College of The Bahamas Lift Problems at Low Reynolds number in Forward Flapping Paper Research Paper Details

The paper must be written on an aerodynamics or performance topic (theory, aircraft design, wing design, etc.). The best source for topics is the table of contents and index of aerodynamics and performance texts. The paper cannot be a topic that has been used in another class and will be submitted to SafeAssign. Only one topic per student (first come first serve) so reserve your topic by emailing me the abstract ASAP but no later than the date shown on the course schedule. If your topic is taken I will advise and you will have to select another. If you have not had a topic approved by the deadline, you will not be allowed to turn a topic in later, and will forfeit the bonus opportunity. Do not assume I received and approved your topic unless you receive an email reply. Save my email approval for your records. If you reserve a topic you may change it ONLY ONCE as long as it is done before the topic due date. You should have several sources actually cited in your work (not just listed at the end). A quality paper normally has one reference per page, so a 1250 word paper would have at minimum 5 different sources.

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Some Guidance on the Optional Bonus Research Paper

First read all the information and guidance about the paper available on Blackboard. In selecting your topic and writing your paper keep in mind the objective of this is to research new information about a topic related to aerodynamics and/or aircraft performance beyond what is already covered in the course and the course text, and beyond what you personally already know. In doing so you are rewarded by getting up to 10 bonus points on your overall score (a full letter grade). A common mistake is to pick a topic off the top of the head without first having researched what sources and information is available about that topic. Sometimes this quick selection of a topic results in one that is far too broad, or one that does not have good external (beyond the course or course text) sources of information.

Keep in mind it must be related to this course (aerodynamics and or aircraft performance) do not write a paper that would be better suited for another course such as a course on aviation history, engines, aviation safety, meteorology, or pilot operational ground school course. Writing a paper suited for a class other than this one would not result in bonus points.

You must go deeper and more detail than what is already covered in the course material and course textbook. Use that as your starting point. So if you selected a topic covered in the course, read the course and text material first (even if we have not yet covered it) but your paper should have new material from sources outside the course. So in that case, start where the course and textbook leaves off and let your resources take you into new material not covered in the course. What you can find in your sources outside the course is what will tell you if it is a good topic or not. I

And the reference should be no less than 6 and one should be the course text book AIRCRAFT PERFORMANCE

FOR

PROFESSIONAL PILOTS

A Technical Based Pilots Performance Reference

SEVENTH DRAFT EDITION

by

James E. Lewis

Contributors:

Melville R. ( By ) Byington, Jr.

David Esser

Don Smith

J. Erin Webb, Editor

Copyright © 2003

by James E. Lewis

All rights reserved. Written permission must be secured from the author to

reproduce any part of this book.

1.1

1.2

PREFACE

This book is designed for use in a college level aircraft

aerodynamics or performance course where students have had a

background in college level algebra, introductory calculus, and

physics. The book attempts to bridge the gap between basic

aerodynamics and performance books written for nontechnically trained pilots and more advanced books written for

technically oriented professionals. Few books address the

aerodynamic and performance needs of an advanced flight

training program without either being too basic or too complex.

This is not an engineering text. It is written for the purposes of

preparing young pilots with limited large airplane experience for

the world of corporate jet or heavy jet transport airplanes.

JEL

1.3

1.4

DEDICATION

This book is dedicated to my oncologist, Dr. Gregory Favis and surgeon, Dr.

Harry Black. Without their outstanding medical care I would not have been able to

finish this project. They have done a wonderful job caring for me during a rather

difficult time in my life.

I would also like to thank my wife Kathy and daughters Erin and Robin for

their infinite patience in dealing with me while I was working for endless hours on

my computer. They also help me correct most of my vast writing errors as I

attempted to bring this work together.

I would also like to thank Boeing Aircraft Corporation for allowing me to

use many of their aircraft performance charts as examples in this book. Their

commitment to aviation education and safety is amazing.

