1) Introduction
I have always been interested in the Great
Pyramid and all of the funny and curious things it is supposed to be showing in
a geometrical way. I have read many books and magazines about, and so far I
have been satisfied with this, assuming that the facts that were told in were
correct and doublechecked. But suddenly one amazing thing came to my mind: it
is vastly said that numbers π and Φ (the golden
number) are in the Pyramid. Well, if they were in two
independent parts of it, I would have been happy with that...but
they are not! The Pyramid is a geometrical figure whose dimensions
follow certain mathematical rules and relationships. (see http://www.omsi.edu/teachers/schools/portland/winterhaven/pyramids/
). For instance, the base of each side’s triangle is equal to the length of
this side. Hence, the distance from the centre to the middle point of any of
the sides equals half of the length of the side. This can be seen on this
figure:
This figure shows that Φ is in the pyramid of Cheops. It is also known that the heigth is related
to the base perimetre in a factor of 2 π.. (taken from Jim Loy’s page http://www.jimloy.com/pseudo/pyramid.htm and slightly modified)
Then, the base of the triangle shown in purple is related to the size of the sides of
the Pyramid. This triangle is were Φ is supposed to be
found:
In the Great Pyramid, the ratio between the length of the apothem and
the distance from the centre is Φ.
And this
triangle is a right triangle. So, the height, the side and the
hypotenuse must follow Pythagoras’ rule: L^{2} = x^{2} + y^{2}.
One more relationship between these values.
It is also said that π is found in the Pyramid as:
In the Great Pyramid, the ratio between the total perimetre of the base
and the height is π.
So, there comes another dependency regarding
the side (the perimetre is the sum of the four sides) and the height.
From a mathematical point of view, we have two
variables (height and side) that depend on two formulae...and Pythagoras’ theorem.
Are π and Φ then related?
That is the question I made to myself, because it does not make any sense!
That is why I am currently doing this study.
2) Building a triangle that has π and Φ, somehow:
In a right triangle like this, H^{2}
= h^{2} + L^{2}, as usual. If we want h and H
to be somehow proportional to L (or if somebody wanted so), let:
h = αL and H = βL
That eventually yields to: α^{2} = β^{2} + 1
Now, if somebody wanted (long ago!) to
have α and β set to something special, we should simply replace these variables
with the desired values, and check if the result is true.
For instance,
2 π h = 8L ó α = 4 / π
and (why not?) β = Φ = (1 + sqrt(5))/2
That yields:
Φ^{ 2} = 16 / π^{ 2} + 1
And given that Φ is a very nice number, and has the following property: Φ^{ 2} = Φ +1
We have:
Φ ≈ 16 / π^{ 2}
Or:
Sqrt
(Φ)*π ≈ 4
Isn’t that funny?
I have computed (with Linux C) the value of Sqrt(Φ)*π and I got 3.9961675861352627291 (which is NOT 4, but it looks very
much like!)
I guess the error comes from assuming it is a
right triangle.
So, instead of doing reverse engineering (what
actually friends of paleoufology do) I
decided to design this nonright triangle using TurboCad (some free CAD
programme) with a standarized L = 1. As you can see on the picture, a
nice modern programme has the same errors than the pyramid: it makes a right
triangle! And, by the way, with the famous angle of 51,8º...
Anyway,
3.9961675861352627291 was, is and will always be different than 4.
If we
solve instead mathematically the intersection of those two
circles, so we may be as much accurate as we’d wish to be, let the first circle
(of radius Φ) be
centered at (0,0) and the second (radius = 4 / π ) at (1,0). Hence,
x
= ( 1 – 16/π^{2} +
Φ^{2} ) / 2 = 0,9984475252363
On the
first circle, since cos(α) = x / Φ, we have:
α
= acos(x/Φ) = 51,89718723793º
On the
second, cos(β) = π
(1 – x)/4, then:
β = acos(π (1 – x)/4) = 89,93013861832º ≠ 90º : this is NOT a right triangle!
Then, the
third angle is:
γ = 180  51,89718723793 
89,93013861832 = 38,17267414375º
3) “Circling” a square, regarding perimetre:
Now, let’s try and construct the following
figure:
In a square of side 2L (whose perimetre
is 8L), let’s find the lines starting at its centre and ending at the
very square, whose length R is such that they “behave” as a circle, that
is:
2 π R = 8L. Or, R = 4 L / π
That leads to 8 segments with 8 angles, the
main (between 0 and 90º) ones being 38,2º and 51,8º. This is some
kind of “circling the square”, instead of
“quadrating the circle”. The figure shows this:
This
shows again the shape of the section of the pyramid (for
instance, left upper blue triangle,
whose dimensions are L, 4L/π and (4L/π)cos(51,8º)
).
