INDEX

S. No Topic Page No.

Week 1
1 Introduction , course content and classification of UAVs 1
2 Measurement of Flight Velocity and Standard Atmosphere 20
3 Anatomy of Airplane and Airfoil Nomenclature 45
Week 2
4 Examples, Pitot and static tube and differential pressure sensor. 66
5 Generation of Lift and Drag 79
6 Aerodynamic center and center of pressure, Various wing planform 94
7 Lifting line theory, NACA airfoil nomenclature 109
Week 3
8 Airfoil and Finite wing, Various wing planform 124
9 Interpreting airfoil data, Cl vs Alpha and drag polar, selection of airfoil 142
10 Introduction to Airplane performance, Equation of motion 166
Week 4
11 Thrust required and Power required 185
12 Calculation of Performance parameters and selection of power plant 211
Week 5
13 Climb Performance, Engine Sizing and Power Plant selection 229
14 Weight Estimation , Common propulsion systems 245
15 Weight Estimation contd., Electric propulsion, Battery Sizing 267
Week 6
16 Iterative weight estimation and Wing sizing 287
Wing Planform selection and sizing and Flight test of Cropped delta
17 wing UAVs 306
18 Effect of variation of CG location and Static Stability 322
Week 7
19 C.G. location and Longitudinal Static stability 339
20 Tutorial 1 353
Week 8
21 Contribution of tail in static stability and Neutral point. 395
22 Tutorial 2 421
23 Tutorial 3 436
 Design of Fixed Wing Unmanned Aerial Vehicles
Dr. Subrahmanyan Sadrela
Department of Aerospace and Aeronautical Engineering
Indian Institute of Technology - Kanpur

Lecture 01
Introduction, Course Content and Classification of UAVs

Good morning friends. Welcome to this course on design of fixed wing UAV. I am
Subrahmanyam Saderla, instructor for this course and here we have 2 of our TA s, Mr. Deep
Pareek and Mr. Salaudeen Khazi. They will be helping us in solving the assignments and; you
can mail them for whatever doubts you have. They will be helping you. Thank you being a part
of this course.

So right now, we heard that registration has clicked about 2500. Now we can no more assume
the students who have registered in this course have performed the prerequisite. So, we have to
redesign the course content in such a way that even the fresher, a student who is out of this
aerospace engineering, should be able to understand whatever we are going to discuss in this
lecture. Let us look at what are we going to discuss in this course. So, we will start with some
basic introduction.

(Refer Slide Time: 01:07)

First few lectures will cover

Introduction about UAVs and their classification

1
 Brief anatomy of aircraft
Basic aerodynamics
Aerodynamic characteristics of wing sections
Aerodynamic characteristics of finite wings
What is aerodynamic centre (xac), one of the important variables
Centre of pressure (xcp)
Relationship between aerodynamic centre (xac) and centre of pressure (xcp).
What is lift and drag, which is in fact the drag polar.
Concepts related to flight path angle
How to measure the velocity during flight, subsequently the standard atmosphere

These are some of the introductory concepts that we will cover.

(Refer Slide Time: 04:37)

Then we will look at some of the basic prerequisites of this course that relates to performance
characteristics. This includes

Study or analysis about level flight

Thrust requirement
Power requirement Range
Endurance
L/D
Wing loading (W/S)
Thrust loading (I/W)

2
So, we use these concepts to figure out what should be the weight of the battery that you would
like to carry to perform a particular mission.

How do we get to know?

By understanding what is the thrust requirement of the system to perform or to execute that
particular mission. Since, most likely, we will be talking about propeller driven aircrafts or
UAVs here, we have to talk in terms of the power requirement.

How to select the propulsion system?

The answer is by performing this performance analysis i.e L/D, W/S.

We will also look at

Climb performance
Rate of climb
Angle of climb.

For the design that you are going to do, what are the parameters that limit your rate of climb and
angle of climb. To know that, we have to study this climb performance. We will look at both,
study rate of climb as well as accelerated rate of climb.

(Refer Slide Time: 08:30)

3
The third chapter will be more about, we can say analytical estimation where you talk about
various stability and control derivatives of the system.

Analytical estimation.

The main emphasis is how to design a system which is stable by itself without any add on, like
control system. How do you make the system stable by itself during its flight? So here we start,
we talk about

Equilibrium
Static stability; Conditions to satisfy the static stability
Longitudinal static stability
Lateral static stability

In Longitudinal static stability, we will talk about

what should be the size of the wing and tail

How far they should be apart for a stable flight

When we say wing, tail, fuselage and all these things in propulsion system, each carries a
weight. So, there exists a cg for this entire system. Once you assemble it, you will figure
out a cg location. Where does this cg location should be? What are the limitations of this
cg travel? So, we will also look for

Most forward and aft cg locations

All these conditions are meant for a stable flight.

We will also look at how big should be your elevator or the control surface if you want to
control your aircraft. Does that sizing involve the distance between wing and tail or the cg
plays any role in that sizing of the elevator. We will address all these questions while
performing

Sizing of control surface.

How the sizing can be performed and what is the role of the cg location as well as the
distance between wing and tail. When I say the distance between wing and tail, we
broadly talk in terms of distance between the aerodynamic centers of these two
components.

4
 Once we do this control surface sizing and at the same time the distances are fixed, then
we need to understand what is the range of angle of attacks for which the aircraft can be
trimmed for. So, we will look at

possible range to trim.

When we say possible range to trim, it talks about control surface deflection.

(Refer Slide Time: 13:39)

For lateral directional case, we will perform

Vertical tail sizing

which are mainly governed by means of some lateral directional parameters, stability and
control parameters. So, we have to design in such a way that the vehicle should be able to
come back to its equilibrium whenever there is a disturbance in lateral directional case.

Rudder sizing

Rudder is a control surface that is located on this vertical tail which is used for directional
control.

Now, we will proceed to talk about

Simulation

5
Why do we need simulation? Whatever you have designed, you have to translate.

What is the output of the design?

A working model with physical dimensions. You should have the plan for geometry as well as
the cross-sectional properties and their respective locations with each other.

Let us say if you have designed the wing with a particular span and chord and you have a tail and
you have located both of them, so the output should be a geometry. Now whether this geometry,
in reality will fly or not. Is there a step that is involved before flying the actual design or the
prototype?

We can do that by means of simulation. In simulation, we convert this physical model to the
mathematical domain and then give various inputs to see how the system behaves. Whatever the
disturbance that you provide, does it vanish with time or increases with time?

All these things you can analyze before actually going for fabrication. Since we will not be
fabricating any model that we are going to design in this course, let us concentrate on simulation.
It is worth spending sometime performing the simulation. So, what we do here is,

o Derivation of a rigid body equations of motion.

Rigid body equations of motion are common but how can you simulate your design by using
that rigid body equations of motion. The answer is

o Modelling external forces and moments.

Whatever the shape or whatever the design that you have performed, it will reflect in terms of
external forces and moments called aerodynamic forces and moments. I call it as

o Aerodynamic model

This aerodynamic model will go as an input, right to the equation of motion and you use a

o Numerical integration algorithm

The above algorithm is to solve the derived rigid body equations of motion which in general
are first order differential equations. And then
o Analysis

6
And finally,

Design Methodology

We will take one or two case studies and we will do the entire process.

Without understanding the performance and stability aspects of this, we will not be able to
design a configuration. If possible, we will also try to include some optimization here, but this
limits the scope of this particular course.

Our TA s will take you to the UAV laboratory here where there are few UAV models. If
possible, we will also include the flight test of those models during this course.

So, you might have heard a general term called drone and how does it differ from an UAV. We
should know about that. We more or less use drone and UAV as a common word. Now we need
to know what is the difference between them.

(Refer Slide Time: 20:34)

Drone:

Drone is also an unmanned system which is meant to perform a mission, more of a pre-
programmed mission. So that means it will not be able to send whatever data that it has acquired
during that particular mission to the ground station unless it come back to its home. So, it is like
an unmanned flight vehicle with limited onboard intelligence and mostly designed to perform
pre-programmed missions.

7
UAV:

UAV stands for an Unmanned Aerial Vehicle, of course there is no onboard pilot. It is more
controlled from the ground station or in an autonomous mode but it can communicate to ground
station. When there exists a communication link, you will be able to change the mission
whenever it is required. During the flight, it is like dynamic mission planning. You can actually
change the mission whenever it is required.

(Refer Slide Time: 23:08)

It is a better onboard intelligence. When I say it can communicate to the ground station, it can
update about the payload data as well as health and the performance of the system. So, it can
update details or the data acquired by the onboard payload as well as data related to sensors
health or health of the system and also the performance of the system. So, in this course, we will
talk about the UAV s.

So how can we classify these UAVs? There are many UAVs starting from 10 cm to 10 m span
UAV. How to classify these UAVs?

(Refer Slide Time: 25:12)

8
First of all, let us look at the mode of operation and how they are nomenclatured.

UAVs can be broadly classified into

Fixed wing-based
Rotary wing based
Flapping wing based
Combination of either of the above (mostly a combination of fixed wind and rotary wing)
and can be called as hybrid UAV.

