The, this figure is taken from Patfield book.
What the kind of maneuver we are talking about
is, see, the helicopter goes in a spiraling
path, but it can have a sideslip also. Now,
if this is the kind of motion, how do you
trim the helicopter? Because it is a very
complicated maneuver, not like our what we
solved earlier, level flight. But if we solve
or if we formulate equations for this problem,
then you will be able to solve any problem,
which will be a simplified version of this.
So, essentially, the helicopter attitude can
be anything and the velocity vector V f is
oriented in space and that vector is also
turning.
Now, the flight speed is given by velocity
vector, you may say, that is a constant, the
magnitude is a constant. And you will specify
a flight path angle, that is, you have to
define what is that flight path angle because
this vector is oriented in space in some direction
and then, you also define one turn rate, the
rate that, which the vertical from the helicopter,
other helicopter is spinning. You can say
turn rate, spin rate, anything, but that is
about a vertical axis.
And of course, one more quantity you specify,
the sideslip. You will define what sideslip
is, the only, these quantity are specified,
they have nothing to do with that type of
equation they are looking at, but if you see
overall for this type of a maneuver, you have
to treat nine equations.
Those nine equations are - three force balance,
three force balance equations and three moment
balance 
and then another, 3 kinematic relations. Now,
force balance, moment balance, you know for
a rigid body because you say f is equal to
a method of Newton’s law and the moment
balance is moment related to angular acceleration,
etcetera.
But the kinematic relation represents, essentially,
the instantaneous angular acceleration or,
sorry, angular velocity instantaneous because
at every instant the aircraft is turning.
So, it will have an angular velocity, which
is a vector, but how that angular velocity
vector is related to the orientation rate
of the helicopter with respect to some earth
fixed coordinate system.
Now, this is, like, orientation of the helicopter
is one, angular velocity is another, how do
we relate this? This is where you have to
have, now, coordinate systems very precisely
marked.
First, you write the body fixed system; body
fixed coordinate system. This is, it will,
you take the helicopter, you, and please note,
I am using slightly different coordinate system
for this case because this is what, generally,
most of the flight dynamic people use, x body
and then, this is y body and then, this is
z body. This is the center of mass of the
helicopter; this is body fixed coordinate
system. That means, when the body turns, this
axis system also turns and this is attached
to the centre of mass.
No, no, z is going down, you can take this
as the dash, dash, dash; this is going in;
this is into the board. So, you can say, x
b, z b is in the plane of the board, this
is the standard thing. Now, how do I get this?
This means you have to first define an earth
coordinate system, earth coordinate system.
You can say this is my x earth, z earth and
then, this is y earth; you can have this any
direction, local.
Now, you see the sequence of rotation from
this to this is actually, you follow yaw,
then pitch, then roll. And yaw is psi, I think,
then it is, theta is pitch and then, phi is
roll; yaw, pitch, roll, this is the sequence,
all counterclockwise. So, that is what I have
shown here. If you look at it, first this
is the z earth, x earth, y earth; I give a
psi rotation counterclockwise.
I get the coordinates system, x 1, y 1 and
of course, z 1 is same as z ea, I am not writing
the transformation matrix fully, then what
you do is you take y 1, x 1, z 1, give a pitch
rotation, then you will get x 2, z 2 and y
2; y2 and y 1 will be same.
Next, you go to the roll, about x 2. Now,
this is the sequence of transformation. If
you follow this sequence, this is the very
well set orthogonal transformation.
I will just finally, write the transformation
between, this is the x b, y b, z b, this is
cosine theta, cosine psi, then cosine theta
sine psi sine theta, then sine phi sine theta
cosine psi minus cosine phi sine psi, then
sine phi sine theta sine psi plus cosine phi
cosine psi and then, this is sine phi cosine
theta. And here, cosine phi sine theta cosine
psi plus sine phi sine psi, this is cosine
phi sine theta sine psi minus sine phi cosine
psi and then, cosine phi cosine theta. And
this is x earth, y earth, z earth, this is
the full because this you may find in text
books also this transformation, because this
is the product of 3 matrices, that is all.
But follow this sequence, that is the most
important thing; yaw, pitch, roll sequence.
Now, what is the angular velocity of the helicopter?
At any instant, you define by angular velocity
of the helicopter as pe xb qe yb plus re zb,
the, that e is the unit vector. Now, this
is the instantaneous angular velocity vector
defined with respect to body axis.
