course integrate and get
that. Another way is what is the deformation
of the spring, deformation of the spring is
we said K beta into beta p this is the moment
due to the deformation and that is acting
at the hub. Now, you see there are two ways
of calculating the hub moment, even shear
also this, if you do vibration you will learn
later. I can get the aerodynamic load, inertia
load integrate the whole thing come to the
root, give that as a moment that is one way,
another way is I look at what is the root
deformation.
I just take that multiple by those spring
concerned in this case, it is much easier.
Suppose, if it is a flexible structure, I
want to know the root bending moment, root
shear force how do you get, I am just throwing
it to you to think about it. But here we take
only this, because there are two ways of theory
like this, I am just giving you I am deviating
a little bit.
You take this as a cantilever beam; you have
learned some loading, I want to know what
is the bending moment and shear at the root.
May be shear force some bending moment, how
do you do, you integrate you will get that
you put it, that is one way that means, you
are integrating the loads. But you also know
bending moment is E I, what this is also and
the sheer force you have another d over d
x of maybe E I is a shear force simple. Now,
if you know w, we can get this also at any
location, you take the slope you take the
moment whatever.
So, this is the another way that means, you
solve for w, you solve the deformation after
that put it, one is take the applied load,
please understand. Now, you apply the load,
you integrate the load and you get the bending
moment and shear at some section. Another
one is you solve for deformation and then
take whatever the standard, in aero-elastic
problem what happens, if the load is a function
of deformation, please understand the load
what applied load, what I put it is a function
of deformation. How much it deforms that means,
what you have to do, you have to solve for
deformation to get the load, but you need
to know the load to get the deformation. So,
this is an iterative process.
So, usually we do not go and calculate bending
moment and shear by this approach, because
of the simple reason you are taking second
derivative all approximate you are doing numerically.
Whenever, you do differentiation you may have
error. So, when you want to get the loads,
you simply integrate the aerodynamic and inertia,
but if you want to get the deformation. Because
you also need to know the deformation, then
you have to solve the full problem. So, it
is like here what we are doing is we are not
integrating the loads like, what we did for
thrust and other things, we simply say what
is the deformation at the root it is easy
for me.
But only thing is I must know beta exactly,
here I know that I am just assuming that beta
is known. So, I take it I take this is the
flap moment acting at the hub at any time
because, beta is a function of psi. So, now
I want to get this is my rotor disk, this
is my helicopter and this is my y axis this
is my x axis the flap moment, when the blade
is here the flap moment is about this axis
right, about the this is x 1 this is y 1 this
is flap moment, which is this. So, I must
take component of that flap moment, along
y and x direction that is the body hub hub
direction. And then I must add the values
due to all the blades and take the mean value
that way I get a hub moment, what we did was
we get all the forces on one blade. Then we
added all the blades assuming that the same
the blade behave the same way, here also it
is a same thing.
Now, let us write this quantity that is the
moment, flap moment. I will write it flap
moment I b omega square omega bar R F square
minus 1 into beta minus beta p. So, I am converting,
the spring stiffness directly in terms of
flap frequency, rotating flap frequency. This
is a modification just because you know that
what is K beta. Now, usineg this, I will go
get my moment, because this is my psi. So,
this angle is also psi I take along x and
y what is the moment.
So, I will have, M y will be my moment here
pitch moment. And m x will be my roll moment;
this is M x this is M y in the hub system.
So, my M y becomes, please understand I am
now doing a M which is the flap moment into.
This is pitch, pitch moment and similarly,
roll moment M x N over 2 pi integral 0 2 pi
M this is roll. Please understand these are
at hub, because you know that a flap moment
how much the spring is deforming, what is
the force this is very similar simple problem.
You also you know that, if you have a spring
match whatever, force excreta, everything
acting here f m k. If you know what is the
force that acts, it is just k x that is all
same way, here is the force here it is the
moment. So, k beta times the deformation and
k beta is replace by this.
Now, you put it here and you know beta is
beta naught 1 c cosine psi plus 1 s sine psi,
because we are truncating rest of the quantities,
we are not considering only first harmonic.
Now, you substitute here and then integrate
the whole thing, you will find beta naught
cosine psi beta naught is a constant, it will
go to 0 and then beta 1 as sine psi cosine
psi that is again sine two psi integral that
is also 0. What you will have is only beta
1 c that M is this expression; you are substituting
for beta this expression. You will find only
beta 1 c will remain for pitch moment. And
for roll moment only beta 1 s that expression
is I will write it here, because is now the
non dimensionalize it may be I erase here.
