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Scan Pyramids, quest for hidden chambers
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Franz Gnaedinger
2015-10-28 07:40:22 UTC
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Denizens of sci.archaeology, does anyone of you know an e-mail address
of someone participating in the Scan Pyramids project?

In the late 1990s I developed my interpretations of the early Egyptian
pyramids here in sci.archaeology, and made predictions about a sun chamber
in the Red Pyramid (44 meters above the basis) and another sun chamber
in the Great Pyramid (90.6 meters above the geometrical basis).

Original numbers of the Red Pyramid: basis 210 by 210 royal cubits (rc),
height 200 rc, slope exactly 290 rc, radius of the imaginary sphere
inscribed in the frame of the geometrical building exactly 84 rc,
hypothetical sun chamber in the center of the imaginary sphere,
floor height of sun chamber 84 rc or 44 meters.

Original numbers of the Great Pyramid: basis 440 by 440 rc, height 280 rc.
This pyramid combines the Golden Pyramid and what I call the Pi-ramid.
Imagine a hemisphere inscribed in the frame of the geometrical building:
its radius equals the golden major of the pyramid's height, 173 rc,
according to the golden sequence 9 16 25 41 66 107 173 280. The hemisphere
symbolizes the sky goddess Nut overarching the Earth (Geb symbolized by
the limestone hill of the basis), and the zenith of the hemisphere her
womb wherein the sun child grows, before leaving the pyramid as the sun.
Now picture a circle whose vertical diameter is given by the height of
the Pi-ramid. It's area equals the area of the building's cross-section.
So the king whose body of stone was the pyramid is transformed into
the circle of the solar disc ... We have this scenario: the king dies,
his soul is reborn in the womb of Nut, or in the sun chamber on the
zenith of the imaginary hemisphere, then leaves the building as the sun,
or at least accompanying Ra, the supreme god manifesting himself in the
sun ... Now the floor of the sun chamber on the zenith of the imaginary
hemisphere would have a height of 173 royal cubits or 90.6 meters above
the geometrical base.

The Scan Pyramids project offers the first chance of having my predictions
tested. More numbers, and several early mathematical techniques including
a systematic method of calculating the circle on the basis of the Sacred
Triangle 3-4-5 can be found on my website

http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
Franz Gnaedinger
2015-10-29 07:59:52 UTC
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Post by Franz Gnaedinger
Denizens of sci.archaeology, does anyone of you know an e-mail address
of someone participating in the Scan Pyramids project?
In the late 1990s I developed my interpretations of the early Egyptian
pyramids here in sci.archaeology, and made predictions about a sun chamber
in the Red Pyramid (44 meters above the basis) and another sun chamber
in the Great Pyramid (90.6 meters above the geometrical basis).
Original numbers of the Red Pyramid: basis 210 by 210 royal cubits (rc),
height 200 rc, slope exactly 290 rc, radius of the imaginary sphere
inscribed in the frame of the geometrical building exactly 84 rc,
hypothetical sun chamber in the center of the imaginary sphere,
floor height of sun chamber 84 rc or 44 meters.
Original numbers of the Great Pyramid: basis 440 by 440 rc, height 280 rc.
This pyramid combines the Golden Pyramid and what I call the Pi-ramid.
its radius equals the golden major of the pyramid's height, 173 rc,
according to the golden sequence 9 16 25 41 66 107 173 280. The hemisphere
symbolizes the sky goddess Nut overarching the Earth (Geb symbolized by
the limestone hill of the basis), and the zenith of the hemisphere her
womb wherein the sun child grows, before leaving the pyramid as the sun.
Now picture a circle whose vertical diameter is given by the height of
the Pi-ramid. It's area equals the area of the building's cross-section.
So the king whose body of stone was the pyramid is transformed into
the circle of the solar disc ... We have this scenario: the king dies,
his soul is reborn in the womb of Nut, or in the sun chamber on the
zenith of the imaginary hemisphere, then leaves the building as the sun,
or at least accompanying Ra, the supreme god manifesting himself in the
sun ... Now the floor of the sun chamber on the zenith of the imaginary
hemisphere would have a height of 173 royal cubits or 90.6 meters above
the geometrical base.
The Scan Pyramids project offers the first chance of having my predictions
tested. More numbers, and several early mathematical techniques including
a systematic method of calculating the circle on the basis of the Sacred
Triangle 3-4-5 can be found on my website
http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
Sorry for the typo. The base of the Red PYramid measure 420 by 420 royal
cubits.

I believe that Hemon planned three pyramids:

A) former cult pyramid of the Red Pyramid: base 100 by 100 rc, hypothetical
height 49 rc, slope practically 70 rc, edge practically 86 rc, radius of
inscribed hemisphere practically 35 rc, diameter practically 70 rc.

B) Red Pyramid: base 420 by 420 rc, height 200 rc, slope exactly 290 rc,
edge practically 358 rc, radius of inscribed sphere exactly 84 rc,
hypothetical sun chamber in the center of the imaginary sphere, floor
height 84 rc or 44 meters.

C) Great PYramid 440 by 440 rc, height 280 rc, slope practically 356 rc,
radius of inscribed hemisphere 173 rc according to the golden sequence
9 16 25 41 66 107 173 280, hypothetical sun chamber on the zenith of
the imaginary hemisphere, floor height 173 rc or 90.6 meters.

The hypothetical sun chambers would have been almost impossible to reach
- high above the base, deep inside the building. I hoped one might one day
search for them with a sophisticated equipment of echo-sounding. Now they
plan to scan the pyramids with infrared sensors. Don't know whether this
works, but we'll see.
Franz Gnaedinger
2015-11-30 07:53:15 UTC
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Post by Franz Gnaedinger
Post by Franz Gnaedinger
http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
Sorry for the typo. The base of the Red PYramid measure 420 by 420 royal
cubits.
A) former cult pyramid of the Red Pyramid: base 100 by 100 rc, hypothetical
height 49 rc, slope practically 70 rc, edge practically 86 rc, radius of
inscribed hemisphere practically 35 rc, diameter practically 70 rc.
B) Red Pyramid: base 420 by 420 rc, height 200 rc, slope exactly 290 rc,
edge practically 358 rc, radius of inscribed sphere exactly 84 rc,
hypothetical sun chamber in the center of the imaginary sphere, floor
height 84 rc or 44 meters.
C) Great PYramid 440 by 440 rc, height 280 rc, slope practically 356 rc,
radius of inscribed hemisphere 173 rc according to the golden sequence
9 16 25 41 66 107 173 280, hypothetical sun chamber on the zenith of
the imaginary hemisphere, floor height 173 rc or 90.6 meters.
The hypothetical sun chambers would have been almost impossible to reach
- high above the base, deep inside the building. I hoped one might one day
search for them with a sophisticated equipment of echo-sounding. Now they
plan to scan the pyramids with infrared sensors. Don't know whether this
works, but we'll see.
Recent georadar analyses carried out in the tomb of Tutenkhamon
under the supervision of a British archaeologist apparently support
the hypothesis of a secret chamber: possible gangways behind the northern
and western walls may lead to the tomb of Nefertiti whose mummy had never
been found. Tutenkhamon died young, at the age of nineteen years, his tomb
was not yet finished, and so he would have hastily been buried in an
antechamber of Nefertiti's tomb. Japanese experts will examine the radar
pictures in the coming four weeks. Mahmoud al-Damati, minister of Egyptian
Antiquities: There is a ninety per cent chance of a secret chamber,
a second tomb behind the one of Tutenkhamon ...

Searching for the hypothetical sun chambers in the Red Pyramid and Great
Pyramid is a more challenging task - high above the ground (84 royal cubits
or 44 meters and 173 rc or 90.6 m respectively), secured by vast masses
of surrounding stone, inaccessible for grave robbers, ancient and modern
ones alike.
Franz Gnaedinger
2015-12-01 08:50:05 UTC
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Post by Franz Gnaedinger
Post by Franz Gnaedinger
http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
Recent georadar analyses carried out in the tomb of Tutenkhamon
under the supervision of a British archaeologist apparently support
the hypothesis of a secret chamber: possible gangways behind the northern
and western walls may lead to the tomb of Nefertiti whose mummy had never
been found. Tutenkhamon died young, at the age of nineteen years, his tomb
was not yet finished, and so he would have hastily been buried in an
antechamber of Nefertiti's tomb. Japanese experts will examine the radar
pictures in the coming four weeks. Mahmoud al-Damati, minister of Egyptian
Antiquities: There is a ninety per cent chance of a secret chamber,
a second tomb behind the one of Tutenkhamon ...
Searching for the hypothetical sun chambers in the Red Pyramid and Great
Pyramid is a more challenging task - high above the ground (84 royal cubits
or 44 meters and 173 rc or 90.6 m respectively), secured by vast masses
of surrounding stone, inaccessible for grave robbers, ancient and modern
ones alike.
Why were the King's Chamber and sarcophagus of rose granite in the Great
Pyramid left undecorated? bare of reliefs, figures, ideograms and
hieroglpyhs?

Their transformatorial magic lies in the numbers.

Jean-Philippe Lauer discovered the Sacred Triangle 3-4-5 in the form of
15-20-25 royal cubits in the King's Chamber: diagonal short wall 15 rc,
chamber length 20 rc, diagonal volume 25 rc. This triangle is the key
to the Egyptian method of systematically calculating the circle (long
before Archimedes).

Earth and heaven are represented by the royal cubit and seven Horus cubits
respectively. The Horus cubits vary around the body length of a kestrel
(beak to tail) and are defined by (relatively) low-number ratios, the most
important one as follows

11 Horus cubits are 7 royal cubits

royal cubit of the Great Pyramid 52.36 cm (rc)
most important Horus cubit 33.32 cm (Hc)

model of Great Pyramid: base 1 rc, height 1 Hc

base Great Pyramid 440 by 440 royal cubits
height 280 ryoal cubits or 440 Horus cubits

If the diameter of a circle measures 1 Horus cubit,
the circumference measures 2 royal cubits

If the radius of a circle measures 1 Horus cubit,
the area measures 1 royal cubit by 2 Horus cubits

6 Horus cubits are the golden minor of 10 royal cubits
(vertical axis of the niche in the Queen's Chamber)

If the side of a square measures 9 rc (or 20 Hc),
the diagonal measures 20 Hc (or 18 rc)

The implicit value for the square root of 2 is 140/99 and appears
in the number column approaching this value; the first lines are
1 1 2, 2 3 4, 5 7 10, 12 17 24, 29 41 58, 70 99 140 ...

How was the base of the Great Pyramid measured out to the stunning
precision reported by Rainer Stadelmann, and around the large limestone
hill still present under the huge building? Combine a wooden square with
a wooden cross, mark the corners of the square = ends of the arms with
four nails, distances of the nails 9 royal cubits (side of square) and
20 Horus cubits (diagonals of square, arms of cross), then leave nail
impressions on a grid of levelled wooden blocks.

Hypothesis: the seven Horus cubits were combined in the sarcophagus
of rose granite (tub and lost lid, four of the seven measures still
present in the tub): outer measurements 4 by 3 by 7 Horus cubits,
inner measurements 3 by 2 by 6 Horus cubits, diagonal volume 7 Horus
cubits (quadrupel 2-3-6-7).

Placing the body of the king into the sarcophagus turned him into a divine
being, while his soul was reborn and raised as the mythical sun child
in the sun chamber on top of the imaginary hemisphere that symbolizes Nut
bending over the Earth, Geb in form of the large limestone hill at the base
of the Great Pyramid. The grown up sun child leaves the pyramid via the
large circle of Ra whose area equals the one of the pyramid (220 rc by
440 Hc) and whose vertical diameter is given by the height of the pyramid
(280 rc or 440 Hc).

Floor height of hypothetical sun chamber above the mathematical center
of the base 173 royal cubits (90.60 m) or 272 Horus cubits (90.63 m)
according to a pair of golden sequences

2 7 9 16 25 41 66 107 173 280 royal cubits

8 8 16 24 40 64 104 168 272 440 Horus cubits

You can't build an Egyptian pyramid let alone a Great Pyramid without
mathematics, without a solid body of simple yet clever numerical methods,
mathematics being the logic of building and maintaining.
Franz Gnaedinger
2015-12-03 08:12:43 UTC
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Post by Franz Gnaedinger
Post by Franz Gnaedinger
http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
Why were the King's Chamber and sarcophagus of rose granite in the Great
Pyramid left undecorated? bare of reliefs, figures, ideograms and
hieroglpyhs?
Their transformatorial magic lies in the numbers.
Jean-Philippe Lauer discovered the Sacred Triangle 3-4-5 in the form of
15-20-25 royal cubits in the King's Chamber: diagonal short wall 15 rc,
chamber length 20 rc, diagonal volume 25 rc. This triangle is the key
to the Egyptian method of systematically calculating the circle (long
before Archimedes).
Earth and heaven are represented by the royal cubit and seven Horus cubits
respectively. The Horus cubits vary around the body length of a kestrel
(beak to tail) and are defined by (relatively) low-number ratios, the most
important one as follows
11 Horus cubits are 7 royal cubits
royal cubit of the Great Pyramid 52.36 cm (rc)
most important Horus cubit 33.32 cm (Hc)
model of Great Pyramid: base 1 rc, height 1 Hc
base Great Pyramid 440 by 440 royal cubits
height 280 ryoal cubits or 440 Horus cubits
If the diameter of a circle measures 1 Horus cubit,
the circumference measures 2 royal cubits
If the radius of a circle measures 1 Horus cubit,
the area measures 1 royal cubit by 2 Horus cubits
6 Horus cubits are the golden minor of 10 royal cubits
(vertical axis of the niche in the Queen's Chamber)
If the side of a square measures 9 rc (or 20 Hc),
the diagonal measures 20 Hc (or 18 rc)
The implicit value for the square root of 2 is 140/99 and appears
in the number column approaching this value; the first lines are
1 1 2, 2 3 4, 5 7 10, 12 17 24, 29 41 58, 70 99 140 ...
How was the base of the Great Pyramid measured out to the stunning
precision reported by Rainer Stadelmann, and around the large limestone
hill still present under the huge building? Combine a wooden square with
a wooden cross, mark the corners of the square = ends of the arms with
four nails, distances of the nails 9 royal cubits (side of square) and
20 Horus cubits (diagonals of square, arms of cross), then leave nail
impressions on a grid of levelled wooden blocks.
Hypothesis: the seven Horus cubits were combined in the sarcophagus
of rose granite (tub and lost lid, four of the seven measures still
present in the tub): outer measurements 4 by 3 by 7 Horus cubits,
inner measurements 3 by 2 by 6 Horus cubits, diagonal volume 7 Horus
cubits (quadrupel 2-3-6-7).
Placing the body of the king into the sarcophagus turned him into a divine
being, while his soul was reborn and raised as the mythical sun child
in the sun chamber on top of the imaginary hemisphere that symbolizes Nut
bending over the Earth, Geb in form of the large limestone hill at the base
of the Great Pyramid. The grown up sun child leaves the pyramid via the
large circle of Ra whose area equals the one of the pyramid (220 rc by
440 Hc) and whose vertical diameter is given by the height of the pyramid
(280 rc or 440 Hc).
Floor height of hypothetical sun chamber above the mathematical center
of the base 173 royal cubits (90.60 m) or 272 Horus cubits (90.63 m)
according to a pair of golden sequences
2 7 9 16 25 41 66 107 173 280 royal cubits
8 8 16 24 40 64 104 168 272 440 Horus cubits
You can't build an Egyptian pyramid let alone a Great Pyramid without
mathematics, without a solid body of simple yet clever numerical methods,
mathematics being the logic of building and maintaining.
Hiram of Tyre made a molten sea for king Solomon (a basin of brass,
1 Kings 7 in the Bible). Its diameter measured 10 cubits while
the circumference measured 30 cubits, yielding 30/10 or 3 for pi.
How come that the master architect of wise king Solomon used such
a poor value for the number of the circle?

Or did he? Maybe a detail fell out of the written tradition.

There might have been two combined measures of nearly the same lenghts,
here called black and red cubit, in reference to the black and red ink
used by Ahmose in the Rhind Mathematical Papyrus

black cubit 21 units
red cubit 22 units

diameter molten sea 10 black cubits or 210 units
circumference 30 red cubits or 660 units
660/210 or 66/21 or 22/7 for pi

The seven hypothetical Horus cubits of Hemon (whom I consider the master
architect of Cheops), combined with the royal cubit, allow simple
calculations, and so do the combined black and red cubit

If the diameter of a circle measures 1 black cubit,
the circumference measures 3 red cubits

If the radius of a circle measures 1 black cubit,
the area measures 3 black cubits by 1 red cubit

If the diameter of a sphere measures 1 black cubit,
the surface measures 3 black cubits by 1 red cubit

If the diameter of a sphere measures 2 black cubits,
the volume measures 2 black cubits by 2 black cubits by 1 red cubit

2 black cubits are the golden minor of 5 red ones

If the side of a square measures 20 black cubits (or 27 red ones),
the diagonal measures 27 red cubits (or 40 black ones)

My formula for understanding early civilization: simple yet complex.
And the one for understanding early mathematics: simple yet clever.
Franz Gnaedinger
2015-12-04 08:44:12 UTC
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Post by Franz Gnaedinger
Post by Franz Gnaedinger
http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
Hiram of Tyre made a molten sea for king Solomon (a basin of brass,
1 Kings 7 in the Bible). Its diameter measured 10 cubits while
the circumference measured 30 cubits, yielding 30/10 or 3 for pi.
How come that the master architect of wise king Solomon used such
a poor value for the number of the circle?
Or did he? Maybe a detail fell out of the written tradition.
There might have been two combined measures of nearly the same lenghts,
here called black and red cubit, in reference to the black and red ink
used by Ahmose in the Rhind Mathematical Papyrus
black cubit 21 units
red cubit 22 units
diameter molten sea 10 black cubits or 210 units
circumference 30 red cubits or 660 units
660/210 or 66/21 or 22/7 for pi
The seven hypothetical Horus cubits of Hemon (whom I consider the master
architect of Cheops), combined with the royal cubit, allow simple
calculations, and so do the combined black and red cubit
If the diameter of a circle measures 1 black cubit,
the circumference measures 3 red cubits
If the radius of a circle measures 1 black cubit,
the area measures 3 black cubits by 1 red cubit
If the diameter of a sphere measures 1 black cubit,
the surface measures 3 black cubits by 1 red cubit
If the diameter of a sphere measures 2 black cubits,
the volume measures 2 black cubits by 2 black cubits by 1 red cubit
2 black cubits are the golden minor of 5 red ones
If the side of a square measures 20 black cubits (or 27 red ones),
the diagonal measures 27 red cubits (or 40 black ones)
My formula for understanding early civilization: simple yet complex.
And the one for understanding early mathematics: simple yet clever.
Robert Bauval had a brillant idea: the Giza pyramids might represent
the belt stars of Orion ... If only he had read Rolf Krauss (instead of
joining a moneymaker on the kooky side of Egytology) he would have gotten
a confirmation that Orion was the constellation of Osiris, learned that
the swaying kha channel was the band of the ecliptic, and then he could
have expanded his pattern as I did on egypt2 (link above).

The wider pattern sees the Dahshur pyramids of Sneferu as the Golden Gate
of Mesopotamian astronomy: Aldebaran and the Pleiads flanking the ecliptic.

The Bent Pyramid stands for the Pleiads - peculiar shape intended right
from the beginning, on a grid involving the number 57 that served for
an astronomical device allowing to measure small angles in the sky,
the arctan 7/57 being very close to 7 degrees; division and doubling
and addition then lead to a scale of 1 to 14 degrees within the easily
constructible angle of 15 degrees.

The Red Pyramid stands for the red star Aldebaran. Sneferu was buried
in this pyramid. A small bark might have been placed in the ante chamber,
allowing Sneferu to join Ra who traveled along the kha channel, from east
to west.

Maybe a small temple between the two pyramids could have marked the harbor
of the kha channel? Someone ready for an archaeological survey of that
region?

While the Great Pyramid combines a pair of mathematically distinct pyramids
(golden pyramid of Nut and the pi-ramid of Ra) the Great Gallery within
combines a pair of triangles

base 80 royal cubits
ascension 39 or 40 royal cubits
slope 89 royal cubits or 140 Horus cubits

The slope of the rectangular triangle 80-39-89 royal cubits would have
aimed at a star in Orion, and the slope of the triangle 80 rc - 40 rc -
140 Hc just below the midwinter sun in the zenith.

A pair of barks looking upward might have been placed in the gallery
- pilot boats guiding the large barks in the long pits found at the base
of the pyramid, leading them toward Osiris in Orion and Ra in the sun.
Franz Gnaedinger
2015-12-05 09:51:18 UTC
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Post by Franz Gnaedinger
Post by Franz Gnaedinger
http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
Robert Bauval had a brillant idea: the Giza pyramids might represent
the belt stars of Orion ... If only he had read Rolf Krauss (instead of
joining a moneymaker on the kooky side of Egytology) he would have gotten
a confirmation that Orion was the constellation of Osiris, learned that
the swaying kha channel was the band of the ecliptic, and then he could
have expanded his pattern as I did on egypt2 (link above).
The wider pattern sees the Dahshur pyramids of Sneferu as the Golden Gate
of Mesopotamian astronomy: Aldebaran and the Pleiads flanking the ecliptic.
The Bent Pyramid stands for the Pleiads - peculiar shape intended right
from the beginning, on a grid involving the number 57 that served for
an astronomical device allowing to measure small angles in the sky,
the arctan 7/57 being very close to 7 degrees; division and doubling
and addition then lead to a scale of 1 to 14 degrees within the easily
constructible angle of 15 degrees.
The Red Pyramid stands for the red star Aldebaran. Sneferu was buried
in this pyramid. A small bark might have been placed in the ante chamber,
allowing Sneferu to join Ra who traveled along the kha channel, from east
to west.
Maybe a small temple between the two pyramids could have marked the harbor
of the kha channel? Someone ready for an archaeological survey of that
region?
While the Great Pyramid combines a pair of mathematically distinct pyramids
(golden pyramid of Nut and the pi-ramid of Ra) the Great Gallery within
combines a pair of triangles
base 80 royal cubits
ascension 39 or 40 royal cubits
slope 89 royal cubits or 140 Horus cubits
The slope of the rectangular triangle 80-39-89 royal cubits would have
aimed at a star in Orion, and the slope of the triangle 80 rc - 40 rc -
140 Hc just below the midwinter sun in the zenith.
A pair of barks looking upward might have been placed in the gallery
- pilot boats guiding the large barks in the long pits found at the base
of the pyramid, leading them toward Osiris in Orion and Ra in the sun.
In the beginning was the darkness Keku Semau, a sea of no dimension,
neither space nor time, nor anything else, just hovering mud particles
(an Ancient Egyptian myth of creation told along modern cosmology by
Erik Hornung). Then it came to pass that the mud particles aggregated
and rose as primeval hill above the primeval sea. The heron Benu Benu
stood on top of the hill, did a loud cry, the hill opened up and released
the sky of Nut and sun of Ra ...

