Now, I just want to mention because there
were some questions last time, because I want
to clarify, because the autorotation, when
we mean by autorotation, it can happen in
the sense, it is a particular condition, that
condition is the power that is given to the
rotor, is 0.
So, the power to the rotor, if you say, that
is, you can say, P shaft, this is what essentially
the climb power. It is a climb velocity, this
is the induced, then you have rho… This
quantity very, very simplistically, only the,
because this is refer to only the main rotor,
because you say, this is climb induced, this
is profile, they will be 0 at a particular
descend velocity because this descend means,
V is negative. V is negative means, this quantity
can be made equal to 0 and that condition
is called autorotation.
Please understand, because you may think,
that these are all, this is a descend state,
this is also descend, windmill break is also
a descend state, everything is descend, descend,
descend, but autorotation, when we mean, it
is that particular state where the power required
to rotate the rotor is basically 0. That is,
engine has failed; engine is not supplying
any power that is all, so, but yeah…
No, no, you will have this also. See, you
have a profile…
In the graph, that is why, now you were asking,
that graph is drawn, see you, only to indicate,
see this is the graph you are talking about.
V plus nu 0 is this axis, essentially. Yes,
it is close to autorotation because if you
neglect this term, that means, the profile
drag, I am not including only the induced
power because usually, the induced power is
much larger, you understand. This is not very
large, but it is there, that is why, the real
operation to mathematical, just an ideal definition
for autorotation, simply say, V plus nu 0
is autorotation because you are considering
only induced power, you are not considering
all the other power. If you really consider
everything, you may have the tail rotor, also
may require and then, there could be some
transmission losses and then the fuselage
drag also can be there.
So, there can be several other factors contributing
to the power of engine, which have to be supplied
to the rotor, but we are basically not including
those. In the real life, it may not happen
at V plus nu 0, it will happen a little down,
T plus nu, a little lower. Because even if
you include this, V plus nu is not 0, V plus
nu is equal to this quantity, you understand,
that is why, autorotation happens at a descend
velocity slightly greater than in the one,
that meets the V plus nu line, it is 0 value,
that is, from real calculations.
Because there was a question last time, where
will it auto rotate? Here also it will generate
power, but that condition you do not call
it autorotation, is it clear? Now, let us
look at the, is it, there are some interesting
physics associated with this autorotation,
I thought I will briefly explain that part
today and then see, one is the, the easiest
because this is very important for you.
The power shaft, so note down this. Forces
on, which I started, on a blade element 
in autorotation because here I am going to
just briefly describe one section and then,
we will analyze the autorotation condition
and the physics associated with that. Later,
you see the complexity, which actually happens
in the real life.
You take an aerofoil, which is acting at the
angle of attack, not Angle of attack, this
is the pitch angle theta and this is omega
r, which is due to the rotational velocity
at that section and you have a, because this
is descending you have a U p, this angle you
call it phi and let us say, this is X because
last time and the resultant and this angle
is phi.
This is a very simplistic problem of an aerofoil,
which has oncoming flow and flow, which is
coming up because it is descending. U p includes
descend velocity as well as induced velocity,
please understand, that is why I use a very
general perpendicular. The flow, that comes
perpendicular to the aerofoil plane of rotation,
that includes in the simple case the descend
velocity and the induced velocity. Induced
velocity is actually down descent velocity,
is the relative.
Now, tan phi, that is the induced angle, is
basically U p over omega r. It is fixed, this
angle is, once you know omega r, once you
phi, sorry, U p, you know this angle, now
if you calculate what are the forces that
act on this element, that force is only the
lift force, which you call it the force in
the Z direction, another one is the X direction.
F Z is L cosine phi plus D sin phi and F X
is, and I will say, D cosine phi minus L sin
phi. Now, let us look at only the F X quantity.
Of course, lift is dynamic pressure into,
you say, it is the unit section. So, you take
chord into lift coefficient, this will be
a dynamic pressure chord drag coefficient.
But this quantity can be positive or negative
because the F X, F X can be positive or negative
or it can be 0, depending on, of course, D
and L contribute depending on the angle phi.
Let us define the angle phi, that is only
the condition where this is 0 because if F
X is 0, please understand, initially the rotor
is rotating because the power is coming from
the engine and that section is rotating with
the omega r initially. Now, engine is shut
off or engine is failed. So, engine is disconnected
from the rotor and this is still rotating.
Now, what will happen? You will have descend;
you will generate an F X, if F X is positive.