Flapping wing are mostly termed as ornithopters. The design is carried out by means of the
aerodynamics associated with the insects, flies and birds, so on.

In case of rotary wings again there are

o Single rotor UAVs

o Multi rotor UAVs

There is single rotor as well as multirotor based test beds that you can develop under this rotary
wing UAVs.

What we will be talking about is fixed wing UAVs.

So, the content that we have presented here is more or less related to fixed wing UAVs. To
further narrow it down, as we know, fixed wing works well in the atmosphere. So, we will be
talking more closely about an atmospheric flight vehicle which in our case is an fixed wing

9
atmospheric UAV design.

The above classification is made by means of principal of operation. Now let us look at the
classification based upon the size, weight and the mode of operation.

(Refer Slide Time: 28:13)

(Refer Slide Time: 28:20)

There are five different classifications in terms of dimension as well as weight of the UAVs.
They are micro, mini, very small, small, medium and large. Very small and mini, they are more

10
or less similar, we cannot exactly differentiate them. But here, the micro UAVs whose weight
class falls below 10 cm, whereas, very small will come under anything between 30-50 cm. The
small UAVs will have a span of 0.5-2 m and medium sized UAVs will be of 5-10 m. The large
size UAVs are 20-50 m, as high as 50 m.

(Refer Slide Time: 29:16)

(Refer Slide Time: 29:30)

11
(Refer Slide Time: 29:40)

(Refer Slide Time: 29:50)

12
(Refer Slide Time: 30:01)

(Refer Slide Time: 30:12)

13
(Refer Slide Time: 30:23)

(Refer Slide Time: 30:34)

14
(Refer Slide Time: 30:45)

(Video Starts 30:57 - Video Ends 31:07)

(Refer Slide Time: 31:08)

15
(Refer Slide Time: 31:24)

(Refer Slide Time: 31:35)

16
(Refer Slide Time: 31:46)

(Refer Slide Time: 31:56)

17
(Refer Slide Time: 32:10)

There is a classification based upon its mode of operation. There is a tactical and combat UAV
which will be used more in a dynamic sense. Those are more active and they are used to drop the
payloads, whatever the payload may be. It can be a warhead or anything. So, these tactical and
combat UAVs are more active. There is a MALE UAV which is abbreviated for Medium
Altitude Long Endurance UAV and the HALE, High Altitude Long Endurance UAV. These two
UAVs are meant to perform surveillance and reconnaissance missions. So, the high-altitude long
endurance UAV can have a range up to 40,000 km and a medium altitude long endurance UAV
can have similar range but at a lower altitude.

(Video Starts 33:03 - Video Ends 36:26)

18
(Refer Slide Time: 33:03)

Dear friends, we are right now near the runway of this flight laboratory of IIT Kanpur. So, flight
laboratory is in front of me. As you can see this is a fixed wing UAV. The 1.5 m class. Overall
takeoff weight is 1.6 kgs and it was designed to have an endurance of 2 hours right now. So, let
me introduce our chief test pilot. He is a test pilot for most of our configurations, Mr. Naveen
and you know Mr. Deep again. He is here.

So, let us quickly have a flight test of this and this has been designed by us and see how this
design behaves right now. So, it is a bit windy as you can see the model is almost waving here.
So, let us see how this performs in the wind. So, you can zoom that wind bag. A bit gusty. (())
(35:01) We can witness a power of flight. The model is gliding and is able to sustain its weight
because of this gust.

19
 Design of Fixed Wing Unmanned Aerial Vehicles
Dr. Subrahmanyan Sadrela
Department of Aerospace and Aeronautical Engineering
Indian Institute of Technology - Kanpur

Lecture 02
Measurement of Flight Velocity and Standard Atmosphere

Dear friends. Welcome back.

In our previous lecture, we saw flight test of a fixed wing UAV. It was a flying wing which weighs
about 1.5 kgs, 1.6 kgs approximately and with a span of 1.5 m.

What do you think is the velocity of that flight? Or in other words, how to measure the velocity of
that flight? What are the sensors that we require to measure the velocity and altitude? What are the
various methods used to measure this velocity?

(Refer Slide Time: 00:45)

20
One simple thing is, install an onboard GPS which have an IMU (Inertial Measurement Unit) that
measures the accelerations, say linear accelerations, and integrate these linear accelerations to get
the corresponding velocity by means of radar.

Consider you have a radar at the ground station with which you can measure the position as well
as the velocity and then we also have some sensor called pitot and static tube. Here, we are not
going to argue about how accurate these measurements are? but these are the various sources from
which we can measure the velocity. Now in this lecture, let us discuss about how to use this pitot-
static tube to measure the velocity.

In the first place, what is this pitot tube and static tube? Consider a hollow cylindrical tube with an
open mouth at one end and the closed mouth at the other end. Now drill a small hole here to tap
the pressure reading and connect this to a pressure sensor. Now let us take this setup and place it
in a wind tunnel. These two lines represent the boundaries of the wind tunnel test section. Now
run the wind tunnel.

So, wind tunnel is a device in which we can produce the required velocities. We can produce
variable velocities of flight. Either you are moving in a static air or you are holding a body and
you are blowing the air, in both the cases, the effect is same. Now let us run this wind tunnel to
produce a velocity, V .

Let V be the free stream velocity here.

Now let us consider point A. Initially, what happens, as soon as you run this wind tunnel, the air
will fill this tube, this entire apparatus. Now once the tube is completely filled, this pitot tube, let
us assume a fluid particle is coming from point A. Since it is in the free stream, the velocity of that
fluid particle will be V .

VA = V

Let us assume this fluid particle is brought to rest at point B isentropically, which is reversible
adiabatic process (no addition of feed or loosing of heat from this setup). During this process, there
is no addition of heat or loss of heat. Now since we brought the molecule to rest, at this particular
fluid element, at this point B, the corresponding velocity here will be 0. Let P0 be the pressure
measured at this point B.

Now from the Bernoulli's theorem, we have,

----------------(i)

21
(Refer Slide Time: 05:35)

which is derived from the law of conservation of energy. Since we are not removing or adding any
energy here, we can apply this Bernoulli's principle across this points A and B, right. So, the
pressure measured at point A be PS,
Total pressure at point A = Total pressure at point B
where P0 be the corresponding pressure measured by this pitot tube at point B and velocity is 0.

---------------(ii)

So, this particular pressure (Ps) corresponds to static pressure and the pressure due to flow or
velocity (1/2 v 2) is known as dynamic pressure where is the density of fluid. Now this is the
total pressure (P0) is the summation of static pressure and the dynamic pressure. Here the pressure
that you measure is ultimately the total pressure.

Now if you want to find the velocity of the flight vehicle, we can rearrange this equation.
Rearranging equation 2, you can measure the velocity by measuring the differential pressure and
the density, right.

22
So, from the above equation, I will get the velocity in m/s, the units of are kg/m3and P is in
Newton/m2, right. Now what are the quantities that I need to know if I want to calculate V ?

They are P0, PS and .

We witnessed that P0 is measured by this pitot tube. How about PS? What exactly is the static
pressure?

What is air? It is a mixture of gasses. We are in an atmosphere which is a mixture of gasses and
gasses are characterized by its state of matter which is loosely packed and characterised by its
random motion. Now due to its random motion, these gasses will exert a normal pressure which is
the rate of change of momentum that happens across the surface.

So, the static pressure, if you want to tap, you need to measure the rate of change of momentum
normal to the surface. Now for this case, at point A, the pressure is known as static pressure which
we need to capture normal to the surface.

(Refer Slide Time: 09:39)

23
Let us say if I take another cylindrical tube which is also hollow but with a closed mouth. And
again, I make a small drill at this end and tap the corresponding pressure by means of this pressure
sensor. Now if I have to capture the static pressure, what I need to do is to drill the holes along the
surface or on the circumference of this tube. Here, I can tap static pressure by using this setup.
This particular tube, that helps you to capture the static pressure, is a static tube (Refer Slide Time:
09:39).

A combination of this pitot and static tube can give you a differential pressure. But we witness that
this UAV is flying at different altitudes. So, it is important to know whether the density is changing
with altitude or not.

Whether the density changes with altitude or not?

The answer is standard atmosphere.

Now, if I can make this pitot and static tube coaxial, it will look something like this (Refer Slide
Time: 09:39). If you attach this setup to your UAV, you will be able to measure the static and total
pressure from which you will be able to find out the corresponding velocity of flight. Either you
can use two separate sensors or you can have a single sensor which can measure the differential
pressure. Here what you get is PS, here P0. This is your V . Now let us have a close look at this
pitot-static tube (Refer Slide Time: 09:39).

(Refer Slide Time: 12:44)

24
See this is one VTOL UAV which we are now working on (Refer Slide Time: 12:44). This was
built by our team members and I do not know whether you are able to see this pitot tube or not,
but I will try to make it clear. This is your pitot tube and there is a coaxial tube here where you
have the holes on the surface with a closed mouth, that is your static tube (Refer Slide Time:
12:44).