The Grear Pyramid above the Nile (that periodically flooded the river plain)
evokes the primeval hill and gives it a geometrical shape

base 440 royal cubits
height 280 royal cubits or 440 Horus cubits

Nut was bending herself over the Earth. In the Great Pyramid she is present
in the imaginary hemisphere whose radius measures 173 royal cubits (90.60 m)
or 272 Horus cubits (90.63 m) according to the golden sequences

2 7 9 16 25 41 66 107 173 280 royal cubits
8 8 16 24 40 64 104 168 272 440 Horus cubits

Ra was the supreme god appearing in the solar disc that is evoked by the
imaginary large circle whose vertical diameter is given by the pyramid's
height and whose area equals the one of the pyramid's cross-section.
The implicit pi value 22/7 was found via the Ancient Egyptian method
of systematically calculating the circle (egypt2, link above, chapter 10,
Long before Archimedes) and also by means of a typical number sequence
that may explain itself. Pi is less than 4 and a little more than 3

4/1 (plus 3/1) 7/2 10/3 13/4 16/5 19/6 22/7 25/8

11 and 7 are the magic numbers defining the Great Pyramid. A close
examination reveals a pair of mathematically and symbolically distinct
pyramids, the golden pyramid of Nut and the pi-ramid of Ra that go along
with Nut and Ra emerging from the primeval hill.

Yet the two pyramids are a close match. Given the base 440 royal cubits,
the exact height of the golden pyramid is 279.844... royal cubits,
and the one of the pi-ramid 280.112... royal cubits, average 279.978...
royal cubits, only 1.125... centimeters short of 280 cubits.

By the way, a recent thermo scan reveals an anomaly in the center
of the eastern base. Mamdouh al-Damaty asks for ideas. Well, 47 royal
cubits or 24.6 meters behind the anomaly is the easternmost point of
the hemisphere that symbolizes the sky of Nut. Maybe there was a secret
passage connecting that point and the ground door of the pyramid temple?
gangway of 'going out in the morning'? Here my perspectivic reconstruction
of the pyramid temple (diameter of the gold disc above the upper door
7 Horus cubits of the sarcophagus, double divine measure, circumference
14 royl cubits) Loading Image...

(Does any Egyptologist visit sci.archaeology?)
Franz Gnaedinger
2015-12-07 08:54:54 UTC
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Post by Franz Gnaedinger
In the beginning was the darkness Keku Semau, a sea of no dimension,
neither space nor time, nor anything else, just hovering mud particles
(an Ancient Egyptian myth of creation told along modern cosmology by
Erik Hornung). Then it came to pass that the mud particles aggregated
and rose as primeval hill above the primeval sea. The heron Benu Benu
stood on top of the hill, did a loud cry, the hill opened up and released
the sky of Nut and sun of Ra ...
The Grear Pyramid above the Nile (that periodically flooded the river plain)
evokes the primeval hill and gives it a geometrical shape
base 440 royal cubits
height 280 royal cubits or 440 Horus cubits
Nut was bending herself over the Earth. In the Great Pyramid she is present
in the imaginary hemisphere whose radius measures 173 royal cubits (90.60 m)
or 272 Horus cubits (90.63 m) according to the golden sequences
2 7 9 16 25 41 66 107 173 280 royal cubits
8 8 16 24 40 64 104 168 272 440 Horus cubits
Ra was the supreme god appearing in the solar disc that is evoked by the
imaginary large circle whose vertical diameter is given by the pyramid's
height and whose area equals the one of the pyramid's cross-section.
The implicit pi value 22/7 was found via the Ancient Egyptian method
of systematically calculating the circle (egypt2, link above, chapter 10,
Long before Archimedes) and also by means of a typical number sequence
that may explain itself. Pi is less than 4 and a little more than 3
4/1 (plus 3/1) 7/2 10/3 13/4 16/5 19/6 22/7 25/8
11 and 7 are the magic numbers defining the Great Pyramid. A close
examination reveals a pair of mathematically and symbolically distinct
pyramids, the golden pyramid of Nut and the pi-ramid of Ra that go along
with Nut and Ra emerging from the primeval hill.
Yet the two pyramids are a close match. Given the base 440 royal cubits,
the exact height of the golden pyramid is 279.844... royal cubits,
and the one of the pi-ramid 280.112... royal cubits, average 279.978...
royal cubits, only 1.125... centimeters short of 280 cubits.
By the way, a recent thermo scan reveals an anomaly in the center
of the eastern base. Mamdouh al-Damaty asks for ideas. Well, 47 royal
cubits or 24.6 meters behind the anomaly is the easternmost point of
the hemisphere that symbolizes the sky of Nut. Maybe there was a secret
passage connecting that point and the ground door of the pyramid temple?
gangway of 'going out in the morning'? Here my perspectivic reconstruction
of the pyramid temple (diameter of the gold disc above the upper door
7 Horus cubits of the sarcophagus, double divine measure, circumference
14 royl cubits) http://www.seshat.ch/home/temple.GIF
(Does any Egyptologist visit sci.archaeology?)
The anomaly in the center of the eastern base of the Great Pyramid,
220 royal cubits east of the mathematical center of the base;
the easternmost point of the imaginary hemisphere (sky of Nut enclosed
in the primeval hill), 173 royal cubits east of the mathematical center
of the base; the hypothetical sun chamber on top of the hemisphere,
173 royal cubits above the center of the base; and the top of the pyramid,
280 royal cubits above the base, are related by an equation

280 x 173 = 48,440
220 x 220 = 48,400

The products are nearly the same and become identical with the exact
golden pyramid. Herodotus reported a similar equation: square of height
= area of a face. Let us check on this equation via the numbers of
the simplified pyramid

base 440 royal cubits, half base 220 rc
height 280 rc, slope 356 rc (136 84 136)
half base = golden major of slope
4 4 8 12 20 32 52 84 136 220 356
radius of imaginary hemisphere 137 rc
golden major of height
2 7 9 16 41 66 107 173 280

280 x 280 = 78,400
220 x 356 = 78,320

A theorem valid for any pyramid connects the above pair of equations:
radius of inscribed hemisphere x slope = area of cross-section
(height x half base)

173 x 356 = 61,588
280 x 220 = 61,600

The base of the imaginary hemisphere is an interesting circle: radius
173 rc, diameter 346 rc, circumference 1086.991... rounded 1087 rc.
The vaule 1087/346 for pi is found in a sequence. Begin with 3 in
the form of 9/3 and 'add' repeatedly 22/7 in the form of 154/49

9/3 (plus 154/49) 163/52 317/101 471/150 (157/50)
625/199 779/248 939/297 (311/99) 1087/346

I found ample indirect evidence for the use of many pi sequences
in the Rhind Mathematical Papyrus (lost original from around 1 850 BC,
copy by Ahmose from around 1 650 BC), for example

4/1 (plus 3/1) 7/2 10/3 13/4 16/5 19/6 22/7 25/8 28/9

3/1 (plus 22/7) 25/8 47/15 ... 157/50 ... 311/99 ... 377/120

9/3 (plus 19/6) 28/9 47/15 66/21 (22/7) ... 256/81

Ra had many names but nobody knew the real name. His hieroglyph was
a circle for the solar disc. Pi sequences generate many values for
for the number of the circle, for the number of Ra, if you like,
none exact (nor is our ever longer decimal fraction of pi) but very
practical: choose a value that comes handy in a given calculation.

Now there are two hypothetical hidden rooms: the sun chamber on top
of the imaginary hemisphere, and a real or symbolical gangway connecting
the easternmost point of the hemisphere and with the anomaly in the center
of the eastern face, a passage of 'going out in the morning'.
Franz Gnaedinger
2015-12-09 07:43:17 UTC
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Post by Franz Gnaedinger
Post by Franz Gnaedinger
http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
Egyptian method of calculating the circle (part 1)

Egyptology is impeded by the dogma of the Greek invention of mathematics.
And there is a strange phenomenon: mathematicians can understand most
complicated relations but fail when it comes to early and very early methods.

In this and the next messages I will shed light on Egyptian mathematics,
beginning with the greatest achievement, a systematic method of calculating
the circle.

Picture a square measuring 10 by 10 royal cubits or 70 by 70 palms or
280 by 280 fingerbreadths. How long are the diagonals? Consult the number
column for the approximation of the square root of 2, here given as a number
sequence. The first lines are 1 1 2, 2 3 4, 5 7 10, 12 17 24, 29 41 58,
70 99 ... If the side of a square measures 70 palms, the diagonal measures
practically 99 palms.

Now imagine a circle around the square. How long is the circumference?
The diameter is given by the diagonal 99 palms. Draw up a suitable pi
sequence, here the first one followed by the most important one

4/1 (plus 3/1) 7/2 10/3 13/4 16/5 19/6 22/7

3/1 (plus 22/7) 25/8 47/15 69/22 91/29 113/36 135/43 157/50
179/57 201/64 223/71 245/78 267/85 289/92 311/99

If the diameter of a circle measures 99 palms, the cicumference measures
practically 311 palms.

And the exact length of the circumference of a circle around the square
measuring 10 by 10 royal cubits? 311.00180... palms, not even one seventh
of a millimeter more than 311 palms!

This calculation shows the advantage of Egyptian mathematics: operate
with number patterns, columns and sequences, choose values that come handy
in the given calculation, the mistakes will even out, sometimes almost
completely, and you get a fine result.

Next time: we shall measure and calculate the circumference of the circle
inscribed in the square 10 by 10 royal cubits or 70 by 70 palms or 280
by 280 fingerbreadths
Post by Franz Gnaedinger
The anomaly in the center of the eastern base of the Great Pyramid,
220 royal cubits east of the mathematical center of the base;
the easternmost point of the imaginary hemisphere (sky of Nut enclosed
in the primeval hill), 173 royal cubits east of the mathematical center
of the base; the hypothetical sun chamber on top of the hemisphere,
173 royal cubits above the center of the base; and the top of the pyramid,
280 royal cubits above the base, are related by an equation
280 x 173 = 48,440
220 x 220 = 48,400
The products are nearly the same and become identical with the exact
golden pyramid. Herodotus reported a similar equation: square of height
= area of a face. Let us check on this equation via the numbers of
the simplified pyramid
base 440 royal cubits, half base 220 rc
height 280 rc, slope 356 rc (136 84 136)
half base = golden major of slope
4 4 8 12 20 32 52 84 136 220 356
radius of imaginary hemisphere 137 rc
golden major of height
2 7 9 16 41 66 107 173 280
280 x 280 = 78,400
220 x 356 = 78,320
radius of inscribed hemisphere x slope = area of cross-section
(height x half base)
173 x 356 = 61,588
280 x 220 = 61,600
The base of the imaginary hemisphere is an interesting circle: radius
173 rc, diameter 346 rc, circumference 1086.991... rounded 1087 rc.
The vaule 1087/346 for pi is found in a sequence. Begin with 3 in
the form of 9/3 and 'add' repeatedly 22/7 in the form of 154/49
9/3 (plus 154/49) 163/52 317/101 471/150 (157/50)
625/199 779/248 939/297 (311/99) 1087/346
I found ample indirect evidence for the use of many pi sequences
in the Rhind Mathematical Papyrus (lost original from around 1 850 BC,
copy by Ahmose from around 1 650 BC), for example
4/1 (plus 3/1) 7/2 10/3 13/4 16/5 19/6 22/7 25/8 28/9
3/1 (plus 22/7) 25/8 47/15 ... 157/50 ... 311/99 ... 377/120
9/3 (plus 19/6) 28/9 47/15 66/21 (22/7) ... 256/81
Ra had many names but nobody knew the real name. His hieroglyph was
a circle for the solar disc. Pi sequences generate many values for
for the number of the circle, for the number of Ra, if you like,
none exact (nor is our ever longer decimal fraction of pi) but very
practical: choose a value that comes handy in a given calculation.
Now there are two hypothetical hidden rooms: the sun chamber on top
of the imaginary hemisphere, and a real or symbolical gangway connecting
the easternmost point of the hemisphere and with the anomaly in the center
of the eastern face, a passage of 'going out in the morning'.
Franz Gnaedinger
2015-12-10 10:22:07 UTC
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Permalink
Post by Franz Gnaedinger
Post by Franz Gnaedinger
http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
Egyptian method of calculating the circle (part 1)
Egyptology is impeded by the dogma of the Greek invention of mathematics.
And there is a strange phenomenon: mathematicians can understand most
complicated relations but fail when it comes to early and very early methods.
In this and the next messages I will shed light on Egyptian mathematics,
beginning with the greatest achievement, a systematic method of calculating
the circle.
Picture a square measuring 10 by 10 royal cubits or 70 by 70 palms or
280 by 280 fingerbreadths. How long are the diagonals? Consult the number
column for the approximation of the square root of 2, here given as a number
sequence. The first lines are 1 1 2, 2 3 4, 5 7 10, 12 17 24, 29 41 58,
70 99 ... If the side of a square measures 70 palms, the diagonal measures
practically 99 palms.
Now imagine a circle around the square. How long is the circumference?
The diameter is given by the diagonal 99 palms. Draw up a suitable pi
sequence, here the first one followed by the most important one
4/1 (plus 3/1) 7/2 10/3 13/4 16/5 19/6 22/7
3/1 (plus 22/7) 25/8 47/15 69/22 91/29 113/36 135/43 157/50
179/57 201/64 223/71 245/78 267/85 289/92 311/99
If the diameter of a circle measures 99 palms, the cicumference measures
practically 311 palms.
And the exact length of the circumference of a circle around the square
measuring 10 by 10 royal cubits? 311.00180... palms, not even one seventh
of a millimeter more than 311 palms!
This calculation shows the advantage of Egyptian mathematics: operate
with number patterns, columns and sequences, choose values that come handy
in the given calculation, the mistakes will even out, sometimes almost
completely, and you get a fine result.
Next time: we shall measure and calculate the circumference of the circle
inscribed in the square 10 by 10 royal cubits or 70 by 70 palms or 280
by 280 fingerbreadths
Egyptian method of calculating the circle (part 2)

Inscribe a circle into the square 10 by 10 royal cubits or 70 by 70 palms
or 280 by 280 fingerbreadths and combine it with the grid 10 by 10. The
circumference of the circle passes through the ends of the horizontal axis
and vertical axis, moreover through eight inner points of the grid that are
defined by the Sacred Triangle 3-4-5. In all the grid offers 12 rational
points of the circle, and divides the circumference into 4 shorter arcs
and 8 longer arcs that measure practically 40 and 90 fingerbreadths
respectively, yielding a total length of

4 x 40 plus 8 x 90 or 880 fingerbreadths or 220 palms

and if we divide the circumference 220 palms by the diameter 70 palms
we obtain 22/7 for pi.

Now let us connect us the dozen points one by one with straight lines.
Thus we obtain a polygon of 4 shorter and 8 longer sides. The shorter
side is given by the square root of 2, and the longer side by the square
root of 2 times the square root of 5. These roots can be approximated
by number columns (here given as number sequences) whose firt lines are

1 1 2, 2 3 4, 5 7 10, 12 17 24, 29 41 58, 70 99 140 ...

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 20, 13 29 65, 42 94 210, 21 47 105, 68 152 340,
34 76 170, 17 38 85, 55 123 275, 178 398 890, 89 199, 445, 288 644 1440,
144 322 720, 72 161 360 ...

By the way, the second number pattern contains a pair of golden sequences,
the so-called Lucas sequence (above) and Fibonacci-sequence (below)

1 3 4 7 11 18 29 47 76 123 199 322 ...
1 1 2 3 5 8 13 21 34 55 89 144 ...

The periphery of the polygon measures

4 x sqrt2 plus 8 x sqrt2 x sqrt5 royal cubits

Considering that the arcs of the circle are slightly longer than the sides
of the polygon we choose values for the roots that are slightly bigger than
the actual roots, 10/7 for the square root of 2, and 9/4 for the one of 5

10/7 x 10/7 = 100/49 (slightly more than 2)
9/4 x 9/4 = 81/16 (slightly more than 5)

Now the periphery of the rounded polygon measures

4 x 10/7 plus 8 x 10/7 x 9/4 royal cubits

4 x 10 plus 2 x 10 x 9 or 40 plus 180 or 220 palms

and if we divide the rounded periphery or the circumference of the circle
220 palms by the diameter 70 palms we obtain again 22/7 for pi.

Next time: calculating the circles of radius 5 25 125 625 ...
Franz Gnaedinger
2015-12-11 09:30:28 UTC
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Post by Franz Gnaedinger
Post by Franz Gnaedinger
http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
Egyptian method of calculating the circle (part 2)
Inscribe a circle into the square 10 by 10 royal cubits or 70 by 70 palms
or 280 by 280 fingerbreadths and combine it with the grid 10 by 10. The
circumference of the circle passes through the ends of the horizontal axis
and vertical axis, moreover through eight inner points of the grid that are
defined by the Sacred Triangle 3-4-5. In all the grid offers 12 rational
points of the circle, and divides the circumference into 4 shorter arcs
and 8 longer arcs that measure practically 40 and 90 fingerbreadths
respectively, yielding a total length of
4 x 40 plus 8 x 90 or 880 fingerbreadths or 220 palms
and if we divide the circumference 220 palms by the diameter 70 palms
we obtain 22/7 for pi.
Now let us connect us the dozen points one by one with straight lines.
Thus we obtain a polygon of 4 shorter and 8 longer sides. The shorter
side is given by the square root of 2, and the longer side by the square
root of 2 times the square root of 5. These roots can be approximated
by number columns (here given as number sequences) whose firt lines are
1 1 2, 2 3 4, 5 7 10, 12 17 24, 29 41 58, 70 99 140 ...
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 20, 13 29 65, 42 94 210, 21 47 105, 68 152 340,
34 76 170, 17 38 85, 55 123 275, 178 398 890, 89 199, 445, 288 644 1440,
144 322 720, 72 161 360 ...
By the way, the second number pattern contains a pair of golden sequences,
the so-called Lucas sequence (above) and Fibonacci-sequence (below)
1 3 4 7 11 18 29 47 76 123 199 322 ...
1 1 2 3 5 8 13 21 34 55 89 144 ...
The periphery of the polygon measures
4 x sqrt2 plus 8 x sqrt2 x sqrt5 royal cubits
Considering that the arcs of the circle are slightly longer than the sides
of the polygon we choose values for the roots that are slightly bigger than
the actual roots, 10/7 for the square root of 2, and 9/4 for the one of 5
10/7 x 10/7 = 100/49 (slightly more than 2)
9/4 x 9/4 = 81/16 (slightly more than 5)
Now the periphery of the rounded polygon measures
4 x 10/7 plus 8 x 10/7 x 9/4 royal cubits
4 x 10 plus 2 x 10 x 9 or 40 plus 180 or 220 palms
and if we divide the rounded periphery or the circumference of the circle
220 palms by the diameter 70 palms we obtain again 22/7 for pi.
Next time: calculating the circles of radius 5 25 125 625 ...
Egyptian method of calculating the circle (part 3)

What if we proceed from the grid 10 by 10 to the finer grid 50 by 50
and the inscribed circle of radius 25 ? The Sacred Triangle 3-4-5
assumes the form 15-20-25 (remember the Sacred Triangle 15-20-25 royal
cubits found by Jean-Philippe Lauer in the King's Chamber of the Great
Pyramid). The number 25 offers one more triple: 7-24-25. The new triple
defines 8 more rational points on the circumference. We now have 4 8 8
sum 20 arcs, and 20 sides of the polygon, 12 shorter sides and 8 longer
ones

shorter side square root of 50 or 5 sqrt2
longer side square root of 80 or 4 sqrt5

For the periphery of the polygon we obtain

12 x 5 sqrt2 plus 8 x 4 x sqrt5

This time we use 17/12 for the square root of 2 and again 9/4 for the
square root of 5

12 x 5 x 17/12 plus 8 x 4 x 9/4 or 5 x 17 plus 8 x 9 sum 157

Divide the rounded periphery or the circumference 157 by the diameter 50
and you get 157/50 or 3.14 for pi.

The first value 22/7 and the second one 157/50 appear in the most important
pi sequence

3/1 (plus 22/7) 25/8 47/15 ... 157/50 ... 311/99 ... 377/120

The method of polygons defined by a sequence of triples can be expanded.
Multiply the radius by a factor of 5, over and over again: 5 25 125 625 ...
There will always be a new triple: 3-4-5, 7-24-25, 44-117-125, 336-527-625
...

3-4-5 or 15-2025 or 75-100-125 or 336-527-625 ...
7-24-25 or 35-120-125 or 175-600-625 ...
44-117-125 or 220-585-625 ...
336-527-625 ...

If you know a triple a-b-c and wish to find the next one, calculate the
following linear terms

plus minus 4a plus minus 3b plus minus 3a plus minus 4b 5c

and choose the positive values that are no multiple of 5 for the long terms.
The polygon has 12 20 28 36 ... sides of two or three different lengths
that are integer multiples of the square root of 2 and/or the square root
of 5 and/or the square root of 2 times the square root of 5. The polygon
is getting rounder and rounder, and the side lenghts can be approximated
to any desired accuracy by means of the number columns that begin with
the lines 1 1 2 and 1 1 5.

So we have a systematic method for calculating the circle, in my opinion
the most valuable treasure contained by the Great Pyramid.
Franz Gnaedinger
2015-12-14 07:49:20 UTC
Reply
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Post by Franz Gnaedinger
Post by Franz Gnaedinger
http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
Egyptian method of calculating the circle (part 3)
What if we proceed from the grid 10 by 10 to the finer grid 50 by 50
and the inscribed circle of radius 25 ? The Sacred Triangle 3-4-5
assumes the form 15-20-25 (remember the Sacred Triangle 15-20-25 royal
cubits found by Jean-Philippe Lauer in the King's Chamber of the Great
Pyramid). The number 25 offers one more triple: 7-24-25. The new triple
defines 8 more rational points on the circumference. We now have 4 8 8
sum 20 arcs, and 20 sides of the polygon, 12 shorter sides and 8 longer
ones
shorter side square root of 50 or 5 sqrt2
longer side square root of 80 or 4 sqrt5
For the periphery of the polygon we obtain
12 x 5 sqrt2 plus 8 x 4 x sqrt5
This time we use 17/12 for the square root of 2 and again 9/4 for the
square root of 5
12 x 5 x 17/12 plus 8 x 4 x 9/4 or 5 x 17 plus 8 x 9 sum 157
Divide the rounded periphery or the circumference 157 by the diameter 50
and you get 157/50 or 3.14 for pi.
The first value 22/7 and the second one 157/50 appear in the most important
pi sequence
3/1 (plus 22/7) 25/8 47/15 ... 157/50 ... 311/99 ... 377/120
The method of polygons defined by a sequence of triples can be expanded.
Multiply the radius by a factor of 5, over and over again: 5 25 125 625 ...
There will always be a new triple: 3-4-5, 7-24-25, 44-117-125, 336-527-625
...
3-4-5 or 15-2025 or 75-100-125 or 336-527-625 ...
7-24-25 or 35-120-125 or 175-600-625 ...
44-117-125 or 220-585-625 ...
336-527-625 ...
If you know a triple a-b-c and wish to find the next one, calculate the
following linear terms
plus minus 4a plus minus 3b plus minus 3a plus minus 4b 5c
and choose the positive values that are no multiple of 5 for the long terms.
The polygon has 12 20 28 36 ... sides of two or three different lengths
that are integer multiples of the square root of 2 and/or the square root
of 5 and/or the square root of 2 times the square root of 5. The polygon
is getting rounder and rounder, and the side lenghts can be approximated
to any desired accuracy by means of the number columns that begin with
the lines 1 1 2 and 1 1 5.
So we have a systematic method for calculating the circle, in my opinion
the most valuable treasure contained by the Great Pyramid.
Egyptian method of calculating the circle (illustrations)

Key figure, grid 10 by 10, square 10 by 10 royal cubits or 70 by 70 palms
or 280 by 280 fingerbreadths, diagonal practically 99 palms (1 1 2, 2 3 4,
5 7 10, 12 17 24, 29 41 58, 70 99 ...), twelve rational points of the
virtual circle, four points given by the ends of tha axes and eight points
defined by the Sacred Triangle 3-4-5 doubled to the rectangle 4 by 3 or
3 by 4 diagonal 5

Loading Image...
Loading Image...

