THE PYRAMID PAPER
G. Nevin, 1996 - 2009
Reverse engineering the Great Pyramid
At most of the high-tech places I have worked, "trouble shooting" was
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
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
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
Pyramid studies are not driven by data, but rather by intuition and
imagination. So in what's follows I will attempt always to
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
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.
Base Lengths And their directions
28" South of due west
30" West of due north
57" South of due west
30" West of due north
The 4 corner angles:
chiseled mark at the bottom of the North face
from North West corner
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
four base lengths, two base diagonals, four face edges, but only one
|(Assumption 1) An obvious choice
then is to look for ratios between the
height and the other line lengths.
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.
Egyptians had a different way of handling fractions and
Whenever a fraction was involved in a math problem, they would convert
it into a sum of fractions, all with a numerator of
some examples: (if you want to check these, it helps if your calculator
has a 1/X
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.
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.
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
With that out of the way, I
can form the first ratio.
2) It is between the North base (230.253m) and the low guess
the height (140m), putting the height in the denominator.
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.
1.644664 - (1/1 + 1/2) =
.1446642, 1 /
.1446642 = 6.91255
230.253/140 = 1 + 1/2 +
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.
1.644664 - (1 + 1/2 +
1/7) = .001807, invert .001807 to get 553.359
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
The first is the South base (230.454m).
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
in the fractions for the South base (longest side) and the North base
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
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:
Base length / height = 1 + 1/2 + 1/X + 1/Y
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
(as was shown above) but the right side of equation 1 is an Egyptian
fraction. So Y
is certainly an integer. From now on
only deal with ratios that convert to integer Egyptian
I'll also start calling Egyptian fractions by their more modern name,
unit fractions, and use the abbreviation UFR for them. So
becomes by some simple rearrangements:
UFR = base / height
So back to the task at hand. That task is to find a set of
fractions that best fit the data. By set I mean four
unit fractions. Each would have the same first three
But the fourth term would be different for each.
One way to approach it is to try them all! Try all unit
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
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
That range will now be changed
to 145 to 148 meters (still a safe guess). The associated
of unit fractions that results is:
1 + 1/2 + 1/13 +
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
using following proportion:
230.391 (east base) /
230.253 (north base) = UFReast /
Then solving for UFReast gives:
UFRnorth x 230.391 / 230.253
Changing to decimal:
1.586923 x 230.391 / 230.253
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
The west is figured in a
similar way. The full set is summarized below.
= 1 + 1/2 + 1/13 + 1/100
= 1 + 1/2 + 1/13 + 1/91
= 1 + 1/2 + 1/13 +
1/88 = 1.588286
= 1 + 1/2 + 1/13 + 1/93
This is the first set. The next set starts with
1 + 1/2 + 1/13 + 1/101.
The one after that is 1
+ 1/2 + 1/13 + 1/102
Then do the same for the unit fractions containing 1/14, 1/15 and
This will produce a list of 14,400 unit fractions
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
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
closer to one is ranked higher than those whose quotient is further
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
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
that if the set of unit fractions perfectly fit the data then the
"error of fit" value would equal zero. Anything less than
fit would give a value other than zero.
One way to figure it is to use the following equation:
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
of unit fractions. Because of the log function, the set with
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,
entered into a computer spreadsheet (Lotus 123) in 3600 rows (one row
for each set). The results are graphed below. The
axis is error of fit and the horizontal axis is height.
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
shown on the graph.
But there is only one set that
has the best fit. This is the one that I'll investigate in
The Best Set
UFRnorth = 1 + 1/2 + 1/14
UFReast = 1 + 1/2 + 1/14
UFRsouth = 1 + 1/2 + 1/14
UFRwest = 1 + 1/2 + 1/14
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
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
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
model is one. This can be shown by rearranging equation 2 to
solve for height:
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
of measurement. Interesting because it suggests the
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
just one height.
(1+1/2+1/14+1/419) = 146.302437m
(1+1/2+1/14+1/300) = 146.302116m
(1+1/2+1/14+1/266) = 146.302539m
(1+1/2+1/14+1/323) = 146.302577m
The small differences in the
height come from the survey data. The data has only three
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.
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
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:
= 479 feet 11 and 15.09 / 16
479 feet 11 and 14.89 / 16
= 479 feet 11 and 15.15 / 16
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
I did not believe
this result at first. I ended up rechecking the calculations
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
1. The data from table one is a hoax perpetrated by J.H. Cole
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
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
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
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.).