So, I will put F X positive, means, your rotor
will start because no other force, only F
X is acting. It will decelerate because F
X positive deceleration 
of the section if F X is 0, that is, autorotation
basically, because there is no force acting,
it will continue to rotate with that omega
r. So, it will, this is the condition for
autorotation; that is why I said, blade element
in autorotation will see what.
And then if F X is negative, negative means,
if this is this side, then your aerofoil will
accelerate. That means, deceleration means,
omega r decreases here, omega r increases
here, omega r constant, that is all, three.
But this, we will relate to, from here in
terms of coefficients rather than D and L
because you can write this, as this will become
F bar, you can take it C D naught because
half rho V square chord is thrown out. So,
C d naught cosine phi C l sine phi, you can
keep this because C D naught is the aerofoil
characteristic; C l is also aerofoil characteristic.
Now, you can take, if F bar X is 0 that is
what is autorotation, I am going to call it
tan phi, is what, C D. This is for autorotation
condition, I am putting a subscript just to
indicate, that this autorotation condition,
tan phi C D naught over C l, it can, yeah,
that we will write it later, that we will
do it later, right now you take it as it is.
Now, you see, if this is greater than 0, that
means what, you divide by tan phi is, what
you do, if this quantity is greater than 0,
this quantity is less than 0, correspond to
this three conditions, right, yes or no? Because
you say, let us write that.
Now, you see, this is always valid, tan phi
is U p over omega r. F X, then this quantity
becomes equal to this value, which is, that
is, phi equals phi a and then you can say,
tan phi is U p over omega r, it is also equal
to C D naught over C l. This is the autorotation
condition; condition for autorotation for
that element, is it clear.
I will always have some value for this. When
this value is exactly equal to C D naught
over C l, then I am in autorotation. Now,
let us look at the other three conditions.
This is deceleration. Well, I do not need
this F Z, this also, this is also not necessary,
I will write here tan phi is…
For this condition this is positive, means
what? This implies C D naught minus C l tan
phi is greater than 0 because C D naught,
I am just dividing.
In other words, this is tan phi is less than
this, is, is 0, this is the phi, I put it
autorotation and this is the other condition,
which is C D naught minus C l tan phi is greater
than, sorry, this is less than 0. In other
words, this implies tan phi greater than C
D naught over C l. Now, you have three.
What we will do is, we will go ahead and you
take an aerofoil, you know, it is a characteristic
with angle of attack. Now, what is my angle
of attack?
Please understand, my angle of attack is basically
alpha, I call that, which is, this is my angle
of attack. Now, I take an aerofoil, any airfoil,
draw the curve. I maybe, I erase this part
because this is right now not required.
We will plot the curve of tan inverse versus
alpha, but keep the axis same; axis in the
sense, x-y same scale I will put it. So, you
plot a same x and y scale because both are
angles, this will be like this, maybe I am
not drawing properly, it may go something
like this.
This is the curve for a given aerofoil, you,
you follow what I am saying because you take
any airfoil, you know the drag value, change
the angle of attack and then plot tan inverse
C D naught over C l like this.
Now, this is very interesting, this curve
is very, very important for, now what is my
angle of attack? Angle of attack is theta
plus phi, now I, this is 0, I take some theta,
this I call it theta. I draw a 45 degree line
from here, this is 45 degrees. Now, this is
same scale, if I take any point here I think,
that point I have noted at a, this is b, maybe
I will put it, this is 45 degrees and some
point C.
If I take this, this is phi because I am taking
actually 45, does not look like 45 because
this is phi and this is also phi, you follow.
Now, if my this phi is greater than this curve
because point a, phi is greater than tan inverse
C D naught over C l. That means phi is greater
than C D naught over C l, follow, which implies
what? It will accelerate. When it accelerates
what happens? Omega is increasing, when it
increases what will happen? phi will decrease,
so it will come to that point b, you understand,
because at b exactly, phi is tan inverse,
that is the autorotation point.
Now, suppose your point is below, that means,
phi is less than tan inverse C D naught, phi
is less than this, that means, it will decelerate.
So, it decelerates means, what happens? Omega
r is already large, so it will start decreasing;
when it decreases, phi will increase. So,
it will go to b, that means, in autorotation
it is a stable point, but what is the pitch
angle you can have maximum?
Please understand, suppose I go and I draw
a tangent to this line 45 degree, this point,
this is the point D, this is theta max, any
point here, this is acceleration, this is
deceleration. That means, I take theta m,
draw the 45 degree exactly, it meets this
curve at a tangent, that means, this is the
point. I can have any disturbance, if I go
beyond, my rotor will start decelerating,
that is all, it cannot take it, you understand.
