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Mathematics AA HL
Mathematics AA HL
Sample Internal Assessment
Sample Internal Assessment

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Table of content
Rationale
Aim
Research question
Introduction
Different types of skyscraper
Process of calculation
Data analysis
Conclusion
Bibliography

Determination of optimized shape of Skyscrapers considering the cost of cleaning the outer wall be minimum.

Determination of optimized shape of Skyscrapers considering the cost of cleaning the outer wall be minimum. Reading Time
11 mins Read
Determination of optimized shape of Skyscrapers considering the cost of cleaning the outer wall be minimum. Word Count
2,092 Words
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Word count: 2,092

Table of content

Rationale

I love heights- hills, mountains, skywalks, skyscrapers and so on. I feel closer to the clouds. My heart feels lighter when I am at a height. This is where my love for travelling started. I have travelled various places but I love those where either there are mountains, hills or skyscrapers. I have read various books about them as they intrigue me the most. Also, I take immense interest in seeing my brother work. He is a civil engineer. So, whenever he gets a skyscraper work, I try following the process. Seeing him work, my curiosity grew. I started researching more about the skyscrapers and gradually acquiring much knowledge about them. Last week, I heard my brother discussing on painting a skyscraper. Hearing the discussion, a thought kept ringing in my mind. I wondered if the cost of painting had a relationship with the structure of the skyscraper.

 

Till then whatever my questions were, I would research and find the answers to all my doubts but this was the first time I could not manage to get the answer to my question. I looked for all sources available, but in vain. As a result, I decided to find it out myself.

 

This IA is about the same. In this IA, I have tried to derive the relationship between the cost of painting the outer surface of a building and the shape of the building.

Aim

This IA aims at discovering the optimal shape of a conventional skyscraper, considering the holding capacity or the number of individuals it may accommodate and height be comparable, such that the cost of painting, polishing or cleaning the outer wall be minimum.

 

We would be obtaining the desired results through the use of concepts like calculus and integration for the calculations of surface areas. This will enable us to determine the optimal shapes and designs of skyscrapers to be constructed in the future as per our requirements.

Research question

What should be the optimized shape of skyscrapers considering the cost of cleaning the outer wall be minimum?

Introduction

Skyscrapers are marvels of architecture. Standing tall in the midst of the cityscape, they not only prove the limits of modern technologies but also set outstanding examples for further improvement. Once constructed, the large amount of money spent in erecting the structure, gets further increased when the cost of maintenance is added periodically. Although the internal buildup of these skyscrapers entails regular and invariant costs, the surface on the outside can considerably reduce the cost of cleaning these surfaces.

 

In this IA I have started by finding the various surface areas of some differently shaped, notable skyscrapers of tentatively equal height and capacity such that the volume or internal capacity of the skyscrapers remains the same. I have worked on four such skyscrapers and using the Desmos software, I have found out the equations for the curve of the surfaces of these skyscrapers. Finally, by integrating these equations using appropriate limits, I have been able to find out the areas under the graphs and by summing up the sectional areas, I have been able to find the total surface area of the chosen buildings. With this data, I have determined the optimal shape of a skyscraper such that it will have a minimal surface area facilitating the reduction in the cost of cleaning. This result will be beneficial to many in the future so that stable skyscrapers can be built while ensuring minimum cost of cleaning or polishing the outer surface.

Different types of skyscraper

While a skyscraper is defined as any building over 40 floors or over 150m, currently the term is clearly used to define high-rises of over 100 floors having a height of several hundreds of meters. Although the shape and size of skyscrapers vary considerably, points like stability, resistance to strong winds or earthquakes, and durability form the basis of any high-rise structure. Moreover, every skyscraper is decided to have such a shape that will help it maintain these points so choosing the shape that will minimize cost of cleaning can be done by finding the shape with minimum surface area.

Figure 1 - Different Types Of Skyscraper
Figure 1 - Different Types Of Skyscraper

Several of these supertall structures have already been erected along with 132 in construction currently. In this IA, I have taken the following skyscrapers and calculated their surface areas:

  • The John Hancock Center:

It is a 100 story, 1128 ft tall building situated at 875 North Michigan Avenue, Chicago. Its name was changed to 875 N. Michigan Ave. on 12th February 2018.

  • The Jin Mao Tower:

It is an 88 story, 420.5 metres tall landmark building in Shanghai. It contains shopping malls, offices and the Hyatt Shanghai Hotel, which at the time of completion was the highest hotel in the world.

