tower final

8/20/2019 Tower Final

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Tower

Mario Nona & Merna Sana

GAT

Mr. Acre

May 27, 2014


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Introduction

With no offense intended, but we think you are completely crazy. We mean, what kind of

person would want a 14 sided building and roof. Although all the shenanigans, it will be our

pleasure to design this tower for you, Mrs. Copeland. uilding it though is a different story.

!n our opinion, aside from all the nonsense, this does seem like a "ery uni#ue building.

$o the thought of, %where would this type of building be built&' came up. (o our best

knowledge, we belie"e a special tower like this would look great in a great big city already with

hundreds of towers) *ew +ork City Although, of course, it is still always up to you.

$o enough of the nonsense. +ou are gi"ing us the option of designing this tower for you,

also with a good price. +our standards are clearly e-pressed through your letter. We see that you

want a 14 sided tower, built upon a / by / ft s#uare plot. +ou also state that you want %the

ma-imum size possible, whilst remaining within the boundaries of buildable space and

remaining perpendicular in some way shape or form to the a"ailable plot.' +ou also want %an

a#uarium built into the flooring, strong foundations, and walls one foot in thickness.' (here

were many other little details included as well that will not be ignored.

(his tower you ha"e presented to us, Mrs. Copeland, is an intimidating duty. !t will

re#uire much effort by both people designated to design this tower. 0"ery ounce of sweat and

tears will be poured into making this tower assignment of yours as perfect as it can be, and we

sincerely hope you are pleased.

The 14 Sided Polyon Ma!i"i#ed on the Plot

(he polygon used for my tower is a 14sided polygon. ! ha"e a plot that is / ft - / ft,

but ! am not allowed to build within 2 ft of the edge, per local rules. (he tower will be built on


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foundations that are the same polygonal shape as the tower. (he foundations must be the

ma-imum size possible on the plot, without breaking any laws. (hen, the tower will be 1 foot in

from the foundation3s edge, and its walls will be 1 foot thick. astly, the foundations will end 1

foot inside of the wall3s inner edge.

When drawn out, this setup will create four concentric polygons, each e-actly 1 foot

further in than the last.

5igure 1. Aerial 6iew of 7olygons

8!n this image, the scaling factor was 91, so all side lengths and areas should be doubled.:

5igure 1 shows the four necessary polygons, within the proper confines of the s#uare.

(he original plot size was / ft - / ft, and nothing can be built within 2 ft of the edge of the

plot. (herefore, the %working place' is ft - ft 8/ minus 2 from each side9 / ;:.


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$eeing as how the polygons are 14sided, the central angle must be 82;?@14:.

(hat means that each angle measure must be 1


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of the %working area,' since "ertices are touching. (he smallest angle in the right triangle is

1./, because it is e-actly half of .=142>. ! used sine to find the appro-imate length of the

side of 7olygon 1, which is .4< ft.

cos81./


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5igure ;. $econd Butermost 7olygon Measurements

5igure ; shows the second outermost polygon, or 7olygon , as well as the two triangles

necessary for finding its area.

-1 11D 8cos 1./


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Area(riangle F D 811D 8cos 1./


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5igure 11 shows how to find the triangle height of 7olygon 2. Again, we Hust needed to

subtract 1 from the pre"ious height.

tan 1./


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5igure 14. !nnermost 7olygon Measurements

5igure 14 displays the innermost polygon, or 7olygon 4, as well as the two triangles

necessary for finding its area.

-2 11D 8cos 1./


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Area(riangle 4 F D 811D 8cos 1./


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5igure 1/ displays Hust the foundations of the tower. *ow, we ha"e to find the "olume of

them. (o do that, we ha"e to find out the area of the ring that the foundations are made inK the

ring that stretches from 7olygon 1 to 7olygon 4. (hen the "olume will easily be found by

multiplying that area by 2.


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6olume5ooting ;1/./@=

6olume5ooting .yd2

5igure 1. Con"erting Cubic 5eet 6olume to Cubic +ard 6olume

(here are e-actly = ft2 for e"ery 1 yd2, so we Hust di"ided the "olume of the foundations

8in ft2: by = in order to get the "olume of the foundations in yd2. (his "olume turned out to be

about .2yd2, which e#uates to 2 bags of $uper5astJryingConcrete 8since we can3t buy

part of a bag:. (herefore, the cost of the concrete can now be found.

Cost 2D11<

Cost L;4<

5igure . Cost Analysis of the 5oundations

y multiplying the number of bags of $uper5astJryingConcrete that we3ll need by the

cost of one bag, we found the total amount of money the foundations will cost. (hey will sot

L;4


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5igure 2. 7le-iglas Jiagram and Measurements

5igure 2 shows the 7le-iglas floor. !t is e-actly 4 in thick, or ft thick. $ince it takes up

7olygon 4, it has the same measurements as 7olygon 4. We want to find the "olume, and we ha"e

all the necessary measurements.

6olume7le-iglas 84':D Area of 7olygon 4

6olume7le-iglas 8:D81?.;


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Cost of 7le-iglass 11??D;

Cost of 7le-iglass ;;??L

5igure ;. Cost of 7le-iglass

5igure ; abo"e shows the cost of the ple-iglass that will be needed for Mrs. Copeland

floor. A sheet of ple-iglass cost 11??L, and there are ; pieces of ple-iglass needed for the

flooring. $o, the number of sheets needed, ;, multiplied by the cost of one sheet, 11??L, e#uals a

total of ;;??L for the making of the floor.

5igure =. Jiagram and Measurements of A#uarium

!n figure =, a diagram has been made of the a#uarium that goes under the flooring. (he

a#uarium will also be placed within the concrete, making the height of the a#uarium 2.< ft. (he

a#uarium will only be filled with =


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ha"e the same dimensions as the halfpolygon abo"e the door, e-cept that they will be full

polygons.