That means, theta m is the maximum pitch angle
you can have. Otherwise, if you go to a pitch
angle, which is beyond this, every point is
deceleration. So, the rotor will simply decelerate
from omega r, it will stop and it will start
rotating in opposite direction, that is very
dangerous.
Now, it is not that you can keep initially
when you are hovering because with the power
pilot will be operating is collective pitch
or whatever pitch angle at some value. Now,
it may be more than this, you can be more
than this, there is nothing because it is
the flight condition, the weight, everything
matters. He may be operating at a pitch angle,
which is larger than this, usually that is
what happens.
The moment engine fails, the pilot is told,
they use the word dump the collective, that
means, reduce the collective angle, that means,
decrease immediately, otherwise if you hold
it, there rotor will stop and immediately,
he has to reduce the collective. But how much
he will reduce, that is another question,
usually reduces?
Now, the interesting part, there is again
one more interesting thing, you know, that
tan phi is U p over omega r. So, what is the
maximum omega r you can have?
When you have maximum omega r, phi must be
minimum, minimum means, the minimum point
on this curve is the point where you will
have maximum omega r. So, that is why, if
you decrease your collective from this point
or anywhere, what will happen? Omega r will
increase, but if you decrease further again,
that will start decreasing. So, you will have
a maximum omega r in the rotational velocity,
will happen only at the point where you have
this is minimum.
Now, these are all drawn for an aerofoil because
you can draw the, because this is purely aerofoil
characteristic, which is done in the wind
tunnel. You will find, that normally most
of the industries, of course, there are several
reasons. Autorotation condition is one choice
of aerofoil and then they will also have,
know, knows down pitching movement, there
are several reasons.
So, you will find, that industry develops
their own cross-section of the aerofoil for
rotor blade and then, they use it, it has
evolved over years of their testing. So, they
would not, you know, reveal all the details.
Now, this is very interesting physics.
Now, if you translate it to, what we did for
one section, you translate your actual rotor.
The actual rotor will be like this because
I am and this is the direction of rotation.
So, you have the blade, it is like this, this
is, now you see, phi is large because omega
r is small near the route. That means phi
is large.
Blade may stall, that is why we drew the diagram
here, Stall. Then, you have another region
where phi s becomes smaller, because from
large I am decreasing. Now, here in that zone,
I may have, this is where power from air to
rotor, it is air to rotor, here it is actually
rotor to the air, that s, it is like a normal
operation. This is from rotor to, sorry, rotor
to air.
Now, drag force will be here, acting here.
Also, drag force will be there, I do not,
I am not showing it. Now, you see, this is
stall, this is pulling it because phi, you,
you will always accelerate. phi X is still
because initially, this angle is large, you
are decreasing it. phi X is attached, flow
phi X is actually accelerating, this may be
somewhere, at one point it may be 0, F X may
be 0, then if you keep on decreasing phi because
what, decreasing phi means what? Omega r is
increasing, as you go towards the tip omega
r is increasing.
Then omega r increases phi becomes small,
when phi can be small, you are actually decelerating,
that means, r is actually dragging the rotor.
So, this is, this is how the loading will
be. So, it will be drawn something like this,
something.
You see, the integrated value of this, this
is a very, very interesting problem for autorotation,
what we should not have any problem. And integral
of this drag force, one is integral to drag
force is total force, but integral of the
r into the drag force, that is the moment,
that is, this is you take it as r.
Integral F X d X, that is 0 to 1, you take
it and this is the total drag force. Another
one is integral 0 to may be r, capital R,
I am sorry; maybe I should use d r, r F X
d r. This is the torque, now what, what should
be the condition for autorotation? You have
drag force; you have torque, then what about
force? It is a very tricky stuff, technically
both must be 0 because drag force must be
0, torque must be 0.
Now, how do you achieve that, that too with
that kind of a distribution? You have a distributed
force, now they have to get it that is why,
it is, it is a tricky problem if you really
want to solve it. How my forces will vary?
Because you have to do it, make sure, that,
that is 0, maybe you, it should be possible,
it is possible to satisfy, but whether that
force is like, because the force is aerodynamic
drag, it is not, that you are applying the
force. If you can apply the force, that is
ok, but this is an aerodynamic force. You
may find one of them is 0, another one maybe
a little bit there. Because I have not done
the calculation, but I thought that this is
a very, very interesting problem in the real
life situation if you want to do.