  • The Shanghai World Financial Center:

It is a 101 story, 494m tall skyscraper situated at 100 Century Ave Lu Jia Zui, Pudong, Shanghai, China. It is the second tallest skyscraper in the city.

  • The Ping’an International Financial Center:

It is a 115 story 599m tall in Futian District, Shenzen, Guangdong, China. The Ping An Insurance commissioned the building and the American firm Kohn Pedersen Fox Associates designed it.

  • The One World Trade Center:

It is a 94 story, 1776ft tall skyscraper situated at 285 Fulton St, New York United States. It was built in the place of the Twin Towers which were destroyed during the 9/11 attack.

Figure 2 - John Hancock Center, Jin Mao Tower, Shanghai World Financial Center, One World Trade Center (From Left To Right)
Figure 2 - John Hancock Center, Jin Mao Tower, Shanghai World Financial Center, One World Trade Center (From Left To Right)

Process of calculation

Prior information / certain considerations

I have used the formula of surface area of cylindrical bodies despite there are two buildings which are not cylindrical in shape. This is because of a certain reason which I have explained below:

 

Firstly, In this IA, two skyscrapers are cuboidal in shape and the other two are having an irregular shape. These two buildings which are having an irregularly shaped outer surface are having an approximately circular base. Moreover, most of the region of the outer surface is curved in nature. Thus, unlike simple formula of surface area of cuboid, it is not possible to simply multiply the dimensions of each face of the skyscraper to find the surface area. Rather, tracing the curve is necessary due to the irregularity in finding the surface area. If surface area of two buildings are found using the formula of surface area of cylinder, it is necessary to find the surface area of the other two buildings using the same formula to maintain a uniformity in the magnitude of the values. Secondly, as there is no comprehensive data on dimension of every edge, face, curve, corner etc. of the buildings, it is mandatory to make certain assumptions. In this IA, we have considered the buildings to be cylindrical in shape to nullify the error to the maximum extent as the two major buildings are curve shaped. On the other hand, any building is not perfectly cuboid due to manufacturing of sharp edges in case of high-rise towers is not possible. Thirdly, in case of any optimization study, the trendline of the dependent variable is more important than the exact magnitude of the dependent variable. This is because with a change in procedure of determination of surface area, there will be difference in errors but that will be proportional as well which will not affect the nature of the graph or trendline of the dependent variable, which is, in this case, the outer surface area of the skyscrapers.

Calculation

In this IA, I am going to calculate the total surface areas of each of the above-mentioned skyscrapers. To achieve this, I have placed the image of the Skyscrapers in the Desmos web application to obtain curves of the surfaces of the buildings using Hit and Trial Method. It is apparent that since the surfaces are uneven, we have to consider separate curve equations and integrate them by using proper limits. At the end we compare the summation of partial surface areas calculated by integration and find out the minimum surface area.

The john hancock center

Figure 3 - We Place An Image Of The John Hancock Center In Desmos
Figure 3 - We Place An Image Of The John Hancock Center In Desmos

From the graph we observe that the equation for the curve within the above-mentioned limits, i.e., from x = 0 to x = 0.1

 

y = 0x + 0.85

 

We can say that y = f(x) so we use the formula for Surface Area as:

 

Surface Area = \(\displaystyle\int^\limits b_a2πydB\)

 

\(dB=\sqrt{1+(\frac{dy}{dx})^2}\ dx\)

 

Thus, by differentiating y with respect to x,

 

\(\frac{dy}{dx}=0\)

 

\(dB=\sqrt{1+(0)^2}dx\)

 

\(∴Surface\ Area=\displaystyle\int\limits^{0.1}_02\pi(0x+0.85)(\sqrt{1+(0)^2}dx\)

 

∴Surface Area ≈ 0.5341

 

I have used an online integration tool – Integral Calculator for the calculation of integrals:

Figure 4 - Integration Calculation
Figure 4 - Integration Calculation
Figure 5 - We Will Follow The Same Steps That We Have Followed In Section I, For Section II
Figure 5 - We Will Follow The Same Steps That We Have Followed In Section I, For Section II

From the graph, we find that the equation for the curve within the above-mentioned limits, i.e., from x = 0.1 to x = 7.31, is:

 

y =- 0.001x- 0.019x + 0.55

 

Thus by differentiating y with respect to x we get:

 

\(\frac{dy}{dx}\) = -0.002x - 0.019

 

\(∴dB=\sqrt{1+(-0.002x-0.019)^2}dx\)

 

∴ Surface Area = \(\ \displaystyle\int\limits^{7.31}_{0.1}2\pi(-0.001x^2-0.019x+0.55)\)

