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The Tower Project Andrew Zannetti GAT 9B Mr. Acre May 30, 2012 Zannetti 1 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. Zannetti 2 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 Zannetti 3 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. Zannetti 4 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 Zannetti 5 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’ Zannetti 6 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. Zannetti 7 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 Zannetti 8 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 Zannetti 9 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 Zannetti 10 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 Zannetti 11 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…’. Zannetti 12 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. Zannetti 13 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 Zannetti 14 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 Zannetti 15 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 Zannetti 16 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 Zannetti 17 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. Zannetti 18 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 Zannetti 19 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. Zannetti 20 Works Cited Adams, Alonzo. Los Angeles Lakers' Kobe Bryant (24) Shoots against Oklahoma City Thunder Guard Thabo Sefolosha (2) during the First Quarter of Game 5 in Their NBA Basketball Western Conference Semifinal Playoff Series. 2012. Photograph. Oklahoma City. Connecticut Post. Hearst Communications Inc., 22 May 2012. Web. 26 May 2012. <http://www.ctpost.com/sports/article/Thunder-surge-past-Lakers-106-90-to-win-Westsemis-3575405.php#photo-2968070>. Andrew. Lakers Logo on Purple Background. 2012. Photograph. Sports Geekery. 6 Jan. 2012. Web. 26 May 2012. <http://www.sportsgeekery.com/2173/los-angeles-lakers-desktopwallpaper-collection/>. Belt Buckle.com. Miami Heat Logo. Photograph. Belt Buckle.com. Web. 26 May 2012. <http://www.beltbuckle.com/nba/miami-heat-logo>. Creamer, Chris. San Antonio Spurs Logo. Photograph. Chris Creamer's Sports Logos. Web. 26 May 2012. <http://sportslogos.net/team.php?id=233>. Elsa/Getty Images North America. Philadelphia 76ers v Boston Celtics - Game One. 2012. Photograph. Boston. Zimbio. Zimbio Inc., 11 May 2012. Web. 26 May 2012. <http://www.zimbio.com/photos/Rajon+Rondo/Philadelphia+76ers+v+Boston+Celtics+ Game+One/GuJ8TdxxSeX>. Zannetti 21 F.Dynamo1986. Boston Celtics Logo. Photograph. F.Dynamo1986. Web. 26 May 2012. <http://www.fdynamo1986.com/?attachment_id=122>. Hui, Kin Man. Spurs Point Guard Tony Parker, Taking a Shot in Front of Clippers Counterpart Chris Paul, Is Averaging a Team-high 35 Minutes a Game in the Playoffs. He Scored 17 Points Sunday. 2012. Photograph. Los Angeles. Spurs Nation. Hearst Communication Inc., 21 May 2012. Web. 26 May 2012. <http://blog.mysanantonio.com/spursnation/2012/05/21/spurs-notebook-popovich-usingbigger-rotation-in-these-playoffs/>. NBA Playoffs Logo. Photograph. 4Quarters with Mike Vamosi. By Mike Vamosi. 28 Apr. 2012. Web. 26 May 2012. <http://fourquarterssports.wordpress.com/2012/04/28/2012-nbaplayoff-predictions-sure-to-go-wrong/>. Ogrocki, Sue. Oklahoma City Thunder Guard Russell Westbrook (0) Shoots over Los Angeles Lakers Guard Kobe Bryant (24) in the First Quarter of Game 5 in Their NBA Basketball Western Conference Semifinal Playoff Series. 2012. Photograph. Oklahoma City. Connecticut Post. Hearst Communications Inc., 22 May 2012. Web. 26 May 2012. <http://www.ctpost.com/sports/article/Thunder-surge-past-Lakers-106-90-to-win-Westsemis-3575405.php#photo-2968086>. Zannetti 22 Shaffer, Tim. Andre Iguodala. 2012. Photograph. Philadelphia. Yahoo! Sports. ABC News Network, 23 May 2012. Web. 26 May 2012. <http://sports.yahoo.com/photos/nbaplayoffs-celtics-vs-76ers-slideshow/philadelphia-76ers-small-forward-andre-iguodaladunks-boston-photo-024316054.html#crsl=%252Fphotos%252Fnba-playoffs-celtics-vs76ers-slideshow%252Fphiladelphia-76ers-small-forward-andre-iguodala-dunks-bostonphoto-024316054.html>. The 700 Level. Philadelphia 76ers Logo. Photograph. Uwishunu. Greater Philadelphia Tourism Marketing Corporation, 23 June 2009. Web. 26 May 2012. <http://www.uwishunu.com/2009/06/the-sixers-rebranding-with-use-of-classic-logo/>. Wade, Jared. Oklahoma City Thunder Logo. Photograph. Both Teams Played Hard. 21 Aug. 2009. Web. 26 May 2012. <http://www.bothteamsplayedhard.net/2009/08/21/the-nbalogo-ranking-project-27-oklahoma-city-thunder/>.