Mathematics HL

8/20/2019 Mathematics HL

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Mathematics HL IA

When Will the Skies

Light Up?Determining when a passenger on an airplane sees day-night transformation

through 3D Cartesian Coordinate, Polar Coordinate and Vector Analysis

Skyler Shi

5/30/2015

Instructor: Mr.Ma

School: Shanghai High School International Division


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Contents

1 Instigation ................................................................................................................................ 1

2 Assumptions-Setting up the model .......................................................................................... 2

3 Overview of Model .................................................................................................................. 5

4 Determining Day- Night Terminator’s Equation ...................................................................... 5

5 Translating International Geographic Coordinates into Polar Coordinates ............................. 7

6 Solving for Time .................................................................................................................... 10

6.1 Determining Cartesian Coordinates of A and B .......................................................................... 10

6.2 Finding the equation for the great circle of AB : ........................................................................ 11

6.3 Solve for Ct .................................................................................................................................. 12

6.4 Finding the Special Ct .................................................................................................................. 13

7 Flying Westwards .................................................................................................................. 15

8 Applying the Model: Flying from Shanghai to Vancouver ................................................... 15

9 Conclusion-Advantages and Limitations of the Model ......................................................... 21

10 Appendix ............................................................................................................................ 22

1 Instigation

I’ve always loved plane rides and there are good reasons to this. On plane rides, I get theopportunity to observe the mechanisms of the planes, and more importantly, I get a beautiful

view of the skies above the clouds.

On multiple plane rides, I’ve taken notice of the skies transitioning from nighttime to daytime or

from daytime to nighttime. In a constant state of flux, these skies are particularly beautiful.

However, the deal with these type of skies is that they only last a short amount of time and

happen at different times on different flights. More than once I have missed out on the chance to

observe such a sky because I slept through it.

Thus, in this paper, I will develop a model that derives the time elapsed after take-off a passenger will see a day-night sky on any flight . I want to use this model to be sure that I don’t

miss out on a day-night sky on my next flight. While inFlight trackers on airplanes do give

passengers a 2D view of the night skies and day skies, they do not tell passengers when a day-

night sky appears. I will utilize knowledge of Cartesian coordinates, polar coordinates, and

vectors to develop such a 3D model.


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2 Assumptions-Setting up the model

There are several aspects of the Earth’s rotation that needs to be pointed out for the establishment

of this model. The Earth rotates around the sun, and the Earth rotates around its own North-South

Pole axis. However, because the Earth is tilted so that its North-South Pole Axis is not

perpendicular to its plane of rotation around the Sun, the Day-Night Terminator 1 (an imaginary

plane) determines the earth’s daylight hemisphere to its nighttime hemisphere, not the plane

containing the North-South Pole axis. (See Diagram 2.1)

Diagram 2.1- 2D View of Model

Diagram 2.1 illustrates the concept of the Day-Night Terminator as determined by the Sun. From

this point on, the sun will be excluded from diagrams. The effect of the Sun on the earth can be

represented simply with the Day-Night Terminator.

In this paper, the Day-Night Terminator axis/line refers

to the line perpendicular to Earth’s plane of rotation

around the Sun. The Day-Night Terminator plane refers

to the plane perpendicular to Earth’s rotation around the

Sun.

Assumption 1:

The Earth of radius 6378 R km 2 is a counter-clockwise

rotating (viewing from the top) sphere rotating around z-

axis (North-South Pole Axis) (Refer to Diagram 2.2)

Justification:

1 http://nova.stanford.edu/~vlf/IHY_Test/Tutorials/DayNightTerminator/DayNightTerminator.pdf

2 http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html

Diagram 2.2 – 3D Cartesian Setup of Model

http://nova.stanford.edu/~vlf/IHY_Test/Tutorials/DayNightTerminator/DayNightTerminator.pdfhttp://nova.stanford.edu/~vlf/IHY_Test/Tutorials/DayNightTerminator/DayNightTerminator.pdfhttp://nova.stanford.edu/~vlf/IHY_Test/Tutorials/DayNightTerminator/DayNightTerminator.pdfhttp://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.htmlhttp://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.htmlhttp://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.htmlhttp://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.htmlhttp://nova.stanford.edu/~vlf/IHY_Test/Tutorials/DayNightTerminator/DayNightTerminator.pdf


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The Earth rotates counter-clockwise (viewing from the top) and has an average radius of 6378

km in reality based on NASA’s measurements. Although the Earth is not a perfect sphere in

reality, this abstraction paper will simplify it into a sphere for easier calculations.

