Today, we will start with the section, which
is a helicopter trim or you say, equilibrium,
that is, if the helicopter is flying at a
particular flight condition, what should be
the control input, that is what you call the
trimming of the helicopter. Now, for that,
pilot input will be evaluated theoretically.
How will you get that trim from what we have
learnt till now? Because we have learnt inflow,
first inflow and then inflow was related to
thrust, so this is one equation. Rotor inflow,
of course thrust is their key, you have forward
flight condition, attitude, everything is
there and the 2nd one is flap response. For
this your input is pitch angle, blade pitch
and the 3rd set, which we obtained are actually
the hub loads, rotor hub loads, but the hub
loads are in terms of flap, in terms of blade
pitch and inflow, everything is there in the
hub loads.
Now, if you want to trim the helicopter, that
means, you must know what are the all the
external loads, that act on the helicopter.
You will have, in terms of loads if you say,
it has several components, one is of course,
main rotor loads, main rotor that means, load
means, all, all the six components. And then,
you have tail rotor and 3rd one is, if you
have any vertical fin, so any vertical fin,
that load and then, horizontal surface, these
are all fixed. Then, fuselage, fuselage aerodynamic,
fuselage loads you can say, fuselage means
one is aerodynamics. So, you may say fuselage
drag because we normally take only drag, but
if you want to do, you have to put internal
to take a particular point and evaluate the
loads at different orientation, different
mark number. So, the fuselage loads is 1,
aerodynamic loads and then, the last one is
the weight also, that is gravity.
Now, what we obtain is only for main rotor
hub loads, we have not; we do not have any
of these things, yeah. The shaft tail rotor
can be mounted on a vertical fin that means,
it is like actual conventional aircraft and
the fin can give its own load. So, that is
why, but if fin is not there, then you neglect,
that horizontal surface, you will find, you
know this is a very, some helicopters have
full big ones, some will have very small one,
some may not have horizontal surface. And
of course, fuselage drag you have to do from,
then gravity, gravity the CG.
Now, once you have the loads you have to transfer
all the loads to the center of mass, but that
means, you must know the location of center
of mass for a particular configuration, means,
you have the weight of that, how much fuel,
how much, everything. And then, write the
six equations of equilibrium, three force
equations, three moment equations, but they
will be in particular coordinate systems,
whether you want to write the equation in
a earth fixed coordinate or it is the body
fixed coordinate because you have to be very
clear about that. All these loads, they are
obtained in the body fixed coordinate system,
so it is always preferable to use body fixed
coordinate system, transfer all these loads
to that and then write the equation. So, you
will have essentially six equations, force
equations.
Now, the helicopter, you know, that it can
be flying in the hub loads, where obtain,
then it is flying straight as a rotor, but
the rotor can fly in any direction, that means,
the component of velocity you must take proper
wing. Now, what will you do in this is because
my fuselage may be like this, rotor may, helicopter
may be going in this direction, lateral, you
can have all velocity components, but what
we have obtained is only one velocity. So,
you will restrict our self to only forward
flight steady condition and how these entire
equations are solved. Now, the question is
what are my unknowns? What are my known quantities?
That is very important before you really start
off the entire sequence of this problem. You
must know what are my unknowns and what are
my known quantities.
If you look at that, the known quantities
if you say because let me put it, the unknowns,
I will write unknowns, no one will worry about
that later. Unknown is a theta naught, 1c,
1s and theta 0, tail rotor, then pitch attitude,
roll attitude. We normally use the symbol…
Pitch and roll attitude of the helicopter
we do not know, these are the six quantities
and then of course, power.
Sometimes you can do a particular case where
power becomes the equation, but usually that
is not done. What you do is, you calculate
power after trimming the helicopter, evaluate
the power, keep on evaluating, may have power
available. It is not, that if you want to
see what is my, I want to use all the power
what should be in my flight condition, that
is a bit tricky, that is why, best is you
prescribe the flight condition and then you
want this, calculate what is the power required.
Now, when we prescribe the flight condition,
first we take the very simplest case. Simplest
case is I will put it known quantities, your
mu, which is the advance ratio. This you say,
I know it, which is, please note this alpha
is attitude of the, basically alpha is that
pitch, attitude pitch, theta alpha. We do
not know this to start with, that is why you
always use advance ratio I am using. Assume,
that this quantity is known; then, weight,
this is the weight of the helicopter. Of course,
CG location must be known for the given weight,
you can have forward CG, half CG, etcetera.
Then, that f, which is the fuselage drag,
related to fuselage drag that f area; then,
of course, the location. Then, you need to
know omega bar R F square flap frequency.