 

\((\sqrt{1+(-0.002x-0.019)^2}dx\)

 

∴ Surface Area ≈ 20.91

 

I have used an online integral calculator for calculating the integrals:

Figure 6 - Integration Calculation
Figure 6 - Integration Calculation
Figure 7 - We Will Follow The Same Steps That We Have Followed In Section I
Figure 7 - We Will Follow The Same Steps That We Have Followed In Section I

From the graph, we find that the equation for the curve within the above-mentioned limits, i.e., from x = 7.31 to x = 7.43, is:

 

y = 0.18

 

Thus by differentiating y with respect to x we get:

 

\(\frac{dy}{dx}=0\)

 

\(∴dB=\sqrt{1+(0)^2}dx\)

 

∴ Surface Area = \(\ \displaystyle\int\limits^{7.43}_{7.31}2\pi(0.18)(1)dx\)

 

∴Surface Area ≈ 0.135

 

I have used an online integral calculator for calculating the integrals:

Figure 8 - Integration Calculation
Figure 8 - Integration Calculation
Figure 9 - We Will Follow The Same Steps That We Have Followed In Section I
Figure 9 - We Will Follow The Same Steps That We Have Followed In Section I

From the graph, we find that the equation for the curve within the above-mentioned limits, i.e., from x = 7.43 to x = 9.43, is:

 

y = 0.04

 

Thus by differentiating y with respect to x we get:

 

\(\frac{dy}{dx}=0\)

 

\(∴dB=\sqrt{1+(0)^2}dx\)

 

\(∴Surface\ Area=\displaystyle\int\limits^{7.43}_{7.31}2\pi(0.04)(1)dx\)

 

∴ Surface Area ≈ 0.502

 

I have used an online integral calculator for calculating the integrals:

Figure 10  - Integration Calculation
Figure 10 - Integration Calculation

Therefore, total surface area of this building = 0.534+20.91+0.135+0.502 = 22.081 sq. unit

 

Height of John Hancock Building = 457 m

 

Therefore, Total Surface Area = 22.081 × \(\frac{457}{9.43}\) = 1070.09 sq. metre

Shanghai world financial center

Figure 11 - We Place An Image Of The Shanghai World Financial Center In Desmos
Figure 11 - We Place An Image Of The Shanghai World Financial Center In Desmos

From the graph, we find that the equation for the curve within the above-mentioned limits, i.e., from x = 0 to x = 9.4, is:

 

y = - 0.011x+ 0.018x + 0.79

 

Thus by differentiating y with respect to x we get:

 

\(\frac{dy}{dx}\) = - 0.022x + 0.018

 

\(∴dB=\sqrt{1+(-0.022x \ +\ 0.018)^2}dx\)

 

\(∴Surface\ Area=\displaystyle\int\limits^{9.4}_{0}2\pi(-0.011x^2\ +\ 0.018x\ +\ 0.79 )\)

 

\((\sqrt{1+(-0.022x\ +\ 0.018)^2})dx\)

 

∴Surface Area ≈ 32.710

 

I have used an online integral calculator for calculating the integrals:

Figure 12 - Integration Calculation
Figure 12 - Integration Calculation

Therefore, Total Surface Area of Shanghai WFC is 32.710 sq. units

 

Height of Shanghai WFC = 494 m

 

Therefore, Total Surface Area = 32.710 × \(\frac{494}{9.4}\) = 1719.01 sq. metre

Jin mao tower

Figure 13 - We Place An Image Of The Jin Mao Tower In Desmos
Figure 13 - We Place An Image Of The Jin Mao Tower In Desmos

From the graph, we find that the equation for the curve within the above-mentioned limits, i.e., from x = 0 to x = 8.4, is:

 

y = 0.71

 

Thus by differentiating y with respect to x we get:

 

\(\frac{dy}{dx}=0\)

 

\(∴dB=\sqrt{1+(0)^2}dx\)

 

\(∴Surface\ Area=\displaystyle\int\limits^{8.4}_{0}2\pi(0.71)(1)dx\)

 

∴Surface Area ≈ 37.47

 

I have used an online integral calculator for calculating the integrals:

Figure 14 - Integration Calculation
Figure 14 - Integration Calculation
Figure 15 - We Will Follow The Same Steps That We Have Followed In Section I
Figure 15 - We Will Follow The Same Steps That We Have Followed In Section I

From the graph, we find that the equation for the curve within the above-mentioned limits, i.e., from x = 8.4 to x = 8.89, is:

 

y = - x + 9.08

 

