The Solar Rectangle that mimics Giza

[Don Barone]

Hi all. I was about 2 lines away from completion when I somehow closed my Firefox. It was all lost. I need to regroup and I will try again.

db

[Don Barone]

Okay to start again. It usually is never as good as the original post but I will try. As some of you may have noted I am trying to put all my ideas down in book form so it will not be lost. As I have often noted this always gives me trouble for when I see an old post it jump starts my mind and I start looking deeper and eventually end up revising the old image. So now to the case in point. I had started to put down all my ideas on the 9 b y 11 triangle and suddenly realized I had never really closed it off. I had never really plotted the 8 planets of our solar system in a clear and concise way on The Giza Rectangle. I decided to go back to the thread where I first introduced the idea and follow along as Nick L. and I built it up together. Sadly all the images were on my old servers but I have all the images saved and it was an easy though lengthy job to find them all. Here is the thread in case you are curious. The Solar Giza Rectangle The first thing I am going to have to do is to agree with Nick L. and admit that the rectangle we need to do the job of plotting our planets is not the 9 by 11 rectangle but is, as he suggested back in 2010, the sq rt of 2, sq rt of 3 and sq rt of 5 right angled triangle.

So we begin anew:

Our Solar Giza Rectangle:

How I discovered this next point is lost in the sands of time but I somehow figured out that the semi major axis of Mars when divided by the semi major axis of Mercury is equal (or almost equal) to sq rt of 3 x 1000 then divided by 440. Thus if we label G1 as Mercury, Mars becomes the full length of the vertical 1732.05, Here is that diagram: (NOTE: I have shown all calculations and proof in the diagram)

 

The honor of the next one I believe goes to Nick L. who came up with the idea of using the sq rt of 2 base to draw and arc to intersect the yellow sq rt of 5 diagonal and then to mark and circle that spot or intersection. After doing the math we found that this intersection lay 821.85 cubits from the north east corner of G1. The calculations are on the page but perfection would have yielded us 822.18. Still extremely close as you will see in the image:

 

Earth was determined when I decided to go 880  cubits down (2 x 440) and from that point draw a horizontal line until it intersected the yellow diagonal. This time when we checked the measurement it yielded us 1136.01 cubits. Perfection would have been 1136.6666. Again the image below is quite revealing:

As you can see we just drew a circle with radius 440 cubits and the diagram speaks for itself:

We have already shown Mars so really the only one left to show is Ceres. Before doing this I had no idea how Ceres would fit in but it turned out to be the most important number other than 440 and Mercury. Unbelievably Ceres and Mars  are in an almost perfect sq rt of 3 (Mars) and sq rt of 2 (added on to get Ceres) that one can imagine. Thus knowing this I present the final image for the inner planets for your viewing pleasure:

We will analyze these ratios in more detail a little later.

I will save this and then move on to the outer planets.

Cheers

db

September 10th, 2024

[Don Barone]

Okay that was nice. Now for the outer planets: Jupiter, Saturn, Uranus and Neptune.

Jupiter is extremely easy to plot as is shown in this image by Bartlett:

The Great Pyramid also shows this as 280 / 3.414214 = 82.01 (cubits) and the floor of The King's Chamber is precisely this ( 1691.5 inches)

and this one going with The Giza Rectangle:

The remaining 3 were not so easy. I had used Ceres as base but was using a number way too large to make it reasonable to fit into a diagram and then I re-read what Nick Lievense had posted on that first thread. Here is his post:

I guess I didn't understand the post or realize what Nick was trying to explain to me. Reading it these 14 years later, it all finally made sense and here is the revised diagram(s)

The first thing I finally realized is that he was telling me to use 250 as a base representing Ceres. But I didn't think there was anyway I could measure 250 cubits and then a couple of days ago it dawned on me. The distance between the south side of G1 and the north side of G2 is precisely 250 cubits. Even nicer than that the distance from the east side of G2 to the center of G1 is precisely 433.012 (sq rt of 3 / 4 then times 1000) The diagonal of course is 500 cubits. Here is the diagram for that:

What Nick also said was something about a square and 4 times and then finally I saw it too. In the next diagram the square of 4 sides of 433.012 is precisely the long sides of The Giza Rectangle and equal to 1732.05 cubits. See the diagram below:

and this image which is showing the entire Giza Rectangle with Uranus the long side on the right and Saturn simply one half of Uranus. Thus Uranus is equal to the square of four sides of 433.012 as well as the long side of 1732.0508075 and Saturn is simply 1 / 2 of 1732.0508075 or 866.025.

FYI: NASA Uranus Data

Here is a little better explanation form a forum I still post to:

It is based on semi major axes and then Saturn = Ceres x 2 x sq rt of 3 and Uranus = Ceres x 4 x sq rt of 3 OR SIMPLY 2 times Saturn

Thus if we make Ceres = 250 cubits on our Giza Rectangle then Saturn = 250 x 2 x 1.7320508075 = 866.025 WHICH IS ONE HALF THE LONG SIDE OF THE GIZA RECTANGLE

Thus if we make Ceres = 250 cubits on our Giza Rectangle then Uranus = 250 x 4 x 1.7320508075 = 1732.05 WHICH IS THE LONG SIDE OF OUR GIZA RECTANGLE

Note that when Ceres = 250 then Uranus is the perimeter of the square defined by 433.012 or sq rt of 3 x 250 x 4

Thankfully the builders were kind enough to make sure 250 cubits was built into the design of The Giza Rectangle as from the south side of G1 to the north side of G2 is 250 cubits !