Then, if you have the blade with the twist,
you must know theta twist and theta; I will
write what are these? This is the twist of
the blade, twist and theta F P, which is the
flight path, flight path angle, I will call
it, I will, later I will describe.
In addition to this you have to know other
geometric condition, where is the tail rotor,
what is the distance from CG, the tail rotor
location and the tail rotor dimension? If
you want to take what we derived for main
rotor expression, completely you can use it
for tail rotor, absolutely no problem, only
thing is you have to use the dimensions, operating
conditions only for tail rotor. But tail rotor
has only collective pitch, there is no cyclic.
We have simplified our problem by considering
only flap frequency that is why I put what
is the rotating flap frequency. And of course,
if you want geometric, what is the precone
beta naught, then you must specify that also,
geometric values. Now, given these quantities,
I will go and get those quantities and what
is the power required for that particular
flight condition, this is the basic question.
We will draw one diagram and then I will we
will describe the diagram is…
And you may have your CG somewhere here, this
is weight and the horizontal, this is your
longitudinal. This is the horizon, in the
sense, the helicopter is flying in that direction,
this angle is theta F P, that is the flight
path angle with respect to horizon, that is
all; horizon is normally earth bound. So,
horizontal, with respect to horizontal, what
is my velocity vector? That is the flight
path, this can be 0, means, you say level
horizontal flight and this angle, I call it
as alpha; that is the angle because this is
what is required. This alpha comes here because
V cosine alpha in the hub.
Now, you may say, this is perpendicular to,
if the shaft is perpendicular to the body
fixed coordinate system, then it is fine,
that this will be 90 degrees because you can
have fuselage access system and that will
be same as hub access system. On the other
hand, if the hub is tilted a little bit with
a respect to the fuselage, then you have to
take that into account in the transformation,
usually in, sometimes it is tilted by 2 degrees,
something like that. The shaft itself, it
will not be perpendicular, that will be slightly
kept. So, these are small, small details,
which one has to consider when you are writing
the equilibrium equation.
Now, given these quantities and what is my
pitch attitude? Pitch attitude is basically,
theta is alpha minus theta flight path, but
I will use some minus sign because the rotations
are counterclockwise, clockwise, positive,
etcetera. This is the angle pitch attitude
of my helicopter
Now, this reference line, so I will fix two
coordinate systems, one is earth coordinate
system, which we may have. X, Y and Z these
are earth system, you may, all X ea… With
respect to this, this is the body, capital
X, capital Y and capital Z, you may call it
b, b, b. That means body system you have to
get the transformation from this, this to
this, that is, by rotation. But you assume
initial because more complicated motions will
take little later, right now we assume, that
the helicopter is only flying in, flying in
the plane X-Z earth axis system, but you may
find it can fly in another direction.
Then, you may have to say what is the because
the heading can be something like this, because
if it is making a maneuver, that means, earth
is a fixed coordinate system, but the helicopter
heading can change, you follow. On the other
hand, you can say, my heading of the helicopter
is same, but my velocity vector is this way,
then you have a side slip, you follow, because
you are also moving side. So, all these flight
conditions there becomes a little complicated.
That part is a most general, that I will just
describe, if time is there at the end of the
class I will show that, otherwise because
that is the complicated thing, which of course,
we are developing it. Right now let us say,
this is the motion, the helicopter is flying
only in the X-Z plane, but the center of.
I will draw one more diagram by the side,
that is, the view from rear 
and weight. So, there is a tilt about, because
if you say with respect, this is my Y body,
Z body and this angle, with respect to this
is my roll. And this distance, you may call
this distance as Y, Y hub. Here, you may say
this is X hub and this is Z hub, then you
can take to tail rotor. This can be X tail
rotor, this is Z tail rotor; you can have
all. That means, essentially, you have to
locate every point. I have not taken any horizontal
surface, nor the fin. I neglected those things
in my simple formulation.
Now, you know that, because these are helicopter
maneuvering, anything; these rotations are
not small angles. So, you have to follow Euler
angle transformation. The Euler angle transformation,
I will write between this, you find out the
sequence of rotation. First you rotate about
x-axis, that means roll and then, after the
rotation of x, you rotate about y-axis, so
that you get the body access system, you understand.
From the earth access you want to go to body
access, so the transformation sequence, you
first say roll, then pitch. So, this sequence
if you follow, it will be like this. First
is, what is it, this is x, y, z, this is the
body, but the rotations are counterclockwise
positive I am taking, that is why, there will
be a minus sign I have to introduce there,
multiplied by 1 0 0 0 cosine phi phi 0 x m,
this is the trap. That means, earth access
system, you first rotate about phi you get
the intermediate access then that intermediate
you rotated about the pitch and then, but
these data are counter clockwise rotations.