Assumption 2:

The Day-Night Terminator axis/plane is 23.5°3 to the

plane containing the z-axis and the y-axis on solstices

(June.21 and Dec. 22). Throughout one year, the Day-

Night Terminator plane will oscillate between the two

angles on solstices as the Earth rotates around the sun.

(Refer to Diagram 2.3)

Justification:

Research has shown that the maximum tilt (tilt on

solstices) of the Day-Night Terminator Axis of the

Earth has varied from 22.1° to 24.5°, but this has

happened over a span of time as long as 40,000 years4,

enabling me to ignore the variations in the tilt for this

model.

The reason why the Day-Night Terminator plane oscillates throughout the year requires 3D

imagination, and can be better understood through a GIF on Wikipedia:

https://en.wikipedia.org/wiki/Terminator_(solar)#/media/File:XEphem-sunset-animation.gif

Assumption 3:

The sun is always shining at the Earth from the

Earth’s right side in this model (viewing from the

front). (Refer to Diagram 2.4)

Assumption 4:

The Earth is placed into a 3D Cartesian coordinate

system. The positives and negatives of the axis from

the front view are shown in Diagram 2.2.1. Any point

on Earth can be modeled by the coordinates (x, y, z).(Refer to Diagram 2.3 or Diagram 2.4)

3 http://www.universetoday.com/47176/earths-axis/

4 http://earthobservatory.nasa.gov/Features/Milankovitch/milankovitch_2.php

Diagram 2.3- Illustration of Day-Night

Terminator

Diagram 2.4- Orientation of Model

https://en.wikipedia.org/wiki/Terminator_(solar)#/media/File:XEphem-sunset-animation.gifhttps://en.wikipedia.org/wiki/Terminator_(solar)#/media/File:XEphem-sunset-animation.gifhttp://www.universetoday.com/47176/earths-axis/http://www.universetoday.com/47176/earths-axis/http://www.universetoday.com/47176/earths-axis/http://earthobservatory.nasa.gov/Features/Milankovitch/milankovitch_2.phphttp://earthobservatory.nasa.gov/Features/Milankovitch/milankovitch_2.phphttp://earthobservatory.nasa.gov/Features/Milankovitch/milankovitch_2.phphttp://earthobservatory.nasa.gov/Features/Milankovitch/milankovitch_2.phphttp://www.universetoday.com/47176/earths-axis/https://en.wikipedia.org/wiki/Terminator_(solar)#/media/File:XEphem-sunset-animation.gif


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Assumption 5:

The Earth will also be placed into a 3D Polar coordinate

system. Any point on Earth can be modeled by the polar

coordinates (R, θ, φ) (Refer to Diagram 2.5). It is

important to note that φ is different than longitudinalangles of the international Geographic Coordinate

System. φ is dependent on the time of a location. For

instance, at 8:00, point X may be (R, θX, φX) , but at

8:01, point X will be at (R, θX, φX+0.25). This will be

explained in further detail later. The positives and

negatives of the angles are defined and shown in

Diagram 2.6. (Refer to Diagrams 2.6.1 & 2.6.2)

Justification:

These positives and negatives are adopted so that translation from polar coordinates into

Cartesian coordinates will ensure the right signs for the Cartesian coordinates.

Assumption 6:

Diagram 2.5- 3D Polar Setup of Model

Diagram 2.6.1- Positive & Negatives of φ Diagram 2.6.2- Positive & Negatives of θ


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The Day-Night Terminator oscillates at a uniform rate, so its oscillation rate can be found:

94(23.5 23.5 ) (365 2) /

365day

I define as the angle between the Day-NightTerminator plane and the Z-axis.

Thus, can be found for different dates on which

airplanes take-off by Equation 4.1:

If Jun.21 < Month.date < Dec.21

9423.5 (Month . date Jun .21) /

365day If

Dec.21 < Month.date < Jun.21

9423.5 (Month .date .21) /

365 Dec day

I realize that space agencies track Day-Night Terminator position down to the minute, but my

calculation of the Day-Night Terminator down to a day is sufficient for my model. Determining

the Day-Night Terminator down to the minute will make my model more accurate, but at the

same time, my model will be greatly complicated. The Day-Night Terminator plane will have to

be modelled by a different equation each minute of the airplane’s flight. This is not desirable, soI only model the Day-Night Terminator down to a day.