Now, you will see, if you want to know what
is the angular velocity vector, but defined
in terms of time, rate of change of these
angles because when the body is rotating,
which means, they are also rotating with respect
to the earth axis. But you will find the same
omega, you can write it as, I am putting it,
psi dot e z earth because this is with respect
to the earth and then, plus theta dot e y1
because look at this, theta dot is y 1 plus
phi dot e x. These are time rate of change
of orientation of the vehicle, as defined
by my yaw, pitch, roll sequence, because yaw
is about the z ea, this is about the rotator
y 1.
Now, what you do is you make a transformation
of each one of them to body axis because you
will find each one is the intermediate position,
you can get the transformation from z a to
x b, y b, z b; y 1 also, you can transform,
but this is actually the body, phi dot is
along the body because that is x 2, which
will be the body axis.
If you do the transformation, you will get
this relationship, here is this part here.
I will write the relationship is finally,
you convert everything into x b, y b, z b;
you will get p equals phi dot minus psi dot
sine theta and q equals theta dot cosine phi
plus psi dot sine phi sine, sorry, cosine
theta and then, r is minus theta dot sine
phi plus psi dot cosine phi cosine theta.
So, this is the relationship between instantaneous
angular velocity to rate of change of the
orientation angle. But please remember, this
relation is not that, they also depend on
what is the theta and phi. That means, you
have to solve for the orientation angle, this
is a non-linear relationship because theta
dot is there, then there is also theta, so
this is where, solving this problem, this
is a, this is what navigation, even the satellite
communication navigation, everything this
is what is used.
So, now, you have a relationship between instantaneous
angular velocity to orientation of the vehicle
because orientation of the vehicle can be
with respect to earth axis. In any orientation,
you always want to know what orientation it
is there, so particularly, if you are designing
autonomous, any vehicle, you have to know
all this, yeah, and the estimation of this.
Now, you can always solve for this in terms
of p, q, r also, that is like you take because
psi dot, phi dot, theta dot, you can get it
in terms of p, q, r, but again, theta will
be sitting, theta and phi. So, this is like
a just transformation of one to the other
coordinate, we will not use that.
Now, what I want is the body, what I specify
is my velocity vector, I will specify velocity
vector of the helicopter, which is actually,
I define my velocity vector as u e xb plus
v e yb plus w e zb. That means, this is my
u, v, w and you have p, q and r.
So, at, at a time, I have a velocity of the
body and the angular velocity of the body
defined with respect to body fixed system.
Now, if I want, yeah
Yes, as we have defined here, that is why,
that sequence you follow. What sequence, that
is the most important thing; this is the sequence
we are following. Suppose, if you change the
sequence, you will find the different thing,
usually, that is why I said, yaw, pitch, roll
sequence is followed, so that the convention
everybody understand, and the coordinate system
also, they try to use the same thing.
Now, the question is, I have the velocity
and the angular velocity, translational angular,
all, now I can go and calculate, what is my
acceleration of the centre mass. So, maybe,
I erase this part because this transformation,
I use it as a transformation from this coordinate
to this coordinate, that is all.
Now, my velocity is, velocity of the centre
of mass is u e xb v e yb plus w e zb. Now,
what is my acceleration of the centre of mass,
time derivative of this and then, these vectors
are also changing their orientation, but they
change with the angular velocity. So, omega
cross V will go. So, you will have, u dot
e xb plus v dot e yb plus w dot plus omega
cross V. Now, you put, the omega is there;
p, q, r, you use p, q, r; do not use this
omega because that we will keep it as separate
relationship, always use p, q, r. Take a cross
product, you will have the acceleration of
the vehicle, vehicle. I will write it here,
this is u dot minus rv plus qw e xb plus v
dot minus pw plus ru e yb plus w dot minus
qu plus pv e zb. So, I have my acceleration,
only thing is, I have to now apply Newton’s,
this is the absolute acceleration of the.
So, what are the loads that act on the helicopter?
You have a main rotor load, tail rotor load,
fuselage aerodynamics and then, any external
bodies, aerodynamic surfaces, then gravity,
all. So, what you do is you put, take all
the loads, transfer them to the CG. So, you
will write, f x equals mass of the helicopter
times, this f y is mass of the helicopter,
f z is this; that means, you have 3 force
equations.
So, you have F x is u dot minus r v plus q
w, but what normally done is, the gravity
load you write it separately, so you will
get, I am putting it, because please note,
that gravity load acts along on z earth. So,
you transform the z earth to body and that
will come.
