Thanks very much ... here is a bit more ...

The main thrust of my research was to show a couple of things. One that "The Creator" had used simple geometric tools and rules to shape our reality and that it was being shown to us by "The Ancients" in their major megalithic constructions.

Here is a continuation on the sq rt of 2, sq rt of 3 and sq rt of 5 along with 1 and 2 and of course Phi (sq rt of 5 + 1 then divided by 2)[ I will get back to Atlantis soon]

An Analysis of The Red Pyramid at Dashur
Mar 30, 2018 at 12:10pm

The base of The Red Pyramid is 420 cubits. The question is why ? Well let's do some math ...

420 cubits is equal to 420 x 20.62 = 8660.4 inches divided by 10,000 is equal to .86604 and is virtually exactly 1/2 the square root of 3 ! and the perimeter of this pyramid would be 4 x 8660.4 = 34641.6 / 10,00 or 3.46416 or TWICE THE SQ ROOT OF 3 OR SQ ROOT OF 12 ! and the circle enclosing this pyramid is ... get ready for it .... 2.720699046 and The Megalithic Yard !!!!!!

The picture in lieu of the thousand words ...

[link to imagizer.imageshack.com (secure)]

and then we have this finale ...

[link to imagizer.imageshack.us (secure)]

[link to imagizer.imageshack.us (secure)]

The diagonal is equal to the square root of 0.86604 squared then times 2 and we get 1.2247655 (sq rt of our 1.5) and 1.22447655 x sq rt of 2 or 1.41421536 and it equals, not surprisingly I should think ... 1.7308 or the square root of 3 and lending more and more support of the sq rt of 3 by sq rt of 2 theory for The Giza Rectangle ... Note as well that the diagonal appears to be simply the square root of the height when used in a ratio to the diagonal ...

Moving on with a couple of new images before I correct the extremely large error I have made in one of my earlier drawings. (I have no idea how I could have made such a serious error) For now though ... forward.

[link to imagizer.imageshack.com (secure)]

In this image you will notice that instead of 200 I have chosen as the height 197.9898 cubits. The reason I have done this is to keep in tune with our theme of all things square root of 2 or square root of 3 with 140 x sq rt of 2 = 197.989899. This distance is also the distance between where the shafts of the so called King's Chamber exit the Great Pyramid. As shown below ... but before we go there I have to post the horrible error I made ... Oops just double checked my drawings as I am apt to say ... "I thought I was wrong .. but I was mistaken" so let's carry on. Now what you are about to see is possibly what Pythagoras was playing at when the soldiers stormed his home and he cried ... do what you will with me but leave my circles alone for I think I may have inadvertently solved for The Pyramids but most importantly I think I have shown in the series of some 16 images I am going to be posting that all the pyramids are tied into each other and can be shown to do so from simple, and I repeat here, simple geometry ! And the cherry on top will of course be when I beautifully tie it all into our solar system and the planets for "The Creator" was a very simple creator and used only the simplest of ratios and such and thus it was easy for The Ancient Builders to emulate "it's" work for remember what is written in Enoch where "The Creator" said he would make it so simple anyone could understand it ... and so we begin ...

We will start with Gary Osborn's copyrighted drawing of The Great Pyramid to scale ...

[link to imagizer.imageshack.com (secure)]

Our starting point will be the fact that like the height of The Red Pyramid the distance between where The King's shafts exit the Great Pyramid IS ALSO 197.989899 cubits (most have it at 198 cubits exactly) ... so let's show that ...

[link to imagizer.imageshack.com (secure)]



Apr 1, 2018 at 12:50pm

Moving on with a couple of new images before I correct the extremely large error I have made in one of my earlier drawings. (I have no idea how I could have made such a serious error) For now though ... forward.




In this image you will notice that instead of 200 I have chosen as the height 197.9898 cubits. The reason I have done this is to keep in tune with our theme of all things square root of 2 or square root of 3 with 140 x sq rt of 2 = 197.989899. This distance is also the distance between where the shafts of the so called King's Chamber exit the Great Pyramid. As shown below ... but before we go there I have to post the horrible error I made ... Oops just double checked my drawings as I am apt to say ... "I thought I was wrong .. but I was mistaken" so let's carry on. Now what you are about to see is possibly what Pythagoras was playing at when the soldiers stormed his home and he cried ... do what you will with me but leave my circles alone for I think I may have inadvertently solved for The Pyramids but most importantly I think I have shown in the series of some 16 images I am going to be posting that all the pyramids are tied into each other and can be shown to do so from simple, and I repeat here, simple geometry ! And the cherry on top will of course be when I beautifully tie it all into our solar system and the planets for "The Creator" was a very simple creator and used only the simplest of ratios and such and thus it was easy for The Ancient Builders to emulate "it's" work for remember what is written in Enoch where "The Creator" said he would make it so simple anyone could understand it ... and so we begin ...

