Post by Don Barone on May 4, 2017 18:26:40 GMT -5
Hi again all ... Well now that I no longer work in a store my research has to consist of mind games and so it was today as I was driving around town, getting wet I may add as it rained almost all of the day and delivering my auto parts I started to speculate if there might be a simpler tie in between the many 3rd and IVth Dynasty pyramids is Egypt, I speculated to myself that perhaps a clue could be hidden within the decimal value of the angles or ratios of the several main pyramids and so I started to ponder on this ...
Khafre we can surmise is 3, 4 5 or .75 (1.33333 and using 3 and 4) or 0.6 (1.66667 and using 3 and 5) and 0.8 (1.25 and using 4 and 5)
While for Menkare I had guessed it was 4,5 and 6.403 or 1.6008 (using 6.403 and 4) or 1.2806 (using 6.403 and 5) and 0.8 (using again 4 and 5)
But then where did The Great Pyramid fit in ? Before moving on I would like you to take note of the slope of Menkare. It is to be totally precise ... the square root of 41.
Now before I go on I would like you to look at this page on The Giza Palteau layout from John Legon
There is one image and notation I would like to bring to your attention. Here is the image and the comment that Giza was built from a 1:2 or 250 x's 500 cubit rectangle.
Yes I know this is all old news however having worked on Giza for many years I had memorized what all facets of a 250 and 500 cubit triangle would yield but in other places it was 500 as the hypotenuse and 250 as one side. And of course a number I had calculated hundreds of time was The sq rt of [250,000 (500 x 500) - 62,500 or sq rt of 187500 or 433.0127. Okay you are probably saying to yourself ... well yeah then so what !
Well when I moved over to The Great Pyramid to figure out the ratios of the sides I got 7 / 5.5 and got 1.2727273 and inverse of this gave me 0.7857142857 and did not seem to fit in and so I did something that believe it or not I had not done in 18 years of research .... I decided to use 7 AS THE HYPOTENUSE !!!!!!!!!!! AND SO IT WAS THAT I DECIDED TO FIGURE OUT THE OTHER SIDE ... do I have to tell you what it equals ?
Well if 7 is the long side and 5.5 in one of the other two sides then calling on Pythagoras again we get 49 - 30.25 = 18.75 and yielding precisely the same number as the other side in a 5 (hypotenuse) and 2.5 being one of the other sides ...
So ....
7 sqd - 5.5 sqrd = 4.33012702 squared ... AND ALSO ...
5 sqrd - 2.5 sqrd = 4.33012702 squared
So there is a nice tie from the simple 1,2 and sq rt of 5 right angled triangle to the equally simple 1, sq rt of 1/Phi and sq rt of Phi (approximate) triangle ( 1, .618, 1.272 )
Again another neat development in the mathematics at Giza and for Egypt in general.
Cheers
Don Barone
.
Khafre we can surmise is 3, 4 5 or .75 (1.33333 and using 3 and 4) or 0.6 (1.66667 and using 3 and 5) and 0.8 (1.25 and using 4 and 5)
While for Menkare I had guessed it was 4,5 and 6.403 or 1.6008 (using 6.403 and 4) or 1.2806 (using 6.403 and 5) and 0.8 (using again 4 and 5)
But then where did The Great Pyramid fit in ? Before moving on I would like you to take note of the slope of Menkare. It is to be totally precise ... the square root of 41.
Now before I go on I would like you to look at this page on The Giza Palteau layout from John Legon
There is one image and notation I would like to bring to your attention. Here is the image and the comment that Giza was built from a 1:2 or 250 x's 500 cubit rectangle.
Yes I know this is all old news however having worked on Giza for many years I had memorized what all facets of a 250 and 500 cubit triangle would yield but in other places it was 500 as the hypotenuse and 250 as one side. And of course a number I had calculated hundreds of time was The sq rt of [250,000 (500 x 500) - 62,500 or sq rt of 187500 or 433.0127. Okay you are probably saying to yourself ... well yeah then so what !
Well when I moved over to The Great Pyramid to figure out the ratios of the sides I got 7 / 5.5 and got 1.2727273 and inverse of this gave me 0.7857142857 and did not seem to fit in and so I did something that believe it or not I had not done in 18 years of research .... I decided to use 7 AS THE HYPOTENUSE !!!!!!!!!!! AND SO IT WAS THAT I DECIDED TO FIGURE OUT THE OTHER SIDE ... do I have to tell you what it equals ?
Well if 7 is the long side and 5.5 in one of the other two sides then calling on Pythagoras again we get 49 - 30.25 = 18.75 and yielding precisely the same number as the other side in a 5 (hypotenuse) and 2.5 being one of the other sides ...
So ....
7 sqd - 5.5 sqrd = 4.33012702 squared ... AND ALSO ...
5 sqrd - 2.5 sqrd = 4.33012702 squared
So there is a nice tie from the simple 1,2 and sq rt of 5 right angled triangle to the equally simple 1, sq rt of 1/Phi and sq rt of Phi (approximate) triangle ( 1, .618, 1.272 )
Again another neat development in the mathematics at Giza and for Egypt in general.
Cheers
Don Barone
.