It is
amazing that: cos(51,8º)
≈ Φ – 1 = 1 / Φ
Actually, acos(Φ – 1) = acos(1 / Φ) = 51,82729237299º.
Besides, 4
sin(51,82729237299) = 3,14460551103 = 4
/ sqrt(Φ) ≈ π. (Well, this obviously comes from the fact that Φ^{ 2} = Φ + 1)
So, since:
sin(51,8º) = π /4 =>
cos (51,8º) = sqrt(1 – (π /4)^{2})
≈ Φ – 1
or
(π /4)^{2} ≈ Φ – 1 = 1 / Φ
we have
the same result than above (1):
Yet,
let’s take a look at the left upper blue
triangle, whose dimensions are L, 4L/π and (4L/π)cos(51,8º), and also at the blue segment that ends at the circle.
The ratio between the radius
(which is also the long side of the triangle) and the base side is 1/cos(51,8º),
which is almost Φ.
The ratio between the base side
and the segment (whose length is (4L/π)(1  cos(51,8º)) ) is cos(51,8º)/(1  cos(51,8º)), which
is also almost Φ.
It sems
as if this angle leads us back to the
golden ratio!
Again,
let’s solve this mathematically, as much accurate as we wish:
From the
above’s figure we can see that the two angles are the intersection of the
circle centered at (0,0) whose radius is 4L/π, with respectivelly, the lines x = L and y =
L.
Hence, sin(α’) = π/4 and cos(γ’) = π/4. Or:
α’ = asin(π/4) = 51,75751851602º
γ’
= acos(π/4) = 38,24248148398º
Obviously,
α and γ are now complementary
angles, so now β’ = α’
+ γ’ = 90º. For these accurate values, let’s
find again the previous ratios:
The ratio between the radius
(which is also the long side of the triangle) and the base side is 1/cos(α’) = 1,615532655169 ≈ Φ
The ratio between the base side
and the segment is cos(α’)/(1  cos(α’)) =
1,624609176462 ≈ Φ
Those angles are not the same than the ones
found on the previous figure (1) ...but they are very similar!
Let’s
then find the absolute errors (difference) of each angle:
E(α)
= α – α’ = 51,89718723793  51,75751851602 =
0,13966872191º
E(β) = β –
β’ = 89,93013861832  90 = 0,06986138168º
E(γ)
= γ – γ’ = 38,17267414375  38,24248148398 = 0,06980734023º
(We all know that: E(α) + E(β) + E(γ) = (α – α’) + (β – β’) + (γ – γ’) = (α + β + γ) – (α’+ β’+ γ’) = 180 – 180 = 0)
Let’s now
compare α and α’ with acos(1 / Φ):
α
– acos(1 / Φ) = 51,89718723793 
51,82729237299 = 0,06989486494º
α’
– acos(1 / Φ) = 51,75751851602 
51,82729237299 =
0,06977385697º
These “errors” are equivalent to E(β) or E(γ) !!!!
(Just for curiosity purposes:
If we extend this analisys to Nsided
polygons with side 2L , we eventually get to: Θ_{1} = acos ( (π/N) / tan (π/N) ) and Θ_{2} = asin ( (π/N) / tan (π/N) )
Θ_{1} and Θ_{2} being the main angles of the segment line whose
length R is such that 2πR =
2NL. Obviously, when N is VERY big (almost a
circle), Θ_{1} is 0.
This figure shows how Θ_{1} (pink ) and
Θ_{2} (blue) (in
degrees) behave as a function of N:
But it seems as if all of the Nsided polygons
have a number of lines (actually 2N in total) of length R that comply
with 2πR = 2NL )
4) “Circling” a square, regarding area:
Another
very curious image is shown when the criteria for finding the lines is the area
instead of the perimetre.
In this
case, R = 2L/sqrt(π), and the image is as follows:
Now there is nothing related to Φ (as far as I may have found), but the blue
lines configure a nice standard Templar/Maltese cross.
5) Actual Great Pyramid data:
So far,
we have been dealing with theoretic ideal figures, but now, let’s go to the
real Pyramid facts:
(Most
data has been taken from:
Terry
Nevin’s http://www.aloha.net/~hawmtn/pyramid.htm ,
http://wwwgroups.dcs.stand.ac.uk/~history/HistTopics/Egyptian_mathematics.html and
http://www.primenet.com/~kjohnson/quickgp.htm
.