So, phi is a counterclockwise rotation positive,
then if you use that, this data because this
is here as per diagram, I put to the access
system clockwise rotation theta; I put it
as alpha minus theta f p. So, actually, I
must put a minus sign, that you look at, you,
you look at, if you want I will describe it.
See, this is Y earth, Z earth, X earth is
back, so you rotate about phi. So, I call
it X, sorry, Y some intermediate prime, Z
prime, this angle is phi. So, Y earth to Y
prime cosine minus sine; this is this matrix
if you put x 1, y 1, z 1, you follow. Here,
x prime, y prime, z prime. Now, you take here,
here is X ea, Y ea, 1, 0, 0, 0. Here, you
will have cosine phi, the same thing it will
give you, minus sine phi. And here, this is
sine phi, in the Z ea sine phi cosine phi,
this is fine. Now, you got x 1, y 1; you got
x prime, y prime. Now, you are in the prime
system, you are giving a rotation about y
prime.
So, you will have, this is my Z prime and
X prime is here. This is actually my Y prime,
X prime is coming out of the board, you are
giving a rotation about Y prime. Now, what
it will be? This is Z prime, this is X prime,
this is the Z body, X body. Now, you will
see, that you will get that, do not just blindly,
that is why, in these transformations all
of them, how they are given. Because now,
you see what is the transformation, because
here you, what X prime, Y prime is 1, Z prime,
X prime cos theta and X prime sine theta.
Whereas, if you go to Z, Z prime cos theta,
Z prime, it will become minus sine theta.
So, that is why, the minus sign there.
Now, with this you have the relation between
earth to body. Why we have to use this is,
because please note, that I still do not know
this and this angle, but I use, I would not
make any small angle approximation because
that is why, please understand. The moment
you have large angle, they are not vectors,
you have to follow this transformation. The
weight is, acts along the earth access system,
that is why you have to transfer the weight
into body system and then, you can write all
your equilibrium equations in the body system.
That is the reason you need to have this transformation
properly done.
Now, you got this, only thing is theta, you
have to put up minus sign because as per the
diagram, that is the key. Now, please understand,
that is, the phi is rotated about earth access,
whereas, theta is not rotated about earth
access, theta is rotated about the rotated
access; it is not about the fixed access systems
you are doing the rotation. So, every rotation
is about the rotated access after you get
a new position about which you rotate, this
is the sequence you follow. Now, if you want,
there is another transformation, which is
there, we have used all those things in our
work, you rotate only about a fixed access
system. What you said is, because you say,
should I rotate theta with respect to Y prime?
These are, if you have a heading, that also
will come.
Not exactly, these are Euler angles; do not
say earth coordinate systems.
Because jairo will give you rate, that I will
come to later. Jairo will give you angular
velocity; angular velocity is a vector because
instantaneous angular velocity, it will give
you about three axis. If you have a three
axis jairo, you will give rotation about the
three axis. Now, the rate of change of attitude,
please understand, there is a lot of difference
between angular velocity vector of the body
at an instant to rate of change of attitude
of the body, there is a relationship. If the
angles are small, you say angular velocity
is same as rate of change of angle; if not,
see because in the airlines, you have to do
one more, that is kinematic relation, that
data we will do it later, if it is interesting,
if need it to know from flight dynamics point
of view, because if you really want to solve
that problem you have, know little bit more,
that is a real life, real life problem, you
have to handle that, I will describe that
part.
This angle is the rotation about earth, I
can find out how much roll rate is happening,
roll rate with respect to earth, phi dot.
How much theta dot is happening here, but
theta dot is about the y prime axis, yeah,
already tilted, well, phi dot is about X ea,
theta dot is about y prime. So, when I want
to get the angular velocity vector, what I
will do is, I use the same equation, I will
put phi dot here and then get this.
What is the body fixed? What is the phi dot?
But then, when I come here I throw this off,
I will put here theta dot y prime axis and
then I will calculate what is the effect of
that, you follow, because you see y prime
is nothing 
but y body. So, basically, theta dot becomes
basically pitch rate, whereas the phi dot
is not exactly the role rate experienced by
the jairo on the vehicle, that depends on
what is the orientation of the aircraft at
that instant, that is why, you need to solve
even in the navigation of any… What is that
navigation, satellite navigation, navigation,
everything you have to follow this because
these angles are not small angles, they can
be large angle. So, you solve the angle, you
get the attitude, you have to solve for that.
Now, let us take as, it is this one and we
will write our, so I thought I will show directly,
yeah, you can do, there is nothing wrong.
