Download Introduction - Andrew G. Zannetti

The Tower Project
Andrew Zannetti
GAT 9B
Mr. Acre
May 30, 2012
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Introduction
I am a successful architect who specializes in making towers. Recently I was approached
by a millionaire who requested that I build a tower for their dog. I was given a twenty foot by
twenty foot plot of land and was told to make the tower have twenty sides as well. The process of
finding the dimensions of the tower took nine separate steps. This following paper will include
blueprints and explanations on how the dimensions of the tower were found. This way the
processes used can be referred to for the future construction of towers. The actual measurements
are all in feet but a scaling factor was used for the blueprints. I used a scaling factor of half a
centimeter would equal one foot in the blueprints.
Part Two: A 20 Sided Polygon Maximized on a 20’ by 20’ Plot of Land
F
H
F'''
L
H'
M
LM = 1’
F’’’H’ = 14’
FH = 20’
FG = 20’
F’’’G’ = 14’
G'
G
Figure 1. Plot and Bases of Tower
Figure 1 shows all four polygons maximized on the twenty foot by twenty foot plot. In
the city the millionaire lives in structures must be three feet away from the edge of the property
on all sides. This made the plot that I could work with fourteen feet by fourteen feet. The outer
polygon represented by the purple and everything inside of it is where the footing would start.
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The colored spaces in between each polygon are 1 foot long. The plot was maximized by placing
a side or face on each side of the plot.
Measure of Central Angle = 360/n
MCA = 360/20
MCA = 18º
Figure 2. Formula and Substitutions to Find Central Angle
Figure 2 shows the formula used to find the measure, in degrees, of the polygon’s central
angle. In the formula n stands for the number of sides. The polygon has twenty sides therefore n
equals twenty. All four polygons have the same number of sides so they all have a central angle
of 18 degrees.
Polygon 1
A
B
D
C
Figure 3. Polygon 1
Figure 3 shows the outermost and largest polygon, Polygon 1. Triangle ACD was used to
find the area of the polygon.
Height (BD) = (Length of one side of the original plot – 6) / 2
Height (BD) = (20-6) / 2
Height (BD) = 14 / 2
Height (BD) = 7’
Figure 4. Formula and Substitution to Find Height of Triangle ACD
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In Figure 4 the height of Triangle ACD was found. In the formula the 6’ is there because
the plot is shrunk by 3’ on all sides. This means that the distance across the entire polygon is the
plot size (20’) minus 6’ which equals 14’. Then the height of one triangle is half the distance
across the polygon. That means 14’ has to be divided by 2 which will give the answer 7’.
tan(measure of angle ADB) = (AB / DB)
tan(9º) = AB / 7
7 * tan(9º) = AB / 7 * 7
1.10…’ = AB
AB * 2 = AC
1.10… * 2 = AC
2.21…’ = AC
Figure 5. Formulas and Substitutions to Find One Side Length of Polygon 1
Figure 5 shows the formulas that were used and the substituted to find the length of one
side of Polygon 1. Two congruent right triangles were made so that trigonometry could be
utilized. The tangent of angle ADB was used to find the length of AB. Angle ADB measures 9º
because the height of the Triangle ACD cuts Angle ADC in half. After AB was found it was
multiplied by 2 to find the entire length of one side of Polygon 1.The length of AC is 2.21…’.
Area of Polygon 1 = (1/2b * h) * 20
AP1 = ((1/2) * 2.21… * 7) * 20
AP1 ≈ 155.22 feet2
Figure 6. Formula and Substitutions to Find the Area of Polygon 1
The formulas and substitutions in Figure 6 were used to find the volume of Polygon 1.
The area of one of the twenty triangles that Polygon 1 is split up into was found. It was then
multiplied by 20 to find the total area. The area of Polygon 1 is about 155.22 feet2.
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Polygon 2
K
L
M
J
Figure 7. Polygon 2
Figure 7 is a sketch of Polygon 2. Triangle KMJ was used to find the area of Polygon 2.
Height (JL) = (Plot – 8) / 2
Height (JL) = (20 – 8) / 2
Height (JL) = 12 / 2
Height (JL) = 6’
Figure 8. Formula and Substitutions to Find the Height of Triangle KMJ
In Figure 8 the formula and substitutions used to find the height of Triangle KMJ are
shown. The original plot size was subtracted by 8’ because Polygon 2 moves in 1’ on all sides
from the original plot. This would then come to be 12 which when divided by 2 becomes 6’.
tan(measure of Angle KJL) = KL / JL
tan(9º) = KL / 6
6 * tan(9º) = KL / 6 * 6
0.95… ‘ = KL
KL * 2 = KM
0.95… * 2 = KM
1.90…’ = KM
Figure 9. Formulas and Substitution to Find One Side Length of Polygon 2
The formulas and substitutions shown in Figure 9 were used to find the length of one side
of Polygon 2. The measure of Angle KJL is 9º because angle KJM which measures 18º is split in
half by the height which is JL. Trigonometry can be used because the height creates two
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congruent right triangles. After KL was found it was multiplied by 2 to find the total length of
the side of Polygon 2. The length of KM is 1.90…’.
