THE PYRAMID PAPER

Egyptian fractions
A safe guess
A model height
The inch connection
What Herodotus said?
The indents
Done
The Face Areas

Reverse engineering the Great Pyramid

     At most of the high-tech places I have worked, "trouble shooting" was somewhere in the job description. I have always been pretty good at it. I was even good at, what I call, blind trouble shooting. This is where (as an example) I would be led to a big piece of equipment and told that nobody know how it works, they lost the manual and would I please fix it. I griped allot but I usually accepted the challenge.

       Today this type of thing is usually called reverse engineering. When I started to look into the Great Pyramid I soon saw that it was just a big reverse engineering job. But I never thought when I started, that it would take me over 10 years to figure it out.

     Before starting a reverse engineering project you have to collect all the data you can.   There is a valid complaint that most Pyramid studies are not driven by data, but rather by intuition and imagination. So in what's follows I will attempt always to show how each step comes either from the data or a clearly stated assumptions about how to manipulate the data.

Math Starting Point

The purpose of this section is to construct a rational, mathematical three-dimensional model of the exterior of the Great Pyramid at Giza. The starting data is from a 1925 survey first proposed by Ludwig Borchart of the German Institute of Archeology in Cairo, and conducted by J.H.Cole (The first modern surveyor to un-cover the cornerstones). The data is summarized below.

Table 1 Base Lengths And their directions
North	East	South	West
230.253 meters	230.391 meters	230.454 meters	230.357 meters
2' 28" South of due west	5' 30" West of due north	1' 57" South of due west	2' 30" West of due north

The 4 corner angles:
North West	North East	South East	South West
89 59' 58"	90 03'02"	89 56' 27"	90 00' 33"

The chiseled mark at the bottom of the North face
115.090m from North West corner	115.161m from the North East corner

     A major goal now is to find integer ratios associated with the data. The first question is, ratios between what?

     Figure 1 shows some of the line lengths to consider. There are four base lengths, two base diagonals, four face edges, but only one height.

Figure 1

(Assumption 1) An obvious choice then is to look for ratios between the height and the other line lengths.

Egyptian fractions

But the actual height has been lost, and the number of ratios involved could be quite large. There is a way to solve both of these problems at the same time, but just a little background in Egyptian fractions is necessary first.

The early Egyptians had a different way of handling fractions and ratios. Whenever a fraction was involved in a math problem, they would convert it into a sum of fractions, all with a numerator of one. Here are some examples: (if you want to check these, it helps if your calculator has a 1/X key)

The first three are from a papyrus document written about 1700 B.C. The last is one, I made up for practice. These are conversion examples starting with the integer fractions. I'll take another approach, using the survey data and guesses about the height to form ratios, and then converting the ratios to the nearest equivalent Egyptian fraction.

The procedure used for converting is essentially the same as the one provided by J.J. Sylvester, a British mathematician (1814-1897).

A Safe Guess

     Instead of one guess for the height of the Pyramid, I guess a safe low height of 140 meters (much too low!) and a high of 150 meters (much too high!). The actual height is most definitely somewhere within this range.

     With that out of the way, I can form the first ratio.

(Assumption 2) It is between the North base (230.253m) and the low guess for the height (140m), putting the height in the denominator.

     Thus:

230.253 / 140 = 1.644664

     This decimal value will now be converted to a four term Egyptian fraction.

         The first term is obviously 1/1 or 1.

         The best choice for the second term is 1/2.

     The third terms denominator (the numerator is one, of course) is found by adding the first two terms, then subtracting that sum from the original decimal value. Then put one over (invert) the value found to get the third term's denominator.

     So:

1.644664 - (1/1 + 1/2) = .1446642,      1 / .1446642 = 6.91255

     Thus:

230.253/140 = 1 + 1/2 + 1/6.91255

     To get the fourth term, take the integer part of the denominator of the third term; add one to it to get 7 (note that this is different from rounding up). Seven becomes the new denominator of the third term. I then add the three terms and proceed as before.

