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Post by Don Barone on Dec 20, 2007 16:28:03 GMT -5
Hi all ... Follow with me as we go in search of how "They" built Giza. Step 1)  Image supplied by Jon B. --------------------------------------------------------------------------------- Step 2)  ----------------------------------------------------------------------------------- Step 3)  -------------------------------------------------------------------------------- Step 4)  ---------------------------------------------------------------------------------- Step 5)  ---------------------------------------------------------------------------------- Step 6)  ----------------------------------------------------------------------------------- Step 7) Square the circle.  ---------------------------------------------------------------------------------- Step 8) Confirming some measurements  ----------------------------------------------------------------------------------- Step 9) Determining a point  ------------------------------------------------------------------------------------ Step 10) Just adding some colour   ----------------------------------------------------------------------------------- Step 11) Drawing some intersection lines for Pyramid 2  ----------------------------------------------------------------------------------- Step 12) Completing Scott's Triangle  and adding some colour ....  Step 13) Everywhere we look there is 1,2 and square root of 5  ----------------------------------------------------------------------------------- Unexpected Bonus: One half the bottom line of triangle is ... at least by scaling it .... and "sensing it" ... is ..... and I again am in awe of what is found at times .... the same as the diagonal of half the blue rectangle ... which is unbelievably ... square root of 5.  Thus it would appear that the side of Scott's Triangle could equal the radius of the large circle. It could be even more perfect that I even dreamed. Amazing !!! cheers Don Barone |
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Post by ariston on Dec 21, 2007 4:43:30 GMT -5
Dear Don,
Great insight, all the roots are repetative aswell and the consistency is sublime. Ideally, I would like to see a Vesica Piscis highlighted also, for not only is the Vesica defined by the sq. root of 3, you can also construct the Seal of Solomon(2 overlapping triangles) from a double Vesica, from the intersections that comprise the base and head of the Piscis. This is also consistent with a 2 to 1 rectangle that can also be made from the same points in the Vesica, root 5 and 2 squares in the rectangle giving root 2. However, a Vesica may be incorporatd into how the design was perceived, rather than designed.
The design consistencies and geometric correspondence make this a probability, rather than a possibility, Beunos Trabahando!!!!!
Kind Regards
Latona
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Post by Don Barone on Jan 17, 2008 11:37:50 GMT -5
Hi all ... Things are (mind) melding very nicely lately ... Reworking my diagrams and noted that the top (light blue line) line that defines the top of the circle enclosing 4 of the 5 points of The Giza Pentagram is at the 1/5th mark. Very, very, very nice discovery ... Now The Sphinx (actually I now believe it is The "Khafre" Valley Temple which defines it all) is at the 1/5th mark of square root of 5 and the circle defined by a line at the 1/5th mark of square root of 20. (sq rt of 5 x 2 = sq rt of 20 Cheers makes the drawing of it extremely easy now ...  Amazing builders to be sure .... Best Don Barone |
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Post by PMacG on Jan 18, 2008 8:48:43 GMT -5
Hi Don. This is getting very very close indeed to what I have. The Valley Temple is the key to the smaller circles based on G3 IMHO, and it falls as on 1/3rd divisions between the two centres of G3 and its imaginary placement on a diamond of 51 degrees. (Gary gave you first first bit and you sent to me). But this is another story so firstly I'll try and give you a simple one that can easily be seen from you drawing the Hexagram.
You have the right dimensions for the radius of the intersecting circles of Vesica Piscis based on G1 centres, as the central line of the rectangle or intersection between these circles is drawn on the East/West line of G3 and its opposite partner on the diamond 51 degrees. The centre of both circles are based on the radius between the centre of G1 and (as you now have it in the lines of your defined central rectangle at the points where the triangles of Heaven and Earth cross-over).
If you see this you will see the rest of what I have been speaking of, but I will attempt to send you a drawing of what I mean very soon - Paul.