Here you can take the static output as well as the total output and you can connect it to the
corresponding sensor. If you have to accurately calculate the velocity of flight which I can use it
at a later stage to control the UAV, I need to know what is the corresponding density at that
particular altitude. Now let us look at how to find density at a particular altitude.

(Refer Slide Time: 14:11)

What is standard atmosphere?

It is a mathematical model to estimate variation of pressure, density and temperature with altitude.
Now we have defined something called altitude here. So, there is a need to define it more precisely.

What we use as altitude in standard atmosphere?

A geometric altitude.

Let us consider an UAV, flying at a height h, which can be directly measured from the UAV to

25
the surface by means of the tape (Refer Slide Time: 14:11). The altitude that can be measured
directly from the aircraft to the mean sea level or to the surface is known as the geometric altitude.

So, there is another altitude as well. Let us say, there is a satellite or another planet. If you want to
find out the distance between the two planets, then we have to talk in terms of their centre of
masses. Let us say there is a satellite revolving around the earth or a moon revolving around the
earth. Now if you want to talk about the distances between them, we have to talk about the
distances between their centre of masses. Let R be the radius of earth and the distance between
these center of masses is ha, which is known as absolute altitude (Refer Slide Time: 14:11).

Researchers have also performed sounding rocket as well as hot air balloon experiments to
measure the temperature by varying the altitude.

(Refer Slide Time: 18:26)

So, this typical plot represents how the temperature varies with altitude and it continues (Refer
Slide Time: 18:26). So, it is up to say 11 km. Initially the x-axis here is temperature in Kelvin. Y-
axis here is altitude, h or hG, in km (Refer Slide Time: 18:26). Initially it is observed that there is
a temperature change with respect to altitude. After a while, there is a constant temperature region
and again there is a gradient region.

26
So, as you can see this is like change in temperature with respect to altitude here (Refer Slide
Time: 18:26). Although the altitude was supposed to be plotted on the x-axis, but altitude was apt
to plot along the z axis vertically. So that is the only reason why it has been plotted in this way.
So, the slope of this curve is defined as lapse rate which is again dT/dh (in terms of Kelvin/km or
m). There are some regions where the temperature is changing with altitude. Those regions are
known as gradient layers. The regions where the temperature remained constant with altitude are
known as isothermal layers. As this goes up, first gradient layer is up to 11 km and the first
isothermal layer is up to 25 km and then up to 47 km to 55 km and so on and so forth (Refer Slide
Time: 18:26).

But we are not interested in this. Why because we are talking about atmospheric flight vehicle and
we know even the commercial aircraft will fly up to 11 km altitude. So, what we are more
interested is in this region. Here, the slope is defined as a which is also known as lapse rate (dT/dh)
and the units are Kelvin/km.

At sea level, it is 288.16 Kelvin and this corresponds to 216.66 Kelvin (Refer Slide Time: 18:26).
With this information, can we reconstruct or can we estimate density and pressure variation with
respect to this altitude?

That is the idea of the standard atmosphere, right. So, we will be able to estimate.

Now the basis of the standard atmosphere is a hydrostatic equation.

(Refer Slide Time: 22:55)

27
Consider an infinitesimal fluid element with rectangular faces in unit cross sectional area (Refer
Slide Time: 22:55). Each face is having unit cross sectional area. Let us label this as A and the
top face be B (Refer Slide Time: 22:55). Say these two faces has a unit cross sectional area and
these two faces are separated by distance dhG. So further, since it is a fluid element, there will be
forces acting on it.

Because of the other fluid elements present nearby, the force here is pressure P. Let us say on
bottom face it is P x cross-sectional area = 1, i.e. P x1. What about this? Let the force acting on
the face B is P+dp. Further assume that this fluid element is at rest. When we say it is at rest, it is
at equilibrium. This is your z-axis and this is your x-axis. So, the force is acting along z is positive
and those acting opposite to this is negative. So, what is this Fz here?

There is a weight of this fluid element which is acting downwards. So that weight is (mass x
acceleration due to gravity) and since we are talking about a fluid element, it is apt to talk in terms
of density and volume. Let be the density and dV is the volume of this fluid element which is (1
x 1 x dhG).

-----------------------(1)

Say this is my equation 1. But if you observe here, what we are interested in? Like coming back
to the question like why we are doing this? We want to know how the density is varying with
altitude, right. So, if I integrate this equation, I will get to know what is the variation of pressure
with respect to altitude given is a function of altitude. But we should not miss that g is also a
function of altitude. So, in order to ease this integration, let us take one more step.

We have equation of state. What is equation of state, which is valid everywhere.

--------------------(2)

Divide these (1&2) equations.

------------------(3)

Now if I integrate this, I will get variation of pressure with respect to altitude but the issue is, there
is a variation of temperature with respect to altitude. That we can figure it out from this gradient

28
layer and isothermal layer. But how about these two parameters (Refer Slide Time: 22:55).
Is g also a function of altitude here?

Now we have to define new altitude called geopotential altitude. In order to ease this integration,
let us now define a new fictitious altitude called geopotential altitude. So, what is that?
Rewriting equation (1) as equation (4).

-----------------(4)

This also represents the same pressure at that point. So, equation 1 and 4 represents the pressure
at the same point.

But what is the difference here? In order to ease the integration, we have defined a new altitude
called geopotential altitude, dh, where h is the geopotential altitude which physically does not
exists. It is a fictitious altitude. Since (1) and (4) represents the pressure at the same point, we can
equate these two equations. To figure out the relation between real altitude and the fictitious
altitude, let us equate (1) and (4). From equation (1) and (4), we have

----------------------(5)

Now what is g/g0?

(Refer Slide Time: 30:25)

29
Let us go back to the definition of absolute altitude. Say we have our moon here. This is our earth
and moon or you can also assume a satellite (Refer Slide Time: 30:25). There is a satellite
revolving around the earth. Now we have defined absolute altitude, ha, which is a sum of geometric
altitude and the radius, hG, right.

So according to Newton universal law of gravitation, the force of attraction between these two
bodies is directly proportional to the product of their masses, where m be the mass of the satellite
and M be the mass of the earth and inversely proportional to the square of the distance between
them.

Now let us bring the satellite on to the surface of the earth. What it becomes?

-------------------(6)

---------------(7)

(Refer Slide Time: 33:02)

30
So, if you divide this equation (6) and (7), what you have is,

Now substitute this g/g0 in equation (5). What you have is

Now integrate between your altitudes of interest. Say from 0 to h, and the corresponding limit for
right hand side is 0 to hG.

(Refer Slide Time: 34:02)

By integrating, what I have is

------------------(8)

So, this is the relation between fictitious altitude and real altitude.

Now why we are deriving this? Since we want to ease this integration, we have converted the

31
actual hydrostatic equation which is in terms of geometric altitude to a geopotential altitude or to
an equation with geopotential altitude.

From equation (3), we have

-----------------(9)

Now g0 is constant, right. It is on the surface of the earth which is 9.81 m/s 2, the average value of
g0. Now if I integrate the equation (9), I will be able to figure out how the pressure varies with
altitude and then we will see how the density varies with altitude. Now let us consider the layer,
gradient layer.

(Refer Slide Time: 36:26)

So, for gradient layer, integrating (dp/P)

Now from the gradient layer, we have lapse rate which is defined as dT/dh. So, what is dh?

32
So corresponding integration limits will also change to T 1 to T2, where T1 be the temperature at
altitude 1, T2 be the temperature measured at altitude 2.

(Refer Slide Time: 37:38)

Now in above equation, a is constant for a particular gradient layer. R is constant which is 287
J/kg-K. The average value of a for the first layer -6.5 K/km. This is the change in temperature. i.e.
if you travel 1 km vertically up, then there is a drop of 6.5 K.

So now let us integrate this equation.

--------------------(10)

33
Now we got the variation of pressure with altitude. So, if you know the lapse rate, you will be able
to find out what is T2. If you know P1 i.e let us say if you are taking sea level condition, P is 1 atm,

what is STP?

Standard Temperature and Pressure (Ps) = 1 .1325 x 105 N/m2 or Pascal; s =1.225 kg/m3 and
temperature is 288.16 K.

(Refer Slide Time: 40:22)

(Refer Slide Time: 41:10)

34
We know the equation of state, at an altitude h1.

-------------------(11)

And we also have equation of state, at altitude h2.

------------------(12)

Now dividing these 2 equations, we have

So, with the help of the equation (10),

-----------------(13)

Equation (13) is your gradient layer relationship for density variation with altitude whereas
equation (10) is your pressure variation with altitude.

(Refer Slide Time: 43:24)

35
Now for isothermal layer, coming back to the equation (9)

In isothermal layer, we observe that temperature remains constant with altitude. So, we can
consider it as a constant. T here is no more a variable. Integrating equation (9)

So, by using

where T2=T1 in this case, that means

for isothermal layer.