See that is where, what is the sequence I
should follow? There is some convention that
is it; you can follow another sequence, absolutely
no problem. But so long as whatever sequence
you follow, the result will correspond to
that, only thing is you cannot compare his
that number with the result, which is following
another sequence. Numbers may not mean, but
the orientation will be fine, that is, this
problem cropped up when people were deriving
the equation for the rotor blade. Whether
you use flap, lag, torsion sequence or lag,
flap, torsion or these are so many and in
the 70s, that was a big debate. Finally, when
we do rotor blade, I will explain that part
right now, you will go into the equation for…
So, you got the, now let us look at what are
the equations, which we need to solve.
So, I said, first is inflow equation in forward
flight, uniform in flow I have taken, please
understand, this equation is one, then I have
trust equation. When I say vertical force,
this is along hub. So, this is, this is the
thrust, this is T, this is my capital H, this
is my, because T H Y, then you have m, x then
you have m y and you have m g, all the loads
at the hub.
You transfer these loads to the CG and then
of course, tail rotor I have used, only thrust
and tail rotor is giving, only thrust in this
y direction that is all I have neglected.
All other loads, which is, which is reasonably
all right for most of the analysis, unless
you want to complicate more for a good solution,
that is enough. That is why you see the drag
force D, which is the fuselage drag, that
is, along this, along the flight path. This
is the D that is why you see this angle, that
horizontal angle, this angle is alpha. So,
D cosine alpha, D sine alpha; that is why,
weight is acting here. When the earth system,
you put weight, put minus w because this is
acting in the negative z direction. So, minus
w you convert it, find out along what is the
force along the z body because this product
you can. The product is actually and this
is a, this is essentially a cosine theta sine
theta sine phi minus sine theta cosine phi
0 cosine phi sine phi and sine theta minus
cosine theta sine phi and cosine theta cosine
phi, x ea, y ea, z ea. Now, you see weight
is acting along minus w because that is minus
Z-axis z earth. So, that will be along the
body, you will put this; that will become
minus w cosine theta cosine phi. Theta is
minus alpha, so you will put cosine that is
why, here I have put cosine alpha minus theta
flight path cosine phi. And D is here, D acts
along the velocity direction and the velocity,
we took it although it is in the X-Z plane.
Please understand, that is why, I said velocity
is in the X-Z plane. Suppose, if the velocity
you have a side, the drag will not be X-Z
plane, drag also will give a side force, you
follow. So, all those, that is why, when you
do vectorially, you have to do everything
properly.
Now, you have the vertical force equation,
similarly horizontal force. Horizontal force,
you have the hub load because tail rotor,
you say it is not giving any vertical force.
Suppose, if the tail rotor is given sometimes
in the design, the tail rotor fin may be like
this, something like this, you may give a
slight tilt of the tail rotor, it is not perfectly
the shaft. This is the tail rotor; looking
from back you may have, there is the fuselage
is going you, see this is given a small angle,
maybe in this they can also give it in the
front direction, this is called Cant angle,
few degrees.take a little bit, they can do,
why?
Now, a component of thrust, you can use it
in supporting the weight, a component; that
is all. But usually, cant is not, you know,
arbitrarily you do not do because not that
everybody goes and gives the cant the advantage
or disadvantage, whatever it may be. The advantage
is, they say, that I can use the tail rotor
force a component in the vertical direction;
that is all.
No, you cannot change, fixed, not that helicopter,
no way to introduction in the design configuration;
that is all. But not all helicopters have,
not all the helicopter have no random means,
what do you mean by random?
Zero; most of the design they will have 0.
Some helicopters have some angle and I have
to be very honest, I only know the effect
of this is, it will give some component because
you need to have a tail surface always. If
the helicopter has to fly, it is slightly
higher speed, we need to have a tail surface,
why? It means, if you do not have, you will
not get equilibrium, moment equilibrium; we
will not be able to achieve as a result, which
means you cannot fly the helicopter at that
speed. Engine may have power, but pilot cannot
fly it will not be in equilibrium. So, that
is why all the fixed surfaces they introduce.
Now, you may say why they have to have a vertical
fin because you have vertical fin. What happens
is, that vertical fin, you will not be at
the 0 degree, it is slightly given 1 or 2
degree to the longitudinal line of the helicopter
at, there is at an angle of attack. Now, what
it will do is, it will try to relieve the
tail rotor load in forward spin a little bit.
That is all nothing more because tail operates
in a very complex environment. Industries
will have a test wing, they will do some test,
otherwise all simple calculation. You take
the thrust, whatever we have, for the main.
