Final Report Test5

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Aerial Navigation: The Mathematics Behind It

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The Mathematics Behind It

This section covers the topics of:

The history behind satellites Early types of navigationThe problem this world faces

History of the Satellite

It is phenomenal how much of an impact one person can make on this world.

If Maria Mykolayivna Moskalenko, a Ukrainian mother, had not given birth to SergeiPavlovich Korolev in Zhytomyr (A city in modern-day Ukraine) on January 12 th,

1907, it is highly probable that our world would be technologically handicapped.

Korolev developed the first artificial satellite, the Sputnik 1, fifty years after his birth.

In 1957, the Soviet Union launched Korolevs life-long concoction, initiating a

nascent era known as the space race. Within this era, both the United States andSoviet Union were scientifically transcending each other by further improving

Korolevs development. By combining the research and development that the Soviet

Sputnik Program and the United States Explorer Programboth did on Korolevs firstsatellite during the International Geophysical Year (1958), a stalwart foundation for

modern satellite technologies was formed.

Todays society is highly dependent on satellites. Communications,

transportation, and information are all everyday necessities for the average personwhich rely on satellites. An artificial satellite is any man-made object launched into

space that orbits the Earth in either a circular or elliptical motion. The overall

purpose of artificial satellites is to simply relay information between two or morepoints. There are three different types of artificial satellites: fixed, mobile, andscientific, all with different orbital paths: geosynchronous, asynchronous, and polar.

Various satellites with several orbital paths are used for a number of different

applications such as telecommunications, radio broadcasting, geographical imaging,scientific research, and global positioning.

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Aerial Navigation by Jerry Lawlor

Early Types of Navigation

Prior to 1940, if a person desired to travel between two places, it was mandatory to know the

exact directions in order to arrive at the correct destination. Satellites were viewed as fiction ratherthan a future reality in this era. Primarily a method called dead reckoning was used to navigate

between geographical points. Dead reckoning is a system of navigation in which an objects current

location is estimated by advancing the original location of the object according to directionalmovements and speeds. For example, the public library is located exactly 10 miles east from the

school and a student is trying to get from school to the library. They would have to drive in theeastward direction at a constant speed of 30mph (miles per hour) for 12 minutes in order to

successfully arrive there. This method of navigation has been used by many cultures for thousands of

years. It is a method based off of common logic and basic mathematics. Although an ideal method of

navigation for land, it is not practical for oceanic and aerial navigation. In the 1600s whenexploration was the revelation of a new era, dead reckoning, along with celestial navigation, steered

explorers thousands of miles away from their desired destination. Celestial navigation is based off of

the trigonometric and geometric properties of a sphere. Sometimes errors lead them to completelydifferent continents. This evidence proves dead reckoning is not accurate when it comes to aerial and

oceanic navigation. With indebted gratitude towards Josh Stone who invented the 1st

radio directionfinder, pilots and sailors were finally able to navigate through long distance trips without mistakes.

Ideal for semi-long-distance routes, radio direction finders could not withhold a worldwide range.Later pronounced an evolution for radar detection, radio direction finders soon became obsolete

when the first satellite navigation system, Transit, came out in the 1960s.

Transit, also known as NAVSAT (Navy Navigation Satellite System), was the first type of

satellite navigation system. With its first successful launch in 1960, Transit 1B sparked a majoradvancement in the space race era. After years and years of constant development and refinement,

the Transit system soon transformed into a global positioning system, or as most people know it, a

GPS.

Global Positioning Systems are currently embedded into modern life, as people have become

undeniably dependent on these systems. A GPS works with many satellites calculating andcomputing the exact position of an object in relation to space and time. The information is then

transmitted back to a receiver where it is displayed on a screen by either graphics or directional

output commands. These systems are solely dependent on the mass amount of artificial satellites

simultaneously orbiting the Earth every second of every minute of every day.


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The Problem this World Faces

Developers are constantly thinking and innovating new ways to continuously improve GPS

technologies, such as the current development of the Galileo, an upcoming GPS system for Europe. But itseems that there is no desire to try and further develop navigational methods that are not solely dependent

on satellite communications. Ever since Korolev invented the first artificial satellite for the Soviet Union,

there has been an international obsession with satellite technologies. In the twenty-first century, a lifewithout television, cell phones, the internet, and GPS systems is unimaginable. This world has become so

accustomed to satellite technologies that the majority of people would deny the possibility of losingsatellite communications; or people would label this possibility highly improbable. Well it is actually

more probable than most people think.

