Giza and 9/11: A new proof of intent

[Don Barone]

The Giza Pyramids or monuments if you will appear to be haphazard but they nicely form a 9 by 11 rectangle. The measured sizes are 1417.5 cubits allowing 20.62 inches per cubit by 1732.5 cubits.




Now I have also claimed that The Great Pyramid is an exact image of The Giza Plateau but in a smaller (1 to 5.4) scale. I will post proof of that later but it is to The Great Pyramid that we will now go to show that 9 by 11 on the plateau in our rectangle is no fluke or co-incidence.

Most who know anything about the pyramids know that the accepted size of The Great Pyramid is base of 440 cubits and height of 280 cubits or if we use 40 as our unit of measure we get 11 (base) by 7 (height) and there most research stops allowing these elements to continue but I have I think come up with something original and it is quite extraordinary. Firstly here is a great image copyright to Gary Osborn of The Great Pyramid to scale.




As I said most researchers, including myself have always used 40 and arrived at 11 by 7 or 1/2 base or 5.5 by 7 in height but I was looking at this image of John Legon and had a thought. Here is the image.




However here were John Legon measurements:



Now the reason what I have discovered has not been found before is because everyone including myself knew that the measurement from the center of The Queen's Chamber to the center of The King's Chamber was 21 cubits and we always used that and it yielded nothing of consequence and then suddenly just the other day I saw this image again and decided to do this to it. I decided to call the distance 22 cubits instead of 21 and things just began to fall into place.



Once we establish the center distances as 22 cubits IT NOW BECOMES OUR BASE UNIT ! So the unit of measure we base everything else on now is the distance (or so I thought) between the centers of The King's and Queen's Chambers and from that one thought comes this.

The base now becomes 198 (220 - 22) and 242 (220 + 22) but it also becomes very cleverly 9 and 11 ! For 198 is 9 x 22 and 242 is 11 x 22. Very clever. Observe ...



But that is just a scratch on the surface for if we measure the distance from where the shafts would exit the pyramid we get a distance also equal to 198 cubits and we have this diagram:




But then we draw a (blue) line joining the two 198 cubit lines and we create a parallelogram. Observe what happens when we draw one more line:




Okay let's draw a few more lines and finish an astounding image:




Okay and just a couple more ...




Okay this image also shows us that where the King's Chamber shafts meet is precisely half way between the pyramid floor base and a line drawn where the shafts would exit the chamber. Observe ...




Okay seeing everything fit so perfectly I was convinced that the distance between the two centers had to be 22 cubits but the only source I could find was Petrie and this is what he had to say:

The position of the King's Chamber in the Pyramid is defined thus: N. wall at base 330.6 ± .8 S. of centre of Pyramid; S. wall 537.0 ± .8 from centre; E. wall (284.4 ± 20.7) = 305.1 ± 3.0 E. of centre; W. wall 107.7 ± 3.0 W. of centre. Base of walls 1686.3 to 1688.5 ± .6 above pavement; actual floor 1691.4 to 1693.7 ± .6 above pavement; ceiling 1921.6 to 1923.7 ± .6 above pavement.

So from wall to wall divided by two would yield our center distance and here it was ... or so I thought for 330.6 + 537 = 867.6 and divided by two gives us 433.8 inches and given 20.62 inches per cubit we get, according to Petrie, 21.04 cubits and with the allowable error precisely 21 cubits NOT 22. but I was convinced that my design was correct so where was I going wrong and then I saw the light. I once again managed to see again how unbelievably clever the builders were and what a great sense of humor they had for I had forgotten the main theme of this thread and a lot of my research and that was 9 and 11 and then the obvious solution revealed itself to me and that will be for my next installment.

Can anyone guess the solution ?

Cheers

[Don Barone]

Well tragically the solution I thought I had did not pan out so we have to figure out why they went 21 cubits instead of 22. Any guesses ?

[Don Barone]

Hi while going through some of my old notes and websites I had found in the past I found my solution in the work of Jin Allison I will post it tomorrow ... no I will post it now. The solution lies in the fact that the shafts do not meet directly in the center of The King's Chamber but precisley 1 cubit south of center and thus gives us the 22 cubits needed to show my diagrams to be true and accurate and very neat and tidy. Here are two diagrams of mine taken directly after Jim Allison's diagram which follows.



then we get this ...



and thanks to Jim Allison for this image which sets it all in place oh so nicely.



Cheers
Don Barone
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[Charlotte]

Hi Don,

I was wondering what you were up to lately, just can't let go of Giza, can you the reason being apparent. I never doubted "intent" the first time I saw and felt the Pyramids and the sacred atmosphere, and can transport myself there even with eyes open. Even if our ways of apprehending Giza is different, they still converge in meaning that it is a place of utmost importance and superior knowledge, to say the least, giving a glimpse into the Mind of the Architect.

[Don Barone]

Hi all just a comment on what I had psoted above. I should know better than to sue other people's data without checking it out thoroughly first. Case in point the shafts

[Don Barone]

Hi all just a comment on what I had posted above. I should know better than to use other people's data without checking it out thoroughly first. Case in point the shafts do not exit the pyramid at the 154 cubit mark but about a cubit and a half lower. Because of that the entire fabrication based on 154 is meaningless and we must move on. As Sherlock Holmes was apt to say once we have ruled out everything else that which remains must be the truth. Or at least he said something like that. so these failures only bring us closer to the truth. So allow me to post this from my new blog post:

Hi all. Well to say I have been busy would definitely be an understatement. I have been working on a few projects such as figuring out how to tell angles using just the pyramids (it is quite easy once you know how) and figuring out the connection between the pyramid fields at Giza and at Dahshur which is what this posting is about. Now this is extremely convoluted and I will try to make it as easy as possible and for those who are mathematically and geometrically challenged I apologize for this is quite a lot to handle but also quite elegant and beautifully simple.

