CHAPTER 13

Units of Measurement Used in the Great Pyramid

One of the matters of interest about the Great Pyramid, not fully identified, is the unit of measurement used by the ancient Egyptians for the construction of their buildings. Some Egyptologists sustain that they used cubits (distance from the elbow to the end of the center finger in the hand). However, the reality is that in the same way this measurement varies in the hands of human beings, it varies in length in the reference books. Its value goes from about 20 to 25 inches. In accordance with its length, it is known as sacred cubit, royal cubit, profane cubit, etc. It is also known that its measurement varied according to the region or country in which it was used.

The well-known physicist and mathematician, Sir Isaac Newton (1643 -17227), was one of the experts who dedicated time to determine and calculate the unit of measurement used by the Egyptians to build the Great Pyramid. Newton computed from the dimensions of the King’s Chamber a value for the cubit, equivalent to 20.63 British inches. Many years afterwards (1881), the engineer and Egyptologist, Sir Flinders Petrie, after an extensive topographical and careful measuring work at the Great Pyramid, verified the value of 20.632 more or less (.004) inches for the Egyptian cubit.

Other investigators affirmed that the unit of measurement used was the pyramidal inch, equivalent to 1.001 English inches [Ref. #47, Smyth, Piazzi]. The truth is that there is no recognized and confirmed unit of measurement used during the pyramid’s building epoch.

In a certain occasion, in my research, I tried to conceive in my mind a situation that could have happened in ancient times, before the pyramid’s construction. The Egyptian engineers wanted to build a wooden scaled ruler to measure distances in their construction work. The ruler had to be divided in 12 units, as the 12 constellations of the Zodiac. In the same way it occurred to me the idea of using the product of 153 multiply by p, expressed in feet units, to establish the height of the Great Pyramid, in this occasion I had the idea that I should used the product of 153 multiply by f to create the ruler expressed in inches. This was only an idea in my meditation, but to satisfy my curiosity, I made the mathematical calculation. The result indicated a ruler’s length of 247.55920 inches. Dividing the ruler’s length in 12 units, each unit measures 20.6299 inches. This number is so close to 20.63 inches that it can be assumed as 20.63 inches with no harm. As a matter of fact, I did not know if Newton, and Petrie, did the same approximation to the number. The result really shocked me, since both lengths, coincidentally, could be considered equivalent.

This meant that according to Newton and Petrie’s calculations, my ruler would measure exactly 12 cubits. I want to make clear, that I do not pretend to demonstrate that this is the way to establish the value of the cubit, or that is the method used by Newton and Petrie with that purpose. I present these facts only as a feeling I had, and that led me to find this coincidental equivalency in the numbers.

Nevertheless, since my ruler was equivalent to 12 cubits, I considered it was correct. Besides, the equivalencies in the numbers permitted me to continue the development of my idea.

I needed to divide the ruler in 12 equal parts, but I wanted to do it using a geometrical method. In this way, all pyramid's supervisors could prepare their rulers following a geometrical procedure. I needed the length of 153 inches, or its equivalence in any unit of measurement, to design my ruler. The geometrical method I used, is equivalent to mathematically multiplying a number by the value of f, and then, dividing the product in 12 units, or equal parts. The method was valid to be used in their time, and is valid, even in ours.

As shown in figure 210, trace a square (ABCD) having its sides equal to the number desired to multiply by f, in this case, 153 inches, or as been mentioned, a length in any unit of measurement equivalent to 153 inches. From the midpoint between D and C (point E), and radius equal to EB, trace an arc from point B to intersect the extension of the line DC. Mark point F at the intersection. The distance between points D and F represents the product of 153 inches multiplied by f, which would be equivalent to the length of the designed cubit ruler. The next step would be to divide the length in 12 units. Figure 210

Figure 211 shows my geometrical method used to set the 12 units for the ruler. The distance DF in figure 210 is equal to DF, in figure 211. To divide the ruler, proceed as follows: With center in the midpoint between D and F (point O), trace a semicircle from point D to point F. Trace a vertical line through point O and another through point F. Trace the line FQ, which is equivalent to Ö2 (R). Figure 211

From F, project the length of the line FQ, to the vertical line through point F. FQ will be equivalent to FN. Trace OS equal to FN.

The length of FS will be equal to Ö3 (R). Mark point T in the intersection of line FS with the circle circumference. Trace a vertical line from point T to intercept line DF, and mark the intersection M. The segment MF of the line DF is equivalent to one third (1 / 3) of line DF. That is, DF/3. Mark the segment of line MV equal to MF. Segment VD will be also equal to MV and DF. If each of these three segments is divided in four equal parts, which can be easily done, the 12 cubits for the ruler will be obtained, each one equivalent to 20.63 inches, or its equivalent in feet, 1.71916'. Figure 212

I found that the basic unit of 153 inches has its origin in the measurement of 153 feet. If 153 feet is divided in 12 units, and expressed in inches, that is: 153 (12) / (12) = 153 inches /unit. In figure 211, I used the same geometric system as in figure 210, to define the 12 units of 153 inches. To simplify it, what it is been done is to divide 153 feet in 12 equal parts, each one having 153 inches. I designed a ruler where the proportion of its length to 153 inches will be equivalent to p. Therefore; the length of the ruler would be equivalent to 153 f = 247.5592 inches. To be consistent, the ruler is further divided in 12 units, where each unit represents a cubit. That is, 153 f  / 12 units = 247.5592" / 12 = 20.63". In other words, the value of the cubit is 20.63", or equal to 20.63"/12 = 1.719167 feet.