     So:

1.644664 - (1 + 1/2 + 1/7) = .001807, invert .001807 to get 553.359

     Finally:

230.253 / 140 = 1 + 1/2 + 1/7 + 1/533.359

     The numbers after the decimal point in the denominator of the fourth term have to be there, of course, to keep both sides of the expression equal. They also serve as a fit indicator. They give a way to indicate how well the particular Egyptian fraction fit the ratio that is being tried.

     For now I will look at other ratios without going through the full procedure. The first is the South base (230.454m).

Hence:

230.454 / 140 = 1 + 1/2 + 1/7 + 1/308.370

     This fraction shows why four terms are needed for the data being used. If only three terms were used there would be no difference in the fractions for the South base (longest side) and the North base (shortest side).

     Next I check the ratios of the North and South bases with the 150m heights.

North 230.253 / 150 = 1 + 1/2 + 1/29 + 1/1861.361
South 230.454 / 150 = 1 + 1/2 + 1/29 + 1/532.697

     So now I have tried both height guesses and have checked the longest and shortest base lengths. This gives enough information to make a general statement.

     If there is a set of four Egyptian fractions that equal ratios between respective base lengths and the height, they follow this pattern:

Equation 1

Base length / height = 1 + 1/2 + 1/X + 1/Y

With X between 7 and 29
Y would be different for each base length
But X would be the same for all four base lengths.

Equation 1 is true; of course, if we allow Y to be a decimal expression (as was shown above) but the right side of equation 1 is an Egyptian fraction. So Y is certainly an integer. From now on I'll only deal with ratios that convert to integer Egyptian fractions. I'll also start calling Egyptian fractions by their more modern name, unit fractions, and use the abbreviation UFR for them. So equation 1 becomes by some simple rearrangements:

Equation 2

UFR = base / height

     So back to the task at hand. That task is to find a set of unit fractions that best fit the data. By set I mean four different unit fractions. Each would have the same first three terms. But the fourth term would be different for each.

One way to approach it is to try them all! Try all unit fractions from:

1 + 1/2 + 1/7 + 1/?   To   1 + 1/2 + 1/29 + 1/?

     With the procedure I've been using this amounts to over 40,000 unit fractions to check. Actually it would be four times that number, because I am looking for a set of four to fit the four base lengths.

     The numbers are so large because of the 140 to 150 meters guess done earlier. It's time for a more realistic approximation for the height.

     That range will now be changed to 145 to 148 meters (still a safe guess). The associated range of unit fractions that results is:

1 + 1/2 + 1/13 + 1/100    to   1 + 1/2 + 1/16 + 1/1000

     I will not go into the reasons behind the choice for the fourth terms. For now accept that the values 1/100 to 1/1000 for the fourth term are appropriate for our data and these unit fractions.

     Now I'll make the first "set" of unit fractions. Beginning with:

UFRnorth = 1 + 1/2 + 1/13 + 1/100 = 1.586923

     The important starting assumption is that UFRnorth represents the length of the of the north base. This lets us find UFReast by the using following proportion:

230.391 (east base) / 230.253 (north base) = UFReast / UFRnorth

Then solving for UFReast gives:

UFReast = UFRnorth x 230.391 / 230.253

Changing to decimal:

UFReast = 1.586923 x 230.391 / 230.253
UFReast = 1.587874

Finally convert this value to the nearest unit faction.

UFReast = 1 + 1/2 + 1/13 + 1/91 = 1.587912

For UFRsouth I set up a proportion as before:

230.454 / 230.253 = UFRsouth / UFRnorth

Solve for UFRsouth and convert to nearest unit fraction:

UFRsouth = 1 + 1/2 +1/13 + 1/88

     The west is figured in a similar way. The full set is summarized below.

UFRnorth   = 1 + 1/2 + 1/13 + 1/100    = 1.586923
UFReast    = 1 + 1/2 + 1/13 + 1/91    = 1.587912
UFRsouth     = 1 + 1/2 + 1/13 + 1/88    = 1.588286
UFRwest    = 1 + 1/2 + 1/13 + 1/93    = 1.587675

     This is the first set. The next set starts with UFRnorth = 1 + 1/2 + 1/13 + 1/101. The one after that is 1 + 1/2 + 1/13 + 1/102 etc.