1.5

1.6

Table of Contents

Chapter

Subject

Page

1

Basic Physical Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1

2

The Standard Atmosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.1

3

Airspeed Computations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.1

4

Basic Lift and Drag Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.1

5

Thrust ( or Drag ) and Power Required . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1

6

Thrust and Power Available . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1

7

Takeoff Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1

8

Large Airplane Takeoff Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1

9

Determine V1, VR, V2, and Balanced Field Length . . . . . . . . . . . . . . . . . . . 9.1

10

Climb Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1

11

Temperature Ram Rise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12

Jet Airplane Range and Endurance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1

13

Optimum Range and Endurance Performance . . . . . . . . . . . . . . . . . . . . . . 13.1

14*

Large Aircraft Weight and Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.1

15*

Using Jet Takeoff Charts . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1

16*

Using the Planned Takeoff Data Message . . . . . . . . . . . . . . . . . . . . . . . . 16.1

*NOTE: Chapters being revised at time of this printing

1.7

11.1

AS-310: Aircraft Performance

Chapter 1: Basic Physical Concepts

Purpose: Over the years, as we have taught aerodynamics and aircraft performance, we have

noticed that students often do not retain certain key information from their introductory physics

courses, which makes it difficult for them to do well in aerodynamics and performance. Often,

they have not been exposed to aviation related applications of physics, so they do not understand

why the subject is important to them. In reality, understanding airplane aerodynamics and

performance is Applied Physics. We take certain principles learned from physics and make use

of them in order to more fully understand real-world flight related situations. There was a reason

to learn physics! Such study was not simply a sophisticated form of mental torture. We want you

to know a few fundamentals before we begin our adventures in aerodynamics and performance.

Learning Objectives: After completion of this lesson you should,

1. Understand the four fundamental concepts of Basic Units, Basic Equation Basic

Units, Unit Conversion, and Proportional Thinking.

2. Using the British Gravitational System (BGS), know the definitions and basic units for

the most commonly used physical terms which apply to the subject of applied

aerodynamics or aircraft performance.

3. Be able to fully explain the key physical principles from introductory physics that

apply to the technical understanding of the subjects of aerodynamics and aircraft

performance.

Basic Concepts

Basic and Derived Units The concept of Fundamental, which we call Basic Units

versus Derived Units. is important to grasp if you are going to successfully work problems using

physical quantities. You must know the definitions of certain physical quantities and their

fundamental or basic units. We will use the British Gravitational System (BGS) in this course

since it is widely used in aviation. In this system all units are based upon a distance unit of the

foot, a force unit of the pound, and a time unit of the second. Many other physical quantities we

will use are derived from these three basic units (the foot, the pound, and the second).

Table 1.1 on the next page lists the basic (fundamental) and derived mechanical units one

should know for a more advanced study of aircraft performance. The derived units are defined in

terms of our three basic units, then expressed in British Gravitational System (BGS) units. Next,

the symbols used in this book are shown. It is probably worthwhile to take the time to reacquaint

yourself with these units. We will be using them quite often in this course.

1.8

Quantity

Definition

(symbol)

Fundamental or

Basic Unit

Force (F)

Force

BGS Unit

BGS Symbol

pound

p

Length (X)

Length

feet

ft

Time (t)

Time

second

s

Area (S)

Length 2

feet 2

ft 2

Volume

Length 3

feet 3

ft 3

Velocity (V)

Length / Time

feet / sec

ft / s

Acceleration (a)

Velocity / Time

feet / sec 2

ft / s 2

Mass (m)

pound / feet / sec 2 (which

is called a slug)

pound / feet / sec 2

p / ft / s 2 = p-s 2 / f = sl

Density (!)

Force /

Acceleration

Weight /

Acceleration

Mass / Volume

Work (W)

Force x Distance

pound foot or foot pound

ft-p

Energy (PE or

KE)

Power (P)

Force x Distance

pound foot or foot pound

ft-p

Work / Time

foot pound per second

ft-p / s

Momentum (M)

Mass x Velocity

pound second

p-s 2 / f x ft / s = p-s

Impulse (I)

Force x Time

pound second

p-s

Derived Units

Mass (m)

slug / feet

3

W/g

where, g = 32.174 ft / s 2

p-s 2 / f / f 3 = sl / ft 3

Table 1.1: Basic and Derived Units – British Gravitational System (BGS)

Example 1.1 Without using Table 1.1, derive the units for impulse?

