## Monday, October 21, 2013

### √3, Phi, Pi our 9 by 11 triangle and The Pyramids at Giza and their story

Hi all this may be a bit of a long post or I may break it into sections I don't know yet. In a previous post I showed how simple it was to layout the inner three planets using a simple 9 by 11 triangle and the height and base of The Great Pyramid but now we are going to expand on this and go to another level of sophistication. What follows was discovered by a gentleman named Nick L. as he and I searched for answer on The Giza Plateau but what follows is his and it is beyond remarkable and sadly it has gone sorely underestimated and under appreciated but I will post it here for you to enjoy. As I said it is amazing.

To begin we will start with the discovery of mine that the ratio between The Great Pyramid and G2 or the one commonly referred to as Khafre's Pyramid is in the exact proportion of the square root of 3 to the value Phi or The Golden Ratio. In mathematics and the arts, two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities, i.e. their maximum. The figure below illustrates the geometric relationship.

So this translates to:

a = 1.00000
b = .618033988
a + b = 1.618033988

And Phi is equal to 1.618033989 while Phi squared = 2.618033989 and 1 divided by Phi = 0.618033989 and lastly 1 divided by 0.618033989 = 1.618033989

So according to me The Great Pyramid or 440 cubits is equal to the square root of 3 or 1.7320508075688772935274463415059 while G2 or Khafre's Pyramid is equal to 411.04 cubits and Phi or 1.6180339887498948482045868343656. So let's do some division and see if I am correct in my discovery.

G1 / G2 = ?
440 / 411.04 =  1.0704554301284546516154145581938

But we have decided that G1 / G2 is equal to or is the same as √3 / Phi

√3 / Phi =

1.7320508075688772935274463415059 / 1.6180339887498948482045868343656 = 1.0704662693192697958259095291383

The degree of accuracy is 1.0704554301284546516154145581938 / 1.0704662693192697958259095291383 = 0.99998987 almost a perfect match. Close enough however to conclude this is indeed what was meant.

Okay now for a very quick recap. Here is the triangle when we last saw it.

What Nick did next was the stuff of genius. He decided that to solve for our solar system he needed to use all three pyramids at Giza and this he did in a most amazing and profound way. For by using all three pyramids he solved not only the correct placing of Venus (where I had only been approximate) but manged to also place Mars correct to a degree of accuracy that again goes far beyond the realms of coincidence.

Fist of all let me list the three sizes for the three pyramids at Giza.

First and largest is The Great Pyramid and the base of it is 440 cubits
Second is G2 or Khafre's Pyramid and it is 411.04 cubits
and third is Menkare's pyramid and it is 201.46 cubits.

We have used The Great Pyramid's height and base to find Mercury and Earth and an approximation for Venus but Nick took it up a notch when he devised this simply but profound drawing by using G2 in a most remarkable and simple way. Observe:

Nick simply lined up G2 on the hypotenuse or long side twice and it set the point of Venus almost perfectly. A most ingenious discovery and setting the stage for finding Mars and the rest of the planets as well.

Nick then found something even I was mesmerized by,  by the sheer genius of it. He managed to find Mars as well and to do this he brought in the smallest of Giza Pyramids and used it in a way only someone on a roll would think of doing. It is remarkable again in it's simplicity and for the fact that he found it. But first a bit about G3 and the smallest of the pyramids. Firstly and foremost is the fact that the first quarter of it from the ground up was encased in RED GRANITE. Was this perhaps a bit of a hint that G3 was meant to show us Mars, the red planet ? Also interestingly the west face of G2 or the middle pyramid is precisely 353.5534 cubits from the west face of G3 and 353.5535 is precisely 1000 times the square root of 2 DIVIDED BY FOUR !!! or 1/4 or the square root of 2 !!!

And now let's continue with the genius of Nick L.

As I mentioned the base of G3 or Menkare's Pyramid is 201.46 cubit's and working with that nothing was able to be found but then Nick did something quite clever he decided to calculated THE DIAGONAL OF G3 and this we find is √201.46² x 2 or √40586.1316 x 2 or √81172.2632 or 284.91. Nick then had the idea to subtract 284.91 from the bottom base of our triangle and got 880 (2 x 400) - 284.91 or 595.09 cubits. Nick then decided to add this to our hypotenuse which we had calculated to equal 1137.01363 cubits or √9² + 11² = √202 or 14.212670403551895496970929487628 and by the way 1137.01363 cubits = 440 x 2.584122 Here is the final drawing for the four inner planets of our solar system.

## Wednesday, October 16, 2013

### Introducing 9/11

Hi all the main theme of all of this so far has been that there may be a very simple formula for the layout of our solar system. With a major contribution by Nick L. we came up with this design. First a very quick image to show you where we get the 9 by 11 rectangle at Giza.

Here is the same diagram but setting up the "9" portion of our next lesson.