The sun is an extremely large spherical plasma consisting of mostly the hydrogen isotopes deuterium

and tritium, and helium. It is interwoven with electromagnetic field caused by the pulsing of constant

nuclear fusion reactions. Quite often, accelerated particles interact with plasma causing twoelectromagnetic fields to subjoin to create a solar flare. A solar flare is an extreme burst of energy from

the sun. To give a generalization of the magnitude of these solar flares, one flare has the capacity to

power this whole world for years since it emits approximately 6 x 1025

Joules of energy. Some of thesesolar flares release clouds, formally known as coronal mass ejections, composed of electrons andpositively charged ions (byproducts of a fusion reaction). In specific conditions, these clouds have the

ability to cause worldwide blackouts, increase polar radiation levels, and even inhibit satellite signals. On

average these clouds collide with Earths ionosphere once every three days and can disrupt long-rangeradio communications. If a solar flare has a peak flux range of 10-4 Watts per square meter or greater, it is

highly probable that all communications could be inhibited. A solar flare in such magnitude can be

compared to an EF5 tornado, but consider these solar flares critical because flares of this magnitudeactually occur quite frequently.

It is true that solar flares alone engender more than enough risk towards the loss of satellite

communications, but there are several other constituents which conspicuously add to the overall risklevel. Foreign countries strengthen international tension with the endeavor to target and even launch

intercontinental ballistic missiles (ICBMs) at the United States artificial satellites in high hopes tocripple the nation. Aside from the physical threats our nation faces, cyber threats also convey an immense

amount of risk to our nations satellite communications. It has been stated and confirmed that foreignhackers have successfully penetrated and bypassed our nations cyber security, gaining undiminished

access to our satellites. In the worst case scenario, the United States would lose all satellitecommunication capabilities. A catastrophic event like this would seriously blind our nation into a

sustained isolation and cause an unimaginable amount of economic distress and tension.


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If Worst Comes to Worst

Regardless of how advanced modern-day GPS systems are, if all satellite communication

capabilities were temporarily or even permanently disabled, then such GPS systems would meannothing. Radio direction finders would also be useless because they are dependent on long-range

radio waves, which in most scenarios, would be inhibited and disabled like satellite communications.

With the absence of satellite navigation systems and radio direction finders, this world is left withtwo forms of long-range navigation: celestial navigation and dead reckoning.

Dead reckoning may be doable for automotive and land navigation, but it is seemingly useless for

oceanic and aerial navigation. Celestial navigation techniques are mainly used for oceanic navigation

and may give a generalization of where to go when sailing; but again, its techniques supply minimal

aid for aerial navigation. Not many, if any, non-satellite navigation techniques or methods provideaccurate directions and guidance for aerial devices.

What about Aerial Navigation?

Regardless of how advanced modern-day GPS systems are, if all satellite communicationcapabilities were temporarily or even permanently disabled, then such GPS systems would meannothing. Radio direction finders would also be useless because they are dependent on long-range

radio waves, which in most scenarios, would be inhibited and disabled like satellite communications.

With the absence of satellite navigation systems and radio direction finders, this world is left withtwo forms of long-range navigation: celestial navigation and dead reckoning.

The problem with dead reckoning methods navigating aerial devices is that dead reckoningmethods are based off of two points, the origin and the destination. With only two usable points and

so many mathematical factors involved, dead reckoning is as effective as a compass for aerial

navigation, providing general directions but nothing more. Celestial navigation is impossible as an

aerial navigation technique. Although celestial navigation is based off of the trigonometric propertiesof a sphere, the mathematical variables that affect flight pathways are not included in this method of

navigation making celestial navigation an inadequate method of navigation for aerial devices.


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This section covers the topics of:

The purpose The hypothesis The system of equations

Whats the Purpose?

The purpose of this experiment was to try and develop a mathematical system that could

successfully navigate an aerial device without sustained satellite communications. Inspiration for this

experiment arose from the recognition of how satellite dependent this world has become and how thefailure of satellite communications could seriously handicap this world. The mathematical system of

equations formulated in this experiment could further improve this world by providing an accurate aerial

navigation system that is not satellite dependent. Such navigation systems would be vitally important in

emergency situations such as the loss of satellite communications due to solar flares.