The Bent Pyramid at Dahshur

Most researchers including myself have spent their lives studying The Giza Pyramids and have basically ignored the pyramid field at Dahshur which includes The Red Pyramid and The Bent Pyramid. I think it is partially because data for these two pyramids are not as easily found yet I think I have finally deciphered the true measurements and it is quite elegant. First of all I would like to state that I have had 9/11 on the brain and kept thinking that The Bent Pyramid had to be in an 11 by 9 ratio, that is the top was 11 units and the bottom 9 units. However there was one thing that kept nagging at me as I went over and then worked on overhauling my theories on The Bent Pyramid and this was the single fact that at Giza the distance from the north face of The Great Pyramid to the south face of G2 was not 1100 cubits as one would have expected but was 1101 cubits ! John Legon used this fact to conclude that the top portion of The Bent was not 11 units but actually 1101 or not 110.0 cubits but 110.1 cubits. I keep thinking he was wrong but the more I looked at it the more I came to realize that yes this was indeed the same measurement and was probably meant to be the same and then I wondered well maybe there would be other matching similarities as well and so for the very first time I decided to try to tie in my measurements at Giza to what I knew of Dahshur and The Bent Pyramid. It was an amazing ride. I found it hard to believe that to solve for The Bent I had to start at Giza but that was how it turned out. Firstly I think I need to post a map of The Giza Plateau with the measurements we will be using. It is an abridged version of Petrie's diagram with adjustments made to make the south side of The Great Pyramid run directly east-west. I will post both for you to compare.

Petrie:



And now the corrected version:



Now I had in past research realized that the angle formed from a line between the centers of G1 and G2 gave us almost the same angle as at The Bent (upper portion) but I had been unable to do anything with it. This was to change quite dramatically this time around. I am not going to post a bunch of references but suffice it to say that the consensus of the angles for The Bent have the top portion at around 43 degrees 22 minutes while the bottom ranges from about 54.3 to 55 degrees. Below I have joined the centers for you and have marked in the angles for you to see. Observe below: We get 43.3637 and 46.6363 as the two angles and that is the beginning of a very strange voyage we are about to take and when it is over I hope you will see the pyramid builders in a completely new light.




Now what I am about to present, in my opinion goes way beyond the bounds of coincidence and has to be proof it was planned from the outset, and that includes all three pyramids at Giza and probably all the Old Kingdom IVth dynasty pyramids as well. Stay tuned .... Nov 6th, 2014 @12:59 AM

Okay it is Thursday morning Nov. 6th and onward we go. Now what led to this latest round of insights was the fact that I chanced upon a diagram I had made around a year ago and although I thought it unique I did not pursue it. Here is the diagram from around September of 2013.



So I took this diagram below and did the same thing.




Now some might wonder how I figured out the angles on the diagram above but it is really quite simple geometry and trigonometry. In any right angled triangle the small side divided into the large side (not the diagonal) will yield a ratio or a number and this is called the "tan" of the angle in question so dividing 13162 by 13936.1 will give us the tan of the angle and the ratio turns out to be 0.9444536 and this is precisely what the distance is from the center of G2 to G3 along the horizontal multiplied by 10,000 ! Now I mean really think about that for a moment. Of all the measurements that randomly could have resulted from placing G3 on the plateau what do you think the odds are of placing it precisely 10,000 times the ratio of the distances between G1 and G2. Now we have met 10,000 here at Giza before when we realize that The Great Pyramid is representing the square root of 3 and the height of this pyramid in inches is precisely, or was projected to be, the square root of 3 divided by 3 times 10,000 and we get 5773.50 inches. But after coming to terms with this latest coincidence one then would need to try to explain the following. What follows was pointed out by me last year but it is within a better framework that I offer it this year.

There have been many arguments on whether The Ancient Builders knew Pi and/or decimals and all I can say is how the heck could they have left the distance between G2 and G3 the exact DECIMAL VALUE or ratio of the distances between G1 and G2 if they did not know decimals ? I mean really how long can "they" keep yelling ... COINCIDENCE !

There are 4 different measurements at The Great Pyramid for the sides and as I have suggested in the past this is because I feel that there is probably at least 4 different solutions But let's deal with just one scenario. Let us assume that The Pi angle was meant and used here at The Great Pyramid how would you prove to visitors to your pyramids that you had used Pi especially when 5.5 by 7 and Phi were so close. So how would you do it ? Well I am going to show how The Master Builders showed us !

In earlier posts I showed my reasoning for calling The Great Pyramid the square root of 3 and it is from that base that we branch out. The Pi angle is the ratio of Pi divided by 4 and so if the angle intended at Giza for G1 was indeed The Pi angle then 1/2 the base would be [(Pi / 4 ) times 5773.50] inches and we get 0.7853981634 times 5773.5026919 and we get 4534.4984 inches and multiplied by 2 gives us 9068.9968 or rounded off to 9069 inches well with the distances allowed at Giza which are as follows:



So we will stick with 9069 inches which is within one half of one inch to what is measured by Petrie. Now we can not forget about G3 which was miraculously placed precisely at the tan of the angle between centers of G1 and G2 but what of it's size ? Well here is what Petrie has for G3 ...

