The equation of one blade is always written
in the rotating frame you understand, because
that is easy to identify what is the motion.
Suppose, if it is in the non-rotating frame
it is difficult because after sometime blade
may be their somewhere else. You write the
motion or the degree of freedom beta with
reference to a rotating coordinate system,
but you get the absolute values are you clear
about that hub loads are in the fixed frame,
because you want to transfer it to the fuse
large. Fuse large is not rotating please understand
there is one non-rotating frame there is a
rotating frame in the rotating frame the blade
is moving.
So, you are observing the blade motion that
is later when we go you cannot look at the
particular blade, but you see a rotor rotating
suppose, you are lifting standing on the ground
you see some rotor is spinning. You do not
know which blade is coming where, but you
see a disk, then you are looking at the motion
of the disk you will come to that later. Because
you are observing a flap motion from a non-rotating
frame you understand, but right now the equation,
which we have written is observing the motion
of one blade as it goes round and round is
it clear and it does the same motion please
understand.
That is why we wrote beta equal beta naught
does beta 1 C sign psi. So, we wrote this
that this is a and psi you know this is psi
and this is actually your capital x and this
is my this are a hub fixed this is you have
x 1 and you had y 1 and in this x 1 y 1 the
blade is moving up and down. So, you were
writing the equation of motion in the rotating
frame x 1 y 1, because in that frame only
you have the blade is flopping up and down.
How that is beta that is this quantity you
understand, but this particular this itself
this is a function of time time means, I can
put it non-dimensional time you agree? But
how I am representing it is in this passion,
this psi is non-dimensional time only you
follow.
So, the blade what happens is it has a constant
flapping agree that is called the conning
plus the value depends on beta depends on
where the location list, but we do not know
these quantities. You do not know what they
are that is what that equation is because
that depends on what is the aerodynamic load
that acts on the blade that f z and r cross
that moment you took it.
That is how you get the flap equation, flap
equation is a simple dynamical system equation
which has a like your m x double dot plus
e x dot plus k x equals some f of t. Only
thing is m c k they are not constants time
is constant, but c and k are not constants
that is all they are also varying with time.
So, please understand the difference and we
are looking at steady state response, steady
state response is essentially if you have
a fixed harmonic input that is this you are
giving see this is my equation of motion and
M F I have written below, that moment due
to flap and of course, lap number is here.
Input is theta I theta I we shown as theta
naught theta 1 c plus theta this is a non-dimensional
time. So, if my input is varying how my output
will vary this is a periodic input, because
you must have d1 some at least spring mass
damper system. One if you give an initial
condition actually, spring mass damper in
electronic it is a l c r circuit that is all.
If you give a initial condition how the response
will look like you have under damp over damp
step response you must have study very well
we will get back to later another one is if
you have a harmonic input.
That is if I have this equation 0 I put x
of sorry x of 0 some x naught and x dot of
0 as x dot 0 this is the initial condition
I C. Again analyze how the system will behave,
if this is 0 I give only f dot that is like
a pulse impulse you are giving if this is
0 you give you are giving a initial displacement
and leaving it how the system will respond.
The next part is forced vibration force vibration
this is free vibration with damping. In forced
vibration you would have studied same equation
K X is F of t, but F of t if you study harmonic
motion you may write F naught some sinee omega
t or cosinee omega t something like that where
omega is the excitation frequency, input frequency
it is nothing with natural frequency.
You may that equal to that now, what we are
looking at here is? You convert this into
non-dimensional you may have what x double
dot c over mu x dot plus omega n square x
here you may F over, which could divide by
M from you can do like this and this you may
call it some f naught some other. This frequency
is natural frequency of the system k root
of k over m this is input frequency, if these
two are equal that is what you call it resonance.
In this problem what we are analyzing is your
input is psi, which is 1 per rev because the
variation is 1 in a revolution. And you see
this is also one, because second derivative
this one damping term is on the right side
that I can transfer it left side please understand.
That damping term which is here this can come
back here, but in forward flight you also
have additional stiffness term that is why
we do not write it immediately some frequency,
but if it is an hover no external force then
you say what is the flap frequency that means,
mu is zero and you will have one. So, your
input frequency is also one if it is in hover,
but in forward flight you will have all the
harmonics please understand, because if I
put theta I equal theta naught plus theta
1 c theta 1 s sine psi cosinee psi this is
getting multiplied by with another sine psi.