You non-dimensionalize your moments, that
is C M y like c t we will put M y over rho
phi R square omega R 
and M y, you will get what that is minus N
over 2 I b omega square, because here I am
just substituting this quantity and integrating.
You will have omega bar R F square minus 1
beta 1 c over rho pi R square omega square
R square R. Now, this quantity is 
you knock out this terms and here N by 2.
What is that, what you have there will be
minus N I b over what is that 2 rho pi R square
R square R into omega bar R F square minus
1 beta 1 c what you do is you pi R square
you put N C multiplied by R C R in the numerator
denominator you multiply by again C R.
So, you will have to pi R square, you will
have C R and then what that rho C right, no
there is a when R square R right am I right,
sorry right. Now, what you do is sorry, I
will put the rho this sigma. So, you any way,
you will have this term, you will have minus
sigma rho C R power 4 right. That is I yeah
sorry, rho C R to the power 4, you multiplied
by left curve slope a and divide by a, then
this factor is rho a C R to the power 4 over
I b this is gamma. So, you will have minus
sigma a over 2 gamma sigma a is the this is
gamma and 2 into you will have omega bar R
F square minus 1 beta 1 c. Let me simplify
and write it.
So, your C M y becomes minus sigma a over
2 omega bar R F square minus 1 divided by
gamma over 8 beta 1 c that is all sorry sorry
not 8 I am sorry gamma. Similarly, you can
write for M x here what will happen is beta
sine psi. So, you will kind only 1 s will
come. So, that I you will just write the expression.
So, your beta C M x becomes sigma a over 2
omega bar R F square minus 1 over gamma beta
1 s, these are my hub moments. Now, there
is a little thing further to this, we got
the hub load all the 6 components this is
what I gave in my notes that two page.
If you know the value of these are all rotor
configuration, omega bar R F square flap frequency
is a rotor characteristic, gamma is the rotor
only thing is depending on your beta 1 c or
beta 1 s how much movement, you are generating
at the rotor hub due to the rotor. Now, how
much you generate at the fuselage c g, because
fuselage c z you will get the hub forces,
they will also get transformed into the fuselage
c g hub moment. Also will get directly, transformed
what is the relative order. If you look at
that particular part there is a slight derivation.
Because that is again highly approximated
please understand, because you know the c
t, you know c h, you know c y, assuming that
wherever is the c g in your fuselage very
very simplified form I will show that diagram.
Moment about the hub center, because these
are all for controls, flying, you need to
know that value. You assume that the C G is
here rotor hub is at a height h just I am
putting it right on shaft, it need not be
straight on shaft, it can be anywhere a little
displaced those things, will come when you
do the actual problem. Now, you have a hub
H longitudinal force M x which is the roll
moment. Similarly, you have y which is the
side force, which is the M y which is the
pitch moment. And thrust and the torque, these
are the hub loads, hub loads due to the blade
operating condition.
But what happens to this, if you look at this
because I have written some form to come to
this. See if you want to know the pitch moment,
the thrust is what directly, passing through
the C G, because I am saying C G is right
on the shaft. And thrust is always I defined
on the perpendicular, through the plane of
the hub. And it is through it is not going
to give me any moment. But what will give
me moment is the H and the Y side force and
the longitudinal side force, capital H into
this height, that will give me a pitching
moment. Similarly, the y force and to this
height will give me a roll moment.
So, at the fuselage C G, I generate moment
due to two sources, one direct moment another
one is due to the forces, but this particular
thing which I have written here. In hover
see it is very interesting, in hover we can
reduce it, but we make again assumptions.
That is what I am telling lot of assumptions
are made, in hover if you look at what is
my longitudinal force coefficient C H, because
we got that expression last and that long
one, which I gave you that sheet. You can
say approximately, that is thrust into beta
1 c approximate, because you have to replace
theta 1 c is beta 1 s that kind of a regulation
you have to use it to get that approximation.
It is again I am telling this is obtained
from your that page, which I gave you in that
printout.
Now, if you use these expressions and then
calculate net moment at the C G, please understand
net moment at C G. One is hub moment another
one is due to thrust, because you see it is
converted that H and Y are converted in terms
of. This is the in-plane hub loads. Now, if
you use the omega R F is 1.1 or 1.05 whatever
or 1.15 is a just slightly higher value. You
will find this value is much higher than this
value, if you use. So, this is direct hub
moment this is through tilt of the thrust.