Thus by differentiating y with respect to x we get:

 

\(\frac{dy}{dx}=-1\)

 

\(∴dB=\sqrt{1+(-1)^2}dx\)

 

\(∴Surface\ Area=\displaystyle\int\limits^{8.89}_{8.4}2\pi(-\ x\ +\ 9.08)(\sqrt2)dx\)

 

∴Surface Area ≈ 1.89

 

I have used an online integral calculator for calculating the integrals:

Figure 16 - Integration Calculation
Figure 16 - Integration Calculation
Figure 17 - We Will Follow The Same Steps That We Have Followed In Section I
Figure 17 - We Will Follow The Same Steps That We Have Followed In Section I

From the graph, we find that the equation for the curve within the above-mentioned limits, i.e., from x = 8.89 to x = 9, is:

 

y = 0.19

 

Thus by differentiating y with respect to x we get:

 

\(\frac{dy}{dx}=0\)

 

\(∴dB=\sqrt{1+(0)^2}dx\)

 

\(∴Surface\ Area=\displaystyle\int\limits^{9}_{8.89}2\pi(0.19)(1)dx\)

 

∴Surface Area ≈ 0.131

 

I have used an online integral calculator for calculating the integrals:

Figure 18 - Integration Calculation
Figure 18 - Integration Calculation
Figure 19 - We Will Follow The Same Steps That We Have Followed In Section I
Figure 19 - We Will Follow The Same Steps That We Have Followed In Section I

From the graph, we find that the equation for the curve within the above-mentioned limits, i.e., from x = 9 to x = 9.11, is:

 

y = 0.14

 

Thus by differentiating y with respect to x we get:

 

\(\frac{dy}{dx}=0\)

 

\(∴dB=\sqrt{1+(0)^2}dx\)

 

\(∴Surface\ Area = \displaystyle\int\limits^{9.11}_{9}2\pi(0.14)(1)dx\)

 

∴Surface Area ≈ 0.096

 

I have used an online integral calculator for calculating the integrals:

Figure 20 - Integration Calculation
Figure 20 - Integration Calculation
Figure 21 - We Will Follow The Same Steps That We Have Followed In Section I
Figure 21 - We Will Follow The Same Steps That We Have Followed In Section I

From the graph, we find that the equation for the curve within the above-mentioned limits, i.e., from x = 9.11 to x = 9.45, is:

 

y = 0.12

 

Thus by differentiating y with respect to x we get:

 

\(\frac{dy}{dx}=0\)

 

\(∴dB=\sqrt{1+(0)^2}dx\)

 

\(∴Surface\ Area=\displaystyle\int\limits^{9.45}_{9.11}2\pi(0.12)(1)dx\)

 

∴Surface Area ≈ 0.256

 

I have used an online integral calculator for calculating the integrals:

Figure 22 - Integration Calculation
Figure 22 - Integration Calculation

Total Surface area of Jin Mao Tower = 37.47 + 1.89 + 0.131 + 0.096 + 0.256 = 39.843

 

Height of Jin Mao Tower = 420 m

 

Therefore, Total Surface Area 39.843× \(\frac{420}{9.45}\) = 1770.8 sq. metre

World trade center

Figure 23 - We Place An Image Of The World Trade Center In Desmos
Figure 23 - We Place An Image Of The World Trade Center In Desmos

From the graph, we find that the equation for the curve within the above-mentioned limits, i.e., from x = 0 to x = 1, is:

 

y = 0.72

 

Thus by differentiating y with respect to x we get:

 

\(\frac{dy}{dx}=0\)

 

\(∴dB=\sqrt{1+(0)^2}dx\)

 

\(∴Surface\ Area=\displaystyle\int\limits^1_02\pi(0.72)(1)dx\)

 

∴Surface Area ≈ 4.52

 

I have used an online integral calculator for calculating the integrals:

Figure 24 - Integration Calculation
Figure 24 - Integration Calculation
Figure 25 - We Will Follow The Same Steps That We Have Followed In Section I
Figure 25 - We Will Follow The Same Steps That We Have Followed In Section I

From the graph, we find that the equation for the curve within the above-mentioned limits, i.e., from x = 1  to x = 7.4, is:

 

y = - 0.05x + 0.77

 

Thus by differentiating y with respect to x we get:

 

\(\frac{dy}{dx}\) = - 0.05

 

\(∴dB=\sqrt{1+(-0.05)^2}dx\)

 

\(∴Surface\ Area=\displaystyle\int\limits^{7.4}_{1}2\pi(-0.05x\ +\ 0.77)(\sqrt{1+(-0.05)^2)}dx\)