Getting multiply with sine square; that means,
I am having second harmonic third harmonic
everything comes as an input. So, my response
also will have all the harmonics.
So, we are looking at steady state values
and I am not looking at the second that is
sine 2 psi or sine 3 psi, because I restrict
my response only up to first harmonic. I am
not adding the other terms I have actually,
neglected this you can have sine cos 2 psi
cos 3 psi everything you can keep on writing
that, but I am neglecting those. Now, if I
say I want to get these coefficient see this
is like a solution of the problem how do you
solve differential equation with this as the
right side and this as the input theta I as
the input how you will get the better. One
is I will go and do a numerical integration
that is the easy way, but you can write closed
form solution for this simple problem close
form means you can get an expression closed
expression and that is what I have done it.
In the sense I assumed my output is in this
form there are two ways of doing you I told
you 1 is the harmonic balance, harmonic balance
means simply substitute everything in that
equation. Collect the constant terms, collect
the cosinee psi terms collect the sine psi
terms you will have equations that is what
harmonic balance is each harmonic collects
it separately write it as one equation left
side equals right side. Another way is called
the operator method, which is simply you integrate
how you get a coefficient of Fourier series,
you integrate a F of t some integral 0 to
2 pie 1 over 2 pie equal to 0 that is for
constant.
Then you will multiply by a sinee psi that
is for first cosinee to get the second that
is an operator same thing you can do for operating
method it really does not matter both are
balancing only. One is doing integration and
get the equation another one is collect all
the terms multiply everything then write when
you do that this is what you get, because
I am not doing all those algebra.
So, I am only saying when you do that part
you will get the first if you make the assumption
that beta naught beta 1 c cosine psi beta
and I do the harmonic balance. When I do harmonic
balance please understand I am only looking
at constant term cosinee psi term sign psi,
if I have on the right hand side some 2 sinee,
sinee 3 psi etc I am just throwing it out
I do not consider them I neglect all of them
you understand. I will get these three equations
and three equations I know theta 0 theta 1
c theta 1 s these are my input theta twist
I know and what is the forward speed I know
it. Only thing is gamma I know gamma is lock
number if you know theta naught theta 1 c
theta 1 s and the operating conditions I can
get from these three equations.
Because beta naught directly because it is
independent of there is no coupling here beta
1 s depends on beta naught. So, I have to
solve this substitute here, then I can get
beta 1 s and then here is the beta 1 c, beta
1 c you can directly get it, but it requires
lambda please understand you need to know
lambda inflow. So, if you are given theta
naught theta 1 c theta 1 s mu lambda gamma
you can get beta naught beta 1 c beta 1 s
that means, if you are specified the operating
condition of the rotor I can get the flap
response. Steady state flap response please
understand steady state means, the blade will
keep on doing the same thing and that is what
you have getting from this equation. You follow,
because why I said is if you treat flap equation
al1 as an independent equation without bothering
about the helicopter then, if I want the response
I have to know all these quantities.
Then you may ask who gives me my lambda, lambda
should be known before you come to solving
these three equations please understand. Before
you solve these three equations to get beta
naught beta 1 s beta 1 c you must know what
is the lambda what is the input pitch angle
what is the operating conditions? At least
that part you must know. Because I do not
know that than I will do that part we will
come to the later that is what the trim analysis
is the entire trim analysis is related to
iterating between flap motion and then fuselage
equilibrium.
So, you have to keep going back and forth
that is why it is a complicated analysis,
even the solution procedure is little complicate
this is a very simplest that is what I am
showing. Now, I will go to a slightly highly
restricted condition that is hover same equation.
This is just a blade flap dynamics in the
study state response, if I said mu 0 just
for because I am powering. If I said mu equal
0 this is what I will have from the from these
three equation I can get what is the beta
naught beta naught is given by gamma theta
naught over 8 some twist lambda by 6. That
means, to an extend if you see this is the
lock number which is aerodynamic force by
inertia force i b row a c r power 4 over i
b, i b is the mass movement of inertia of
the blade about the flap. And then of course,
collective pitch this term looks somewhat
like c t somewhat it is not exactly, because
c t is sigma a by 2 because you know that
c t is sigma a over 2 theta naught by 3 minus
lambda by 2 in hover.