North - unavailable

East = 4149.2 inches
South = 4157.8 inches
West = 4153.9 inches

Mean is 4153.6 inches

So let us start with the mean and round up to 4154 inches. If we add this to G1 we get G3 + G1 = 4154. + 9069.0 = 13223 inches and would it not be somewhat convincing if this number in some way showed us Pi given the numbers we have already been using well how about this. 4154 / Pi = 1322.26 this checks to 4/10ths of one single inch and has convinced me that they intentionally encoded their understanding of this ratio. But as always at Giza there is always one more proof and so I offer this. One of the sacred measurements used at Sakkara for their arches or curves was the ratio 0.7 or 1.4 ratio and so what do we get when we use 4154 and 1.4 with a little help from our friend Pi ... do I really need to do the math ... well it is none other than [(4154 / Pi) / 1.4 = 944.47 and again showing us our angle tan decimal of 9445. So what we have is G1 + G3 equal to 1/Pi x 10 or 3.183098862. If we allow this then G3 is equal to precisely 1 unit and G1 is equal to [(1/Pi x 10) - 1] or 2.183098862 and that is indeed what the numbers work out to. Are these just more coincidences ? I think not. In order to be a perfect fit we would only need 2/10th of an inch or 9069 / 2.183098862 or 4154.19. Proof I would suspect of intent. But we are not nearly done yet. Here is a diagram that illustrate this concept a bit better using 9068.8 and 4154.



But wait this post is supposed to be about The Bent Pyramid at Dahshur so maybe we should head on over there. see you there in a while ... to be continued.

[Don Barone]

Oh dear ... I have just discovered something rather interesting and it has totally upset the apple cart. I am not sure what led me to the page, probably a vague recollection that I had seen a number before but decided to check the ratios of the mass of the planets. I had come across 104.71 and I was sure I had seen that somewhere as a ratio of the mass of a planet or something so off I went to Google. Well imagine my total surprise when I discovered that 1 Solar Mass = 1047.56 Jupiter masses (gosh it seems that Ptah is popping up all over the place these days). Now you are probably wondering what this has to do with anything well it all goes back to this listing of John Legon and the measurements of Petrie and Dorner. Allow me to make it a little clearer. Here are the measurements ...

As you can plainly see Dorner has the height of The Bent Pyramid equaling 104.71 meters and this equals 199.92. but come on now really would you build a pyramid 199.92 of your units well no neither would I,  I would probably build it 200 units or 200 cubits if I was an Egyptian so why 199.92 and not 200.0 or is 200 correct.  Well first let us consider this. There are some who might suggest that the battle between Horus and Set was actually symbolically telling us about the battle for control of our solar system between The Sun (Horus) and Jupiter (Ju P(i)tah) or Set. Well up until a few minutes ago it was just a thought but then I managed to find these ratios and once again I do not think i can ever look at The Bent Pyramid or The Red Pyramid the same way again for within the ratio of the solar masses of The Sun and Jupiter another door has been opened for us. First let us once again list the ratio of solar mass to Jupiter masses. It is 1 Solar mass = 1047.56 Jupiter masses.  This ratio is as important as the other one of 1.070466 or sq rt of 3 / Phi for you will be amazed at what it reveals. 

Firstly let us assume that we have 104.756 cubits this would give us 2160.07 inches (simply 104.756 x 20.62) Now I am not sure if that number jumps out at you but it should as it is a most interesting number. Perhaps if I post this it might help:

Equatorial radius of The Moon = 1738.14 km (0.273 Earths)[3] 1738.14 x 2 to get diameter and we get 3476.28 kilometers and believe it or not 3476.28 kilometers is equal to 2160.07 AN EXACT MATCH !!!!!!!!! So let's see what this gives us 1/10 the ratio of solar masses to Jupiter masses times 20.62 which just happens to be the number of inches in a cubit and we get EXACTLY and I mean PRECISELY the diameter of The Moon in miles. 
 
[3] - Williams, Dr. David R. (2 February 2006). "Moon Fact Sheet". NASA (National Space Science Data Center). Retrieved 31 December 2008.

So now if anyone ever asks you gee how many Jupiter masses in The Sun's mass simply tell them to take the diameter of The Moon in miles and divide by the number of inches in a cubit or 20.62 and then multiply by ten. It is quite remarkable and suggests that maybe Jupiter was a major influence in the reason for the sizes of the planets.  But back to The Bent. We were wondering why 199.92 instead of 200.0 or was it supposed to be 200.00 well up until a minute ago I was all for 199.92 but now I am 100% convicned it is 200.00. Basically I now think that it is both or maybe a couple of others for I think all the sides are different and all the angles are different as well and Petrie has already proven that the sides change angles as they rise upward in the lower portion but allow me to show you why I am now a total and firm believer that the height was 200 cubits. Well it is back to Jupiter and The Sun and they are telling us it is 200 cubits they are doing this by telling us that 104.756 IN METERS THIS TIME which is 1/10 the ratio of Solar to Jupiter masses times 39.37 inches in a meter to get 4124.24 inches which just happens to equal 200.01 cubits at 20.62 inches per cubit. So it is back to the drawing board I go as this new information needs to be processed.

Cheers
Don Barone
Nov 6th, 2014 @ 6:19 PM EST

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[Don Barone]

In other words ...

Cubit
--------  x 200 = Jupiter constant or number of Jupiter mass in a solar mass.
Meter

Now can you beat that ?

Now we find the same ratio as Sun to Earth and Earth to Mars shown in the number of Jupiter masses in a solar Mass.

Most interesting.

[Don Barone]

Just more wizardry from Giza. Base of G1 or The Great Pyramid is 9069 inches.