We will start with Gary Osborn's copyrighted drawing of The Great Pyramid to scale ...



Our starting point will be the fact that like the height of The Red Pyramid the distance between where The King's shafts exit the Great Pyramid IS ALSO 197.989899 cubits (most have it at 198 cubits exactly) ... so let's show that ...



Now before we move on I just would like you to think upon the fact that using 197.989899 as the diameter gives us a circumference of 622.00 and is precisely the diagonal in a pyramid that yields the Pi base for The Great Pyramid. We will examine that more later, for now just think about it ...

However from earlier work on The Red Pyramid I discovered that viewed on the 45 degree angle the base to height was in the relationship 1 to 1.5 so to get 1/2 the base we simply add 1/2 of 197.989899 to it and this is the easiest way to accomplish this ...

[link to imagizer.imageshack.com (secure)]

Now let's draw in the line and set the height properly ... NOW THE THING TO KEEP IN MIND IS ALL WE ARE USING IS A COMPASS, A STRAIGHTEDGE AND A MARKER !

[link to imagizer.imageshack.com (secure)]

Okay now let's draw in The Red Pyramid as seen on edge !

[link to imagizer.imageshack.com (secure)]

Okay so we have nicely drawn The Red Pyramid precisely to scale and is shown by our blue triangle !

Now how easy was that ? So let's draw another pyramid. Let's do The Pyramid of Khafre which is in the sizes 3, 4 and 5 and happily our 1/2 base of our blue triangle of //2 base of The Red Pyramid viewed on angle of 45 and since that can be said to be 3 x .5 the height would simply need to be 4 x .5 and is simply 2 and give us the following series of diagrams ...

[link to imagizer.imageshack.com (secure)]

and drawing in the triangle which is in the precise proportions to The Pyramid of Khafre or G2 ...


[link to imagizer.imageshack.com (secure)]

However there will be those who will claim that this is not what was intended so I decided to go a little further with this drawing and drew a circle with diameter of "6" or diagonal of The Red Pyramid ...

[link to imagizer.imageshack.com (secure)]

and then decided to draw the square using 297 ( 1/2 base of blue triangle) and arrived here ...

[link to imagizer.imageshack.com (secure)]

Now there has been much speculation on why the "Queen's" shaft to the left ends where it does. It has been speculated that it ends at 43.3 degree and matches The Red Pyramids angle and I thought that made a lot of sense but I now think the makes more sense for since we have embarked on a path of sq rt of 3 and sq rt of 2 it seemed logical that the pyramid builders would stick with their theme of sq rt of 2 and make the angle 45 degrees and so I moved the diamond (now a triangle) down ... and got this ...

[link to imagizer.imageshack.com (secure)]

... and noted that it would appear that it 'COULD" be that the angle of our black pyramid hits the end of the shaft of The "Queen's Chamber" as shown in this diagram where I have projected the line ...

[link to imagizer.imageshack.com (secure)]

... and then I got the intersection of the two other shafts and got this diagram ... note here that so fat that intersection point has not yielded me anything yet ...

[link to imagizer.imageshack.com (secure)]

... and last but certainly not least I drew in our 45 degree triangle but with base of The Great Pyramid and got the image below ...

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I believe I promised to tie this all into our solar system and so I shall ...

For that we have to turn 90 degrees and we will begin anew ... but that will be our next post ...

To find any correlation to our solar system I have to once again show the solar system and put the distances on it ... For now we will stick with Mars and Earth ... observe ...

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and now the distances using the semi major axis of both Mars and our Earth

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So here are the ratios we have ...

227,939,100 / 149,598,261 = 1.5236747972625163069241827617234

227,939,100 / 78,340,839 = 2.9095820635773379960865622079947

and lastly we have

149,598,261 / 78,340,839 = 1.9095820635773379960865622079947

Now I have already shown that a meter at 39.37007874015748031496062992126 inches (10,000 / 254 ) divided by the cubit at 20.62 inches gives us 1.9093151668359592781261217226605 and checking to within 0.99986

[link to imagizer.imageshack.com (secure)]

[link to imagizer.imageshack.com (secure)]

But this is old hat my regulars cry out and you are right but I need to set it up for those who have not traveled here with us before.

Now according to me the correct base length for The Great Pyramid based on 9068.9968 inches is 9068.9968 / 20.62 = 439.816 while using 280 as a confirmed and fixed height and using The Pi Angle we get 439.823 and using 197.989899 x Pi as a diagonal we get also exactly the same 439.823 (rounded). So we will try both ...

Now it is further my contention that the base of The Red Pyramid WAS NOT meant to be exactly 420 cubits but was in fact meant to be sq rt of 3 / 2 then times 10,000 and is 8660.254 and using 20.62 inches as a cubit this gives us 419.993 cubits.