Area of Polygon 2 = (1/2b * h) * 20
AP2 = ((1/2)*1.90… * 6) * 20
AP2 ≈ 114.04 feet2
Figure 10. Formula and Substitutions to Find the Area of Polygon 2
The area of Polygon 2 was found using the formula and substitutions shown in Figure 10.
The area of Triangle KMJ was found first using the formula A = (1/2b) * h. Then it was
multiplied by 20 because Polygon 2 is made up of 20 triangles congruent to Triangle KMJ. The
area of Polygon 2 is about 114.04 feet2.
Polygon 3
K
L
M
N
Figure 11. Polygon 3
The sketch in Figure 11 shows what Polygon 3 looks like. The triangle KNM was used to
find the area of Polygon 3.
Height (NL) = (Plot – 10) / 2
Height (NL) = (20 – 10) / 2
Height (NL) = 10 / 2
Height (NL) = 5’
Figure 12. Formula and Substitutions for Finding the Height of Triangle KNM
The height of Triangle KNM was found using the formula and substitutions shown in
Figure 12. Polygon 3 is one more foot in than Polygon 2 is, so its 5 feet in on all sides. This is
where the plot minus 10’ comes in because that is how much is taken off of each side overall. 10’
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is then the distance across Polygon 3. It then has to be divided by 2 to get half of the length
which is the height of Triangle KNM. The height of Triangle KNM is 5’.
tan(measure of Angle KNL) = KL / NL
tan(9º) = KL / 5
5 * tan(9º) = KL / 5 * 5
0.79…’ = KL
KL * 2 = KM
0.79… * 2 = KM
1.58…’ = KM
Figure 13. Formulas and Substitutions to Find the Length of One Side of Polygon 3
Figure 13 shows the formulas as well as the substitutions used to find the length of one
side of Polygon 3. The measure of Angle KNL is 9º because Angle KNM equals 18º and the
height cuts KNM in half. Trigonometry can then be used because two right triangles were
created by the height. After KL is found it has to be multiplied by 2 to get the total length which
is KM. The length of KM is 1.58…’.
Area of Polygon 3 = (1/2b * h) * 20
AP3 = ((1/2)1.58… * 5)
AP3 ≈ 79.19 feet2
Figure 14. Formula and Substitutions to Find the Area of Polygon 3
In Figure 14 the formula and substitutions used found the area of Polygon 3. First the
area of Triangle KNM was found. That was then multiplied by 20 because Polygon 3 is made up
of 20 congruent triangles. The area of Polygon 3 is about 79.19 feet2.
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J
K
L
Polygon 4
M
Figure 15. Polygon 4
Figure 15 shows what the fourth and innermost polygon looks like. The triangle, JML
was used to find the area of the entire polygon.
Height (KM) = (Plot – 12) / 2
Height (KM) = (20 -12) / 2
Height (KM) = 8 / 2
Height (KM) = 4’
Figure 16. Formula and Substitutions to Find the Height of Triangle JML
The height of Triangle JML was found using the formula and substitutions shown in
Figure 16. Polygon 4 is yet another foot in on each side from Polygon 3. This means that all
sides are pushed in by 6 on each side making each length actually minus 12’. The 8’ that are left
which is the distance across Polygon 4 is divided by 2 because that is the height of Triangle
JML. The height of Triangle JML is 4’.
tan(measure of Angle JMK) = JK / MK
tan(9º) = JK / 4
4 * tan(9º) = JK / 4 * 4
0.63…’ = JK
JK * 2 = JL
0.63… * 2 = JL
1.26…’ = JL
Figure 17. Formulas and Substitutions to Find the Length of One Side of Polygon 4
The length of one side of Polygon 4 was found using the formulas and substitutions used
in Figure 17. The measure of Angle JMK is 9º because the measure of Angle JML is 18º and the
height of the triangle splits Angle JML in half. The height also creates two congruent right
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triangles and allows trigonometry to be applied. After JK was found it had to be multiplied by 2
to find JL which is the total length of one side. The length of one side of Polygon 4 is 1.26…’.