But usually, they will say torque is 0, good
enough or you say drag force is 0, that is
also torque is 0, because you are everything
relating power. So, I will say directly, torque
is 0. So, I do not bother about, but dynamically
if you want to look at it as a full problem,
it is a rotor blade, there should be no force
on that. That means, F X must be 0 integral,
the moment also be 0, satisfying both is very,
very in real life situation, but you will
find, may be a little F X may be there. I
am not sure what really happens in the actual
rotor system.
Usually, this is the because I thought that
it is quite interesting from Physics point
of view and steady, please understand all
assumptions. Now, you will say, you may ask
lot of questions, how do I know what is my
U p because U p consists of descend velocity
at an inflow. I do not have clear definition
for inflow in the turbulent wake state; I
have in the other state, windmill break state,
but not in the turbulent wake state. So, these
are problems.
Now, you slowly understand, that even a simple
descent thing is not very easy problem to
analyze because the, because we showed, that
one minute, I will show this diagram also,
this is just a dark patch. So, you have to
make an approximation to the climb and I say
that is my inflow that inflow may not be exactly
at every point. You do not know what, but
you have to make an assumption, that my inflow
is constant over the whole rotor disk, which
is not true.
Now, you see, the amount of assumptions you
make, which are really not correct, but still
you have to get an answer because you cannot.
So, usually the power is 0, they take it and
use that condition, come to autorotation,
that is all. It is a very, very, from mathematical
point of view, it is a complex problem, but
what is normally done, it, take it as the
simply curve, show that C D naught by L, ok
minimum, this operates, fine and I will have
my rotor loads and you take typical one section
and then say, if that section is somewhere
around, I do not know what, because the whole
thing goes in a particular fashion.
That is why this is shown to the pilot usually
because this is aerofoil characteristic C
D naught, because you are not doing anything
autorotation, you are just plotting that and
minimum point, you say, ok, hey the pilot
should come immediately, dump the collective
below. But then, autorotation is a stable
rotor, will continue to rotate, but what rpm
it will achieve? That rpm depends on what
C value you are operating on, is it clear.
I think with this I close your, all what you
have learnt in climb; climb is a very, very
smooth thing, whereas descent is most complicated
problem. Even today, if you go to industry,
they will have some thumb rules; just follow
that; that is all.
And autorotation, another important aspect
is the inertia of the rotor because the rotor
should have sufficient inertia. Inertia means
mass into r square because it should be, because
if it has a large inertia, it will not decay
faster because pilot takes some time to react.
You are operating at a higher pitch angle
and engine fails, immediately drag force will
try to stop. So, how do you take the inertia?
What inertia you should take, so that the
rotor should have that sufficient inertia?
Because if it is very large, then it is aircraft
is going to be very heavy, the blade is going
to be very heavy. But if it is very light,
blade will stop. So, you should have sufficient
inertia and there are some guidelines at least,
that much you should have that inertia of
the rotor. So, that is another aspect, this
is purely from aerodynamics. We are not looking
at the inertia aspect because this is a very
complex problem simply because you do not
know what the inflow is, that is all.
Dynamics you can fairly clear, yeah…
What reaction?
Net force, no this is, we have plotted this
for a section…
Which one, this one? This is net force, yeah…
That force is acting on the system on the
helicopter. See, if I isolate the blade because
every, see dynamically you have to look at
it. I can have 0 reaction force, what is the
problem?
See, what you are saying is, you are holding
it, just because you are holding it so there
is a force you are saying.
No, but then, but there is a force acting…
Wait, this is a very tricky thing. You take
this, this is the blade; you say, if I isolate
this, isolate, you say there is a force acting
somewhere, I do not know whether this way
or this way. If force acts what will happen
to this body from basic dynamics?
This is net; if I isolate a body you say there
is a reactor. Yes, I agree, if you have a
force, see this is the other body, right,
this will be like this.
Which reaction?
That the net aerodynamic force can, you said,
that it cancels, cancel means what? This is,
it should be 0. See, you take the aerodynamic
force, this is a rigid body, you are having
aerodynamic force acting on that, so the forces
act on this are, I am just drawing. This will
be resisted by this you may say, maybe other
way round, whatever you may put it, but then,
this entire force has to be what? At the root
you are going to have a moment, see total
force that acts on a body, it has to be 0
and you say moment is 0.