Main rotor is much more complex in terms of
pitch angle, etcetera; tail rotor is more
complex in the aerodynamic environment.
And, and the distance these are all important.
This is mounted because first of all, the
vertical fin you need because if you want
to mount the tail at a slightly higher position
from the, because you may have why I have
to shift it up, that is like a configuration
because the main rotor, the wake, you do not
want to go and hit the tail rotor. So, you
try to keep this center along this height
rotor hub height, so that tail rotor is not
affected by the flow from the main rotor.
If it gets affected, you do not know what
is going on, you can say, well I will put
it this here, then you put it. Whatever you
get you have to take it because you do not
know how they will interact with the wake
from the main rotor, the tail rotor rotation.
So, best is, you put it up even in tandem
helicopter, you see the back one will be up,
the front one will be a little lower. This
is to avoid because you do not know honestly
and, but some helicopters you may say, it
is there, our model has not that lifted, it
is right on that so, but we take it, that
is all.
Now, the horizontal force, you have again
the hub load, the drag and the component of
the weight because the component of the weight
multiplied with this will give a horizontal
force in the body system. So, you use that
term and then side force, you have the y,
you have the tail rotor and you have again
the w in the Y sin phi minus sign because
this is a w becomes, so that side force has
again the weight. That is why, you see, the
weight comes in all directions in the body
fixed system.
Then, similarly, you do the moment balance,
roll moment, pitch moment and yawing moment.
So, you have the roll moment M x from the
main rotor, side force into Z h, this is the
height; side force. And similarly the tail
rotor, tail rotor thrust into Z height tail
rotor. And then, you will have main rotor
thrust into Y h because this distance, the
thrust is this way that will give a moment
if the CG is not on the longitudinal line.
No, I will send you these things, this just
for explanation. And then, you have pitch
moment, again rotor hub and this is due to
thrust and that is due to again side force,
sorry, the longitudinal force will give, and
tail rotor I have taken nothing.
Now, you see, in the pitch moment I do not
have any horizontal surface. If I put a horizontal
surface here, like a small aerodynamic surface
flat plate kept at an angle that will give
a pitching moment. So, usually, big helicopters
they have that, if it is a small weight class,
they do not have. But some time you will find
some helicopters have only one half of the
tail, not on both sides, it will be having
only one half; these are all just for equilibrium.
And you can have a vertical plate also, like
you have horizontal surface you can have a
vertical plate, which is inclined at a slight
angle like your vertical fin; you can have
like this, they will give a force, these are
all to relieve little bit.
And yawing moment, again the main rotor and
the side force will give a yawing moment because
of this distance side force. And of course,
tail rotor will give and your horizontal force
also will give because of this distance. Horizontal
force will give lateral shift in the CG.
So, these are my seven equations, of that
inflow is the first equation. So, six load
equation, you may call it equilibrium. Now,
how do we really solve, that is the question
you have to because where is the flap equation
coming in between, because flap equation is
sitting somewhere because all these thrust
H, Y, M x, M y, they all are dependent on
flap response.
Now, flap response we approximated it as though
first harmonic beta naught, beta 1c, beta
1s, that is all, rest of them we neglected.
So, all these expressions require flap. So,
you have to now introduce the flap equation
in between; that is why, now you see your
equilibrium.
This part we will come to that later, this
is an aeroelastic problem, is part of trim
problem because you cannot decompose aeroelastic
problem in helicopters. Aeroelastic problem
is response of the blade to air, that will
be sitting in between because you cannot just
like that solve equilibrium equations separately,
because you must know blade response. Then,
blade response depends on what type of model
you are having for the blade. Here, we have
included only flap, please understand. In
real helicopter you have to have flap, lag,
torsion, all must be included and it will
be much more complex, I will show those results
a little bit later.
Now, how do you start because it is an iterative
procedure, you do not know inflow. So, first
is you assume my thrust is equal to weight
and always start these analysis from hover
case, do not straightaway go and then start
solving, I want at mu 0.3; straightaway go
and start means, sometimes it will not converge.
So, you start from hover, which is the simplest
case. You take, that equilibrium quantities
and then slightly change those quantities,
then go to next forward speed, which is point
0.5 mu because you have to assume. First assumption
is, assume alpha, alpha is theta F P, maybe
there flight path may be 0, may not be 0,
that is a given quantity, please understand,
given quantity flight path is there. I may
have flight path 0 that means the FP is 0,
means, alpha I do not know. So, I will assume
alpha, you may say I assume 0, no problem
because hover I may take it 0. That is why,
first case hover, whatever result you get
in the alpha, if it is 0, when you go to first
forward flight of point 0.5 mu, you assume
0. Then, thrust is equal to totally weight,
just approximately thrust is weight. Then,
you solve for lambda, inflow because inflow
equation I showed you, that requires only
C T and alpha, mu you will know, there is
no problem, mu is given, that is a flight
condition. So, you first iterate within this,
get a lambda that means what you have done?