Area of Polygon 4 = (1/2b * h) * 20
AP4 = ((1/2)1.26… * 4) * 20
AP4 ≈ 50.68 feet2
Figure 18. Formula and Substitutions to find the Area of Polygon 4
The area of Polygon 4 was found using the formula and substitutions in Figure 18. First,
the area of Triangle JML was found. That was then multiplied by 20 because Polygon 4 is made
up of 20 congruent triangles. The area of Polygon 4 is about 50.68 feet2.
Part 3: Volume of the Concrete Footing, Plexiglas Floor and Aquarium Water
L
LK = 3.5’
K
Figure 19. Footing of Tower
Figure 19 shows the footing of the tower that is made of concrete. It starts at Polygon 1
and ends at Polygon 4. The footing extends 3.5’ into the ground.
Volume of Big Footing Prism = Area of Polygon 1 * height
VBFPr = 155.21 * 3.5
VBFPr = 543.25… feet3
Figure 20. Formula and Substitutions to Find the Volume of the Big Prism
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The volume of the big prism was found using the formula and substitutions in Figure 20.
The area of Polygon 1 was multiplied by the height of the prism to get the volume. The volume
of the big footing prism is 543.25… feet3.
Volume of Small Footing Prism = Area of Polygon 4 * height
VSFPr = 50.68… * 3.5
VSFPr = 177.39… feet3
Figure 21. Formula and Substitutions to Find the Volume of the Small Prism
In Figure 21 the formula and substitutions shown were used to find the volume of the
small prism. The base of the prism is Polygon 4 so the area of that polygon is multiplied by the
height, 3.5’. The volume of the small footing prism is 177.39… feet3.
Volume of Footing = VBPr - VSPr
VFooting = 543.25… – 177.39…
VFooting = 365.86… feet3
Figure 22. Formula and Substitutions to Find the Volume of the Footing
Figure 22 gives the formula and substitutions used to come up with the volume of
footing. To find the volume the volume of the big prism is subtracted by the volume of the small
prism. The volume of the footing is 365.86… feet3.
Cost of Concrete for Footing = (Volume of footing / 27) * 115
CConcrete = (365.86… / 27) * 115
CConcrete = 13.55… * 115
Round 13.55… up to 14
CConcrete = 14 * 115
CConcrete = $1610
Figure 23. Formula and Substitutions to Find the Cost of Concrete for the Footing
The formula and substitutions in Figure 23 were used to find the cost of the concrete. The
volume of the footing was divided by 27 because that is a cubic yard. It costs $115 for every
cubic yard of concrete that is used. The volume of the footing is 13.55… cubic yards so it must
be rounded up to 14 cubic yards. This is then multiplied by 115 to find the final cost of $1610.
Volume of Water in Aquarium = Volume of Small Prism * 0.75
VAW = 177.39… * 0.75
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VAW ≈ 133.04 feet3
Figure 24. Formula and Substitutions to Find Volume of Water in Aquarium
Figure 24 shows the formula and substitutions that were used to find the volume of water
in the aquarium. The small prism is the aquarium and is filled 75% with water. This means that
the volume of the small prism has to be multiplied by 0.75 to get the amount of space taken up
by the water. The volume of the water in the aquarium is about 133.04 feet3.
UV = 4 inches
U
V
Figure 25. Floor
The floor of the tower is made of 4” thick Plexiglas. The floor is the inside of Polygon 4.
Volume of Floor = Area of Polygon 4 * ⅓
VFloor = 50.68… * ⅓
VFloor ≈ 16.89 feet3
Figure 27. Formula and Substitutions to Find the Volume of the Floor
The volume of the floor is the area of Polygon 4 multiplied by 4” which is ⅓ of a foot.
The volume of the floor is about 16.89 feet3.
Sheets of Plexiglas = Area of Polygon 4 / (4’ x 8’)
PlexiglasSheets = 50.68… / 32
PlexiglasSheets = 1.58… feet2
Round Sheets of Plexiglas up to 2
Cost of Plexiglas = PlexiglasSheets * $1100
CostPlexiglas = 2 * 1100
CostPlexiglas = $2200
Figure 28. Formula and Substitutions to Find the Cost of the Plexiglas for the Floor
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The total cost of the Plexiglas for the floor is $2200.
Part 4: One Lateral Face of the Outer Prism Base
K
L
M
J
KM = 1.90…’
Figure 29. Base of Outer Prism
The base of the outer prism of the tower is shown in Figure 29. This base is also Polygon
2 from Part 2.