In the previous posting I showed how easy it was for you to remember and lay out our solar system remembering just a few numbers, 12, 19, 31 and the square root of 505 AND DRAWING A FEW CIRCLES.  This next exercise is going to show you the same thing but in a totally different way. For this exercise we are going to use a triangle !  However I have to emphasize here that our base unit of 440 and 9 were given to us large as life from The Giza Plateau where the height of The Great Pyramid is 280 cubits and thus will be 3.5 of our 9 units and 440 will be the other 5.5 units remembering of course the ratio is  7/5.5 or 3.5 (280 cubits) / 2.75 (220 cubits or 1/2 440)

Here is our first tentative line drawing of our next lesson. We are going to draw the 9 by 11 triangle using what Giza has given us. 440 cubits (base of The Great Pyramid) and 280 cubits (height of The Great Pyramid) and 5.5 (440) and 3.5 (280) and 11 (2 x 440 or 2 x 5.5)

Now one of the things I have advocated over and over again is that not only is The Great Pyramid representing Earth but it is surely representing Mercury as well. More proof of this will follow in later posts. So in keeping with the simplistic view we are building of our solar system I am going to allow the base of The Great Pyramid or 440 cubits to represent the distance to Mercury. I have already shown a direct link between G2 or center pyramid and Venus and G1 or Great Pyramid as Mercury by showing that 1/2 of 440 or 220 x the ratio of Mercury to Venus or 1.8686  = 411.08 and the base of G2 but that is for another lesson. Suffice it to say that there is some powerful evidence to suggest using Mercury as 440 cubits and The base of The Great Pyramid. So let's relabel our diagram with this in mind.

Now before going on we need a recap I think in ratios and the planets so here they are again

Mercury = 57,909,050 kilometers but we use 1.00000000 units
Venus = 108,208,930 kilometers but this is 1.868601 times Mercury or 1.868601 units
Earth as we know (or hopefully we remember) is 31/12 or 2.58333333 units or 149,598.261 kilometers
and Mars is 227,939,100 kilometers and is 3.93615678 units or 3.93615678 times Mercury

And now that we have learned the way to figure out the third side of a triangle that is right angled let's solve for our solar system triangle. This one is pretty straight forward it is simply the square root of 9 squared plus 11 squared or square root of 81 + 121 or square root of 202 or 14.212670403552 units and here is the diagram labelled.

Now interestingly if we divide 1137.0136 cubits which is the long side or hypotenuse of our 9 by 11 triangle by 440 cubits which we have also designated at the distance to Mercury we find that we get 1137.0136 / 440 =  2.58412181818182 and if we further multiply 57,909,050 which is the semi major axis to mercury we find that we get 57,909,050 x 2.58412181818182 or 149,644,040 kilometers, This shows an accuracy of 149,598,261 (actual semi major axis of Earth)  / 149,644,040 = 0.9997 So amazingly with this very simple diagram we have plotted Mercury and Earth to a degree of accuracy of 99.97 % ! We find that we get the following:

Well that is nice but it would be better if we could find Venus as well and so I tried this:

Now wouldn't it be nice if we could find Venus in this image.  I figured the logical place for Venus would be where the diagonal lines cross and so I set about calculating the distance to it. Here is that image.

Now I have to admit to scaling these distances as for the life of me I couldn't think of a way to calculate it exactly and here is what I got. The lime green diagonal calculates out to be 3.9874427 units or 318.9954 cubits. This was solved by first scaling to get 202 cubits or about 10.1 squares on the diagram where each square is equal to 20 cubits then using what we learned about right angled triangles and tan I divided 11 by 9 and got 1.222222 and then multiplying this by 202 we arrive at 246.888 cubits or 3.08611 units (2.525 x 1.2222) for 202 by 246.888 is just a mini version of 9 by 11.
Then to get the long side or hypotenuse we just use what we already learned about a² + b² = c² and we get c = √a² + b² or c = √2.525² + 3.08611² or c = √6.375625 + 9.5240749321 or c = √15.8996999321 or c = 3.9874427810 So now let's go back to our overview image and fill in the details and the sizes:

Okay now to see if it fits to Venus. Well as we know by now that ratio of Mercury to Venus is 1.868601 or 108,208,930 (semi major axis of Venus) / 57,909,050 (semi major axis of Mercury) so to be exact we would need 440 or 5.5 units x 1.868601 and this would give us 822.18444 or 10.2773055

We have 818.02 and 10.23 for an accuracy rating of 99.49 not one of my better efforts but as I said I had to scale the distances. Still it certainly would be close enough if you were going to draw the circles now using these distances. To work out exactly we would need diagonal to equal 3.9353649 instead of 3.987427 this would make our other distances  3.045804 and 2.49502

So is it close enough to draw in Venus ? Well I think so. Now is there an easier way to draw our solar system ? I would think not.

And now let's draw our circles and complete this lesson.

I mean really could it be any easier ?

Cheers
Don Barone

## Thursday, October 10, 2013

### A Possible Masonic Secret ... and it may not be as easy as 3, 4, 5

Hi all ...