 

∴Surface Area ≈ 22.54

 

I have used an online integral calculator for calculating the integrals:

Figure 26 - Integration Calculation
Figure 26 - Integration Calculation
Figure 27 - We Will Follow The Same Steps That We Have Followed In Section I
Figure 27 - We Will Follow The Same Steps That We Have Followed In Section I

From the graph, we find that the equation for the curve within the above-mentioned limits, i.e., from x = 7.4  to x = 7.6, is:

 

y = 0.39

 

Thus by differentiating y with respect to x we get:

 

\(\frac{dy}{dx}=0\)

 

\(∴dB=\sqrt{1+(0)^2}dx\)

 

\(∴Surface\ Area=\displaystyle\int\limits^{7.6}_{7.4}2\pi(0.39)(1)dx\)

 

∴Surface Area ≈ 0.490

 

I have used an online integral calculator for calculating the integrals:

Figure 28 - Integration Calculation
Figure 28 - Integration Calculation

Total Surface Area of World Trade Centre = 4.52+22.54+0.49 = 27.55 sq. units

 

Height of World Trade Centre = 541 m

 

Therefore, Total Surface Area = 27.55 × \(\frac{541}{7.6}\) = 1961.12 sq. metre

Data analysis

In this section of the IA, I will establish an analytical reasoning about how the surface area depends on shape and how can it be optimised based on the data that has been deducted and calculated in the previous section.

 

The table given below comprises the outer surface area of the skyscrapers as obtained from the process of calculation along with the year of its construction.

Year of construction
Building
Surface Area (in square metres)
1968
John Hancock Center
1070.09
1997
Shanghai WFC
1719.01
1994
Jin Mao Tower
1770.80
2012
One WTC
1961.12
Figure 29 - Table On Surface Area And Year Of Construction For The Skyscrapers Chosen
Figure 30 - Bar Graph Showing The Surface Area Of The Sky Scrapers Taken
Figure 30 - Bar Graph Showing The Surface Area Of The Sky Scrapers Taken

From the graph, we can see that the outer surface area of the building is evidently increasing over the years of construction though the height  is approximately equal or lies within a particular range. Furthermore, from the skyscrapers, we can assume that the surface area is more dependent on the intensity of curvature of the building than it’s holding capacity. The skyscrapers with cuboidal shape are of lesser surface area in comparison to the skyscrapers having a curved outer surface. On the other hand, those with more curves in the building.

Conclusion

In this IA,  the outer surface area of four skyscrapers from different countries have been investigated. In this case study of analysing surface area of skyscrapers or architecture, the holding capacity or the number of individuals the buildings can accommodate without any haste as well as it’s height, is very important. An assumption was made that the holding capacity of all the four buildings are constant in order to compare the surface area. The four skyscrapers taken are also having comparable height.

 

However, the analytical study has shown different inferences. From the above study, we have seen that the skyscrapers with more curved outer surface has comparatively more surface area than that of the skyscrapers with cuboidal shaped surface. This doesn’t satisfy the surface to volume ratio of curved objects. Surface area to volume ratio of any curved object is always less than that of the objects with no curvature or less curvature in the outer surface. But, it is seen to be contradicted in this case. One World Trade Centre and the Shanghai World Financial Centre are two skyscrapers that are considered in this IA. Both of the skyscrapers’ outer surface area has come out to be more than that of the John Hancock Centre or the Jin Mao Tower which are cuboidal in shape. As the height of each skyscraper are almost constant and the holding capacity being assumed to be constant, and the outer surface area must vary with respect to the shape of the skyscrapers. However, in this particular mathematical exploration, it is found that the outer surface area is varying with respect to the time of construction.

 

Jin Mao Tower, having a cuboidal outer surface is having greater surface area than that of John Hancock Centre which is again a cuboidal shaped skyscraper of approximately equal height and holding capacity. This statement also validates the assumption of the previous paragraph that the surface area is dependent on time of construction because Jin Mao Tower was built in the year of 1994 which is 26 years after the construction of John Hancock Tower.

 

Similarly, surface area of One WTC is more than Shanghai WTC despite of the fact that both are curved shaped skyscrapers with similar type of curve.

 

Hence, I can suggest that surface area is increasing with passing years and does not have direct or inverse relationship with the shape of the skyscrapers. It is possible that with passing years, the amenities and building unnecessary structures have increased significantly in the skyscrapers resulting in an increase in the overall structure, i.e., volume of the skyscrapers without increasing the holding capacity of the building.

Bibliography