So, if you know the initial conditions, change in the altitude and the corresponding temperature of
the isothermal layer, you will be able to find out what is the pressure here. But we are interested
in the gradient layer, let us solve few problems figuring out the velocity as well as pressure for this
gradient layer.

36
(Refer Slide Time: 45:47)

Let us rewrite these equations for gradient layer and isothermal layer.

For gradient layer, we have

And for isothermal layer, we have

So, what are these T1, T2 for gradient layer and h1, h2 for isothermal layer?

The sea level conditions are P1, T1 and 1 for gradient layer. For each isothermal layer or each
gradient layer, the end point of this first gradient layer will be the starting point of the isothermal
layer. So, you have to consider P1, 1, T1 for isothermal layer at 11 km i.e.if you want to calculate

37
these thermodynamic properties at say 20 km altitude, then you have to consider P1, 1, T1 at 11
km altitude for isothermal layer.

(Refer Slide Time: 47:57)

Now let us quickly solve 1 or 2 examples.

Example 1: Consider the fixed wing UAV flying at an altitude of 5 km with a velocity of 30 m/s.
Find the corresponding total pressure measured by pitot tube?

Solution:

Here there is a fixed wing UAV flying at 5 km altitude and cruising at a speed of 30 m/s.

Now we have to find out the corresponding total pressure measured by this pitot. How do we do
this?

As we know 5 km corresponds to gradient layer here. Now what is the total pressure?

Total Pressure (P0) = Static pressure (Ps) at 5 km + dynamic pressure with density ( at 5 km
altitude and the velocity (V) of 30 m/s.

38
Given V = 30 m/s,
h = 5 km = hG

Now first we need to find out what is the static pressure at 5 km and density at 5 km altitude.

We have to convert this hG to h. How we are going to do that?

So, what is the difference between this hG and h at 5 km altitude?

(Refer Slide Time: 52:41)

So, the difference is not much. It is just 4 m of difference. And so, the result will also be not much
affected by whether you convert this to h or hG.

Now by using the gradient layer equations,

39
How to find T2?

We have lapse rate for this first layer which is,

So, what is T2 in this case,

So now substituting this value,

(Refer Slide Time: 56:40)

40
and the corresponding density at this altitude is,

Now you can directly solve this since you have P5 and T5.

Otherwise, you can also use

You can solve from either of these equations to get density at 5 km

Now that you have the density as well as the static pressure at 5 km altitude, by substituting these
two values, for a given velocity, you can find out the corresponding total pressure measured by the
pitot tube.

(Refer Slide Time: 58:38)

So, what do we get,

41
So, this is the pressure measured by the pitot tube which is installed in this UAV, which is flying
at 5 km altitude at a speed of 30 m/s.

Now take another example.

Example 2. Determine the cruise velocity of a wing alone UAV flying at an altitude of 10 km?

Solution:

The pitot tube measures total pressure of 26.37 KPa.

In the first question, we need to find out the total pressure for the corresponding velocity. Here if
we have the total pressure, what will be the velocity of flight?

But the difference here is, we are flying at 10 km and still we are in gradient layer.

(Refer Slide Time: 01:02:13)

So here

42
Now again consider the gradient layer equations where

Okay. So, we got the pressure, static pressure at 10 km. Now let us find out the density at this
altitude, right.

(Refer Slide Time: 01:05:00)

We have from the equation of state,

We can easily find out what is the density at 10 km from here.

43
Now by substitute these two values,

So that is the velocity at which this UAV is flying. If the pitot is measuring 26.37 KPa and UAV
is flying at 10 km, the cruise velocity of this UAV is 30 m/s.

44
 Design of Fixed Wing Unmanned Aerial Vehicles
Dr. Subrahmanyan Sadrela
Department of Aerospace and Aeronautical Engineering
Indian Institute of Technology Kanpur

Lecture - 03
Anatomy of Airplane and Airfoil Nomenclature

Good morning friends welcome back. In our previous lecture, we were talking about how to
find the velocity of UAV right.

(Refer Slide Time: 00:25)

So, we derived that

If you know the differential pressure i.e., the total pressure (P0) and static pressure (Ps) and
also if you know the density ( ) you will be able to determine the velocity. So, we measure this
total pressure (P0) with the help of pitot tube and corresponding pressure sensor and static
pressure (Ps) is measured with the help of static tube connected to a pressure sensor. To know
the density ( ) at that particular altitude we have derived standard atmosphere equations.

We observed that there are some altitudes at which the temperature changes with altitude and
the pressure and density also changes with altitude. Standard atmosphere is a plot that
represents the variation of this temperature with altitude.

45
So, let us limit our self to the first two layers. This is a gradient layer and the slope is -6.5 K/km
(Refer Slide Time: 00:25). The gradient layer extends up to 11 km altitude which is on Y-axis
and temperature variation is on X-axis. The sea level temperature is 288.16 Kelvin. It is
observed that from 11 to 25 km the temperature remains almost constant and the corresponding
layer is termed as isothermal layer. At 11 kilometers it is observed that the temperature is
216.66 K.

To find out the temperature pressure and density at different altitudes we have derived the
equations for gradient layer.

(Refer Slide Time: 03:32)

The above two relationships stand for gradient layer where if you know the altitude you can
find the corresponding temperature at that particular altitude by the definition of the lapse rate
which is

46
If you know the altitude of your flight, you will be able to find out the corresponding
temperature (T2).

Hence using the temperature (T2) and the sea level temperature (T1) and known sea level
density ( 1) you will be able to find out the density ( 2) at the required altitude as well as the
pressure (P2) at the required altitude.

What are these STP conditions?

(Refer Slide Time: 05:00)

Standard Temperature Pressure (STP) conditions:

Pressure at Sea level, P = 1 atm = 1.01325 x 105 Pa

Density, =1.225 kg/m3

Temperature, T = 288.16 K

Universal gas constant, R = 287 J/kg K

g0 = 9.81 m/s2

Earlier, we have solved few examples in which we assumed that the altitude is known and the
corresponding velocity is obtained by means of the measured either pitot pressure or the static
pressure. But now to know the altitude or to determine at what height your UAV is flying, let
us take some examples.

47
(Refer Slide Time: 06:57)

In some cases, we can also have differential pressure sensors. Instead of measuring pressure
from pitot tube and static tube separately, we can connect to a differential pressure sensor. One
can be P0, the other can be Ps (Refer Slide Time: 07:55). You can have a pitot tube which is
connected to the total pressure port and then you can have a static tube at the same time. Here,
we will get Ps from here, P0 from here (Refer Slide Time: 07:55). Hence, we can use a
differential pressure sensor directly.

(Refer Slide Time: 07:55)

48
(Refer Slide Time: 09:15)

Example: 3 Consider the differential pressure measured by the pressure sensor of a UAV
cruising at 30 m/s is 409.05 Pa. Find the corresponding altitude of flight?

Solution:

We have

P0-Ps = 409.05 Pa

Cruise velocity = 30 m/s.

let us consider for the gradient layer till 11 km

(Refer Slide Time: 11:05)

49
If I want to find out h, either I need to know what is the density at that altitude or the static
pressure at that altitude. So, let us see what we can find from this given data.

Now we know density at height, h or let it be the density at the corresponding altitude of the
flight. Now using this relationship,

1 = Sea level density

T1 = Sea level temperature

(Refer Slide Time: 13:49)

50
By using the definition of lapse rate, you can find what is the h

Since you have dT you can find dh.

h1 = 0 at Sea level condition

Now, geopotential altitude = 3 km

Converting to geometric altitude

(Refer Slide Time: 15:47)

51
Radius of earth,

The difference between geometric altitude and geopotential altitude is hardly 1.5 meters.

Assuming that there is no on-board GPS, only source where you can get the altitude of flight,
and with the absolute pressure, whatever you are going to measure, how to find out the
corresponding altitude. Next problem will address that situation. So, let us take another
example.

(Refer Slide Time: 17:31)

Example 4: The static pressure sensor of the UAV measures a pressure of 53.75 kilopascals.
Find the altitude of flight of this UAV?

(Refer Slide Time: 19:12)

52
Solution:

Here,

P at a particular altitude

Ph = 53.75 kilopascals

Now we need to find out the corresponding altitude of flight

By using the relationship

(Refer Slide Time: 20:51)

Now the temperature at the altitude h,

Th = 255.66 Kelvin

From the definition of lapse rate,

53
We have,

the corresponding geometric altitude, hG = 4.996 km

So, what you have done here?

In the question, it is mentioned that we have a static pressure data from the sensor which is
about 53.75 kilopascals. By using this static pressure, we found the corresponding temperature
at that altitude by using gradient layer equation. With the help of definition of the lapse rate,
we were able to find out the corresponding altitude of flight and the geopotential altitude turned
out to be 5 kilometers. With the help of the relationship between geopotential altitude and the
geometric altitude, we figured out geometric altitude to be 4.996 kilometers.

Now let us look at an aircraft.