There are two conditions. I have isolated
this body; this body is initially, though
it is rotating with a steady omega, that means,
you will have only acceleration towards the
center, there is no acceleration in the tangential
direction. But if I isolate the body, if I
have a resultant force, that means what? It
has to have an acceleration or deceleration
depending on whatever direction you are having.
Now, deceleration will stop, acceleration
may take it up, I may have torque 0, you understand.
Then, it is like a rigid body, I do not have
any moment, but I have a resultant force;
can you have like that, yes or no? See, what
I am saying is, if you neglect, if you have
a leftover force, that force will be resisted
by the reaction here, you understand.
See, basic mechanics of solids, what you do,
you take a beam, you apply some load. If you
apply just this load, will it be stable? No,
because you have to have a reaction at the
root end, the reaction depends on what you
apply here, agreed or not? Because you should
have a reaction force, you will have some
moment or whatever it may be, you will have
something. This is depended on this, this
is purely static condition, agree?
Now, this is the net body force, but I want
the moment 0 because the torque, the rotor
drag is going to come, it will try to stop
the rotor, I do not want the rotor to stop,
that means, what? That means, what this must
be; hey, balance means what? This is balanced,
always this is balanced in static condition,
I want torque to be 0, but when I make torque
0, am I making the force also 0? Not necessarily,
that is what I put, this is, torque is 0,
this is the force, it, it can be, need not
be, I do not know, you follow. Now, if you
take it as a problem, that I need to get both
of them 0, you know just a mathematical problem.
There can be a slight thing because autorotation,
they do not keep on going all over the place
as usually, it dumps the collective rotor,
rpm will come to some stabilized value because
now, you see the minimum phi, which we said
is maximum omega. Whether your maximum omega
is what you have designed, because please
understand, you rotate the rotor at a particular
omega because that is a fixed omega, are you
going to achieve that omega or you are going
to some other omega, which is higher than
the design value or is going to be less than
the design value? So, these are other issues.
If your omega increases more than your centrifugal
load, everything because the root will get
tremendously structurally strained.
Static equilibrium is, once static equilibrium,
this is 0 you say, understood, but our situation
is this and you have a dynamics also it is
there.
So, you will have your dynamics is what? The
mass, you may, suppose this is, this is, that
is why it is little bit complex. Problem there
is a mass center; mass center will have acceleration,
agreed. So, it will have, because this is
rotating, it will have acceleration towards
the center, it can have a tangential acceleration,
agreed. That is the mass center, and then,
this body is also, there can be an angular
acceleration, agreed.
Your angular acceleration must be 0, then
torque is 0, but your tangential acceleration,
is it 0? Because you want it rotate at constant
omega, constant omega if it rotates means,
this must be 0, because there is no tangential
acceleration of the center of mass. Because
this is basic dynamic you know, that this
is, this is r, this is r alpha, you know that
from basic dynamics, rigid body dynamics.
Tangential acceleration r is fixed, constant
r and that is alpha, now this should be 0.
So, you need to have, every condition should
be whether you will satisfy, usually torque
is set 0, that means, your rpm is take it.
But I do not know, whether the forces are
really balanced to 0, but it may come very
close to 0. See, there will be some redistribution
of the loads because you need to maintain
because you do not want the rotor to, that
is why pilot dump the, may not hold it at
the same value, he may keep changing a little
bit, that only you have to ask the pilot.
But as Physics of the problem for a 2- D aerofoil
it is explained, but then real situation is
much more complex because you will find different
sections of the rotor blade operate in different,
different conditions. Some will be accelerating,
some will be decelerating, the condition of
these effects, is it clear. So, I have confused
you enough. So, autorotation is otherwise
simple; if you do not want to get into this
business, autorotation is, power is 0, that
what, what and that is good enough, yeah…
No, no, this is alpha, this, this is X-axis
is, yeah, yeah…
Yeah, yeah, all these points are stable, will
be stable here, every point is stable.
Yeah, yeah, see this is done for a section,
you understand. But that is why I said, section
means, which section you are exactly balancing
this. That means, somewhere F X must be 0,
they may be here and here I do not know where
they are, but other places you are having
the force in those places, you are actually
operating faster.
See, this is for a section, this is for the
rotor, every section cannot operate here,
it is not possible to operate, you understand.
This is shown only two-dimensional aerofoil
characteristic; that is all, is it clear,
am I… You integrate over the blade…
Yeah, yeah, see some point you may be here,
some point you may be below. So, some points
you are accelerating, some points you are
decelerating, integrate, that is what? Some
points you are accelerating, you are decelerating,
net is, net is 0, that is why you have to
integrate the along the blade knowing the
characteristic. Otherwise, mostly in the industry,
they may take in my opinion a typical section.