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
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.
= 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
/ (7*419) =
/ (7*300) =
/ (7*266) =
/ (7*323) =
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
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 ...
... two'. Taylor interpreted this as; the area
of a face of
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:
Area face = height2
A Pyramid face is a triangle so its area is:
Area face=1/2 * base *
The altitude of a face is
called the apothem. Combine the two equations to eliminate
= 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 *
Solving for apothem gives:
Apothem = 2 / base
Using the fractions from Table
3 and Equation 7, I found the four apothems.
|2933 / 2308
||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
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
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
of fit analysis produces some very pleasing results (for me).
Taylor's height squared relationship really seems to tie everything
together. I believe J.H. Cole could easily have fudged his
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.
critical piece of information is missing. And that is, what
on the area of the base is the peak of the Pyramid directly
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
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
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
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
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
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
the four Corners from that spot (this spot I'll refer to as
Cent_eq-area). The lengths are listed below.
I then converted the values
above to the nearest unit fractions. Hence:
1 + 1/9 + 1/
1 + 1/9 + 1/ 432
1 + 1/9 + 1/ 474 =
1 + 1/9 + 1/ 600 =
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
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
The first thing
to note in figure 5 is the two
triangles on the South and East
sides. Their side lengths are all integer
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.
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
the dashed line with arrow) until their ends met. The meeting
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
the model for the full pyramid would be done.
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
Since for a triangle:
Area= base * altitude / 2,
Solving for altitude gives:
South triangle's altitude
East triangle's altitude
I then noticed that the apothem was the hypotenuse of a right triangle
whose other sides were the height and this altitude. But
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
there is a choice between apothems. Equation 6 and 7 seem
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).
(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
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.
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).
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
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
sure what to try next. The picture of the south side
shows the problem.
triangle with side of h, ap
can be thought to pivot left
or right and thus moving the place where the indent appears.
saw that the lines marked lft
in the picture is the
key. But how do I choose their respective lengths?
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.
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
thought about this guy's job and the problem of finding the lengths
it came to me.
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
for the lengths of lft
are lengths that produce equal
slopes on the left and right side. The picture below
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
so that the base angles marked R and L are equal.
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
not show up no mater how big the drawing was.
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.
||1 + 1/2 + 1/14 + 1/
||1 + 1/2 +
1/14 + 1/
||1 + 1/2 + 1/14 + 1/ 266
||419 / 266
+ 1/2 + 1/14 + 1/ 323
||3560 / 2261
||1 + 1/9 + 1/
||2003 / 1800
1/9 + 1/
1/9 + 1/
Some other relationships are:
= 2 / UFRa
= 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
do is divide all the lengths below by 146.3023 (the
you wanted to make a simple 6 foot high pyramid with
four 2x4s with a tenth of a foot accurancy. The table
|| / height
|| = new edge
The tables shows, take the 4 edges nnw
divide by height to get the model edge lengths. Multiply
lengths by the new height (6 feet) to get the new edge length.
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
height is six feet. Stake the edges to the ground and its
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).
picture, for example, 71 indicates row 7 col. 1 value)
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
model as actually two models, a big one and a small one. The
one is simply the same model with no concave faces (no indents) and the
has the concave faces as in figure 10.
I well discuss face area first. Equation 6 says that face
equals height squared. This means of course that each face
should equal one (12 =1). I checked
first the face areas of
big model (no concave faces). The result are shown below:
But the face areas of the small model amazed me. This is
adding the area of the left and right triangle for each face.
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
Each year for thousands of years before the Pyramid the Egyptians had
to redraw all the farm boundaries after the Nile flood. The
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
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:
|| South base
And the small model (this area is the sum of two triangles):
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....
by the Greeks) has something to do with the above areas. The
areas from the big model are about 0.3% larger than the
And the four areas from the small model are roughly 0.3% smaller than
the constant. This suggests that the Egyptians knew
constant to better than 0.15%. And considering the precision
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
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
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
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:
(area of base)*height / 3
Since the height of the model
is one, if the base had slightly more area the formula could be:
4*GR / 3
That would make the face
areas, height, base area and volume all have special exact values.
THE END FOR NOW
Terrance G. Nevin, 1996 - 2009