So, you have two sources due to tilt of the
thrust vector, the moment you get is some,
but direct hub moment is some sometime two
to four times greater than the effect due
to tilt. Because C T value, you can substitute
sigma, you will substitute a you substitute
H over R you take you will find.
Now, you see if pilot wants to do a good maneuver,
which moment is good. Suppose, if you make
my flap frequency is equal to 1 that means,
this term is done, I do not get any hub moment.
He has to fly only with this C T only by thrust
tilt that is why in helicopter, you do not
go to 0 thrust condition. That is the inverted
loop when you go 0 G, 0 G means 0 thrust that
means, he cannot generate any moment that
means, he will lose control over the vehicle.
That is why helicopters particularly, they
are two bladed where it centrally hinged;
you do not have any moment. You would do not
fly low G, because you do not have the moment,
that is why they try to put some spring. So,
that increase this the control. But then,
if you keep on increasing the flap frequency
it will become highly sensitive, because any
small change immediately it will start rolling.
Now, this is the trade off, what you would
like to have whether, you want a sluggish,
but stable vehicle or you want highly sensitive,
but highly sensitive means again it is subjective.
So, when they go through the flight they always
have the pilot. So, pilot's opinion plays
a major role in then actually, drew some diagrams
in which all the pilots, different pilot then
they come oh this is convenient. This is I
can handle it, but then experienced pilot
will be able to his opinion will be different
from a new pilot.Once as you gain experience,
it is a I can handle it, but new pilot will
say it is very difficult. So, these are all
the tradeoffs, which one has the finally,
they will says all right. Let us decide on
some combination of in the sense the flap
frequency gamma how do we play.
Flap frequency is most dominant, usually it
is kept you will be surprised omega bar R
F for articulated, articulated rotors it maybe
1.02 to 1.03 somewhere around there 0 2 0
3 maybe 0 4 is a slightly higher, for hinge
less omega bar R F is 1.07 to 1.1 somewhere
around them. Now, you see articulated rotor,
because you have a hinge, but you do not have
a spring. So, the model is little different
that we that part, when we do the next section
we will learn. You see the flap frequency
difference this is the high flap frequency,
that is why I am saying I have written of
two two decimal points, industry will calculate.
You will be surprised; they will want to go
to at least three decimals two to three decimals
in flap frequency.
Why because if I square it 1.07 becomes what
1.14 approximately. You subtract 1 that means,
one is gone the leading term is gone. So,
what is left is only 0.14 so whether it is
1.15 1.05 1.07 those numbers become that one
is not dominant now. So, even if you change
your flap frequency little bit, your hub load
is change by substantial. So, you will see
very interesting, the change in flap frequency
here what is the it is almost 5 in 100 right,
that is what 5 in 100.5 percent right here.
Whereas, if you substitute the square and
then get the hub moment, that is one point
what 0 2 square minus 1, this factor will
be this is what .014 to 0.4 right 0.04 approximately.
Whereas here 1.07 square minus 1 this is 0.14
three times they changed whereas, flap frequency
changes, you see only new 5 percent, I have
changed my flap frequency by one percent two
percent, but I changed my hub loads like,
three four times this is a very very critical
change in the helicopter another goes down.
So, you have to integrate from minus r to
plus r that is why that stabilizer bar part
you have to do that.
This is for each blade is independent; it
is not the other blade is dependent on the
motion of the one blade, if you do that that
equation will slightly change. So, is it clear
because you find flap frequency is a very
very dominant parameter, really influence.
So, industry spends enormous time, because
in deciding and then going through the iterative
thing, several design. Now, imagine you are
actually, making a blade compulsive blade
with the hub design. You must know the flap
frequency to the third decimal, usually they
will say 1.092 and 93 something like that,
you may wonder what is the third decimal going
you know in the engineering thing, you will
say it is all right. Even 1.09 or 1.1 it makes
a lot of difference.
So, this is 09 do not round it off to the
next decimal, only for flap that is why I
am saying in the helicopter, certain things
are really critical, then the criticality
of each feature gets changed or reduced just
for the blade alone. So, the design you will
be surprised, they may do the blade frequency
calculation almost every day like, you are
an engineer you keep doing change, some small
stiffness change go read run the whole program
and then get the flap frequency. So, everyday
your job will be get the flap frequency, flap
frequency and on force lead lag keep plotting
that is all. Because that is very very important,
you may find what is that I am doing everyday
same program, I have running change some small
number get it, but it has to be done that
way.