Typical section in a helicopter is about 70
percent or 75 percent, but that is, I do not
know for the autorotation, whether they take
that value, it could be even 60, I do not
know, but you may be able to tell me. For
that section you satisfy this, after that
rest of, then take over. But here you see,
please understand, you have not considered
the elastic twist of the blade, nothing is
considered. This is like a rigid blade because
I keep a pitch angle and then only the inflow
angle is taken, you follow.
But in real life it is not just the inflow
alone, you are going to have even elastic
twist of the blade and then you will also
have blade flapping, please understand. So,
that will also give velocity. So, it is a,
it is a more complex thing, that is why, what
industry will do? Autorotation is, you have
the code, you say power 0, the condition for,
which you keep on solving the problem for
different descent velocities, you solve the
problem, the trim of the helicopter, you solve
the entire problem assuming some inflow velocity
because that you have to take this curve approximation.
You say, if I design at this velocity, this
is my inflow, calculate every section, what
is the drag force, integrate over the whole
thing, you will know what is the power.
So, they will keep on drawing the power curve
with respect to climb and descent, wherever
it meets 0 or you extrapolate to 0, that is
the descent velocity that is all, they go
for power evaluation not by section. But for
explanation from Physics point of view, this
is done because it depends on the aerofoil
characteristic, but he is told reduce. That
is why, why pilot is, whenever they say you
decrease the collective, do not, see, there
is very, very, see the general tendency is,
if you want to go up what you do?
You increase the collective, but provided
the rotor is attached flow, you increase the
collective lift, will go up. Sometimes if
your angle is already large, you increase,
you may be going down further. Then, in that
case it defines your logic, hey I have to
reduce the collective, so that get it attached
flow after that climb. But usually, the instinct
says, hey if I reduce the collective, I am
going to go further down because in autorotation
what is going to happen is going to come down,
the helicopter is going to come down, engine
fails because there is no power, it is going
to come down.
At that time what we have to do? We have to
decrease it, you follow, and you decrease
it to, because if you are here, that is all,
your rotor will stop and it will start rotating
backwards. So, you decrease it such that you
are in a stable zone, that is why he is told
decrease, but how much he should decrease,
that may be way to do vehicle or something.
But this is drawn for aerofoil, not for the
integrated value of the rotor because integrated
if you have to draw, that is difficult task;
yeah…
No, no lag hinge may be there. See, lag hinge,
yes, there is, because please understand,
moment you have to come, a force will come.
See, lag-hinge is not right at the center
of the shaft, you understand. How rotor is
rotated? How power is given to that, how do
you give? Because if you have a hinge, suppose
you say, this is a hinge, by rotating this
will you rotate this, how no hinge? See, this
is the shaft, I put a hinge, if I rotate this
what nothing will happen? So, I cannot have
centrally lag-hinged rotor, I, yeah…
From the hinge it will go. How it will go
is, this is what, you take this, this is the
or in this diagram, let us take, this is the…
You have a hinge in this direction and you
may have your blade starting here. Here is
the hinge, this is free to move back and forth
about this point, but then a force will come
here and that force gets transferred as a
force and moment and that is how you will
be able to rotate the helicopter, otherwise
if I directly put the hinge right here, if
I rotate the shaft, blade will not rotate.
So, you do not have that situation always.
You will find, that transfer of power is there
by the offset, but this offset is most of
the rotor they will have.
That is all transfers to the hinge; it will
go, transfer, it will get transferred. Because
otherwise, without, see hinge is, is a good
design because you do not transfer the moment,
but if you do not transfer anything at all,
then how will you rotate in the first place
because if I put a ball bearing, attach the
rotor and in the inner rays I attach the shaft.
What will you do if you rotate the shaft?
Blade will just stay as it is, it will not,
blade will not rotate, follow. You have to
have an offset and of course, the offset come
because of physical constrain also. So, you
need to transfer the rotation, is this clear
or to an extent?
Because this looks a little interesting, I
thought I will give the physics because there
are people from industries. So, it will be
interesting to know how things are really
happening in actual, it is a more complex
problem. Now, let us, this is just a brief
thing, what you call the ground effect because
this is just for information to you because
there is no calculation or anything like that.