Hint: Use Newtons 2nd Law (F = ma)

F = ma = mV/t = M/t

(M or m V we call Momentum)

s

F t = Impulse = m V = sl (f / s) = (p s 2 / f) (f / s)

1.9

=

ps

Basic Concepts

“Basic Equation Basic Units”. When dealing with unfamiliar relationships, it is

sometimes difficult to know what units should be used. Many times in numerical calculations

keeping track of the proper units is half the battle. This task is easier if we always make it a

practice to think in terms of fundamental or basic units. If you look at an equation and it does not

contain any unusual conversion factors (e.g. no number other than 2, ½, 3, or 1/3), then you most

likely can use any basic system of units (British Gravitational or Metric) as long as you are

consistent and be relatively comfortable with your results. To solve this sort of basic equation

simply use basic units with your numeric values. You can check this by substituting the units into

the equation. The units to the left of the equal sign will exactly equal the units on the right.

Example 1.2 Considering the relationship, F = m a, show that the units on the

right side of the equal sign are exactly equal to the units on the left.

Note: This expression does not contain any unusual conversion factors;

therefore, this is a basic equation so basic units should work. The

“slug” is not the basic unit for mass. The derived unit (slug) should

be converted to its basic form (p-s 2 / ft) before continuing.

F=ma

or

p = (sl) (ft / s 2) = (p s 2 / ft) (ft / s 2)

or

p=p

Note: The units on both sides of the equation turn out to be equal. This

key point can be helpful to know when trying to figure out what

units were used in an equation.

Example 1.3 Consider the Lift Equation from Basic Aerodynamics, show that

the units on the right side of the equal sign exactly equal the units

on the left.

V = [295 W / (CL ” S)]

Note: Here we see a “strange” number 295. This is a clue to tell us that

someone has converted this relationship to some non-basic system of

units. The 295 is a conversion factor. Unless we derived the equation, we

may not know the units involved. They could be light years per cubit.

Cannot solve the equation

without more information.

1.10

1.11

Basic Concepts

Unit Conversion Converting from one system of units to another is a lot easier if you

are aware of the process also called Dimensional Analysis (big word describing a simple process).

Lets learn what this means by example.

Example 1.4 Consider the situation where you want to convert from knots to feet

per second, in order to use a basic equation (without a conversion

factor).

We can do this by carefully analyzing the units involved.

Nautical Mile

6,076 feet

1 hour

1 Knot = ————- x ——— x ———— = 1.6878 ft / s

hour

Nautical Mile 3,600 seconds

Example 1.5 Consider a situation where you want to convert from RPM to radians

per second (the basic unit for rotational motion is the radian per

second or radians / s) in order to use a basic equation without a

conversion factor.

Revolution 2 Radians

1 Minute

1 RPM = ———– x ————- x ———— = 0.10472 radians / s

Minute

Revolution

60 Seconds

Next is an example of using this conversion factor,

Example 1.6 Convert 2,400 RPM to basic physical units.

2,400 Rev / min x 0.10472 radians / sec / min

= 251 radians / s

Proportional Thinking Take the expression for kinetic energy, KE = ½ m V 2. This

expression states that KE is directly proportional to mass (m) and velocity (V 2) or KE ? m and

KE ? V 2. The symbol ? means is proportional to. If either mass or velocity are increased,

the kinetic energy (KE) will increase. This type of relationship is called a Direct Proportion. If

only one factor in this expression is varied at a time, the direct proportion can be expressed as

follows,

KE = K1 m ,

where K1 = ½ V2

or

1.12

KE = K2 V2 , where K2 = ½ m

Basic Concepts

Proportional Thinking

Varying one factor at a time is a handy procedure called completing a Parametric Study. This sort

of study is often used to examine the impact of one factor in a relationship on the resulting answer.

In this case, how much will kinetic energy (KE) change if only mass (m) is increased or how much

will kinetic energy change if only velocity (V) is increased?

The expression KE = K1 m can be rewritten as, KE / m = K1 (a constant)

Comparing two situations, Case 1 where m = m 1 and Case 2 where m = m 2

KE 1 / m 1 =

KE 2 / m 2

or

KE 2 = KE 2 (m 2 / m 1)

The given direct proportion has now been expressed as a Ratio. See the following examples

for clarification of the importance of proportional thinking.

Example 1.7 Say that an airplanes mass (m) is increased by 50%. By how much

will its kinetic energy (KE) change?