Finally, they freeze this is the number, but
now you see that is a theoretical calculation,
then you do experiment of course, experiment
what you get you take it, because you cannot
do much about it you cannot change anything.
So, in the case of of course, a l h it is
around 1.093 or 9 1 something like that flap
frequency which is high why it is high they
said.
Now, you find that when i have my flap frequency
high, I get a good control moment, but that
the same time that means, I am transferring
lot of loads from the rotor to my fuselage.
So, I will transfer more vibration also. So,
if you have less small flap frequency, the
load transfer is less vibration is also may
be less and you know, but control is also
less agility is less. Now, what is the trade
off. This is where the design goes into several
iterations and once industry has gained experience,
in a particular design they will stick to
it, because they know that this a this kind
of design, will do this effect and I know
the sequence all right. Now, with this I close
the fla, simple flap motion next starts the
trim.
But I will start that in the next class because,
I want to show one small demo, how see you
have a blade of course, it is a centrally
hinged. How it will respond, because we said
if I change some body motion it is rotating.
How fast or how slowly it will respond, that
is why I thought I will show you a small video
clipping, it is not video clip sorry actual
model, we have a model and that will be able
to see seeing is believing. So, I will remove
this, what is it here this is better. Now,
you rotate what you can. So, now this is spinning,
if I tilt you see it takes slowly, that see
this line the rotor disk, see rotor disk is
perpendicular to the shaft. If i tilt the
shaft it will not immediately come you, see
it takes time to come and this particular
delay, please understand see if I tilt see
this delay takes time.
This does not have any aerodynamic, only two
masses and just like a stick, which is spinning
this is called the stabilizer bar. Now, you
see this is has a inertia it keeps rotating.
So, even I tilt my shaft it takes some time
for it to come to the other position, this
gives a feedback in my now, you have to put
the. Now, you put the just a blade you see
the response, no only one of them it will
be very fast there it takes some time for
it to come.
Now, you are actually it is only inertia in
that case, now you add aerodynamics that is
like you are having the aerofoil, if I tilt
you see it will immediately come the time
it takes for the rotor to respond. It is very
quick, particularly in flap motion that is
why even though there is a lot of dynamics
involved in this. The pilot will usually,
feel that he gives an input, it does not mean
that, it takes lot of time to tilt or there
is no that sluggishness will not be there,
you will feel that it is acting quickly response
is very good. It is a very interesting that
is why because this is like a gyroscope, please
understand right. Now, what we have done is
we just spinning it.
Now, what we did was we had put two aerofoil
shapes, just they generate lift now you see
you rotate it, see if I tilt now see it is
very quick whereas, in the other case, it
took some time to the aerodynamics alone influences.
The response you can actually, write a equation
for this type of dynamics, just a rod you
are just rotating and it has do this response
what is the delay. But if I put to just aerodynamic
lifting surface, very simple model whatever,
we have done how quickly it response. And
the same thing happens in the rotor blade,
because it is a rotor blade you give an input
it quickly response.
So, it does not take lot of time, pilot gives
it will do slowly it will be pretty fast.
Now, we were I will just briefly mention about
the I think this is enough the trim part.
So, I will describe the trim in the next class,
what is that procedure it is only the procedure
and final results I will show for a general
helicopter system, after that the subject
can take in different directions. Because
it can be only blade dynamics, please understand
not flap dynamics blade dynamics, that means,
you do flap you do lead lag, you do torsion.
And then you do combined coupled, that is
a one direction of work.
Another one is you take only the flap, but
include fuselage motion that means, coupled
blade fuselage dynamics, you follow and that
has it is own set of problems, like you have
flap motion coupling with the fuselage, lead
lag motion coupling with the fuselage, each
one has it is own stability problems. And
then flight dynamics is rotor and the fuselage.
Now, you see you can pursue your course or
the lectures in various directions, because
everything is going to be complicate, because
it is not that I can write an equation. Only
for blade isolated flap I can write, isolated
torsion I can write, that I will cover after
the trim part. Then we will it is pretty much
like, what are the various types of problems
you face in helicopters, that is all it is
mostly descriptive with some mathematics,
but you cannot derive the equations, because
it is too much for us.