The total length of the ruler, expressed in feet, would be equal to (153")(f ) = 247.5592", or 247.5592" / 12 = 20.63 feet. Note that we have 20.63 inches as the length of the cubit, while the 20.63 feet refers to the total length of the ruler. As comparison: to measure a distance equivalent to the height of the Great Pyramid, it will take a litlle more than 23 straight-line measurements with this ruler.

In the reference book "Cómo se construyeron las Pirámides, Ediciones Tikal, 1989, pages 46-50, Peter Hodges indicates as reasonable, to measure long distances in the construction of the Great Pyramid using two wooden rulers, with metal ends, and having a length of approximately 6 meters. To take the measurements, the rulers are alternated in straight-line positions, until the measurement is completed. He explained that this method would be more accurate than using a cloth tape or the use of only one ruler. The equivalency of a 6 meters ruler, in feet units, would be equal to 19.68', that is, less than one foot, in relation to the 20.63' length of my designed 12 cubits ruler.

Let’s examine some interesting equivalencies in relation to the mentioned numeric relations, in reference to the Great Pyramid.

* The length for the Grand Gallery ceiling is equal to:

153' = (144 / f ) (1.71916) feet = 89 cubits.

* The height of the Great Pyramid would be equal to:

480.66' = 153' (p) = (144) (p / f) (1.71916) feet = 279.59 cubits.

* The apothem measurement is 611.41' = (Öf)(153 p)

= 355.65 cubits

* The base’s sides are equal to 755.75' = 2 (153 p) / Öf

= 439.60 cubits.

* The displacement of 24 feet in the central axis for the corridors, to the east of the face’s center, represents 13.96 cubits.

* The width of the King’s Chamber is 10 cubits, equivalent to 17 feet and 2 inches, while its length of 20 cubits, is equivalent to 34 feet and 4 inches.

Figure 213 shows the King Chamber measurements. From another point, the chamber is rectangular and its width and length are in the ratio 1:2. Therefore, the area of the floor (or ceiling) can be divided in two triangles, whose sides, contrary to the hypotenuse also keep the ratio 1:2. These data indicate that the value of p is present in its geometric configuration, according to my indications in Appendix B (Properties of the right triangle, with sides opposite to the hypotenuse, in the ratio 1:2).

It can be verified that if the length of the diagonal distance between the floor’s opposite corners (or ceiling), is added to the width of the chamber and the result is divided by the chamber’s length, the value of f will result. This is, (38.3858 + 17.1667) / 34.3333 = f =1.618. From another view, if half the width is subtracted from the product between the width and f, it will give the chamber’s height. That is, h = (17.1667)( f) - (17.1667) / 2) = 19.19’.

The chamber height is also equivalent to half the diagonal measurement between the floor’s opposite corners. This is shown in the figure with the distance (gh) = (Ö5)(w)/2) = (Ö5)(17.1667) / 2 = 19.1929’.

Another interesting fact in this configuration is that the inclined plane between the opposite ceiling’s sides and those of the floor creates two triangles that are in the proportion of 3, 4 and 5 in their sides. That is, the sides of the triangles (gbc) and (gcf) in the figure are in the proportion 3, 4 and 5.

DIMENSIONS

Width = w = 17' - 2"

Height = (Ö5 / 2) (w) = 19.1929’

Length = 2w = 34' - 4"

Diagonal bg = 3w /2 = 25.75’

Diagonal cg = 5w /2 = 42.92’

KING CHAMBER MEASUREMENTS Figure 213

Before I finish this chapter, I want to illustrate for the benefit of those interested, how to trace geometrically a distance within a circle which approximates the value of p, that is, when p =1.2 f² (in the Appendix D it is also shown another method to set this approximation) between these two constants.

As shown in figure 214, setting two adjacent and unitary squares, ABCD and DCFG, the value of f² can be easily set. Using the midpoint of side DG (point E) as center, trace an arc from point F to intercept the extension of line AG, at a point identified as O. The distance DO will be equal f = 1.61603, while the distance AO will be equal to (1 + f), which is equivalent to f². Figure 214

If point O is used as the center of a circle and the distance OA = (f²) as its radius, as shown in figure 214 its diameter would be equivalent to f² + f² = 2 f², number which is also equivalent to (f³ + 1). It can be calculated that 0.6 (D), when D = 2 f², is equal to 0.6 (2 f²) = 1.2 f² = 3.141640787, number which differs only by 0.000048133 from the value of p (= 3.141592654...)

To set geometrically a length equivalent to 0.6 of the diameter (D), trace two tangent circles at point B. Trace their horizontal and vertical diameters. Set a line from point K in the circumference of the first circle, to point M in the circumference of the second. Set point W at the intersection of line KM and the circumference of the circle with center L. From point K, using KW as its radius, trace and arc to intercept point C in the same circumference, and trace a line from point C to point D. The length of the line CD will be equal to 0.6 of the diameter of the circle. Therefore, when the diameter is equivalent to 2 f², the length of the line CD will represent the value of 3.141640787, which is an approximate value to the value of p. Figure 215

It can be established that the distance CD = 0.6 (D), by using the Pythagorean theorem (see figure 215). Triangle QDK is rectangular, based on the theorem that states that if the diagonal of the triangle is equal to the diameter of the circle and its sides intersect in a point located in its circumference, the triangle is rectangular. Triangle QDM is also rectangular by complement angles. Therefore, since (QK) = D, then, D² - (KD)² = (QD)² = (QM)² - (DM)². Solving the equation (2 f²)² - (KD)² = [(Ö10)(f)] ² - [(Ö10)(f) - (KD)]². Besides, calculating from the proportional triangles OM/OK = ND/NK = 3f²/f², it is found that ND = 3 (NK), and substituting in the above formula, the length of CD = 0.6 (D) = 0.6 (2 f²) = 3.141640787, which represents a very close approximation to the value of the constant p.