A good morning, last class we discussed about recall it was Steady side slip and how do
you define a steady side slip, suppose a plane is moving like this straight like this then
factory angle is 0 and if the airplane is instead moving like this we have a positive
side slip that is if I draw a diagram. If I flying such that relative air speed is coming
from the right that is I am moving like this so relative air is coming from the right if
I see from like this lady where should come from the right okay.
. So this is a positive side slip angle, and
we have seen to execute a positive side slip angle I need to give a null deflection I need
to give another deflection and we have seen how their reflections are related, and also
we have seen that I can use studies actually maneuver to estimate some aerodynamic parameters.
Especially C and beta right, and you can do it for estimate a few other parameters as
well. By doing experiments by using some subset of internal data the basic philosophy is to
in this course is to educate you how to conduct explain how to think.
The second part today what we will be doing is basically we will be doing steady coordinated
turn; steady coordinated turn that is what to do will be giving a brief. What is the
steady coordinate turn see earlier we are discussing about the airplane let us say having
a cruise it is going like level cruise is going like this right. For a steady sight
sleep airplane is going like this but for the coordinated turn steady coordinator turn
I am turning like this and that I can do if I have some bank and I am using some other
deflection. .
I can as well do it like this just give another deflection but the problem is as I get on
a deflection you could see this left wing will see a larger velocity because of this
turn so there will be more left or it will try to bank so you use both rudder as well
as a lid on and you turn it like this so that beta is zero all the time. That is very important
in a steady cornea terms beta is 0 and of course you are ensure that you are not losing
the altitude, but if that is agreed that I am executing a maneuver using a lid on and
others and some throttle setting. Such that I am turning, such that beta is
equal to 0 then how the sixth of equation will get modified the first thing we will
see right because when you will be doing experiment and will be putting their data I should know
what is a mathematical model I am going to use which will be a subset of a 6 12equations
of motions right. So let me take up that for the steady turn.
. Let the FY equation from a six off you are
aware it is half Rho v2 is to Cy into p b / 2 v plus cy r into r b /2 v plus cy beta
into beta cy delta A into Delta A Cy Delta r into Delta r this is the aerodynamic force.
Which are coming because of chill in the sixth off if they roll it is a unit if there is
a beta if there is a alone deflection greatest another deflection. Of course the FY will
have some component because of the bank will be compared of the weight right that tell
me that mg generator with mg cos theta sine five.
. So this is I think knew I have written this
is part of the one of the equations of the sixth of model you should understand before
you do an experiment that this part is aerodynamic force and this part is because of the gravity
and theta is the pitch angle okay. Now the question is we are talking about study coordinator
turn I like to investigate which of these terms will contribute for it steady coordinated
ton .What is CY please let us understand what is the CY be it is that side force let us
say if I take the axis system which conventionally follow this is X this is Y and this is Z.
. So if I use this as if I assume this aircraft
then if I am moving in this direction you know x axis is this y axis is towards view
and z is downward right X Y Z said right and now what is beta positive beta positive is
the wind is coming from the right so this is the beta positive direction and if the
wind is coming from the right and there is a vertical tail which way the vertical tail
will experience a force it will be in the opposite of y direction is it clear.
If I see it like this is X this is y and this is z and positive beta wind coming from here
and this is a vertical tail so vertical tail experience force towards me and this direction
is opposite to positive y direction ok. So for a positive beta CY which is a non-dimensional
coefficient of FY will be negative that is I say CY beta will be negative well, right
ok. And by diagram wise if I write this is my Hooks law this is the tail this is the
wing if this is positive beta let me draw an effort if this is positive beta.
I know my convention this is X this is Y and then Z is inside the board for positive beta
this vertical tail which is here it will also see closer to positive beta, so there will
be a reinforce in this direction and this direction is opposite to the positive y direction.
. So this FY because of the beta is negative
always say d cy / d beta which says per unit beta how much CY will change direction wise
that is negative. We understand it back; now let us talk about CYP. What is CYP ,CYP is
suppose this is the airplane what is p+ p positive is right wing going down that is
the right wing going down this P+ so if I roll this airplane positive like this now
imagine with a vertical tail here so as it rotates the vertical tail also rotates in
same direction it pushes their towards positive Y so it also experiences the force in the
negative y direction is it clear. So they repeat again Let us assume I am repeating
this for your clarity let us say this is a vertical tail all the vertical tail as I will
be positive p right wing going down this tail is pushing there so there will be a reaction
on the tail in this direction and this direction of force is opposite to the direction of positive
Y, so CY because of P also will be negative as the major contributor is the vertical tail.
So I said this negative, sometimes you sign my changes because of sweepback of the wing
kind of high-wing aircraft because of the fuselage.
But I am not giving you a general idea how to find that component I see like be negative
and actually meaning this contribution is primarily because of vertical tail right normal
aircraft you may find because of sweep because of a fuselage effects design may not be what
I am Telling your numbers may not be what I am telling. Similarly Cyr if you see what
is Cyr and if you recall your earlier class are r means you are you are it that is yours
is positive is right wing going like this she or it positive.