And then, F y because this is the standard
way of writing, v dot minus pw plus ru minus
mg sine phi cosine theta. And then F z is
m w dot qu plus pv minus mg cosine phi cosine
theta. So, these are my three force equations.
Can you look to the transformation relation,
you will find, that this is very obvious,
all right.
Now, you have to get the rotation equations;
rotation equations are a bit messy because
you know, if you have done rigid body dynamics,
you know, that angular momentum, angular momentum
is given by I omega. So, I am writing here
H x, H y, H z because angular momentum is
a vector, this is symmetric matrix, I yz minus
I zx I zy I zz, sorry, the p, q, r, this is
my angular momentum of the body H vector.
So, H x, this is along the body axis. Please
note that because this can be derived, I am
not going to the derivation of this because
this you must have done in the rigid body
dynamics.
Angular momentum and the axis system is the
body axis system, that is why, in any aircraft,
anything, you first decide your axis system
and then I xx, I yy, all these quantities
are defined with respect to that axis system.
Now, imagine, if you put a load, if you put
some equipment, your initial change, that
is, the reason for using body fixed axis system
is only to simplify the moment equation, otherwise
it will become really missing because these
quantities, they do not change, you say with
the body fixed system. Of course, if fuel
is expanding, it is changing, but you do not
bother about that.
So, this inertia, this is the inertia tensor,
I hope you know, that this is a 3 by 3, it
is known as inertia tensor because this follows
the tensor transformation, that is all. Now,
what I want is that is the moment, this is
defined with respect to my axis system in
the body p, q, r. Now, I want moment, moment
is essentially H dot, this is about the C.G.,
this is about the C.G. If you want, I can
write it the C.G. H dot moment, about the
C.G. plus omega cross H because these vectors,
they are also changing with time. Therefore,
when I take a derivative of that, I will get
omega cross H, so this is my moment equation.
So, I may put C.G., that is why, always you
take moment equation about the centre of mass,
it makes simpler. Now, what you do is, this
is the moment, due to all the external loads,
anyway, the gravity passes through the C.G.,
so it will not create any moment. You have
to take a dot product of this, which means,
you expand the whole thing, take a dot, dot
means p dot, q dot, r dot, that will come.
Then omega cross H, omega is p, q, r; H is
also a function of p, q, r, so you have to
take the cross product, write the full equation,
that is the correct exact way. But usually,
what happens is they make certain assumptions.
The assumption is, if you write the full stuff
it becomes a little messy, so I leave that
part, I will make the simplified assumption,
that cross product of inertia, that is, these
quantities.
If it is a symmetric body, if it is a sphere,
perfect sphere, then the cross product is
0, you will have I xx, I yy, I zz, but if
it is a symmetric body about 1 plane, which
is the longitudinal plane, if you take then
what will have? For every plus y, you will
have a minus y, so wherever y is there, they
will go off. That means, these quantities,
they will go to 0, but this will stay I xz,
because they say, that number for help is
for aircraft or anything is substantial. Even
though, this may not be 0, I xy, I yz, you
neglect it, but if you do not want to neglect,
carry the entire term, it is not very difficult.
If you have computational aspect, you can
carry it, it is not a problem.
But for ease of understanding, they try to
throw away lot of things and then try to get
a simplified version. Now, I will write generally
all cross products, except I xz. That means,
these are neglected, that is because you say,
but in the aircraft case, yes, there is symmetry,
but in the helicopter it is not exactly symmetric.
Because you have tail rotors sticking around
one side, but you say, that it is all right.
Let me make that assumption, then it becomes,
I can write it in small, so I will have my
M x. M x is moment of the all the loads, that
is, the rotor loads, tail rotor, horizontal
surface, every aerodynamic load, whatever.
You get inertia, aerodynamic load, everything,
transfer it to the C.G., that is the along
the x direction is the M x.
That is why we get c m x, c m y, c m z, all
those things. The rotor, they transfer to
fuselage centre of mass, this becomes p dot
minus I xz because I have made the assumption
r dot plus pq minus qr into I yy minus I zz
and then, M y is I yy q dot minus I xz r square
minus p square minus rp I zz minus I xx. Then,
M z I zz r dot minus I xz p dot minus q r
minus p q I xx minus 
I yy, these are my, normally you see, I xz
is comparable to I xx magnitude wise.