F
T
Q J
U
QR = 1.90...'
TJ = 2.5'
JL = 1.5'
TU = 0.23...'
QF = 3.80...'
L
R
Figure 30. Lateral Face of Outer Prism with Door
The lateral face of the outer prism is about 3.80’ tall and about 1.90’ wide. The door
measures 2.5’ tall and 1.5’ wide. The dimensions of the face are known because the outer prism
is Polygon 2. This makes the width the length of one of Polygon 2’s sides which equals 1.90…’.
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The height of the face is double the length of the width so it equals 3.80…’. The dimensions of
the door had to be scaled down to somewhat of a doggie door because that is who will be living
in this tower. The half window above the door had its dimensions found using trigonometry.
W
V
G H
WV = 1.90...'
TU = 0.23...'
VR = 3.80...'
GH = 0.23...'
R
Figure 31. Lateral Face of Outer Prism with a Window
Figure 31 shows a face of the outer prism and window with the dimensions labeled.
Area of Door = (b * h) + (1/2b * h) * 10
ADoor = (1.5 * 2.5) * (1/2 * 2 * (0.75 * sin(9º)) * (0.75 * cos(9º))) * 10
ADoor = 3.75 + 0.86…
ADoor = 4.61… feet2
Area of Windows = 4 * ((1/2b * h) * 10)
AWindows = 4 * (((1/2)(2 * (0.75 * sin(9º)) * (0.75 * cos(9º))) * 10)
AWindows = 3.47… feet2
Lateral Surface Area of Outer Prism = (b * h) * 20 – ADoor – AWindows
LSAOPr = (1.90… * 3.80…) * 20 - 4.61… - 3.47…
LSAOPr ≈ 136.39… feet2
Figure 32. Formulas and Substitutions of the LSA of the Outer Prism
In Figure 32 the lateral surface area of the outer prism minus the door and windows was
found. The total lateral surface area of the outer prism is 136.39… feet2.
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Part 5: Volume of the Inner Prism
O
F
OP = 1.58...'
P
Figure 33. Base of Inner Prism
Figure 33 shows the inner prism’s base with the side length labeled.
Q
OQ = 3.80...'
OP = 1.58..'
O
P
Figure 34. Lateral Face of Inner Prism
The face of the inner prism has the same height as the outer prism but has a different
width.
Volume of Inner Prism = Area of Polygon 3 * h
VIPr = 79.19… * 3.80…
VIPr = 301.02… feet3
Figure 35. Formula and Substitutions to Find the Volume of the Inner Prism
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The volume of the inner prism of the tower was found in Figure 35. The volume of the
inner prism is 301.02… feet3
Part 6: Dimensions of the Outer Pyramid
N
M
I
J
G
IN = 8.27...'
IG = 6'
NG = 5.70...'
Figure 36. Drawing of Outer Pyramid Base and One Lateral Face
Figure 36 shows the height of the outer pyramid as well as the slant height of a lateral
face. The height of the outer pyramid is 3 times the length of one side of Polygon 2. That means
that it would be 1.90…’ times 3 which is 5.70…’.
Slant Height (IN)2 = a2 + b2
Slant Height (IN)2 = 62 + 5.70…2
Slant Height (IN)2 = 68.51…
Slant Height (IN) = sqrt(68.51…)
Slant Height (IN) = 8.27…’
Figure 37. Formula and Substitution to Find the Slant Height of the Outer Pyramid
In Figure 37 the Pythagorean Theorem was utilized to find the slant height of the outer
pyramid. The slant height of the outer pyramid is 8.27…’.
Measure of Angle GIN = tan-1(NG / GI)
MGIN = tan-1(5.70… / 6)
MGIN ≈ 43.54º
Figure 38. Formula and Substitution to Find the Measure of Angle GIN
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In Figure 38 trigonometry is used to find the angle of measure between the prism base
and the outer pyramid’s lateral faces. The angle between the outer pyramid and the prism is
43.54º.
Part 7: Lateral Faces of the Outer Pyramid
D
DC = 8.27...'
AC = 0.95...'
DB = 8.33...'
AB = 1.90...'
A
C
Measure of Angle DAC ≈ 83.45º
Measure of Angle DBC ≈ 83.45º
Measure of Angle ADC ≈ 6.55º
Measure of Angle BDC ≈ 6.55º
Measure of Angle ADB ≈ 13.1º
B
Figure 39. One Lateral Face of the Outer Pyramid
Figure 39 shows the lengths of each side and the height of one lateral face of the outer
pyramid. It also has each angle measurement labeled.