9069 x Pi = 28491.1

28491.1 squared = 811742994.34

811742994.34 / 2 = 405871497.17

Square root of 405871497.17 = 20146.3

20146.3 inches is equal to base of G3 in cubits x 100

So scale is 1 cubit of G3 = 100 inches of this

Always something interesting

[Don Barone]

Just keeping my notes and thoughts here for others to read:

The length of each side of the Red Pyramid is 220m and its original height was 104m. The German team have recently found remains of the pyramid’s capstone, or pyramidion, which has now been reconstructed and placed on the pyramid’s eastern side. However, the angle of slope of the pyramidion differs from the Red Pyramid, suggesting that it was not originally intended to be placed on this structure. The entrance to a 60m long sloping passage is on the northern side of the pyramid, 28m above the ground and leads to the first of three chambers. The first two rooms are tall antechambers with high corbelled roofs and separated by a narrow rectangular passage and the roof of the second chamber coincides with the vertical axis of the structure. The antechambers are very similar in size and construction with projecting vaulted ceilings of large blocks of fine white limestone. In the second antechamber and the passage to the burial chamber many early explorers have left graffiti, reminders of their presence inside the pyramid.

360 / Pi = 114.59155902616464175359630962821 cubits
114.59 x 20.62 inches = 2362.88
2362.88 / 39.37 = 60.017 Checks very close to 60 meters

Base of The Red Pyramid is sq rt of 3 x 10,000 / 2 = 8660.254 inches
8660.254 inches / 12 = 721.688 feet
8660.254 / 39.37 inches = 219.97 meters
8660.254 / 20.62 = 419.993 cubits

Ratio of mass of Mars and Earth = 5,972,190,000,000,000,000,000,000 / 641,693,000,000,000,000,000,000 = .10744684948067626783474738747428
13162 (from Giza and east-west measureent) x .10744684948067626783474738747428 = 1414.2154328646610372409451139364 squared = 2000005.2905525805893781812741103 Checks to 2000000 to accuracy of 0.9999974
13162 / 15176 = 0.86729 (as a tangent) = 40.93478 degrees x 10 = 409.3478 x 1.0744684948067626783474738747428 = 439.8314

I call this mental doodling.

Cheers
Don

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[Don Barone]

The Square Root of The Square Root .... of .... 3 !



The Square root of The Square root of ... Pi !

Let us explore these shaft and their exit points more fully. As you can see in the diagram above I kept wanting to use 23333 and 33333 but in a lot of my work (can't find them right now but I will) I kept getting 133.13 or 13313 or around those numbers and could not see the light until just today. I was playing around with my calcualtor at work and happend to hit the sqaure root of Pi and then for some strange reason (well there was a reason which I will explain later), perhaps providence I decided to hit square root again and got ... 1.3313353638. Now I must have done this a million times in the past but this time, maybe because I was just looking at this diagram a day or two ago it suddenly jumped out at me 1.331335 or times 10,000 to get 13313.35. So the square root of the square root of Pi PLUS 1 would eqaul 2.331335 and times 10,000 equals 23313.35. Now these numbers are not contrived they have been published long before I came along and so we have it that the square root of 23313.35 is 152.687 and if we call these cubits we have in inches 152.687 x 20.62 and equaling 3148.41 which equals to within 1/100th of an inch the height to the top of course 103 or bottom of course 104. (Please see Petrie's diagram above) But there is a bit more to this solution but let us recap so far. We have the height of The Great Pyramid as 5773.5027 and we have found that the sq rt of the sq rt of Pi + 1 and then square rooted itself yields EXACTLY the height of the bottom of the shaft were it to exit the pyramid or precisely where the 104th block is. See diagram of courses by Petrie above.

and the link ... The Square Root of The Square Root of ...

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[Don Barone]

Hi all.

Here is an image of the Giza Plateau with two sets of measurements on it. The black are from Petrie in 1881 and the red are my adjusted measurements to prove a point. for now I will just leave the image with an explanation of why I chose these to follow in a day or two. See if you can figure out why these were chosen. I have also adjusted the diagonal measurements by an inch or two. No more. First Petrie notes:



... and my adjusted figures (in red)





Punchline to follow ...

Cheers

.

[Don Barone]

Hi all here is a revised image with the correct measurements on it. I have closed this problem to 5/100ths of an inch and for now that is good enough.



Meaning of these numbers tomorrow ...

Cheers

[Don Barone]

Sigh the diagram above is incorrect ... I can't seem to get it to close so that may explain why it reads 19168.4 instead of my 19167.58511 ... I will work on it a bit longer but I think it is not solvable.


Cheers

[Charlotte]

Cheers Don, "that is good enough for now", right or wrong, the quest for knowledge and understanding makes life interesting.

[Don Barone]

Nov 19, 2014 6:52:34 GMT -5 Charlotte said:

Cheers Don, "that is good enough for now", right or wrong, the quest for knowledge and understanding makes life interesting.

Indeed it does.

Trying to prove the following:

Distance between center of G1 and G2 is exactly the distance between The Sun and The Earth allowing distance to equal number of miles in cubits and then in inches. Allow me to illustrate.

Semi major axis (or averaged distance from The Sun) For Earth this is 149598261 kilometers.

1 mile = 1.60934 kilometers EXACTLY by convention.

So 149598261 / 1.60934 = 92,956,281 miles / 100,000 = 929.56281 and let us call these cubits on our diagram. So since 1 cubit = 20.62 inches we arrive at this. 929.56281 x 20.62 Petrie has this distance as 19168.4 inches

Step 1:



Second step:

If we allow base of G3 to equal 4154.128 inches then 4154.128 / Pi then x 10 = 13223 inches. If we take our base of G3 out of this we are left with 9068.87 which is precisely the mean base length of The Great Pyramid. This tells us two things:

1) Pi was encoded between G1 and G3
2) It is very safe to assume that the distance or mean length for G3 was 4154.128 inches or 201.46 cubits.



More to follow a bit later today.

Cheers
Don

[Don Barone]

Let's step up to a Step Pyramid.

It is really interesting that as I start to look at the other Old Kingdom Pyramids how they all seem to be following a set plan. And yes 9 by 11 is involved.

First the agreed upon data of The Step Pyramid.