This is the ground effect; what ground effect
means? Your rotor is operating close to the
ground rather than far away. Because if you
look at it just from what happens, the rotor
inflow comes, it hits the grounder, that means,
the normal velocity should go to 0 at the
ground level.
Now, this particular thing is idealized very
simplistically. Your rotor inflow, you put
a mirror image here because this is a method
of mirror image. You put another, as though
fictitious rotor is rotating beneath the ground,
such that both of them are giving an inflow.
This inflow will come up like this, that inflow
goes down, both cancel out and leaving earth
as the, ground as 0.
Now, the effect of this will be to reduce
the inflow there because you are pushing the.
So, what happens is, to lift the same weight
the power required to hover near the ground
is less. Because you have induced basically
power, induced power, your induced velocity
you require less, therefore the power is less.
So, if you are hovering near the ground, you
have, you require less power, but in other
words the same thing is plotted in terms of
thrust. You, if you give the same power, that
means, you can lift the more weight near the
ground.
So, that is why, that particular thing is
shown as a diagram, that is, T over T infinity,
T is the thrust. Rotor thrust over T infinity
is when it is far away from the ground and
the curve actually starts from 1 to 2. Here,
Z over R is the height of the rotor with respect
to the ground.
Now, you see, as you come near, take this
curve, that green, as you come near to the
ground. Your lifting capability increases
for the same power because this is drawn for
same power. So, you see, I can lift more weight
if I am near the ground. So, this particular
thing is used because if you want to carry
a little they also go a little forward, so
that the power required becomes less. So,
you carry a little extra load and then you
can take off and fly, but of course, this
effect is highly dependent on forward speed,
which I meant is cross wind. Suppose, you
are hovering near the ground, suppose there
is a cross wind, what will happen? The wake
will, instead of coming down like this, if
we swift, then suddenly rotor will lose thrust,
that is why, this is also dependent on the
forward velocity, the effect of ground.
Depends on if the forward velocity is to induced
velocity, if it is more than 2, you see, is
has no effect T by T infinity it has. The
whole wake is pushed backwards and the effect
of ground is lost. It is quite sensitive to
side winds. Usually, the thumb rule is, if
you go beyond 1 rotor diameter, ground effect
is, you can take it as 0, there is no ground
effect; 1 rotor diameter, that is the general
thumb rule. And there are some expressions
given some earlier because we are not going
to use this. Basically, this is for information
because if it is near the ground, you can
develop more thrust that is all, for the same
power because this is called the ground effect
of the helicopter rotor. From now on, it is
forward flight. The entire course will go
because this is much more complex problem,
because I have indicated the complexity in
hover.
Then, of course, climb in forward flight,
why it is complex? Because we split the entire
rotor disk, assume, because later, we will
define all the velocity components motion,
etcetera, everything will be defined. You
say, this is a rotor disk you are viewing
from the top. Now, here in this course, I
normally take counterclockwise rotation is
positive and my rotor is rotating in the counterclockwise
direction, but usually you may ask, which
direction the rotor rotates? Europeans rotate
clockwise, Americans rotate counter clockwise,
why it is anybody’s guess, it does not matter
as far as the rotor. Only thing is, pilot
has to get adjusted to that little bit the
direction of rotation, otherwise all the calculations
everything is same. Somehow American helicopters,
they always rotate counterclockwise, whereas
all these European, the Germans, everybody,
they, they rotate clockwise.
So, we are going to take the counterclockwise
rotation positive. So, that is why, I put
psi and psi is basically the azimuth location,
which is omega T and omega is the rotor angular
velocity. And if you take kth blade, it will
become psi k. Now, you see this rotor, in
the earlier case it was just stationary, now
it is moving forward.
So, the forward direction I put it up. So,
there is a relative wind V, which is coming
towards the rotor. Now, this is also rotating,
this is having omega r, so you will find...
So, this is, you will have this is omega r
and you have V and this is psi, that means,
in this half, the relative air actually acts.
That means, the rotor section experiences
a larger velocity of wind, whereas in this
half, this is actually moving this way; that
wind is also coming, so there is a reduction
in the velocity. So, this is advancing side,
this is retreating side, this is briefly
Now, you immediately know because my velocity
is now varying, oncoming flow is time varying
because if you take a cross-section or basic
aerodynamics says, what is the normal velocity,
but that normal velocity is no longer a constant,
it is varying with time. So, now, you know.
In your aerodynamics we have a situation where
my oncoming velocity is time varying. This
is the first complexity between fixed wing
and rotor wing because you do not do. In fixed
wing theory the time varying, oncoming flow,
it is all fixed.