     Then do the same for the unit fractions containing 1/14, 1/15 and 1/16. This will produce a list of 14,400 unit fractions (3,600 sets).

     The thing to find now is a way to rank the sets according to how well they fit the data. The corner angles in the data from Table 1 can help here. Begin by splitting the base into two triangles.

     The values of UFRnorth, UFReast, UFRsouth and UFRwest are taken from a set and combined with the corner angles in trig formulas to find the values of Diag.1 and Diag.2.

     An example using the Cosign law to find diagonal 1:

     If the triangle's sides were originally laid out with this unit fraction set and these corner angles, then Diag.1 would equal Diag.2.

     In fact, the quotient between Diag.1 and Diag.2 (it should equal 1) is what can be used for ranking.   The set whose quotient is closer to one is ranked higher than those whose quotient is further from one.

     But there are two more triangles that can be made. Just split the base in the other direction and use the other two corner angles to find two new diagonals Diag.3 and Diag.4

So if I compute all four diagonals and combine them right, I can come up with an "error of fit" value for the set. Error of fit means that if the set of unit fractions perfectly fit the data then the "error of fit" value would equal zero. Anything less than perfect fit would give a value other than zero.

One way to figure it is to use the following equation:

Equation 3

Error of fit = log (abs (Diag.1 / Diag.2 - Diag.3 / Diag.4))

In words, subtract Diag.3 divided by Diag.4 from Diag.1 divided by Diag.2; take the absolute value and then the log of this difference. This value is called the "error of fit" for the set of unit fractions. Because of the log function, the set with the largest negative value would indicate the best fit; -5 would indicate a better fit than -4.

So the calculations;
(1) Figuring the set of unit fractions,
(2) The four diagonals and
(3) The error of fit,
Were entered into a computer spreadsheet (Lotus 123) in 3600 rows (one row for each set). The results are graphed below. The vertical axis is error of fit and the horizontal axis is height.

Graph1

Graph 1 shows that there are indeed some sets that have a close fit with the data. The first unit fraction of the four best sets is shown on the graph.

But there is only one set that has the best fit. This is the one that I'll investigate in detail.

The Best Set

UFRnorth = 1 + 1/2 + 1/14 + 1/419	= 1.574761905
UFReast = 1 + 1/2 + 1/14 + 1/300	= 1.573815206
UFRsouth = 1 + 1/2 + 1/14 + 1/266	= 1.575187970
UFRwest = 1 + 1/2 + 1/14 + 1/323	= 1.574524547

This set has some very interesting features that I will get into shortly but first some words on modeling.

A model height

Now I come to the question of the height of the Pyramid. When I made the first "set" of four unit fractions an important assumption was that the four base lengths are proportional to the four unit fractions. Because of this, the sets themselves can be considered as part of the specifications of a model.

The best set of unit fractions from graph 1 can thus be thought of as actual base lengths on a pyramid model. The height of this new model is one. This can be shown by rearranging equation 2 to solve for height:

Equation 4

Height = basea / UFRa

     But since the unit fractions are considered as actual base lengths, UFRa would equal basea length so:

Height = UFRa / UFRa = 1

     So from here on when the model is mentioned it is assumed that the model is one high. One what? The interesting answer is one unit of measurement. Interesting because it suggests the possibility that the basic unit of measurement of the Pyramid is the height itself!

     The Pyramid's actual height can also be found by using equation 4 and the data. But the following summary shows that they do not give just one height.

Basea / UFRa = height	mm diff.
230.253 / (1+1/2+1/14+1/419) = 146.302437m	0.019 North
230.391 / (1+1/2+1/14+1/300) = 146.302116m	0.301 East
230.454 / (1+1/2+1/14+1/266) = 146.302539m	0.122 South
230.357 / (1+1/2+1/14+1/323) = 146.302577m	0.160 West

146.302417m Average

     The small differences in the height come from the survey data. The data has only three decimal digit accuracy, while the unit fractions values are known exactly.