Now maybe it is only me and I haven't been paying attention but I do not believe that anyone ever mentioned before that twice around the Step Pyramid equals once around the Great Pyramid. that is that the perimeter (according to above article) is (231 + 209 + 231 + 209) x 2 =  880 x 2 = 1760 cubits while we all know that the perimeter of The Great Pyramid is 1760 ... I mean how could I not know this ? How could this not have been discussed ? Strange and I didn't even get it from the cubits I solved it from the meters which I now say equals  9068.88 / 39.37 to get meters or  230.349552 and divide by 40 and times 21 = 120.9335 meters and times 19 = 109.4160 Now what we have to realize is that nobody ever, and I mean ever checks their math. It is supposed to be 231 cubits by 209 cubits for a total of 440 cubits but if one adds the meters up we get 230.9 and that only yields 9070.06 inches NOT 9072 which it would have to if it was 440 x 20.62 inches. I however get 120.9335 + 109.4160 x 2 = 460.699 x 2 = 921.398 meters / 4 = 230.3495 (per side)  x 39.37 = 9068.86 and checking with the mean of The Great Pyramid to 1/100th of an inch. So now according to  Egyptologists  we are supposed to  believe that the Ancient Builders made the exact  same error at The  Step Pyramid and at Giza with The Great Pyramid.  Sure they did.

And last but not least the height is said to be 59.90 meters however I should think that it was in reality 60.01 meters and thus 10 meters per layer but more importantly 60.01 meters OR 114. 59156 cubits which just happens to be 360 / Pi or  2362.878 inches or  60.017 cubits

I mean after all it was Imhotep who was building this thing ... wasn't it ?

,

[Don Barone]

Arrrrgggggggghhhhhhh. I just spent about 30 minutes making a huge post and then the computer jammed. When I tried to refresh ... it was gone. Sigh I will try again but it is never as good the second time.

Okay where was I ? Oh yes ...

In this post I will point out the beauty and the harmony and the reasons for The Giza Plateau and as in all things we must start at the beginning. To understand Giza is to understand math and our solar system for the two are inseparables. But to the beginning. The first thing we need to conquer Giza was an understanding of the height of The Great Pyramid and that came when DUNE and I (there is some argument who used it first ... not really that important) both figured out that the height was the square root of 3 x 10,000 then divided by 3 or 5773.50 INCHES ! YES THE SECRET IS STUDYING IT IN INCHES. So if we use as a height 5773.50 and divide by 20.62 inches in a cubit we get 279.995 cubits or close enough to 280.00 to keep everyone happy. For myself however there was a more compelling reason for thinking square root of 3 and that was because I had determined that G1 and G2 were in the precise ratio of the square root of 3 to Phi ! As a proof I offer this. The mean of G1 is 9068.88 inches and the ratio of square root of 3 to Phi is 1.732050808 / 1.618033988 or 1.07046627 so 9068.88 / 1.07046627 = 8471.9 and wouldn't you just know that the north side of G2 is precisely and I mean exactly 8471.9 making this an exact match and beyond doubt if you ask me. Here is Petrie's data:



But what of Pi ? Many would have you believe that the builders knew Pi. Could I prove they did ? I think I can. G1 and G3 and in a most unusual ratio to each other and as a matter of fact they are in an exact 1/Pi relationship. As proof I offer this. Mean of G1 is 9068.88 inches and mean of G3 is 4154.128 (surmised) and together they give us 4154.128 + 9068.88 or 13223.008 inches. Now 1/Pi x 10 = 3.1831 so if we divide 13223.008 by 3.1831 we go right back to G3 and we get 4154.128 (4154.13). Showing us that G1 and G3 are in an exact 1/Pi relationship and I guess coincidence #2 to the non believers but I challenge anyone to find another example of the ratio anywhere. But there is still much more.

How were we going to determine how big to make our Giza Plateau ... well shall we call in the sq root of 3 again ? No better still let's go to the square root of the square root of 3 this time and use 1.316074 but multiply by 10,000 (same number times as in our height of G1 ) and we get 13160.74 and let's use this to separate G1 and G2. So let it be said, so let it be done. Okay so now we have one distance for our plateau and we shall carry on. But what of my claim of a solar system connection, Well I have claimed (as have others) that G1 is representing The Earth so if G2 is The Sun as I and others have suggested wouldn't it be neat to code the distance from The Sun to Earth in a distance on the plateau and so let's do it. But how would we do it ? Well semi major axis ( simply average distance of The Earth from the Sun) is 149598261 kilometers so let's make it miles by dividing by 1.60934 (1 mile = 1.60934 kilometers by convention) and we get 149598261 / 160934 = 929.562808 miles (x 10,000 ) and let's assume these are cubits (1 cubit = 10,000 miles) we get 929.562808 x 20.62 or 19167.585 inches. So we now have 2 distances between G1 and G2. we have 19167.58 inches and we have 13160.74 inches and we can solve for the other one so let's see what we have after using basic trig to find the third side



Weird things are happening.  Segments are vanishing and well it is starting to bug me. Anyway the 9445 comes from multiplying 4154.128 by the number of cubits in G1 / 1000 or .43982 x 4154.05 and we get 9445 and that is our distance form G2 to G3 So now we have our distance east-west and it is 1/2 of 13223 or 6611.5 + 9445.0 + 13160.74 = 29217.24 which may be a slight problem as most would have the east-west distance 29228 inches. We shall see but that is all for tonight.



Cheers

Don
.

[Don Barone]

Well once again I overlooked something in my rush to put it all down. Those diagrams still do not close on themselves properly as 9445 is supposed to be a product of 13160.74 / .9945 and does not equal 13935.25 but 13934.08 so once again my drawing does not close for about a single inch. Very frustrating ... I am starting to wonder if I can close it using 19167.58, 36856.7 and 17874.25 or whether I will need to use 19168.4, 17873.1 and 36856.7. If I make it check to .9445 then I can not get it to close to 36856.7 and then if I get that right then it will not be the correct number for 17874.25 which is supposed to be 36856.7 x 10 and then divided by 20.62 = 17874.25

Hmmm ... I may have to consult that Geometry Problem site again and put it in their hands to tell me if it will ever close.