Hypothesis

By using different mathematical laws

and functions, it is predicted that it will be

possible to formulate a mathematicalsystem of equations that can accurately and

effectively navigate an aerial device from

point A to point B without sustained

satellite communications. This is based on

different trigonometric functions andproperties as well as various topological

equations and methods.

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Jerry Lawlor

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The System of Mathematical Equations

Determine the Mean Radius of Earth

The value for the mean (average) radius of Earth is calculated by adding the value of the polarradius to double the value of the equatorial radius and dividing the sum by three. According to the

International Union of Geodesy and Geophysics (IUGG), the Earths equatorial radius is equivalent

to 6,376.1370 kilometers (km) and the Earths polar radius is equivalent to 6,356.7523 km. After allof the calculations, the value for the mean radius of the Earth is equivalent to 6,371.0088 km. It is

important to calculate the mean radius of the Earth due to the fact that the Earth is not a perfect

sphere; instead, it is considered to be an ellipsoid.

IUGGs equation for calculating the mean radius ofEarth Substituting the IUGGs values for the equatorial and polar radii

The mean radius of the Earth

Plot the Origin and Destination on the Geographical Polar GridDetermine the coordinates of the origin F(1 , 1) and the destination D(2 , 2) and plot them on

the geographical polar grid.

Step One: Calculate the Central Angle

between the Origin and DestinationDetermine the angular distance (also known

as the central angle) between the origin andthe destination. The angular distance is the

angle between two polar coordinates on asphere or ellipsoid. Such angle is calculated

by the trigonometric properties of half angles

incorporated in the angular distance formula.


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Step Two: Convert F(1 , 1) and D(2 , 2) to Cartesian Coordinates

Subtract the value of the central angle () from 180 and divide the difference by two in order tocalculate . Finding the difference between the central angle and 180 is important for graphing the twopoints on a Cartesian plane because 180 - and are supplementary angles, meaning that the sum ofthe two values equals 180 (It is important to note that the value ofcannot exceed 180 due to theEarths spherical and epileptic properties).

Next, use the trigonometric functions sine and cosine and the properties of a right triangle to

determine the (x , y) values of the Cartesian coordinates corresponding to (1 , 1) and (2 , 2). These

coordinates will correspond given the fact that 2 + = 180 (meaning that they are supplementary).

Step Three: Determine the Distance between F and D

Convert to radians and then multiply by the value of the mean radius to calculate the distance(km) between point F and D. The distance is represented by the constant z. This is derived from theformula s = r.


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Step Four: Graph point BPoint B is located d1 kilometers closer to the destination from the origin. To graph this point,

subtract double the value of d1 from z and then divide by the value of the mean radius of Earth. Next,

convert the quotient into degrees to get ( + ). After that, use the sine and cosine functions tograph point B.

+ Step Five: Graph point E

Point E is located d2 kilometers closer to the origin from the destination. To graph this point,

subtract double the value of d2 from z and then divide by the value of the mean radius of Earth. Next,

convert the quotient into degrees to get (

+

). After that, use the sine and cosine functions to

graph point E. +

Step Six: Graph point GUse the trigonometric sine and cosine functions and the properties of the right triangle according

to Euclidian Geometry to graph point G. Point G is described as the geographic marking point atwhich the aerial device needs to reach maximum altitude.

Step Seven: Graph point I

Use the trigonometric sine and cosine functions and the properties of the right triangle according

to Euclidian Geometry to graph point I. Point I is described as the geographic marking point at

which the aerial device needs to begin its descent from maximum altitude.


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Step Eight: Formulating an Equation for Since is a line with a constant inclination/declination (slope), it is a linear function with the base

equation y = mx + b. The first step of formulating an equation for the line is determining the slop ofthe line. Linear slopes can be found by dividing the y by the x. Next, the (x , y) for either G or F need

to be entered into the equation y = mx + b to solve for b. Step Nine: Formulating an Equation for

Since is a line with a constant inclination/declination, it is a linear function with the base equationy = mx + b. The first step of formulating an equation for the line is determining the slop of the line.Linear slopes can be found by dividing the y by the x. Next, the (x , y) for either I or D need to be

entered into the equation y = mx + b to solve for b.

