A Mathematical Analysis of Erosion Rates in Giza's Ancient Pyramids Using Geometry and Calculus.

My interest in this exploration of erosion rates in Giza's ancient pyramids started during my visit to the pyramids last summer. At that time, I found myself in an archaeology and history phase, deeply intrigued by everything related to these structures including their historical significance and the preservation challenges faced. Through this exploration I hope to uncover factors contributing to differences in erosion rates and highlight the practical applications of my mathematics knowledge in archaeology by employing mathematical tools like calculus and geometry to analyse erosion rates. This exploration will not only deepen my understanding but also contribute to insights to the broader field of cultural heritage preservation, helping in the development of practical strategies for conserving ancient structures against environmental and anthropogenic challenges. I chose to write this IA instead of my other options as it has strong ties to the real world and can contribute to solving a major challenge of preserving the pyramids.

The aim of this exploration is to employ mathematical tools, specifically calculus and geometry, to conduct an analysis of erosion rates in Giza's ancient pyramids. By focusing on the calculation of volume and surface area, this exploration seeks to provide a deeper understanding of the erosion processes affecting these structures. The primary objective is to compare the erosion rates between two pyramids of Giza; the pyramid of Khafre and the pyramid of Khufu, investigating potential reasons for any variations observed. Through mathematical analysis, this exploration aims to shed light on the preservation challenges faced by these ancient wonders.

This study not only emphasises the practical applications of mathematics in archaeology and historical preservation but also aims to contribute valuable insights to the ongoing discourse on the conservation of cultural heritage. The insights gained from this exploration can inform targeted preservation strategies, guide restoration projects, and contribute to the development of sustainable practices for safeguarding cultural heritage in the face of environmental challenges. Additionally, as cultural heritage extends beyond national borders, the findings of this exploration may serve as a valuable resource for international collaborative efforts aimed at preserving and understanding ancient structures globally.

The Giza pyramids are the biggest and most well-known pyramids in the world. They were built around 2686-2181 BC, during the Old Kingdom, as a tribute to Egypt's Fourth Dynasty pharaohs. (Hemeda, 2020) The environmental conditions surrounding the pyramids, such as climate, weather patterns, and geographical location, contribute to erosion rates of the limestone that was used to build them. (Hemeda, 2020) One estimate suggests that the erosion of the pyramids is approximately 0.1 centimetres per decade. (Thompson, 2023) Studies examining the nature and sustainability of the construction materials reveal that the pyramids' building material display remarkable resilience against weathering and decay, despite having been constructed nearly 5,000 years ago. (Hemeda, 2020)

In this exploration, I will employ mathematical tools, specifically geometry and calculus, to analyse erosion rates in Giza's ancient pyramids. Geometry will be utilised to calculate the volume and surface area of the pyramids, providing essential quantitative data for my analysis. Calculus will then be applied to determine the rate of erosion for each pyramid by examining changes in volume and surface area over time. Geometry serves as the basis for understanding the geometric properties of the pyramids, such as their volume and surface area. By accurately measuring these parameters, I can quantify the extent of erosion affecting the structures.

Calculus enables me to analyse how these geometric properties change over time, allowing the calculation of erosion rates by considering the rates of change in volume and surface area. By integrating geometry and calculus, I can gain deeper insights into erosion dynamics and compare erosion rates between different pyramids at Giza. Can calculus be used to predict future erosion rates based on current data, and if so, how reliable are these predictions? This mathematical analysis will not only enhance our understanding of the preservation challenges facing these monuments but also contribute to the development of effective conservation strategies. I will be using multiple forms of representing mathematical data such as formulas, diagrams and tables to enhance the comprehensibility and visual representation of the exploration.