Imagine this airplane is taking a you are at about center of gravity it is turning like
this so again this tail is pushing the air this direction so it will experience a force
in this direction and incidentally now for the positive or eight the force experienced
by vertical tail is in the positive y direction so Cyr because of vertical tail if it is predominant
which is generally true will be positive right .Now I come to see Y Delta R for time being
I neglect this I said he is not much if I talk about CY Delta R let us see what is the
sine of CY Delta R . Again I come here Delta it means other this
is the rudderless say if I deflect it towards left which is a positive radar deflection,
I repeat towards left when I look from the top is a positive rather deflection by convention
so if I have deflecting towards left and moving forward then the force will come in this direction
which is the positive direction. So CY Delta R will be this sign will be positive this
much understand you need to have or you are doing experiment at least times we should
know from the first principle that is why I thought I will explain you this then we
come back to our experiment. So this is the expression of a FY which are
essentially I return the forces and the kinematics part if I now they could we see the signal
equation this term whole of this term will be equated to M(v.+ru-pw) v dot is a component
of V along from total v which is half Rho v 2 capital u2 number that V along the local
y direction. .
And u is the component of total V along X direction local X direction and local z direction
is w that is aircraft people is moving like this and then suppose this is a relative air
speed he had a component along X along Y along Z that same aircraft is moving like this unless
the velocity is still the exits in the parallel to the ground then the component will change
so that is why I am saying but you know that VW u VW are the component of total velocity
along local body axis XYZ direction right and V dot is of course you understand is the
rate is acceleration along that direction. Okay this is the part which I am taking out
from the sixth topic of motion which you have already done you know one of us postage and
you are aware of it. Now I put some atoms I am assuming that theta is zero that means
what I am conducting the experiment so that the pitch angle is very small pitch attitude
angle please under this theta like very small ice that is why I am taking zero and also
assuming that the bank angle is also very small where was in 10 degrees.
. If I do this some Back angle side greater
than equal to 15 degrees I would say 10degrees when I do the experiment. Once I do this I
make an approximation that this term mg cos theta sine Phi I can now easily write it to
be mg Phi. No objection, right. .
Now once I am doing a steady coordinator turn that which I know that there is no beta these
gentlemen those for steady coordinator run because in a steady coordinate time I am Telling
like this every time we tied 0 latitude is aligned towards the velocity vector okay,
similarly when I am doing steady coordinator turn what is the value of P it is not doing
anything like that it is only turning like this so p is also 0 so I drop this term so
long steady coordinate kind of proclamations what about R can I drop it you cannot drop
it because R is there it is turning like this right. For this stage and then delta R is
there this also stays and we have any approximation that see y delta is weak.
. So I am not adding that into the expression,
so finally what happens if I do this assumptions then my equation becomes M G Phi plus half
rho v2 s into cy r into r b / 2 v +CY Delta R into Delta R this is equal to M. Now V dot
it is a steady coordinator return so we got automatically goes there is no role p so the
speed of loose on these goals because steady coordinator turn there is no role for these
also goes. .
So what is remaining now is m and U into I right and it will easily understand when I
am doing a static or a turn like this and I will not be very much in corrective file
at V equal to U and I am moving like this every time align to the velocity vector an
exception align so I can replace this U / V second right mg phi plus half rho v2 square
and to CY RB / 2 B plus CY Delta R into Delta R equal to MVR.
. Where our location wise I am using this so
let me or is I or it. Now see what is this MVR, MVR is angular rate MB omega relate and
mv2 upon X centripetal acceleration okay and you guys true if they have the plane turning
like this it is like a stone being turned by a string so some force must be making it
happen to give a centripetal acceleration in the circular path. So that is the effect
MVR which is a centripetal acceleration who is causing it, for a stone and rotating a
stone is the string the tension the spring causes that so somebody must be causing this
acceleration who is causing it primarily mg Y component of the weight and a dynamic force
this, this is clear right. So now what we will do will do further little
approximation engineers are always in famous for doing approximations many a times or sometimes
something very realistic comes out if I do an approximation he let the aerodynamic force
contribution is negligible anyway you could see that ladders are designed to give moment
is a large momentum not force, forces from the wing side so this is a force component
from the runner and this is there because of cross because of rate the side forces coming.
So we cannot a bad idea to see what happens if I say gentlemen this force component is
not very appreciable competitive back to the force coming because of the weight component
which is at a bank angle of Phi so if I neglect this if I neglect this then what I get is
minus I am getting is sorry right mg phi is equal to MVR or R whatever additional writing
nor I can rank mmk gets cancelled so I am writing that minus VR plus b phi equal to
0 or proceed G equal to vr/ five. .
So how do I use this information and get some enrichment to experiment what I will be doing
I will actually go for a steady coordinator I will take stopwatch I tell the pilot take
a turn of 90 degree on 180 degree depending upon mutual agreement. I have a stop watch
and I know how much time it has taken to turn from there I will get the value of R. It will
be how the total angle turn divided by total time so I know R we I know from a speed indicator
phi I know from the turn back indicators so I will measure accurate those data and find
this ratio and if it counts we are not between nine point seven to nine point eight something
like that and which should remain constant. .