Step Ten: Determine the Distance between F and G

Use the analytical geometric distance formula (xy-plane) to calculate the distance (km) between

points F and G. This formula is specifically designed to calculate the distance between two Cartesian

points.

Step Eleven: Determine the Distance between I and D

Use the analytical geometric distance formula (xy-plane) to calculate the distance (km) betweenpoints I and D.


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Step Twelve: Calculate the Tangential Speed at 2The tangential speed is the speed (km per hour) at which the Earth rotates at a given latitude. In

order to calculate this, it is necessary to calculate the equatorial speed of the Earth first. According to

the World Geodetic System (GWS), the Earth rotates at a speed of 7.2921150 E-5

radians per solar

second. Overall, to find the Earths equatorial speed, 7.2921150E-5 radians per solar second has to be

converted into kilometers per hour. The equatorial speed of the Earth is

.

To find the tangential speed at a given geographic latitude, multiply the equatorial speed by the

cosine of the geographic latitude.

Step Thirteen: Determine the New Longitude of Destination

Due to the Earths constant rotation, a new longitude for the destination has to be calculated. This

new polar coordinate will compensate for the Earths rotational factor and direct the aerial device tothe point at which the original destination would be after the total time of flight. The equation for

this formula is based off of the fact that distance divided by velocity equals time. Since the time the

Earth rotates during flight is equal to the time the aerial device is airborne, the distance of the flightpath divided by the average velocity of the aerial device is equal to the distance the Earth rotation

divided by the tangential speed at the latitude 2. Before the longitude of the new destination can be determined, four separate calculations arerequired first. The four calculations are: the distance of the flight path, the mean velocity of theaerial device, the distance that the Earth rotated, and the tangential speed of the Earth at latitude 2.

Since the tangential speed of the Earth has already been calculated and the mean velocity of the

aerial device has already been stated, only two main calculations need to be made.


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A variation of the angular formula used in step one is used to calculate the distance of the flight path.Instead of using the known polar coordinates of the original destination, the polar coordinates of the new

(compensated) destination will be used. The latitude of the new destination is equal to the latitude of the

origin based off of the fact that the earth rotates on an axis running through the latitudinal coordinates 90and -90. This means that the axis in which the Earth rotates on is perpendicular to the latitudinal lines of

the Earth. On the other hand, the longitude of the destination will be substituted by the longitude of the

new destination because they hold different values.

Next, the central angle between the two points has to be converted into radians. Then the sum of the

radians corresponding to the distance parameters d1 and d2 are converted in to radians and subtracted

from the radians between the two points. This step calculates the radians that the aerial device will be at

a sustained altitude.

After that, the radians that the aerial device will be at a sustained altitude is multiplied by the mean

radius of the Earth with the altitude of sustained flight incorporated into it (altitude = kilometers above

sea level). This step is based off of the arc length formula. When the arc length is finally calculated, the

distances dfg and did are added to the arc length to get the distance of the flight path (dfg is equal to thedistance of the flight path during inclination and d id is equal to the distance of the flight path during

declination.

The distance that the Earth rotated is also calculated by using the angular formula. It is used to find

the angle between the original destination and the new destination. The angle calculated is then

converted into radians and multiplied by the value of the mean radius of Earth. This is, again, based offof the arc length formula.


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Since all of the necessary components have been calculated, solve for each individual ratio of

distance over velocity.

Cross multiply each ratio to get an expression of equality and eliminate fractions.

Simplify the expression with kconstants.

After all of the kconstants are substituted and the equation is simplified down to an easilycomprehensible level by using properties of sine and arcsine functions , the final expression comes

out to be:

Due to the fact that the variable appears on both sides of the equation, the only way to solve for would be to use estimated guesses or write a computational program. Once a value for isentered that makes both sides of the equation equal, store that value as the longitude of the new

destination. It is possible to completely isolate the variable on a single side of the equation, butthat just increases the amount of unnecessary complexity and chances of computational error.

*Graphing from this point on will use the (Xn , Yn) system.


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Step Fourteen: Convert Fn(1 , 1) and Dn(2 , 3) to Cartesian Coordinates

Subtract the value of the central angle () from 180 and divide the difference by two in order tocalculate . Finding the difference between the central angle and 180 is important for graphing thetwo points on a Cartesian plane because 180 - and are supplementary angles, meaning that thesum of the two values equals 180.