Cheers but now off to The Salt Mines

Don

[Don Barone]

Okay this seems to be a correct solution.



19167.58 squared = 13160.74 squared + 13935.25 squared

17874.25 squared = 9444.21 squared + 15175.50 squared

36856.7 x 10 = 368567 / 20.62 (inches in a cubit) 17874.25

and 13160.74 / .9444.21 = 13935.25

and lastly 36856.7 squared = 29110.75 squared + 22604.95 sqaured

All would seem in order and this closes to about 7/100ths of an inch and given the fact that I have rounded off could be the reason.

However what of those who would claim that the plateau is 1417.5 x 1732.5 cubits and 29228 inches by 35724 inches ... well they are wrong for if they are correct then Petrie is out on his long diaognal by almost 10 inches for  we have figured out that base of  G3 + G1 = 13223 inches thus we need to take 1/2 this off of both of these measurements and we get 29228 - 6611.5 = 22616.5 and 35714 - 6611.5 = 29112.65. However the long diagonal then becomes sq rt of 22616.5 squared  + 29112.65 squared and this gives us  sq rt of 511460840.25 + 847546390.0225 = sq rt of 1359007230.2725 which equals 36864.72 BUT THIS IS IMPOSSIBLE UNLESS YOU WANT TO CLAIM THAT PETRIE WAS OUT 10 INCHES for the diagonal would never change. No matter what angle we claimed for true north the distance from G1 to G3 would always have to be 36856.7 inches and thus anyone who claims other wise is claiming that Petrie is wrong and I do not think he is.

Here is Petrie's data:



Cheers
Don

.

[Charlotte]

Imhotep "The Most Sacred One", is thought to be the architect of Zoser. An inscription refers to Imhotep as "The Sunbearer of the King of Egypt, one who is near the head of the King, director of the Great Mansion, royal Representative, High Priest of Heliopolis, Imhotep, the Carpenter and Sculptor.

It reads "Sunbearer of the King of Egypt, near the head of the King", not one particular King, and since he was "High Priest of Heliopolis", City of the Sun, the Most Sacred One near the head of the Kings, could have well illuminated the Architects. How else could "the thing" be achieved.

[Don Barone]

Just a bit of a side trip.

Always fascinated by how closely connected Phi and Pi and now the sq rt of 3 are and here is some mental doodling I did at work today.

Firstly we must always go back to my early discovery that the sq rt of 3 / Phi is EXACTLY IDENTICAL to the ratio of the base of G1 and G2

√3 / Phi = 1.73205... / 1.618033988... = 1.07046626932

Base of G2 = 8471.9 x 1.07046626932 = 9068.8832 THE EXACT MEAN FOR G1

But now for the stuff I found today.

360 (degrees) / Pi = 114.591559
√114.591559 = 10.70474469692

10.70474469692
-------------------- 1.070474469692
         10

1.07046626932 / 1.070474469692 = 0.999992

So this means we have this equation:

√360/Pi         √3   
--------- = --------- (to a degree of accuracy of 0.999992)
    10              Phi

But this bring with it a new set of problems. How will we ever determine whether the builders meant √360/Pi/10 or √3 / Phi ...

As an example let us use √360/Pi
                                    --------- = 1.070474469692
                                        10
Base of G2 again is 8471.9 x 1.070474469692 = 9068.95 within 7/100ths of an inch and finally we have this ...

360 / √3 = 207.8461 and x 2 = 415.69 and times 10 = 4156.9

Bases for G3
N - ?
East - 4149.2
South - 4157.8
West - 4153.9


Almost forgot one of my favourites.

Phi / Pi = 1.618033988 / 3.14159265 = 0.5150362148

This could translate to 51° 50' 36.22" or 51.84339444°

Pi angle = 51.8540°
Phi angle = 51.8273°
5.5/7 = 51.8428

[Don Barone]

More diversions and side tracking.

Is this yet another clue toward Pi ?

Here is a chart taken from this website:



For some strange reason I decided to add all the "tans" together to see if they gave us any further insights and I did not get very far for by adding 0.714288 and 1.058703 I found that the total was 1.772991. Now anyone who has fooled around with the numbers as much as me immediately would recognize this number as being very close to the square root of Pi or 1.772453851. You will notice that this chart uses the angle 43° 22' 00" instead of what Petrie has which is 43º 22' 52" and equals 43.381111... and give a tan of 0.9450287 the reciprical being  1.0581689225472 - Now if you have read earlier posts I have suggested that the second angle of The Bent Pyramid incorporated tan of 1.4 or 54.46232221º or 54º 27' 44.36" RECIPRICAL OF 1.4 IS 0.71428571428 ... NOW WHEN WE ADD THESE 2 TOGETHER WE GET:

0.7142857142857
1.0581689225472
-------------------------
1.77245463683289
===============

Square root of Pi is 1.77245385090552

So we have 1.77245385090552 / 1.77245463683289 = 0.9999996

So we have the "tan" of the two reciprical angles of The Bent Pyramid equaling square root of Pi to a degree of accuracy of 4/10,000,000ths

Conclusion has to be that they were trying to show us Pi !

Best
Don

[Don Barone]

Continuing to bounce around like a ping pong ball we have this.