(Refer Slide Time: 23:23)

Let us say this is the UAV which I am interested in (Refer Slide Time: 23:23). Why do the
aircraft fly? One simple answer can be there is some force which is opposing the weight. Let
us term this force as lift. Since, we are talking about a fixed wing UAV, we need to move the
aircraft at a required velocity. Who is moving this aircraft? Thrust is the force that is helping
the aircraft to move at the required velocity. There is some penalty at the same time for

54
generating lift. As we are moving in a fluid, there can be friction because of which there is an
opposing force called drag.

These are the four forces that we need to understand here before we start to design.

How lift is generated?

Let us look at the anatomy of the aircraft.

(Refer Slide Time: 24:45)

Let us assume this as an aircraft (Refer Slide Time: 24:45). The major components are wing,
horizontal stabilizer or tail, vertical stabilizer or vertical tail, propulsion system i.e., engine
(Here it will be on either side but you can have a single engine as well in some of the aircrafts),
fuselage and undercarriage.

(Refer Slide Time: 27:47)

55
For a conventional aircraft, there are fixed components as well as moving components. The
fixed components are wing, horizontal tail, vertical tail and fuselage. We also have some
moving components. But why do we need them? To control the orientation of the aircraft.

(Refer Slide Time: 28:34)

These moving components are also known as control surfaces. They are So we have something
moving surface on horizontal tail called elevator which helps to control the nose up and nose
down motion and the moving component on the vertical tail called rudder which helps you to
turn left and right. There are control surfaces on wings known as ailerons which helps you to
roll.

It is worth mentioning about various motions that are possible with an aircraft. An aircraft is
considered as a rigid body in space with six degrees of motion.

What it can do? It can move, it can translate.

(Refer Slide Time: 29:57)

56
In total an aircraft has six degrees of freedom, three translational and three rotational. The
rotational orientation of the aircraft can be controlled by the control surfaces.

(Refer Slide Time: 30:33)

Assume the point as the CG of the system and be the origin of the coordinate frame whatever
I am considering (Refer Slide Time: 30:33). Let us consider a right-handed coordinate system
where x cross y is z here (Refer Slide Time: 30:33). The translational movement along x, y
and z axes are the other three degrees of freedom. The linear translation and an angular rotation
are possible here.

Let us start with x axis, the rotation about x axis is known as roll. What is the positive roll? We
have to define the convention as well. It follows the right-hand thumb rule. Stretch your free
thumb along the positive x-axis, the curl of your fingers will give the positive rotation about
the respective axis. The right wing can be called as starboard and the left wing as port side. The
right wing going down is a positive roll. Similarly, the roll is controlled by the ailerons as well.
You can control the rolling motion by deflecting the corresponding control surfaces called
ailerons here.

The rotation about y axis is known as pitch. The positive pitch will be nose up. If the nose is
going up, it is a positive rotation about y-axis. If the nose is going down, it is a negative rotation
about y-axis.

Similarly, you have yaw. If the right wing coming back, it is a positive yaw. The pitch rotation
about y-axis can be controlled by elevator. We will see about that at a later stage. At the same
time, the yaw can be obtained by means of rudder deflection.

57
(Refer Slide Time: 33:11)

The moving surfaces that we have with wing are ailerons which are used for control and the
flaps or high lift devices are used to enhance the lift. The lift is an essential force that opposes
the weight. If you want to enhance the lift at a lower velocity, we need to operate these flaps.
At the same time, we have spoilers which can be used as a control surface as well. The basic
aim of the spoiler is to reduce the lift, disturb the flow and increase the drag.

(Refer Slide Time: 33:24)

How about horizontal tail?

Here, we have is an elevator, it can be a split elevator as well. You can have four such split
elevators to have the differential control. You also have a vertical tail chord and the
corresponding control surface known as rudder.

58
How to control the speed?

One variable that we have here is thrust. The thrust is often considered as a control input.

We just now saw wing is one of the major components of the aircraft.

(Refer Slide Time: 35:47)

Let us cut this wing. Say this is your center line or fuselage reference line (Refer Slide Time:
35:47). What happens if I cut this wing? We will often find a profile similar to the wing. That
profile is known as airfoil which is a 2D wing or infinite wing. This airfoil is responsible for
generation of lift.

How efficient the lift will be?

There are many ways to generate lift. You can also use a flat plate. The shape speaks a lot about
how efficiently you can generate the lift. Let us look at the nomenclature of this airfoil first.

(Refer Slide Time: 37:30)

59
Let us consider an arc or a line whose nomenclature is Mean camber line. What is this mean
camber line? Now if I want to draw an airfoil with the help of this mean camber line what
should I do? I can build the entire airfoil with the help of this mean camber line.

Let us say I know the thickness distribution at each and every point. Thickness is measured
perpendicular to the mean camber line at that particular point. I have series of points along this
camber line and varying these points, I will get the thickness distribution. With that
information, I will be able to plot upper and lower boundaries. These upper and lower
boundaries are my upper surface and lower surface of airfoil.

The thickness is the distance measured from the upper surface to the lower surface
perpendicular to the mean camber line. Here the thickness distribution is symmetric about the
mean camber line. This mean camber line will be the midpoints of the thickness at that
particular location. So mean camber line can be defined as locus of midpoints of thickness of
an airfoil.

(Refer Slide Time: 39:42)

Hence, here we have upper surface and lower surface, thickness which is the distance between
upper surface and lower surface and measured perpendicular to the mean camber line and the
locus of midpoints of thickness. The starting point of this mean camber line (MCL) is a leading
edge and the after point of the MCL is trailing edge.

A straight line joining this leading edge and the trailing edge is known as chord line.This is my
chord (Refer Slide Time: 41:26).

60
(Refer Slide Time: 40:52)

(Refer Slide Time: 41:00)

(Refer Slide Time: 41:26)

61
(Refer Slide Time: 41:36)

(Refer Slide Time: 41:56)

Now comes the camber, the distance between the mean camber line and chord line is known
as camber. What happens if the camber is zero? The mean camber line coincides with the
chord line. In that case the thickness distribution will be symmetric about chord line.

When there is camber the thickness, distribution is always symmetric about mean camber line
and it is not symmetric about the chord line. But when the camber is zero, the thickness
distribution is symmetric about the chord line. It is like the upper surface is a mirror image of
bottom surface about the chord line. In that case this is how an airfoil look like (Refer Slide
Time: 41:56).

In the above case, the chord line and the mean camber line both coincides with each other and
we have equal thickness distribution. This particular airfoil is known as symmetric airfoil. Here
we have leading edge and trialing edge.

62
The airfoil having a camber is known as cambered airfoil. If this camber is above the chord
line then it is known as positively cambered airfoil. If this camber is below the chord line, then
it is known as negatively cambered airfoil. Hence, there are two airfoils known as positively
cambered and negative cambered airfoils.

If I make a wing out of the symmetric airfoil, I get a symmetric wing and if I make a wing out
of the cambered airfoil, I get a cambered wing.

Let us see how to plot an airfoil. What do you require? You require upper surface and lower
surface and something called leading edge radius.

(Refer Slide Time: 45:46)

The curvature which is tangent to the upper and lower surface right known as leading edge. If
you draw a circle with a radius such that the circle becomes tangent to the upper and lower
surface. The radius of that circle is known as leading edge radius.

Now how can I plot the airfoil? When I say plot here, I need the coordinates of upper and lower
surface. How can I plot it?

If you want to design a new airfoil what all you need to do?

The camber is varying with respect to the mean camber line which is in turn varying as the
chord varies. Let us say you have a chord here. At each and every location of the chord there
are corresponding coordinates for the mean camber line.

(Refer Slide Time: 47:10)

63
(Refer Slide Time: 47:53)

If I have the coordinates of the mean camber line,

This is my y-axis and this is my x-axis (Refer Slide Time: 47:53). Now let us assume the
chord is along this x axis. If I know the equation of the mean camber line, equation that talks
about the distribution of thickness at each and every x, the corresponding thickness distribution
about mean camber line and the slope at each and every point of the mean camber line, I will
be able to plot the airfoil. What do I require here?

Let us say this is my chord of this airfoil. Now for different locations of chord what I require
is what will be the y of camber line at corresponding x, yc(x), thickness at that corresponding
x, yt(x) and theta of the camber line at that x, c(x).

If I have this information, then I will be able to figure out what are the coordinates of upper
and lower surface. Let us see how?

Now let us consider point x on the chord. Now what will be the corresponding point on the
camber line? From yc(x) you will get to know what is the corresponding point on the camber
line right.

Now for the slope here,

If I draw a tangent, theta of the camber line at that particular x, c(x) = (dyc/dx).

Thickness is measured perpendicular to this mean camber line. If I have to draw the thickness,
I need to know the reference. The reference here is the tangent. So perpendicular to this tangent,
I have this yt(x). Consider this is a symmetric thickness distribution.

Now what are these coordinates then (Refer Slide Time: 47:53)?

64
If I draw a local horizontal here which is parallel to x-axis, the corresponding angle to the
perpendicular component, yt, will be theta here. This angle will be equal to c. If I have a cos
component of the angle, you will get the height yt and you know the height yc.