So, this has to be, that is why weights and
C.G. group in any aircraft industry or aircraft
or aerospace, any company, they have to get
these values. Before the aircraft is build,
it is like, you make estimate from the drawing
and then you try to calculate the complete
I xx, I yy, I zz, all these numbers for a
given coordinate system because if you make
a mistake, there your flight dynamics is all
wrong. So, this is done, that is why some
time it is a routine job, but one has to do
it in an aircraft industry. As a matter of
fact, all your dynamics is only this, but
of course, I made an approximation. If I do
not make approximation, that is, all the full
dynamics of a rigid body, there is nothing
more beyond all problems, whatever. You have
learnt in physics, one on one, or dynamics
you can solve with this, with this set only,
only thing is you must be very clear.
Now, you see, how do I solve, I have nine
equations, how do I solve for my problem because
what are my unknowns? problems, first you
say, hey I am not, I am going to fly a steady
maneuver and subsequently, I will say, if
there is a perturbation about the steady state,
what is the stability? Only these two are
solved, using these equations otherwise, this
is a fully non-linear equation. Because you
see p square, r square, p q, everything is
non-linear, so solving it is not that straight
forward. So, how do they attempt to solve
this problem? In the case of helicopters,
I will go to the helicopter thing, see what
is done is, first I say I am doing a particular
flight condition. That means, I am not accelerating;
my u, v, w are fixed; p, q, r are also fixed
Then, I will write my u is, u equilibrium
plus delta u, which is a, and v, which is
the v equilibrium and w, w equilibrium delta
w time. And then, p, p equilibrium plus delta
p; q 
and r, r equilibrium plus delta r time t.
Then, you need to have also your theta equilibrium,
phi equilibrium because theta and phi, that
is the orientation of the vehicle. So, I will
have, theta is theta equilibrium plus maybe,
delta theta. And phi, which is the phi equilibrium
plus delta phi except psi because there is
no psi equilibrium because the orientation
can change; as the helicopter turns, my angle
psi, you can keep on turning, there is no
equilibrium for that. Please remember, that
is why, that turned allow because it is like
a yaw rate, pretty much it can keep on turning.
But the vehicle orientation is theta equilibrium,
phi equilibrium there, so what you do is you
substitute these quantities now in this equation.
And then, you say, my deltas are small, product
of deltas I neglect, and I will collect all
the terms with the equilibrium quantity separately,
and collect all the terms with the delta separately.
Now, I will have one set of equations, I call
it as the steady state equilibrium equation
and another is the perturbation equation.
Perturbation equation I need to use it for
my stability, whereas equilibrium I use it
for my trim. This is how it is done, otherwise
you will go nuts. This is highly complicated
formulation, you just cannot get anything;
first you split, when you split you will write
the equilibrium equation. So, I will just
write that equilibrium force equation.
So, you, you understand this, you substitute
here, then take. So, when I take u dot, that
means, it is only delta u dot; u equilibrium
is constant value.
Now, I can write my equations. This is force
equation and moment m minus r equilibrium
v equilibrium plus q equilibrium w equilibrium
plus m g sine theta equilibrium is X equilibrium;
that is the force along X directions equilibrium
force; that means, the steady state force,
that is what it means. m minus p equilibrium
w equilibrium r equilibrium u equilibrium
minus, sorry, I should put a minus m g sine
phi equilibrium cosine theta equilibrium is
Y… Then the Z equation, minus q e u e plus
p equilibrium v equilibrium minus m g cosine
phi equilibrium cosine theta equilibrium is
Z…
Then, you write the moment equation also,
3 moment equations, that is, moment, you will
have minus I xz p equilibrium q equilibrium
I yy minus I zz q equilibrium r equilibrium
and this the standard notation L e roll, this
is roll equation. And then, minus I xz r square
minus p square equilibrium minus I zz minus
I xx r e equilibrium p equilibrium is M, we
may say equilibrium, this is pitch. And then,
the yaw equation is I xz q equilibrium r equilibrium
minus I xx I yy p equilibrium q equilibrium,
which is N equilibrium, which is the yaw.
And then, you have to change this equation
also, in this equation because you know that
these are dots. So, dot means that is the
time derivative.
So, what will happen is, p equilibrium is
minus psi dot equilibrium theta equilibrium;
q equilibrium will be sine phi eq cosine theta
eq; and then, r eq becomes psi dot equilibrium
cosine phi equilibrium cosine theta equilibrium.