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Post by Don Barone on Jan 18, 2008 11:39:34 GMT -5
Hi Paul et al This just off the drawing board ... Adding to my earlier "re-discovery" of "The Pentagram at Giza" I can't help but be amazed at how seamlessly it fits into my latest "re-discovery" of "The Root 10 Circle at Giza" a concept I had not even dreamed about when I first found The Pentagram. Here are a few images to enjoy and ponder on and try, along with me, to figure how in blazes it could have possibly been laid out on such a huge scale. NOTE: Top line that defines the circle that encloses The Pentagram appears to be ... 1/5th of the diameter of this circle ... measured down from the top of the circle. Image 1 - An Overview  ---------------------------------------------------------------------------------------------- Image 2 - Marking some points  ---------------------------------------------------------------------------------------------- Image 3 - Listing the co-ordinates of these points  ----------------------------------------------------------------------------------------------- Image 4 - Marking a few more points and giving 1 distance  ----------------------------------------------------------------------------------------------- Image 5 - A conclusion  ------------------------------------------ And now just a final interesting thing about this design which really makes it all that much easier to draw it with a minimum of measuring ... the largest square root of 5 side that can be obtained in this image (3162.275 and 6324.55) runs and is tangent to the circle defining The Giza (Wood) Pentagram's four inside corners.  ------------------------------------------------------------------------------------------- More as it becomes available ... And this to close with ... In this largest of right angled 1,2 and square root of 5 triangle where sides 1 and 2 are square root of 10 x's 1000 (3162.275) and 2 x's square root of 10 x's 1000 (6324.555) the long side is square root of 10,000,000 plus 40,000,000 which equals: Square root of 50,000,000 = 7071.0678118654752440084436210485 or 10,000 times the square root of 2 and EXACTLY 5 times length from E-W centerline of P1 to E-W centerline of P3. (1414.21 x 5 = 7071.0678) So we have sq rt of 5 = 5 x's sq rt of 2, Alice can be seen to be giggling hysterically somewhere I am sure .. Thanks Giza I did not know that. Cheers Don Barone |
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Post by PMacG on Jan 18, 2008 16:36:03 GMT -5
Hi Don.
I think the way these roots tie together in the GPs is amazing, and I think your the last leg of the puzzle now, but whoever laid out the complex was indeed complex, a great feat of engineering.
Kindest regards - Paul.
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Post by Don Barone on Jan 18, 2008 18:53:54 GMT -5
Hi Paul then I think you are going to like this one. Gives us a very big hint and also almost proves beyond a doubt that feet and inches were based on cubits. Here is a post of mine from another board ... Hi Stephen not too many people have found this one I bet ... Distance on a circle to the point to square the circle by perimeter with radius square root of 10 x's 1000 = 3162.275 x's 2 x's Pi = 19869.176531592202468867101340641 divide by 8 = 2483.6470664490253086083876675801 Now the good part says Alice, "I got this one from the hookah smoking catepillar ... " 2483.6470664490253086083876675801 - sq rt of 2 = 2483.6470664490253086083876675801 - 1414.2135623730950488016887242097 = 1069.4335040759302598066989433704 cubits .... The catepillar suggested Alice compare this to the diagonal of Pyramid 1 in feet and Alice gasped in amazement when she realized it was [square root of X = 756 squared x 2} = 1143072 = 1069.1454531540598568940766755025 This means if we take 1000 square roots of 2 and add it to a diagonal of a square of 756 cubits we get the distance to square the circle to within 0.288 cubits or about 6 inches ... getting closer ... This also means that if 440 cubits = 756.20368 feet or 20.623736802550407226066190749117 inches per cubit The Ancient Builders have indeed shown us how to square this circle of ours. Alice thought this was worth mentioning .  Cheers Don barone |
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Post by Charlotte on Jan 18, 2008 21:24:32 GMT -5
Love out loud, Don, the grinning Cheshire, again, the step by step geometry, the newest images and your never ending praise of the Builders in which I join. Anyway, I looked at your post this morning and shut off the comp, turned on the TV and heard a Sesame Street character say exclaim: "Triangles everywhere, see, triangles everywhere". Then they talked about numbers and even sang a song about triangles. Thought I, I should sit down and begin learning with little kids  unless you could do a Canada-California mind-meld  I so regret not having been taught even the basics. What about the Burlington story? Cheers Charlotte |
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Post by PMacG on Jan 19, 2008 9:07:26 GMT -5
Hi Don.
Therefore your also saying that the key to this is in the 8th division, and the differences you have found can be determined by 288.
I got a list of differences between Cubits, Imperial, and the Prof. Thom's Megalithic Yard that also fit the working values of the hexagrams - does this idea fit and make sense to you? As they were all measures of the same thing.
Bringing another point in that got ditched but in light of how it this works in the realms of time kind of hang around to be looked at. I first looked at in respect that 440 cubits x 1.72 gave 756.8, and firstly I thought this is really close to the actual dimension of base that this must be more than just a coincidence. Anyway a miss by even a few inches is a miss but still it was very close.
So I then went about this a different way as 440/1.72 = 255.8139535, now if that is taken as 2.558139535 + 440 = 442.55813935 x 4 = 1770,23252 etc x 2 = 3540,46504/12 = 295.0387533. If we look at this value the correct value of the moon should be 29.538 days per lunar month. In fact if we take the highly accurate value of 384/13 =29.53846154 except for the extra 0 in 29.5038 the value would be bang on. Now because of the extra 0 I have also keep it on back burner, but it is very interesting is it not.
Regards - Paul.
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unless you could do a Canada-California mind-meld 
I so regret not having been taught even the basics.