In this case, m 2 = 1.5 m 1

(Representing a 50% increase in mass)

Expressing the fact that KE ? m, as a ratio,

KE 2 = KE 1 (m 2 / m 1) = KE 1 (1.5 m 1 / m 1)

It can now be said that,

KE 2 = 1.5 KE 1 or Kinetic energy has increased by 50%

Example 1.8 Say that a airplanes velocity (V) is increased by 50%. By how

much will the kinetic energy (KE) change?

In this case, V 2 = 1.5 V 1

(Representing a 50% increase in velocity)

Using the fact that KE ? V 2 and expressing it as a ratio,

KE 2 = KE 1 (V 2 / V 1) 2 = KE 1 (1.5 V 1 / V 1) 2

Therefore,

KE 2 = 2.25 KE 1 or Kinetic energy has increased by 125%

( 2.25 times KE1 is a 125% increase)

1.13

Basic Concepts

Proportional Thinking

Example 1.8

Note

(continued)

Do you understand why increasing velocity (V) has a greater impact

on kinetic energy (KE) than increasing the mass (m)?

Consider the uniformly accelerated rectilinear motion expression relating time to distance

and acceleration, t = (2 X / a). In this case t is directly proportional to X and inversely

proportional to a. If distance (X) increases, the time (t) will increase but if acceleration (a)

increases, the time (t) will decrease. This second type of relationship is called an Inverse

Proportion. Again, if only one factor is varied at a time, these proportions can be expressed as

follows,

t = (2 X / a) or

K1 X , where K1 = (2 / a)

(Direct proportion)

and

t = (2 X / a) or K2 (1 / a),

where K2 = (2 X)

(Inverse proportion)

Again, varying one factor at a time a parametric study is performed. This sort of study

can again be used to examine the impact of varying one factor on the resulting answer. In this

case, focusing on the inverse relationship, how much will time (t) change if acceleration

(a) is increased?

The expression t = K2 (1 / a) can be rewritten as, t a = K2

(a constant)

Comparing two situations, Case 1 where a = a 1 and Case 2 where a = a 2

t1 a1 = t2 a

2

The given inverse proportion can also be expressed as a ratio,

t 2 = t 1 (a 1 / a 2)

1.14

Basic Concepts

Proportional Thinking

Example 1.9 An airplane is experiencing uniformly accelerated rectilinear motion.

By how much will the time (t) change for the airplane to move a

given distance (X) if acceleration (a) is increased by thirty (30)

percent?

In this case, a 2 = 1.3 a 1

(Representing a 30% increase in acceleration)

Expressing the fact that t ? 1/ a as a ratio,

t 2 = t 1 (a 1 / a 2) = t 1 (a 1 / 1.3 a 1)

It can now be said that,

t 2 = t 1 / 1.3 = 0.877 t 1 or Time has decreased by 12.3 %

Example 1.10 Repeat Example 1.9 except decrease acceleration (a) by thirty (30)

percent.

In this case, a 2 = 1 – 0.3 = 0.7 a 1 (A 30 % decrease in acceleration)

Again, expressing the fact that t ? 1/ a as a ratio,

t2 =

t 1 (a 1 / a 2) = t 1 (a 1 / 0.7 a 1)

It can now be said that,

t 2 = 1.195 t 1 or Time has increased by 19.5%

1.15

Important Physical Principles

Newton’s First Law (The Law of Inertia) A body left to itself has constant velocity

(speed and direction).

Newton’s Second Law (The Law of Momentum) The acceleration (a) of a body is

directly proportional to the resultant force (F) acting on the body, is inversely proportional to the

mass (m) of the body, and has the same direction as the resultant force (F). The resultant force

(F) acting on a body can be the vector sum of the many forces.

Re-arranging the terms slightly, this principle can be expresses as,

1.1

Where,

F = ma

F

m

a

(Newtons Second Law – also called Newtons Law of Momentum)

= force (p)

= mass (sl)

= acceleration (ft / s 2)

Note: To change somethings velocity we must apply a force.

Example 1.11 What would be the acceleration on a 130,000 pound airplane

subject to a net unbalanced force of 30,000 pounds?