Very first step, you got the inflow, after
knowing the inflow, then you go to... So,
that is why I said, use equation one there
is, solve for lambda. Once you get lambda,
I have given you in your class notes the flap
frequency not equal to 1; equal to 1 I have
both, beta naught, beta 1c, beta 1s. Solve
for beta naught, beta 1c, beta 1s assuming
theta naught, theta 1c, theta 1s; please understand
you have to assume something.
Now, how do we assume? This is from previous
flight condition, that is why, you take the
hover case first, solve for that because hover
is easy to converge. Assume these values,
then substitute here because here this requires
a lambda, please understand. Lambda and theta
twist, everything, you will get beta naught,
beta 1c, beta 1s. Now, I have lambda, I have
all these quantities, then go to the six equilibrium
equations. What you do is, what are the unknowns?
For that case you say, I know lambda, beta
naught, beta 1c, beta 1s, but I do not know
the six quantities of theta naught, 1c, 1s,
alpha, phi and tail rotor. Tail rotor you
can take it as a tail rotor thrust itself,
later you can, once the thrust get converged
you can actually go and the use the simplified
path to get the tail rotor pitch angle.
Now, when you solve these six equations, six
unknowns, they are all algebra equations;
please understand, they are all algebraic
equations, but none can be used it may be,
but after getting theta naught, theta 1c,
theta 1s and alpha and phi, you again go back
to step 1, because now you know alpha, you
go back, you can again and you may get the
thrust also from the problem. You can calculate
because all these angles are known, you can
get what is the thrust. Know the alpha, know
the thrust, again get a nu lambda. Once you
get the nu lambda, again go here, solve and
this whole procedure you iterate, till equations
are converged. Now, the conversions is itself
is a very interesting thing, you converge
here, then whatever, you come here, you take
it, but the key problem, which is spaced is,
you will find the thrust, which is a dimensional,
it is like a 4000 kg helicopter or 5000, that
is a very big number. Whereas a side force,
they are all very small numbers. Then, when
you try to take equations, you will find,
that the big number will drive the result,
but it may not converge properly.
So, always, all of them are non-dimensionalized
such that they converge. They are all given
equal importance and then, you get the converged
solution and once you converge in the pitch
angles, then you say your result is over,
but you will never get everything equal to
0 beautifully, some small number, that is
ok. If it is 4000 kg, if it comes to 3950,
it is alright, already 980 because otherwise,
it will keep on accelerating the result, you
keep on shunting. And these iterations if
you go, real complicated thing, it takes about,
the problem, which we solved, it takes 14
hours for one flight condition because that
is much more complicated, much more complicated.
This you can do because you have everything
in the closed form solution, you can write
a code, that is why, I am saying, you can
write. I have given every expression to you
in the class derived, this can be actually
exercise for the student, you see, how to
solve the problem. Now, you see all these
equations are there, you write a program,
it will not work initially, it will give you
garbage. Then, you have to find out what error
because even the solution technique has to
be proper, it is all algebraic equation.
Now, I am just extrapolating the complexity.
We made several assumptions, one is in the
inflow. Uniform inflow in it is not uniform,
it is variable. The 2nd one is flap, who told
you only 1st harmonic, I should have all the
harmonics. Then, I should have lead lag motion,
I should have torsion motion. That means,
your problem gets complicated and you should
have stall, you have reverse flow effects.
When you start adding everything the problem
becomes sometime intractable, it is a complicated
problem. So, what industry was doing, initially
they do lot of approximations, get a result
first, after that they will go, that is why,
here when I use flap, so if you take only
flap, you call it flap term.
And another thing is some of the research
work, they do not take all the six equations
because I do not know tail rotor, I am not
interested. So, they will say in most of the
research publications, when they are analyzing
the aeroelastic response of rotor blade, only
blade. Then, they will take side force equation
and yaw moment equation, they neglect it that
means, you have thrown away two equations
that means. What are the two unknowns you
will throw? You basically throw roll angle,
you throw from that and the tail rotor is
thrown, that means, two unknowns are thrown,
remaining is theta naught, theta 1c, theta
1s and alpha, that is all. But that is, they
have for a research publication, where you
want to study more about only blade response,
blade characteristic influence of certain
geometry and the aeroelastic thing, they do
not do full analysis of the helicopter.
