Hi all. I will be using this a lot in the near future since my usual sounding board, Graham Hancock Message Forum, has issued me with an "indefinite suspension" after deleting a series of post (Noah's Great Rainbow) that were moved from "Mysteries" to "Inner Space". Since that is not where I posted them and did not agree that that is where they should have been I deleted them and boy oh boy like all those in power they do not like people disagreeing with them so they kicked me off the board. Oh well ...

I continue to try to find rhyme or reason in the distances of the planets from The Sun and while trying a few things with the outer planets came up with this little gem. But first we must understand that Kepler found a neat little formula that explains the ratio of the planets. It is a formula that tells us where they are but not why they are there but maybe we can find that one day. Here is his third law:

The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

Earth has it's orbital period 365.2564 days and we get for Mercury at 365.2564 / 87.969 (days in Mercury year) or the orbital ratio period of Mercury as 4.152103582. Thus if we square this we get 17.2399641551892 and now if we find the cube root of this we get 2.5833234495 and if we check the semi major axis we get Mercury at 57,909,050 and times 2.5833234495 we get 149,597,806.8. Now the semi major axis of Earth is 149,598,261 so this checks for 149,597,806.8 / 149,598,261 = 0.999997 or basically an exact fit. But then we have to remember Giza and this is what I came up with.

Here is a chart that I think will be helpful:

In the chart I made above we have the speeds of the planets as they rotate around The Sun using Mercury as a base of 1 (the quickest) and the ratio of the speeds of the other planets as a percentage of the quickest and then we have how many times their speed can be divided into the speed of Mercury. [example: Mercury = 47.87 / 35.02 (Venus) = 1.366933 ] and the last column is the ratio of the semi major axis' with Mercury of 57,909,050 being 1 as a given. Now column 1 and 2 are simply reciprocals of each other so that for Earth 1/0.622102 is equal to 1.607455

So where does this leave us well we have this formula now:

Ratio of the ORBITAL PERIOD of any given planet in relation to Mercury and then squared will yield us the ratio of those same two planets semi major axis' CUBED !.

Example: Jupiter = 4332.59 days to orbit Sun divided by 87.969 days for Mercury to orbit sun) = 49.2513 and now squared and we get 2425.6932 and cube rooted = 13.43627 and Jupiter is 778,570,200 kilometers from The Sun so 778,570,200 / 57,909,050 = 13.444706829071 or 0.9993

But let's concentrate on Earth and Giza. For a very long time now I have made the claim that The Great Pyramid is representing Earth

__Mercury (it is also really representing the entire solar system but that will be another posting) I have also pointed out (can't remember where) that the ratio of the speeds of Earth and Mercury of 0.622102 is strangely similar to the diagonal at Giza and The Great Pyramid of 440 cubits squared x 2 and square rooted to get 622.254 as you can see very close to 0.622102__

**AND**Now as I mentioned I was doing work on the outer planets specifically Neptune and noticed that Neptune was ABOUT 30.08 AU (astronomical unit = 149,597,870.7 kilometers) from The sun on it's semi major axis. So what you might say well I was playing with cube roots and noticed this little gem ... cube root of 30.08 is ... 3.109992. Now because I have worked these numbers to death I immediately knew what this meant. If we multiply by 100 we get 310.992 and this is a very important number. But first back to The Great Pyramid. So now we go with a couple of IF's"

__the height of The Great Pyramid was meant to be exactly 280 cubits and__

**IF**__the angle involved was supposed to be Pi__

**IF**we would get this:

Pi / 4 = 3.1415926535897932384626433832795 / 4 = 0.78539816339744830961566084581988 x 280.00 and we get for half the base of The Great Pyramid 219.91148575128552669238503682957 and times 2 we get 439.8229715 for the base. So what ? Well I am getting to the punch line. If we now calculate the diagonal of the perfect Pi measurements at Giza for The Great Pyramid we get ... Square root of [(439.8229715 squared) x 2) and we find it equals 622.0036 and now we find a couple of things.

1) 622.0036 is almost identical to the ratio of the speeds of Mercury and Earth but 1000 times larger

2) One half this number 622.0036 = 311.0018 and is virtually exactly (if we divide by 100 to get 3.110018) identical to

**THE CUBE ROOT OF NEPTUNE IN AU'S FROM THE SUN !**So 3.11018 cubed = 30.080755 and if we multiply this times 149,597,870.7 we get for the semi major axis of Neptune 4,500,016,888.4 and the number given for the semi major axis of Neptune ranges from 4,498,542,600 km to 4503443661 km (depending on site checked and my gut feeling that is was supposed to be 4,500,000,000 EXACTLY) and thus we have basically scored a bulls eye and have shown that

**the orbital period and distance of Neptune is****directly related to the ratios of Earth and Mercury**

**AND THE MEASUREMENTS FOUND FOR THE GREAT PYRAMID !**Now there is another clever discovery I have made which I will post a bit later on a new blog. Cheers and I hope you enjoyed this and the other of my blogs.

EDIT ADDED ON OCT. 10TH, 2014 @ 2:27 PM EDST.

This is from a post I made on a certain message board on Feb 20th, 2012:

*As most of us here know in order for The Great Pyramid to work out to Pi exactly the dimensions would have to be 439.82. Using this measurement as the base allows all angles and everything else to equal Pi.*

Now imagine my surprise as I was playing around with Bennett's numbers and decided to multiply 0.8660254 his distance from Venus to Earth if we allow Mercury's distance as 1. Total distance would then be 1.8660254 and if we divide this by 3 we get 0.622008468

Now I guess it is because I have worked with these numbers so much but I immediately recognized that number and came up with this.

0.622008468 x 1000 = 622.008468

If we allow this as a diagonal of a right angled 1,1 to sq rt of 2 triangle we get the sides to equal astonishingly as:

622.008468 squared = 386894.5342

386894.5342 / 2 = 193447.2671

Square root of 193447.2671 = 439.82641

Unbelievable ...

So the equation we have is this:

{[Sq rt of 3 divided by 2] + 1}

-------------------------------- x 1000 = Diagonal of G1 in it's perfect Pi form.

3

Nice and neat and accurate.

Now imagine my surprise as I was playing around with Bennett's numbers and decided to multiply 0.8660254 his distance from Venus to Earth if we allow Mercury's distance as 1. Total distance would then be 1.8660254 and if we divide this by 3 we get 0.622008468

Now I guess it is because I have worked with these numbers so much but I immediately recognized that number and came up with this.

0.622008468 x 1000 = 622.008468

If we allow this as a diagonal of a right angled 1,1 to sq rt of 2 triangle we get the sides to equal astonishingly as:

622.008468 squared = 386894.5342

386894.5342 / 2 = 193447.2671

Square root of 193447.2671 = 439.82641

Unbelievable ...

So the equation we have is this:

{[Sq rt of 3 divided by 2] + 1}

-------------------------------- x 1000 = Diagonal of G1 in it's perfect Pi form.

3

Nice and neat and accurate.

Now to this we can add sq rt of 3 divided by 2 and then add 1 to get 1.86602540378 and if we divide by 3 we get 0.622008468 but this is precisely the ratio of the speed of Earth to the speed of Mercury of 29.78 (average speed of Earth in km/sec) divided by 47.87 (average speed of Mercury of 47.87 IF WE USE THE SEMI MAJOR AXIS OF 57,909,050 AS THE RADIUS OF A TRUE CIRCLE) or 0.622102 accuracy 0.99984 (0.6220036 / 0.622102 )

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