Now, you see, this is my full equation, nine
equations here, here, here. Now, this I xz,
no minus 1. Now, you see these 9 equations
you have to solve. What are the quantities,
which I have to know? Now, I will show this
because I would like to come here, which will
be the nine, sorry, not nine.
The unknown quantities of the problem, what
are the unknown quantities, because when I
want, these are 9 equations, what are things
I do not know? You say, mass of the helicopter
you know, please note that, that is, is very
important, I know the mass of the helicopter;
I know the inertia properties the tenser.
External loads, these are my rotor loads and
fuselage aerodynamic loads, any horizontal
surface, any lifting surface, everything,
they provide those loads here.
I do not know what is u, u e, v e, w e. I
do not know, that angular velocity P e, Q
e, R e, I put a capital because for equilibrium
I just used a capital here, that is all. Otherwise,
if the p equilibrium, q equilibrium, all the
e is equilibrium, then this is the turn rate,
psi dot and orientation angle, theta equilibrium,
phi equilibrium and then control input 4.
So, I have thirteen quantities, which need
to be noted, but I have nine equations. Therefore,
you have to prescribe 4 quantities before
you really can solve the problem. Now, what
are those four quantities you will specify?
You will not specify four from here that is
the problem.
The four quantities, which you specify are
flight speed and flight path angle. I will
tell you what the flight path angle is, then
turn rate and then, you will say what is the
side slip angle.
The definition for side slip angle sine beta
e is v e by v fe. Please understand, now the
quantities, which are mentioned there are
different from, except this, this is different,
this is different, I do not have any, I do
not have v fe defined there, you follow, I
do not have gamma f e defined there, I have
only u e, v e equilibrium quantities.
But what I specify is in some other four quantities.
Now, you have to relate this 4 from here to
some of those quantities, that is again a,
I will briefly describe that, we will, may
be go what, to the first figure. If it is
there, I will specify, that what we need is,
this is their z earth axis. Body axis is x
b, y b, z b and velocity vector of the body
is v bar fe. The components of this are u
e, v e, w e, they are equilibrium quantities,
components of the velocity vector along the
body fixed system, please understand, is u
equilibrium, v equilibrium, w equilibrium.
Now, the flight path angle is defined, that
this is the z earth, x earth, and y earth;
the velocity vector can be in any orientation,
can be up, down, anywhere. What you do is,
you project this velocity vector on to a horizontal
plane, on to the horizontal plane; project
this vector on to the horizontal plane. And
then, the angle between the vector, velocity
vector and the horizontal plane is the flight
path angle and then you project it and the
angle between the projected vector and the
x earth, you call it as a chi; just put some
because that will come later.
Now, these are your quantities, I have the
velocity of the vehicle, I do not know what
is the chi is, but flight path angle I specify,
velocity I know; that means, I know this and
I know this, but I do not know this. But from
here, if you look at it, I know omega e, that
is the turn rate, but I do not know u e, v
e, w e, none of these quantities I know.
Now, what you do is you take the velocity
vector, transform them along the earth fixed
system first. That means, I have the velocity
vector, I am going to get components along
the earth fixed system because this is a little
interesting. Finally, you will not use that.
The velocity vector in the earth bound system
is U earth phase V fe cosine gamma fe, this
is the flight path, and cosine chi, this is
not psi, this is chi. And then, V ea, V fe
cosine gamma fe sine chi and then, w earth
is V fe sine gamma fe.
This you can get, only thing you do not know,
these quantities are mathematically, I can
write it, but I know this, given this sine
gamma fe, I know w ea. Now, what you do is
you go back, transform this in the body fixed
because you know the transformation relation
between this and the body.
When you do that, you will get, now I am,
say I am writing for, I will write it for
one quantity, U e is v fe cosine gamma fe
cosine theta equilibrium cosine chi cosine
psi equilibrium plus sine chi sine psi, this
is psi equilibrium minus sine gamma fe sine
theta equilibrium. You will find this is the
exact transformation in u e or u equilibrium,
sorry, I, we are using equilibrium. What you
will find is you will have a term cosine,
cosine plus sine, sine. This is nothing but
you call it by a chi e is chi minus… This
you call it track angle.