Many students of Masonry have come to the conclusion that the secret of the Masons was the 3, 4 and 5 sided right angled triangle. Although many good arguments have been offered up I was never convinced that this is what it was. I always felt it had to relate to my ongoing research into the solar system and the ratios found there.  As some may know who have read my previous posts I have made some interesting discoveries lately on our solar system and eventually ended up with this diagram.

I then turned it 90 degrees to get this.

Please note again that the angles are EXACTLY 1/2 of each other or 2 times the other and then I made a chance discovery on Google of a Knight's Templar apron from The Civil War era and it was this ...

Now I am sure even you can see the obvious similarity

Here is the overlay: an apparent exact match.

I next decided to extend the line where the smaller triangle hits and join it to the place where it meets Earth orbit as below and then I simply squared it.

I then decided to draw a line from where the top red line hits the red square and Earth's orbit and got this diagram:

THE STARLING RESULTS:

Now as far as I can scale it it turns out that the red square exactly matches the circumference of Earth ... in other words the solar system is squaring the circle for us. And to top it all off the angle we get by drawing the green diagonal gives us The Phi angle or 51.827 degrees.

It is magic ... And it is all at Giza !!!!!!!! Just look a little deeper.

Seems there was more than one secret being hidden...

NOTE: I turned the angle with Swish and it could be anywhere from 51.827 to 51.86 ... so it could be either The Pi or The Phi angle.

either way we are definitely on The Square now :o)

Cheers

### Part IV: Finding other mathematical things in the layout of Our Solar System

Back again. To begin this exercise I am going to take the diagram we ended with and turn it 90 degrees and we will have this:

And then next I am going to duplicate on the left what we have drawn on the right and we will get this configuration.

Okay at this point a very quick lesson on trigonometry and incorporating what you learned about before. Right angle triangles have a second guiding principle and that is that if the sides are in ratio to each other the angles will always be the same and the tan or the ratio between side A and side B will be the same as well. For example we are going to use side 19 and side 12. The ratio between 12 and 19 is 12/19 or 0.631578947368. the neat thing about right angled triangles is that any ratio that agrees with this for example 24 / 38 or 48 / 76 or even  18 / 28.5 will yield the ratio of 0.631578947368 and this we call "the tan " of the angle. A "tan" of 0.631578947368 can and does yield only 1 possible angle. So even though all these ratio have different numbers the ratio remains the same and then so does the angle. So in the old days we would have to either figure out the angle or consult a book but now most calculators have the function built in so we would simply key in 12 / 19 up would come 0.63157894736842105263157894736842 you would then press inverse tan and we would get an angle of 32.2756443 degrees. It is that simple.

Now another thing we have to know is that all triangles when their interior angles are added together ALWAYS EQUAL 180 DEGREES So in the triangle of 12, 19 and 22.47220505 we have an angle of 90 (by definition of right angled triangle) and 32.2756443 so this would mean the other missing angle HAS TO EQUAL 180 - 90 - 32.2756443 or 57.724355685422368135731924931312. Again forgive the long number but it is imperative to prove a point. Since 90 degrees is always a given it is quicker and the norm to simply take the one angle from 90 degrees for example 90 - 32.2756443 = 57.7243556... So to sum up a tan (or cotangent) is simply the ratio of the small side into the medium side (or reverse) Every angle has a tan and an inverse or cotangent and they are always different except for the only exception and that is 45 degrees where they are the same.

So now let's solve for our solar system diagram and calculate the angles involved in the diagram. I am first going to label the angles as so:

Angle A would simply be tan of 19 /12 or 1.5833333 or angle of 57.724355685422368135731924931312

Angle C is  90 - 57.724355685422368135731924931312 or 32.275644314577631864268075068688

Okay now let's solve for B. Angle B would be tan of 19 / 34.472205054244231864598140445491 or 0.55116868706548588761042844449951 and we find that this relates to angle 28.862177842711184067865962465656

Angle D is  90 - 28.862177842711184067865962465656 or 61.137822157288815932134037534344

On the surface this doesn't seem overly important but it is showing us an amazing thing. The two angles off of the main line of 34.7422 are EXACTLY in the ratio of 1 to 2. That is that angle 57.724355685422368135731924931312 is EXACTLY double the angle of 28.862177842711184067865962465656 2 x 28.862177842711184067865962465656 = 57.724355685422368135731924931312

THEY ARE EXACTLY DOUBLE OR 1/2 OF EACH OTHER. SO IN OUR LITTLE DIAGRAM OF OUR SOLAR SYSTEM WE HAVE MANAGED TO FIND A WAY OF SHOWING HOW TO DRAW ONE HALF OF ANOTHER ANGLE.

But I then found this ratio and diagram in the most unusual of places and that will be for our next posting.

Cheers
Don

### Part III: Looking for Mars - Teaching a 9 year old to build The Solar System and The Giza Plateau

Hi again all. Today we are going to try to find Mars. This is going to be the beginning of an interesting exercise and it may be a bit advanced for the average reader but I will try to make it as simple as possible. At the moment we have:

Mercury = 12 units (exactly)
Venus =  √505 or 22.47221
and Earth = 31 units (exactly)

What will Mars equal ?