Now, another aspect is you can do wind tunnel.
In a wind tunnel what happens is, it is kept
at a specific shaft angle, that means, alpha
is fixed, that is called a wind tunnel trip.
There you do not go and balance the thrust
because you know, it is held. Whatever force,
that comes, it is not, that it is equal to
weight, so thrust equation is out. You may
directly solve what is the developed force,
developed only. You can say, there is no equilibrium
equation wind tunnel. What people do is, they
give prescribed collective, prescribed cyclic,
prescribed theta naught, 1c, 1s, now find
the response of the blade. But for that you
need to know inflow. So, they will give a
shaft angle, that is all, but you do not know
thrust, you follow. So, you have to get the
thrust and then get the inflow and then get
the all the response quantities. So, depending
on the situation, you use that particular
set of equations and solve them.
I hope you have, I have conveyed the message
here. Now, let us look at some of the, how,
see once you have converged, you go back and
calculate power, what is the power required
for this because power equation, you know
it earlier I had, we have derived in the class.
Now, I am going to show few results, which
I have taken from the book and some, which
we have generated. You see, in general what
happens is, beta naught does not vary so much
with respect to forward speed. It does not
vary a lot up to some 0.3, some 0.3, 0.35.
So, very small variation because the thrust
vector, it is almost supporting the weight
of the helicopter. So, it just, only with
change in weight it may change.
Whereas, beta s and 1c, 1s I have shown here
like this, but actually this is not the, this
is some theoretical calculation. 1c they increased,
this is also increased with, but this is not
correct. This was a main point of research
in 80s, later I will show some other results.
Some of the experiment show, that at low value
of mu, the lateral flap, lateral flap, it
is about 2 degrees, 3 degrees that is all;
it goes up and then it decreases. These are
all various theories, this line, I wanted
to say this line is my uniform inflow theory,
this is the line.
Now, I will come to our results, which we
have, this was the, we published, just last
year it came, I think last year, yeah, 2009,
yeah, 2009.
Here is the, we have, this is a complicated
problem we have solved, same thing, flap,
lag, torsion completely, but full 6 degree
of freedom helicopter analysis. This is where
it took about 14 hours I was telling you with
various aerodynamic models. So, if you look
at the collective, collective it will decrease
and then it will increase.
But you see, as you go to high speed, that
is, around 0.35 mu, you see the slope of this
curve, it is very steep, the program will
not converge so easily, that means, even if
the vehicle has power, enough power. But the
helicopter cannot be converged, in the sense,
trimmed properly because of, it requires very
large, it is not restricted. Now, how do you,
why this curve shoots up like this, that is,
because of the fuselage drag, if you reduce
the drag of the fuselage, but how would you
reduce it, that is the problem?
Now, that is why, that f, which I showed you
equivalent flap plate area for the helicopter,
that has to be, if it is brought down somehow,
then the curve may tilt. You can go further
if it is there, you just cannot trim. The
drag will shoot very high, that is one of
the key things and of course, you see, theta
1s, that is the longitudinal cyclic. You see,
this is a negative and that keeps increasing,
these are with different aerodynamic models.
Now, here is the theta 1c, which is the lateral
cyclic. With different aerodynamic models
I am getting different, different curves.
This is the uniform inflow curve, the lower
that continuous line and these are, I have
included some dynamic stall model and dynamic
wake, some additional more complex models,
you see they change drastically just in equilibrium,
this was a point of.
Why the helicopter it is like this? Pilot
wants to fly forward, usually the general
tendency is, what if you want to go high speed,
you give more angle, but here, initially you
give more angle, then as you increase this
speed, you start decreasing the angle. So,
from hover you go to very large and then you
start coming down, this is, it defies the,
you know, traditional. You want to go high
speed, what you do? You increase the throttle.
It is not that you go to high throttle for
low speed and then come to low throttle for
high speed, it does not happen that way.
But here, this was why it happens; this was
a question a lot of people… Of course, this
is due to inflow modeling, but inflow and
stall put together it came, whereas model,
which I have mentioned to you, that gives
slightly this result, but we need to have
stall also. And of course, the tail rotor
follow similar trend as a main rotor and these
are roll and pitch, but roll attitude we cannot,
I do not have data to compare with experiment
because I do not have data for this with a
real helicopter. So, any way HUL, once we
develop, they will be using this code for
them because this is, you cannot measure it
and then you have some other mechanism.
What is a rotor tilt, tilt angle? You should
measure it and then plot. Because the tilt
angle depends on where you put your tail what
happens, etcetera. So, and this is the pitch
attitude. So, you see, the angle values, that
is, what I am telling you approximately, this
is about 10 degrees, it comes to 7 goes up
cyclic, that is theta 1s, which is directly
related to beta 1c they, go up to minus 10,
minus 20. So, the pitch angle can go very
high values in the retreating blade.