So, it is some crude way of theta by 3, but
here theta naught by 8 there is it is not
that same expression and lambda by 6, but
some close thing it looks similar theta minus
lambda. So, you know that flap coning beta
naught is related to what is the rotor loading,
if the rotor is loaded more your coning will
be more if it is less it will be less. And
then when you come to the cyclic response
that is 1 s and 1 c this is 
you see this is somewhat directly related
to this angle theta naught coning is related
to theta naught theta naught is related to
your loading.
So, you say my beta naught is somewhat related
to how much my helicopter is loaded my rotor
is supporting, if you put more weight it will
go up a little bit same operating condition
please understand it is not that my r p m
is changing fixed r p m. But when you look
at this 1 c, 1 c is related to and as from
this equation beta 1 s is related to theta
1 c.
That means, when I change my pitch angle if
I take this as my rotor this is my x this
is my y and my psi, if I change here pitch
angle gets changed here, but I get a because
theta 1 c is pitch if I increase my pitch
angle when psi is 0. That means, when the
blade is here if I increase my pitch angle
the blade will flap up here, it will go up
their not that immediately go up please understand.
Similarly, beta 1 c is related to theta 1
s theta 1 s is given here and the blade will
have beta 1 c here or you can call it beta
1 c you follow that means, when I give an
input here if I increase my pitch angle beta
1 c will become minus, minus means what it
will go up.
Because beta 1 c positive means 0 is up 180
is down whereas, I want if beta are you may
say if my theta 1 s is negative, negative
means I am reducing my pitch angle at 90 degree
then what beta 1 c is positive that means,
the blade will go up at the back at the 180
it will come down. So, there is a phase shift
of 90 degree between input to the output.
But please note this 90 degree is only for
this simple case of hover plus centrally hinged
blade it will vary the moment I have a different
flap model, if I put a spring it will not
be 90 degree it will change. If I put something
else hinge offset and a spring it will change
then how it changes depends on that is what
we will derive.
The phase shift between input and output now,
you know if the helicopter has to rotor disc
is to tilt forward down you must change the
pitch angle when the blade is here not on
the zero this is the key important there is
a phase shift of 90 90 phase shift comes because
it is a resonance. Because if you have studied
little bit more carefully your vibration part
of a m x w dot problem how the phase varies
with frequency all no matter what damping
you have. All the phase curves will cross
omega over omega and 1 they will cross us
90 that means, when the input frequency is
equal to the natural frequency. The phase
difference between input and output is 90
degree that means, if you give an input now
effect will be shown after 90 degree then
you convert it to time.
Here we know then 90 degree is basically if
I in this disk if I change the pitch angle
here the response will be here. That I can
show it in the lab with a when we rotate the
blade, because this you can see clearly only
the stabilizer bar you will not be able to
see the rotor blade rotor blade is basically
attached so, but you can get those from experimentally.
So, this is the one of the important results
of input to output. Now, if I change my model
please understand 90 will not be there 90
will change 90 will shift and in forward flight
this is in hover condition only please understand.
Now, in forward flight will it be 90 no it
will not be 90 it will be different, but then
because pilot please understand now you were
slowly pilot is having a stick in his hand
he will move the stick forward. But what he
is doing? He is only giving a pitch input
he is not giving flap response he gives a
pitch input, because of the pitch cyclic variation
the blade flaps and net result is the rotor
disk tilts forward he gets a thrust in that
direction. Now, this is also a you can take
this two terms on the left side and then you
can combine and then try to give a geometrical
interpretation. It is a little interesting
geometrical interpretation I taught I will
just show you that is all this for only centrally
hinged, because when you say centrally hinged
immediately you must know flap frequency is
1 per rev that means, flap frequency flap
natural frequency is rotor r p m itself nothing
else.
Now, let us look at because we define earlier
some different planes suppose, you put that
this is the shaft and 
this is the tip path plane this is my hub
plane. Now, what is this angle? If this is
my x direction this is beta 1 c, because this
is x and y is down and this is E direction
tip path plane is this, but we also define
the no feathering plane no feathering plane
it is a plane in which there is no change
in the pitch angle. If that is theta 1 s now
look at it because plane is like this 
no feathering plane means, I am giving a theta
1 s because in this I am looking at it this
way in this y direction.
The pitch angle I gave when the blade is at
90 degree I give a theta 1 s which is positive
you assume that you are giving a positive
pitch angle and then when it comes to 270
you give minus, because sinee 270 is negative.