The Day-night Terminator plane has vectors v and u

where

sin

0

cos

v

0

1

0

u

Their vector product will give the vector perpendicular

to the Day-night Terminator plane:

Diagram 4.2- Defining

Diagram 4.3- Day-Night Terminator Plane

Vectors


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sin 0 0 cos cos

0 1 0 0

cos 0 sin 0 sin

Term

Term

Term

A

v u B

C

Thus, the equation for the Day-night Terminator plane is:

1 1 1Term Term Term Term Term Term A x B y C z A x B y C z

Where (x1,y1,z1) is an arbitrary point on the plane

cos sin cos 0 0 1 sin 0 x z

Equation 4.2: cos sin 0 x z

5

Translating International Geographic Coordinates into Polar

Coordinates

Given: Time of airplane take-off, A ( A LAT , A LONG ), B ( B LAT , B LONG )

Find: Polar coordinates of location A (R, θA, φA), B(R, θB, φB)

The airplane flies from location A that has latitude A LAT and longitude A LONG to location B

that has latitude B LAT and longitude B LONG .

The difference between the International Geographic Coordinates and Polar Coordinates in thismodel is that Polar Coordinates depend on time whereas International Geographic Coordinates

are independent of time. For example, location M in International Geographic Coordinates at

8:00 and 8:01 will both have coordinates (LATM, LONGM). But location M will have Polar

coordinates of (R, θM, φM) at 8:00 and (R, θM, φM+0.25°) at 8:01. This difference is caused by

Earth’s self -rotation around the Z-axis (North-South Pole Axis).

θ is equal to the latitude of a location. As the Earth rotates around the Z-axis, the latitude of a

location remains unchanged. Thus θ=LAT.

Referring to Diagram 5.1, the latitudinal plane that

contains A can be represented with the equation:

sin A

z R LAT .

Solve the three equations:

- Equation of latitudinal plane

- Equation of sphere (Earth’s surface)

Diagram 5.1 Latitudinal Plane


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- Equation of Day-Night Terminator plane

Equations 5.1:

2 2 2 2

sinLAT

cos sin 0

A z R

x y z R

x z

The roots (the intersection of the two planes and the

sphere) are the sunrise and sunset locations on the

latitudinal plane that contains A. (Refer to Diagram 5.2)

I will then translate these derived Cartesian coordinates of sunset and sunrise locations into polar

coordinates. I do this because it is easier to derive polar coordinates of locations A with polar

coordinates of the sunset and sunrise locations.

However, the sign of the angles need to be taken into account, so the translation of Cartesian

coordinates into polar angles needs to be divided into four cases.

Equations 5.2: To translate the sunrise and sunset

locations into polar coordinates:

/Sunrise set A LAT

1. 1 //

/

tan ( )Sunrise set Sunrise set Sunrise set

x

y if /Sunrise set x ,

/Sunrise set y are positive

2. 1 //

/

180 tan ( )Sunrise set Sunrise set Sunrise set

x

y if

/Sunrise set x is positive, /Sunrise set y is negative

3. 1 ///

tan ( )Sunrise set Sunrise set Sunrise set

x

y if /Sunrise set x is

negative, /Sunrise set y is positive

4. 1 //

/

180 tan ( )Sunrise set Sunrise set Sunrise set

x

y if /Sunrise set x is negative, /Sunrise set y is negative

Diagram 5.2 Sunrise Sunset Locations

Diagram 5.3 Four Cases


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Thus, sunrise/set location:/ /

( , , ) sunrise set sunrise set

R

Once the sunrise and sunset locations are found, then

location A can be found by backtracking. If the airplane

takes off at location A during daytime, then it can be

inferred that location A has rotated from the sunrise

location and will rotate to the sunset location as time

passes. If the airplane takes off at location A during

nighttime, then it can be inferred that location A has

rotated from the sunset location and will rotate to the

sunrise location as time passes.

Knowing this, the polar coordinates of A can be found

by deducting the angular distance A is to travel to either

the sunset or sunrise location from the polar coordinates

of either the sunset or sunrise location.