The diagram I have started has some unusual properties which we will discover later but for now let's do another one of those right angled triangle calculations. This time we are going to use the distance from The Sun to Mercury added to the distance from The Sun to Venus. This gives us 12 (Mercury distance) plus √505 or 22.47221 (Venus distance) and we get 34.47220505. This will be our "B" side. Our "A" side will be 19 units so we will have this configuration: Our "B side is in red and our "A" side is in baby blue.

So our long side is remembering our formula of c = √a² + b² or C = √34.47220505² + 19² so doing the math we find that our C side is equal to √1549.33292130186 or 39.361566550403727100264412305514. Forgive the long number here but it is imperative to show the accuracy of my next point. Here is the diagram:

Okay now it is back to NASA for the correct distance to Mars. Checking their website (NASA's) we find that the semi major axis (remember just the average of furthest and closest) of Mars is equal to 227,939,100 km. Now a by product of this diagram is a very interesting relationship. We find that if we take 5/6ths of Mercury's inches or 10 inches we find this:

Mars is 227,939,100 km
Mercury is 57,909,050 km
The ratio ( or the number of times Mercury's distance can be divided into Mars distance) is  3.9361567837842272

And now if we divide 10 (our new distance for Mercury) and divide it into our pink line above of 39.3615665504037271 we get 3.93615665504037271

On checking for the accuracy of this we find this:

3.9361567837842272  / 3.93615665504037271 = 0.999999967 or 33 billionths or a virtual match !

Okay but what can we find still using Mercury as 12 inches. Well really the only thing I could come up with to find a close match was to take the distance from The Sun to Earth or 31 inches and add 51 divided by Pi to it. Why it works I don't know at the moment but we really are just teaching Grade 3 here. So 51 / Pi = 16.2338042 and if we add this to Earth or 31 inches we get 47.2338042 inches. if we divide this by our Mercury base of 12 we get 47.2338042/ 12 or 3.936150349614444. since we already know what it should be from above (The ratio ( or the number of times Mercury's distance can be divided into Mars distance) is  3.9361567837842272)  we get a ratio or accuracy rating of:
3.936150349614444 / 3.9361567837842272 = 0.99999837 or 1.7 millionths ! So now we have the four inner planets easily drawn and easily remembered. They are as follows:

Mercury = 12 units or inches
Venus = √505 units or inches
Earth = 31 units or inches
Mars = Earth + 51/Pi or 31 + 16.2338042 or 47.2338042 units or inches

And we arrive at this image:

I hope this has made it easy for you and you should now be able to remember and draw our solar system with no trouble at all.

But this diagram shows us so much more and this we will discuss in our next posting.

Cheers
Don Barone

## Wednesday, October 9, 2013

### Part II: Teaching a 9 year old to build The Solar System and The Giza Plateau

Hi Tommy and those who have decided to follow along. Well we have found a very easy way to remember the distances to Mercury and Earth.  Distance to Mercury from our Sun is 1 Giant's Foot. 12 Giant's inches or simply 12 units while Earth works out to be EXACTLY 31 units. That is very interesting but is there anyway we can find the other inner planets (inner planets being Mercury, Venus, Earth and Mars with Ceres being on the border) ?  Well I thought there might be a way so I started to try a few things. The first logical thing to try was to see if the distance from Mercury to Earth worked into anything and so I took this distance (Mercury to Earth) or 31 inches - 12 inches or 19 inches and tried to see if I could do anything with it. The first step was to draw a 19 inch or 19 unit line 90 degrees to our original line as in the diagram below: (Please remember that clicking on any of the images will give you the larger and true size.)

Now before we go on we have to remember a very simple yet very important mathematical oddity. if you have watched the Wizard of Oz you will have heard The Scarecrow say, on receiving his brain or diploma, the square of the hypotenuse is equal to the sum of the squares of the other two sides. This shows symbolically how important this one little thing is and that is that in a right angle triangle, a right angled triangle being a triangle with one angle exactly 90 degrees, that if you square the length of side "A" and then add it to the square of side "B" you will get the square of side "C". It is clearer in this simplest example of  this triangle the 3, 4 and 5 unit triangle where 3 x 3 or 9 plus 4 x 4 or 16 is equal to 5 x 5 or 25.

.Another way of saying this is that Side C (or 5 in our diagram) is equal to the square root of 25 -  (square root simply being that number which when multiplied by itself gives us the number in question: eg. square root of 25 would be 5 ( 5 x 5) and square root of 16 would be 4 ( 4 x 4 ) and the square root of 9 would be (3 x 3 ). In any right angled triangle the small side squared plus the middle side squared WILL ALWAYS EQUAL THE LONGEST SIDE SQUARED ! So in the above diagram 3 would be equal to our 12 while the 4 would equal our 19 unit side. So we need to look at the image below ...