So, your hub plane to no feathering plane
it will be because you are giving f which
angle like this that is a theta naught we
have increased one side you have decreased
other side. That means, in this this is 90
degree you have increased theta naught plus
theta 1 s 270 degrees you made it theta naught
minus theta 1 s that mean the plane should
tilt a little this way. Because then you are
actually decreasing the theta naught is same
theta 1 is g1 up that means, you are trying
to reduce it is angle so, you were no feathering
plane will be I mean draw like this no feathering
plane with this angle that is theta 1 s.
With respect to the hub plane please understand,
all are referred in the hub when you look
at a blade pitch angle will be what here theta
naught when you go to 90 degree theta naught
plus theta 1 s 270 theta naught minus theta
1 s. So, you want to make both of them equal
because that means, you tilt the plane a little
up as a result when you are look at that the
angle through which you tilt is theta 1 s.
That means, you are actually subtracted from
here you have added here, then my pitch angle
is theta naught in the no feathering plane
the pitch angle is only theta naught is it
clear not clear. See what happens is? You
see this side this angle is theta naught plus
theta 1 s that is at 90 at 270 this is theta
naught minus theta 1 s this is less I want
this angle to increase means, the plane must
come down here it should be so, this to line
you join that plane is this display.
So, in the no feathering plane the pitch angle
is constant in the tip path plane there is
no flapping motion all tip is going. Now,
if you look at the angle between T P P and
no feathering plane that is what beta 1 c
plus theta 1 s, this is the angle between
tip path plane and no feathering plane. Because
that is the angle beta this is the hub plane
please understand this is hub plane these
two are parallel both side theta nine minus
theta 1 s y axis x axis is here, x is this
this is x. See here aerofoil comes here I
am looking at it then what will happen if
you look at it will be like this is it clear.
Now, you see the angle between tip path plane
and no feathering plane is this in hover you
look at it beta 1 c plus theta 1 s is 0 that
means, these two planes must be parallel that
is this angle should be zero this angle should
be zero means, what both plane must be parallel.
So, what and if you look at the other this
is tip path plane and no feathering plane
in longitudinal direction, because you are
looking at x and z, if you look at the tip
path plane no feathering plane in the y z.
You will get the opposite in the y z this
is x z plane that is longitudinal plane and
in the lateral plane you will have that is
y z plane T P P and N F P.
This will be, because you are giving what
beta 1 s if you look at the y z plane this
is my hub plane and this is the z axis and
this my y direction this will be beta 1 s
this is beta 1 s. And theta 1 c will be because
you are looking this way here you would have
added theta naught plus theta 1 c here it
is theta naught minus theta 1 c. So, you will
tilt the plane like this because it is more
pitch angle. So, you will tilt it like this
which means this will be your N F P with this
angle theta 1 c. So, the angle between tip
path plane and no feathering plane in this
lateral is minus theta 1 c now, this is also
0 for hover, but please understand in forward
flight they will not be 0.
Forward flight you will have some value sitting
there and that is a function of forward speed
how much is the rotor disk loading, because
please understand this is I am again putting
it hover. The moment I go to forward flight
these two planes are different, because forward
flight I have to solve if you want I can show
the I have it no it is not here I have to
write it. If the c z is not on the shaft what
will happen is if the suppose, you have the
helicopter this is the shaft c z is somewhere
here if you have then what will happen? So,
you have to give cyclic such that what will
happen is ultimately the resultant this will
come below so, that the helicopter attitude
also will change slightly.
Shifting in the sense because this should
be right below this that is all so, you orient
this suppose orient it means how do I, because
I want this point to be below suppose, if
I rotate it like this what will happen? I
am the diagram so, the c z has come here.
Now, the shaft is like this I may be I should
put it like this I it is a little to too much
the shaft is here I go like this, this is
on 
the may be c z is here tip path plane is horizontal
shaft is tilted here. That means, hub plane
is like this which means I have to give a
cyclic input depending on the c z location
that is what happens usually, became depending
on the c z location he has to give a cyclic
that will.
If it is right on shaft no problem thrust
is you take it a hub plane in the sense it
is you will get a horizontal comp1nt that
is why in hover, if the c z is right on shaft
if you give a cyclic helicopter will start
moving you cannot be in hover. And at the
same time I do not want the helicopter to
move, but I will give a cyclic no, but if
you do clamp it that means, you do not allow.