     This situation means that the data that lead me to this best set can be left for a while and a new set of model data can be used. First a little about inches.

The inch connection

     A very interesting thing happens when the heights from above are changed from meters to feet and inches. The old conversion formula is:

Length in Meters / .30479974 = Length in Feet

     First the average height:

146.302417 / .30479974 = 479.9952159 = 479' 11 and 15.08/16"

     Then the rest of them:

North = 479 feet 11 and 15.09 / 16

East = 479 feet 11 and 14.89 / 16

South = 479 feet 11 and 15.15 / 16

West = 479 feet 11 and 15.18 / 16

     So this best set indicates that the height of the Pyramid is one sixteenth of an inch less than 480 feet. The first of those most interesting features.

     I did not believe this result at first. I ended up rechecking the calculations and trying three other variations of the "error of fit" equation (equation 3). For me in the end, it came down to one of the following two conclusions:

1. The data from table one is a hoax perpetrated by J.H. Cole or
2. The Egyptian used feet and inches.

     On a trip to Washington D.C. I went to the Congressional Library to try and investigate both of the above. All I could find about J.H. Cole's survey was that he was criticized because his data on the north side mark (115.090 + 115.161) did not match the data for the north base (230.253).

     And on the foot, I found a lot of authors claimed it was a very ancient unit of measurement but none were really sure of how ancient it was or where it really had come from (but most agreed that the foot of today was derived from the British Imperial Yard). So none of this is conclusive. It was only after studying the unit fractions of the "best set that I started to lean toward conclusion 2.

Unit fractions to normal fractions

     The Egyptians used normal fraction also (2/3, 8/7 etc.). But whenever they had to do a calculation with a normal fraction they would change it to a unit fraction first. There turn out to be a simple formula to change the unit fractions (from the best set) to normal fractions (or vice versa).

     I'll start by converting the unit fraction 1+1/2+1/14+1/X to a normal fraction.

1+1/2+1/14+1/X = 11/7+ 1/X = [7+(11*X)]/7X

Hence the unit fraction is equal the normal fraction:

UFRa = [7 + (11*X)] / 7*X

  This means that each of the denominators of the fourth term in the unit fractions of the "best set" can be substituted in turn for X in equation 4 to find the normal fractions that equals it. These calculations are summarized in Table 3

Table 3

Unit fraction		Normal fraction
1+1/2+1/14+1/419 =	(7+11419) / (7419) =	4616 / 2933 North
1+1/2+1/14+1/300 =	(7+11300) / (7300) =	3307 / 2100 East
1+1/2+1/14+1/266 =	(7+11266) / (7266) =	2933 / 1862 South
1+1/2+1/14+1/323 =	(7+11323) / (7323) =	3560 / 2261 West

First note the South fraction's numerator matches the North's denominator.

Before I go further I will explore and expand on a relationship first put forward in English by John Taylor (1781-1864). He said he got it from reading Herodotus (484-425 B.C.) a Greek historian who visited the Pyramid about 440 B.C.

Just what did Herodotus say anyway?

He said something roughly like this 'the faces rise as ... peak ... two'. Taylor interpreted this as; the area of a face of the Pyramid equals the square of the height of the Pyramid (some Egyptologist and math historians say Herodotus never said anything that would translate to anything near what Taylor said). I reserve judgment. Putting it into equation form:

Equation 6

Area face = height²

A Pyramid face is a triangle so its area is:

Area face=1/2 * base * altitude
The altitude of a face is called the apothem. Combine the two equations to eliminate area:

Height² = 1/2 * base * apothem

But I am working with the model dimensions now so the height is equal one and the four base lengths are the fractions from table 3. So:

1² = 1 = 1/2 * base * apothem

Solving for apothem gives:

Equation 7

Apothem = 2 / base

Using the fractions from Table 3 and Equation 7, I found the four apothems.

North ap.	East ap.	South ap.	West ap.
2933 / 2308	4200 / 3307	3724 / 2933	2261 / 1780

The following illustration shows that the number 2933 goes in what could be called a flow between the South and the North (like the Nile).