Next, use the trigonometric functions sine and cosine and the properties of a right triangle to

determine the (x , y) values of the Cartesian coordinates corresponding to (1 , 1) and (2 , 3). These

coordinates will correspond given the fact that 2 + = 180.

Step Fifteen: Determine the Distance between Fn and Dn

Convert to radians and then multiply by the value of the mean radius to calculate the distancebetween point Fn and Dn. The distance is represented by the constant zn. This is derived from the

formula s = r.

Step Sixteen: Graph point Bn

Point Bn is located d1n kilometers closer to the new destination from the origin. To graph this point,

subtract double the value of d1n from zn and then divide by the value of the mean radius of Earth. Next,convert the quotient into degrees to get ( + ). After that, use the sine and cosine functions tograph point Bn.

+ Step Seventeen: Graph point EnPoint En is located d2 kilometers closer to the origin from the new destination. To graph this point,

subtract double the value of d2n from zn and then divide by the value of the mean radius of Earth. Next,

convert the quotient into degrees to get ( + ). After that, use the sine and cosine functions tograph point En.


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+ Step Eighteen: Graph point G

n


to Euclidian Geometry to graph point Gn. Point Gn is described as the geographic marking point atwhich the aerial device needs to reach maximum altitude.

Step Nineteen: Graph point In


to Euclidian Geometry to graph point In. Point I is described as the geographic marking point atwhich the aerial device needs to begin its descent from maximum altitude.

Step Twenty: Formulating an Equation for

Since is a line with a constant inclination/declination (slope), it is a linear function withthe base equation y = mx + b. The first step of formulating an equation for the line isdetermining the slop of the line. Linear slopes can be found by dividing the y by the x. Next, the

(x , y) for either Gn or Fn need to be entered into the equation y = mx + b to solve for b.

Step Twenty-One: Formulating an Equation for

Since

is a line with a constant inclination/declination, it is a linear function with the base

equation y = mx + b. The first step of formulating an equation for the line is determining theslop of the line. Linear slopes can be found by dividing the y by the x. Next, the (x , y) for either Ior D need to be entered into the equation y = mx + b to solve for b.


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Step Twenty-Two: Determine the Distance between Fn and Gn

Use the analytical geometric distance formula to calculate the distance between points Fn and Gn.

Step Twenty-Three: Determine the Distance between In and Dn

Use the analytical geometric distance formula to calculate the distance between points In and Dn.

Step Twenty-Four: Graph P1nLine P1n is the derivative of the circle representing Earth at Fn. In other words, it is the perpendicular

line to at point Fn. To graph this, use the negative reciprocal of the slope of as the slope forP1n and solve for the y-intercept using the y-intercept equation.

Step Twenty-Five: Transpose

to Contain Point Gn

Formulate and equation for a line containing point Gn and is parallel to . This is important inorder to find the angle of take-off because it forms a right triangle containing F n and Gn. Step Twenty-Six: Find the Intersection of P1n and

To find the intersection, set the equation of P1n equal to the equation and solve forx. The values of the two x points are equivalent given the fact that they are located at the intersection ofthe two lines. After the x value has been determined, calculate the y value.


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Step Twenty-Seven: Determine the Distances between and Fn , GnUse the analytical geometric distance formula to calculate the distance the between points.

Step Twenty-Eight: Find the Angle of Take-Off

By utilizing the trigonometric properties of tangent and arctangent, determine the angle of take-off.

Step Twenty-Nine: Graph P2n

Line P2n is the derivative of the circle representing Earth at Dn. In other words, it is the

perpendicular line to at point Dn. To graph this, use the negative reciprocal of the slope of as the slope for P2n and solve for the y-intercept using the y-intercept equation.

Step Thirty: Transpose to Contain Point InFormulate and equation for a line containing point In and is parallel to . This is important inorder to find the angle of take-off because it forms a right triangle containing D n and In.

Step Thirty-One: Find the Intersection of P2n and To find the intersection, set the equation of P2n equal to the equation

and solve

for x. The values of the two x points are equivalent given the fact that they are located at theintersection of the two lines. After the x value has been determined, calculate the y value.


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Step Thirty-Two: Determine the Distances between and Dn , InUse the analytical geometric distance formula to calculate the distance the between points.

Step Thirty-Three: Find the Angle of LandingIn order to find the angle of landing, calculate the absolute value of an arctangent function.