So where is all of this going you must be asking yourselves by now and I will tell you.

Firstly I noted using our new 1 to 1.5 ratio of The Red Pyramid that if we allowed 220 cubits to represent the base of The Red Pyramid then the height would be 220 / 3 and then times 2 or 220/3 or 73.33333 x 2 or 146.6666666 and there it might have stood if I had not decided to divide it into the height of The Great Pyramid and got this ...

280 / 146.66666 = 1.90909091 or 21 / 11 and close to our above ratios of 1.9095820635 and 1.9093151668 but I felt it might be closer than this so I decided to use the exact measurements I felt were their and those were 419.993 for base of The Red Pyramid and 439.823 for base of The Great Pyramid ... what would these values yield ? These values were not very good so I went back to the drawing board and decided to reverse engineer it and start with 419.993 for the Red and divide in by our solar system ratio of 1.90958206358 and got 219.94 or base of 439.88 and times 20.62 = 9070.315 inches and equaling 230.386 meters all very extremely close so let's put up one last diagram and maybe we can all see it visually ...

[link to imagizer.imageshack.com (secure)]

So what are our final conclusions after all of this ... well they are simply this ...

The base of The Red Pyramid represents the distance from The Sun to The Earth !
The base of The Red Pyramid represents one meter !

One half the base of The Great Pyramid represents the distance between Earth and Mars !
One half the base of The Great Pyramid represents one cubit of 20.62 inches

Well since I seem to be on an Easter roll I might as well keep going ...

I was thinking what could the ratio be between Mars and Ceres and knowing I had come to a dead end on this many times before I tried yet again to divide Ceres semi major axis distance of 414,001,000 by Mars semi major axis distance at 227,939,100 and got 1.816279. Now only because I had just recently been fooling around with square root of 1.5 which is in fact simply the sq rt of 3 / the sq rt of 2 and it comes to 1.2247448714 but the reciprocal is the key for it gives us 0.816496581 and of course I immediately noticed the close similarities to 1.81628 so I decided to press on and got this ...

I decided that the distance from The Sun to Ceres shall equal sq rt of 2 / sq rt of 3 then plus 1 or simply 1.81649658092772603273242802490196. This should make Sun to Mars equal to 1 and Mars to Ceres equal to 0.816496581 ( sq rt of 2 / sq rt of 3) or thereabouts ... shall we see how we have done ?

Okie dokie ... :)

The most up to date figure we have for the semi major axis of Ceres (remember they just sent a probe there ) is 414,001,000 kilometers and Mars as we all know (or should know) by now is 227,939,100 so let's do the math ...

414,001,000 / 1.81649658092772603273242802490196 (remember this should give us Mars) = 227,911,797 and divided by actual is 227,911,797 / 227,939,100 and checks for 0.99988 ... I guess close enough for my tastes and remembering after all the planets are probably not where they were at "The Creation" :) So how about another diagram or two ... ?

[link to imagizer.imageshack.com (secure)]

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[link to imagizer.imageshack.com (secure)]

[link to imagizer.imageshack.com (secure)]

Okay well the fact that Mars = 1 ties in so very nicely with other things I had noted.

So now we have ....

The Sun to Mars = 1

The Sun to Ceres = 1 + sq rt of 2 divided by sq rt of 3

AND FROM PREVIOUSLY DISCOVERED THINGS

Jupiter = 1 + (sq rt of 1 + 1 or 2.41421356 ) = sq rt of 2 + 2 or 3.4142135623731

Here is an image that is worth way more than a thousand words ...

(very large image) [link to imagizer.imageshack.com (secure)]

So it would seem that the solar system is reasonably simple ... at least up to Jupiter ..

Well shall we keep on going ? Why not ?

Next stop Saturn and amazingly the distance from The Sun to Saturn is the semi major axis of Mars x 2 x Pi or 1 432,183,604 kilometers and here is the image I have made ...

[link to imagizer.imageshack.com (secure)]

So in this image the red of The Martian Circle is equal to the straight path from The Sun to Saturn ...

And from the peanut gallery I hear ... why ?

Well I am still working on that :)

But keep all this in mind for we will be getting back to the pyramids and our Giza Rectangle eventually ...

Okay got another one for you. In this diagram the grey-blue line is equal to the circumference of Saturn (Saturn x Pi) but it is also equal to (Mars x 2 x Pi) then times Pi again or simply Mars x 2 x Pi squared or 227,939,100 x 2 x 9.8696044011 = 4,499,337,489 = Neptune semi Major axis ...

And of course the image ...

[link to imagizer.imageshack.com (secure)]

Uranus is Mars x 2 x 2Pi = 227,939,100 x 2 x 6.28318530718 = 2,864,367,208 (image will follow later, much later)

So it would appear that Mars is yet another unit of measure in our solar system.

But the real question is ... why ?

Much more to come ...

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