The shorter arcs of the circle indicated by the dozen points measure
practically 40 fingerbreadths each, and the longer ones 90 fingerbreadths,
yielding a circumference of 880 fingerbreadths or 220 palms, and if we
divide the 220 palms by the diameter 70 palms we obtain 22/7 for pi.

Drawings of the first four polygons that have 12 20 28 36 shorter and
shorter sides

Loading Image...
Loading Image...
Loading Image...
Loading Image...

12 20 28 36 points of the grid evoke the circle

Loading Image...
Loading Image...
Loading Image...
Loading Image...

The corners of the rounder and rounder polygon are generated by a turning
angle of arctan3/4 (36.86989... degrees), mirror each point around the
axes (y = 0, x = 0) and the oblique cross (y = +-x)

Loading Image...

Another way of generating the triples is to draw up a number column
that uses a factor of minus four

1 1 -4
2 -3 -8 3 8 3-4-5
-1 -11 4
-12 -7 48 7 48 7-24-25
-19 41 76
22 117 -88 117 88 44-117-125
139 29 -556
168 -527 -672 527 672 336-527-625
-359 -1199 1436
-1558 237 6232 237 6232 237-3116-3125
and so on

Many more drawings can be found via egypt2 and egypt3 (links above),
here for example the hemisphere of Nut (below) and circle of Ra (above)

Loading Image...




http://www.seshat.ch/home/
Eric Stevens
2015-12-14 21:03:58 UTC
Reply
Permalink
On Sun, 13 Dec 2015 23:49:20 -0800 (PST), Franz Gnaedinger
Post by Franz Gnaedinger
Post by Franz Gnaedinger
Post by Franz Gnaedinger
http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
Egyptian method of calculating the circle (part 3)
What if we proceed from the grid 10 by 10 to the finer grid 50 by 50
and the inscribed circle of radius 25 ? The Sacred Triangle 3-4-5
assumes the form 15-20-25 (remember the Sacred Triangle 15-20-25 royal
cubits found by Jean-Philippe Lauer in the King's Chamber of the Great
Pyramid). The number 25 offers one more triple: 7-24-25. The new triple
defines 8 more rational points on the circumference. We now have 4 8 8
sum 20 arcs, and 20 sides of the polygon, 12 shorter sides and 8 longer
ones
shorter side square root of 50 or 5 sqrt2
longer side square root of 80 or 4 sqrt5
For the periphery of the polygon we obtain
12 x 5 sqrt2 plus 8 x 4 x sqrt5
This time we use 17/12 for the square root of 2 and again 9/4 for the
square root of 5
12 x 5 x 17/12 plus 8 x 4 x 9/4 or 5 x 17 plus 8 x 9 sum 157
Divide the rounded periphery or the circumference 157 by the diameter 50
and you get 157/50 or 3.14 for pi.
The first value 22/7 and the second one 157/50 appear in the most important
pi sequence
3/1 (plus 22/7) 25/8 47/15 ... 157/50 ... 311/99 ... 377/120
The method of polygons defined by a sequence of triples can be expanded.
Multiply the radius by a factor of 5, over and over again: 5 25 125 625 ...
There will always be a new triple: 3-4-5, 7-24-25, 44-117-125, 336-527-625
...
3-4-5 or 15-2025 or 75-100-125 or 336-527-625 ...
7-24-25 or 35-120-125 or 175-600-625 ...
44-117-125 or 220-585-625 ...
336-527-625 ...
If you know a triple a-b-c and wish to find the next one, calculate the
following linear terms
plus minus 4a plus minus 3b plus minus 3a plus minus 4b 5c
and choose the positive values that are no multiple of 5 for the long terms.
The polygon has 12 20 28 36 ... sides of two or three different lengths
that are integer multiples of the square root of 2 and/or the square root
of 5 and/or the square root of 2 times the square root of 5. The polygon
is getting rounder and rounder, and the side lenghts can be approximated
to any desired accuracy by means of the number columns that begin with
the lines 1 1 2 and 1 1 5.
So we have a systematic method for calculating the circle, in my opinion
the most valuable treasure contained by the Great Pyramid.
Egyptian method of calculating the circle (illustrations)
Key figure, grid 10 by 10, square 10 by 10 royal cubits or 70 by 70 palms
or 280 by 280 fingerbreadths, diagonal practically 99 palms (1 1 2, 2 3 4,
5 7 10, 12 17 24, 29 41 58, 70 99 ...), twelve rational points of the
virtual circle, four points given by the ends of tha axes and eight points
defined by the Sacred Triangle 3-4-5 doubled to the rectangle 4 by 3 or
3 by 4 diagonal 5
http://www.seshat.ch/home/key1.GIF
http://www.seshat.ch/home/key2.GIF
The shorter arcs of the circle indicated by the dozen points measure
practically 40 fingerbreadths each, and the longer ones 90 fingerbreadths,
yielding a circumference of 880 fingerbreadths or 220 palms, and if we
divide the 220 palms by the diameter 70 palms we obtain 22/7 for pi.
Drawings of the first four polygons that have 12 20 28 36 shorter and
shorter sides
http://www.seshat.ch/home/poly1.GIF
http://www.seshat.ch/home/poly2.GIF
http://www.seshat.ch/home/poly3.GIF
http://www.seshat.ch/home/poly4.GIF
12 20 28 36 points of the grid evoke the circle
http://www.seshat.ch/home/polyg1a.GIF
http://www.seshat.ch/home/polyg1b.GIF
http://www.seshat.ch/home/polyg1c.GIF
http://www.seshat.ch/home/polyg1d.GIF
The corners of the rounder and rounder polygon are generated by a turning
angle of arctan3/4 (36.86989... degrees), mirror each point around the
axes (y = 0, x = 0) and the oblique cross (y = +-x)
http://www.seshat.ch/home/poly5.GIF
Another way of generating the triples is to draw up a number column
that uses a factor of minus four
1 1 -4
2 -3 -8 3 8 3-4-5
-1 -11 4
-12 -7 48 7 48 7-24-25
-19 41 76
22 117 -88 117 88 44-117-125
139 29 -556
168 -527 -672 527 672 336-527-625
-359 -1199 1436
-1558 237 6232 237 6232 237-3116-3125
and so on
Many more drawings can be found via egypt2 and egypt3 (links above),
here for example the hemisphere of Nut (below) and circle of Ra (above)
http://www.seshat.ch/home/pyramid2.GIF
All of this is very interesting but, did the ancient Egyptians have an
arithmetical system capable of processing the calulations you have
demonstrated?
Post by Franz Gnaedinger
http://www.seshat.ch/home/
Error 403 "Access forbidden".
--
Regards,

Eric Stevens
Franz Gnaedinger
2015-12-15 07:58:36 UTC
Reply
Permalink
Post by Eric Stevens
All of this is very interesting but, did the ancient Egyptians have an
arithmetical system capable of processing the calulations you have
demonstrated?
Of course, the Egyptians had a decimal system for integers, and unit
fractions series. The novelty of my approach is that you can carry out
even highly demanding calculations with integers only. An example.
How long is the diagonal of the square 10 by 10 royal cubits or 70 by 70
palms? We calculate the square root of 2 and get an ever longer decimal
fraction: 1.41421356237309... Then we multiply 10 royal cubits or 70 palms
by that number. The Egyptian way was different. Draw up the number column
for the square root of 2

1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
and so on

Now pick up the value that comes handy, namely side 70 diagonal 99.
So if the side of the square measures 70 palms, the diagonal measures
practically 99 palms. Another question. How long is the diagonal
of the cube 10 by 10 by 10 royal cubits or 70 by 70 by 70 palms
or 280 by 280 by 280 fingerbreadths? Draw up another number column

1 1 3
2 4 6
1 2 3
3 5 9
8 14 24
4 7 12
11 19 33
30 52 90
15 26 45
41 71 123
112 194 336
56 97 168
and so on

Choose the number 56 for the edge of a cube and the number 97 for the
diagonal of the volume. Multiply them by a factor of 5 and you obtain
the edge 280 and diagonal of the volume 485. So we have these numbers
for the square and the cube

side of square and edge of cube 10 royal cubits or 70 palms or 280 f
diagonal of square 99 palms
diagonal volume 485 fingerbreadths

Another example. What if the side of the square and edge of the cube
measures 41 fingerbreadths? Consulting the above number columns you
find the diagonal square 58 f and the diagonal volume cube 71 f.

All the calculations I presented can be carried out with integers only.
The ancient methods I am reconstructing since 1979 work with integers.
Mistakes even out in the long run, often miraculously well, Eric,
we discussed this in around 2003, as I recall. I demonstrated to you
how the small mistakes in measuring out a long distance with a square
and cross even out.

Later on I will demonstrate how astronomical calculations work with
integers. Here an example. The Maya had phantastic values for the
Venus year. How did they achieve their Venus calendar? A Venus year
is between 583 and 584 days, closer to 584 than 583 days. So draw up
this number sequence

583/1 (plus 584/1) ... 7007/12

You get a sequence of values and can again choose the one that comes handy.
A most handy value for the Mayan number and calendar system are the above
7007 days for a dozen Venus years.

Entropy increases not only toward the future but also toward the past.
We need ever more imagination to reconstruct early ways. I developed
an experimental math-history in analogy to experimental archaeology.
Moreover, mathematical examinations can be a non-invasive method.
Also I profit from the insights gained by invasive methods, but then
I draw conclusion that reach way farther, owing to experimental early
mathematics, methods based on addition - what is the summing up of some
numbers compared to the piling up of millions of bricks and stone blocks?
Peter Jason
2016-02-15 00:38:43 UTC
Reply
Permalink
Instead of all the headache-inducing mathemetics, the Egyptians
probably used developement by incribing the plan on the ground and
projecting this up onto the rising pyramid by means of theodolites
etc.
Loading Image...
Loading Image...
etc.
See Romer:
Loading Image...
Franz Gnaedinger
2016-02-15 08:38:07 UTC
Reply
Permalink
Post by Peter Jason
Instead of all the headache-inducing mathemetics, the Egyptians
probably used developement by incribing the plan on the ground and
projecting this up onto the rising pyramid by means of theodolites
etc.
http://www.davidfurlong.co.uk/images/pyramid_scan6.jpg
http://www.destinypictures.net/masterplan2.jpg
etc.
http://ecx.images-amazon.com/images/I/31FBcUzPJYL._AC_UL320_SR244,320_.jpg
The biggest headache would be to measure out the square of a pyramid
base with the methods proposed by academe. Recently seen in a tv
documentary, professors explaining how the pyramids were built:
begin with the square of the base, take two long ropes (some 220 meters
in the case of the Great Pyramid) and a wooden equèrre as we used it
in school for the right angle, and voilà ... Typical desk top half-idea,
would never work in the field, could never explain how the base of the
Great Pyramid was measured out to the phantasmagorical precision reported
by Rainer Stadelmann in the second edition of his book on the Egyptian
Pyramids (he was responsible for the pyramids before Zahi Hawass took
over), and this around a large limestone hill still extant under the
building, visible at a corner. Mark Lehner proposes a wooden triangle
3-4-5 which poses the same problem: you can't simply prolong a straight
line of, say, one meter, to 220 meters, keeping it perfectly straight.
Might be done with a laser pointer, but never with a rope. I have practical
experience, three years of measuring out and drawing sports fields in
my younger days. Professors lack such experience, the methods they propose
work for a small NOVA pyramid, but never for a real pyramid. And then,
what is the adding of integers to the piling up of millions of blocks,
each weighing tons? All the mathematical methods I reconstructed are
simple yet clever additive methods, including the systematic method
of calculating the circle, a long time before Archimedes. The big problem
are not such methods but the dogma issued by academe: the Egyptians had
no real mathematics. Everybody can see that a pyramid is a geometrical
hence a mathematical building - but if, as the professors declare,
the Egyptians had no mathematics, then the pyramids, or at least the
Great Pyramid, must have been built by aliens or Atlantians (big topic
in sci.archaeology in the late 1990s). My compatriot Erich von Daeniken
wrote a book on the Great Pyramid wherein he says that the alien who
built this masterwork is sleeping (still sleeping) in a hidden chamber
within. EvD can publish anywhere anytime; I can't publish my insights
on pre-Greek mathematics.
Peter Jason
2016-02-16 01:36:05 UTC
Reply
Permalink
On Mon, 15 Feb 2016 00:38:07 -0800 (PST), Franz Gnaedinger
Post by Franz Gnaedinger
Post by Peter Jason
Instead of all the headache-inducing mathemetics, the Egyptians
probably used developement by incribing the plan on the ground and
projecting this up onto the rising pyramid by means of theodolites
etc.
http://www.davidfurlong.co.uk/images/pyramid_scan6.jpg
http://www.destinypictures.net/masterplan2.jpg
etc.
http://ecx.images-amazon.com/images/I/31FBcUzPJYL._AC_UL320_SR244,320_.jpg
The biggest headache would be to measure out the square of a pyramid
base with the methods proposed by academe. Recently seen in a tv
begin with the square of the base, take two long ropes (some 220 meters
in the case of the Great Pyramid) and a wooden equèrre as we used it
in school for the right angle, and voilà ... Typical desk top half-idea,
would never work in the field, could never explain how the base of the
Great Pyramid was measured out to the phantasmagorical precision reported
by Rainer Stadelmann in the second edition of his book on the Egyptian
Pyramids (he was responsible for the pyramids before Zahi Hawass took
over), and this around a large limestone hill still extant under the
building, visible at a corner. Mark Lehner proposes a wooden triangle
3-4-5 which poses the same problem: you can't simply prolong a straight
line of, say, one meter, to 220 meters, keeping it perfectly straight.
Might be done with a laser pointer, but never with a rope. I have practical
experience, three years of measuring out and drawing sports fields in
my younger days. Professors lack such experience, the methods they propose
work for a small NOVA pyramid, but never for a real pyramid. And then,
what is the adding of integers to the piling up of millions of blocks,
each weighing tons? All the mathematical methods I reconstructed are
simple yet clever additive methods, including the systematic method
of calculating the circle, a long time before Archimedes. The big problem
are not such methods but the dogma issued by academe: the Egyptians had
no real mathematics. Everybody can see that a pyramid is a geometrical
hence a mathematical building - but if, as the professors declare,
the Egyptians had no mathematics, then the pyramids, or at least the
Great Pyramid, must have been built by aliens or Atlantians (big topic
in sci.archaeology in the late 1990s). My compatriot Erich von Daeniken
wrote a book on the Great Pyramid wherein he says that the alien who
built this masterwork is sleeping (still sleeping) in a hidden chamber
within. EvD can publish anywhere anytime; I can't publish my insights
on pre-Greek mathematics.
Yet they must have had some feedback procedure by which they could
check progress & accuracy. Of course Aliens could have had a laser
dummy pyramid that could be filled with stones without any checking at
all.
But for simple humans, only external reference points would be
essential.
Post holes & trenches for just this purpose have been found south of
the Cheops pyramid.
Franz Gnaedinger
2016-02-16 08:33:33 UTC
Reply
Permalink
Post by Peter Jason
Yet they must have had some feedback procedure by which they could
check progress & accuracy. Of course Aliens could have had a laser
dummy pyramid that could be filled with stones without any checking at
all.
But for simple humans, only external reference points would be
essential.
Post holes & trenches for just this purpose have been found south of
the Cheops pyramid.
Here is my procedure of aligning the pyramid and measuring out the base
and leveling it at the same time

Loading Image...

text in

http://www.seshat.ch/home/egypt3.htm

and mathematical methods in

http://www.seshat.ch/home/egypt2.htm

All you need for leveling and measuring out the base is a grid.
Place a wooden block on each point of the grid. Then check the heights
of two neighboring blocks with a device depicted in my large drawing.
If one of the blocks is too low, add a piece of wood on its top.
Go on until you have leveled all blocks on the grid. Then use the
wooden frame-and-cross with four nails. Place the nails on four blocks
and make an impression. Then turn the frame-and-cross around an angle
of 90 degrees and check whether the nails fit in the small holes made
before. If they do, fine, you can proceed. By and by you get a grid
of reference points. You don't need to level the entire plateau,
just the tops of the wooden blocks, an ideal grid of 50 by 50 blocks
in the case of a large pyramid, but way less in the case of the Great
Pyramid, because of the large limestone hill (that accounts for one
third of the pyramid's volume according to Rainer Stadelmann).

Now for the wooden frame and cross: royal cubit of the Great Pyramid
52.36 centimeters, side of the square 9 royal cubits, diagonal 20 Horus
cubits of 33.32 centimeters (most important of the seven Horus cubits,
defined as follows: 11 Horus cubits equal 7 royal cubits). These numbers
rely on the ratio 99/70 and 140/99 for the square root of 2 which are
found in the basic number column for calculating the square.

Here again, for Jason and his Archaeonauts: how to calculate a square.

Practical experience will sooner or later lead to a definition of the
saquare: four sides of equal length, and a pair of diagonals, both
equally long. But what is the ratio of side and diagonal? Lay out
a square in a field, marking the corners with pebbles, then measure
the side and the diagonal with your feet, or steps, and you may found
this formula

if the side of a square measures 5 paces
or a multiple thereof
the diagonal measures practically 7 paces
or a multiple thereof
and if the side measures 7 paces
or a multiple therof
the diagonal measures 2 x 5 = 10 paces
or a multiple thereof

I have indirect evidence that this formula was already known 35,000
years ago in Africa. Then, perhaps in Ancient Egypt, a mathematician
asked: what if I make the side 5 plus 7 paces long? will the diagonal
measure 7 plus 10 paces, together 17 paces? Yes, it does. And this
insight led to a number column: add neighboring numbers, and double
the first number of each line

1 1 2
2 3 4
5 7 10 (original formula)
12 17 24 (used by Imhotep at Saqqara)
29 41 58
70 99 140 (used by Hemon at Giza)

I found ample indirect evidence for the use of this and further number
columns in the Rhind Mathematical Papyrus from around 1650 BC, copy
of a lost original from around 1850 BC. The great advantage of those
number columns, and of additive number sequences, are that you can
operate with integers - pluck the number that comes handy in a given
calculation, the mistakes even out, often amazingly well. And so do
the small mistakes when you measure out a grid for the base of a large
pyramid (I carried out an error calculation years ago in sci.archaeology,
discussing this topic with Eric Stevens).
Franz Gnaedinger
2016-02-24 07:37:50 UTC
Reply
Permalink
Post by Franz Gnaedinger
Here is my procedure of aligning the pyramid and measuring out the base
and leveling it at the same time
http://www.seshat.ch/home/base.GIF
text in
http://www.seshat.ch/home/egypt3.htm
and mathematical methods in
http://www.seshat.ch/home/egypt2.htm
All you need for leveling and measuring out the base is a grid.
Place a wooden block on each point of the grid. Then check the heights
of two neighboring blocks with a device depicted in my large drawing.
If one of the blocks is too low, add a piece of wood on its top.
Go on until you have leveled all blocks on the grid. Then use the
wooden frame-and-cross with four nails. Place the nails on four blocks
and make an impression. Then turn the frame-and-cross around an angle
of 90 degrees and check whether the nails fit in the small holes made
before. If they do, fine, you can proceed. By and by you get a grid
of reference points. You don't need to level the entire plateau,
just the tops of the wooden blocks, an ideal grid of 50 by 50 blocks
in the case of a large pyramid, but way less in the case of the Great
Pyramid, because of the large limestone hill (that accounts for one
third of the pyramid's volume according to Rainer Stadelmann).
Now for the wooden frame and cross: royal cubit of the Great Pyramid
52.36 centimeters, side of the square 9 royal cubits, diagonal 20 Horus
cubits of 33.32 centimeters (most important of the seven Horus cubits,
defined as follows: 11 Horus cubits equal 7 royal cubits). These numbers
rely on the ratio 99/70 and 140/99 for the square root of 2 which are
found in the basic number column for calculating the square.
Here again, for Jason and his Archaeonauts: how to calculate a square.
Practical experience will sooner or later lead to a definition of the
saquare: four sides of equal length, and a pair of diagonals, both
equally long. But what is the ratio of side and diagonal? Lay out
a square in a field, marking the corners with pebbles, then measure
the side and the diagonal with your feet, or steps, and you may found
this formula
if the side of a square measures 5 paces
or a multiple thereof
the diagonal measures practically 7 paces
or a multiple thereof
and if the side measures 7 paces
or a multiple therof
the diagonal measures 2 x 5 = 10 paces
or a multiple thereof
I have indirect evidence that this formula was already known 35,000
years ago in Africa. Then, perhaps in Ancient Egypt, a mathematician
asked: what if I make the side 5 plus 7 paces long? will the diagonal
measure 7 plus 10 paces, together 17 paces? Yes, it does. And this
insight led to a number column: add neighboring numbers, and double
the first number of each line
1 1 2
2 3 4
5 7 10 (original formula)
12 17 24 (used by Imhotep at Saqqara)
29 41 58
70 99 140 (used by Hemon at Giza)
I found ample indirect evidence for the use of this and further number
columns in the Rhind Mathematical Papyrus from around 1650 BC, copy
of a lost original from around 1850 BC. The great advantage of those
number columns, and of additive number sequences, are that you can
operate with integers - pluck the number that comes handy in a given
calculation, the mistakes even out, often amazingly well. And so do
the small mistakes when you measure out a grid for the base of a large
pyramid (I carried out an error calculation years ago in sci.archaeology,
discussing this topic with Eric Stevens).
XQ The Super School Project (hors concours)

Being Swiss I can't participate in the XQ Super School Project,
and having neither a Facebook nor a Twitter account I can't present
my idea, so I post a message here.