     Furthermore, the South normal fraction can be reduced, hence:

2933/1862 = (7 * 419) / (7 * 266) = 419/266

And since:

UFRnorth = 1 + 1/2 + 1/14 + 1/419
UFRsouth = 1 + 1/2 + 1/14 + 1/266

     This ties the South and North unit fractions together. With the North being derived from the South (this matches how the nation of Egypt was set up at the time of the pyramid).

     So here is what it comes down to. Using Cole's data and my error of fit analysis produces some very pleasing results (for me). But Taylor's height squared relationship really seems to tie everything together. I believe J.H. Cole could easily have fudged his survey to make the 480-foot connection. But he could not have simultaneously made the mathematics so interesting.

     Now back to the task of making a mathematical model. Unfortunately even knowing mathematical expressions for the bases, apothems and the exact height does not give me enough information to even start constructing a model.

     One critical piece of information is missing. And that is, what point on the area of the base is the peak of the Pyramid directly over? Until this is known with the same exactness as the bases and the height are now known, no model can be built.

     Equation 6 ( Area of face = height² ) can help here also because it implies that the areas of all four faces of the Pyramid are equal (because there is only one height). This gave me a method to find that exact spot on the base (if equation 4 is correct) where the peak was over by using a step and repeat procedure. The procedure goes something like this.

1. Pick any place on the base and position the peak over it
2. Construct lines from the four corners to the peak (defining the faces)
3. Compute the area of each face
4. If they are different, move the peak to some other point on the base
5. Keep moving the peak until all face areas are equal
I automated the procedure on a computer and plotted the result in a 3d graph.

I found that spot that produces a Pyramid with four faces all of equal area. I then computed the lengths of the four lines that go to the four Corners from that spot (this spot I'll refer to as Cent_eq-area). The lengths are listed below.

NW corner	to Cent_eq-area	1.11414
NE corner	to Cent_eq-area	1.11343
SE corner	to Cent_eq-area	1.11322
SW corner	to Cent_eq-area	1.11278

I then converted the values above to the nearest unit fractions. Hence:

NW	1 + 1/9 + 1/ 330.171 =	1.1141398
NE	1 + 1/9 + 1/ 432 =	1.1134259
SE	1 + 1/9 + 1/ 474 =	1.1132208
SW	1 + 1/9 + 1/ 600 =	1.1127777

The unit fraction for the length NW to Cent_eq-area never did work out right. I began to suspect that this length is actually a derived one. It can be derived because the three other unit fractions help to form two exact triangles that can be used to find the NW to Cent_eq-area length.

     The first thing to note in figure 5 is the two triangles on the South and East sides. Their side lengths are all integer expressions (so they can be exact parts of the model) also their placement determines the exact spot of the Cent_eq-area. These two triangles can also determine the rest of the base.

     The way to do it is to consider the North base length (UFRnorth) to be hinged at the NE corner and the West base length (UFRwest) to be hinged at the SW corner. These two lengths could then be rotated as shown (by the dashed line with arrow) until their ends met. The meeting of these lengths would then determine the NW, SW and NE corner angles.

     This also would create the length from the NW corner to the center (dashed line with question mark) and explain why it cannot be written as an easy integer expression (just what I found). I was happy with the above model for the base. It looked like just four more lines and the model for the full pyramid would be done.

The indents

     But while rechecking the math I found a discrepancy between the apothem and the altitude of the two triangles discussed above.

     I had computed the areas of the two triangles using just the sides for the computation. They were:

South base triangle's area = 0.61938
East base triangle's area = 0.619744

Since for a triangle:

Area= base * altitude / 2,

Solving for altitude gives:

South triangle's altitude = 0.786424
East triangle's altitude = 0.787096

     I then noticed that the apothem was the hypotenuse of a right triangle whose other sides were the height and this altitude. But solving for altitude gives:

South =. 782375
East = .782933

     The triangle' altitude does not match the altitude given by the apothem's right triangle (see figure 6).   It shows that there is a choice between apothems. Equation 6 and 7 seem to lead to conflicting results. I tried converting the various decimal numbers (from figure 6) to unit fractions. But none of them were even close to an exact fit (they have to fit exactly because they are part of a right triangle).