... and try to figure out what our "5" side is equal to. Following our formula of a² + b ² = c² or c = √a² + b² So substituting in our image above we have a = 12,  b = 19 and c is unknown at this time. So now all we do is substitute so we get .c = √a² + b² or c = √12² + 19² or c = √144+361 or c = √505 or
c = 22.47221 units or inches So the white line in our above diagram is equal to 22.47221 "inches". But having used certain numbers very often I immediately recognized that this number was significant for in a year on Venus there are, wait for it, there are 224.7 Earth Days, This means that Venus orbits or circles The Sun in 224.7 Earth days. I found this very nice that the length equaled the Venus Year in Earth days but could it relate to our drawing of the solar system. Well for this we have to visit NASA where it is said that Mercury is 57,909,050 kilometers on it's semi major axis (remember semi major axis is simply the average distance that a planet is away from The Sun when you take it's closest and furthest approach. Mercury being 69,816,900 km (furthest) and 46,001,200 km (closest) for a total of 115,818,100 and divided by 2 to give us 57,909,050.) Now Venus is 108,208,930. The ratio between them or the number of times Mercury's distance can divide into Venus' distance is simply 108,208,930 / 57.909.0505 and we get 1.868601367. Now since Mercury is our diagram is 12 units Venus would simply be 12 x 1.868601367 or 22.423216406. Not exactly correct but very close. Accuracy would equal 22.423216406 / 22.47221 = 99.78 % Not as close as I would like it to be but heck Tommy is only in grade 3 and I am sure he would get a passing grade. So as a recap now we have:

Mercury = 12 "inches"
Venus =  √505 inches
Earth = 31 inches

So now Tommy (or anyone for that matter) can draw the solar system TO SCALE by using just these three simple numbers.

We will find Mars and a whole lot more next.

## Monday, October 7, 2013

### Teaching a 9 year old to build The Solar System and The Giza Plateau

Hi all today I am going to teach little Tommy how to build the solar system. You can listen in if you want.You might find it interesting. Are you ready Tommy ? Great ! Let's begin.

Okay Tommy before we begin I need to ask you a question. "What is the closest planet to The Sun ?"

"Mercury ?"  "Yes that is correct. Good job !"

"Now how far is it away from The Sun ?"

"Excuse me what was that you asked ? What measurement system do I mean ? Well now isn't that a loaded question. It could be miles, it could be kilometers, it could be meters, it could be feet and it could be inches. All of them would be correct but which one should I use for this exercise ? Well Tommy which one would you like to use ?

Feet you want to use feet ? Gosh that would give us a pretty large number. Wait what do you mean I don't understand I thought you said you wanted to use feet ? No that is not what you said. You want to use FOOT ? YOU WANT THE DISTANCE FROM THE SUN TO MERCURY TO EQUAL 1 FOOT ? Oh dear you say I have it wrong again. You did say FOOT but you suggest we use giant inches instead. Okay so let me get this straight you want the distance between The Sun and Mercury to equal 12 super giant inches or one giant foot, is that correct ? Okay why not.

Now folks the point of this is that distances really are only designed to be relative. Whether it is 10 miles or 10 million inches really doesn't matter as long as when we compare the distances between two things we use the same measurement system. For example let us say from Town A to Town B is a set distance, for now not important. From Town B to the Next town is double the distance from town A to town B.

Se we would have this simple example. From A to B is simply 1 unit. It could be any measurement we decided to use. Be it inches, feet , meters or any new measuring system you cared to devise. For a joke I am going to say that this 1 unit is equal to 62 Barones. It really doesn't matter what a Barone is as long as when I relate it to a second distance I use the unit measure of "The Barone" So since the next distance is twice the distance from A to B we have B to C equaling 124 Barones and total distance is equal to 186 Barones. Problem is however you are not familiar with the unit measure "The Barone" as you use something else. You use a measurement called "The Mile" and it turns out that from town A to town B is exactly 31 of your miles. So by simple reasoning we can see that if I have 62 units of my measurement and you have 31 units of your measurement then 1 of your units equal 2 of my units. So the ratio is 1 mile = 2 Barones. So you are correct by saying 31 miles and I am correct by saying 62 Barones and Little Tommy would be just as correct as saying the distance from A to B IS NOT 31 MILES OR 62 BARONES BUT IS IN FACT SIMPLY 1 GIANT'S FOOT OR 12 GIANT INCHES and the ratio or as called when using this on a map the scale would be 1 Giant's Foot = 31 miles and 1 Giant's Inches would simply be 31 miles divided by 12 or 2.5833333. So 1 Giant's Inch equals 2.58333333 miles.

So now let's use that simple reasoning to our solar system as suggested by Little Tommy. Let's use 12 giant inches to equal distance to Mercury and we would have a diagram something like this:

And below see the orbit of this. The radius of Tommy's Mercury is simply 12, 12 Giant's Inches or ... well let's call them Tommys for lack of a better name for now.