Now, what happens because of that? Because
of that if you use directly the formula, lift
is what? Half rho dynamic pressure chord C
l lift per unit length, this is varying, time
varying and square, that means, if it is 1
harmonic, it will go to 2 harmonic also. So,
your loads are now having time varying, they
vary with time. The moment you have any load
lift is a lift holds acting if aircraft is
steady flight, please understand. This is
a steady velocity, V is steady, omega is steady.
Even if both of them are steady, I am having
my lift force as a result all the aerodynamic
forces changing with time. When they change
with time, because this is like a beam, I
am having a load, which is changing, that
means what? That blade is going to respond
because it is a flexible blade, it will respond.
Now, when the blade responds, so you see higher
velocity in advancing side, then retreating
side. Now, this is the first complexity.
Now, blade responds this periodicity because
you are varying cause. This is the root cause
for periodic aerodynamic loads, please understand.
Periodic, I did not say harmonic. It is periodic
because you will have, all harmonics will
come because you, you will be, you will see
later, is a highly complex.
You, even if you take first harmonic because
this is 1 per revolution you are putting square.
When you put a square, if a sine psi is there
it is going to become sine square psi; sine
square psi you can put it in cosine to psi
form. That means, you are getting second harmonic.
Now, you will find, slowly you will keep on
adding all the harmonics, harmonics means
time varying.
The moment you have time varying, what you
are going to have? The blade is going to respond;
blade is going to vibrate. Now, blade response
you cannot neglect it in forward flight. You
cannot say, my rotor is rigid, it does not,
no matter I may oscillate my load, but my
blade is not going to respond at all, that
is why, that is our assumption you can make.
But then, you cannot fly because then it is
like a propeller and the propeller is very
stiff, whereas in helicopter, the blades are
not stiff, they are very, very flexible. So,
response of the blade you have to include,
this is what the aero-elasticity. Now, you
see, aero-elasticity cannot be separated from
helicopter dynamics in forward flight I need.
In hover, you please understand, we never
talked about blade motion, only blade pitch
angle, inflow, nothing else. We did not even
mention about flap, lag or torsional deformation,
nothing. We treated as though blade is rigid,
we did not even mention about its flexibility,
whereas in forward flight, you cannot do that,
you have to have the blade response.
Now, when the blade starts moving up and down,
now this is adding another complexity. My
load is changing; as a result, my blade is
oscillating. Now, my blade is oscillating,
therefore what will happen? My load, also
because my angle of attack, everything will
change, as a result, it is an aero-elastic
problem. You cannot split these two because
what we normally study in aerodynamics? There
is a steady flow, the angle of the aerofoil
is captured, some pitch angle you find, the
lift you find, the drag wonderful and you
find the pitching moment; this is a steady
case.
Now, here, one is, the V is changing; because
of the V, this load is changing; because of
the load change, this is going to oscillate
up and down; now, when this oscillates up
and down, that is going to change the angle
of attack, everything. So, my load is, this
is the simple aero-elastic problem, but it
is complicated.
Now, how do I solve? I do not have any theory,
which says, that this is an unsteady aerodynamic
situation, but I still use whatever you have
learnt in basic aerodynamics steady flow.
We will use the same condition and that is
what is, of course it is complicated, in research
level you complicate the whole thing, but
here for the course, you know, we will do
it very, very simplistic.
And then of course, you see, because vibration,
because we have this is oscillating, so you
have vibration. You also have these two phenomena,
which is stall, the blade will stall. Why
the blade should stall you may ask? The stalling
is your angle of attack, is going up beyond
the, whatever the stall angle here, advancing
side your velocity is more, that means what?
You will generate more lift, please understand.
Retreating side, velocity is less. If you
keep the pitch angle same, then one side you
generate more lift, other side you generate
less lift assuming inflow is same; inflow
is another problem. Then, what will happen?
The helicopter will roll. So, what is done
is, pitch angle at the advancing side you
reduce it, on the retreating side you increase
the pitch angle such that you balance both
and this is done through the cyclic control,
which you had. Now, when you increase the
pitch angle you may also stall somewhere that
happens at some forward speed. So, you will
have dynamic stall as part of aero-elastic
problem.
Then, the other one I mentioned, reverse flow.
You know, that this flow is coming, this is
omega r. If this is small, omega r is small
near the root, the flow will be coming from
trailing edge to the leading edge, this is
called the reverse flow. So, you have all
problems related to aerodynamics in the forward
flight.
