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Topic: Ariadne's thread
Replies: 86   Last Post: Jan 28, 2013 5:19 AM

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Franz Gnaedinger

Posts: 330
Registered: 4/30/07
Re: Ariadne's thread
Posted: Nov 25, 2011 2:52 AM
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The Maya method of calculating the circle, perhaps
encoded in the Venus Pyramid (Citadel), is based
on the number 13 and the starting triple 5-12-13.
The Babylonian method, I believe, was based on
the number 17 and the starting triple 8-15-17.
And the Egyptian method, encoded in the Great
Pyramid, was based on the number 5 and the
Sacred Triangle 3-4-5 or 15-20-25 or 75-100-125
or 375-500-625 ... Jean-Philippe Lauer recognized
this triangle in the so-called King's Chamber of
the Great Pyramid: diagonal of a short wall 15
royal cubits, length of the chamber 20 royal cubits,
diagonal of the volume 25 royal cubits.

Every subdivision of the radius, from 5 to 25 to 125
to 625 ... makes a new triple emerge, from 3-4-5
to 7-24-25 to 44-117-125 to 336-527-625 ... If you
know a triple and wish to find the next one, calculate
these terms

+-3a +-4b / +-3b +-4a / 5c

and choose the positive values for the pair of long
terms that are neither zero nor divisible by 5 (in the
case of the Maya method you have to calculate the
terms +- 5a +-12 b and +-5b +-12a and 13 c, and
choose the positive values for the long terms that
are neither zero nor divisible by 13).

The arms of the cross mark 4 points on the circle,
the triples 8 and 8 and 8 and 8 ... more points.
The resulting polygon has 12 20 28 36 ... sides
of two or three different lengths (never more) that
are whole number multiples of the square roots
of 2 and/or 5 and/or 2x5 (2 and/or 13 and/or 2x13).

The square root of 2 is approximated by the number
column whose first lines are 1 1 2 / 2 3 4 / 5 7 10 /
12 17 ..., and the square root of 5 by the number
column whose first lines are 1 1 5 / 2 6 10 / 1 3 5 /
4 8 20 / 2 4 10 / 1 2 5 / 3 7 15 / 10 22 50 / 5 11 25 /
16 36 80 / 8 18 40 / 4 9 ...

The polygon inscribed in the circle of radius 25
and diameter 50 is defined by the triples 15-20-25
and 7-24-25. It has 20 sides of two different lengths,
12 sides of 5 times the square root of 2, and 8 sides
of 4 times the square root of 5

12 x 5 x sqrt2 plus 8 x 4 x sqrt5

Considering that the circumference of the circle is
a little longer than the periphery of the inscribed
polygon we may choose values that are a little above
the actual roots, 17/12 and 9/4

12 x 5 x 17/12 plus 8 x 4 x 9/4 equals 157

Divide the periphery 157 of the numerically rounded
polygon by the diameter 50 and you get 157/50 = 3.14
for pi, from the sequence

3/1 (plus 22/7) 25/8 47/15 69/22 ... 157/50 ...

The base of the Great Pyramid measured 440 and
the height 280 royal cubits. Imagine a circle whose
vertical diameter is given by the height of the pyramid.
Its area equals the one of the pyramid's cross-section
(implicit pi value 22/7). As if the pyramid, symbol of
the Primeval Hill, released the solar disc of Ra ...


