## Thursday, October 3, 2013

### Part V - Giza continues and then putting on some "logs"

Hi Nick what am I missing ?

... Well the ratio between the semi-major axes of Neptune and Jupiter is 2798.3101568091 million miles divided by 483.76680187701 which = 5.7844195714788.

I get semi major axis of Jupiter as 778,279,101.84 + or -

How do you get 483.76680187701 ?

Best
Don Barone

-------------------------------------------------------------------

Author: Nick L (82.4.204.---)
Date:   19-Jul-10 03:04

>>I get semi major axis of Jupiter as 778,279,101.84 + or -

How do you get 483.76680187701 ?<<

Hi Don

I am working solely in miles here, Wikipedia (which is where I took all my data from for consistency) gives 778,547,200 km for Jupiter's semi-major axis. Converting this to miles gives 483766801.877 miles or 483.76680187701 x 1 million.

Best
Nick

-----------------------------------------------------------------------

Hi Nick yeah thanks sometimes switching back an forth you miss it HOWEVER it does bring up an interesting point on why I missed it.

The distance from The Sun to Neptune IN MILES (2,798,310,156.81) equals almost exactly the distance from The Sun to Uranus IN KILOMETERS (2,876,679,082)

I guess if you were going to create something and show two different measurement system this would be an ideal way of doing it. Make the last planet the same distance as the second last planet when converting to the two different measurement system you wanted the people to use or the two measurement systems you used in "The Creation"..

Most interesting indeed.

Best
Don Barone

Edit October 3rd, 2013: Actually the ratio is not close at all. Really about 97.25 %

--------------------------------------------------------------------------
Hi Nick playing around with that theme, that theme being that the difference in the measurements was deliberately encoded into the Solar System we can come up with this:

If we allow the distance to Neptune as 4,50,000,000 kilometers PRECISELY we would get 2,796,170,365.07 miles

If we take the square root of this we get 52878.827 if we divide by 10 we get 5287.8827. If we allow this to equal inches we get 440.65 very nearly the base of The Great Pyramid in cubits.

So let us now work backwards from 440 even. Multiply by 12 gives us 5280

5280 x 10 = 52800

52800 squared = 2,787,840,000 (Neptune = 2,798,310,156.8)

However if we allow 440 cubits or as Petrie states 9068.4 inches we can arrive at this. Simply divide by two to get 4534.2. If we then multiply by a million we get 4,534,200,000 - Is this the true distance that Neptune is from The Sun in kilometers ?

Not sure but close enough maybe to warrant further investigation.

Regards
Don Barone

---------------------------------------------------------------------------------
Author: Nick L (82.4.204.---)
Date:   21-Jul-10 16:31

>>Nick L "there can only be one rectangle which encloses the 3 pyramids"

Not true actually.
You have the far NE corner and the far SW corner as starting points.
With these several rectangles are possible.
I favor root 2 x 1735<<

According to Legon's interpretation of Giza which is the template I work from there is only one viable option. If you have found something else that works for you then that's cool.

>>What evidence is there of knowledge of the English mile in AE times ?<<

That depends upon what you consider "evidence." Personally I am happy enough at the moment just to point out the patterns and relationships which emerge from studying Giza, which to my mind constitute evidence - at least once an accumulation thereof begins to challenge the notion of "coincidence." This public study has only scraped the surface of these issues. However, whether or not intent by the AE or pyramid builders can ultimately be proven is neither here nor there really, since there is a far more important story which emerges from all of this which can in fact be verified by anyone with a calculator. I am talking here about the proportions of the solar system. If you look closely at the table below you may detect some patterns, ie. the first 5 prime numbers, consistent multiples of the number 5 with square roots and pi, and consistent accuracies of no less than around 99.92% to current data. The precise values for the square roots and pi are employed here, no approximations.

The base length of G1 is 40 x 11 cubits and Legon found multiples of the first 4 prime number square roots with 250 in the spacings between the pyramids. Ultimately that is irrelevant though, as this table stands perfectly by itself without any reference to Giza required, as in fact do the solar system maps. However the fact that the maths and geometry of this table and the solar system maps can be seen to tie in rather neatly with Legon's and indeed Clive Ross' interpretation of Petrie's Giza survey should give any conscientious thinker pause for thought in my opinion.

And here's what we get when the first planet of the outer solar system is assigned the number 11:

-------------------------------------------------------------------------------------------------

Author: Graham Chase (82.14.33.---)
Date:   21-Jul-10 18:02

Nick L : "According to Legon's interpretation of Giza which is the template I work from there is only one viable option"

You are working from a wrong interpretation as Legon assumes that the rectangle is root 3 by root 2 - which it isn't, as well you know, as have stated so above. From false premise all that follows is wrong.

The table above that approximates the orbits using square roots is not consistent in its method. It just demonstrates that it is possible to approximate to any number using square roots and other factors.

Graham Chase

--------------------------------------------------------------------------
Author: Graham Chase (82.14.33.---)
Date:   21-Jul-10 18:24

I don't think that my comment about the table has been addressed.

The table above that approximates the orbits using square roots is not consistent in its method. It just demonstrates that it is mathematically possible to approximate to any number using square roots and other factors.

Graham Chase

-------------------------------------------------------------------------------------
Hi Nick I had seen that before. It starts with an error by assuming that the satellite pyramids are in a straight line which they are not. After such a simple error I did not take a great deal of notice on what else it said.

Anyway I have read a bit more of Clive Ross' book: "1o6: The Dawn of Man" and he has introduced A FOURTH MEASURING SYSTEM or at least another way of measuring.