Equations 5.3:

Airplane takes-off in daytime:

A( , , ) ( , , ) A A LATA Sunset

R A R

Airplane takes-off in nighttime:

A( , , ) ( , , ) A A LATA Sunrise R A R

where

/ /

/ /

( ) 360 24 60 ( ) 0.25 / min sunrise sunset take off sunrise sunset take off Earth self rotationspeed

timeofsunrise set timeoftake off timeofsunri se set timeoftake off

T T T T

(If / 180Sunrise set , then /A( , , ) ( , , 360 ) A A LATA Sunrise set R A R (This is to account

for the discontinuity of at 180 and 180 .

Note: the sunrise/sunset times of different locations can be found online at:http://sunrisesunset.com/calendar.asp

And once the polar coordinates of A are determined, the polar coordinates of B can be easily

determined through the relationship of the latitudes and longitudes of A and B.

Equation 5.4:

Diagram 5.4- Finding Location A

http://sunrisesunset.com/calendar.asphttp://sunrisesunset.com/calendar.asphttp://sunrisesunset.com/calendar.asp


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(R, , 0.25 t)t A A

A

(R, , 0.25 t)t B B B

(R, , ) ( , , )t A A t At At At A A x y z

cos sin( 0.25 )

cos cos( 0.25 )

sin

At A A

At A A

At A

x R t

y R t

z R

(R, , ) ( , , )t B B t Bt Bt Bt B B x y z

cos sin( 0.25 )

cos cos( 0.25 )

sin

Bt B B

Bt B B

Bt B

x R t

y R t

z R

6.2

Finding the equation for the great circle of AB :

This section of the paper outlines the method of determining the equation for the great circle of

AB at a given time t after the plane takes-off.

At a given time, there are vectors a and b where

Equations 6.1:

cos sin( 0.25 )

cos cos( 0.25 )

sin

At A A

At A A

At A

x R t

a y R t

z R

cos sin( 0.25 )

cos cos( 0.25 )

sin

Bt B B

Bt B B

Bt B

x R t

b y R t

z R

Their vector product will give the vector perpendicular to the AB plane:

Equation 6.2:

AB At Bt At Bt At Bt

At Bt At Bt At Bt AB


A x x y z z y

a b B y y z x x z

z z x y y xC

Diagram 6.3- Vectors a and b


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Thus, the equation for the plane containing AB is:

1 1 1 AB AB AB AB AB AB A x B y C z A x B y C z

Equation 6.3: 0 0 0 0 AB AB AB AB AB AB

A x B y C z A B C

6.3

Solve for Ct

The previous sections have derived the equation of the Day-Night Terminator plane and the

equation of the Great Circle containing AB . Using the below three equations:

- Equation of Day-Night Terminator plane

- Equation of Great Circle of AB

- Equation of sphere (surface of Earth)

Equations 6.4:

2 2 2 2

cos sin 00

40678884 AB AB AB

x z A x B y C z

x y z R

I find roots Ct(xt, yt, zt).

Where Ct is the intersection of the Day-Night

Terminator and the Great Circle of AB on the surface

of Earth at any time.

According to Diagram 6.4, there will be two roots to

the system of Equations 6.4. I will discard the C t that is

not on the shorter great arc of AB (the airplane’s route).

Ct indicates the point on the airplane route that

observes a Day-night sky.

However, Ct is actually the locus of points of the intersection of the Day-night Terminator planeand the route that the plane follows. There are many solutions to C t because both the equation for

the Day-Night Terminator plane and the equation for the Great Circle of AB have variable t in

them. At every instance of time, there is a solution of the system of equations Ct.

However, there is only one Ct out of all the locus of points that I am concerned with- the point

where not only the Day-night Terminator plane intersects the plane route, but also where the

Diagram 6.4- Ct


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airplane happens to be on the point. This is where a passenger on the plane can observe a sunset

or sunrise sky. This is the special Ct. If I find this special Ct, then I will be able to find the

distance the airplane travels before it arrives at this special Ct. This distance divided by the

airplane’s velocity produces the time after take-off a passenger encounters a Day-Night sky.

6.4

Finding the Special Ct I will use a table to derive the Special Ct and hence the time after take-off a passenger encounters

a Day-Night sky.

I will guess times at which produce the special Ct in the table. Each guessed time will generate a

Ct. Using this Ct, I will derive the distance of the flight of the airplane to Ct t t A C

D . I will find the

quotient of this distancet t A C

D and the airplane’s velocity airplanev to find time of travel to reach Ct

airplanet .