Things to remember

360 / sq rt of 3 = 207.8461

Base of G3 varies from 4149.2 (east side) to 4157.8 (south base)

2 x 207.8461 = 415.69 and times 10 = 4156.9 and suggests that height of The Great Pyramid to a point drawn from the corners through where the Queen Chamber shafts end is possibly showing us G3 in a scale of 1 cubit (in Great Pyramid height) = 20 inches (base of G3 in inches)

The diagram: (This by the way opens up many new avenues on how to look at these alleged 3rd and 4th Dynasty marvels one of which is that this angle is the same as the top of The Bent Pyramid and angle of The Red Pyramid as discussed on guardian egypt and discovered by Avry but published by andrew bayuk - long story)

[Don Barone]

Okay so carrying on with this diagram:



So let us assume that the angle is correct and is a match for the angle at The top of The Bent and for The Red Pyramid what does or would that gives us. Allow me to illustrate:

 Here is a diagram extending the point of escape for The King's Chamber downward and showing that it may hit the point where The Queen's Chamber shafts end.



Next we will draw a line from the two ends of The Queen's Chamber and we get this configuration:



Now if we assume that we have The Bent Pyramid hidden here we have to figure out the base and when we assign a value of 11 for the top and 9 for the bottom we find that the base of our Bent within The Great Pyramid as being a line defined by the base of The Queen's Chamber as shown below.



Okay now that we have the base and the height of  the top of The Bent we must figure out where the final bottom base would be and to do this we simply use a protractor and set it at about 54 degrees 30 minutes and we get this configuration:



Now let's do the same for the other side and then extend the lines toward the top and and draw in our "Bent Pyramid" within The Great Pyramid like so to get this configuration:

 
 
Things to note here are the fact that the projected lines of our 54 degree angle hits the very top of the existing pyramid and some have suggested that is as far as they ever built and another very interesting thing to note is that the angle from our baseline and the start of our 54+ degree angle to the exit of our King;s Chamber shafts appears to be precisely 60 degrees as illustrated below.




Since in an earlier diagram I had show that the height of the exit of The Kings Chamber shafts would be:

Course 104

Bottom = 3148.4 and 3149.5 inches
Top = 3174.7 and 3176.0 inches

And 1/tan of 60 degrees = sq rt of 3 / 3 we get .57735 x 3149.5 and we get a distance from the center of the pyramid to the 54+ degree line as being 1818.3647 inches or dividing by 20.62 we get 88.1845 cubits if we use 3148.4 we get 88.1537 cubits. We also noted that at this height the base is 200 cubits or 1/2 of this would equal 100 cubits so the distance to the start of our 54+ line is 100.0 + 88.1845 = 188.1845 and subtracted from 220 we are left with  cubits 31.8155 cubits. If we use exact 1/Pi x 100 we get base as 220.015 cubits



These days it seems to always be sq root of 3 and Pi ....

Cheers
Don

[Don Barone]

Continuing our theme with a post entitled ...

THE 104TH COURSE

Working assumed facts:

Height of The Great Pyramid: 5773.50 inches or sq rt of 3 x 10,000 then divided by 3 and giving us 279.995 cubits if 20.62 inches = 1 cubit
Bases of The Great Pyramid: 9068.88 or 4534.44 for 1/2 base
Angle of slope is The Pi Angle or tan reciprocal of Pi/4 or  0.7853981634 = 38.1460 or 51.854 degrees
So we have this to be exact:
(Projected) Height of Great Pyramid = 5773.50 inches
1/2 base to equal 4534.4963 inches


And now our "Hero" .. The 104th Course (of The Great Pyramid)


Now the first thing to note is that the top of the 104th is 99 cubits from the center line of The Great Pyramid to the slope side while the bottom (of the 104th) is precisely 100 cubits or 2062 inches from the center line while according to Petrie the height to the bottom of the 104th (or top of the 103rd) is 3148.4 inches up from the bottom floor while the top of the 104th is 3174.7 up. Using simple subtraction this gives us the thickness of the 104th as 26.30 inches (as per Petrie) and this leads us to some startling and awe inspiring "co-incidences" as I will show you.

Cheers and more soon ...

[Don Barone]

Okay we are back and let's try a few math examples.

Height of G1 is 5773.50 inches (by definition and logic)
Bottom of 104th course is 3148.4 inches
Difference is 5773.50 - 3148.4 = 2625.1 inches.

But hold on just a second. Does anyone remember the thickness of the 104th course ? Well to save you going back it is 3174.7 - 3148.4 or 26.3 inches and checking to 1/100 th of the height from bottom of the 104th course to the apex of The Great Pyramid. So we have the 104th course as a beacon or marker of sorts to show us precisely what they were doing. They were telling us that the angle is The Pi Angle and that the 104th shows and proves it by its distance of 100 cubits from center line to slope and height 127.31 cubits high (2625.1 inches divided by 20.62) and tan angle of 51.854 or The Pi Angle is 1.27324 x 100 = 127.324 and thus checking to 1/100th of a cubit or 1/5th of an inch.





So I guess you can call all of this a great big giant coincidence but methinks you would be wrong.

And now to ponder the fact that many have used 9 and 11 and 99 and 198 and amazingly they are all right because the bottom of the 104th is 100 cubits and the top is 99 cubits and this 99 allows all the angles of the shafts to equal even multiples of 11 which we will examine again in the next post. So it is the long way around proving it but all the work done showing the angles  as multiples of 11 are also correct. You really have to admire the builders who were clever enough to encode Pi in the pyramid but also encoded 7/11. So it would now appear that within The Great Pyramid is a Phi Pyramid, a 7/11 pyramid and The Bent Pyramid.

I mean how clever can you get ?



Cheers and maybe more later ...
Don


Cheers
Don

[Don Barone]

And very quickly back to this diagram we have this proof of Pi...



The sq rt of the sq rt of Pi is --- sq rt of Pi = 1.772453851 and sq rt of 1.772453851 = 1.331335

In our diagram please note that we have sq rt of 23313.35 and sq rt of 33313.35.