This is something new, track angle, that is
why you do not define it, but later you will
get, the track angle will come, you have to
solve for the track angle. Now, if I write
in this my u equilibrium in the body because
this u is in the body access system, I will
have it in this form v fe cosine gamma fe
cosine theta equilibrium cosine chi. Maybe,
I will put, if you want equilibrium, you can
put equilibrium, equilibrium minus sine gamma
fe sine theta equilibrium.
In a similar fashion, you can write v eq,
I will later tell you what is, why I am using
because in the equilibrium state you say,
that my velocity is constant, flight path
angle remains constant, orientation angles
are constant and this is also constant, they
do not vary with time. But you will find chi
and psi, what is psi equilibrium, then nothing
like psi equilibrium; it is also changing
with time.
So, I cannot put a psi because you said, psi
dot has an equilibrium; that means, psi cannot
have equilibrium because it is turning. So,
usually, what is done is you do not try to
put an equilibrium here, you say these two
quantities are varying with time, but there
difference, it is a constant.
Both of them are varying in a same amount,
the difference is a constant. And now, you
have to get that, now flight mechanics possibly
they may, because the track angle is…, Now,
what you do is, you will have your v e is
v fe cosine gamma fe sine phi equilibrium
sine theta equilibrium cosine chi equilibrium
plus cosine gamma fe because these are all
long expression phi equilibrium sine chi equilibrium
plus sin gamma fe sine phi equilibrium cosine
theta equilibrium.
Similarly, you will have w because the combination,
it comes such a way, here this sine chi, you
will have; if you substitute this, you will
have sine, sine chi cosine psi minus something.
So, that is how the expression will come,
that is why they will come on to be sine chi
and cosine chi.
Now, you will have w and your w equilibrium
will be again, v fe cosine gamma fe cosine
phi equilibrium sine theta equilibrium cosine
chi equilibrium minus cosine gamma fe sine
phi equilibrium and then, sine chi equilibrium,
then you will have one more term, plus sine
gamma fe cosine phi equilibrium cosine theta
equilibrium.
Now, you see, for me to get u e, v e, w equilibrium
quantities, I know flight path, the velocity
I know, the flight path angle is given. What
I do not know in this is, I do not know chi,
I do not know theta, I do not know phi, these
3 equilibrium quantities I do not know; you
follow what I am saying? That means, if I
want to get the velocity along the body axis,
given the quantities of these four, I still
need to know three, chi equilibrium, theta
equilibrium, phi equilibrium.
So, what do I do is, I first go and solve
for, how I define my chi equilibrium. What
is given here is sine beta e because side
slip angle is specified, you have to specify
what is side slip; side slip angle is what?
This is v equilibrium in the body system,
whether you are moving along the y body, along
the y axis of the body system, if you have
a velocity there, the ratio of that velocity
to the total velocity resultant; that is my
side slip. That means, side slip angle is
specified; that is a given quantity. Now,
you know from here v equilibrium, v by v fe,
this is nothing but sine beta, which is the
cyclic.
So, I write sine beta e, this entire expression,
that is, sine beta e is cosine gamma fe sine
phi equilibrium sine theta equilibrium cosine
chi equilibrium, then plus cosine gamma fe
cosine phi equilibrium sine chi equilibrium,
maybe I should put it, plus sine gamma fe
sine phi equilibrium cosine theta equilibrium.
Please note, that here, what I will do is,
I know this quantity, this quantity I know,
I can form a quadratic equation in sine of
them, call this as k 1, k 2, k 3 and then,
take this term sine square cos square, something
chi, you solve that quadratic equation, you
can write it as k 1 cosine chi e equilibrium
plus k 2 sine chi equilibrium minus k 3 equal
0. Because you bring this term here, you can
say this is k 1, this is k 2, and k 3 is,
you bring this term and then put a minus sign.
Now, what you do is, you convert to either
sine to cos or cos to sine and then take this
term on this side, this term on the other
side, square both, you will get cosine square.
Then, this is sine square, sine square can
be or cosine square can be written as 1 minus
sine square. So, you will have an equation
in sine chi. So, finally, you will write sine
chi equilibrium is given by k 2 k 3 plus minus
root of k 1 power 4 k 1 square k 2 square
minus k 3 square k 1 square over k 1 square
plus k 2 square. This is what you will get.
That means, there are two routes, you always
pick the meaningful route, that is what, in
that, what you know in this is, knowing beta
e side slip angle and you have to assume phi
equilibrium, theta equilibrium and then, these
are assumed, gamma fe is given. That means,
when you are starting, you specify those 4
quantities; first thing is I will go and get
the track angle.