If the time of travel of the airplane airplanet is equal to the guessed time, then the calculated C t is

the special C t and the guessed time is the time after take-off a passenger encounters a Day-Night

sky. In this case, the airplane takes the same amount of time to fly to Ct as the Day-night

terminator plane takes to rotate to Ct. The airplane coincides with the Day-night terminator plane,

and passengers aboard observe a day-night sky.

If 0airplanet t t , then I need to guess another time and confirm t again.

This process is repeated until 0airplanet t t

A table is made:


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t (time of

travel)( , y ,z )t Ct Ct Ct C x

t t A C

(angle of arc t t A C ) t t A C D (distance of

arc t t A C )

airplanet (time it

would take for plane to travel

to Ct)

airplanet t t

0 min Solve the set of equations:

2 2 2 2

cos sin 0

0

40678884

AB AB AB

x z

A x B y C z

x y z R

sint t

At Ct

At Ct

At Ct

A C At Ct

At Ct

At Ct

x x

y y

z z

x x

y y

z z

22

t t

t t

t t

A C

AC

A C

D R

R

t t

t t

A C

airplane

airplane

A C

airplane

Dt

v

R

v

Since t is taken

at intervals of1, there might

not be a time

where airplanet isequal to t .

Therefore the

objectiveshould be to

minimize t .

1 min

2 min

3 min

….

I will use Wolfram Mathematica to perform the calculations in order to complete the table.


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9423.5 (Aug .29 Jun .21) /

365day

9423.5 69 / 5.73

365day

Equation 4.2(Equation for Day-Night Terminator):

cos sin 0 x z

cos( 5.73 ) sin( 5.73 ) 0 x z

Equations 5.1:

2 2 2 2

sin

cos sin 0

LATA z R

x y z R

x z

2 2 2 2

6378 sin

40678884

cos( 5.73 ) sin( 5.7

31

)

2

0

.2

3

z

x y z R

x z

Roots (Sunset/Sunrise Locations on Latitudinal Plane):

The system of equations was solved by Wolfram

Mathematica (refer to Appendix 8.1):

The sunrise location is the root with the positive y value.

The sunset location is the root with the negative y value.

(Refer to Diagram 7.1)

Sunrise: Sunset:

x1=-331.719 x2=-331.719

y1=-5444.263 y2=5444.263

z1=3305.880 z2=3305.880

Since the flight takes off at 11:00, which is in Daytime,the location of A(Shanghai) will be calculated through the Sunset location of Shanghai.

Equations 5.2 (converting Cartesian Sunset coordinates into relative Polar coordinates):

/ 31.22Sunrise set LATA

Diagram 7.1 Sunset and Sunrise


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1 1//

/

331.719180 tan ( ) 180 tan ( ) 176.513

5444.263

Sunrise set Sunrise set

Sunrise set

x

y

Thus, the sunrise location is (R, 31.22 , 176.513 ).

The sunset time for Shanghai is 18:23 on Aug. 29.9

Equation 5.3 (ascertaining polar coordinates of A(Shanghai)):

Because / 180Sunrise set ,

/A( , , ) ( , , 360 ) A A LATA Sunrise set R A R where

/

/

( ) 0.25 / min (18 : 23 11: 00) 0.25 / min 443 0.25 / min 110.75 sunrise sunset take off

timeofsunrise set timeoftake off

T T

Thus, A( , , ) (6378,31.22 , 176.513 360 110.75 ) (6378,31.22 ,72.737 ) A A R A A

Equation 5.4 (ascertaining polar coordinates of B(Vancouver)):

( , , ) ( , , ( )) B B LATB A LONGB LONGA

B R B R

( , , ) (6378, , 72.737 ( 123.12 121.46 ) 360 ) B(6378, 48.25 ,188.157 )48.25 B B B R B

Note: B(6378,48.25 ,188.157 ) B(6378,48.25 , 171.843 )

(6378,31.22 ,72.737 ) A (Shanghai)

B(6378, 48.25 , 171.843 ) (Vancouver)

Equations 6.1 (Determine vectors a and b):

cos sin( 0.25 ) cos 31.22 sin(72.737 0.25 )

cos cos( 0.25 ) cos31.22 cos(72.737 0.25 )

sin sin 31.22

At A A

At A A

At A

x R t t

a y R t t

z R

9 http://sunrisesunset.com/calendar.asp

http://sunrisesunset.com/calendar.asphttp://sunrisesunset.com/calendar.asphttp://sunrisesunset.com/calendar.asphttp://sunrisesunset.com/calendar.asp


