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 double-checked. 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: L2 = x2 + y2. 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, H2 = h2 + L2, 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: α2 = β2 + 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.
2 π h = 8L ó α = 4 / π
and (why not?) β = Φ = (1 + sqrt(5))/2
Φ 2 = 16 / π 2 + 1
And given that Φ is a very nice number, and has the following property: Φ 2 = Φ +1
Φ ≈ 16 / π 2
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 non-right 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/π2 + Φ2 ) / 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)
sin(51,8º) = π /4 => cos (51,8º) = sqrt(1 – (π /4)2) ≈ Φ – 1
(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 N-sided polygons with side 2L , we eventually get to: Θ1 = acos ( (π/N) / tan (π/N) ) and Θ2 = asin ( (π/N) / tan (π/N) )
Θ1 and Θ2 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), Θ1 is 0.
This figure shows how Θ1 (pink ) and Θ2 (blue) (in degrees) behave as a function of N:
But it seems as if all of the N-sided 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 ,
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 internet-published 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 4-pointed 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 North-East 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.
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!
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 ):
Cole Survey’s data
89º 59' 58" = 89,9994º
90º 3' 2" = 90,0505º
89º 56' 27" = 89,9408º
90º 0' 33" = 90,0092º
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 m2
N_A = (114,6757 * 229,1353) / 2 = 13138,1255 m2
NE_A = (81,8015 * 162,048) / 2 = 6627,8847 m2
SW_A = (81,813 * 161,9534) / 2 = 6624,9468 m2
S_A = (114,5138 * 229,1353) / 2 = 13119,577 m2
SE_A = (81,9417 * 162,0167) / 2 = 6637,9619 m2
Total Area = 52790,6509 m2
After above’s data, we can obtain the following values:
(1) Total base perimetre = BP = 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 = Ab = 52790,6509 m2
(4) Total Volume ≈ (Area * height) / 3 = (52790,6509 * 146,3023) / 3 = 2574464,5484 m3
(5) Ratio BP / 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)
(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