Step Thirty-Four: Graph P4n

Line P4n is the derivative of the circle representing Earth at Gn. In other words, it is the perpendicular

line to at point Gn. To graph this, use the negative reciprocal of the slope of as the slope forP4n and solve for the y-intercept using the y-intercept equation. Step Thirty-Five: Plot (Xgpn , Ygpn)

(Xgpn,Ygpn) is a point located on . It is designed to give tcn the properties of an isoscelestriangle. With such properties, the value of angle tcn can be determined. The steps in doing this require aset value for the temporary (non permanent) constant i. This method was derived from the distance

formula.


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Solve the equation until both sides are equal. After that, plug in the values of x in to the equation:

( )

Step Thirty-Six: Plot (Xpgn , Ypgn)

(Xpgn,Ypgn) is a point located on . It is designed to give tcn the properties of an isoscelestriangle. With such properties, the value of angle Nn can be determined.


( )

Step Thirty-Seven: Determine the Distance between ( ),( )Determine the distance between ( ) and ( ) by using the distance formula.

Step Thirty-Eight: Find the Midpoint between ( ),( )Find the midpoint between points ( ) and ( ) by dividing the sum of the two xand y values by two and storing the quotients as the coordinates for the midpoint.

( )Step Thirty-Nine: Determine the Distance between ( ) and GnDetermine the distance between midpoint ( ) and Gn by using the distance formula.


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Step Forty: Find the Angle of Adjustment at Maximum AltitudeFind the measure of the angle of adjustment at maximum altitude. This is the angle at which the aerial

device has to adjust by when it reaches maximum altitude in order to stay within the flight parameters.

The angle of adjustment at maximum altitude can be found be doubling the negative arctangent of the

distance between ( ) and ( ) divided by twice the value of the distance between(

) and Gn.

Step Forty-One: Graph P3n

Line P3n is the derivative of the circle representing Earth at In. In other words, it is the perpendicular

line to at point In. To graph this, use the negative reciprocal of the slope of as the slope for P3nand solve for the y-intercept using the y-intercept equation.

Step Forty-Two: Plot (Xpin , Ypin)

(Xpin,Ypin) is a point located on . It is designed to give tn the properties of an isosceles triangle.With such properties, the value of angle tn can be determined. The steps in doing this require a set valuefor the temporary (non permanent) constant i. This method was derived from the distance formula.


( )

Step Forty-Three: Plot (Xipn , Yipn)

(Xipn,Yipn) is a point located on

. It is designed to give tn the properties of an isosceles triangle.

With such properties, the value of angle tn can be determined.


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( )

Step Forty-Four: Determine the Distance between (

),(

)

Determine the distance between ( ) and ( ) by using the distance formula.

Step Forty-Five: Find the Midpoint between ( ),( )Find the midpoint between points ( ) and ( ) by dividing the sum of the two x

and y values by two and storing the quotients as the coordinates for the midpoint.

( )

Step Forty-Six: Determine the Distance between ( ) and InDetermine the distance between midpoint ( ) and In by using the distance formula.

Step Forty-Seven: Find the Angle of Declination from Maximum AltitudeFind the measure of the angle of declination from maximum altitude. This is the angle at which

the aerial device has to adjust by when it starts its declination from maximum altitude in order tostay within the flight parameters. The angle declination from maximum altitude can be found be

doubling the negative arctangent of the distance between ( ) and ( ) divided bytwice the value of the distance between ( ) and In.


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What about Aerial Navigation?

Imaginary Vertical Asymptotes

Graphing the Flight Path

Navigational Statement

If then route in the eastward direction (represented by )If then route in the westward direction (represented by )If = then a system error is present because the slope is undefinedIf then route in the northern direction (represented by ) *prefix inIf then route in the southern direction (represented by ) *prefix deIf

=

then route in the no direction (represented by

) *disregard the

input because no value for

is existentCommanding Statement

Proceed in the direction at a in respect to the geographicalcoordinate system. Take off at a(n) angle with a speed of kilometers per hour for kilometers. At an altitude of kilometers above sea level, adjust the aerial device at a(n) angle toconstantly stay parallel with the earths curvature and stay within the parameters of the flight path. After

an accumulative distance of kilometers, adjust the angle of flight to from the currentposition and continue flight for

kilometers. Once an accumulative distance ofz kilometers has been

reached, adjust the directional motion of the aerial device by for landing.