Measure of Angle ADC = tan-1(AC / DC)
MADC = tan-1(0.95… / 8.27…)
MADC ≈ 6.55º
Measure of Angle DAC = tan-1(DC / AC)
MDAC = tan-1(8.27… / 0.95…)
MDAC ≈ 83.45º
Measure of Angle ADB = 2 * MADC
MADB = 2 * 6.55
MADB ≈ 13.1º
Measure of Angle DBC = MDAC
Measure of Angle CDB = MADC
Figure 40. Formulas and Substitutions to Find the Angles of One Lateral Face
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Figure 40 above shows the formulas and substitutions used to find the angle
measurements one face of the outer pyramid. The angles of the face of the outer pyramid are
about 13.1º, 83.45º and 83.45º.
Area of One Lateral Face = 1/2b * h
ALF = ½ * 1.90… * 8.27…
ALF = 7.86… feet2
Figure 41. Formula and Substitutions to Find the Area of One Lateral Face
The formula and substitutions in Figure 41 were used to find the area of one lateral face
of the outer pyramid. The area of one lateral face of the outer pyramid is 7.86… feet2.
Lateral Surface Area of Outer Pyramid = ALF * 20
LSAOPy = 7.86… * 20
LSAOPy = 157.31… feet2
Figure 42. Formula and Substitutions to Find the Lateral Surface Area of the Outer Pyramid
In Figure 42 the lateral surface area of the outer pyramid was found to be 157.31… feet2.
Part 8: Height and Volume of the Inner Pyramid
H
G
E
F
HC = 4.75...'
C
Figure 43. Inner Pyramid with One Lateral Face
Figure 43 shows a drawing of the inner pyramid with the height labeled.
Volume of Inner Pyramid = ⅓ (Area of Polygon 3) * Height
VIPy = ⅓ (79.19…) * 4.75…
VIPy = 125.42… feet3
Figure 44. Formula and Substitutions to Find the Volume of the Inner Pyramid
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The formula and substitutions used in Figure 44 helped to find that the volume of the
inner pyramid is 125.42… feet3.
Part 9: The Tower
Figure 45. Outer Tower
Figure 45 shows the outside of the tower with the walls that cannot be seen represented as
dotted line segments.
Total Surface Area of Tower = LSA of Outer Prism + LSA of Outer Pyramid
TSA = 136.39… + 157.31…
TSA ≈ 293.71 feet2
Figure 46. Formula and Substitutions to Find the Total Surface Area of the Tower
The formula and substitutions utilized in Figure 46 helped find the total surface area of
the outside of the tower.
Total Volume of Tower = Volume of Inner Prism + Volume of Inner Pyramid
VTotal = 301.02… + 125.42…
VTotal ≈ 426.46 feet3
Figure 47. Formula and Substitutions to Find the Total Volume of the Tower
Figure 47 shows the formula and substitutions used to find that the total volume of the
tower is about 426.46 cubic feet.
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Conclusion
Overall it was not very difficult to find the dimensions, volume and surface area
of the tower itself. However, there were a few problems that I encountered while finding the
dimensions of the tower. One problem that was encountered was the door that I usually use for
towers was too wide and tall to fit on the side of the tower. I had to make the door half the size of
what the doors usually are. Also there were a few rounding errors made while finding
measurements. These were checked and corrected to make sure that all of the measurements
were correct. I also built a scale model to show the millionaire what the tower will look like.
While building the model I encountered a few problems. At first I had decided to use foam board
to create the scale model. I figured a 1cm = 1 ft scale would be good so I made that the scale.
After I cut out a few of the foam board pieces I realized the cuts were not accurate. I discarded
that idea. After the foam board did not work I went on to use wood for my model. The first scale
had to be changed so that the model would be big enough to work with. The new scale is 2 cm =
1 ft. While cutting the footing for the model I could not find wood with the right dimensions so I
had to use two pieces and glue them together to create the footing. Later on in the building of the
model I tried to use wood to make my pyramids but did not have the right tools to do so. I
decided that I would use the foam board and cut the pyramid faces using that. I hope that this
paper including blueprints and the formulas used to find the measurements of this tower will
prove useful to others and me in building towers in the future. The millionaire is a big time fan of
professional basketball so I have decided to theme their dog’s tower with basketball teams. The
millionaire believes that the Miami Heat will be the NBA Champs so the inside of the tower is
Miami Heat themed. On the outside there are five other teams that are also in the playoffs this
year. The ground outside of the tower is a basketball court with two hoops. I also hope that the
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millionaire and their dog will be very pleased with this basketball themed tower. Maybe they will
one day watch basketball games inside of it together.
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