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cos sin( 0.25 ) cos 48.25 sin( 171.843 0.25 )

cos cos( 0.25 ) cos 48.25 cos( 171.843 0.25 )

sin sin 48.25

Bt B B

Bt B B

Bt B

x R t t

b y R t t

z R

Equation 6.2 (Determining the cross product of vectors a and b):

AB At Bt At Bt At Bt



A x x y z z y

a b B y y z x x z

z z x y y xC

cos 31.22 cos(72.737 0.25 ) sin 48.25 sin 31.22 cos48.25 cos( 171.843 0.25 )

sin 31.22 cos48.25 sin( 171.843 0.25 ) cos 31.22 sin(72.737 0.25 ) sin 48.25

cos 31.22 sin(72.737 0.25 ) cos 48.25 cos( 171

t t

t t

t

.843 0.25 ) cos31.22 cos(72.737 0.25 ) cos48.25 sin( 171.843 0.25 t t

Equation 6.3 (Equation of the great circle of AB ):

0 AB AB AB

A x B y C z

(for the sake of simplicity, I won’t ex pand AB

A , AB

B and AB

C

Equations 6.4 (systems of equations to determine C t):

2 2 2

cos( 5.73 ) sin( 5.73 ) 0

0

40678884

AB AB AB

x z

A x B y C z

x y z

Table:

The airplane takes off at 11:00, and Shanghai experiences sunset that day at 18:23. It takes 7

hours and 23 minutes for the airplane to experience the day-night sky without flying. Thus, a

good approximation of the time the airplane meets the day-night sky when flying would be 200

minutes – 300 minutes. I will then narrow down the time


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t (time

of

travel)

0 AB AB AB

A x B y C z

( , y ,z )t Ct Ct Ct C x

t t

A C (angle of arc t t A C ) t t A C

D (distance

of arc t t A C )

airplanet (time it

would takefor plane to

travel to Ct)

airplanet t t

0 min

100min

200

min300

min

….

Solve the set of

Equations 6.4

sint t

At Ct

At Ct

At Ct

A C At Ct

At Ct

At Ct

x x

y y

z z

x x

y y

z z

22

t t

t t

t t

A C

A C

A C

D R

R

85

6

t t

t t

t t

A C

airplane

airplane

A C

airplane

A C

Dt

v

R

v

R

Since t is

taken atintervals of 1,

there might

not be a timewhere airplanet

is equal to t .

Therefore theobjective

should be to

minimize t .200min

0.16293308523456076 AB

A

0.8298728828467303 AB

B

0.514319834283114 AB

C

(-546.746, -3269.602,

5448.822)

30.401 =0.5306 rad 3384.167 km 238.882 min 38.882

300min

0.498386959128417 AB

A

0.6832617587126657 AB B

0.5143198342831141 AB

C

(-527.515, -3572.51,

5257.17)

36.153 =0.6310 rad 4024.419 km 284.077 min -15.923

250

min0.33868828692328046

AB A

0.7749364078744585 AB

B

0.514319834283114 AB

C

(-541.994, -

3348.03,5401.46)

33.017 =0.5763 rad 3675.641 km 259.457 min 9.457

275

min0.421040143232046

AB A

0.7334585126205393 AB

B

(-536.246, -

3439.64,34.503 =0.6022 rad 3840.725 km 271.110 min -3.89


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20

0.5143198342831141 AB

C 5344.17)

265min

0.38864639533214784 AB

A

0.7511259357416193 AB

B

0.5143198342831141 AB

C

(-538.863, -3398.37,

5370.25)

33.891 =0.5915 rad 3772.649 km 266.305 min 1.3046

270min

0.40493963394924837 AB

A

0.7424689116760677 AB

B

0.514319834283114 ABC

(-537.61, -3418.2,

5357.77)

34.194 =0.5968 rad 3806.341 km 268.683 min -1.317

(Refer to Appendix 8.2 for Mathematica Script)

As t is minimized for 265 min and 270 min, the airplane that flies from Shanghai to Vancouver at 11:00 on Aug.29 will

encounter a Day-Night sky 265-270 min after takeoff.