My concern are math lessons on the level of primary school and highschool.
I saw a lot of my comrades suffer in those lessons, way back in my
school years, and remember well how upset I was when my teacher began
geometry with Euklid's parallel-axiom: how can he know that parallels
never meet? Looking back on my early years in school I find that math
lessons begin on too high a level, owing to academe's dogma of the Greek
origin of mathematics. My studies of several decades reveal a mathematical
cosmos below the level of Greek mathematics, consisting of simple yet clever
additive methods, including a systematic calculation of the circle, long
before Archimedes. Playing with such methods would allow an easier access
to mathematics, a discipline feared and even hated by many, and would
also be a step toward a prospering global society that requires a fair
history of civilization including mathematics, logic of building and
maintaining: a fair history of civilization acknowledging all contributions
and encouraging well-meaning people all over the world to participate
in the development of our common civilization.

Readers in German might look up my brochure

http://www.seshat.ch/home/mathe.pdf
Franz Gnaedinger
2016-03-02 07:32:11 UTC
Reply
Permalink
Post by Franz Gnaedinger
XQ The Super School Project (hors concours)
Being Swiss I can't participate in the XQ Super School Project,
and having neither a Facebook nor a Twitter account I can't present
my idea, so I post a message here.
My concern are math lessons on the level of primary school and highschool.
I saw a lot of my comrades suffer in those lessons, way back in my
school years, and remember well how upset I was when my teacher began
geometry with Euklid's parallel-axiom: how can he know that parallels
never meet? Looking back on my early years in school I find that math
lessons begin on too high a level, owing to academe's dogma of the Greek
origin of mathematics. My studies of several decades reveal a mathematical
cosmos below the level of Greek mathematics, consisting of simple yet clever
additive methods, including a systematic calculation of the circle, long
before Archimedes. Playing with such methods would allow an easier access
to mathematics, a discipline feared and even hated by many, and would
also be a step toward a prospering global society that requires a fair
history of civilization including mathematics, logic of building and
maintaining: a fair history of civilization acknowledging all contributions
and encouraging well-meaning people all over the world to participate
in the development of our common civilization.
Readers in German might look up my brochure
http://www.seshat.ch/home/mathe.pdf
Super School Project (postscript)

My second primary school teacher, by then a young man whom I liked well,
began the very first geometry lesson by explaining Euclid's axioms,
ending with the parallel axiom. This one upset me: how can my teacher
know how those lines behave at an infinite distance? Years later I learned
that the parallel axiom caused a big problem, as it can't be deduced
from the other axioms. I pride myself on having sensed that gap
intuitively. The incoherence between the parallel axiom and the other
axioms led to non-Euclidean geometries, elliptic and hyperbolic ones
where parallels intersect or drift apart respectively.

Again many years later I learned that there had been two or even three
Euclids who were no single personalities but groups of mathematicians
who gathered the mathematical knowledge from Ancient Egypt and
Mesopotamia and other regions of the ancient world.

Beginning with Euclid and Greek geometry is too much of a challenge
for many pupils, who subsequently fear and often hate mathematics
for all their life. Beginning with the simple pre-Greek methods
I reconstructed will ease their access to that discipline.

What makes me wonder is that mathematicians understand the most
complicated theories but fail when it comes to early methods.

If I were to found a mathematical discipline, then the quest for
the simplest solution to a given problem.
Franz Gnaedinger
2016-03-12 09:28:18 UTC
Reply
Permalink
Post by Franz Gnaedinger
Super School Project (postscript)
My second primary school teacher, by then a young man whom I liked well,
began the very first geometry lesson by explaining Euclid's axioms,
ending with the parallel axiom. This one upset me: how can my teacher
know how those lines behave at an infinite distance? Years later I learned
that the parallel axiom caused a big problem, as it can't be deduced
from the other axioms. I pride myself on having sensed that gap
intuitively. The incoherence between the parallel axiom and the other
axioms led to non-Euclidean geometries, elliptic and hyperbolic ones
where parallels intersect or drift apart respectively.
Again many years later I learned that there had been two or even three
Euclids who were no single personalities but groups of mathematicians
who gathered the mathematical knowledge from Ancient Egypt and
Mesopotamia and other regions of the ancient world.
Beginning with Euclid and Greek geometry is too much of a challenge
for many pupils, who subsequently fear and often hate mathematics
for all their life. Beginning with the simple pre-Greek methods
I reconstructed will ease their access to that discipline.
What makes me wonder is that mathematicians understand the most
complicated theories but fail when it comes to early methods.
If I were to found a mathematical discipline, then the quest for
the simplest solution to a given problem.
from the lunisolar calendar of the Göbekli Tepe region as of 12,000
years ago to the one of the Horus eye in Ancient Egyt (more numbers,
but easy to understand for an interested reader)

Stone Age way of counting lunations or synodic months

30 29 30 29 30 29 30 ... days for 1 2 3 4 5 6 7 ... lunations

(also 29 30 29 30 29 30 29 ... days, less accurate)

15 and 17 lunations counted in the 30 29 30 mode yield 443 and 502 days
respectively

17 15 17 15 17 or 17 32 49 64 81 lunations

502 443 502 443 502 or 502 945 1447 1890 2392 days

1,890 days correspond to 63 lunations and are 63 continuous periods
of 30 days, allowing to formulate a lunisolar calendar of the Göbekli
Tepe region as of 12,000 years ago

a month has 30 days,
12 months are a basic year of 360 days,
add 5 and occasionally 6 more days
and you get a regular year of 365 days
and an occasional leap year of 366 days,
while 63 continuous periods of 30 days
are 1,890 days anc correspond to 64 lunations
(mistake per lunation less than one minute,
or half a day in a lifetime)

I reconstructed this calendar in late 2004 and immediately recognized
the connection to the Horus eye of Ancient Egypt.

The left eye of the Horus falcon was the moon, and his right eye the sun.
Seth destroyed the lunar eye. Wise Thoth healed it by reassembling the
fragments '2 '4 '8 '16 '32 '64, whereupon he called the restored lunar eye
The Whole One. However, the numbers don't really add up to one, a tiny part
is missing. Why then the whole one? This term does not refer to the whole
moon, the eye of the night sky, but to a whole lunation, the whole lunar
cycle. Multiply the 30 days of an Egyptian month by the series of the
Horus eye '2 '4 '8 '16 '32 '64 and you get 29 '2 '32 days for a lunation,
or 29 days 12 hours 45 minutes, exact value 29 days 12 hours 44 minutes
2.9 seconds (modern average from 1989 AD).
Franz Gnaedinger
2016-03-26 09:07:06 UTC
Reply
Permalink
Post by Franz Gnaedinger
from the lunisolar calendar of the Göbekli Tepe region as of 12,000
years ago to the one of the Horus eye in Ancient Egyt (more numbers,
but easy to understand for an interested reader)
Stone Age way of counting lunations or synodic months
30 29 30 29 30 29 30 ... days for 1 2 3 4 5 6 7 ... lunations
(also 29 30 29 30 29 30 29 ... days, less accurate)
15 and 17 lunations counted in the 30 29 30 mode yield 443 and 502 days
respectively
17 15 17 15 17 or 17 32 49 64 81 lunations
502 443 502 443 502 or 502 945 1447 1890 2392 days
1,890 days correspond to 63 lunations and are 63 continuous periods
of 30 days, allowing to formulate a lunisolar calendar of the Göbekli
Tepe region as of 12,000 years ago
a month has 30 days,
12 months are a basic year of 360 days,
add 5 and occasionally 6 more days
and you get a regular year of 365 days
and an occasional leap year of 366 days,
while 63 continuous periods of 30 days
are 1,890 days anc correspond to 64 lunations
(mistake per lunation less than one minute,
or half a day in a lifetime)
I reconstructed this calendar in late 2004 and immediately recognized
the connection to the Horus eye of Ancient Egypt.
The left eye of the Horus falcon was the moon, and his right eye the sun.
Seth destroyed the lunar eye. Wise Thoth healed it by reassembling the
fragments '2 '4 '8 '16 '32 '64, whereupon he called the restored lunar eye
The Whole One. However, the numbers don't really add up to one, a tiny part
is missing. Why then the whole one? This term does not refer to the whole
moon, the eye of the night sky, but to a whole lunation, the whole lunar
cycle. Multiply the 30 days of an Egyptian month by the series of the
Horus eye '2 '4 '8 '16 '32 '64 and you get 29 '2 '32 days for a lunation,
or 29 days 12 hours 45 minutes, exact value 29 days 12 hours 44 minutes
2.9 seconds (modern average from 1989 AD).
astronomical number 57 (Bent Pyramid)

Angles of 60 and 30 and 15 degrees are easily constructed, while a simple
device allows very close approximations to small angles of one half degree
up to 7.5 plus 7.5 sum 15 degrees: take a rod 57 fingers long and fix a short
one 15 fingers or 30 Ra marks long to one end of the long rod in such a way
that you get a sighting device in form of a T with a long shaft

Loading Image...

Now the astronomical number 57 determines the grid of the Bent Pyramid
with temenos and lower and upper part of the building

grid 579 by 570 royal cubits

subdivision 105 66 228 66 105 royal cubits

Loading Image...

The grid contains multiples of the triples 3-4-5 and 7-24-25 and 11-60-61
while the height measures 99 plus 108 sum 207 royal cubits, lower slope
and edge practically 119 and 136 rc, upper slope and edge practically
157 and 194 rc respectively (close approximations). In all there is
a beautiful mathematical model that speaks for a well intended shape

Loading Image...

The Bent Pyramid may represent the Hyads and the Red Pyramid the red star
Aldebaran, both near the swaying kha channel (band of ecliptic, Rolf
Krauss), while the soul of king Sneferu would have been waiting in the
hypothetical sun chamber of the Red Pyramid (64 rc or 44 m above the center
of the base), waiting to join the solar barque of Ra ...
Franz Gnaedinger
2016-04-14 08:13:53 UTC
Reply
Permalink
Post by Franz Gnaedinger
astronomical number 57 (Bent Pyramid)
Angles of 60 and 30 and 15 degrees are easily constructed, while a simple
device allows very close approximations to small angles of one half degree
up to 7.5 plus 7.5 sum 15 degrees: take a rod 57 fingers long and fix a short
one 15 fingers or 30 Ra marks long to one end of the long rod in such a way
that you get a sighting device in form of a T with a long shaft
http://www.seshat.ch/home/device.GIF
Now the astronomical number 57 determines the grid of the Bent Pyramid
with temenos and lower and upper part of the building
grid 579 by 570 royal cubits
subdivision 105 66 228 66 105 royal cubits
http://www.seshat.ch/home/grid57.GIF
The grid contains multiples of the triples 3-4-5 and 7-24-25 and 11-60-61
while the height measures 99 plus 108 sum 207 royal cubits, lower slope
and edge practically 119 and 136 rc, upper slope and edge practically
157 and 194 rc respectively (close approximations). In all there is
a beautiful mathematical model that speaks for a well intended shape
http://www.seshat.ch/home/bentp.GIF
The Bent Pyramid may represent the Hyads and the Red Pyramid the red star
Aldebaran, both near the swaying kha channel (band of ecliptic, Rolf
Krauss), while the soul of king Sneferu would have been waiting in the
hypothetical sun chamber of the Red Pyramid (64 rc or 44 m above the center
of the base), waiting to join the solar barque of Ra ...
Golden Gate of Mesopotamian astronomy (Maidum Dahshur)

The Pleiads and Aldebaran were the Golden Gate of Mesopotamian astronomy,
framing the band of the ecliptic, seen as kha channel in Ancient Egypt
(Rolf Krauss), so if the Red Pyramid symbolizes the red star Aldebaran
and the Bent Pyramid the Hyads, the Maidum Pyramid could well symbolize
the Pleiads, and the Maidum Pyramid together with the Dahshur Pyramids
the Golden Gate of the kha channel, while the Lisht pyramid could have
marked the earthly correspondence to the heavenly kha channel, maybe
its harbor, where the soul of Sneferu had joined Ra in his solar bark ...

The Maidum Pyramid was planned to have seven steps that would correspond
to the seven stars of the Pleiads visible to the naked eye. Huni died
before it was completed. Sneferu added one more step - symbolically
placing his father as an eighth star among the Pleiads? Unfortunately
the pyramid mantle collapsed, and now the building stands as a strange
structure of great beauty in the desert.

I propose an archaeological survey of a sufficiently wide East-West band
in the desert on the geographical height of Lisht.
Franz Gnaedinger
2016-05-09 06:51:47 UTC
Reply
Permalink
Post by Franz Gnaedinger
Golden Gate of Mesopotamian astronomy (Maidum Dahshur)
The Pleiads and Aldebaran were the Golden Gate of Mesopotamian astronomy,
framing the band of the ecliptic, seen as kha channel in Ancient Egypt
(Rolf Krauss), so if the Red Pyramid symbolizes the red star Aldebaran
and the Bent Pyramid the Hyads, the Maidum Pyramid could well symbolize
the Pleiads, and the Maidum Pyramid together with the Dahshur Pyramids
the Golden Gate of the kha channel, while the Lisht pyramid could have
marked the earthly correspondence to the heavenly kha channel, maybe
its harbor, where the soul of Sneferu had joined Ra in his solar bark ...
The Maidum Pyramid was planned to have seven steps that would correspond
to the seven stars of the Pleiads visible to the naked eye. Huni died
before it was completed. Sneferu added one more step - symbolically
placing his father as an eighth star among the Pleiads? Unfortunately
the pyramid mantle collapsed, and now the building stands as a strange
structure of great beauty in the desert.
I propose an archaeological survey of a sufficiently wide East-West band
in the desert on the geographical height of Lisht.
Keeping my thread alive.

New examinations of the two hypothetical even probable chambers behind
Tut-enkh-Amun's tomb had been announced for the end of April. What has
become of that? and of the scanning of the Great Pyramid?

Another archaeological story. Traces of 7,000 years old Swiss cheese
had been identified via finger prints on lumps of animal fat, not
necessarily the one with big holes ;-) For the time being a speculation,
but would make sense. Many people had been lactose intolerant in those
times, whereas fermented milk, yoghurt and cheese, contain little amounts
of lactose, and milk is well preserved in the form of cheese.
Franz Gnaedinger
2016-05-26 06:57:02 UTC
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Post by Franz Gnaedinger
Keeping my thread alive.
New examinations of the two hypothetical even probable chambers behind
Tut-enkh-Amun's tomb had been announced for the end of April. What has
become of that? and of the scanning of the Great Pyramid?
Another archaeological story. Traces of 7,000 years old Swiss cheese
had been identified via finger prints on lumps of animal fat, not
necessarily the one with big holes ;-) For the time being a speculation,
but would make sense. Many people had been lactose intolerant in those
times, whereas fermented milk, yoghurt and cheese, contain little amounts
of lactose, and milk is well preserved in the form of cheese.
Chephren's pyramid (and a surprise)

Chephren's pyramid is defined by a Sacred Triangle 3-4-5

base 411 royal cubits, height 274 rc, height : base = 2 : 3

3 times 68.5 rc - half base
4 times 68.5 rc - height
5 times 68.5 rc - slope

A pyramid of the same type is mentioned in the Rhind Mathematical Papyrus:
base 12 royal cubits, height 8 royal cubits. While the slope measures
exactly 10 royal cubits, according to the Sacred Triangle 3-4-5. So far
so good. But there is a surprise waiting for you if you play with the
numbers and shapes of this model pyramid.

Picture a wooden model of the base 12 fingers, height 8 fingers, and slope
10 fingers. What is the volume? 384 cubic fingers. And the surface of the
model including the square of the base? 384 square fingers.

Picture a sphere inscribed in the model. How long is the radius?
Exactly 3 fingers, diameter 6 fingers.

If there is a secret chamber hidden in Chephern's pyramid it might be
in the center of the imaginary inscribed sphere evoking the sun inside
the primeval hill, 102.75 royal cubits above the center of the base.

A famous problem in the Rhind Mathematical Papyrus equates the area of
a square of side 8 with the one of a circle of diameter 9, implicit value
of pi 256/81, a value found via one of the many pi sequences

4/1 (plus 3/1) 7/2 10/3 13/4 16/5 19/6 22/7 25/8 28/9

9/3 (plus 19/6) 28/9 47/15 66/21 ... 256/81

Now there is a surprise in form of a three-dimensional equivalent of
the famous formula: a model of the Chephren pyramid of height 8 and base
12 has practically the same volume as a sphere of diameter 9, implicit
pi value 256/81.

Picture a pair of stone models, one of the pyramid and the other of the
sphere: they have practically the same weight, hence the same volume.
Franz Gnaedinger
2016-06-08 06:52:45 UTC
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Post by Franz Gnaedinger
Chephren's pyramid (and a surprise)
Chephren's pyramid is defined by a Sacred Triangle 3-4-5
base 411 royal cubits, height 274 rc, height : base = 2 : 3
3 times 68.5 rc - half base
4 times 68.5 rc - height
5 times 68.5 rc - slope
base 12 royal cubits, height 8 royal cubits. While the slope measures
exactly 10 royal cubits, according to the Sacred Triangle 3-4-5. So far
so good. But there is a surprise waiting for you if you play with the
numbers and shapes of this model pyramid.
Picture a wooden model of the base 12 fingers, height 8 fingers, and slope
10 fingers. What is the volume? 384 cubic fingers. And the surface of the
model including the square of the base? 384 square fingers.
Picture a sphere inscribed in the model. How long is the radius?
Exactly 3 fingers, diameter 6 fingers.
If there is a secret chamber hidden in Chephern's pyramid it might be
in the center of the imaginary inscribed sphere evoking the sun inside
the primeval hill, 102.75 royal cubits above the center of the base.
A famous problem in the Rhind Mathematical Papyrus equates the area of
a square of side 8 with the one of a circle of diameter 9, implicit value
of pi 256/81, a value found via one of the many pi sequences
4/1 (plus 3/1) 7/2 10/3 13/4 16/5 19/6 22/7 25/8 28/9
9/3 (plus 19/6) 28/9 47/15 66/21 ... 256/81
Now there is a surprise in form of a three-dimensional equivalent of
the famous formula: a model of the Chephren pyramid of height 8 and base
12 has practically the same volume as a sphere of diameter 9, implicit
pi value 256/81.
Picture a pair of stone models, one of the pyramid and the other of the
sphere: they have practically the same weight, hence the same volume.
No more information about the ongoing campaigns in Egypt, and still
no interest in new ideas and approaches of the non-invasive kind
(numbers and hermeneutics combined).
Franz Gnaedinger
2016-06-28 07:52:28 UTC
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Post by Franz Gnaedinger
Post by Franz Gnaedinger
Chephren's pyramid (and a surprise)
Chephren's pyramid is defined by a Sacred Triangle 3-4-5
base 411 royal cubits, height 274 rc, height : base = 2 : 3
3 times 68.5 rc - half base
4 times 68.5 rc - height
5 times 68.5 rc - slope
base 12 royal cubits, height 8 royal cubits. While the slope measures
exactly 10 royal cubits, according to the Sacred Triangle 3-4-5. So far
so good. But there is a surprise waiting for you if you play with the
numbers and shapes of this model pyramid.
Picture a wooden model of the base 12 fingers, height 8 fingers, and slope
10 fingers. What is the volume? 384 cubic fingers. And the surface of the
model including the square of the base? 384 square fingers.
Picture a sphere inscribed in the model. How long is the radius?
Exactly 3 fingers, diameter 6 fingers.
If there is a secret chamber hidden in Chephern's pyramid it might be
in the center of the imaginary inscribed sphere evoking the sun inside
the primeval hill, 102.75 royal cubits above the center of the base.
A famous problem in the Rhind Mathematical Papyrus equates the area of
a square of side 8 with the one of a circle of diameter 9, implicit value
of pi 256/81, a value found via one of the many pi sequences
4/1 (plus 3/1) 7/2 10/3 13/4 16/5 19/6 22/7 25/8 28/9
9/3 (plus 19/6) 28/9 47/15 66/21 ... 256/81
Now there is a surprise in form of a three-dimensional equivalent of
the famous formula: a model of the Chephren pyramid of height 8 and base
12 has practically the same volume as a sphere of diameter 9, implicit
pi value 256/81.
Picture a pair of stone models, one of the pyramid and the other of the
sphere: they have practically the same weight, hence the same volume.
No more information about the ongoing campaigns in Egypt, and still
no interest in new ideas and approaches of the non-invasive kind
(numbers and hermeneutics combined).
RMP 33 (mathematical wit)

Imagine a cube measuring 41 by 41 by 41 fingerbreadths or simply fingers.
How long is the diagonal of a face? and the diagonal of the volume?

Let us draw up the number columns approximating the square roots of 2 and 3

1 1 2
2 3 4
5 7 10
12 17 24
29 41 58 *

If the side of a square measures 41 fingers, the diagonal measures
practically 58 fingers.

1 1 3
2 4 6
1 2 3
3 5 9
8 14 24
4 7 12
11 19 33
30 52 90
15 26 45
41 71 *

If the edge of a cube measures 41 fingers, the diagonal of the volume
measures practically 71 fingers.

In problem 33 of the Rhind Mathematical Papyrus, Ahmes divides 37 by
1 ''3 '2 '7 and obtains 16 '56 '679 '776.

This calculation turns the above cube into a wooden chest measuring
41 by 41 by 41 fingers on the outside, while the boards are 2 fingers thick,
and the inside is a hollow cube measuring 37 by 37 by 37 fingers (2 37 2
sum 41). How long is the the diagonal of the hollow volume in palms?
and what is the capacity of the wooden chest in cubic cubits?

Let us prolong the second number column that approximates the square root
of 3

15 26 45
41 71 123
112 194 336
56 97 168 *

The edge of the hollow inner cube measures 37 fingers, and the diagonal
of the hollow volume 37 times 168/97 fingers or 37 times 42/97 palms
or 16 '56 '679 '776 palms, number obtained by Ahmes. The capacity equals
37/28 times 37/28 times 37/28 or practically 1 ''3 '2 '7 cubic cubits,
divisor used by Ahmes - a numerical joke for the mathematically inclined,
rewarding the pupils of Ahmes and confirming a modern interpreter of the
RMP.
Franz Gnaedinger
2016-08-09 07:25:03 UTC
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Post by Franz Gnaedinger
RMP 33 (mathematical wit)
Imagine a cube measuring 41 by 41 by 41 fingerbreadths or simply fingers.
How long is the diagonal of a face? and the diagonal of the volume?
Let us draw up the number columns approximating the square roots of 2 and 3
1 1 2
2 3 4
5 7 10
12 17 24
29 41 58 *
If the side of a square measures 41 fingers, the diagonal measures
practically 58 fingers.
1 1 3
2 4 6
1 2 3
3 5 9
8 14 24
4 7 12
11 19 33
30 52 90
15 26 45
41 71 *
If the edge of a cube measures 41 fingers, the diagonal of the volume
measures practically 71 fingers.
In problem 33 of the Rhind Mathematical Papyrus, Ahmes divides 37 by
1 ''3 '2 '7 and obtains 16 '56 '679 '776.
This calculation turns the above cube into a wooden chest measuring
41 by 41 by 41 fingers on the outside, while the boards are 2 fingers thick,
and the inside is a hollow cube measuring 37 by 37 by 37 fingers (2 37 2
sum 41). How long is the the diagonal of the hollow volume in palms?
and what is the capacity of the wooden chest in cubic cubits?
Let us prolong the second number column that approximates the square root
of 3
15 26 45
41 71 123
112 194 336
56 97 168 *
The edge of the hollow inner cube measures 37 fingers, and the diagonal
of the hollow volume 37 times 168/97 fingers or 37 times 42/97 palms
or 16 '56 '679 '776 palms, number obtained by Ahmes. The capacity equals
37/28 times 37/28 times 37/28 or practically 1 ''3 '2 '7 cubic cubits,
divisor used by Ahmes - a numerical joke for the mathematically inclined,
rewarding the pupils of Ahmes and confirming a modern interpreter of the
RMP.
RMP 31 (granary on a ring)

A fair history of civilization is more important than ever. I go on
revealing marvels of Egyptian mathematics and geometry. This time
a stunning theorem on the area of a ring in relation to the side
of the inscribed complete or incomplete polygon.