Let us say the point at the local horizontal is yc, xc. The height here is yc, and the horizontal
component is xc. I get yc from yc(x), i.e for a particular x, I know what is y of the mean camber
line.

Now there is a point A and point B (Refer Slide Time: 47:53) where I have (xa, ya) & (xb, yb).

From the above equations, I can get to know what is the coordinate. As I am calculating the
entire length, the second half of the length is acquired by subtraction which leads to a negative
quantity that is why we have negative y-axis here.

Assuming the chord line as your x-axis, you will get the negative y-coordinate. Now, you have
(xa, ya) and (xb, yb). At each and every point you can find out the corresponding upper and
lower surface coordinates for a given point from the chord.

Are we done? What about leading edge radius? Consider this is my leading edge and you know
the slope. At that particular point draw a tangent to the mean camber line with the slope equal
to the slope at that particular chord location. Now measure the radius along this chord line and
draw a circle by using that point as a center so that it should become tangent to the upper and
lower surface. This becomes the leading-edge radius.

Hence, to define the airfoil what we need is i) equation of mean camber line ii) equation for
thickness distribution as function of chord iii) theta of the slope of the camber line as a function
of chord and iv) the leading edge radius ( ).

65
 Design of Fixed Wing Unmanned Aerial Vehicles
Dr. Subrahmanyan Sadrela
Department of Aerospace and Aeronautical Engineering
Indian Institute of Technology Kanpur

Lecture - 04
Examples, Pitot and Static Tube and Differential Pressure Sensor

Dear friends welcome back. In our previous example problems, there was a small mistake that
we have made, so let us solve these example problems again. I will quickly go through these
examples again.

(Refer Slide Time: 00:27)

Example 1:

Consider a fixed wing UAV flying at an altitude of 5 km with a velocity of 30 m/s. Find
corresponding total pressure measured by the pitot tube?

Solution:

The information given here is

Geometric altitude, hG = 5 km
Velocity of the flight vehicle, V = 30 m/s

First, we need to find the total pressure measured by the pitot tube. How to find the total
pressure?

66
At 5 km, we know that the aircraft is flying in the gradient layer. So, we need to find the
corresponding static pressure at 5 km by using gradient layer equations.

(Refer Slide Time: 02:02)

This is a relationship between pressure and temperature at different altitudes where

g0 = 9.81 m/s2
a = -6.5*10-3 K/ m
R = 287 J/kg-K

Now, if I want to find static pressure at 5 km, I need to know what is the corresponding
temperature at 5 km. We know the static pressure at sea level and the corresponding
temperature at sea level. So, if I can find out T5, I will be able to calculate P5.

(Refer Slide Time: 02:50)

67
To find out T5, I use the definition of lapse rate,

First, we need to convert the geometric altitude to the geopotential altitude,

r = radius of earth = 6400 km

See the difference is very less. At lower altitudes, the geopotential and geometric altitudes are
almost same. Now what is temperature T2?

So now to find out the total pressure at the particular altitude I need to find corresponding
density at that particular altitude.

Where

68
(Refer Slide Time: 05:06)

Now let us look at example 2.

(Refer Slide Time: 08:41)

Example 2: Determine the cruise velocity of a wing alone UAV flying at an altitude of 10 km.
The pitot tube measures a total pressure of 26.723 kilopascals. What will be the velocity of
flight?

Solution:

We need to determine the velocity of flight at 10 kilometers altitude.

hG = 10 km
Total pressure, P0 = 26.723 kpa

69
Now if I need to find the velocity what I need is a differential pressure.

To find the velocity at a particular altitude we need to know the differential pressure which is

(Refer Slide Time: 10:05)

We need to convert hG to h

If we want to know what is the static pressure at 10 km, we need to find temperature at that
particular altitude. Hence, the corresponding temperature at that altitude is

Now pressure at 10 km altitude by using gradient layer equation,

70
So, static pressure at 10 km is 26.537 kilopascals. To find the velocity, we need density at that
particular altitude. By using the equation of state,

The density at sea level is,

See the density is decreased by more than half.

So, this is the velocity at which this wing alone UAV is flying at 10 kilometers altitude.

(Refer Slide Time: 13:30)

Let us revisit example number 3.

71
(Refer Slide Time: 15:48)

Example 3: Consider the differential pressure measured by the pressure sensor of a UAV
cruising at 30 meters per second is 0.4 kilopascals. Find the corresponding altitude of flight?

(Refer Slide Time: 16:03)

Solution:

We have here

Differential pressure at a particular altitude, (P0-Ps) h = 409.05 Pascal

Corresponding velocity of flight, V = 30 m/s

72
Say by some means you got to know some other instruments you got to know what is the
velocity of this flight right. Now we have to find the corresponding altitude of this flight.

Since we know velocity and the differential pressure, we can find the density of that particular
altitude

Now since most of the flights are in gradient layer, we assume that for this current course the
flight happens to be in a gradient layer. I have the definition of lapse rate where I have dT/dh.
If I know this dT, I can find the dh since the slope for any of these gradient layers is constant.

Now I know what is the temperature at this particular altitude h. By using definition of lapse
rate,

Now the corresponding geometric altitude is

73
So, the UAV is approximately flying at 3006 meters. Hence, the altitude of flight is
approximately 3 km, so the difference between geometric and geopotential altitude is very,
very less.

(Refer Slide Time: 17:40)

(Refer Slide Time: 20:04)

Let us solve this final example again.

74
(Refer Slide Time: 22:23)

Example 4: The static pressure sensor of a UAV measures a pressure of 53.75 kilopascals. Find
the altitude of flight of this UAV?

(Refer Slide Time: 22:37)

This is the practical way to find out the altitude by using the static pressure sensor. You have
the static pressure sensor and that particular sensor measures the static pressure at that particular
altitude. Now you need to find out the corresponding altitude of flight. So, we have static
pressure Ps at h as,

75
Here, what I need to find is altitude. If I have to find the altitude, I need to know what is the
corresponding temperature at that altitude. From the gradient layer equations,

(Refer Slide Time: 24:42)

So you have the temperature. Now by using the definition of lapse rate,

We have geometric altitude. Now converting to corresponding geopotential altitude,

76
Hence, hG is the altitude at which this UAV is flying when it is measuring the static pressure
of 53.75 kilopascals.

(Refer Slide Time: 27:06)

Now what is the difference between geometric and geopotential altitude?

The difference is hardly 3.9 m.

Now, let us look at some of the sensors that can measure the total pressure and static pressure.

(Refer Slide Time: 27:58)

Let us look at this pitot-static tube. The holes that you can see on this periphery are meant for
static pressure and you can see one hole which is along the longitudinal axis of this tube that is
meant for total pressure (Refer Slide Time: 27:58). These holes are along the circumference
of the outer tube. You can consider there are two coaxial tubes, one with the closed mouth, the
other with the open mouth.

77
This particular outlet is for static pressure and the one which is straight here along the
longitudinal axis of this tube is for total pressure (Refer Slide Time: 27:58). Let us look at the
pitot-static tube and the corresponding pitot-static sensor pressure sensor. The bottom inlet is
meant for static pressure to measure the static pressure and the top one is for the total pressure.
The output we get from here is a differential pressure.

(Refer Slide Time: 29:04)

This is a differential pressure sensor and the scale of this sensor is less than a centimeter and
the total sensor size is about 2 cm (Refer Slide Time: 29:04). The differential pressure sensor
size is about 2 cm including the PCB which helps you to acquire the data and it weighs hardly
15 grams.

(Refer Slide Time: 29:42)

We have another sort of sensor that is mounted on the Navio2 Autopilot. See what you have
here is a static pressure sensor. For the static pressure sensor, you will not be able to connect
any tube from the static port rather wherever you mount the sensor, you need to allow the
particular board to interact with the atmosphere. All you need to do is make a small slit, so that
the air in that cabin interacts with the surrounding atmosphere.

78
 Design of Fixed Wing Unmanned Aerial Vehicles
Dr. Subrahmanyan Sadrela
Department of Aerospace and Aeronautical Engineering
Indian Institute of Technology Kanpur
Lecture - 05
Generation of Lift and Drag

(Refer Slide Time: 00:13)

Good morning friends, welcome back. In our previous lecture, we discussed about how to plot
an airfoil given slope of the camber as a function of chord and the corresponding camber
thickness distribution about the mean camber line as a function of chord. We also require the
leading-edge radius to complete the airfoil.

What exactly this airfoil does when you place it in a flow?

In other words, if I take the airfoil in my hand and start running, what happens to that airfoil?
The effect acquired by holding the airfoil in air while running can be duplicated or simulated
by holding the airfoil and allowing the flow at the required velocity which can happen inside a
wind tunnel. Let us say we place the airfoil inside a wind tunnel.