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9 Conclusion-Advantages and Limitations of the Model

This model incorporates simpler logic than other models because it chose a 3D geometry

approach instead of a 2D geometry approach. Other models online have modelled the Day-Night

Terminator into a wave function on a 2D map. Whereas this may make calculations easier, the

knowledge of translating a 3D plane’s intersection with a sphere onto a 2D wave function is

beyond the math course I’m taking.

This 3D approach is simpler in logic and it compartmentalizes the calculations so that they are

staged out step-by-step.

However, this model is limited. I designed this model to calculate the time an airplane will meet

a day-night sky. The calculations took quite some time, proving the model not as practical as I

thought it would be. Furthermore, 3D graphs are inherently limited, and they might not appear to

be as straightforward as 2D graphs. I found it difficult to convey a lot of information with the 3D

graphs. Finally, my table approach to ascertain the time needed requires a system of non-linearequations to be solved. This is very calculation intensive and most definitely not something that

can be done on a napkin offered on an airliner ’s flight.

Nevertheless, this model does offer a great deal of accuracy and information. It shows that an

airplane taking off on not only different times in the day but also different dates in the year

affects the results. A 2D model would be much more complicated if it were to consider the

differences of an airplane taking off on different dates. There would be a need for many different

equations to represent the Day-night terminators. (Refer Figure 9.1) In this 3D model, a single

equation with a variable determining the tilt of the Day-night terminator is sufficient. (Recalling

equation 4.2) In this sense, the 3D model is more simplistic and comprehensive.

I’m thrilled to have incorporate IB math skills of polar coordinates, Cartesian coordinates and

vectors to solve this problem that has frustrated me for quite some time. In section 5, I first

calculated the cartesian coordinates of the sunset/sunrise locations. It became evident to me that

a conversion of cartesian coordinates into polar coordinates would not only make the

calculations but also the logic henceforth easier. This showed me the importance of

understanding the connections between system of knowledge in math.

Figure 9.1- 2D Modelling of Day-Night Terminator


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10 Appendix

Section 8.1:

Solve[^2 + ^2 + (6378 ∗ Sin[31.22Degree])^2 == 40678884&& − Cos[−5.73Degree] ∗ + Sin[−5.73Degree] ∗ 6378

∗ Sin[31.22Degree] == 0, {, }, Reals]

{{ → −331.7190224116311, → −5444.2631535811315}, {

→ −331.7190224116311, → 5444.2631535811315}}

Section8.2:

Syntax::bktmcp: Expression "[Cos[31.22Degree]Cos[72.737Degree+0.25Degree*#]Sin[48.25Degree]-

Sin[31.22Degree]Cos[48.25Degree]Cos[−171.843Degree+0.25Degree*#]],{200,300}]" has no closing "]".

Syntax::bktmcp: Expression

"Map[[Cos[31.22Degree]Cos[72.737Degree+0.25Degree*#]Sin[48.25Degree]-

Sin[31.22Degree]Cos[48.25Degree]Cos[−171.843Degree+0.25Degree*#]],{200,300}]/@{200,300}]" has

no closing "]".

Function[Cos[31.22Degree]Cos[72.737Degree+0.25Degree*#]Sin[48.25Degree]-

Sin[31.22Degree]Cos[48.25Degree]Cos[−171.843Degree+0.25Degree*#]]/@{200,300,250,275,265,270}

{-0.162933,-0.498387,-0.338688,-0.42104,-0.388646,-0.40494}

Function[Sin[31.22Degree]Cos[48.25Degree]Sin[−171.843Degree+0.25Degree*#]−Cos[31.22Degree]Sin[

72.737Degree+0.25Degree*#]Sin[48.25Degree]]/@{200,300,250,275,265,270}

{-0.829873,-0.683262,-0.774936,-0.733459,-0.751126,-0.742469}

Function[Cos[31.22Degree]Sin[72.737Degree+0.25Degree*#]Cos[48.25Degree]Cos[−171.843Degree+0.2

5Degree*#]−Cos[31.22Degree]Cos[72.737Degree+0.25Degree*#]Cos[48.25Degree]Sin[−171.843Degree

+0.25Degree*#]]/@{200,300,250,275,265,270}

{-0.51432,-0.51432,-0.51432,-0.51432,-0.51432,-0.51432}


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Solve[−Cos[−5.73Degree]*x+Sin[−5.73Degree]*z==0&&(-0.16293308523456076)*x+(-