Ahmes carries out a division:

33 divided by 1 "3 '2 '7 equals 14 '4 '56 '194 '388 '679 '776

Beginners learn how to work with unit fraction series, while advanced
learners are given a more demanding task.

Imagine a regular hexagon whose side measures 66 fingers. Inscribe and
circumscribe a pair of circles. The two circles form a ring. The radius
of the outer circle measures 66 fingers. How long is the radius of the
inscribed circle in palms?

Consult the number column for calculating the diagonal of the cube,
the equilateral triangle and regular hexagon:

1 1 3
2 4 6
1 2 3
3 5 9
3 5 9
8 14 24
4 7 12
11 19 33
30 52 90
15 26 45
41 71 123
112 194 336
56 97 168

The side of the hexagon measures 66 fingers. Multiply 66 fingers by 168/97
and you obtain the diameter of the inscribed circle in fingers. Multiply
33 fingers by 168/97 and you obtain the radius in fingers. Divide 33 fingers
by 97/42 = 1 "3 '2 '7 and you obtain the radius of the inscribed circle in
palms:

33 f divided by 1 "3 '2 '7 equal 14 '4 '56 '97 '194 '388 '679 '776 palms

The area of the ring is given by the difference

area circumscribed circle minus area inscribed circle

The area of the circumscribed circle is found as follows:

radius x radius x pi

4/1 (plus 3/1) 7/2 10/3 13/4 16/5 19/6 22/7

3/1 (plus 22/7) 25/8 47/15 69/22 ... 311/99

Choose the value 311/99 for pi

66 fingers x 66 fingers x '99 x 311 = 13,684 square fingers

Now for the area of the inscribed circle. It measures

14 '4 '56 '194 '388 '679 '776 palms
times
14 '4 '56 '194 '388 '679 '776 palms
times
pi

Is anyone prepared to carry out that multiplication ???

Ahmes would smile and offer a much simpler solution based on a fine theorem:

Imagine a regular polygon of 3, 4, 5, 6, 7 ... equal sides. The circumscribed
circle and the inscribed circle form a ring. Draw a circle around a side of
the polygon. Its area equals the area of the ring.

The side of the regular hexagon measures 66 fingers, the radius of the circle
around a side measures 33 fingers, and the area of the ring measures

33 fingers x 33 fingers x '99 x 311 = 3,421 square fingers

The area of the outer circle measures 13,684 square fingers, the area of the
ring measures 3,421 square fingers, and the area of the inner circle measures
13,684 - 3,421 = 10,263 square fingers. Comparing these areas reveals the
following proportions:

area inner circle / area ring / area outer circle = 3 / 1 / 4

Build a granary on the ring. If the height measures 5 '2 royal cubits or 154
fingers, the volume of the wall measures practically 24 cubic cubits, and
the capacity 72 cubic cubits or 108 khar or 540 quadruple-hekat or 2160 hekat.

Theorem: The area of a ring equals the area of a circle around the side
of the inscribed polygon, whether it is a complete polygon (equilateral
triangle, square, pentagon, hexagon, and so on) or an incomplete polygon.
You can transform the area of any ring into a circle of the same area.
The diameter of that circle is given by a secant of the outer ring-circle
that is at the same time a tangent of the inner ring-circle.

(It will help if you make drawings.)
Franz Gnaedinger
2016-09-19 06:41:33 UTC
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Post by Franz Gnaedinger
RMP 31 (granary on a ring)
A fair history of civilization is more important than ever. I go on
revealing marvels of Egyptian mathematics and geometry. This time
a stunning theorem on the area of a ring in relation to the side
of the inscribed complete or incomplete polygon.
33 divided by 1 "3 '2 '7 equals 14 '4 '56 '194 '388 '679 '776
Beginners learn how to work with unit fraction series, while advanced
learners are given a more demanding task.
Imagine a regular hexagon whose side measures 66 fingers. Inscribe and
circumscribe a pair of circles. The two circles form a ring. The radius
of the outer circle measures 66 fingers. How long is the radius of the
inscribed circle in palms?
Consult the number column for calculating the diagonal of the cube,
1 1 3
2 4 6
1 2 3
3 5 9
3 5 9
8 14 24
4 7 12
11 19 33
30 52 90
15 26 45
41 71 123
112 194 336
56 97 168
The side of the hexagon measures 66 fingers. Multiply 66 fingers by 168/97
and you obtain the diameter of the inscribed circle in fingers. Multiply
33 fingers by 168/97 and you obtain the radius in fingers. Divide 33 fingers
by 97/42 = 1 "3 '2 '7 and you obtain the radius of the inscribed circle in
33 f divided by 1 "3 '2 '7 equal 14 '4 '56 '97 '194 '388 '679 '776 palms
The area of the ring is given by the difference
area circumscribed circle minus area inscribed circle
radius x radius x pi
4/1 (plus 3/1) 7/2 10/3 13/4 16/5 19/6 22/7
3/1 (plus 22/7) 25/8 47/15 69/22 ... 311/99
Choose the value 311/99 for pi
66 fingers x 66 fingers x '99 x 311 = 13,684 square fingers
Now for the area of the inscribed circle. It measures
14 '4 '56 '194 '388 '679 '776 palms
times
14 '4 '56 '194 '388 '679 '776 palms
times
pi
Is anyone prepared to carry out that multiplication ???
Imagine a regular polygon of 3, 4, 5, 6, 7 ... equal sides. The circumscribed
circle and the inscribed circle form a ring. Draw a circle around a side of
the polygon. Its area equals the area of the ring.
The side of the regular hexagon measures 66 fingers, the radius of the circle
around a side measures 33 fingers, and the area of the ring measures
33 fingers x 33 fingers x '99 x 311 = 3,421 square fingers
The area of the outer circle measures 13,684 square fingers, the area of the
ring measures 3,421 square fingers, and the area of the inner circle measures
13,684 - 3,421 = 10,263 square fingers. Comparing these areas reveals the
area inner circle / area ring / area outer circle = 3 / 1 / 4
Build a granary on the ring. If the height measures 5 '2 royal cubits or 154
fingers, the volume of the wall measures practically 24 cubic cubits, and
the capacity 72 cubic cubits or 108 khar or 540 quadruple-hekat or 2160 hekat.
Theorem: The area of a ring equals the area of a circle around the side
of the inscribed polygon, whether it is a complete polygon (equilateral
triangle, square, pentagon, hexagon, and so on) or an incomplete polygon.
You can transform the area of any ring into a circle of the same area.
The diameter of that circle is given by a secant of the outer ring-circle
that is at the same time a tangent of the inner ring-circle.
(It will help if you make drawings.)
Reading 'The Great Mathematical Problems' by Ian Stewart (original 2013
translation 2015) makes me wonder once again that mathematicians and
brillant scholars understand the most complex and complicated modern
theories while being blind for the marvels of early and very early
mathemaitcs.
Franz Gnaedinger
2016-09-22 07:43:36 UTC
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Post by Franz Gnaedinger
Reading 'The Great Mathematical Problems' by Ian Stewart (original 2013
translation 2015) makes me wonder once again that mathematicians and
brillant scholars understand the most complex and complicated modern
theories while being blind for the marvels of early and very early
mathemaitcs.
Prediction regarding the recently discovered Etruscan stele from Poggio Buco,
dedicated to the supreme god Tinia, Etruscan Zeus, and his wive Uni, goddess
of fertility: the long inscription might refer to a ritual union of the god
and goddess, perhaps enacted by a priest and a priestess, a symbolic union
making and keeping the land fertile ...

My prediction relies on a linguistic experiment of meanwhile nearly a dozen
years, and on calendar reconstructions. I can't render the linguistics
involved here, but give an idea of a hypothetical Etruscan religious
calendar.

It would have been a calendar of ten periods alternatingly lasting 36
and 37 days, the periods of 36 days being consecrated to emanations
of Uni, and the periods of 37 days to emanations of Tinia, first half
year beinning on the winter solstice, December 21 in our modern calendar,
in all 182 days, and the second half year beginning on the summer solstice,
June 21, in all 183 days, together 365 days of a regular year

36 37 36 37 36 37 36 37 36 37 sum 365 days

Picture a circle of the radius corresponding to four lunations or synodic
months (modern average of one lunation 29 days 12 hours 44 minutes 2.9
seconds) and a decagon inscribed in the circle (a regular polygon of ten
sides) - the side of that polygon will correspond to 73 days, combined
period of the female 36 and male 37 days. A temple on the basis of such
a decagon might have celebrated the cosmic union of Uni in her lunar aspect
and of Tinia in his solar aspect.

I hope that we shall soon get an online transliteration and translation
of the long inscription.
Franz Gnaedinger
2016-10-23 07:48:50 UTC
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Post by Franz Gnaedinger
Prediction regarding the recently discovered Etruscan stele from Poggio Buco,
dedicated to the supreme god Tinia, Etruscan Zeus, and his wive Uni, goddess
of fertility: the long inscription might refer to a ritual union of the god
and goddess, perhaps enacted by a priest and a priestess, a symbolic union
making and keeping the land fertile ...
My prediction relies on a linguistic experiment of meanwhile nearly a dozen
years, and on calendar reconstructions. I can't render the linguistics
involved here, but give an idea of a hypothetical Etruscan religious
calendar.
It would have been a calendar of ten periods alternatingly lasting 36
and 37 days, the periods of 36 days being consecrated to emanations
of Uni, and the periods of 37 days to emanations of Tinia, first half
year beinning on the winter solstice, December 21 in our modern calendar,
in all 182 days, and the second half year beginning on the summer solstice,
June 21, in all 183 days, together 365 days of a regular year
36 37 36 37 36 37 36 37 36 37 sum 365 days
Picture a circle of the radius corresponding to four lunations or synodic
months (modern average of one lunation 29 days 12 hours 44 minutes 2.9
seconds) and a decagon inscribed in the circle (a regular polygon of ten
sides) - the side of that polygon will correspond to 73 days, combined
period of the female 36 and male 37 days. A temple on the basis of such
a decagon might have celebrated the cosmic union of Uni in her lunar aspect
and of Tinia in his solar aspect.
I hope that we shall soon get an online transliteration and translation
of the long inscription.
Lunisolar calendars encoded in Terrazzo Buildings 1 and 2 at Nevali Cori
(Chori) in the region and from the era of the Göbekli Tepe.

Both temples form rectangles, along the peripheries 13 and 12 stone pillars
of the Göbekli Tepe type respectively, in the center of each temple a pair
of larger pillars, again in the Göbekli Tepe style. The second building
replaced the first one. They encode the following calendars:

Terrazzo Building 1

13 pillars along the periphery for 13 months of 28 days,
space between pair of larger central pillars for 1 or 2 more days,
yielding a regular year of 365 days and occasional leap year of 366 days
while 135 continuous periods of 28 days are 3,750 days
and correspond to 128 lunations or synodic months

Terrazzo Building 2 replacing Building 1

12 pillars along the periphery for 12 months of 30 days,
space between pair of larger central pillars for 1 or 2 more days,
yielding a regular year of 365 days and occasional leap year of 366 days
while 63 continuous periods of 30 days are 1,890 days
and correspond to 64 lunations or synodic months

Regular year of Buidling 1

Dec 22 - Jan 18 New Year following winter solstice
Jan 19 - Feb 15
Feb 16 - Mar 15
Mar 16 - Apr 12
Apr 13 - May 10
May 11 - Jun 7
Jun 8 - Jul 5 summer solstice in the middle
Jul 6 - Aug 2
Aug 3 - Aug 30
Aug 31 - Sep 27
Sep 28 - Oct 25
Oct 26 - Nov 22
Nov 23 - Dec 20
December 21 winter solstice, New Year's Eve


Regular Year of Building 2

Dec 22 - Jan 20 New Year following winter solstice
Jan 21 - Feb 19
Feb 20 - Mar 21 month ending on spring equinox (Mar 21)
Mar 22 - Apr 20
Apr 21 - May 20
May 21 - Jun 19
June 20 21 22 summer festival, solstice in the middle
Jun 23 - Jul 22
Jul 23 - Aug 21
Aug 22 - Sep 20
Sep 21 - Oct 20 fall equinox near beginning (Sep 23)
Oct 21 - Nov 19
Nov 20 - Dec 19
December 20 21 winter festival, solstice at the end, New Year's Eve

Here is how to calculate the lunisolar aspects in the Stone Age way.
One lunation or synodic month lasts 29 days 12 hours 44 minutes 2.9 seconds
(modern average from 1989 AD). Count lunations as follows

30 29 30 29 30 29 30 ... days for 1 2 3 4 5 6 7 ... luntions

15 and 17 lunations counted that way are 443 and 502 days respectively.

17 15 17 15 17 or 17 32 49 64 81 lunations
502 443 502 443 502 or 502 945 1447 1890 2392 days

63 continuous periods of 30 days are 1,890 days and correspond to 64
lunations; mistake less than one minute per lunation, or half a day
in a lifetime. 135 continuous periods of 28 days are 3,750 days and
correspond to 128 luntaions (same numerical definition, same small
mistake).

Calendar 2 was taken over in the Egyptian Horus eye calendar.
The eyes of the Horus falcon were sun and moon (right eye the sun,
left eye the moon). Seth destroyed the Horus eye. Wise Thoth healed
it, assembling the parts '2 '4 '8 '16 '32 '64, and called the restored
lunar eye The Whole One. However, the parts don't really add up to one.
Why then 'the whole one' ? This refers to one whole lunation. A whole
month had 30 days, a whole lunation 30 days times the Horus eye series
'2 '4 '8 '16 '32 '64 yielding 29 '2 '32 days, or 29 days 12 hours
45 minutes.

A further variation is the rosette of eight petals found on beautiful
Kamares ware from Middle Minoan Crete, and in the center of the Tiryns
side of the Phaistos Disc: each petal for a long month of 45 days,
in all a basic year of 360 days, while the small circle indicates
five and occasionally six more days, in all a regular year of 365 days
and an occasional leap year of 366 days, while 21 continuous periods
of 45 days are 945 days and correspond to 32 lunations (a relation
found again via the above number pattern).
Franz Gnaedinger
2016-11-03 09:07:15 UTC
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Post by Franz Gnaedinger
Lunisolar calendars encoded in Terrazzo Buildings 1 and 2 at Nevali Cori
(Chori) in the region and from the era of the Göbekli Tepe.
Both temples form rectangles, along the peripheries 13 and 12 stone pillars
of the Göbekli Tepe type respectively, in the center of each temple a pair
of larger pillars, again in the Göbekli Tepe style. The second building
Terrazzo Building 1
13 pillars along the periphery for 13 months of 28 days,
space between pair of larger central pillars for 1 or 2 more days,
yielding a regular year of 365 days and occasional leap year of 366 days
while 135 continuous periods of 28 days are 3,750 days
and correspond to 128 lunations or synodic months
Terrazzo Building 2 replacing Building 1
12 pillars along the periphery for 12 months of 30 days,
space between pair of larger central pillars for 1 or 2 more days,
yielding a regular year of 365 days and occasional leap year of 366 days
while 63 continuous periods of 30 days are 1,890 days
and correspond to 64 lunations or synodic months
Regular year of Buidling 1
Dec 22 - Jan 18 New Year following winter solstice
Jan 19 - Feb 15
Feb 16 - Mar 15
Mar 16 - Apr 12
Apr 13 - May 10
May 11 - Jun 7
Jun 8 - Jul 5 summer solstice in the middle
Jul 6 - Aug 2
Aug 3 - Aug 30
Aug 31 - Sep 27
Sep 28 - Oct 25
Oct 26 - Nov 22
Nov 23 - Dec 20
December 21 winter solstice, New Year's Eve
Regular Year of Building 2
Dec 22 - Jan 20 New Year following winter solstice
Jan 21 - Feb 19
Feb 20 - Mar 21 month ending on spring equinox (Mar 21)
Mar 22 - Apr 20
Apr 21 - May 20
May 21 - Jun 19
June 20 21 22 summer festival, solstice in the middle
Jun 23 - Jul 22
Jul 23 - Aug 21
Aug 22 - Sep 20
Sep 21 - Oct 20 fall equinox near beginning (Sep 23)
Oct 21 - Nov 19
Nov 20 - Dec 19
December 20 21 winter festival, solstice at the end, New Year's Eve
Here is how to calculate the lunisolar aspects in the Stone Age way.
One lunation or synodic month lasts 29 days 12 hours 44 minutes 2.9 seconds
(modern average from 1989 AD). Count lunations as follows
30 29 30 29 30 29 30 ... days for 1 2 3 4 5 6 7 ... luntions
15 and 17 lunations counted that way are 443 and 502 days respectively.
17 15 17 15 17 or 17 32 49 64 81 lunations
502 443 502 443 502 or 502 945 1447 1890 2392 days
63 continuous periods of 30 days are 1,890 days and correspond to 64
lunations; mistake less than one minute per lunation, or half a day
in a lifetime. 135 continuous periods of 28 days are 3,750 days and
correspond to 128 luntaions (same numerical definition, same small
mistake).
Calendar 2 was taken over in the Egyptian Horus eye calendar.
The eyes of the Horus falcon were sun and moon (right eye the sun,
left eye the moon). Seth destroyed the Horus eye. Wise Thoth healed
it, assembling the parts '2 '4 '8 '16 '32 '64, and called the restored
lunar eye The Whole One. However, the parts don't really add up to one.
Why then 'the whole one' ? This refers to one whole lunation. A whole
month had 30 days, a whole lunation 30 days times the Horus eye series
'2 '4 '8 '16 '32 '64 yielding 29 '2 '32 days, or 29 days 12 hours
45 minutes.
A further variation is the rosette of eight petals found on beautiful
Kamares ware from Middle Minoan Crete, and in the center of the Tiryns
side of the Phaistos Disc: each petal for a long month of 45 days,
in all a basic year of 360 days, while the small circle indicates
five and occasionally six more days, in all a regular year of 365 days
and an occasional leap year of 366 days, while 21 continuous periods
of 45 days are 945 days and correspond to 32 lunations (a relation
found again via the above number pattern).
Stonehenge (former ring of 56 bluestones)

Mike Parker Pearson believes that a straight part of the parallel lines
of the Stonehenge causeway were a natural phenomenon caused by melting
waters of the last Ice Age. Those lines would have baffled people: in one
direction indicating the setting midwinter sun, in the other direction
the rising midsummer sun ...

The original monument would have dated to around 5 000 BP and would have
comprised the circular ditch, the wall inside of the ditch, and a ring of
56 bluestones along the wall. Each bluestone marked a burial pit wherein
the ashes of worthy members of an aristocratic elite had been embedded.

Still according to Mike Parker Pearson, the iconic moument we see today
was built later on, the bluestones removed from the ring and placed within
the new monument.

Now let us have a look at the lunisolar aspect of the number 56 that may
have invoked the seeming regenerative powers of moon and sun.

Count lunations or synodic months as follows

30 29 30 29 30 29 30 ... days for 1 2 3 4 5 6 7 ... lunations

15 and 17 lunations counted in the 30 29 30 mode yield 443 and 502 days
respectively

15 17 sum 32 lunations or 443 502 sum 945 days

945 days for 32 lunations
doubled 1,890 days for 64 lunations (Göbekli Tepe formula)
doubled 3,780 days for 128 lunations
doubled 7,560 days for 256 lunations

7,560 days are 56 times 135 days. Count 135 days for each bluestone in
the original ring of bluestones along the wall and you obtain the equivalent
of 256 lunations or synodic months; mistake less than one minute per
lunation, or half a day in a lifetime (same mistake as in the Göbekli Tepe
formula of 63 continuous periods of 30 days being 1,890 days, equivalent of
64 lunations).

Moon and sun set on the western horizon and rise from the eastern horizon.
The moon wanes and waxes again. The sun follows lower and lower trajectories
toward midwinter, then higher and higher trajectories toward midsummer.
Apparently they have regenerative powers. And those powers might have
been invoked by the lunisolar aspect of the number 56 of the original ring
of bluestones - may the worthy dead be reborn in the beyond ...

Interesting are the numbers 56 and 135 and 256 for they are composed of
small factors

56 = 2 x 2 x 2 x 7

135 = 3 x 3 x 3 x 5

256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

Small numbers working together form big numbers of a cosmic dimension.
In a similar way we human beings, tiny under the huge heavenly vault,
can achieve big things when we co-operate.
Franz Gnaedinger
2016-12-09 09:50:00 UTC
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Post by Franz Gnaedinger
Stonehenge (former ring of 56 bluestones)
Mike Parker Pearson believes that a straight part of the parallel lines
of the Stonehenge causeway were a natural phenomenon caused by melting
waters of the last Ice Age. Those lines would have baffled people: in one
direction indicating the setting midwinter sun, in the other direction
the rising midsummer sun ...
The original monument would have dated to around 5 000 BP and would have
comprised the circular ditch, the wall inside of the ditch, and a ring of
56 bluestones along the wall. Each bluestone marked a burial pit wherein
the ashes of worthy members of an aristocratic elite had been embedded.
Still according to Mike Parker Pearson, the iconic moument we see today
was built later on, the bluestones removed from the ring and placed within
the new monument.
Now let us have a look at the lunisolar aspect of the number 56 that may
have invoked the seeming regenerative powers of moon and sun.
Count lunations or synodic months as follows
30 29 30 29 30 29 30 ... days for 1 2 3 4 5 6 7 ... lunations
15 and 17 lunations counted in the 30 29 30 mode yield 443 and 502 days
respectively
15 17 sum 32 lunations or 443 502 sum 945 days
945 days for 32 lunations
doubled 1,890 days for 64 lunations (Göbekli Tepe formula)
doubled 3,780 days for 128 lunations
doubled 7,560 days for 256 lunations
7,560 days are 56 times 135 days. Count 135 days for each bluestone in
the original ring of bluestones along the wall and you obtain the equivalent
of 256 lunations or synodic months; mistake less than one minute per
lunation, or half a day in a lifetime (same mistake as in the Göbekli Tepe
formula of 63 continuous periods of 30 days being 1,890 days, equivalent of
64 lunations).
Moon and sun set on the western horizon and rise from the eastern horizon.
The moon wanes and waxes again. The sun follows lower and lower trajectories
toward midwinter, then higher and higher trajectories toward midsummer.
Apparently they have regenerative powers. And those powers might have
been invoked by the lunisolar aspect of the number 56 of the original ring
of bluestones - may the worthy dead be reborn in the beyond ...
Interesting are the numbers 56 and 135 and 256 for they are composed of
small factors
56 = 2 x 2 x 2 x 7
135 = 3 x 3 x 3 x 5
256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
Small numbers working together form big numbers of a cosmic dimension.
In a similar way we human beings, tiny under the huge heavenly vault,
can achieve big things when we co-operate.
Here is a message I posted to another group and find also relevant for
sci.archaeology (although the single message out of a wider context may
rise more questions thant it can answer - feel free asking me if you are
interested in a specific topic).

humanity's diary (another way of reading the Bible)

Carel van Schaik, evolutionary anthropologist, and Kai Michel, historian,
wrote a book on the Bible as 'humanity's diary'. Agriculture changed human
life dramatically, and the Bible tells in many stories how people tried to
cope with numerous challenges posed by their new life. I warmly welcome
the book, and can confirm that the beginning of the Bible goes back to
the invention of agriculture in the region of the Göbekli Tepe, between
12,000 and 10,000 years ago. The authors take the Bible seriously, neither
literally nor dismissing it in the way of Richard Dawkins.