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Pin Point MethodThe Pin Point Method is a method I designed to further improve the accuracy of aerial navigation.

It is when several geodesic coordinate points are selected and stored as x n variables (n is the number

of x points that exist). Then the central angle between each point xn and the aerial device z (*locatedat origin F at first, but then moves further towards destination D as flight progresses)(z is a function

of distance and time) is calculated using the formula:

*assume z is on the geographical surface of Earth

Next, calculate the arc lengths for each of the angle measures and store each value in set FX[]. Do

not forget that since the location of z is constantly moving, the values in set FX[] will vary as afunction of time and distance. For the next step, the central angle between each point x n and the

destination D point must be calculated. From these values, each arc length needs to be calculated and

stored in set DX[]. Now please keep in mind that numbers in set DX[] have constant values

while number in set FX[] vary with time. Once the , the destination has beenreached.


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To Sum things up

Overall, the system of equations developed in this project completely satisfies and even surpassesthe standardized goal set in the beginning. After studying numerous mathematical books and proofs, an

independent system of equations was formulated that can successfully navigate an aerial device between

two points. This mathematical system has the capabilities of accurately navigating an aerial device

between two points without the aid of satellite communications. During the developmental phase manymathematical obstacles were encountered dealing with fundamental mathematics, unknown variables,

and elegance. This system of equations has an extraordinary potential for non-satellite aerial navigation

systems. If a time ever comes where a solar flare inhibits radio capabilities, or our nations satelliteshave been overtaken by foreign terrorist groups, then this system of equations can accurately navigate

aerial devices between two geographical points without the use of satellite or radio technologies.

Although this project is well developed, room for improvement still exists. In the future, I plan onincorporating the mathematical factors of flight deviation due to directional wind forces and imbedding

the gravitational factors with respect to altitude into the system of equations developed in this project.

These factors are extremely important because they can help determine the rates and values ofacceleration during flight to give a more accurate value for the velocity of the aerial device. This would

increase the accuracy and precision of the navigation system by replacing the value of a mean velocity

of flight with a far more accurate value of a predicted velocity of flight utilizing factors of acceleration.Such calculations could be made by optimizing Newtons Second Law of Gravitational Forces into asummation that could calculate the average gravitational force during flight in respect to distance and

time. From this, vector calculus can be used to project the compensating angles of inclination anddeclination in respect to distance and time. Using the properties of the definition of a resultant force, f =ma, the acceleration of flight can then be calculated. Another factor that could be incorporated into this

project in the future is air density in respect to altitude and temperature in order to calculate frictional

factors of force and repulsions against gravity and the aerial devices directional forces. All in all this

system of equations can prevent the crippling of aerial navigation in times of satellite failure byaccurately navigating aerial devices without sustained satellite communications.

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Jerry Lawlor

Page 24

BibliographyAlfred Rosenthal (1968). Venture Into Space: Early Years of Goddard Space Flight Center.

NASA. pp. 15.

Alicia Chang. "50th anniversary of first U.S. satellite launch celebrated". Associated Press.Retrieved 2011-12-08.

Bob Brewin (2003). "Homemade GPS jammers raise concerns".Computerworld. Retrieved 2012-

02-07.Daphne Burleson (2005). Space Programs Outside the United States. McFarland & Company.

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David S. F. Portree; Joseph P. Loftus, Jr (1999). "Orbital Debris: A Chronology".Lyndon B.

Johnson Space Center. pp. 18. Retrieved 2011-10-21.

"First time in History".The Satellite Encyclopedia. Retrieved 2011-11-06.

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2011-10-25.NSSDC Master Catalog. Note: In searching all satellites ("Spacecraft Name" = blank), a few

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"Orbital Debris Education Package". Lyndon B. Johnson Space Center. Retrieved 2012-01-02."Workshop on the Use of Microsatellite Technologies". United Nations. 2008. pp. 6. Retrieved

2011-10-16.

Tariq Malik. "SpaceX Successfully Launches Falcon 1 Rocket Into Orbit". Space.com. Retrieved2011-10-23.

Picture and Graph Citationshttp://sv.wikipedia.org/wiki/Fil:GPS-24_satellite.png

http://www.fordesigner.com/imguploads/Image/cjbc/zcool/png20080526/1211793593.pnghttp://www.wi-skyinflight.com/images/airplane_intro.png

Jerry Lawlor
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