The helicopter to move clamp what we are doing
as an experiment you give a cyclic then you
will see the tip path plane tilting it doing
like this depending on what input you give,
if you give a longitudinal then it may go,
but then that is only for centrally hinged
we can demonstrate that in the experiment.
Now, is this clear because the face shift
90 degree essentially, what the pilot does
is? Pilot does swash plate plane which is
actually if there is no other coupling etcetera
that is the control plane, which is also no
feathering plane if you do not have any other
some flap pitch coupling etcetera. So, he
is tilting only this plane no feathering plane
he tilts that is all and because of the blade
dynamics the tip path plane tilts he does
not tilt the tip path plane he tilts only
the no feathering plane that is the swash
plane. He just tilts lateral at any orientation
and as a result, because you know in hover
they were highly simplified case they are
parallel so, wherever it tilt that also goes
there.
Now, the question is you can ask what is the
time lag between is input to that tilt you
can do that if there is a time lag then you
will find pilot will give what is an may slowly
that will come that part we will study. Because
the flap is very quick it does within almost
1 quarter of a revolution substantial thing
we will do the time constant and other thing
then you will know that that is why pilot
will not feel that I give an input the helicopter
rotor takes lot of time to tilt. But that
part if you really want to see we have a small
experiment may be next class we can bring
it, but with the aerofoil and without the
aerofoil that I can demonstrate here by showing
you just a simple model we have.
I will just tilt it you will see with aerodynamics
it will respond very quickly without that
it will come very slowly. So, that seeing
is believing so, next class we will show a
video, video means you have to take the video,
but we will demonstrate that motion then you
will see this is what really happens.
Now, the major part of the flight the relation
between input flap is clear for a very simplified
case. Now, if you complicate the problem because
all this is that is why usually industry will
say what is the input face shift 90 degree,
but 90 degree is valid only for centrally
hinged with flap frequency 1 per rev that
is 1 omega. Suppose, if it is not that which
is true actual helicopter do not have 1 then
the face will change and is it same in forward
flight will the face shift remain constant
in forward plane no it will not it will change
slightly.
So, now, we will go for a slightly more I
would call it an improved model here what
I have d1 is I put 1 more spring, because
this 1s we understand this part then we can
there are lot more things even here. I put
a instead of a hinge I put a root spring which
is k beta this is the flap spring that will
also give a moment when I move it, it will
give that means, in my equation of motion
I must include this effect the effect of the
root spring. And one more thing I have added
in this diagram that I put a red line and
I put beta sub p which I call it precone that
means, my rotor blade in the undeformed position
is not horizontal it is slightly kept it angel
when you attach it itself in the manufacturing
in the assembly of the blade you are giving
a coning angle precone it is a set geometric
angle.
If you put these two what the what really
happens, we will just study that part I will
not again derive the entire equation, please
understand that is why I said once you get
one equation position vector you take again
rotate it get the acceleration get the aerodynamic
load only load what I have added is a spring.
Now, the spring will give a spring moment
which is essentially depends on how much is
the deformation that is I call it beta is
the flap angle please, understand beta is
measured from horizontal plane hub plane.
But the blade is already kept that means,
how much elastic deformation I give into the
spring is beta minus beta precone and that
much of the moment it will act on the blade.
So, my equation I will just write that equation,
because the flap equation we wrote inertia
is I b into beta double dot this is the derivative
with respect to time plus omega square we
make that approximation sine beta is beta
cosine beta this is the inertia. And aerodynamic
was 0 to R r F z d r this was the aerodynamic
moment we said inertia plus external now,
you have to put the hub moment this moment
is clockwise whereas, the spring because the
aerodynamic lift is this way. So, if you have
that because the lift is this way f z and
this is my y axis y 1 this is x 1 and this
is flap this is r. So, this moment from y
axis is a clockwise that is minus, but when
the spring deform this will give a spring
moment K beta into beta minus beta p, that
will be counter clockwise which is a positive
so, I add K beta into this is 0 that is all.
So, this is my may be I have I should put
it a little clear you will have a spring moment
this is my flap equation that is all. Now,
what happens? I have to add all of them and
then non-dimensionalize. So, if I do I b I
take it out I bring this term here that will
be beta double dot plus omega square plus
K beta into beta equals you have K beta into
beta p plus integral 0 to R r F z d r this
is my equation sorry I have to put the I b.