Okay now comes the only memorization that is required. I have not figured out exactly why it is this but it is. The distance to Earth using this Giant's Scale is 2 feet and 7 inches or 2 and 7/12ths or 2.583333. as I said I am not sure why it is but again as I said ... it simply is.

So basically now for Mercury we have 12 inches and for Earth we have 31 inches. and as I have shown above the ratio or the number of times Mercury's distance from the Sun divides into Earth's distance from The Sun is 2.583333 times or 2.58333 x 12 or 31 Giant's Inches.

But of course we do not use Giant's Inches so we have to convert and check to see if this system has worked. The actual value of Mercury from The Sun is 57,909,050 kilometers or 12 Giant's Inches or 1 MERCURY UNIT OR 1 GIANT'S FOOT ! The actual distance to Earth from The Sun is 149,598,261 kilometers or 31 Giant's Inches or 2.5833333 Mercury Units or 2.583333 Giant Feet.

So checking we simply multiply 57,909,050 x 2.58333 and if all is well we should get 149,598,261. Doing the multiplication we find that 57,909,050 x 2.58333 = 149,598,379 and dividing the answer we should have got by this answer will give us how close we are. So 149,598,261 / 149,598,379 = 0.9999992 or 8/10,000,000ths or 8 ten millionths. Virtually an exact match. So from now on when you think of Mercury and Earth just think of 12 and 31. 12 the number of zodiacs or disciples or inches in a foot and 31 ... well how about 12 + 1 and then reversed or 13 reversed or 31 :o)

Mercury and Earth ... 12 and 31. It is that simple.

It is as simple as 1 2 3 (1+2) or 3 4 (3+1)

We shall find Venus and Mars in our next exercise.

Cheers

## Sunday, October 6, 2013

### Just a very short addendum to previous post.

We have calculated for the base a distance of 9068.9968211710892529703912882108 inches with 1/2 of that being 4534.4984105855446264851956441054. We have also determined that this equals the quarter distance of the circumference or in this case what we are now calling The Megalithic Yard of 2.720699046. So we have  2.7206990463513267758911173864632 = 4534.4984105855446264851956441054 so what would 1 equal ? Well let's divide and we get surprisingly 1666.6666666666666666666666666667 inches. Very nice. Also ( and please forgive the very long numbers it helps to emphasize accuracy) let's calculate the circumference in inches if we assume a height of 1 / √3 x 10,000 or 5773.5026918962576450914878050196 C = 2Pi x r and in this case it equals 36275.987284684357011881565152843. If we now divide this by the value of our megalithic yard of 2.7206990463513267758911173864632 we get 13333.333333333333333333333333333 again very nice.

## Saturday, October 5, 2013

### The Great Pyramid: Squaring the circle and showing us that it really is square root of 3 and so much more

Hi again.

This time I hope to present a show that Barnum would have been proud of. First I am going to square the circle using The Great Pyramid. Second I will show how The Great Pyramid is indeed representing the square root of 3. We will then show how The Megalithic Yard (simply 1/4 of the circumference of a Pi circle) is derived and how important it is. And last (for now ) I will show (and prove) why and how G2 or the middle pyramid is equal to Phi or 1.618033989. The fact that they also represent Mercury and Venus will be for the next post.

SQUARING THE CIRCLE AT GIZA

There have been many attempts to show how the fact that  2 times Pi and all that equal The Great Pyramid's base but it is a bit more elegant than that. First here is a diagram I found on a site that dealt with The Megalithic Yard.

And then I drew the red and the blue line you see in the next diagram.

In the Perfect Pi triangle (pyramid) that which we have labelled √3 would be 4 and the blue line would have a value of Pi or 3.14159265 so the resulting ratio would equal 0.785398163397 if we use the inverse which would give us the larger angle or 1/tan then the angle for Perfect Pi is equal to 51.85397401278. From this we can also calculate the distance for the blue line. It is simply 0.785398163397 times the square root of 3 or  0.785398163397 x 1.732050807569 =  1.360349523176 or PRECISELY 1/4 OF THE BASE OR 1/2 OF THE MEGALITHIC YARD OF  2.720699046352

The next step  was obvious to me and that was to draw a square using the blue line as 1/2 of one of the sides or 1/8th of the perimeter of a square and I got the next diagram.

Now let's make the circle a little clearer.

Now I suggest we label the sizes but I think most can already see where this one is going.

THEY ARE IDENTICAL ! THIS MEANS THAT THE BLUE SQUARE BOX SQUARES THE RED CIRCLE !!!

Also please note that our new radius is precisely or exactly 1/2 of The Megalithic Yard or GB in our diagram of 2.72069903.

This is a neat little example of a perfect Pi triangle and now I am going to show you that The Great Pyramid is that perfect Pi triangle (pyramid).