The discrepancy (for me) was resolved when I took an overhead view. It was then that I realized that if I use the shortest apothem (1.269689 from figure 6), it would cause the face to be slightly concave. See figure 7.

I say resolved because I had done enough research to know that the faces of the Pyramid actually are slightly concave! This was discovered around 1940 from an aerial photograph.

Fig. 8

This gave me confidence that the model that I am pursuing is right. I went through all the faces and found that equation 7 caused all of them to be slightly concave. It also meant that instead of one big base triangle associated with each face there are two smaller ones (additionally this means that each face is split into a right and left side).

I then made a simple assumption and it took me a long time to realize it was a mistake. The assumption I took was that if I looked at an overhead view of the Pyramid, the line of an apothem would cross the base line at a right angle. When I started to doubt this I wasn't sure what to try next. The picture of the south side shows the problem.

sp1

The fixed triangle with side of h, ap and dtc can be thought to pivot left or right and thus moving the place where the indent appears. I saw that the lines marked lft and rht in the picture is the key. But how do I choose their respective lengths?

sp2

     The solution began to come to me when I heard about one small finding from the archeological diggings at the so called "workers village" at Giza. The finding was that they had found a tomb and the inscription on the walls proclaimed that the man buried there was the boss or lead man of the west side of the Pyramid.

     I started to think about this man and the tremendously hard job he had. He probably had to make sure that all the stones were cut and dressed right, so they would fit correctly in the west side. And when I thought about this guy's job and the problem of finding the lengths of lft and rht, it came to me.

     What I realized is that this man's job would be twice as hard, if he had to have slightly different shaped stones for the left and right side of the west face. And so it just seemed obvious that the proper choices for the lengths of lft and rht are lengths that produce equal slopes on the left and right side. The picture below illustrates this.

After some figuring, I found an easy way to make the slopes of the left and right sides turn out equal. It was simply to adjust the lengths of lft and rht so that the base angles marked R and L are equal.

Done

At first I thought there had to be more unit fractions to complete the model. But it turned out that using just the seven I had found, the height squared relationship, and the indent positioning, is enough to fully specify a model.

I made a 3D drawing using a CAD program, and learned the model has some subtle features. The first was that the concave faces would not show up no mater how big the drawing was.

Fig. 9

In fact it appears that the only way that you could see the concave effect is with something like the sun's shadow (as in figure 8).

For this reason I will use figure 10 (the concave face is exaggerated) to discuss some components of the model.

Fig 10

Summary of fractions:

Name	Unit Fraction	Normal Fraction
UFRnorth	1 + 1/2 + 1/14 + 1/ 419	4616 / 2933
UFReast	1 + 1/2 + 1/14 + 1/ 300	3307 / 2100
UFRsouth	1 + 1/2 + 1/14 + 1/ 266	419 / 266
UFRwest	1 + 1/2 + 1/14 + 1/ 323	3560 / 2261

UFRA	1 + 1/9 + 1/ 600	2003 / 1800
UFRB	1 + 1/9 + 1/ 432	481 / 432
UFRC	1 + 1/9 + 1/ 474	1583 / 1422

     Some other relationships are:

ap_a    = 2 / UFR_a    (equation 7)

dtc_a    = SQRT (ap² - 1)    (distance to center)

     And of course:

Height = 1

     The model is composed of 16 triangles (8 on the base & 8 on the faces). I went about finding the full lengths and angles for those. I hope everyone, by now, understands if you would want the decimal numbers for my model all you have to do is divide all the lengths below by 146.3023 (the height).