Date Subject Author
11/17/11
Read Ariadne's thread
Franz Gnaedinger
11/17/11
Read Re: Ariadne's thread
Milo Gardner
11/18/11
Read Re: Ariadne's thread
Franz Gnaedinger
11/18/11
Read Re: Ariadne's thread
Milo Gardner
11/19/11
Read Re: Ariadne's thread
Franz Gnaedinger
11/19/11
Read Re: Ariadne's thread
Milo Gardner
11/20/11
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Franz Gnaedinger
11/20/11
Read Re: Ariadne's thread
Milo Gardner
11/20/11
Read Re: Ariadne's thread
Milo Gardner
11/21/11
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Franz Gnaedinger
11/22/11
Read Re: Ariadne's thread
Franz Gnaedinger
11/22/11
Read Re: Ariadne's thread
Milo Gardner
11/23/11
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Franz Gnaedinger
11/24/11
Read Re: Ariadne's thread
Franz Gnaedinger
11/24/11
Read Re: Ariadne's thread
Franz Gnaedinger
11/24/11
Read Re: Ariadne's thread
Franz Gnaedinger
11/24/11
Read Re: Ariadne's thread
Milo Gardner
11/25/11
Read Re: Ariadne's thread
Franz Gnaedinger
11/26/11
Read Re: Ariadne's thread
Franz Gnaedinger
12/2/11
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Franz Gnaedinger
12/2/11
Read Re: Ariadne's thread
Milo Gardner
12/3/11
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Franz Gnaedinger
12/4/11
Read Re: Ariadne's thread
Franz Gnaedinger
12/4/11
Read Re: Ariadne's thread
Milo Gardner
12/5/11
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Franz Gnaedinger
12/5/11
Read Re: Ariadne's thread
Milo Gardner
12/7/11
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Franz Gnaedinger
12/8/11
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Milo Gardner
12/10/11
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Franz Gnaedinger
12/12/11
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Franz Gnaedinger
12/12/11
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Milo Gardner
12/13/11
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Franz Gnaedinger
12/13/11
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Milo Gardner
12/15/11
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Franz Gnaedinger
12/15/11
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Milo Gardner
12/15/11
Read Re: Ariadne's thread
Milo Gardner
12/16/11
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Franz Gnaedinger
12/16/11
Read Re: Ariadne's thread
Milo Gardner
12/18/11
Read Re: Ariadne's thread
Franz Gnaedinger
12/18/11
Read Re: Ariadne's thread
Milo Gardner
12/19/11
Read Re: Ariadne's thread
Franz Gnaedinger
12/20/11
Read Re: Ariadne's thread
Franz Gnaedinger
12/20/11
Read Re: Ariadne's thread
Milo Gardner
12/21/11
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Franz Gnaedinger
12/22/11
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Franz Gnaedinger
12/23/11
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Franz Gnaedinger
12/24/11
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Franz Gnaedinger
12/29/11
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Franz Gnaedinger
1/2/12
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Franz Gnaedinger
1/3/12
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Milo Gardner
1/4/12
Read Re: Ariadne's thread
Franz Gnaedinger
11/28/11
Read Re: Ariadne's thread
Velev, Petyr
1/6/12
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Franz Gnaedinger
1/6/12
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Milo Gardner
1/9/12
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Franz Gnaedinger
1/17/12
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Franz Gnaedinger
1/19/12
Read Re: Ariadne's thread
Franz Gnaedinger
1/19/12
Read Re: Ariadne's thread
Milo Gardner
1/27/12
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Franz Gnaedinger
2/10/12
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Franz Gnaedinger
2/28/12
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Franz Gnaedinger
3/2/12
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Franz Gnaedinger
3/23/12
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Franz Gnaedinger
3/24/12
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Milo Gardner
4/9/12
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Franz Gnaedinger
4/10/12
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Franz Gnaedinger
4/13/12
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Franz Gnaedinger
4/17/12
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Franz Gnaedinger
4/18/12
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Franz Gnaedinger
4/18/12
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Franz Gnaedinger
5/5/12
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Franz Gnaedinger
5/7/12
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Franz Gnaedinger
5/7/12
Read Re: Ariadne's thread
Milo Gardner
5/8/12
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Franz Gnaedinger
5/8/12
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Milo Gardner
5/8/12
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Franz Gnaedinger
5/8/12
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Franz Gnaedinger
5/9/12
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Franz Gnaedinger
5/10/12
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Franz Gnaedinger
8/14/12
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Franz Gnaedinger
1/13/13
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Franz Gnaedinger
1/19/13
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Franz Gnaedinger
1/23/13
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Franz Gnaedinger
1/23/13
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Franz Gnaedinger
1/24/13
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Franz Gnaedinger
1/26/13
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Franz Gnaedinger
1/28/13
Read Re: Ariadne's thread
Franz Gnaedinger

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