Please
take careful note that my goals and aims in this study have nothing to do with
other people’s, especially above’s mentioned T. Nevin, from whose web pages I simply
borrowed the information I needed, not the ideas.
So, I am nor for nor against any other
internetpublished study about.)
These are
the actual dimensions of the Pyramid of Cheops:
This
second and third figures show how the base of the Pyramid is not a
square, but some kind of 4pointed star.
As well
as this photo, taken from a plane on 1940:
Using
TurboCad I have drawn the base of the pyramid, using above’s values (note that DIRECTIONS
are meant CLOCKWISE, rather than the way the standards suggest:
otherwise East would be at the left of North!).
For
making this picture I simply drew the lines ndtc,
wdtc, edtc and stdc with
their corresponding lenghts and angles. Then I made the circle centered at the
top of ndtc with radius ln,
and the circle centered at the right of edtc
with radius re. Obviously, their intersection is the NorthEast corner.
And so forth for the remaining corners...
The resulting values for nw, ne, sw and se
segments match data provided by above’s figure, so I can assume I did
not make any mistake or misunderstanding.
DIRECTIONS:
This
figure shows the calculated DIRECTIONS for the
N, W, S and E lines (from East):
Comparing
to the ones taken from the web, which are relative to North and clockwise:
Ndtc = 90º  (359,9586º  360º) = 90,0414º
Edtc = 89,9083º + 270º = 359,9083º
Sdtc = 179,9673º  90º = 89,9673º
Wdtc = 90º  (269,9587º  360º) = 180,0413º
This
figure shows the calculated DIRECTIONS for the NW,
NE, SE and SW lines (from East):
Note that sw is very close of being 135º!
ANGLES:
This
figure shows in red the big inner angles:
By the way,
the
angular results are different than the ones from Cole Survey in 1925 ( http://www.primenet.com/~kjohnson/quickgp.htm
):

NorthWest 
NorthEast 
SouthEast 
SouthWest 
Cole
Survey’s data 
89º 59'
58" = 89,9994º 
90º 3'
2" = 90,0505º 
89º 56'
27" = 89,9408º 
90º 0'
33" = 90,0092º 
Calculated 
89,4025º 
89,4527º 
89,3436º 
89,4112º 
Nevertheless,
if we draw straight lines (that is, we
neglige the small angle on each side) from each corner to the other, we find
now the same angles than in Cole Survey, as in the figure:
The
reason is very simple: Cole Survey happened on 1925, whilst the discovery
of the small bending on the sides ocurred 15 years later, on 1940,
due to above’s photo!
They
simply assumed in 1925 that the sides were straight lines! And then the numbers
match…
This
figure sows in blue the detailed inner angles:
Note that the angle is exactly 90,3029º on W and E faces, and 90,294xº on N and S!
AREA of the BASE:
For
calculating the area of the base of the Pyramid, we divide it into six triangles:
Whose areas are:
NW_A = (81,9255 * 162,1511) / 2 = 6642,155 m^{2}
N_A = (114,6757 * 229,1353) / 2 = 13138,1255 m^{2}
NE_A = (81,8015 * 162,048) / 2 = 6627,8847 m^{2}
SW_A = (81,813 * 161,9534) / 2 = 6624,9468 m^{2}
S_A = (114,5138 * 229,1353) / 2 = 13119,577 m^{2}
SE_A = (81,9417 * 162,0167) / 2 = 6637,9619 m^{2}
Total Area = 52790,6509 m^{2}
VALUES:
After