This equals sq rt of the sq rt of Pi + 1 and sq rt of the sq rt of Pi + 2

sq rt of 23313.35 = 152.687 and this in cubits changed into inches = 3148.41 EXACTLY what Petrie got for distance to bottom of course 104 from floor of The Great Pyramid,

Magic ... or designed

Best
Don

[Charlotte]

Hi Don,

People I meet in daily life believe the Pyramids are tombs for Kings. I stopped reading accepted Egyptology years ago, so I don't know what current theories are, and don't care. It used to be tombs, treasures which have been stolen, maybe also Khufu's body, and some person who apparently has never been inside the GP imagined colorful pictures on the walls of the GG. Someone here in LA told me that he was led by an old guardian of the Pyramids to a chamber, maybe the elusive tomb of Osiris, and when he looked upon the image, he became Osiris, asserting also that anyone looking upon the image would become the god. Don't know whether this is true or not, and as outlandish as it might sound, it got my attention considering the writ in the Book of the Dead - Book of coming forth by Day, inscriptions, spells, etc., would suggest as much, even if such experience is subjective, in which case it would be even more "real". A mysterious process on which no words can be imposed.

Supposedly, there is a Book of Records hidden somewhere under the Plateau. Considering your work and many others exploring the Pyramids and Plateau thusly, why could the place itself not be the "Book of Records", seeing all that is recorded in the Pyramids and the Plateau entire. The Rosicrucians demonstrate future events recorded in the GP, which I doubt not reflecting on what I have learned since.

I think for a person who simply wants to see the Pyramids, knows nothing of what conventional Egyptology relates, head not full of preconceived notions, looks upon the Pyramids and feels the Atmosphere with a sense of wonder, is most fortunate because the GP, the first in view, allows no thought of tomb, treasures and images, only look, you are standing before an enormous triangle of stones in eternal time. Such was my never to be forgotten experience and I remembered the two dreams which compelled me to visit Giza. Not that it matters too much, but I said all this before.

[Don Barone]

Well thanks to DUNE (Dennis Payne) over at Graham's another possible major break through in understanding The Great Pyramid. Now this could all be a coincidence but really since this makes about 1000 coincidences I am just a bit curious when it will be admitted that it was all designed this way. Anyway onward ...

DUNE came to the realization that the center of the shafts as they enter The King's Chamber were 1732.05 inches above the base. I am still trying to confirm this as data on the shafts and where they enter the chamber are few and far between. However from this simple observation come many important realizations. Observe ...

1732.05 is of course 1000 x sq rt of 3 and we have been exploring the fact that it may all be about this obviously very important measurement and we have concluded (as Dennis did) that the height of The Great Pyramid is 5773.50 or sq rt of 3 x 10,000 and then divided by 3. This also makes the center of the shafts EXACTLY 3/10ths of the way up the pyramid (5773.50 /10 = 577.35 x 3 = 1732.05). If we allow 20.62 inches to equal a cubit then 1732.05 equals 83.99858 cubits. Remember this number.

If all of this is correct then we have this interesting observation. If we take 1732.05 and divide by 44 we get 39.36479 and if we subtract this number from 1732.05 we get ... 1692.6860 and if we divide by 20.62 we get 82.0895 and this is EXACTLY what Petrie got for the floor of The King's Chamber ... actual floor 1691.4 to 1693.7 ± .6 above pavement;

So what do we have in all of this. Well we have the floor of the King's Chamber being 43/44ths of the height of the center of the shafts and the difference between the floor and the center of the shaft is 1732.05 - 1692.686 = 39.365 inches. If we divide 39.365 by 20.62 we get 1.9090587 basically the ratio between the cubit and the meter and between Mars and The Sun as a ratio to the distance between Earth and Mars.



But there is one more very neat thing about all of this and I will show you after you look at this diagram.



We now have a triangle with base of 2062 inches or 100.00 cubits, height of 1732.05 inches or 83.998 cubits and slope or hypotenuse of 2692.9243 or 130.60 cubits. Now interestingly when we divide 2062.00 / 1732.05 we get 1.190496255 and true to the genius of the builders of The Great Pyramid this when taken as a reciprocal or 1 / 1.190496255 gives us 0.839985 PRECISELY WHAT THE HEIGHT TO THE CENTER OF THE SHAFTS ARE IN CUBITS DIVIDED BY 100 or .8339985 x 100 or 83.9985 cubits.

and last but not least the angle of our slope in the triangle of 2062 by 1732.05 by 2692.92 is that of 1.190496255 or 1/1.190496255 or 0.839985  and this is tan of 40.0298 degrees and I might be inclined to suggest that the angles that are being called ...

Queen's Chamber ...

 Northern shaft Both shafts are approximately 3 (1.57) above floor

Distance from eastern wall 5.54 (2.9)

Height 0.4 (0.21) , width 0.4 (0.21)

Total length unknown, horizontal length 3.68 (1.93)

Angle varies from 33° 18' to 40° 6'
Southern shaft Distance from eastern wall 5.5 (2.88)

Height 0.4 (0.21) , width 0.4 (0.21)

Total length 113.5 (59.44), horizontal length 3.74 (1.96)

109.8 (57.5) at 39° 36' 28"

King's Chamber:

Southern shaft Distance from eastern wall 4.75 (2.49)

Height 0.27 (0.14) , width 0.34 (0.18)

Total length 102.25 (53.55), horizontal length 3.28 (1.72) ,

4.2 (2.2) at 39° 12', 8.3 (4.35) at 54° 32' 24"

and 86.5 (45.3) at 45°

... are in fact 40.0298 degrees or at least one of them might have been.


These builders were genius'  beyond belief and I always surprised when we find something new..