Several assumptions are made regarding erosion rates in Giza's ancient pyramids. Firstly, it is assumed that erosion occurs uniformly across the pyramid surfaces, despite the potential influence of various factors like weather, environment, and human activities, which may lead to non-uniform erosion patterns. Secondly, it is assumed that the geometric properties of the pyramids remain constant over time, disregarding potential changes due to structural instability or historical modifications, which could affect the accuracy of calculations. Thirdly, uniform material properties throughout the pyramids, such as density and durability, are assumed, neglecting possible variations in construction materials or degradation over time that could result in differing erosion rates across pyramid sections. Lastly, linear erosion rates are assumed over the analysis time frame, simplifying calculations, but ignoring potential nonlinear variations due to environmental fluctuations or preservation/restoration efforts. These assumptions are made in order to simplify the data retrieved and use standard mathematics to tackle the analysis, facilitating easier computation and interpretation of results. However, they may oversimplify the complex reality of erosion processes, potentially leading to inaccuracies in predictions and requiring careful consideration of their limitations.

Additionally, there are a few mathematical assumptions that are made as well it is assumed that the height of the stone layers for both the pyramids is the same and are equal for each layer hence using the original height of the pyramid of Khufu and dividing it by the number of stone layers the height of the stone layer is found and assuming the same height is for the pyramid of Khafre the height of the pyramid originally is divided by the height of the stone layer to find the number of stone layers in the pyramid of Khafre. The calculations are provided further ahead in the exploration.

Finding volume assuming ideal shape using geometry

The volume of an ideal pyramid is given by

\[V=\frac{1}{3} l w h\]

Using values from table 2

\[\begin{gathered}V=\frac{1}{3} \times 230 \times 230 \times 137.5 \\V=2.43 \times 10^{6} \mathrm{m}^{3}\end{gathered}\]

This gives a good reference however it is not the ideal shape as the pyramids of giza are not perfectly triangular but rather have step-like structures on each row.

- I assumed the height of each layer on the real pyramid are equal and is h.
- I also assumed the width of each layer on the real pyramid is equal and is d.

The volume of the highest row is equal to

\[h \times \text{area of row} = h \times(d)^{2}\]

The length of the second row from the top is equal to 3d as seen in the diagram
The area of the second row is equal to (3d)²
Thus, the volume of the second row from the top is equal to

\[h \times \text{area of row} = h \times(3d)^{2}\]

Using the same logic, the volume of the third row from the top is equal to

\[h \times \text{area of row} = h \times(5d)^{2}\]

In order to find the volume of the nth row, it is important to find the length of the side of the nth row.

The above sequence is an arithmetic one with U₁=d and common difference =2d
Hence, the

\[\begin{gathered}U_{n}=d+(n-1)2d \\=d+2dn-2d \\=2dn-d=d(2n-1)\end{gathered}\]

Area of the nth row from the top is equal to [d(2n-1)]²
Volume of nth row from the top is equal to h×[d(2n-1)]²
The volume of the pyramid is equal to

\[\begin{gathered}hd^2+h(3d)^2+h(5d)^2+h(7d)^2+...+h\times[d(2n-1)]^2 \\=h\{d^2+(3d)^2+(5d)^2+(7d)^2+...+[d(2n-1)]^2\} \\=hd^2\{1+3+5+7+...+(2n-1)^2\} \\=hd^2\{\sum_{n=1}^{n}(2n-1)^2\}\end{gathered}\]

\[\begin{aligned}&=hd^2\{\sum_{n=1}^{n}(4n^2-4n+1)\} \\&=hd^2\{4\sum_{n=1}^{n}n^2-4\sum_{n=1}^{n}n+\sum_{n=1}^{n}1\} \\&\sum_{n=1}^{n}n^2=\frac{\lceil n(n+1)(2n+1)\rceil}{6} \\&\sum_{n=1}^{n}n=\frac{n(n+1)}{2} \\&\sum_{n=1}^{n}1=n \\&4\sum_{n=1}^{n}n^2-4\sum_{n=1}^{n}n+\sum_{n=1}^{n}1=4(\frac{[n(n+1)(2n+1)]}{6})-4(\frac{n(n+1)}{2})+n \\&\Rightarrow=\frac{4[n(n+1)(2n+1)]-12(n(n+1))+6n}{6} \\&\Rightarrow\frac{2n[2(n+1)(2n+1)-6(n+1)+3]}{6} \\&\Rightarrow\frac{n[2(n+1)(2n+1)-6n-3]}{3} \\&\Rightarrow\frac{n[2(n+1)(2n+1)-3(2n+1)]}{3} \\&\Rightarrow\frac{n(2n+1)[(2n+2)+3]}{3} \\&\Rightarrow=\frac{n(2n+1)(2n-1)}{3}\end{aligned}\]