However, civilization did not start with the Göbekli Tepe. Who can imagine
that the marvellous paintings in the Lascaux cave had been carried out by
a bunch of local artists? No, the very best painters came from a wide region,
probably from various parts of the Franco-Cantabrian space that was a refuge
in the last Ice Age (as revealed by mitochondrial DNA) and would have been
the home of a proto-society ruled by wandering arch shamans and arch
shamanesses represented by megaceroi (giant stag and hind respectively).

The lying red ocher birdman in the pit of Lascaux can be seen as river map
of the Guyenne

Loading Image...

The five swimming stags crossing a river in a freeze of the nave are
visiting shamans, while the roaring megaceros in the axial gallery is
an astronomical arch shaman explaining one of the two Lascaux calendars.

The composite animal seen from the ancient entrance as if through the left
eye socket shows the face of a bearded man, a pair of lances growing as horns
out of his front, the mottled hide of a feline, the strong body of a bull,
and the swollen belly of a pregnant mare

Loading Image...

This can be read as message to aspiring tribal leaders who might have been
initiated in that cave: make a wise use of your weapons; be patient and then
act in a decided manner as a lion; be strong as a bull; and care for your
own like a mare for her foal ...

The glorious rotunda symbolizes midsummer and life, the niche at the rear
end of the axial gallery midwinter, the axial gallery connecting them a year,
and the longer way from the rotunda to the pit the career of a supreme tribal
leader who, if worthy, may be reborn in the region of the Summer Triangle
Deneb - Vega - Atair ...

The Göbekli Tepe culture of the Neolithic did not emerge out of nothing,
and religion played an important role for the civilizatory process already
in the Paleolithic.
Franz Gnaedinger
2017-01-12 07:48:03 UTC
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Permalink
Post by Franz Gnaedinger
humanity's diary (another way of reading the Bible)
Carel van Schaik, evolutionary anthropologist, and Kai Michel, historian,
wrote a book on the Bible as 'humanity's diary'. Agriculture changed human
life dramatically, and the Bible tells in many stories how people tried to
cope with numerous challenges posed by their new life. I warmly welcome
the book, and can confirm that the beginning of the Bible goes back to
the invention of agriculture in the region of the Göbekli Tepe, between
12,000 and 10,000 years ago. The authors take the Bible seriously, neither
literally nor dismissing it in the way of Richard Dawkins.
However, civilization did not start with the Göbekli Tepe. Who can imagine
that the marvellous paintings in the Lascaux cave had been carried out by
a bunch of local artists? No, the very best painters came from a wide region,
probably from various parts of the Franco-Cantabrian space that was a refuge
in the last Ice Age (as revealed by mitochondrial DNA) and would have been
the home of a proto-society ruled by wandering arch shamans and arch
shamanesses represented by megaceroi (giant stag and hind respectively).
The lying red ocher birdman in the pit of Lascaux can be seen as river map
of the Guyenne
http://www.seshat.ch/home/menhir6i.GIF
The five swimming stags crossing a river in a freeze of the nave are
visiting shamans, while the roaring megaceros in the axial gallery is
an astronomical arch shaman explaining one of the two Lascaux calendars.
The composite animal seen from the ancient entrance as if through the left
eye socket shows the face of a bearded man, a pair of lances growing as horns
out of his front, the mottled hide of a feline, the strong body of a bull,
and the swollen belly of a pregnant mare
http://www.seshat.ch/home/menhir6f.JPG
This can be read as message to aspiring tribal leaders who might have been
initiated in that cave: make a wise use of your weapons; be patient and then
act in a decided manner as a lion; be strong as a bull; and care for your
own like a mare for her foal ...
The glorious rotunda symbolizes midsummer and life, the niche at the rear
end of the axial gallery midwinter, the axial gallery connecting them a year,
and the longer way from the rotunda to the pit the career of a supreme tribal
leader who, if worthy, may be reborn in the region of the Summer Triangle
Deneb - Vega - Atair ...
The Göbekli Tepe culture of the Neolithic did not emerge out of nothing,
and religion played an important role for the civilizatory process already
in the Paleolithic.
My compatriot Erich von Daeniken can publish anything anytime. He wrote
a book on the Great Pyramid, which, he claims, was built by an alien
who still sleeps in a hidden chamber ... The Egyptians, obviously, were
not able to build that monument on their own. I can show that they were,
using simple yet clever methods, but I have no chance to publish my work.

Near the end of last year, EdV wrote in a newspaper column that woodhenges
were built as messages for aliens, whereas calendars are their most boring
explanation. I never interpreted a woodhenge, given the uncertainty of
numbers, but for example the original ring of 56 bluestones of Stonehenge
(in a previous message) as a lunisolar calendar, on top of the calendar
interpretation of the younger iconic monument

http://www.seshat.ch/home/stonehen.htm

with a parallel to the bronze needle calendar from Falera in the Swiss Alps.

Calendars, in my opinion, mark the beginning of mathematics and turned
earth and sky, which they united, so to say, into the house of humankind.
Franz Gnaedinger
2017-02-06 08:36:50 UTC
Reply
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Post by Franz Gnaedinger
a book on the Great Pyramid, which, he claims, was built by an alien
who still sleeps in a hidden chamber ... The Egyptians, obviously, were
not able to build that monument on their own. I can show that they were,
using simple yet clever methods, but I have no chance to publish my work.
Near the end of last year, EdV wrote in a newspaper column that woodhenges
were built as messages for aliens, whereas calendars are their most boring
explanation. I never interpreted a woodhenge, given the uncertainty of
numbers, but for example the original ring of 56 bluestones of Stonehenge
(in a previous message) as a lunisolar calendar, on top of the calendar
interpretation of the younger iconic monument
http://www.seshat.ch/home/stonehen.htm
with a parallel to the bronze needle calendar from Falera in the Swiss Alps.
Calendars, in my opinion, mark the beginning of mathematics and turned
earth and sky, which they united, so to say, into the house of humankind.
how long lived Methuselah? (variable time scale)

Genesis 5 surprises with high ages. Adam begot Seth when he was
130 years old, and lived 930 years ... Enoch begot Methuselah when
he was 65 years old, and lived 365 years. Methuselah begot Lamech
when he was 187 years old, and lived 969 years ...

Maybe the Biblical year varied?

While English hour has a precise meaning (one hour 60 minutes,
24 hours one day) Greek ho:ra (omega, long o) is open for a bunch
of meanings: period of time, season, spring, harvest, climate, year,
daytime, day, hour, moment, age, youth.

In the Göbekli Tepe calendar a month had 30 days, a season 90 days,
and a basic year 360 days.

Time is long when we are young, short when old. So let us consider
a season of 90 days a young man's 'year', and a month of 30 days an
old man's 'year'

Adam begot Seth when he was 130 seasons or 11,700 days or
32 years old, and reached an age of 930 months or 27,900 days
or 76 years

Enoch begot Methuselah when he was 65 seasons or 5,860 days
or 16 years old, and lived 365 months or 10,950 days or practically
30 years

Methuselah begot Lamech when he was 187 seasons or 16,830 days
or 79 years old, and lived 969 months or 29,070 days or 79 years

The modified numbers are in the range of plausibility, while the
actual numbers are all composed of factors, which may symbolize
fertility, like the big number in Exodus

2 x 2 x 2 x 2 x 2 x 3 x 5 x 5 x 5 x 5 = 60,000

A group of Israelites followed Moses and Aaron into the Sinai,
multiplied, and became a people.

(short version, some linguistic lines omitted)
Franz Gnaedinger
2017-02-07 08:14:46 UTC
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Post by Franz Gnaedinger
how long lived Methuselah? (variable time scale)
Genesis 5 surprises with high ages. Adam begot Seth when he was
130 years old, and lived 930 years ... Enoch begot Methuselah when
he was 65 years old, and lived 365 years. Methuselah begot Lamech
when he was 187 years old, and lived 969 years ...
Maybe the Biblical year varied?
While English hour has a precise meaning (one hour 60 minutes,
24 hours one day) Greek ho:ra (omega, long o) is open for a bunch
of meanings: period of time, season, spring, harvest, climate, year,
daytime, day, hour, moment, age, youth.
In the Göbekli Tepe calendar a month had 30 days, a season 90 days,
and a basic year 360 days.
Time is long when we are young, short when old. So let us consider
a season of 90 days a young man's 'year', and a month of 30 days an
old man's 'year'
Adam begot Seth when he was 130 seasons or 11,700 days or
32 years old, and reached an age of 930 months or 27,900 days
or 76 years
Enoch begot Methuselah when he was 65 seasons or 5,860 days
or 16 years old, and lived 365 months or 10,950 days or practically
30 years
Methuselah begot Lamech when he was 187 seasons or 16,830 days
or 79 years old, and lived 969 months or 29,070 days or 79 years
The modified numbers are in the range of plausibility, while the
actual numbers are all composed of factors, which may symbolize
fertility, like the big number in Exodus
2 x 2 x 2 x 2 x 2 x 3 x 5 x 5 x 5 x 5 = 60,000
A group of Israelites followed Moses and Aaron into the Sinai,
multiplied, and became a people.
(short version, some linguistic lines omitted)
(correcting the last paragraph)

The modified numbers are in the range of plausibility, while the
actual numbers are all composed of factors, which may symbolize
fertility, like the big number of the Exodus mentioned in the
Book of Numbers

2 x 2 x 2 x 2 x 2 x 2 x 3 x 5 x 5 x 5 x 5 x 5 = 600,000

A group of Israelites followed Moses and Aaron into the Sinai,
multiplied, and became a people. The twelve factors evoke
the twelve tribes of Israel. Note also the twelve ancestors of
Noah in Genesis 5.
Franz Gnaedinger
2017-02-10 08:16:35 UTC
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Post by Franz Gnaedinger
(correcting the last paragraph)
The modified numbers are in the range of plausibility, while the
actual numbers are all composed of factors, which may symbolize
fertility, like the big number of the Exodus mentioned in the
Book of Numbers
2 x 2 x 2 x 2 x 2 x 2 x 3 x 5 x 5 x 5 x 5 x 5 = 600,000
A group of Israelites followed Moses and Aaron into the Sinai,
multiplied, and became a people. The twelve factors evoke
the twelve tribes of Israel. Note also the twelve ancestors of
Noah in Genesis 5.
(one more correction) There are only nine ancestors of Noah mentioned
in Genesis 5.

Hebrew shanah 'year' might once have had a wide range of meanings,
comparable to Ancient Greek ho:ra. Especially a month of 30 days and
a season of 90 days. (Astronomical relation: 63 continous months or 21
seasons are 1,890 days and correspond to 64 lunations or synodic months; lunisolar formula of the Göbekli Tepe. Harran or Haran forty kilometers
south of the Göbekli Tepe was consecrated to the moon god Sin.)

Adam begot Seth when he was 130 years old. 130 seasons are 11,700 days
or 32 years. Adam lived 930 years. 930 months are 27,900 days or 76 years.
The 800 years in between may have been added for the purpose of redundancy,
preventing mistakes in a copy, and as a number of many factors indicating
fertility.

Noah mentioned at the end of Genesis 5 begot Shem and Ham and Japheth
when he was 500 years old. 500 seasons are 45,000 days or 123 years -
which is only possible if there were three Noahs taken together.

The first Noah might have left the Göbekli Tepe region for the Jordan Valley
where he begot Shem who became the father of the Semites. The second Noah
might have left the Göbekli Tepe region for Sumer (Uruk of the cattle
enclosure, Erech and the ark in the Bible) and begotten Ham who became
the father of the Hamites. And the third Noah might have left the Göbekli
Tepe region for Armenia (at the base of Mount Ararat) and then Central Asia,
and would have begotten Japheth Greek Japetos who became the father of
the Proto-Indo-Europeans, Japhetic being an old term for their language
(Proto-Indo-European).

In a later chapter we read that Noah lived 950 years. 950 seasons are
28,500 days or 78 years, a proud age for those times. This Noah would
have been the father of Shem alone, no longer the triple Noah.

There is a middle way between taking the Bible litterally and dismissing it
as propaganda of the Jahwist: trying to understand the symbols and many
combined layers of meaning.
Franz Gnaedinger
2017-02-17 07:23:59 UTC
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Post by Franz Gnaedinger
(one more correction) There are only nine ancestors of Noah mentioned
in Genesis 5.
Hebrew shanah 'year' might once have had a wide range of meanings,
comparable to Ancient Greek ho:ra. Especially a month of 30 days and
a season of 90 days. (Astronomical relation: 63 continous months or 21
seasons are 1,890 days and correspond to 64 lunations or synodic months;
lunisolar formula of the Göbekli Tepe. Harran or Haran forty kilometers
south of the Göbekli Tepe was consecrated to the moon god Sin.)
Adam begot Seth when he was 130 years old. 130 seasons are 11,700 days
or 32 years. Adam lived 930 years. 930 months are 27,900 days or 76 years.
The 800 years in between may have been added for the purpose of redundancy,
preventing mistakes in a copy, and as a number of many factors indicating
fertility.
Noah mentioned at the end of Genesis 5 begot Shem and Ham and Japheth
when he was 500 years old. 500 seasons are 45,000 days or 123 years -
which is only possible if there were three Noahs taken together.
The first Noah might have left the Göbekli Tepe region for the Jordan Valley
where he begot Shem who became the father of the Semites. The second Noah
might have left the Göbekli Tepe region for Sumer (Uruk of the cattle
enclosure, Erech and the ark in the Bible) and begotten Ham who became
the father of the Hamites. And the third Noah might have left the Göbekli
Tepe region for Armenia (at the base of Mount Ararat) and then Central Asia,
and would have begotten Japheth Greek Japetos who became the father of
the Proto-Indo-Europeans, Japhetic being an old term for their language
(Proto-Indo-European).
In a later chapter we read that Noah lived 950 years. 950 seasons are
28,500 days or 78 years, a proud age for those times. This Noah would
have been the father of Shem alone, no longer the triple Noah.
There is a middle way between taking the Bible litterally and dismissing it
as propaganda of the Jahwist: trying to understand the symbols and many
combined layers of meaning.
everlasting memorials (playing with numbers)

Each of the nine patriarchs preceeding Noah in Genesis 5 would have lived
A seasons of 90 days plus B variable periods, in all A plus B months of
30 days. One period lasts 30 days minus 60 days times A/B. Mathematically
minded readers may check on the following relations:

Adam lived 130 seasons plus 800 periods, in all 930 months. Four Adam
periods are 81 days. If the diagonal of a cube corresponds to five weeks
or 35 days, the edge measures practically one Adam period. 35 Adam periods
are 24 lunations in the Göbekli Tepe definition.

Seth lived 105 seasons plus 807 periods, in all 912 months. If the diagonal
of a 2x3 rectangle corresponds to 80 days, the rectangle measures practically
2 by 3 Seth periods.

Cainan lived 70 seasons plus 840 periods, in all 910 months. One Cainan period
lasts 25 days. 13 Cainan periods are 325 days or 11 lunations counted in the
30 29 30 mode.

Enos lived 90 seasons plus 815 periods, in all 905 months. If the diagonal
of a square corresponds to 33 days, the side measures practically one Enos
period.

Mahalaleel lived 65 seasons plus 830 periods, in all 895 months. If the
diagonal of a 1x3 rectangle corresponds to 80 days (diagonal of the Seth
2x3 rectangle), the 1x3 rectangle measures practically 1 by 3 Mahalaleel
periods.

Jared lived 162 seasons plus 800 periods, in all 962 months. If the circum-
ference of a circle corresponds to eight weeks or 56 days, the diameter
measures practically one Jared period.

Enoch lived 65 seasons plus 300 periods, in all 365 months (number of days
in a regular year). One Enoch period equals 17 days. 33 Enoch periods
(Enos number 33) are 561 days or 19 lunations counted in the 30 29 30 mode.

Methuselah lived 187 seasons plus 782 periods, in all 969 months. If the
diagonal of a double square corresponds to five weeks or 35 days (diagonal
of the Adam cube), the rectangle measures practically 2 by 4 Methuselah
periods (360/161 for the square root of 5 - 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 20 ...
72 161 360).

Lamech lived 182 seasons plus 595 periods, in all 777 months. If the diagonal
of another 2x3 rectangle corresponds to six weeks or 42 days, the rectangle
measures practically 2 by 3 Lamech periods.

The authors of Genesis 5 might have said: 'Let us make the memory of the
patriarchs live on forever, in form of unbreakable hence eternal if only
approximated geometrical relations, and/or in lunar cycles, and let us
carefully render the pair of neccessary numbers (A, B) and their sum for
control ...' The modified times (when a patriarch begot a son, how long
he lived) are in the range of plausibility but of course not historical,
instead immaterial and indestructible memorials for the patriarchs who
combined the human sphere and divine sphere (what we now call rational
and irrational numbers).

Compare the different meanings of year with the pair of different cubits
in Solomon (diameter of the 'molten sea' measuring 10 cubits of 21 units,
in all 210 units, while the circumference measures 30 cubits of 22 units,
in all 660 units, yielding 660/210 or 22/7 for pi), and consider the vision
of a heavenly Jerusalem composed of astronomical cycles (wheels turning
within wheels) in Ezekiel http://www.seshat.ch/home/calendar.htm
(scroll toward the end).
Franz Gnaedinger
2017-02-21 07:53:50 UTC
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Post by Franz Gnaedinger
everlasting memorials (playing with numbers)
Each of the nine patriarchs preceeding Noah in Genesis 5 would have lived
A seasons of 90 days plus B variable periods, in all A plus B months of
30 days. One period lasts 30 days minus 60 days times A/B. Mathematically
Adam lived 130 seasons plus 800 periods, in all 930 months. Four Adam
periods are 81 days. If the diagonal of a cube corresponds to five weeks
or 35 days, the edge measures practically one Adam period. 35 Adam periods
are 24 lunations in the Göbekli Tepe definition.
Seth lived 105 seasons plus 807 periods, in all 912 months. If the diagonal
of a 2x3 rectangle corresponds to 80 days, the rectangle measures practically
2 by 3 Seth periods.
Cainan lived 70 seasons plus 840 periods, in all 910 months. One Cainan period
lasts 25 days. 13 Cainan periods are 325 days or 11 lunations counted in the
30 29 30 mode.
Enos lived 90 seasons plus 815 periods, in all 905 months. If the diagonal
of a square corresponds to 33 days, the side measures practically one Enos
period.
Mahalaleel lived 65 seasons plus 830 periods, in all 895 months. If the
diagonal of a 1x3 rectangle corresponds to 80 days (diagonal of the Seth
2x3 rectangle), the 1x3 rectangle measures practically 1 by 3 Mahalaleel
periods.
Jared lived 162 seasons plus 800 periods, in all 962 months. If the circum-
ference of a circle corresponds to eight weeks or 56 days, the diameter
measures practically one Jared period.
Enoch lived 65 seasons plus 300 periods, in all 365 months (number of days
in a regular year). One Enoch period equals 17 days. 33 Enoch periods
(Enos number 33) are 561 days or 19 lunations counted in the 30 29 30 mode.
Methuselah lived 187 seasons plus 782 periods, in all 969 months. If the
diagonal of a double square corresponds to five weeks or 35 days (diagonal
of the Adam cube), the rectangle measures practically 2 by 4 Methuselah
periods (360/161 for the square root of 5 - 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 20 ...
72 161 360).
Lamech lived 182 seasons plus 595 periods, in all 777 months. If the diagonal
of another 2x3 rectangle corresponds to six weeks or 42 days, the rectangle
measures practically 2 by 3 Lamech periods.
The authors of Genesis 5 might have said: 'Let us make the memory of the
patriarchs live on forever, in form of unbreakable hence eternal if only
approximated geometrical relations, and/or in lunar cycles, and let us
carefully render the pair of neccessary numbers (A, B) and their sum for
control ...' The modified times (when a patriarch begot a son, how long
he lived) are in the range of plausibility but of course not historical,
instead immaterial and indestructible memorials for the patriarchs who
combined the human sphere and divine sphere (what we now call rational
and irrational numbers).
Compare the different meanings of year with the pair of different cubits
in Solomon (diameter of the 'molten sea' measuring 10 cubits of 21 units,
in all 210 units, while the circumference measures 30 cubits of 22 units,
in all 660 units, yielding 660/210 or 22/7 for pi), and consider the vision
of a heavenly Jerusalem composed of astronomical cycles (wheels turning
within wheels) in Ezekiel http://www.seshat.ch/home/calendar.htm
(scroll toward the end).
Ark of the Covenant (lost, but not entirely)

We may assume that the Ark of the Covenant as described in Exodus 25 was
a later version made for a temple, in a time when the relatively small group
of Israelites who had left Pi-Ramesse for the Sinai had become a people of
600,000. The ark is lost, but not entirely; we still have the numbers,
and we can get a lot of information from them.

The basic chest measured 2.5 by 1.5 by 1.5 cubits. What is the surface of
such a right parallel-epiped? Practically the same as the one of a sphere
whose diameter measures 2.5 cubits. An imaginary sphere evoking the
medieval "Deus est sphaera" - God is present in the perfect shape of
the sphere.

Double the numbers 2.5 and 1.5 and you have 5 and 3, in between 4, together
the numbers of the Sacred Triangle 3-4-5 which is the key to the Egyptian
systematic method of calculating the circle via a series of circles of the
radii and diameters 5 25 125 625 3125 ... and 10 50 250 1250 6250 ... ever
smaller units respectively - all ten numbers are factors of 600,000 -
and their slowly rounding inscribed polygons based on the Sacred Triangle
and a series of triples derived from it by means of a linear plus minus
algorithm

http://www.seshat.ch/home/aim.pdf

The second polygon of the diameter 50 units relies on the triples 3-4-5
or 15-20-25 and 7-24-25. It has twenty sides, a dozen sides measuring
practically 7 units (exactly the square root of fifty) and eight sides
measuring practically 9 units (exactly the square root of eighty), in all
a periphery of practically 156 units. Divide the this measurement by the
diameter 50 units and you get 156/50 or 78/25, implicit value of pi in
the above transformation of the surface of the right parallel-epiped
into the one of the imaginary sphere.