Now, I just want to show this result because
this is what this is also we presented it
in it is published in the journal see this
is the taken from a book this is the flight
test this is the code which they have used
this is a paper if the pat field book I actually
took permission from him to put it in my publication
in our publication I would say because the
laxman student did I took his permission.
So, that I can reproduce this in our journal
paper just to say because I do not have practical
data because usually even if you have one
practical data and then show the quality of
the paper will go drastically very high.
So, I just wanted to show to the people look
here this is what the experiment or flight
test qualitative this is how my my data for
the given helicopter test. So, different models
if I use see this line looks somewhat like
this this is only lateral cyclic please understand
rest of them I do not have any data and this
is you see goes up to three point something
here also it shoots up; that means, the rise
t price in the angle for trimming this is
only trim problem please understand we have
not gone to stability other things this itself
has then we said this is what that is why
we put qualitative comparison of variation
of lateral pitch angle with forward speed.
Now, I just want to show this result, because
this is what, this is also we presented it
in, published in the journal. See, this is
taken from a book, this is a flight test,
this is the code, which they have used, there
is a paper in the Patfield book, I actually
took permission from him to put in my publication,
in our publication I would say because Laxman
student did, I took his permission so that
I can reproduce this in our journal, paper,
just to say, that I do not have practical
data because usually, even if you have one
practical data and then show, the quality
of the paper will go drastically very high.
So, I just wanted to show to the people, look
here, this is what the experiment qualitative,
this is how my, my data for the given helicopter
does.
So different models if I use, see, this line
looks somewhat like this, this is only lateral
cyclic, please understand, rest of them I
do not have any data, and this is you see,
it goes up to three point something, here
also it shoots up. That means, the rise, steep
rise in the angle for trimming, this is only
trim problem, please understand, we have not
one to stability, other things, this itself
has, then we share, this is what, that is
why we put qualitative comparison of variation
of lateral pitch angle with forward speed.
So, this is a present study, this is reproduced
from some reference, that is, the Patfield
book, what I am saying is, this we just published
last year AHS, this is the helicopter society
journal.
This is a flight test; this is a flight test,
but flight test for some helicopter, please
understand. But this, why I took it is, we
all know, everybody knows this is what happens
in flight, but how would you get that, what
model affects your trim result, simple. So,
ours is purely a theoretical study, but we
gave various aerodynamic models just to see,
hey if you change this model to another model
what happens to the trim, you follow and this
is only a qualitative comparison. Now, if
you want quantitative, then we should take
HAL that is all; HAL should supply the data,
see they also know that.
Now, there is a collaboration to make it our
own Indian developed code for the analysis
of helicopter thing because please understand,
these things take 15 years, it is not one
day's job and a lot of companies, they are
also working in US, they have put lot of millions
of dollars for developing analysis code. Today,
they are doing maneuver, we are now doing
maneuver for the next thesis, after this is
the maneuver.
Now, I will just show you one diagram, it
is an interesting, maybe I will show this
first, this is what I took it from the paper.
See, these are all experimental data, this
is the one, which I took because the other
things are all, everybody gets the trend,
there is nothing special, only this is the
key. I just want to show how the angle, pitch
angle varies.
I have put the numbers here, this is 0.05,05
these are nothing, but the midpoint of the
rotor. And these small numbers 2, 3, 4, 4.55,
5.6, these are effective angle of attack on
the rotor disk at various radial locations
because this we took it as 0.23 as the hub
for a advance speed 0.01, very low speed,
everything is within the stall angle, that
is, 6 degrees is the maximum near, somewhere
near the tip.
But when you go to high speed, high speed,
this is 0.35. You see here, these angles are
16 degrees, 15 degrees; that means the blade
starts stalling here and on the advancing
side, there about 1.2 here, 3, 4, 5 here,
there is no stall, all the stall goes here.
And this is the reverse flow region, very
high angle, you can even have, I think we
have put minus 180 degrees. That means, the
aerodynamic data we have for the full 360
degrees, the rotor airfoil, we took the data
from a text book only.
Now, that is why, at high speed you start
stalling on the other side and that will give
you vibrate. We did vibrate, load, etcetera,
it is just for you to know the, how do you
do trim problem, but trim problem will give
you the vibration load as a part of solution.
That is why you do not solve trim separately,
aero-elastic load separately, you have to
get both of them simultaneously, but simultaneously
means, at one shot you will not be able to
solve it; iteration.