Because K beta over I b alright now this is
like spring by k by m. So, I am adding additional
stiffness to the blade.
So, I non-dimensionalize now the entire equation
and write a because this part is straight
forward only thing is you have to substitute
for I am not writing the entire substitute
where F z which we know earlier that is a
lift force lift per unit span put that value
and expand all the terms multiply. And then
divide by this dot converted into non-dimensional
time derivative so; you will get a omega square.
So, this I can convert it non-dimensional
because this dot is d by d t so, I put a omega
denominator another omega so, I will get a
omega square. So, when I take out the omega
square I b omega square beta double dot.
So, I will have I b omega square into beta
double dot plus 1 plus I b omega square into
beta here you will have again K beta beta
p plus your integral F z d r. Now, I make
it completely non-dimensional then this will
become beta double dot plus 1 plus K beta
I b omega square beta 
plus gamma M F this will not change M F is
what we wrote earlier. By adding a spring
and a precone I am getting a constant momentum
on the right side this is the aerodynamic,
because gamma is a lap number M F is all those
theta I etcetera, this is my now new natural
frequency in flap which is actually more than
one. Now, I will write it omega bar R F square
rotating flap frequency is 1 plus K beta by
I b omega square please understand I have
to non-dimensionalize with respect to a r
p m this is the omega square.
So, this is my and my non-rotating frequency
N R F non-rotating N R is non-rotating, because
you have a spring you have I b if you non-dimensionalize
that will be k beta by I b and of course,
non-dimensionalize with respect to some reference
omega that is why I put a bar you do not put
zero value there you put whatever. Now, you
see my simple equation tells me rotating flap
non-rotating flap is 1 plus.
So, this is my rotating flap square is 1 plus
omega bar two sorry and this is K beta over
I b omega two that is all. So, if you put
a spring my natural frequency in flap has
increased because become more than one. Now,
this is where the industry real problem how
much more than one you can keep it is a very
very see that is where I said we introduce
now two terms. One is this lap number another
one is the rotating flap natural frequency
this is natural frequency please understand.
These are the two new parameters actually,
you will find the industry in selecting the
rotor system rotor blade this is where the
major effort goes on what should be my flap
frequency it is a real challenging.
Because this is going to affect my performance
control characteristics my loads hub moment
many quantities, that is why the flap frequency
is a very very dominant frequency there is
a first flap natural frequency, first natural
frequency we will come to that influence part
later.
Now, I have modified now i have to get again
beta equal to beta naught plus beta 1 c plus
beta 1 s all those things. I should solve
for beta naught 1 c 1 s again I do the harmonic
balance may be I will write that part.
This is omega bar R F square beta naught equals
minus 1 beta p this I will send you also plus
gamma I have d1 some simplifications here,
theta twist mu square over 60 minus lambda
over 6 plus mu theta 1 s over 6. This term
is identical to I think that earlier term
only thing is I have put some point 8 I added
because that comes, because here this is the
gamma part is almost same because 60 you add
point 8 location. So, you will get 1 plus
mu square you take out in this, because you
add 1 over 6 minus 1 over 6. So, that is how
you get that mu theta 1 s over six minus lambda
by 6, but then the next two terms is omega
bar this is a little different minus 1 beta
1 c is gamma 1 over 8 theta 1 c minus beta
1 s into 1 plus half mu square minus mu over
6 beta naught.
And then the next one is omega bar R F square
minus 1 beta 1 s is gamma 1 over 8 theta 1
s plus beta 1 c into 1 minus half mu square
plus mu over 3 theta 0.75 minus mu over 4
lambda plus mu square over 4 theta 1 s, n
f p is mu square minus 
that expression any mu square theta. Now,
this is just a modified form you have to solve
this equation now, if you are looking for
you will see there is a modification in the
beta naught beta 1 c beta 1 s, because of
change in the flap frequency. Now, you have
to solve this equation to get your theta naught
sorry beta naught beta 1 c beta 1 s.
Now, if you take the hover case because again
I am coming to simple flight condition which
is the hovering flight condition, because
then you said mu is zero. If you do that let
us look at the only the beta naught part that
is the coning beta naught becomes I am because
your mu will go up all the mu will go up.
You will have omega bar R F square, because
from here you get a very interesting result
divided by omega bar R F square into precone,
because this is a precone term is there plus
gamma over omega bar R F square into you will
have theta 0.8 over 8 minus lambda over 6.
