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 ?

Don Barone

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