How do I get track angle? First I solve, I
assume the orientation of the helicopter,
you assume, then you get that track angle.
Once you get the track angle, you have already
assumed these two, track angle you know, gamma
fe you know, you can immediately get u e,
v e, w e. That means, you have all the three
quantities and given psi dot, that is the
input equilibrium turn rate. You have already
assumed theta eq, v eq; that means, I know
p e, q e, r e, you go back the six equations
and then solve. I will just briefly mention
because this procedure is a quite complicated,
that is where the convergence, all the, see
here, I briefly describe the procedure, that
is done.
Trim procedure, I just put it complex. Given
the data, velocity flight path and then turned
rate, side slip. Once I have this, I go back
and solve first the track angle, but assuming
theta eq and phi eq, so that is why, assume.
I say, here these 4 quantities are pilot control
input, I assume these two quantities, which
are eq, compute the track angle. Once you
get the track angle, obtain u equilibrium,
velocity equilibrium, w equilibrium, then
P, Q, R. There, I use capital, here P, Q,
R. Then, you solve rotor inflow, that is,
the, you may assume uniform inflow or any
other inflow, first assume that, go and solve
rotor inflow. Once you get the rotor inflow,
then you solve the blade equations, even if
you have only flap equation just solve them.
But you should write the flap equation based
on a hub, which is having all these motions
because you cannot use that level flight flap
equation. So, your flap equation must contain
the hub motion. That means, all the equations,
you have to develop, then solve them, blade
flap, assuming only flap. If you have lead
lag torsion everything, then you have to solve
the full, get the rotor hub loads and other
aerodynamic loads, transfer them to the fuselage,
get the force and moment equilibrium. You
will find, that they do not match, but you
take the average load. Please understand,
you do not hub loads only, you get, obtain
the rotor hub loads. That means, you take
1 revolution, get the mean values, then transfer
the mean values for balancing the force and
moments, it will not balance this. You have
six unknowns, which are these pilot input
and orientation of the vehicle.
So, six equations, you use these six. Now,
once you get this, you go back here because
for getting the blade nodes, you need the
pitch input, otherwise you cannot get the
blade load. That is why, you assume this once,
you get a new update, go back here again,
compute the track angle again, calculate all
the u e, v e, w e and then, you do this. This
is quite a complicated procedure, sometimes
the program will not converge; it is very,
very tricky problem and this is the complicated
trim procedure.
Pardon.
No, it is not how long, it is actually the
convergence becomes complicated problem. See,
we are going to do this because we have developed
the equations now we plan to do, but initially
we will do a steady turn.
Do not go and complicate all sides of, just
give a, we have done for a flap, we need a
steady turn and we are getting results, which
are quite similar to what are obtained by
some other people.
Now, we want to do more complicated motion,
ultimately the problem is the industry required.
Hey, if I give a control input how my vehicle
will perform, that one is you can go and solve
my linearized, that delta u, delta v, delta
w, that equation, that is a perturbation only,
but that is, will give you only at the instant
what this happening, it will not tell you
over a long time because long time. However,
you would have done the equilibrium about
a hovering condition and a perturbation about
the hovering condition.
But you go to forward flight at some other
speed, then you will have equilibrium under
perturbation, for that, like that you will
have equilibrium, perturbation equilibrium,
perturbation, you cannot say this perturbation
will lead to that perturbation. If you want
to solve the full problem, you have to solve
complete, non-linear problem and this is where
in between you have to get the rotor inflow.
Please understand this is also equally complicated
problem because rotor inflow is related to
rotor thrust, so you can have, you know, various
types of inflow models, dynamic wake model
and we have incorporated something, which
is model, you can put, but that is again a
constant, even though it varies along the
radial and azimuth, but it does not vary with
time. So, you can have different types of
inflow models.
So, usually, these problems, first you try
to solve in a simplified way. Take only flap,
check whether a program is converging because
you will be able to get the mistakes from
that. Once it is done, then you can say, these
are my loads and if you want to include stall
model, please understand, when I get the aerodynamic
load I should have stall model for the blade,
all those things. I cannot have a closed form
expression, everything has to be integrated
in these this part, so that is why the aero-elastic
problem is part of flight mechanics problem.
It is not that this in independent problem
in helicopters and if you are doing this kind
of a maneuver, this is, this is a real challenge
problem. I think with this we close.