Here is what Clive says on Page 268 of "1o6: The Dawn of Man" after talking about gravity and the need to find the velocity and acceleration of a falling body: "... Newton was a very clever mathematician, and he theorized the true motion (acceleration) of one body falling toward another body. To develop the theory he had to use logarithms and this type of mathematics was "developed" by The Greeks in Alexandria, Egypt.

Logarithms sound difficult to understand but in reality it is a simple way to multiply and divide by adding and subtracting. "

Clive goes on to explain how logarithms work in a very condensed version and then makes this brilliant insight after explaining that Newton calculated the force of gravity of 9.8 meter per second per second equals 18.76 cubits per second per second

He then again brilliantly noted that  18.76 is Log 1.2734

And his final in your face observation for now was that "the tallest pyramid on this planet has an angular slope equal to 4/Pi and this ratio equals 1.2734 ! "

One has to wonder just how clever Clive really was ...

Cheers and more later for indeed there is more to come ...

Best
Don Barone

PS: Please note that when I checked it I got 4/Pi = 1.27324

I also got 18.72 as the force of gravity in cubits per second squared. Can anybody figure out the log of 18.72 ?

Log site

And for the exact match we have this ....

And anti log of Pi is ...

1.3854557314e+3 or 1385.4557314 and surprisingly almost precisely what the diameter of The Sun is in kilometers divided by a thousand (1,392,000 / 1000 = 1392.00.

Hmmm ... we shall have to study this more closely ....

However Clive noted that the diameter of Mars was 6 794.4 km (my measurement ! ) and Earth was 12,756.28 km (my measurement) for a ratio according to Clive of 1.876 while I get currently 1.8775 ...

so did "God" use logs ?

Hmmm ....

We will have to re-examine Giza using logs to see what we find. :o)

--------------------------------------------------------------------------------------

Author: Nick L (82.4.204.---)
Date:   23-Jul-10 20:19

Hi Don, Elizabeth, Derek...

That's an excellent observation by Clive there Don, seems very relevant. A good angle to pursue. I have done a fair bit of study with the base of natural logarithm e (2.7182818...) which is a very important number in mathematics. Just in regard to logarithms and Elizabeth mentioning time circles, if we sum the synods of the inner planets as viewed from Earth, ie. Mercury, Venus and Mars, we get 1479.72 days. If we sum the synods of the 4 outer planets we get 1514.2 days.
Taking 1479.72 as a great circle of days, and using 22:7 for pi, the diameter of the circle will be 470.82 days. This is √3 x 271.828 days, or 100 x e. Now taking 1514.2 days as a great circle, again using 22:7 for pi we get a diameter of 481.79090909091 days. This is √pi x 271.821 days, once again approx 100 x e. Of course e minus 1 is 1.7182818, which is a good value for a royal Egyptian cubit in feet. I have a lot of notes on this sort of stuff, after reading Richard Heath's "Matrix of Creation" some years ago which addresses the geometry of time in the solar system I became fascinated by the subject.

I have to run on here, I'm getting away for a couple of days for a bit of fresh air and relaxation. Happy hunting everyone and have a nice weekend.

Cheers
Nick

----------------------------------------------------------------------------
Don posts:

Just starting to try a few things.

Anti Log of Phi to 12 places = 41.498651903383

Anti Log of Phi to 2 places = 41.50

Altitude north of step of Queen's Chamber 41.51 cubits (21.74)

Wow this is fantastic

Height of The Kings Chamber is 82.09 cubits

Log of 82.09 is .... and this is really going to knock your socks off ... I can hardly believe what my eyes are showing me ... Log of 82.09 is .... 1.914

1.914 ... This is the angle (x 100) at Giza between the three pyramids and is the distance in degrees traveled by Mars in it's orbit around the Sun in one Earth year.

This is beyond astounding ... I am spellbound and speechless.

And finally this beauty for now ...

1411.32 is the distance down from center of G1 to center of G3

Log of 1411.32 ( or 1411.25) is 3.1496 ...

THIS IS EXACTLY AND I MEAN EXACTLY ONE HALF THE PERIMETER DISTANCE AT GIZA DIVIDED BY 100  (6299.2 / 2 = 3149.2 / 100 = 3.1496).

This is astounding.

.

The diameter of The Earth is 12,756.28 kilometers.

The diameter of Mercury is 4 879.4 kilometers.

The ratio of these two is 2.61441981

THE ANTI LOG OF 2.61441981 is ....

411.5474 ... This is PRECISELY the base distance of G2 in cubits when we use Perring's measurements of 707.10 feet

Simply astounding !

Eerie ....

Diameter of Mercury = 4 879.4 km

Diameter of Saturn = 120,536 km

Ratio = 1 to 24.7030373

Log of 24.7030373 = 1.392750354

Multiply by a million and we have 1,392,750.354

Reversing back in order to get 1.392 precisely we need 24.660393372 as the ratio or 24.660393372 x 4 879.4 or 120327.92

Diameter of Saturn = 120,536 km

Our value = 120,328 km

Difference = 208 kilometers

Not bad when considering we are measuring the diameter of a gas giant with no real surface to use.

More interesting madness from "The Log House at Giza"

The distance from the south side of G3 to the south side of G2 is equal to about 630.95 cubits
Well what do you suppose the log is of 630.96 ?

Well amazingly it is ... 1/100 of the height of G1 in cubit ... yes that is correct it is 2.80000

If we use 631 even we get ...  2.8000293592 or 280.003

Weird ... truly weird.