Now, there is something called ideal precone
that is 
what should be my how much I can set, because
I said that initially blade I attached it
at that location what angle I would prefer.
If you put at that angle the blade will not
experience any moment at the root even though
I have a spring, but that is set for only
one particular flight condition this is called
a ideal precone why you want to have a ideal
precone?
Because usually all the blades there are not
hinged like this they have a spring spring
means, it is a it is again an idealization
real blade will be different, but it can be
made equivalent to this type of blade. Because
if I give you the flap frequency of a blade
rotating flap frequency you immediately can
find out what should be my K beta, if I know
the mass of the blade if it is centrally hinged
assumed that it is centrally hinged, if you
know the rotating flap frequency you know
what is my K beta. So, you can get that this
kind of equivalence you should be able to
handle it easily now if you give a precone
what is the effect of that precone?
That is let us look at the what is the hinge
moment that comes that is if you have a system
spring mass and F of t. What is the force
that here? The force that axis is only k x
that means; the moment that acts at the hub
please understand this is the flap moment.
Flap moment acting at the hub is I am taking
mean value, because beta naught 1 c 1 s I
am not taking otherwise I have to take K beta
beta minus beta p I mistake like that here
I am taking only the mean part of it 1 c 1
s. So, the mean value mean hub moment is K
beta into beta naught minus beta p, this is
the mean hub moment now I substitute this
beta naught here.
If I substitute that my hub moment mean hub
moment will become that is K beta you will
have omega bar R F square minus 1 by into
beta p 1 minus beta p then of course, the
other term plus gamma over into theta and
this it is a beta p because a beta p over
this beta p only that is all. So, you subtract
this will become K beta minus 
plus this value into theta 0.8 by 8 minus
lambda over 6, because this is minus beta
p. Suppose, if my beta p is equal to gamma
into theta 0.8 over 8 lambda over 6 if, that
means, my moment is zero. Even though I have
a spring my mean hub moment is zero, but I
will have some beta 1 c and 1 s they will
ask later that means, what I am doing is in
my design.
Because if I get root moment my design has
to be properly made, because it has to resist
the sheer force and the bending moment like
a cantilever beam, if you are taking a beam
you design what is my root every section you
design for shear and bending moment. If my
bending moment is there I have to design it
a little thicker I can reduce the mean load
only mean load I can reduce it by giving a
precone now you know why rotor blades are
given slight precone. If it is a only when
you have a if it is a root moment because
the pin there is no that moment is zero centrally
hinged no problem, but the moment you put
a spring which is like it is like a cantilever
first.
So, you want to give a mean moment slightly
bring it down rather than otherwise my blade
has to be designed little thicker at the root.
So, the whole idea of putting the precone
is reducing the mean flap moment acting at
the hub slightly less. And this is called
this angle because please understand you will
not eliminate the moment at all flight condition
because this is valid only for one particular
value. If you start flying at a another speed
forward you will have different flap then
you will get a moment usually, the precone
they give about two degrees 2 or 2.5 something
like that that is all whereas, the flap angle
may be of the order of around 5 degrees.
So, whereas, if it is fully tilting to 5 you
all of you to tilt only through 2 3 degrees
rather than full the 5 degrees, because then
the design that is why all are very very important,
but it has a effect on damping also that part
we will come later. The stability and other
aspects right now the precone is given to
relieve hub mean moment slightly otherwise
you have to design a bigger because the centrifugal
force you will try to bring it down in the
aerodynamic force we will try to take it up.
If it these two moments are equal it is kept
at the same location that is all. But if it
is slightly up that is why you can have if
it is fully up it will try to bring it down,
if it is like this it will take it up.
So, that is why usually you do not look at
flap deformation because it depending on the
flight condition it can come down it can go
up anything. Because you can have a if the
coning is too much that means, your lift is
very large because that is why going up if
the coning is very less means my lift is less.
So, the coning angle is changing with your
weight condition, that is why all this are
just give some two degrees you relieve the
hub moment it is purely from a designed structural
design point of view. Of course, air aeroelasitcally
it will have an effect that part will come
I hope you will be able to do it or otherwise
you will forget aero-elastic effect of beta
p on the blade some stability etcetera.
Now, I think I will leave you now because
this is the next part. The phase shift how
much because now you have change the frequency
what will be the change spaceship that part
we will do the next class.