(The same applies for the pyramid of Khafre up until this step after which different values will be substituted in the equation)

Substituting n=210

\[\frac{210(2(210)+1)(2(210)-1)}{3}=\frac{210\times421\times419}{3}=12347930\]

Finding the height of each step

\[h=\frac{\text{total height of pyramid}}{210}=\frac{137.5}{210}\approx0.65476\mathrm{m}\]

Since there are 210 rows, each row has two steps at each end of the row

\[\text{Number of steps}=2\times210=420\]

\[d=\frac{\text{length of side of base}}{420}\]

Volume of the pyramid is equal to:

\[\left[0.65476\times\left(\frac{230.3}{420}\right)^2\right]\times12347930=2430899\mathrm{m}^3\approx2.43\times10^6\mathrm{m}^3\]

\[\begin{gathered}A=\frac{1}{2}bh\times4 \\A=\frac{1}{2}\times230.3\times137.5\times4 \\\text{Surface area}=63332.5\mathrm{m}^3\end{gathered}\]

Using Figure 2 as reference for the labelled variables.
The surface area is equal to

\[4\times\text{surface area with real shape}\]

Surface area of the real shape is:

\[\begin{aligned}=hd&+h(3d)+h(5d)+h(7d)+...\text{ up to }210\text{ layers.}\\&=hd(1+3+5+7+...)\text{ up to }210\text{ layers}\end{aligned}\]

The above sequence is an arithmetic one with U₁=1 and common difference d=2

\[\begin{gathered}U_n=U_1+(n-1)d \\U_n=1+(n-1)2 \\U_n=1+2n-2 \\U_n=2n-1\end{gathered}\]

\[\begin{gathered}\text{Sum of this sequence is }\sum_{n=1}^{n}(2n-1)=2\sum_{n=1}^{n}n-\sum_{n=1}^{n}1=\frac{2(n(n+1))}{2}-n \\=n^2+n-n=n^2\end{gathered}\]

\[\text{Area of a side}=hd(n^2)\]

(The same applies for the pyramid of Khafre up until this step after which different values will be substituted in the equation)

Using values previously found, h≈0.65476, d=230.3/420

\[\text{Surface area of one side}=0.65476\times\frac{230.3}{420}\times(210)^2=15833.07894\approx15833\]

\[\text{Surface area of the pyramid}=15833\times4=63332.31576\approx63332\mathrm{m}^2\]

To calculate the rate of erosion using calculus, we can utilise derivatives. First, we need to establish an equation using the data from Table 2, including data from calculations made regarding the surface area and volume. By deriving this equation, we can find the rate of change, which represents the erosion rate.

Given that height is directly proportional to the volume of the structure, the rate of change of height will also be proportional to the rate of change of the volume.

\( h \propto V \)

The primary objective is to find the derivative of the height based on the given data, also known as the rate of change. Subsequently, we will utilise this information to find the derivative of the volume through calculus, utilising data from Table 2.

h_original = 146.7

h_current = 137.5

Time since construction = 2560 BC + 2024 (current year) = 4584

\( \frac{dh}{dt} = \frac{(146.7 - 137.5)}{4584} = \frac{9.2m}{4584 \text{ years}} \)

An equation representing the change in height over the change in time has been formulated. Now, let's revisit the formula for the volume of a pyramid, previously derived through integration. We'll differentiate this equation with respect to time to determine the rate of change of volume. This differentiation process will enable us to find the rate of volume change over time.