(Refer Slide Time: 01:13)

79
Now say there is a flow for this airfoil. Whenever we say flow, there are these lines which are
known as stream lines (Refer Slide Time: 01:13). It is assumed that the fluid particle moves
along the stream line. In other words, stream line is the path traced by the fluid particle in which
if you draw a tangent at each and every point, it gives you the direction of velocity at that
particular point.

The fluid particle tends to move in a direction at a particular instant. If you draw a tangent,
which means the fluid particle, at that instant or at that location, will be moving in a particular
direction. The equation of stream line is

We are not going to discuss about it. Let us see, what happens when there is a flow? Also, we
say there is a lift but how the lift is generated?

Although there are many viewpoints about how the lift is generated, there is one fundamental
viewpoint that most people believe and now, we will talk about that viewpoint.

(Refer Slide Time: 03:33)

Let us consider a bunch of stream lines in a stream tube where the flow actually happens. Let
us assume the flow is happening in the form of a stream tube. Now as soon as the stream tube
encounters the airfoil, it will be split into two stream tubes here. So, due to the radius of leading-
edge curvature of airfoil, which is followed immediately after the leading edge, the cross-
sectional area here decreases (Refer Slide Time: 03:33).

Let us say if it is a positively cambered airfoil, the curvature is more above the chord line. The
flow is supposed to touch the airfoil at the leading edge. Now there is an upper stream tube and
the lower stream tube. Due to the leading-edge curvature, the cross-sectional area of the stream

80
tube reduces. Consider an incompressible flow, where the density remains constant throughout
the flow. Using continuity equation, law of conservation of momentum

If the cross-sectional area decreases the velocity has to increase, which implies V is increased
here.

At the same time, if you have a symmetric airfoil, you have equal thickness distribution which
also has a similar reduction in cross sectional area on either side. If you have a cambered airfoil,
you have greater reduction in cross sectional area of the stream tube above the chord line and
when you have symmetric airfoil you have equal reduction in area above and below the chord
line.

Now this will accelerate the flow above the surface.

Let us talk above the surface. The dynamic pressure here increases and
theorem for incompressible flows

When the dynamic pressure increases, there is a drop in the static pressure. This particular
phenomenon is known as suction.

So, there is a drop in static pressure here. On the bottom surface, although the cross sectional
is reduced the acceleration is not as high as that happen in the upper stream tube, because of
which there is a greater pressure drop in upper surface. There is also a pressure drop on the
bottom surface but not as high as the upper surface. This pressure difference will help in
generating lift.

Now, for the symmetric airfoil, it has equal thickness distribution about the chord line, so you
have equal reduction in cross sectional area. Here, the acceleration will remain same on upper
stream tube as well as lower stream tube. Hence there is an equal pressure drop on upper and
lower surface.

Eventually a symmetric airfoil at zero angle of attack does not produce any lift. Now we have
defined a term called angle of attack. Let us say, if this is my reference line, the chord line of
of the symmetric airfoil, the angle made by this reference line with respect to free stream is
known as .

V is the velocity at which I am moving in a stationary air assuming there are no winds and
disturbances. If I am moving in a stationary air at velocity V , that means the air is also moving

81
towards me with the same velocity. Hence, the velocity, V , represents the velocity of your
flight vehicle.

For a wing or an airfoil, we consider the chord line as a reference line. For fuselage, we can
say fuselage reference line or center line. The angle made by this with the reference line is
known as angle of attack.

When there is an angle of attack then you have greater curvature on the upper stream tube in
comparison to that of lower stream tube which eventually results in a greater pressure drop why
because there is a greater reduction in cross sectional area which helps to accelerate the flow
resulting in pressure drop which is also true even for compressible flows.

Consider below Euler equation,

Here, density cannot be negative for a fluid. So, if there is an increase in velocity there is a drop
in static pressure. It is true irrespective of whether it is compressible or incompressible.

Let us see the pressure distribution for an airfoil. We talk about pressure drop, so how the
distribution will look like?

(Refer Slide Time: 10:23)

To review again, there is a pressure drop, but the drop occurring above will be higher for
cambered airfoil even at zero angle of attack. It is higher compared to that of lower surface and
this pressure difference will help in generating lift. In case of symmetric airfoil, you have to
maintain certain angle of attack so that you can attain the pressure difference by which the lift
can be generated.

82
Pressure distribution:

Let us say this is my V (Refer Slide Time: 10:23). This is my airfoil which is placed at an
angle of attack . Now so this is how typically a pressure profile of an airfoil look like (Refer
Slide Time: 10:23). This is a cambered airfoil or you can also have a symmetric airfoil at an
angle of attack. The arrows here pointing away from the surface represents the suction or
negative pressure (Refer Slide Time: 10:23).

So even on the bottom surface we have suction or negative pressure but the difference is the
suction is more on the upper surface. Now we understood that there is a pressure distribution
whenever we place an airfoil or a wing in a flow. Since we are moving in a fluid, there is always
a chance that the body may encounter some friction with the fluid.

So that friction is the shear stress and the shear stress will be tangent to the surface. Hence,
there will be a shear stress acting tangential to the surface due to fluid friction. Let us consider
ds be the small surface, so let n^ with the unit vector normal to the surface and k^ with the unit
vector tangential to the surface.

Now the pressure over the surface, a negative pressure, can be,

Force due to pressure = , a negative pressure

Surface integral force =

So, these are the two forces that result in a resultant aerodynamic force, one is due to friction
and the other one is due to pressure.

Resultant Aerodynamic Force =

(Refer Slide Time: 14:44)

83
This is actually the extension of the chord line (Refer Slide Time: 14:44). This is my V and
this is my angle of attack (Refer Slide Time: 14:44). There is a resultant aerodynamic force,
R which we are able to estimate by integrating the pressure force and shear stress distribution.
Now, a component of this resultant aerodynamic force acting perpendicular to free stream is
lift and the component of this resultant aerodynamic force acting parallel to free stream is drag.

If you say the arrows represent the magnitude, then drag is much smaller than lift. We have
defined lift and drag and they belong to wind. When we are talking in terms of lift and drag,
we are talking about wind axis. So, forces in wind access are lift and drag.

(Refer Slide Time: 16:45)

Now let us define lift and drag.

A component of resultant aerodynamic force acting perpendicular to V in free stream velocity

is lift and the component of resultant aerodynamic force acting parallel to free stream velocity
is drag. The drag is acting opposite to the direction of motion or in the direction of free stream
velocity. Now we have defined lift and drag here.

(Refer Slide Time: 18:38)

84
Let us denote lift, L

= dynamic pressure, where is the density

S = reference area
CL = coefficient of lift

Although you are moving at same speed with the same aircraft at the same CL at different
altitudes, you will end up producing different lifts because of the change in the density, .

(Refer Slide Time: 19:58)

Similarly drag is defined as

CD = drag coefficient.

The units of lift and drag is Newton, where CL and CD are non-dimensional quantities.

Let us now assume that we place the airfoil inside a wind tunnel and vary the angle of attack
and measure the corresponding CL.

85
(Refer Slide Time: 20:43)

Here, CL is the function of alpha ( ), Mach number (M) and Reynolds number (Re). In general
Reynolds number is defined as

L = reference length, ratio of inertial forces to the viscous forces

µ = dynamic viscosity of air

Mach number is defined as velocity of the flight / velocity of sound

(Refer Slide Time: 21:12)

86
(Refer Slide Time: 21:53)

Now, let us vary and see how CL varies. CL is plotted along the y-axis and is plotted along
the x-axis. For a cambered airfoil, this is how a typically the CL vs plot look like (Refer Slide
Time: 21:53). For a cambered airfoil, you have lift coefficient called CLo even at zero angle of
attack. CLo is the y intercept at =0 and you have at which CL=0 (Refer Slide Time: 21:53).

This is negative for a cambered airfoil and the slope see remains linear until certain angle of
attack within this region (Refer Slide Time: 21:53). You can define the slope as dCL/d which
known as lift curve slope and is often termed as CL right. Now, beyond this there is certain
nonlinearity and if you further increase the angle of attack, the CL becomes maximum.

If you further increase the angle of attack beyond the CLmax, the lift suddenly starts dropping.
That particular point of CLmax is known as alpha stall and the corresponding angle of attack at
which CLmax is stall. Beyond stall you will start losing lift and the drag will also increases. We
will see how the drag increases later. In the linear region if I have to model the lift, I have to
write an equation for lift with the variation of angle of attack. Since it is a linear curve up to
certain angle of attack what I can assume is

(Refer Slide Time: 24:20)

87
Since it is a straight line, y = mx+c,

y = CL and
slope, m = CL
x=
y-intercept, c = CLo

I can use the above equation to model the CL as a function of angle of attack up to certain
region.

(Refer Slide Time: 24:54)

We took it granted that there exists a resultant force when there is a flow and we resolved it
along the flow as well as perpendicular to the flow from which we got lift and drag. But where
does the resultant force act, we need to know that. Where does that point exist?

The answer is center of pressure. If you look at here, there is a pressure distribution and the
average of this will act at certain point (Refer Slide Time: 27:50). There is a pressure
distribution on bottom surface as well whose average will be acting at some point here, maybe
the centroid of this particular distribution (Refer Slide Time: 27:50). So the particular point
that lies on the chord is a center of pressure.

