Finding both Phi and Pi within the structure of The Queen's

[Don Barone]

Hi again all. It never fails. Just when I think I am too old and too tired to come up with anything new at Giza something smacks me in the head. Case in point are these websites.

Queen's Chamber #1

Queen's Chamber #2

Queen's Chamber - Niche

The Mathematics of The Giza Plateau

I am not sure if I had seem it before but I was amazed this time that a hexagon fit perfectly within the structure of The Queen's Chamber.  Not only this but it appeared that Phi was easily able to be found.

Phi is outlined in blue.

However what I have "discovered" within the context of all these websites is not on any website I have ever looked at so I am going to presume that it is an original discovery of yours truly.

Every once in a while I find something that makes me giggle with delight at what I have manged to find and this is one of them. As a preface to this I had mentioned this fact I had noted before and many thought it was quite an interesting observation. Others of course ignored it but it has come back with a vengeance to haunt them.  It has to do with The Queen's Chamber and an observation made by me regarding the overall build of the structure. I am sure, like me, many of you have wondered why a particular angle was chosen for the angled large megalithic stones that make up the roof of The Queen's Chamber. It is an amazing observation and I am sure you will never look at this structure and the two participating angles the same way again.

So without fanfare I will simply post the images and quote the posting i made at the Ma'at Forum

The things to note here are how the horizontal lines of the "star" seem to match precisely the stones used in the construction on the sides. Phi also seems to fit from the centre line to the edges of the niche. On this other site it tells us that "water" would be 20 centimeters high to make it fit perfectly. [www.nielsbjerre.dk]

Now 20 centimeters is equal to 20/2.54 = 7.8740157480315 inches. This as a ratio of inches in a cubit of 20..62 inches is equal to 20.62 / 7.8740157480315 and we get 2.61874 almost exactly Phi squared or 2 x Phi. Exact would be 20.62 / 2.618033988 = 7.876139152 (rounded off) (0.9997)

October 11, 2023 07:55PM Registered: 19 years ago
Posts: 7,303

Some interesting work being done.

Here is an interesting site on "The Queen's Chamber".

The person has used Hexacubits which is simply 1/6th of a cubit (the person has used 20.625 inches per cubit) since a hexagon had 6 sides.

The thing I would like to point out is that the angles are, starting in bottom left corner; 90 degrees, 121 degrees, 118 degrees, 121 degrees and finally 90 degrees again totaling 540 degrees.

If we want to calculate the angle at the top we get 121 - 90 and we get 31 degrees.

Now astonishingly the sine of 31 degrees is 0.51503807491005421008163193639814. So what you might ask but this number is, believe it or not, almost the exact value of PHI DIVIDED BY PI or 1.6180339887498948482045868343656 / 3.1415926535897932384626433832795 and equaling 0.51503621480048386500726510157442 and is the sine for 30.999875664461018685262468015873 and checking to 31 degrees to an degree of accuracy of 0.999996 (rounded off). The opposite angle would be 59 degrees with a sine of 0.85716730070211228746521798014476 and the reciprocal of this number is 1.1666333972153304924812531274131 or 1/6th of 7.0 (6.9998...)

A graphic illustration of what I have found:

A second image:

And here is a little larger image:

So amazingly in the design of the roof of The Queen's Chamber both Phi and Pi have been used together to give the angle of the roof.  To me I have to admit even after all that I have discovered this just totally blows my mind.

db

[Don Barone]

I have to be a bit more careful and selective on the sites I go to and take to heart.


Example of what I mean.

Among all measurements not corresponding we have the angle of the roof at 30 degrees 26 minutes quite at bit off from 31 degrees.

Hmmm

[Don Barone]

The angle according to Petrie and others [donbarone.altervista.org] IS NOT 31 degrees. When this fails the house of cards comes crashing down.

If height is 9 cubits and width is 10 cubits and apex is 12 cubits the best we can get is 3/5 = 0.6 and angle that has a tan of 0.6 is 30.964 (rounded off)

Unfortunately Petrie does not agree with any of this. Here is what he says about the angle of the supporting megalithic blocks:

As you can see none of the values check. Petrie has the slope in his first example at 120 inches but that would mean (using 20.62 inches per cubit) that the slope is 5.8196 cubits (rounded off). The second example has the slope length at 119.96 inches. This equals 5.81765 cubits (rounded off).

The diagonal of G2 allowing base and sides to equal 411.035 gives us 581.29127 (rounded) this when divided by 100 and then multiplied by 20.62 gives us a slope length of 2.81.29127 x 20.62 = 119.86 inches and well within the tolerances that Petrie found. Assuming that 9 cubits was correct for the height and 10 cubits was correct for the width to solve for height to apex we need 581.29127/100 or 5.8129127 squared minus 5.0 squared and we get 2.9648 (rounded off) and added to 9 we find apex seems not to be 12 but 11.9648 cubits.

If we want to insist on 31 degrees we get height to apex of 9 + 3.0043 or 12.0043. Slope becomes 120.27 inches. It seems like at lot but for heaven's sake it is only 1/4 or an inch more than the value Petrie got.