\( \begin{array}{c} V = \frac{lwh}{3} \\\\ \frac{d}{dh} \left[ V = \frac{lwh}{3} \right] \end{array} \)

The derivative of \(V\) with respect to time is denoted as \(\frac{d v}{d t}\). For the right-hand side of the equation representing the volume of the pyramid, we can apply the chain rule with respect to \(h\). The derivative of \(\left(\frac{l w h}{3}\right)=\frac{1}{3} l w\), and \(h\) is multiplied by \(\frac{d}{d t}\) outside the square brackets and is also removed when the equation is differentiated because the derivative of \(h^{1}\) is \(h^{0}\).Substituting the values calculated for \(\frac{d h}{d t}\) and the values for \(l\) and \(w\) from table 2 into this equation allows us to determine the value of \(\frac{d v}{d t}\), representing the rate of change of volume with respect to time.

\( \begin{array}{l} \frac{dv}{dt} = \left(\frac{1}{3}l\omega\right) \times \frac{dh}{dt} \\\\ \frac{dv}{dt} = \frac{1}{3}\left(230.3\right)^2 \times \frac{9.2}{4584} \\\\ \frac{dv}{dt} = 35.5 \frac{m^3}{year} \end{array} \)

The same equation derived here will be used ahead in calculating the rate of erosion of the pyramid of Khafre.

The volume of an ideal pyramid is given by

\[V=\frac{1}{3}lwh\]

Using values from table 3

\[\begin{gathered}V=\frac{1}{3}\times210.5\times210.5\times136.5 \\V=2.02\times10^6\mathrm{m}^3\end{gathered}\]

Using the equation derived in the pyramid of Khufu calculations n(2n+1)(2n-1)/3
This was found by assuming both the pyramids have the same height for their stone layers.
Using values from both Table 2 and Table 3.
146.7/210=0.69857m height of each stone layer in the Pyramid of Khufu. (using original height)
143.9/0.69857=205.99≈206 layers of stone in the Pyramid of Khafre. (using original height)

Substitute n=206

\[\begin{gathered}\frac{206(2(206)+1)(2(206)-1)}{3} \\=\frac{206\times413\times411}{3} \\=11655686\end{gathered}\]

Finding the height of each step (using current height)

\[h=\frac{\text{total height of pyramid}}{210}=\frac{136.5}{206}\approx0.66262\mathrm{m}\]

Since there are 206 rows, each row has two steps at each end of the row

\[\begin{gathered}\text{Number of steps}=2\times206=412 \\d=\frac{\text{length of side of base}}{412}\end{gathered}\]

Volume of the pyramid is equal to:

\[\left[0.66262\times\left(\frac{210.5}{412}\right)^2\right]\times11655686=2016100.36\mathrm{m}^3\approx2.02\times10^6\mathrm{m}^3\]

\[\begin{gathered}A=\frac{1}{2}bh\times4 \\A=\frac{1}{2}\times210.5\times136.5\times4 \\\text{Surface area}=57466.5\mathrm{m}^3\end{gathered}\]

Using the equation derived above in the pyramid of Khufu calculations

\[\text{Area of a side}=hd(n^2)\]

Using values previously found, h≈0.66262, d=210.5/412

\[\begin{gathered}\text{Surface area of one side}=0.66262\times\frac{210.5}{412}\times(206)^2 \\=14366.59553\approx14367\mathrm{m}^2\end{gathered}\]

\[\text{Surface area of the pyramid}=14367\times4=57468\mathrm{m}^2\]

Using the equation derived above in the pyramid of Khufu calculations

\( h_{original} = 143.9 \)

hcurrent = 136.5

Time since construction = 2520 BC + 2024 (current year) = 4544

\( \begin{aligned} \frac{dh}{dt} &= \frac{(143.9 - 136.5)}{4544} = \frac{7.4m}{4544 \text{ years}} \\ \frac{dv}{dt} &= \left(\frac{1}{3} l\omega\right) \times \frac{dh}{dt} \\ \frac{dv}{dt} &= \frac{1}{3} \left(210.5\right)^2 \times \frac{7.4}{4544} \\ \frac{dv}{dt} &= 24.1 \frac{m^3}{year} \end{aligned} \)