Seven down and just one to go, Neptune. It is actually one of the easier ones to finish with. It is simply Saturn times Pi and since we have established that 866.025 is Saturn and is one half the long side we simply use this as a diameter and draw the circle. The resulting circle is the semi major axis of Neptune and we never left the comfy confines of The Giza Rectangle. Now how cool is that ? 

Three diagrams to ponder. Take your pick:

Choice #1:

Choice #2:

and finally Choice #3:

So what is the point of all this ? For me it is twofold. The first is that it is possible, yes in fact very easy to plot our solar system using a simple special right angled triangle, a sq rt of 2, sq rt of 3 and sq rt of 5 right angled triangle. The thing we must ask ourselves is why had nobody discovered this before Nick and I ?  The answer is very simple. The triangle on it's own is of little value and the Giza Plateau unless one is looking would never give up it's secret. No the only possible way it can be found or discovered is to believe that The Giza Rectangle is indeed an astronomy book and without the placements of the three pyramids precisely where they are, this could never have been found.  It is the unique combination of the triangle within The Giza Rectangle and it's strategic placing of the pyramids that allowed these things to be found.  There are many other things that we have found along the way that are posted throughout this forum. It all goes back to one thing, no one else found these things  ... because no one had ever looked before. The argument is always but nowhere did they write it down, no that's right ... THEY BUILT IT INTO THEIR PYRAMIDS and to find the answers it is there you have to look.

I really would appreciate some input on these recent postings to see if it is reaching souls out there.

Emhotep

db

[Don Barone]

Earlier I stated that the relationship of sq rt of 3 and sq rt of 2 to Ceres was a very important number configuration and this is part of the reason.

Just to confirm the ratio and it's relationship to Ceres let's do the fact checking.

First Ceres semi major axis is 414001000 km (this number was reported by NASA shortly after they returned from their exploration of this dwarf planet).

Mars was reported on Wiki with an accompanying footnote to a semi major axis being 227,939,100 km. It has changed several times over the years but I am sticking with this distance.

To check the math the ratio must give us sq rt of 3 and sq rt of 2. This means that Sun to Mars = sq rt of 3 and the distance from Mars to Ceres must give us the sq rt of 2. So how do we solve this ?

It is pretty easy. We simply add together sq rt of 3 + sq rt of 2 and when we get the result we divide it into the number of kilometer Ceres is from the Sun. This will give us a value of "1 unit". To find Mars we simply multiply this "1 unit" by sq rt of 3 or 1.7320508075.

Here is the math:

sq rt of 2 + sq rt of 3 = 3.14626437 (rnd)

414,001,000 / 3.14626437 = our "1 unit" of 131,584,937.3483

If our premise is correct then 131584937.4883 * 1.7320508075 should give us Mars' semi major axis.

The math gives us 121584937 * 1.7320508075 = 227,911,797 km. This checks to our value of Mars to 227,911,797 / 227,939,100 = 0.99988 or 99.99%

To check sq rt of 2 we do the same thing. 414,001,000 - 227,939,100 = 186,061,900. So sq rt of 2 * our 1 unit of 131584937 should give us 186,061,900.

Math reveals 131584937 * 1.41421356237 = 186,089,202.

Results reveal that sq rt of 2 checks for 186061900 / 186089202 = 0.99985 or 99.985.

Thus I think it is safe to conclude that the distance of The Sun to Mars was meant to be the sq rt of 3 and the distance for Mars to Ceres was meant to be the sq rt of 2.

This relationship, considering the solar system is supposedly a random occurrence, shouts of "intelligent design". There can be little doubt that our universe was conceived and built on 5 basic numbers:

sq rt of 2
sq rt of 3
sq rt of 5
Pi
Phi

Did The Ancient Builders know this ? Let's look at one of the clues they may have left us. It is called The Bent Pyramid and it reveals these hidden secrets (among many)

Post to be finished at a alter date. I am a bit distracted now.

Cheers
Don Barone

[Don Barone]

Hi all. As a change of pace I am going to go back to my all time favourite painting and of course it is "Et in Arcadia Ego" better known as The Shepherds of Arcadia by Nicolas Poussin. For those who were on a different planet while I was studying this to death here is an image of this painting and my observation of Phi.

As you can see it plainly marks Phi for all to see. And that's basically where I left it in relation to Phi until just minutes ago when I re-read the website of mine and since I have Gary Meisner's Phi program I decided to see if there were any more Phi's to be found. This is kind of neat. Here are the lines I ended up drawing on the painting:

And now for more Phi:

and one more ...

... and all together.

Just thought you might be interested.

Cheers

Don Barone

September 11th, 2024