The GREAT PYRAMID IS THE SQUARE ROOT OF 3

A while ago I made the discovery that I thought that G1 represented the square root of 3 and that G2 represented Phi. This was based on the fact that if one divides the square root of 3 (1.7320508075688773) by Phi (1.618033988749895) one gets a ratio of 1.0704662693192698. And the ratios of the two Giza Pyramids G1 and G2 using the then known supposed bases of them was equal to 440 / 411.04 = 1.07045543012845465 accuracy of 0.99999 to be exact we would need 440 /  1.0704662693192698 or 411.035838. close enough for me to declare it an exact match and there it sat for a couple of years until now. Recently we have been having discussions over at a site I post at about what the height of The Great Pyramid could have really been and most claim 280 cubits while others say 481 feet while still others claim 5773.50 inches. Now this last one is an eye opener and is  the true height because of the logic of it. Allow me to ask you a question. if you were designing The Great Pyramid and wanted to encode the square root of 3 how would you do it so it was obvious to all but the truly blind ? Well one way would be to make it an obvious ratio to another large object, maybe say another full scale pyramid and use that other one as an obvious marker as well so that no other known ratio would or could fit (√3 to Phi) or you could do what the builders at Giza did and you could incorporate the square root of 3 in the actual construction by using it's reciprocal to represent the height. But how would you do that and still make it stand out. Well again you could do what the builders id and you could take the reciprocal of the square root of 3 which is 0.577320 and then to make it large enough to last for all eternity you could multiply it by 10,000 and then using the lowliest of all measurements but really the most important one and the one that man himself is built on, you could take the inch or "the digit" and use 5773.50 of them. AND THIS IS WHAT THEY DID !!!

... says the heckler in the third row. Well okay says I.

Let us go back to our perfect Pi triangle. If The Great Pyramid was that triangle then it must follow our perfect Pi formula and that is that the height is "4 units" while 1/2 the base is "Pi units". Therefore if we divide 5773.50 by 4 and then multiply by Pi or 3.14159265 then we should get the value as measured at The Great Pyramid. but first let's list the measured values of the four sides of The Great Pyramid as measured by Flinders Petrie in 1881. they are:

 Length. Difference from Mean. Azimuth. Difference from Mean. N E S W 9069.4 9067.7 9069.5 9068.6 + .6 – 1.1 + .7 – .2 – 3' 20" – 3' 57" – 3' 41" – 3' 54" + 23" – 14" + 2" – 11" Mean 9068.8 .65 – 3' 43" 12"

So now let us do the simple math. Height is 5773.50 inches so divided by 4 and we get  1443.375673 and now multiply by Pi and we get 4534.49841 but that is only one half of the total base. Now if we multiply by two to get the entire base we get 2 x  4534.49841 = 9068.9968. so this shows that according to my calculations Petrie may have been off by 0.2 of an inch. Pretty convincing I'd say. But now what of my claim that G1 is the square root of 3 and G2 is equal to Phi. Does this help or hinder my hypothesis ? Well let's do some more math. If my hypothesis is true then if we divide 9068.9968 or base of The Great Pyramid in inches by the square root of 3 and then multiply by Phi then we should get the base of G2 in inches. Okay here goes. 9068.9968211710892529703912882108 / 1.7320508075688772935274463415059 = 5235.9877559829887307710723054658 but before we go any further let take a look at the inverse of this number. We get 1.9098593171027440292266051604702e-4. We will come back to this. But now if we multiply 5235.9877559829887307710723054658 by Phi or 1.6180339887498948482045868343656 we get ...  8472.006 inches !

The actual measured north side of G2 as measured by Petrie is ... wait for it ...

 Length Diff from mean Azimuth Diff from mean N. side E. side S. side W. side 8471.9 8475.2 8476.9 8475.5 – 3.0 inches + 0.3 inches + 2.0 inches + 0.6 inches – 5' 31" – 6' 13" – 5' 40" – 4' 21" – 5" – 47" – 14" + 65" Mean 8474.9 1.5 – 5' 26" 33"

Yes that is correct it is 8471.9 and checks to within an almost impossible degree of accuracy. Co-incidence ? Not a chance !

So in conclusion we can say two things now with certainty.

1) The Great Pyramid is a perfect Pi triangle or in this case pyramid

and

2) G1 and G2 are in an exact ratio of  √3 to Phi

It is a masterful display of engineering and goes beyond what even I had hoped to find here. However, and this is a huge question, how did they know about ... the inch ?

And the good part is we have only scratched the surface of what is there to find. We have yet to relate it all to the solar system and reality and that will be our next post.

Cheers until next time.

## Friday, October 4, 2013

### The Giza Solar System - Tidying up a few loose ends

Hi all here are a couple of small things I will bring up here while I have them fresh in my mind. Since I am convinced that our solar system was "intelligently" designed or at the very least follows some very simply ratios I would like to submit these two items for your perusal.

The first has to deal with square roots and circles both at the same time. At Giza the largest or Great Pyramid has a presumed size of 440 by 440 cubits. The cubit in Egypt is thought to have been around 20.62 inches + or - 0.06 inches.

Here is a second image that has subdivided the Great Pyramid into 4 sections and we now get a smaller square of 220 by 220 cubits with the resulting diagonal giving us 311.125 cubits.

