Thursday, November 6, 2014

From Giza to Dashur: The Real Story of The Old Kingdom Pyramids Part II

Please read Part I first by clicking here

Well here we are at Dahshur and for those who have lived in a bubble and do not know what The Bent Pyramid looks like here is a picture of it.

As you can see there appears to be two pyramids and here is the sketch we will be using.

John Legon did a lot of great work on these things in the mid 80's but his main fault I feel is that he insisted that The Giza Plateau was in the ratio of the square root of 2 to the square root of 3 rather than trying a more simple and accurate 9 by 11 rectangle. It is because of this fact that most of his work at Giza is in error and does not fit. If we allow 20.62 inches to a cubit then The Giza Rectangle works out to be 1417.5 cubits wide or 29228.85 inches by 1732.50 cubits or 35724.15 inches. These are the facts and the distances as presented by Petrie and I am not sure why John would have insisted that they were not correct. Maybe I will ask him. Anyway after experimenting with 110 for the top and 90 cubits for the bottom pyramid height I finally decided to take the advice of John Legon and use 110.1 cuibts for the upper height. Here are the break downs according to Petrie and Dorner. Again borrowing from Johm Legon's website. Let us look first at the break downs for the two heights.

 But before we do any of this we have to decide on the angle of the upper portion and so I am going to go with The Master Builders and use the angle they left us at Giza of 43.3637 and the tan ratio of 0.94445 . So to start with let us assume that they copied (oh wait they were first) , that they used for a top height the distance of 110.1 cubit what would that give us. Well it reall is a simple matter. one only has to take 110.1 and divide it by 0.94445 to get 1/2 the base of the upper portion and thsi gives us  110.1 / 0.94445 and equaling 116.576 and tiems 1 we have a base of  233.152 cuibts for the upper base. Here is a diagram to illustrate this.

And finally showing the contained angle.

Just a note here that we match Dorner measurement of 110.11 within 1/100th of a cubit. Keep that in mind as we solve for the bottom angle and distance. of base.

The next stage was to figure out what the base dimension was. Was it 360 cubits as some ahd proposed or was John legon correct with his guess of 362 based on explantions which are available on his website. I felt that the answer like all the answers I was now finding had it's roots at Giza and so began to experiment with various base. First I took 360 / 2 = 180 - 116.576 = 63.424 and divided into 89.9 (as per Johm Legon's plan of Giza) I got  1.41744 and an angle of 54.797 and with a base of 180.75 or 1/2 361.50 I got 64.924 / 89.9 = 1.3847 Neither of these reaslly di anything for me and then I decided to sqaure the base add on or 64.924 and got 4215.13 and that gave me an idea. What if we went back to G3 again and took the square root of the base this time and so I did this and got square root of 4153.6 or 64.4484. Now earlier I had multiplied 944.45 by 1.4 and got  1322.23 but this time I decided to use what we had at Giza and instead of multiplying by 1.4 I decided to multiply by 1.39361 (13,936.1 inches north south between G1 and G2) and got this number 64.448 x 1.39361 equals 89.815 and virtually identical with what Dorner had for the lower height AND what he had for the total height of 199.92 as we get 199.916. So by using the measurements at Giza we have arrived at a probable correct solution for The Bent Pyramid. the only remaining thing we need to do is justify the total base of 116.576 + 64.448 or 181.024 and times 2 we get 362.048. But this is such an odd number. Why not 360 or 362 even why on Earth 362.048 ? Well ... But first a word on the ehight that we arrived at. We got 104.71 meters or 199.92 cubits probably doesn't mean too much to most but an interesting "co-incidence" has the ratio of the mass of Jupiter or Ju P(i)tah [Jupiter = Ptah] equalling:

One solar mass M= 1,047.56 Jupiter mass (MJ)

and woudn't you just know it that 1,047.56 meters = (this is a punch line so get ready) 1,047.56 meters = 2000.12 cubits and just to keep this rolling since some would have The Red Pyramid with a base of 420 cubits ... well wouldn't you just know it that 440 cubits / 1.047.56 = 420.02 cubits so we are muddying the waters again but this time we have Jupiter involved instead of The Sun. But jsut found another one ... 1047.56 x 12 inches_ = I think this might be another punch line ... 104.756 cubits =  2160.07 inches and diameter of Moon in miles exact would have it:

Equatorial radius of Moon = 1738.14 km  (0.273 Earths) x 2 = 3476.28 / 1.60934 = 2160.07 Scoring an absolute direct match. 

... to be continued

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