Fairly for that altitude so you know that yes this ratio has to be closer to the value
of G so then you are understand whatever sixth term equation you have understood you have
applied it correctly and their assumptions are realistic. If this value goes off then
the assumptions are wrong ,for example if you are banking at a60-degree and you are
having a some positive theta values and your rudder is very powerful then you need to be
careful that the value of U lot come that means your experiment who are designing is
not correct because the G equal to Phi will only come when theta equal to 0 and Phi is
very small right and this indirectly also tells you how go do your sig Tau’ s model,
right. So this would be also all of the experiments
and this is a theory of building it up then we will be going for a session will actually
demonstrate how to do this experiment clear then but not forget for a steady coordinate
and you are supposed to measure you are eight but if you tell the pilot I want to take a
turn after 90degrees or 180 degrees you should have a stopwatch as you start taking a turn
ensure that is doing a coordinated turn no loss of altitude the speeder should remain
same. And this is the stop watch you find what is
the value of r you are it you stun bang indicator information to get Phi use airspeed indicators
to get the value of V and offline you see this ratio or there is coming closer to G
or not if it coming closer to G the assumptions are fine because you assume the equation of
motions to be perfect right okay. Now so this is one part for this experiment now let us
see what more information you could get out of this experiment.
Let us watch out for rolling moment coefficient in a rolling moment coefficient I can again
write like cl and expand like Clp and P B / 2 V + Clr into RB / 2V+ CL beta into beta
plus Cl delta A into delta A + CL delta R into delta R.
. Alright so when the airplane is doing the
pilot is executing a steady coordinate turn like this what is the value of cl, cl has
to be zero there is a positive value here it will turn like this or turn like this since
it is just making a coordinate turn so CL is no rolling moment is there for CL is 0
so it must satisfy cl is equal to zero. When I put this zero equal to CLP into P B/ 2 V
and ask myself a question whether this contribution of rolling movement will be there or not.
. I we get to see that when I am doing steady
coordinator in a return there is no p they are not all right so this gentleman goes off.
Then plus clr similar story with clr if I say will you be correct no ,no because there
is the R and I have taking a study coordinator turn there is a you are it so I cannot drop
on this stage what about CL beta into beta for a steady coordinator turn what is the
value of beta we have seen beta is 0.So this man also goes well of course cl and delta
which is coming from Ellen on and plus CL Delta R CL Delta R means.
. When I deflect the rudder if I deflect the
rudder like this it gives a young moment plus they the force in this direction that will
give a rolling moment also that is why you have CL Delta R. So if this is true then what
do I do I write simplified equation 0 equal to CL r into r b / 2 v plus Cl delta a into
delta a + CL delta r into delta r . .
While we are doing further approximation we will see if we study between these two terms
CL delta and into CL delta t which control surfaces primary responsible for giving rolling
moment by design it is the al anon from rudders it comes as a second effect because rudders
are useful direction of motion for turning the plane our should also remember that during
stall rudder is our best friend. But in general designed when you are not going to stall and
all the primary rolling moment comes from the ell anon and that he is designed for aeronautics
for rolling motion. So I again do an approximation as a junior
I said between these two terms this contribution is predominant so I say thank you very much
you also go ,so now you have 0 equal to clr to RB by 2 V the CL Delta A into Delta A do
not forget to bring this equation what are the assumptions we have made. And also you
know that from the earlier equation you have seen R equal to G Phi by V then if I now substitute
this R here what will happen then it will show that CLr into r is defined by V into
B /2 V because I am replacing the expression for R by this, this expression of R came remember
for the study coordinator turn and others assumptions.
. So I am simply in this equation I have substituting
the expression for R this plus cl delta A into delta a is equal to 0 .So CL r into r
is this into CL r into r b / 2. So this is fine now what I do I do some manipulation
I could show that right clr into g phi by V into b/ 2 v same expression plus CL delta
a into delta a equal to 0. So I can write clr into v Phi / 2 is to be by v 2square v
v have taken their 2 have taken a to have taken inside the bracket is equal to minus
CL Delta A into Delta A, no issues this is the expression you understand.
. Now here from here to hear nothing great we
have done the two we have taken inside this g Phi to this V has gone outsides of v2 and
this is .So you could easily see that Delta a by Phi varies with 1 by v2 very extremely
important of the vision Delta a by Phi values with 1 by v2 what is the meaning of this.
As you are going to make a back angle change and we do a later on you will find that the
as I increase the speed is a she will go on decreasing right it goes inversely so this
is easily can check from your experimental results because you know you have the data
available around deflection you have Phi through done by indicator your V from airspeed indicator
and we are from this expression has come it has come from a fixed off and go on adding
those simplifications right. .
And if you get something closer to this you will understand oh my sixth of my understanding
is correct my assumptions are correct a meaningful which will help you in understand the equations
of motion this expression is extremely important and this will tell you how to design when
you are designing the control controller or an airplane to make the pilot have a good
handling qualities it should be very, very carefully the delta A / phi will go inversely
with V2 okay that is extremely important okay thank you very much.