above’s data, we can obtain the following values:
(1) Total base perimetre = B_{P}
= ls + rs + ln + rn + lw + rw + le + re = 921,4667 m
(2)
Average
length of each “segment” = 115,1833 m
(3)
Total
Area (from above´s figure) of the base = A_{b} = 52790,6509 m^{2}
(4)
Total
Volume ≈ (Area * height) / 3 = (52790,6509 * 146,3023) / 3 = 2574464,5484 m^{3}
Ratios:
(5)
Ratio B_{P }/ height = 6,2983746667 = 2 *
3,14918733335 ≈ 2π
(6)
Ratio ln
/ (height*π) = 1/3,994825325092 ≈ ¼ ≈ 1/( π sqrt(Φ)) => ln / height ≈ 1 /
sqrt(Φ)
(7)
Ratio le
/ (height*π) = 1/3,990604221564 ≈ ¼ ≈ 1/( π sqrt(Φ)) => le / height ≈ 1 /
sqrt(Φ)
(8)
Ratio ls
/ (height*π) = 1/3,990372094473 ≈ ¼ ≈ 1/( π sqrt(Φ)) => ls / height ≈ 1 /
sqrt(Φ)
(9)
Ratio lw
/ (height*π) = 1/3,985621161301 ≈ ¼ ≈ 1/( π sqrt(Φ)) => lw / height ≈ 1 /
sqrt(Φ)
(10) Ratio rn / (height*π)
= 1/3,989727823475 ≈ ¼
≈ 1/( π
sqrt(Φ))
=> rn / height ≈ 1 / sqrt(Φ)
(11) Ratio re / (height*π)
= 1/3,989142617825 ≈ ¼
≈ 1/( π
sqrt(Φ))
=> re / height ≈ 1 / sqrt(Φ)
(12) Ratio rs / (height*π)
= 1/3,987218536748 ≈ ¼
≈ 1/( π
sqrt(Φ))
=> rs / height ≈ 1 / sqrt(Φ)
(13) Ratio rw / (height*π)
= 1/3,99533579176 ≈ ¼
≈ 1/( π
sqrt(Φ))
=> rw / height ≈ 1 / sqrt(Φ)
(14)
Ratio nap / ndtc = 1,620559750571 ≈ Φ
(15) Ratio eap / edtc = 1,622148170717
≈ Φ
(16) Ratio sap / sdtc = 1,622863727619
≈ Φ
(17) Ratio wap / wdtc = 1,621748418716
≈ Φ
(18) Ratio nnw / nw = 1,345333653985
(19)
Ratio nne / ne = 1,344111118205
(20)
Ratio sse / se = 1,344221781235
(21) Ratio ssw / sw = 1,344461581837
(22) Product (ln / height) * (nap /
ndtc) = (ln * nap) / (height * ndtc) =
1,274433346339 ≈ sqrt(Φ)
(23) Product (rn / height) * (nap /
ndtc) = (rn * nap) / (height * ndtc)
= 1,276061634365 ≈ sqrt(Φ)
(24) Product (le / height) * (eap /
edtc) = (le * eap) / (height * edtc)
= 1,277031871169 ≈ sqrt(Φ)
(25) Product (re / height) * (eap /
edtc) = (re * eap) / (height * edtc)
= 1,277499769847 ≈ sqrt(Φ)
(26) Product (ls / height) * (sap /
sdtc) = (ls * sap) / (height * sdtc)
= 1,277669511454 ≈ sqrt(Φ)
(27) Product (rs / height) * (sap /
sdtc) = (rs * sap) / (height * sdtc)
= 1,278680041607 ≈ sqrt(Φ)
(28) Product (lw / height) * (wap /
wdtc) = (lw * wap) / (height * wdtc)
= 1,278313395081 ≈ sqrt(Φ)
(29) Product (rw / height) * (wap /
wdtc) = (rw * wap) / (height * wdtc)
= 1,275205185185 ≈
sqrt(Φ)
Note that these ratios (20>27)
are actually the ratios of the areas of the corresponding triangles, as well
as the following:
(30)
Ratio ((ln
+ rn)*nap) / ((wdtc + edtc)*height) =
1,277011190357 ≈ sqrt(Φ)
(31)
Ratio ((ls
+ rs)*sap) / ((wdtc + edtc)*height) = 1,277011710688 ≈ sqrt(Φ)
(32)
Ratio ((lw
+ rw)*wap) / ((ndtc + sdtc)*height) = 1,276710824537 ≈ sqrt(Φ)
(33)
Ratio ((le
+ re)*eap) / ((ndtc + sdtc)*height) = 1,276710195568 ≈ sqrt(Φ)
Angles:
(34)
Angle wdtc^wap
= acos(wdtc/wap) = 51,93038573043º
(35)
Angle sdtc^sap
= acos(sdtc/sap) = 51,96122054885º
(36)
Angle edtc^eap
= acos(edtc/eap) = 51,94144399775º
(37)
Angle ndtc^nap
= acos(ndtc/nap) = 51,89746169016º
(38)
Angle nnw^nw
= acos(nw/nnw) = 41,90966940813º
(39) Angle nne^ne = acos(ne/nne) = 41,92791292208º
(40) Angle sse^se = acos(se/sse) = 41,93316454641º
(41)
Angle ssw^sw
= acos(sw/ssw) = 41,94453971902º