Let us define center of pressure and we call it xcp.

It is the point mostly located on chord at which the resultant aerodynamic force or forces act.
If you consider a moment about center of pressure, it will be 0. Now we have defined something
called moment and where this moment is coming from?

88
(Refer Slide Time: 27:50)

Consider an airfoil, so there is a lift, there is drag and there is V . Earlier, we have defined lift
and drag. Now let us also define moment, M about a point.

where Cm = non-dimensional pitching moment coefficient

= mean aerodynamic chord

We will see what is this mean aerodynamic chord for a wing or for an airfoil. Now in this case,
if I have to write moment about a point and there is a force,

pitching moment coefficient about any point, Cm = moment at zero lift, This is a pure
moment that is acting when there is no lift.

(Refer Slide Time: 29:30)

89
There is lift, there is drag and say there is a point O. Here, the pitch up is considered positive,
pitch down is considered negative. What is pitch up here? Consider y- axis into the blackboard,
stretch your thumb along the positive y where pitch up is considered negative. The curl of
fingers gives you the positive moment, so pitch up is considered positive and pitch down is
considered negative.

Let be the moment when the lift is 0. Let us say the pitching moment is the sum of moment
when there is no lift and moment which depends upon the location your moment reference
point.

= moment at CL=0, when there is no lift

= moment due to lift

Now let us consider the point O or the moment reference point at the leading-edge. The lift is
acting at some point which is xcp. We are considering moment about the leading edge. Now xcp
is somewhere behind, so the lift will contribute a negative moment. CL cannot be negative so
has to be negative. If we consider the most forward point, the has to be negative. If we
consider the moment reference point towards the trailing edge then becomes positive.

As is varying from positive to negative, there exist a point at which = 0. That point is known
as aerodynamic center.

(Refer Slide Time: 32:27)

What is happening when = 0?

To review, when you consider the moment reference point towards the leading edge, you have
as negative why because CL will contribute a nose down moment. If CL is positive, lift is
positive and has to be negative. If you consider the same moment reference point at the
trailing edge, becomes positive and there exists a point when = 0.

90
At = 0, you have,

moment about that point = moment when lift is zero (CL=0)

This particular point is known as aerodynamic center. So, the moment about aerodynamic
center (a.c) is

Now let us define the important variable aerodynamic center. It is the point along the chord
about which pitching moment is independent of angle of attack.

Now we witnessed that, we have modelled CL as a function of angle of attack,

Here, the CL varies with angle of attack. If I take moment reference point as the aerodynamic
center, the pitching moment ( will remain constant. Hence, remains constant
irrespective of angle of attack.

Why Cm exist?

You call the aerodynamic center as xac. For a low speed, it is observed that xac lies at the quarter
chord, almost at 0.25c, but what is this moment about aerodynamic center? Why does it exist
even when lift is 0? When can lift be 0? Let us come back to this figure (Refer Slide Time:
35:50).

(Refer Slide Time: 35:50)

91
Let us say by integration,

resultant force along the upper surface = resultant force along the bottom surface

If you integrate the pressure over the upper surface, you will get a force on the upper surface
(Fu, p). and integration of pressure on the bottom surface will give force on the bottom surface
(Fb, p). If these two forces become equal, then CL becomes 0.

What happens for a symmetric airfoil? The upper surface distribution and bottom surface
distribution becomes equal and CL is 0. The resultant of the integrated force will be acting at a
particular point on the upper surface say along a point on the chord line, xupper surface and the
bottom surface integrated force will also act at certain point, xlower surface.

For symmetric airfoil the pitching moment about aerodynamic center is 0, because the two
points (xupper surface and xlower surface) will coincide. But for a cambered airfoil, even for CL = 0, the
two points will not coincide. Hence, there is a pure couple because two forces are acting at an
offset. The couple is independent of moment reference point. So the Cm about aerodynamic
center is the pure couple. So how should I relate this aerodynamic center and center of pressure?

(Refer Slide Time: 37:35)

Let us consider a point O on the chord line, let this be aerodynamic center and there is a center
of pressure. Let the reference be xcp from the point O and xac be the aerodynamic center from
point O. Also, there is V . There is lift and drag. The same lift and drag can be represented at
the aerodynamic center followed by a moment that is moment about aerodynamic center.

Now let us consider a moment about point O,

92
By considering aerodynamic center, moment about aerodynamic center

Now consider moment about point O with respect to center of pressure

; Cmac remains constant.

This is a relation between center of pressure and aerodynamic center.

Now as angle of attack increases what happens? CL will increase that means xcp becomes close
to xac. So, at higher angles of attack, the center of pressure will start moving towards
aerodynamic center. Let us see why?

(Refer Slide Time: 41:27)

Let us assume the flow is completely attached. As you go to higher angles of attack, the flow
may not remain attached. That means the effective pressure distribution will be for a smaller
area of upper side (Refer Slide Time: 41:27). So, the resultant pressure will act somewhere at
the centroid. Say if this is your resultant pressure distribution at higher angles of attack (Refer
Slide Time: 41:27), then xcp for initial case and this case will vary. The xcp will slowly shift
towards your aerodynamic center. When this can happen? At higher CL s which can be
achieved at have higher angle of attack.

93
 Design of Fixed Wing Unmanned Aerial Vehicles
Dr. Subrahmanyan Sadrela
Department of Aerospace and Aeronautical Engineering
Indian Institute of Technology Kanpur

Lecture - 06
Aerodynamic Center and Center of Pressure, Various Wing Planform

Hello friends. Welcome back. In our previous lecture, we were discussing about center of pressure
aeronautical center, relationship between them. And we have witnessed how CL varies with angle
of attack and how can we model CL variation with alpha mathematically.

(Refer Slide Time: 00:32)

We have

This corresponds to variation of lift coefficient and angle of attack in the linear region of angle of
attack. At the same time, we have defined two important parameters. One is aerodynamic center
(ac) the other one is center of pressure (cp). Center of pressure is a point about which resultant
aerodynamic forces act and above center of pressure the pitching moment is zero. Aerodynamic
center is defined as the point about which pitching moment remains constant with angle of attack.

94
Let us now consider an airfoil and say xac is the distance of aerodynamic center with respect to the
leading edge and xcp is the location of center operation with respect to leading edge here. Now let
us take the moment. Let O be the leading edge here and we are going to derive the moment
expression about this point O for this particular aerofoil (Refer Slide Time: 00:32). Say, we are
testing it at certain .

Let us consider there is a flow with a velocity v overr this airfoil. When there is a flow there is
lift and drag and also there is lift and drag at moment about aerodynamic center. Now let us
consider the moment about point O by considering the forces and moment acting at aerodynamic
center.

Now the pitching moment about point O,

Where, ;

Now if you take the moment about the same reference point by considering the forces acting at the
center of pressure (since center of pressure is defined as the point about which the resultant
aerodynamic forces act), what we have is the moment due to lift because the drag contribution is
very less along the same axis, hence neglecting drag.

So, the pitch down moment due to lift is

Above two equations represent the moment about point O due to lift. The moment should be equal
whether you are considering with respect to aerodynamic center or center of pressure. By equating
those 2 what we have?

95
(Refer Slide Time: 05:06)

This is the relationship between center operation and aerodynamic center. We also address that at
higher angle of attack the CL increases due to which becomes very low and this will move
close towards .

Now let us take up an example.

Example:

NACA 2412 airfoil is tested in a low speed wind tunnel. It is observed that the pitching moment
coefficient about the leading edge at zero lift is 0.02. Also, at =8o the CL is measured to be 0.7
and Cm about leading edge is -0.2. Find the aerodynamic center for the airfoil.

Solution:

Given,

96
Given Cm about leading edge at CL=0 is -0.02.

Let us say this is our leading edge and this is my aerodynamic center (Refer Slide Time: 05:06).
Let this distance be xac (Refer Slide Time: 05:06). V and will have lift and moment about
aerodynamic center. Now, if you write an equation for pitching moment about aerodynamic center
about this leading edge with respect to this aerodynamic center, we have

From the above equation,

the moment about leading edge, at CL = 0, measured to be -0.02 turns out to be the moment about
aerodynamic center.

From the same equation we have CL at 8 degrees and the corresponding Cm.

(Refer Slide Time: 08:41)

By substituting the corresponding values,

97
 is a non-dimensional number.

Hence, for low speed flight vehicles, the aerodynamic center in general lies at about 22 to 26% of
mean aerodynamic .

(Refer Slide Time: 09:31)

It is worth noting that the center of pressure keeps varying with angle of attack. Why? Because CL
keep varying here from this equation (Refer Slide Time: 05:06). At the same time, you know as
angle of attack changes, the pressure distribution changes. Once you have different pressure
distribution, you have different centroid for that distribution. So xcp keep varying whereas xac is
remains constant over a range of velocities whereas angle of attack always remains constant.

(Refer Slide Time: 10:47)

98