I really wonder if we will ever know what was intended by the builders/designers. Using 2.9648 and dividing by 5 we get a top angle of 30.666 degrees. [tan of 30.666 = 2.9648 / 5 = 0.592957 (rounded off) ]

The more I think on it the more I believe Pi and Phi and 31 degrees was the intent with the following measurements:

Deflection angle of roof: 30.99987566446 ( rounded off )

Ratio of slope to added on height: Phi / Pi or 0.5150362148 (rounded off). This is sine of 30.99988 (rounded off) degrees

Ratio of the two sides: 3.0043 / 5 or 0.60086. This is tan of 30.99988

and finally

Slope distance is Pi / 2.692874 (rounded off) x 5 = 1.166632 (rounded) x 5 = 5.83316 (rounded) x 20.62 = 120.28 (rounded off) inches

Yes, this seems to fit the best.

[Don Barone]

Hi all. Well as usual I am unable to let sleeping dogs lie. I decided that the logical thing to do was to check Petrie's measurements since I held him in such high esteem.  What I found was, well, unsettling. I think it is best if i post the relevant section from Petrie's and Birdsall online book. So look below.

The first thing we need to notice is that he states, using the values he got, that the width is, on average, 205.85. Now the major problem with that is that we have been indoctrinated into accepting the width as being 10 cubits wide exactly. If we accept this then the width IF IT IS  10 CUBITS must be made up of cubits that are ONLY 20.585 INCHES. A conundrum to be sure. Compounding this is that he states the length as being (on average is understood) 226.47 and if we divide by 20.62 we get 10.983 reasonably close to 11. However we divide 226.47 by 11 we get, yes you guessed it (didn't you ?) we get the same as the cubit for the width of 20.588.  So it appears that for some strange reason the cubit used in the construction of The Queen's Chamber is not 20.62 or 20.6215 or 20.618034 but is in fact 20.585 plus or minus. (I will have to check with hstr1 to see when he has the cubit at this length) To make the waters even more muddy he has given us the height to the apex point and he states it is 245.1 inches. If we divide 245.1 by 20.62 we get 11.8865.

For now we have to move on to something just a bit different. We have seen that that angle of the sloping roof is somewhere ( and everywhere) between 31 degrees and maybe 30 degrees. So what is the correct value. Well as you I have put up a very logical case of the angle being  30.99988 (rounded off) or almost exactly 31.00 degrees. In order to get that we need a height of 12.0043 cubits NOT 11.8865 cubits so let's divide 245.1 by 12.0043 and see what we get. (Who is betting on 20.585 ?) here is the answer, 245.1 / 12.0043 equals 20.4177 rounded off. Well that sucks ! 245.1 / 20.285 = 11.9067 cubits

This would make the angle 11.9067 - 9 = 2.9067 / 5 = 0.58134 and the angle that has this as it's tan is 30.17. Well that sucks as well.

So is there a solution ?  Well I think I have found one. 9 cubits by the way is the height of where the slope touches the wall. If we stick to the answer we got when we divided by 20.62 or 11.8865 and subtract 9 we are left with 2.8865. This number then is the key to the solution ( or at least a secondary solution). If we use this to calculate the angle we get 2.8865 divided by 5 (1/2 width) to get 0.5773. Now I initially missed the significance of this number but it eventually made itself know to me for when we look for the angle that has the tan of 0.5773 ( and this one has got to be one of my best) we amazingly get 29.998 or as close to 30 degrees as we can get rounding off decimal places. So beautifully this would make the number of degrees involved in The Queen's Chamber (starting at bottom left) of 90, 120, 120, 120, and 90. Now how's that for making geometric sense ?

There is one more astounding result of this solution. For when we calculate the slope distance we get - the square root of 25 ( 5 squared) + 8.33188225 (2.8665 squared and remember these values have been rounded off) So our slope is the square root of 25 + 8.33188225 or square root of 33.33188225 or  5.773377 ( if we don't round off we get precisely 5.7735026918962576450914878050196 which is exactly the square root of 300  divided by 3 OR the square root of 3 divided by 3 then times 10. The square root 3 divided by 3 is of course exactly the tan of 30.00 degrees

However we still have one "minor" problem. We have used two different sizes for the cubit. For the width and length we have used 20.585 and for the height we have used  20.62. What can this mean ?

Well is it possible that on the flat plane 20.585 is used but "in" the sky the royal cubit of 20.62 is used ?

Let's try just one test on G1. Height we agree is 280 x 20.62 =  5773.6 and basically exactly 1000 times the slope distance of The Queen's Chamber. But what of the base ? 9069 / 20.585 gives us 440.56 and, well that sucks as well ! 9069 / 440 = 20.611

Oh well but all the other things work perfectly so I will go out on a limb and say that the angle of the slope of The Queen's Chamber IS BOTH 31 DEGREES AND 30.0 DEGREES !!!

Feel free to comment !!!!!

db

[Don Barone]

And speaking of pentagons .. from the "How do they get around this ?" we have this.

This is in the official rule book of Baseball.

Unfortunately the designer of this base must have been an Egyptologist for it is mathematically incorrect and flies in the face of Pythagoras.

If one squares the equal sides 8.5 inches we get 72.25 + 72.25 for the hypotenuse which totals 144.5. Thus the side is square root of 144.50 and equals 12.02081528 NOT 12 INCHES !!!

I WONDER HOW THEY GET AROUND THIS AS EVERY MAJOR AND MINOR AND PEE WEE IN FACT EVERY BASEBALL TEAM IN THE WORLD IS BREAKING THE RULES.

Just FYI

db