5igure 2?. 7olygon of the Buter 7rism

5igure 2? displays 7olygon , which is the polygon that the walls will be contained

within. 0ach side of 7olygon is appro-imately 4.44 ft.

5igure 2?. ateral 5ace and Joor Jiagram and Measurement


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5igure 2? clearly displays one lateral face of the outer prism of the tower. (he dimensions

of one face are 4.44 ft 8the side of 7olygon : by /.// ft 8the side of 7olygon times , because

each side must be twice as tall as it is wide:.

sin 1./


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5igure 22. ateral 5ace and indow Jiagram and Measurements

(he dimensions of lateral face for the window are the dimensions for the lateral face of

the door, ob"iously. Also, the window has the same dimensions as the top of the door. (his made

further calculations "ery easy.

AreaWindow 14 8F:81.4;:8?.;=:

AreaWindow ≈ 6.85 ft

5igure 24. 5inding the Area of the Window

$ince all the measurements were already found when finding the area of the door, all we

had to do was plug the number in together. Again, we found the area of the polygon in the same

fashion as all the other polygons. (he area of the window is about ;./< ft.

(he last step is to find the lateral surface area of the entire bottom prism of the tower. Bf

course, we also need to subtract the areas for the door and the windows.

$A 1/84.44:8/.//: 81/.4P8;./


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$A =?.; 2.1

LSA ≈ 677.57 ft

5igure 2


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5igure 2=. ateral 5ace Jiagram and Measurements

!n 5igure 2= abo"e, the lateral face diagram and measurements of the inner prisms are

discussed. !ts width is the same as the side length of 7olygon 2 8because that is where the inner

prism lays:, and the height must be the same as the outer prism, otherwise the walls would slant

and unusual things would happen.

6olume!nner 7rism 8/.//:D842.1:

6olume!nner 7rism 1


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*ow it3s time for the roof of the tower. (his will be made of a pyramid, with, again, the

same number of sides as the base polygon. (he height of this outer pyramid must be 2 times the

length of one side of its base. (his outer pyramid will reside in 7olygon .

5igure 2. Buter 7yramid Jiagram and Measurements

5igure 2 displays the base of the outer pyramid, as well as one of the lateral faces of the

pyramid. $ome of the measurements in 5igure 2 area already known, such as the side length

4.44 and the line 7B as in .=2. (hese measurements ha"e already been found when calculating

the measurements of 7olygon . (he "alue of 12.2 can also be easily found by multiplying the

side of the polygon by 2, since the pyramid will be e-actly three times as high as one side of its

base. (he last two "alues, the slant height and angle measure, will take some calculations.

SL = !9.72P12.2:

SL = !94.48 " 117.42#


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SL = 271.90

SL ≈ 16.49 ft

5igure 4?. 5inding the $lant Eeight of the Buter 7yramid

(o find the slant height of the outer pyramid, we used the right triangle that is formed by

the apothem of the base and the height of the pyramid. Qsing the 7ythagorean (heorem, we were

able to calculate the slant height as being about 1;.4 ft.

θ ≈ ta$1812.2@.=:

θ ≈ 53.88%

5igure 41. 5inding the Angle etween the 7rism ase and the 7yramid 5ace

&' ($') all a$gles a$* +a(e ',- +'*el as l'se t' -eal as /'ssible

)a$te* t' g,-e ',t the a$gle bet)ee$ the /-is+s ',te- /-is+ base a$* the

',te- /y-a+i* fae. si$g the -ati' 'f ta$ge$t i$ t-ig'$'+et-y the$ a//lie*

the i$e-se ta$ge$t t' $* the +issi$g a$gle )hih )as al,late* t' be ≈

53.88%.

(ne )ateral 'ace o% the (uter Pyra"id

*ow that we ha"e the needed "alues, we need to find the lateral surface area of the outer

pyramid, as well as a few other things.


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5igure 4. Jiagram and Measurements of Bne ateral 5ace

oth the base and the height of the triangular lateral face ha"e already been found.

Eowe"er, the unknown angle measures should be calculated.

θ ≈ ta$181;[email protected]@::

θ ≈ 82.33%

5igure 42. 5inding the ase Angle of the (riangular ateral 5ace

Qsing in"erse tangent, as well as the measurements we found pre"iously, we were able to

calculate the base angles to be about /.22>.

φ ≈ !ta$188.:@1;.4::

φ ≈ 7.672

φ ≈ 15.33%

5igure 44. 5inding the Angle at the (op of the (riangular ateral 5ace

Bnce again, ! used in"erse tangent to figure out the last angle of the triangular face of the

pyramid. *ow that that3s o"er and done with, it3s time to work out the lateral surface area.

$A 148F:84.44:81;.4:


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Bnce again, the measurements for the base of the pyramid are the same as the

measurements of 7olygon 2. 5inding the height was a easy procedure of multiplying the base by

2.

6olume7yramid 8:811.4:842.1:

6olume7yramid ≈ 967.98 ft2

5igure 4=. 6olume of the !nner 7yramid

Qsing the "olume of a pyramid formula9 6 8:8area of base:8height:, we were able to

find the "olume of the inner pyramid. !t came out to be appro-imately ;=./ ft.

My Tower

$o now e"erything is all done 0"erything is calculated, and all that is left is to put it all

together.


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5igure 4/. (he Completed (ower

!n 5igure 4/, you can clearly see the finished tower in all its glory. *ow it3s time to find

the total surface area and "olume

$A ;==.


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!n conclusion, e"en through all the challenges and difficulties, we would be greatly

grateful if you chose our tower design aside from all the others, still including our pay. (hank

you, from the both of us.

tower final

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