The royal cubit of the New Kingdom of Egypt measured 52.5 centimeters,
28 fingerbreadths of 1.875 cm or 7 palms of 7.5 cm. The acacia chest
was gilded on the outside and inside. Make the boards three fingebreadths
thick and the inside measures 16 by 9 by 9 palms. A square of side 16
and a circle of radius 9 have practically the same area - parallel to
a famous problem in the Rhind Mathematical Papyrus. Make the boards
a little thinner (2.8759... f) and you have the exact equation.

The Rhind Mathematical Papyrus encodes theorems in telling numbers
(for example RMP 32 in aim.pdf). The same spirit is found in the Bible.

When I study the figurines from Beersheba (Jean Perrot) I can't but
compare them with the legacy of Predynastic Egypt, and ask whether
some gifted people from the region of Beersheba contributed to the
civilization of their mighty neighbour? and later on took their scrolls
with them when they left Pi-Ramesse for the Sinai? writings on a par
with the Rhind Papyri?
Franz Gnaedinger
2017-02-28 09:14:38 UTC
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Post by Franz Gnaedinger
Ark of the Covenant (lost, but not entirely)
We may assume that the Ark of the Covenant as described in Exodus 25 was
a later version made for a temple, in a time when the relatively small group
of Israelites who had left Pi-Ramesse for the Sinai had become a people of
600,000. The ark is lost, but not entirely; we still have the numbers,
and we can get a lot of information from them.
The basic chest measured 2.5 by 1.5 by 1.5 cubits. What is the surface of
such a right parallel-epiped? Practically the same as the one of a sphere
whose diameter measures 2.5 cubits. An imaginary sphere evoking the
medieval "Deus est sphaera" - God is present in the perfect shape of
the sphere.
Double the numbers 2.5 and 1.5 and you have 5 and 3, in between 4, together
the numbers of the Sacred Triangle 3-4-5 which is the key to the Egyptian
systematic method of calculating the circle via a series of circles of the
radii and diameters 5 25 125 625 3125 ... and 10 50 250 1250 6250 ... ever
smaller units respectively - all ten numbers are factors of 600,000 -
and their slowly rounding inscribed polygons based on the Sacred Triangle
and a series of triples derived from it by means of a linear plus minus
algorithm
http://www.seshat.ch/home/aim.pdf
The second polygon of the diameter 50 units relies on the triples 3-4-5
or 15-20-25 and 7-24-25. It has twenty sides, a dozen sides measuring
practically 7 units (exactly the square root of fifty) and eight sides
measuring practically 9 units (exactly the square root of eighty), in all
a periphery of practically 156 units. Divide the this measurement by the
diameter 50 units and you get 156/50 or 78/25, implicit value of pi in
the above transformation of the surface of the right parallel-epiped
into the one of the imaginary sphere.
The royal cubit of the New Kingdom of Egypt measured 52.5 centimeters,
28 fingerbreadths of 1.875 cm or 7 palms of 7.5 cm. The acacia chest
was gilded on the outside and inside. Make the boards three fingebreadths
thick and the inside measures 16 by 9 by 9 palms. A square of side 16
and a circle of radius 9 have practically the same area - parallel to
a famous problem in the Rhind Mathematical Papyrus. Make the boards
a little thinner (2.8759... f) and you have the exact equation.
The Rhind Mathematical Papyrus encodes theorems in telling numbers
(for example RMP 32 in aim.pdf). The same spirit is found in the Bible.
When I study the figurines from Beersheba (Jean Perrot) I can't but
compare them with the legacy of Predynastic Egypt, and ask whether
some gifted people from the region of Beersheba contributed to the
civilization of their mighty neighbour? and later on took their scrolls
with them when they left Pi-Ramesse for the Sinai? writings on a par
with the Rhind Papyri?
YHVH (encoded in the Ark of the Covenant)

((linguistic passages omitted))

Jewish scholars used the letters of the Hebrew alphabet as numerals,
possibly long before the Maccabaeans in the second century BC

Y 10 H 5 V 6 H 5

The Ark of the Covenant is mentioned in Exodus 25. The gold-covered acacia
chest measured 2.5 by 1.5 by 1.5 cubits. On it was placed a 'mercy seat'
of pure gold: a gold sheet measuring 2.5 by 1.5 cubits, on it a pair of
golden cherubims facing each other and stretching their wings forward
so they touch each other in the middle while covering the whole length
of the mercy seat, 2.5 cubits, yielding 1.25 cubits for one pair of wings
or one wing

length of mercy seat 2.5 cubits
wings of one cherub 1.25 cubits
breadth of mercy seat 1.5 cubits
wings of other cherub 1.25 cubits

length : wing : breadth : wing = 2.5 : 1.25 : 1.5 : 1.25

2.5 : 1.25 : 1.5 : 1.25 = 10 : 5 : 6 : 5 = Y : H : V : H

The mercy seat would have encoded or rather invited Jahwe in form of YHVH
- may he have mercy with us, may he come down on earth and take place on
the mercy seat on the annual day of atonement.
Franz Gnaedinger
2017-03-07 10:58:55 UTC
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Post by Franz Gnaedinger
YHVH (encoded in the Ark of the Covenant)
((linguistic passages omitted))
Jewish scholars used the letters of the Hebrew alphabet as numerals,
possibly long before the Maccabaeans in the second century BC
Y 10 H 5 V 6 H 5
The Ark of the Covenant is mentioned in Exodus 25. The gold-covered acacia
chest measured 2.5 by 1.5 by 1.5 cubits. On it was placed a 'mercy seat'
of pure gold: a gold sheet measuring 2.5 by 1.5 cubits, on it a pair of
golden cherubims facing each other and stretching their wings forward
so they touch each other in the middle while covering the whole length
of the mercy seat, 2.5 cubits, yielding 1.25 cubits for one pair of wings
or one wing
length of mercy seat 2.5 cubits
wings of one cherub 1.25 cubits
breadth of mercy seat 1.5 cubits
wings of other cherub 1.25 cubits
length : wing : breadth : wing = 2.5 : 1.25 : 1.5 : 1.25
2.5 : 1.25 : 1.5 : 1.25 = 10 : 5 : 6 : 5 = Y : H : V : H
The mercy seat would have encoded or rather invited Jahwe in form of YHVH
- may he have mercy with us, may he come down on earth and take place on
the mercy seat on the annual day of atonement.
('molten sea' (hidden gems of Solomonic wisdom)

((story and etymology left out))

The molten sea of Solomon on the temple mound of Jerusalem was a stone basin
partly clad in hammered brass sheet, diameter 10 cubits and circumference
30 cubits, 5 cubits deep, capacity 2,000 baths.

Replace the single cubit by a pair of cubits of nearly the same lenghts,
a 'black' and 'red' cubit in reference to the black and red inks used
by Ahmose in the Rhind Mathematical Papyrus

'black' cubit of 28 fingerbreadths or 21 units or 52.5 cm (bc)
'red' cubit of 22 units or 55 centimeters (rc)

diameter basin 10 bc or 210 units or 5.25 meters
circumference 30 rc or 660 units or 16.5 meters
660/210 or 22/7 for pi

rim possibly 1 bc or 52.5 centimeters
diameter rim-water-rim 12 bc or 6.3 meters
outer circumference 36 rc or 19.8 meters

The combined measures allow simple formulae

diameter of a circle 1 bc
circumference 3 rc

radius of a circle 1 bc
area 1 bc x 3 rc

diameter of a sphere 1 bc
surface 1 bc x 3 rc

diameter of a sphere 2 bc
volume 2 bc x 2 bc x 1 rc

side of a square 20 bc
diagonal 27 rc

If the round wall of the basin was vertical and 5 bc deep or high, the
capacity was 5 bc x 5 bc x 15 rc, 250 old barrels or 2,000 old buckets
(English bath leaning on Hebrew bat 'bucket'), measurements given in
fingerbreadths, 28 f being 21 u or 1 bc or 52.5 cm

28 by 28 by 44 f 8 baths or ca. 227.4 liters old barrel
28 by 28 by 22 f 4 baths or ca. 112.7 liters
28 by 14 by 22 f 2 baths or ca. 56.8 liters
14 by 14 by 22 f 1 bath or ca. 28.4 liters old bucket
14 by 14 by 11 f 1/2 bath or ca. 14.2 liters
14 by 7 by 11 f 1/4 bath or ca. 7.1 liters
7 by 7 by 11 f 1/8 bath or ca. 3.6 liters old jug
Franz Gnaedinger
2017-03-08 07:35:23 UTC
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Post by Franz Gnaedinger
('molten sea' (hidden gems of Solomonic wisdom)
((story and etymology left out))
The molten sea of Solomon on the temple mound of Jerusalem was a stone basin
partly clad in hammered brass sheet, diameter 10 cubits and circumference
30 cubits, 5 cubits deep, capacity 2,000 baths.
Replace the single cubit by a pair of cubits of nearly the same lenghts,
a 'black' and 'red' cubit in reference to the black and red inks used
by Ahmose in the Rhind Mathematical Papyrus
'black' cubit of 28 fingerbreadths or 21 units or 52.5 cm (bc)
'red' cubit of 22 units or 55 centimeters (rc)
diameter basin 10 bc or 210 units or 5.25 meters
circumference 30 rc or 660 units or 16.5 meters
660/210 or 22/7 for pi
rim possibly 1 bc or 52.5 centimeters
diameter rim-water-rim 12 bc or 6.3 meters
outer circumference 36 rc or 19.8 meters
The combined measures allow simple formulae
diameter of a circle 1 bc
circumference 3 rc
radius of a circle 1 bc
area 1 bc x 3 rc
diameter of a sphere 1 bc
surface 1 bc x 3 rc
diameter of a sphere 2 bc
volume 2 bc x 2 bc x 1 rc
side of a square 20 bc
diagonal 27 rc
If the round wall of the basin was vertical and 5 bc deep or high, the
capacity was 5 bc x 5 bc x 15 rc, 250 old barrels or 2,000 old buckets
(English bath leaning on Hebrew bat 'bucket'), measurements given in
fingerbreadths, 28 f being 21 u or 1 bc or 52.5 cm
28 by 28 by 44 f 8 baths or ca. 227.4 liters old barrel
28 by 28 by 22 f 4 baths or ca. 112.7 liters
28 by 14 by 22 f 2 baths or ca. 56.8 liters
14 by 14 by 22 f 1 bath or ca. 28.4 liters old bucket
14 by 14 by 11 f 1/2 bath or ca. 14.2 liters
14 by 7 by 11 f 1/4 bath or ca. 7.1 liters
7 by 7 by 11 f 1/8 bath or ca. 3.6 liters old jug
heavenly Jerusalem (and a majestic river)

((final message ending a series of messages on water symbols
and telling numbers in the Bible))

Ezekiel, at the end of his book (48:30-35) describes a city called
The Lord is there, a heavenly Jerusalem composed of astronomical
cycles, an imaginary city within a round wall - 4,500 measures in
the North, 4,500 measures in the East, 4,500 measures in the South,
4,500 measures in the West, in all a circumference of 18,000 measures.

Use the equations

30 measures for one lunation
64 measures for 63 days

and the 18,000 measures become 600 lunations or synodic months,
while the diameter of the imaginary city within the big circle of the wall
is practically 5,730 measures or 5,640 days or 191 lunations, based on
600/191 = 3.1413… for pi = 3.14159…, from a pi sequence

6/2 (plus 22/7) 28/9 50/16 72/23 … 600/191 …

The side of a square inscribed in the big circle is practically 135 lunations
or 3,988 days or 4,050 measures. Transform the square into a circle of
the same area. The diameter measures practically 4,500 days - key number
of the heavenly Jerusalem, a number composed of seven small prime factors
indicating divine fertility

2 x 2 x 3 x 3 x 5 x 5 x 5 = 4500

Start with 18,000 measures. Use 20 for 63/pi and 4/5 for the square root
of 2/pi and you get exactly 4,500 days. Use a pocket calculator and you
obtain 4500.1114 days, rounded 4,500 days.

Ezekiel was born around 622 BC in Israel. He began working there,
and continued working in his Babylonian exile in Tell Abib on the Chaber
River near an ancient dam. He might have been an astronomer and
mathematician and maybe also a water engineer at a renowned Chaldean
school. It has even been assumed that he was “one of Pythagoras’ masters”
(on what basis I don’t know). In any case he was a master in handling early
calculating methods, and he achieved a marvel of combined approximations
in his vision of a heavenly Jerusalem, encoded in a single number, 4,500.

The corners of a dodecagon (regular polygon of twelve sides) inscribed
in the big circle of the city wall mark the centers of the dozen gates.
One side is practically 1,460 days, four regular years of 365 days,
and the periphery 17,520 days, 30 Venus years of 584 days, rounded
value of the actual 583,92 days. Number sequence relating side and
radius of a dodecagon

3/6 (plus 14/27) 1733 31/60 45/87 59/114 73/141

18,000 measures are the inner circumference of the wall. Count 2 days
for the wall and the radius of the outer circle was 2,822 days, the side
of the inscribed dodecagon practically 1,461 days, three regular years of
365 days plus one leap year of 366 days, and the periphery 48 years.

Back to the inner circumference of the wall. Give a gate 30 measures
or one lunation. The arc of the wall from one to the next gate would then
have been 1,470 measures or practically 1,447 days, and may have been
divided into 510-450-510 measures or practically 502-443-502 days
or 17-15-17 lunations counted in the 30 29 30 mode each.

30 measures for one lunation correspond to the Goebekli Tepe lunation;
mistake less than one minute, or half a day in a lifetime. 1,447 days
for 49 lunations are even better; mistake two seconds, or half an hour
in a lifetime.

We can now say that Ezekiel’s mysterious wheels turning within each other
- “their work was as it were a wheel in the middle of a wheel” (1:16) -
were the cycles of sun and moon and Venus revolving within each other
in the vision of a heavenly Jerusalem.

And the Gihon spring became the source of a majestic river in Ezekiel 47.

(( http://www.seshat.ch/home/lasco13.htm ))
Franz Gnaedinger
2017-03-17 08:15:21 UTC
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Post by Franz Gnaedinger
heavenly Jerusalem (and a majestic river)
((final message ending a series of messages on water symbols
and telling numbers in the Bible))
Ezekiel, at the end of his book (48:30-35) describes a city called
The Lord is there, a heavenly Jerusalem composed of astronomical
cycles, an imaginary city within a round wall - 4,500 measures in
the North, 4,500 measures in the East, 4,500 measures in the South,
4,500 measures in the West, in all a circumference of 18,000 measures.
Use the equations
30 measures for one lunation
64 measures for 63 days
and the 18,000 measures become 600 lunations or synodic months,
while the diameter of the imaginary city within the big circle of the wall
is practically 5,730 measures or 5,640 days or 191 lunations, based on
600/191 = 3.1413… for pi = 3.14159…, from a pi sequence
6/2 (plus 22/7) 28/9 50/16 72/23 … 600/191 …
The side of a square inscribed in the big circle is practically 135 lunations
or 3,988 days or 4,050 measures. Transform the square into a circle of
the same area. The diameter measures practically 4,500 days - key number
of the heavenly Jerusalem, a number composed of seven small prime factors
indicating divine fertility
2 x 2 x 3 x 3 x 5 x 5 x 5 = 4500
Start with 18,000 measures. Use 20 for 63/pi and 4/5 for the square root
of 2/pi and you get exactly 4,500 days. Use a pocket calculator and you
obtain 4500.1114 days, rounded 4,500 days.
Ezekiel was born around 622 BC in Israel. He began working there,
and continued working in his Babylonian exile in Tell Abib on the Chaber
River near an ancient dam. He might have been an astronomer and
mathematician and maybe also a water engineer at a renowned Chaldean
school. It has even been assumed that he was “one of Pythagoras’ masters”
(on what basis I don’t know). In any case he was a master in handling early
calculating methods, and he achieved a marvel of combined approximations
in his vision of a heavenly Jerusalem, encoded in a single number, 4,500.
The corners of a dodecagon (regular polygon of twelve sides) inscribed
in the big circle of the city wall mark the centers of the dozen gates.
One side is practically 1,460 days, four regular years of 365 days,
and the periphery 17,520 days, 30 Venus years of 584 days, rounded
value of the actual 583,92 days. Number sequence relating side and
radius of a dodecagon
3/6 (plus 14/27) 1733 31/60 45/87 59/114 73/141
18,000 measures are the inner circumference of the wall. Count 2 days
for the wall and the radius of the outer circle was 2,822 days, the side
of the inscribed dodecagon practically 1,461 days, three regular years of
365 days plus one leap year of 366 days, and the periphery 48 years.
Back to the inner circumference of the wall. Give a gate 30 measures
or one lunation. The arc of the wall from one to the next gate would then
have been 1,470 measures or practically 1,447 days, and may have been
divided into 510-450-510 measures or practically 502-443-502 days
or 17-15-17 lunations counted in the 30 29 30 mode each.
30 measures for one lunation correspond to the Goebekli Tepe lunation;
mistake less than one minute, or half a day in a lifetime. 1,447 days
for 49 lunations are even better; mistake two seconds, or half an hour
in a lifetime.
We can now say that Ezekiel’s mysterious wheels turning within each other
- “their work was as it were a wheel in the middle of a wheel” (1:16) -
were the cycles of sun and moon and Venus revolving within each other
in the vision of a heavenly Jerusalem.
And the Gihon spring became the source of a majestic river in Ezekiel 47.
(( http://www.seshat.ch/home/lasco13.htm ))
holy Jerusalem descending from heaven (Revelation of John)

The heavenly Jerusalem in the vision of Ezekiel returns in The Revelation
of St. John the Divine, John of Patmos - "the holy Jerusalem descending
out of heaven from God" (21:10), a city made of jewels and pearls and
precious metals, hovering in a sphere of the circumference 12,000 furlongs
(assuming the foursquare city was protected by a round wall, four towers
marking the cardinal directions N E S W). Replace 20 furlongs by 30 measures
and you have the city called 'The Lord is there' in the vision of Ezekiel.

Chapter 13 of the Revelation is about the beast, a dragon of the number
Six hundred threescore and six. If this was the circumference of the dragon's
den in the depth of the sea from where the beast emerged, we have a counter-
part to the heavenly Jerusalem: circumference 666 units, diameter practically
212 and radius 106 units, according to a pi sequence

3/1 (plus 22/7) 25/8 47/15 69/22 ... 333/106 ...

An inscribed irregular hexagon has the sides of practically 118 118 118 118
118 40 units. 118 is the number of four lunations counted in the old way,
30 29 30 29 sum 118 days. Connect the corners of the irregular hexagon with
five lines of practically 196 units each and you have an open pentagram,
all five lines together 980 units (divisible by the numbers of the seven
heads and ten horns of the dragon), while the opening measures 40 units
(again a set of close approximations).

The pentagram was the emblem of the Pythagoreans, and became a sign of
magicians. Goethe's Faust trapped Mephistopheles with an open pentagram,
while Satan tempted Jesus for 40 days in the desert.

Could the dragon's den go back to a now lost script by Ezekiel?

The heavenly Jerusalem in the Revelation needs neither candles nor the sun
- divine light being woven into the geometry of the Lord's own city, in the
form of closed astronomical polygons, whereas the pentagram of the devil's
den is open, unstable, the dragon weaker than God, bound to lose the final
war.

Could this have been the message of the hypothetical script by Ezekiel
that would not be lost entirely but survive as echo in John's Revelation?

While the Gihon spring became the source of a majestic river in Ezekiel,
the morning star Venus personified by an angel (144 cubits tall) showed
John "a pure river of water of life, clear as crystal" (22:1). John's
Revelation is the hope for a better world governed by God and Jesus
(and, unfortunately, also by misogyny, although the whore and the seven
women on the mountains of the dragon refer to Babylon and political powers).
Water plays again an important role, on the level of symbols - fresh water
vs the salt water of the sea from where the dragon rose - and as a necessity
of life.

((ending on the water issue on the Goebekli Tepe and in the Bible omitted))
Franz Gnaedinger
2017-03-20 07:45:54 UTC
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Post by Franz Gnaedinger
holy Jerusalem descending from heaven (Revelation of John)
The heavenly Jerusalem in the vision of Ezekiel returns in The Revelation
of St. John the Divine, John of Patmos - "the holy Jerusalem descending
out of heaven from God" (21:10), a city made of jewels and pearls and
precious metals, hovering in a sphere of the circumference 12,000 furlongs
(assuming the foursquare city was protected by a round wall, four towers
marking the cardinal directions N E S W). Replace 20 furlongs by 30 measures
and you have the city called 'The Lord is there' in the vision of Ezekiel.
Chapter 13 of the Revelation is about the beast, a dragon of the number
Six hundred threescore and six. If this was the circumference of the dragon's
den in the depth of the sea from where the beast emerged, we have a counter-
part to the heavenly Jerusalem: circumference 666 units, diameter practically
212 and radius 106 units, according to a pi sequence
3/1 (plus 22/7) 25/8 47/15 69/22 ... 333/106 ...
An inscribed irregular hexagon has the sides of practically 118 118 118 118
118 40 units. 118 is the number of four lunations counted in the old way,
30 29 30 29 sum 118 days. Connect the corners of the irregular hexagon with
five lines of practically 196 units each and you have an open pentagram,
all five lines together 980 units (divisible by the numbers of the seven
heads and ten horns of the dragon), while the opening measures 40 units
(again a set of close approximations).
The pentagram was the emblem of the Pythagoreans, and became a sign of
magicians. Goethe's Faust trapped Mephistopheles with an open pentagram,
while Satan tempted Jesus for 40 days in the desert.
Could the dragon's den go back to a now lost script by Ezekiel?
The heavenly Jerusalem in the Revelation needs neither candles nor the sun
- divine light being woven into the geometry of the Lord's own city, in the
form of closed astronomical polygons, whereas the pentagram of the devil's
den is open, unstable, the dragon weaker than God, bound to lose the final
war.
Could this have been the message of the hypothetical script by Ezekiel
that would not be lost entirely but survive as echo in John's Revelation?
While the Gihon spring became the source of a majestic river in Ezekiel,
the morning star Venus personified by an angel (144 cubits tall) showed
John "a pure river of water of life, clear as crystal" (22:1). John's
Revelation is the hope for a better world governed by God and Jesus
(and, unfortunately, also by misogyny, although the whore and the seven
women on the mountains of the dragon refer to Babylon and political powers).
Water plays again an important role, on the level of symbols - fresh water
vs the salt water of the sea from where the dragon rose - and as a necessity
of life.
((ending on the water issue on the Goebekli Tepe and in the Bible omitted))
PS regarding the dragon's den in John's Revelation. When we start from
the circumference 666 units, the other numbers are close approximations:
diameter 212 radius 106, sides of the inscribed irregular hexagon 118 118
118 118 118 40, long line of the open pentagram 198, all five lines together
980, divisible by the numbers of the seven heads and ten horns of the dragon.
If you draw the open pentagram in the circle you can see five 'dragon heads'
along the circumference: imagine five heads of the horned viper into them,
Cerastes gasperettii, extremely venomous, living in deserts and well watered
oases, a sidewinder whose way of moving might have suggested an aquatic origin
(Johannes de Moor), and give them a pair of short horns each, so you have in
all ten horns; furthermore the open pentagram has a pair of larger triangles
or 'dragon heads' in the middle, overlapping each other, so you have in all
seven heads (the larger ones hornless). Modern calculation of the irregular
hexagon: 2 arcsin 20/106 plus 10 arcsin 59/106 = 359.963... or practically
360 degrees. We may assume that the ancient astronomers had reliable tables
of angles allowing them to calculate such a polygon.
Franz Gnaedinger
2017-03-24 08:31:19 UTC
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Post by Franz Gnaedinger
PS regarding the dragon's den in John's Revelation. When we start from
diameter 212 radius 106, sides of the inscribed irregular hexagon 118 118
118 118 118 40, long line of the open pentagram 198, all five lines together
980, divisible by the numbers of the seven heads and ten horns of the dragon.
If you draw the open pentagram in the circle you can see five 'dragon heads'
along the circumference: imagine five heads of the horned viper into them,
Cerastes gasperettii, extremely venomous, living in deserts and well watered
oases, a sidewinder whose way of moving might have suggested an aquatic origin
(Johannes de Moor), and give them a pair of short horns each, so you have in
all ten horns; furthermore the open pentagram has a pair of larger triangles
or 'dragon heads' in the middle, overlapping each other, so you have in all
seven heads (the larger ones hornless). Modern calculation of the irregular
hexagon: 2 arcsin 20/106 plus 10 arcsin 59/106 = 359.963... or practically
360 degrees. We may assume that the ancient astronomers had reliable tables
of angles allowing them to calculate such a polygon.
acorn and oak tree (poetic formula of the Bible)

Big claims in the Bible can be seen as product of a realistic basis and
an inherent potential. For example the Lord dividing the Red Sea for the
escaping Israelites who might have found a way across a swamp owing to
their knowledge of the terrain they had acquired as water engineers and
water workers and saw this knowledge as a gift from above. They were a small
group yet had the potential of becoming a people of 600,000. Or take the
molten sea. Nobody could have made a brass basin that large and heavy -
more than five meters across, more than two and a half meters deep, resting
on a dozen oxen also made of brass, and filled with more than fifty thousand
liters of water - as described in 1 Kings 7 and 2 Chronicles 4. The realistic
basis here is a stone basin carved out of the bedrock on the temple mound
of Jerusalem, partly clad with hammered brass sheet, and the inherent
potential the combination of two measures of length, the common cubit of
28 fingerbreadths or 7 palms or 21 units or 52.5 centimeters (royal cubit
of Egypt in the New Kingdom) and the second cubit of 22 units or 55 cm.
Their combination allowed simple formulae for the calculation of the circle
and sphere, and of large squares in the field. This method might have been
forgotten in the time the Bible was written, yet the importance of the
basin was well remembered, and turned a modest stone basin into a marvel
of brass no metallurgist of the early Iron Age could possibly have achieved.
My interpretation of the molten sea focuses on 1 Kings 7:32 and 2 Chronicles
4:2 and the method of combining two measures of length as inherent potential.