Example :
Say you wanted to make a simple 6 foot high pyramid with just four 2x4s with a tenth of a foot accurancy. The table illustrates the steps.

edge	/ height =	model	x new H	= new edge
219.0291	146.3023	1.49710	6	9.0
218.9514	146.3023	1.49657	6	9.0
218.9291	146.3023	1.49642	6	9.0
218.8809	146.3023	1.49609	6	9.0

The tables shows, take the 4 edges nnw thru ssw, divide by height to get the model edge lengths. Multiply those lengths by the new height (6 feet) to get the new edge length. In this case (with the accurancy specified) its 9 feet.

Take 4, nine foot long 2x4s, and drill holes in the ends of each. Tie together with wire or rope. Erect the edges in a square and measure the height. If it is less than 6 feet, move edges in. If more, move out. Continue adjusting until the edges form a square and the height is six feet. Stake the edges to the ground and its done.

I used the above info and Google "SketchUp" to draw a model showing how a shadow falls on the indent on the west side, like Fig. 8. I changed all the lengths from meters to inches. So instead of being 146.3 meters high, it is 146.3 inches (12 feet 2 5/16 inches).

Base Angles

(in picture, for example, 71 indicates row 7 col. 1 value)

Area

     But the most intriguing aspects of the model have to do with area. To begin a discussion on area it will help to think of the model as actually two models, a big one and a small one. The big one is simply the same model with no concave faces (no indents) and the small one has the concave faces as in figure 10.

     I well discuss face area first. Equation 6 says that face area equals height squared. This means of course that each face area should equal one (1² =1). I checked first the face areas of the big model (no concave faces). The result are shown below:

North face	East	South	West
1.0019666	1.00202155	1.001971222	1.0020280353

But the face areas of the small model amazed me. This is found by adding the area of the left and right triangle for each face. The results are:

North face	East	South	West
1.0000081955	1.0000086766	1.0000081774	1.0000086396

     This means that by shaving just a small sliver of material from each face (making them concave) makes their area equal essentially equal one. My first question was could the Egyptians even make such an exact area.

     Each year for thousands of years before the Pyramid the Egyptians had to redraw all the farm boundaries after the Nile flood. The main task was to give each farmer the same area to farm as he had last season. After a thousand years they probably got very good at it. I would say then, that it is reasonable to suppose that when they planned a gigantic national monument, it's various exact areas would be important to them.

     With a renewed respect for area I next calculated the base triangles area. As before, first the big model:

North base triangle	East base	South base	West base
.620255	.619744	.619383	.619895

And the small model (this area is the sum of two triangles):

North	East	South	West
.617071	.616467	.616195	.616618

     Here again comparing the different values from the big and small models is instructive. Anybody that has ever played around with numbers would probably recognize that the constant .6180339887.... (First mentioned by the Greeks) has something to do with the above areas. The four areas from the big model are about 0.3% larger than the constant. And the four areas from the small model are roughly 0.3% smaller than the constant.   This suggests that the Egyptians knew the constant to better than 0.15%. And considering the precision of the face area values (from above) it seems a reasonable assumption.

     Well that seemed to be the end of my quest for a model. The small model with concave faces has enough features (it's height is equal one and each face area is also equal one) to suggest it as the original plan for the Pyramid. It sure has unity going for it!

A simpler model

     It was after considering a simpler model I had second thoughts. The simpler model is a square pyramid with height of one and face area of one. All the base lengths equals 1.5723028 and it turns out the base triangle are exactly 0.6180339... in area (the constant from above which I will call GR [for golden ratio]).

     If the Egyptian could build a complicated model with face area and height equal one. Then they certainly could have built the simpler square pyramid I have described. And if that were true they would know how to construct an area exactly equal to 0.6180339... (Not just to 0.15%).

      Also when considering the volume of the model pyramid, the formula started me thinking. The formula is:

Volume   =   (area of base)*height / 3

     Since the height of the model is one, if the base had slightly more area the formula could be:

VolModel   =   4*GR / 3       0.8240453...

     That would make the face areas, height, base area and volume all have special exact values.

THE END FOR NOW

THE PYRAMID PAPER

Contents

Reverse engineering the Great Pyramid

Math Starting Point

Table 1

Base Lengths And their directions

North

East

South

West

The 4 corner angles:

The chiseled mark at the bottom of the North face