0.8298728828467303)*y+(-0.514319834283114)*z==0&&x2+y

2+z

2==40678884,{x,y,z},Reals]

Solve::ratnz: Solve was unable to solve the system with inexact coefficients. The answer was obtained by

solving a corresponding exact system and numericizing the result. >>

{{x->-546.7462627332769`,y->-3269.601874049342`,z->5448.821534001661`},{x-

>546.7462627332769`,y->3269.601874049342`,z->-5448.821534001661`}}


0.6832617587126657)*y+(-0.5143198342831141)*z==0&&x2+y2+z2==40678884,{x,y,z},Reals]



{{x->-527.5151690154876`,y->-3572.508155665824`,z->5257.166273208386`},{x-

>527.5151690154876`,y->3572.508155665824`,z->-5257.166273208386`}}

a200={Cos[31.22Degree]Sin[72.737Degree+0.25Degree*200],Cos[31.22Degree]Cos[72.737Degree+0.25Degree*200],Sin[31.22Degree]}

c200={-546.746,-3269.602,5448.822}

Norm[a200*c200]/(Norm[a200]*Norm[c200])

{0.719348,-0.462469,0.518326}

{-546.746,-3269.6,5448.82}

0.506056

a300={Cos[31.22Degree]Sin[72.737Degree+0.25Degree*300],Cos[31.22Degree]Cos[72.737Degree+0.25

Degree*300],Sin[31.22Degree]}

c300={-527.515,-3572.51,5257.17}


{0.456502,-0.723149,0.518326}

{-527.515,-3572.51,5257.17}

0.58994


0.7749364078744585)*y+(-0.514319834283114)*z==0&&x2+y2+z2==40678884,{x,y,z},Reals]



{{x->-541.994,y->-3348.03,z->5401.46},{x->541.994,y->3348.03,z->-5401.46}}



c275={-541.994,-3348.03,5401.46}


http://reference.wolfram.com/mathematica/ref/Solve.htmlhttp://reference.wolfram.com/mathematica/ref/Solve.htmlhttp://reference.wolfram.com/mathematica/ref/Solve.htmlhttp://reference.wolfram.com/mathematica/ref/Solve.htmlhttp://reference.wolfram.com/mathematica/ref/Solve.htmlhttp://reference.wolfram.com/mathematica/ref/Solve.html


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{0.532516,-0.669153,0.518326}

{-541.994,-3348.03,5401.46}

0.564023


0.7334585126205393)*y+(-0.5143198342831141)*z==0&&x2+y2+z2==40678884,{x,y,z},Reals]



{{x->-536.245714613068`,y->-3439.6406933255107`,z->5344.174064752974`},{x-

>536.245714613068`,y->3439.6406933255107`,z->-5344.174064752974`}}



c275={-536.246,-3439.64,5344.17}


{{x->-536.246,y->-3439.64,z->5344.17},{x->536.246,y->3439.64,z->-5344.17}}

{0.532516,-0.669153,0.518326}

{-536.246,-3439.64,5344.17}

0.566443


0.7511259357416193)*y+(-0.5143198342831141)*z==0&&x2+y

2+z

2==40678884,{x,y,z},Reals]



{{x->-538.863,y->-3398.37,z->5370.25},{x->538.863,y->3398.37,z->-5370.25}}


0.7424689116760677)*y+(-0.514319834283114)*z==0&&x2+y2+z2==40678884,{x,y,z},Reals]



{{x->-537.61,y->-3418.2,z->5357.77},{x->537.61,y->3418.2,z->-5357.77}}



c265={-538.863,-3398.37,5370.25}

http://reference.wolfram.com/mathematica/ref/Solve.htmlhttp://reference.wolfram.com/mathematica/ref/Solve.htmlhttp://reference.wolfram.com/mathematica/ref/Solve.htmlhttp://reference.wolfram.com/mathematica/ref/Solve.htmlhttp://reference.wolfram.com/mathematica/ref/Solve.htmlhttp://reference.wolfram.com/mathematica/ref/Solve.html


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{0.561197,-0.645288,0.518326}

{-538.863,-3398.37,5370.25}

0.557615



c270={-537.61,-3418.2,5357.77}


{0.546987,-0.657377,0.518326}

{-537.61,-3418.2,5357.77}

0.561992

Date post:	07-Aug-2018
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