What we are going to do now is draw circles using the side and diagonal we see in the above image.Since the formula for circumference is simply Diameter x Pi we get (220 x Pi) and (311.125 x Pi. ) This will give us 691.15 for 220 x Pi and 977.4343 for 311.125 x Pi. However what we are going to discover next is a neat little thing I found and it is amazing how many seem to think this is just a neat little coincidence but they fail to realize that this meshes the circle and the square root beautifully. I am going to next post a diagram that now after I have explained where I got the number from you should be pretty amazed. Here is the image.

What we see above is an image of how The Giza Plateau looks from a little further out. Please notice that the circumference we used of 220 x Pi or 691.15 just happens to match the distance from the north side of the top pyramid to the north side of the middle pyramid. Gee when I found that I thought that it was pretty neat but you know it gets a whole lot better. First we have labelled this 691.15 (20) as simply 1 unit. But we further not that if we call this 1 unit them the orange line and the line we have labelled is the square root of 2 to the 1 being 691.15 for 691.15 x 1.4142135 = 977.43 which corresponds to our orange line in distance. But now let's post an image that combines the two of them.

This lttle discovery of mine led to a complete solving of The Giza Plateau using this method and I will post it shortly but let's move onto much larger circles. The kinds you might find in the solar system. Firstly let me show you a very interesting relationship between 2 sets of 2 of the planets, them being Mars and Jupiter and the second set being Ceres and Uranus. They are in a very special and easily recognized ratio to each other. The semi major axis of Mars, (semi major axis simply being the average of the furthest and nearest that any planet gets to The Sun) is 227,939,100 kilometers while Jupiter is 778,547,200 kilometers with the ratio being, ratio being simply how many times will something divide into another number, 778,547,200 / 227.939.100 =  3.415593025. This is extremely close to (√2 + 2) or 3.414213562 This allows us to set up a couple of interesting diagrams at Giza knowing this information. Here is one of them:

In the above diagram we have Mars situated at a point where it is in the ratio 1 to 2.14142135 with Mars being in kilometers, 227,939,100 of them away from our Sun. Due to the fact that the green line is precisely 2000 cubits (this has been proven with autocad, and Petrie's notes and triangulation). Because of this fact we have the total line equal to distance to Jupiter of 227,939,100 x  3.415593025 = 778,547,200 kilometers. What we have done is divide 2000 by 3.14142135and arriving at 585.79 and this is the value for Mars.

Now below we have another two examples we are going to look at. Here is the first one:

In the diagram you will note that Mars is 1 unit or the radius or 227,939,100 kilometers.

Since in any circle we have the formula for a circumference as 2 x Pi x r (radius) we get for this circumference simply 3.14159265 x 2 x 227,939,100 or 1,432,183,604 kilometers

The actual distance for the semi major axis of Saturn is according to the latest figures 1,426,666,422 km however many agree it is 1,433,449,370 km. so it is somewhere around here and it shows simply that taking the semi major axis of Mars and multiplying by 2 times Pi will give us the distance to Saturn HOWEVER this is simply the circumference so it can be said that the circumference of Mars' orbit around The Sun is the same as the semi major axis of Saturn. (accuracy = 1,432,183,604 / 1,433,449,370 = 0.9991). So in the above diagram Mars is the radius of the circle and Saturn is the purple circumference. A neat little discovery courtesy of The Giza Plateau.

Would you like one more ? Okay one more.

Above I also mentioned a ratio of another two planets, Ceres and Saturn. The semi major axis of Ceres is 413,690,604 km and for Uranus it is 2,876,679,082 km. when we do all the calculations we get the circumference this time as being 2,600,186,830. So could this have been the distance that Uranus was at before it got knocked into it's present weird axis ?

However we must note here that the semi major axis of Neptune is 4,503,443,66 kilometers

The hypothetical circumference of Uranus should be 2,600,000,000 km with the actual calculations showing 2,600,186,830 km

Is there a relatiobship between the two ? Well remember the square root of 3 ? And how the height of The Great Pyramid was 5773.50 inhces ? Well why don't we try that one. I mean afterall it is pretty prominent on The Giza Plateau. So let us divide the larger by the square root of 3 and we find that 4,503,443,66 (the semi major axis of Neptune) / 1.7320508075689 (square root of three) and we get 260,006,441 interestingly almost precisely 10 times (actual of 10.00471 or accuracy of 0.99995) smaller than the hypothetical circumference of Uranus.

A last diagram to show the above.

Okay so we have semi major axis of Ceres of 413,690,604 x Pi x 2 = possible circumference of Uranus but we also have 4,503,443,66 (the semi major axis of Neptune) / 1.7320508075689 (square root of three) and we get 260,006,441 so let fine tune this equation:

Ceres we will label Ca (Ceres axis)
Uranus we will label Ua (Uranus axis)

Ca x 2 x 3.14159265 = (Ua / √3) x 10

Most interesting.

See you later.

Cheers
Don