Using Newtons Second Law (Equation 1.1), F = m a

a = F/m

Where, m = 130,000 p / 32.174 ft / s 2 = 4040.5 sl

a = 30,000 p / 4040.5 p-s 2 / ft

= 7.4 ft / s 2

Note: In this example the slug was expressed in its basic form, p-s 2 / ft. In

this way the units work out for acceleration as ft / s 2.

1.16

Important Physical Principles

Newton’s Second Law (Newtons Law of Momentum)

Sometimes Newton’s Second Law is expressed as the Time Rate of Change of Momentum.

We know that applying a force for a given amount of time will change an objects momentum.

mV

F = m a = m ( V / t ) = —– = M / t

t

1.2

Where,

F = M / t or

F

M

t

I

Ft = M ,

Where, mV or momentum (M)

Where F t = I or Impulse

= force (p)

= momentum (p-s)

= time (s)

= impulse (p-s)

Note: F t is defined as impulse (I). To change somethings momentum (M) we

must apply an impulse (I) or a force (F) for a given amount of time (t).

Example 1.12 What impulse (I) is required to change an airplanes momentum (M)

by 200,000 p-s?

Applying Newtons Second Law directly, we know that momentum is

changed by applying an impulse,

F t = Momentum (M) = Impulse (I)

= 200,000 p-s

Example 1.13 How long would a 20,000 pound force need to be applied in order to change

an airplanes momentum by 200,000 p-s?

Using Newtons Second Law in terms of momentum (Equation 1.2),

F t = M, solving for t

t = M / F = 200,000 p-s / 20,000 p

1.17

= 10 s

Important Physical Principles

Newton’s Third Law (The Law of Reaction). For every action there is an equal and

opposite reaction. This principle is the basis for airplane propulsion. Mass that is accelerated

rearward behind the airplane results in an equal but opposite force we call thrust. Thrust is what

propels an airplane through the air. The airplanes motion in the same direction as its thrust.

Mechanical Energy: There are two types of mechanical energy that are helpful to know

about when examining aircraft performance: potential energy and kinetic energy. Potential

Energy is the energy an object possesses based upon its position. Kinetic Energy is the energy an

object possesses due to its motion.

Potential energy (PE) can be expressed as an objects weight (W) times its height (H).

We know that the force, W = m g,

1.3

PE = W H

Where,

PE

W

H

m

g

or

mgH

= potential energy (ft-p)

= weight (p)

= height (ft)

= mass (slugs)

= 32.174 ft / s 2 = constant in the BGS

Closely examining the units in terms of basic units,

PE

.

=

(pound x second 2) foot

foot

—————— x —- x —- = foot x pounds

foot

second 2

= ft p

Note: The ft-p unit is the same as we defined earlier in general for energy

Review Table 1.1 if need be.

Example 1.14 What is the potential energy (PE) of a 193 pound sky diver falling

from 1000 feet above the ground?

Using the definition of potential energy (Equation 1.3), PE = W H

PE = 193 p (1,000 ft)

1.18

= 193,000 ft-p

Important Physical Principles

Mechanical Energy

An objects kinetic energy (KE) can be expressed as,

1.4

KE = ½ m V 2

Where,

KE

m

V

= kinetic energy (ft-p)

= mass (sl)

= velocity (ft / s)

Ignoring the ½ for now since we are only talking about the units involved,

KE

KE

=

½ m V 2 = sl (ft / s ) 2

=

( p- s2 )

——————–ft

ft

ft

x ——- x ——s

s

= ft-p

Example 1.15 How much kinetic energy (KE) must the brakes of a 400,000 p

Boeing 747 absorb while stopping an airplane after touching down

at 150 ft / s? You will see this is a LOT of energy.

Using Equation 1.4, KE = ½ m V 2

m = (400,000 p) / (32.174 ft / s2 ) = 12,432 p s2 / ft or 12,432 sl

KE = ½ m V 2 = ½ ( 12,432 p s2 / ft) (150 ft / s) 2 = 140 million ft-p

Note: 140 million ft-p of mechanical energy converts to a lot of heat in

the brakes. Remember, p s2 / ft = slug (sl) by definition

Conservation of Mechanical Energy. Energy can be neither created nor destroyed, but it

may change form. One important application of this concept is with Total Mechanical Energy

(TE),

Tot…

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