You can hold a tree in the palm of your hand - an acorn with the potential
of becoming an oak tree.

(Correction of the previous message. One line of the open pentagram
measures 196 units, while the sum of the five lines was given correctly
as 980 units. I carried out each calculation several times in different ways.
Considering the many numbers I am glad that I mistyped only one of them.)
Franz Gnaedinger
2017-04-27 06:50:14 UTC
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Post by Franz Gnaedinger
acorn and oak tree (poetic formula of the Bible)
Big claims in the Bible can be seen as product of a realistic basis and
an inherent potential. For example the Lord dividing the Red Sea for the
escaping Israelites who might have found a way across a swamp owing to
their knowledge of the terrain they had acquired as water engineers and
water workers and saw this knowledge as a gift from above. They were a small
group yet had the potential of becoming a people of 600,000. Or take the
molten sea. Nobody could have made a brass basin that large and heavy -
more than five meters across, more than two and a half meters deep, resting
on a dozen oxen also made of brass, and filled with more than fifty thousand
liters of water - as described in 1 Kings 7 and 2 Chronicles 4. The realistic
basis here is a stone basin carved out of the bedrock on the temple mound
of Jerusalem, partly clad with hammered brass sheet, and the inherent
potential the combination of two measures of length, the common cubit of
28 fingerbreadths or 7 palms or 21 units or 52.5 centimeters (royal cubit
of Egypt in the New Kingdom) and the second cubit of 22 units or 55 cm.
Their combination allowed simple formulae for the calculation of the circle
and sphere, and of large squares in the field. This method might have been
forgotten in the time the Bible was written, yet the importance of the
basin was well remembered, and turned a modest stone basin into a marvel
of brass no metallurgist of the early Iron Age could possibly have achieved.
My interpretation of the molten sea focuses on 1 Kings 7:32 and 2 Chronicles
4:2 and the method of combining two measures of length as inherent potential.
You can hold a tree in the palm of your hand - an acorn with the potential
of becoming an oak tree.
(Correction of the previous message. One line of the open pentagram
measures 196 units, while the sum of the five lines was given correctly
as 980 units. I carried out each calculation several times in different ways.
Considering the many numbers I am glad that I mistyped only one of them.)
(while the hype around the Pyramid Scan project evaporates I go on revealing
treasures in the mathematical methods used by the pyramid builders)

numbers defined by algorithms (and a piece of mathemagics)

Numbers can also be defined by algorithms, for example the square root of 2
by a number column generated by this algorithm

a b 2a
a+b b+2a 2(a+b)

You can begin with any number pair, for example a = 2 and b =5, and even make
mistakes, the number column still approximates the square root of 2

2 5 4
7 9 14
16 25 32 mistake: 25 instead of 23
41 57 82
98 139 196
237 335 474
572 809 ....

809/572 = 1.414335..., squared 2.000345...

The best (fastest) number column is obtained with a = 1 and b = 2 and has
an equivalent in the continued fraction (1;2,2,2,2,2,2,...), furthermore
the lines invoke alternating triples and quadruples

1 1 2 1 = 0 + 1 triple 0-1-1

2 3 4 3 = 1 + 2 quadruple 2-1-2-3

5 7 10 7 = 3 + 4 triple 3-4-5

12 17 24 17 = 8 + 9 quadruple 12-8-9-17

29 41 58 41 = 20 + 21 triple 20-21-29

70 99 140 99 = 49 + 50 quadruple 70-49-50-99

The quadruples are based on divisions of 2. Let me explain this via problem
32 of the Rhind Mathematical Papyrus.

Ahmes divides 2 by 1 '3 '4 and obtains 1 '6 '12 '114 '228. Beginners learn
how to work with unit fraction series, while advanced students are given
a more challenging task. Imagine a right parallepiped measuring 2 by 1 '3 '4
by 1 '6 '12 '114 '228 units. How long is the diagonal of the volume?

Impossible to find out!, the students exlaim. No no, very simple, Ahmes
smiles, just proceed like this:

1 '3 '4 plus 1 '6 '12 '114 '228 units

1 1 plus '3 '6 plus '4 '12 plus '114 '228 units

2 '2 '3 '76 units

Theorem. Divide 2 by A so that you obtain B. Let a right paralelepiped
measure 2 by A by B units. How long is the diagonal of the volume?
Exactly A plus B units.

2 : 1 = 2 quadruple 2-1-2-3

2 : 1 '3 = 1 '2 multiply the numbers by a factor of 6
and you obtain the quadruple 12-8-9-17

2 : 1 '3 '15 = 1 '3 '15 '35 multiply the numbers by a factor of 35
and you obtain the quadruple 70-49-50-99

While the above triples approximate the square 1 by 1, diagonal sqrt2,
the quadruples approximate the right parallelepiped 2 by sqrt2 by sqrt2,
diagonal volume 2 sqrt2.

The most basic number column deserves more studies. Note for example that
the product of the second and third number in a line appears as first number
of a later line. One might say the richness of the square root of 2 unfolds
from the intitial algorithm.
Franz Gnaedinger
2017-05-02 07:08:19 UTC
Reply
Permalink
Post by Franz Gnaedinger
(while the hype around the Pyramid Scan project evaporates I go on revealing
treasures in the mathematical methods used by the pyramid builders)
numbers defined by algorithms (and a piece of mathemagics)
Numbers can also be defined by algorithms, for example the square root of 2
by a number column generated by this algorithm
a b 2a
a+b b+2a 2(a+b)
You can begin with any number pair, for example a = 2 and b =5, and even make
mistakes, the number column still approximates the square root of 2
2 5 4
7 9 14
16 25 32 mistake: 25 instead of 23
41 57 82
98 139 196
237 335 474
572 809 ....
809/572 = 1.414335..., squared 2.000345...
The best (fastest) number column is obtained with a = 1 and b = 2 and has
an equivalent in the continued fraction (1;2,2,2,2,2,2,...), furthermore
the lines invoke alternating triples and quadruples
1 1 2 1 = 0 + 1 triple 0-1-1
2 3 4 3 = 1 + 2 quadruple 2-1-2-3
5 7 10 7 = 3 + 4 triple 3-4-5
12 17 24 17 = 8 + 9 quadruple 12-8-9-17
29 41 58 41 = 20 + 21 triple 20-21-29
70 99 140 99 = 49 + 50 quadruple 70-49-50-99
The quadruples are based on divisions of 2. Let me explain this via problem
32 of the Rhind Mathematical Papyrus.
Ahmes divides 2 by 1 '3 '4 and obtains 1 '6 '12 '114 '228. Beginners learn
how to work with unit fraction series, while advanced students are given
a more challenging task. Imagine a right parallepiped measuring 2 by 1 '3 '4
by 1 '6 '12 '114 '228 units. How long is the diagonal of the volume?
Impossible to find out!, the students exlaim. No no, very simple, Ahmes
1 '3 '4 plus 1 '6 '12 '114 '228 units
1 1 plus '3 '6 plus '4 '12 plus '114 '228 units
2 '2 '3 '76 units
Theorem. Divide 2 by A so that you obtain B. Let a right paralelepiped
measure 2 by A by B units. How long is the diagonal of the volume?
Exactly A plus B units.
2 : 1 = 2 quadruple 2-1-2-3
2 : 1 '3 = 1 '2 multiply the numbers by a factor of 6
and you obtain the quadruple 12-8-9-17
2 : 1 '3 '15 = 1 '3 '15 '35 multiply the numbers by a factor of 35
and you obtain the quadruple 70-49-50-99
While the above triples approximate the square 1 by 1, diagonal sqrt2,
the quadruples approximate the right parallelepiped 2 by sqrt2 by sqrt2,
diagonal volume 2 sqrt2.
The most basic number column deserves more studies. Note for example that
the product of the second and third number in a line appears as first number
of a later line. One might say the richness of the square root of 2 unfolds
from the intitial algorithm.
The history of the square root of 2 might have begun 25,000 years ago in
Zaire when an African genius defined and measured squares in paces and found
a formula that was encoded in the group of 7 5 5 10 tally marks on the
Ishango bone

if the side of a square measures 5 paces
or a multiple thereof
the diagonal measures 7 paces
or a multiple thereof
and if the side of a square measures 7 paces
or a multiple thereof
the diagonal measures 5 5 sum 10 paces
or a multiple thereof

Later on someone wondered whether the numbers can be added as follows

side 5 plus 7 sum 12
diagonal 7 plus 10 sum 17
side 17 diagonal 12 12 sum 24 ?

Also this formula worked, and led sooner or alter to the number column

1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
169 239 338
408 577 816
985 1393 ....

The Djoser Pyramid used 1//12 and 24/17, the Great Pyramid 99/70 and 140/99;
ample indirect evidence for the use of the number column can be found in the
Rhind Mathematical Papyrus (also higher numbers defining octagons on the
second level, for advanced learners, while the apparent first level was meant
for beginners). The Babylonians used the simple value 17/12 = 1;25 and the
excellent value 1393/985 = 1;24,51,10,3,2,... in the abbreviated form of
1;24,51,10 (clay tablet YBC 7289). Later on the Greeks turned the number
column into a continued fraction; Eudoxus reduced the number column to a
ladder of two numbers per line that are generated by a different algorithm
(1 1, 2 3, 5 7, 12 17, ...); and the average of the mirror values offered by
each line of the original number column and by other number columns as well
(e.g. 7/5 and 10/7, average 99/70) offers a much faster approximation.

If mathematical education begun in the same experimental way and proceeded
from simple to more elaborate formulae, children would get a basic under-
standing of mathematics (that may prevent them form becoming kooks later on
in their life, trying to make sense of what they had to learn on an abstract
level and never really understood - comment for sci.math where this message
appeared first).

Begin with a pleasure walk on a gentle foothill before climbing the mountain.
d***@gmail.com
2017-05-02 14:13:20 UTC
Reply
Permalink
You might of interest Brian Pellar's 2014 paper on origin of Astro-Alphabet, it includes 22/7 ratio via zodiac symbols which were alphabet couplets, eg. Taurus = alef/ox head + bet/body & tail.
- - -
DD'eDeN
Franz Gnaedinger
2017-05-15 06:31:51 UTC
Reply
Permalink
Post by Franz Gnaedinger
The history of the square root of 2 might have begun 25,000 years ago in
Zaire when an African genius defined and measured squares in paces and found
a formula that was encoded in the group of 7 5 5 10 tally marks on the
Ishango bone
if the side of a square measures 5 paces
or a multiple thereof
the diagonal measures 7 paces
or a multiple thereof
and if the side of a square measures 7 paces
or a multiple thereof
the diagonal measures 5 5 sum 10 paces
or a multiple thereof
Later on someone wondered whether the numbers can be added as follows
side 5 plus 7 sum 12
diagonal 7 plus 10 sum 17
side 17 diagonal 12 12 sum 24 ?
Also this formula worked, and led sooner or alter to the number column
1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
169 239 338
408 577 816
985 1393 ....
The Djoser Pyramid used 1//12 and 24/17, the Great Pyramid 99/70 and 140/99;
ample indirect evidence for the use of the number column can be found in the
Rhind Mathematical Papyrus (also higher numbers defining octagons on the
second level, for advanced learners, while the apparent first level was meant
for beginners). The Babylonians used the simple value 17/12 = 1;25 and the
excellent value 1393/985 = 1;24,51,10,3,2,... in the abbreviated form of
1;24,51,10 (clay tablet YBC 7289). Later on the Greeks turned the number
column into a continued fraction; Eudoxus reduced the number column to a
ladder of two numbers per line that are generated by a different algorithm
(1 1, 2 3, 5 7, 12 17, ...); and the average of the mirror values offered by
each line of the original number column and by other number columns as well
(e.g. 7/5 and 10/7, average 99/70) offers a much faster approximation.
If mathematical education begun in the same experimental way and proceeded
from simple to more elaborate formulae, children would get a basic under-
standing of mathematics (that may prevent them form becoming kooks later on
in their life, trying to make sense of what they had to learn on an abstract
level and never really understood - comment for sci.math where this message
appeared first).
Begin with a pleasure walk on a gentle foothill before climbing the mountain.
daud..., thank you for the information regarding Brian Pellar. Unfortunately
I could not read his online pages, for they got blurred. In sci.lang his
thesis of the zodiac having inspired the alphabet convinced nobody, yet
it remined me of an old apperçu of mine wich I took up and developed further
in this message that involves a couple of word reconstructions:

from sunrise to sunrise (Ionic alphabet mirroring a special day)

Milet on the Ionian coast of Anatolia was the home of important mathematicians,
astronomers, and natural philosophers. Imagine that one of them developed the
Ionic alphabet some 2,700 years ago, 24 letters that also had numerical vaules,
from Alpha = 1 via via Thaeta = 8 and Pi = 16 to Omega = 24, and that he tried
to synchronize the 24 letters with the 24 hours of a special day, summer
solstice, June 21 in our calendar, 24 hours from June 21 at 4 o'clock
to June 22 at 4 o'clock

June 21, early morning, 4 o'clock, beginning of hour 1, hour Alpha (a)
June 21, midday, 12 o'clock, end of hour 8, hour Thaeta (th)
June 21, evening, 20 o'clock, end of hour 16, hour Pi (p)
June 22, early morning, 4 o'clock, end of hour 24, hour Omega (long o)

June 21, early morning, 4 o'clock, beginning of hour 1 or hour Alpha.
Just before dawn, bright Aldebaran hovers above the eastern horizon,
main star in Taurus 'Bull', from TOR for bull in motion. Aleph then Alpha
represented the head of an ox (that can still be seen when you turn our A
upside down). This might originally have been a reference to Aldebaran,
and to Baal as golden sun calf rising from the tree of life in the morning.
The Minoan Baal was later identified with Zeus. Aldebaran and the Pleiads
were the Golden Gate of Babylonian astronomy, passed by sun and moon
and planets.

June 21, midday, 12 o'clock, end of hour 8 or hour Thaeta. Now the sun reaches
the highest point in the sky, not only the one of the day but also of the year.
In some cases the Thaeta was given as a tiny circle with a central dot, which
also was the hieroglyph of the supreme Egyptian god Ra who manifested himself
in the solar disc. In most other cases, the early Thaeta was a tiny circle
with an inscribed cross, together known as ringcross, according to my studies
the emblem of the supreme sky and weather god. TYR means to overcome in
the double sense of rule and give, emphatic Middle Helladic Sseyr (Phaistos
Disc, Derk Ohlenroth) Doric Sseus (Wilhlem Larfeld) Homeric Zeus. PAS means
everywhere (in a plain), here, south and north of me, east and west of me,
in all five places, wherefrom Greek pas pan 'all, every' and pente penta- 'five'.
TYR PAS named the supreme sky and weather god who overcame everybody
everywhere in weather and time that rule our lives but are also given to us
so that we make the best of them. TYR PAS would have been visualized by
the ringcross of the Bronze Age. A strongly polished version became French
temps 'weather, time', TYR PAS English time and PAS TYR English weather.
Greek theos 'god' begins with a Thaeta and derived from DhAG meaning able,
good in the sense of able. Zeus was a most able one, supreme god in the
Greek pantheon.

June 21, evening, 20 o'clock, end of hour 16 or hour Pi. The sun disappeared
below the western horizon. One version of the early Pi that survived in the
classical Greek alphabet can be seen as a high narrow gate, so the setting
sun passed an imaginary gate in disappearing. Pi from periphaeres 'circular,
round' wherefrom English periphery named the circumference and then also
the number of the circle, here the circle of the 24 hours that were mirrored
in the Ionic alphabet of 24 letters.

June 22, early morning, end of hour 24 or hour Omega. In the archaic alphabet
from Thera, 7th century BC, the long o was given as a circle with a central
dot, hieroglyph of Ra in Egypt, whereupon the Omega, apparently invented
in Milet, evokes the solar disc rising from the horizon, and the lower case
the solar disc on the horns of Hathor, Egyptian goddess in her emanation
of the heavenly cow (especially on an Egyptian standard). The wife of Zeus
had been cow-eyed Hera, descendant of the Divine Hind from Altamira who
called life into existence, also moon bulls, thus creating time, lunations
or synodic months, periods of 30 29 30 29 30 29 30 29 30 29 30 ... days.
As Divine Hind Woman she appeared in Orion below Aldebaran where the
moon bulls waited to go on their heavenly mission, while the sun passes
the Golden Gate framed by Aldebaran and the Pleiads.

From Alpha to Omega, from dawn to dawn, from sunrise to sunrise,
while the letters of the Ionic alphabet allowed to tell what happens in a day,
and from day to day ... Is it a coincidence that the Iliad and Odyssey have
each 24 books?

Now let us consider the numbers 1 8 16 24. As beginning of hour 1 (early
morning on the summer solstice) and end of hour 8 (midday) and end of
hour 16 (later evening) and end of hour 24 (early morning again) they
divide a cycle of 24 hours into 8 plus 8 plus 8 hours. Another arrangement
of the same numbers makes them grow beyond a day in a pattern that
can be prolonged

1 = 1 x 1
1 + 8 = 3 x 3
1 + 8 + 16 = 5 x 5
1 + 8 + 16 + 24 = 7 x 7

We have then a circular movement in the hours of a day mirrored in the
Ionic alphabet, and a linear one in a sequence of days and a written text.

As for the early morning of the summer solstice, also the glorious rotunda
of Lascaux represents that moment of the year, the red mare rising above
the ledge symbolizing the midsummer sun rising above the horizon, and
the proud white bull by her side a full moon occurring at the same time,
ideal start of an eight-year period in the lunisolar calendar of Lascaux.
Franz Gnaedinger
2017-06-02 07:01:16 UTC
Reply
Permalink
Post by Franz Gnaedinger
daud..., thank you for the information regarding Brian Pellar. Unfortunately
I could not read his online pages, for they got blurred. In sci.lang his
thesis of the zodiac having inspired the alphabet convinced nobody, yet
it remined me of an old apperçu of mine wich I took up and developed further
from sunrise to sunrise (Ionic alphabet mirroring a special day)
Milet on the Ionian coast of Anatolia was the home of important mathematicians,
astronomers, and natural philosophers. Imagine that one of them developed the
Ionic alphabet some 2,700 years ago, 24 letters that also had numerical vaules,
from Alpha = 1 via via Thaeta = 8 and Pi = 16 to Omega = 24, and that he tried
to synchronize the 24 letters with the 24 hours of a special day, summer
solstice, June 21 in our calendar, 24 hours from June 21 at 4 o'clock
to June 22 at 4 o'clock
June 21, early morning, 4 o'clock, beginning of hour 1, hour Alpha (a)
June 21, midday, 12 o'clock, end of hour 8, hour Thaeta (th)
June 21, evening, 20 o'clock, end of hour 16, hour Pi (p)
June 22, early morning, 4 o'clock, end of hour 24, hour Omega (long o)
June 21, early morning, 4 o'clock, beginning of hour 1 or hour Alpha.
Just before dawn, bright Aldebaran hovers above the eastern horizon,
main star in Taurus 'Bull', from TOR for bull in motion. Aleph then Alpha
represented the head of an ox (that can still be seen when you turn our A
upside down). This might originally have been a reference to Aldebaran,
and to Baal as golden sun calf rising from the tree of life in the morning.
The Minoan Baal was later identified with Zeus. Aldebaran and the Pleiads
were the Golden Gate of Babylonian astronomy, passed by sun and moon
and planets.
June 21, midday, 12 o'clock, end of hour 8 or hour Thaeta. Now the sun reaches
the highest point in the sky, not only the one of the day but also of the year.
In some cases the Thaeta was given as a tiny circle with a central dot, which
also was the hieroglyph of the supreme Egyptian god Ra who manifested himself
in the solar disc. In most other cases, the early Thaeta was a tiny circle
with an inscribed cross, together known as ringcross, according to my studies
the emblem of the supreme sky and weather god. TYR means to overcome in
the double sense of rule and give, emphatic Middle Helladic Sseyr (Phaistos
Disc, Derk Ohlenroth) Doric Sseus (Wilhlem Larfeld) Homeric Zeus. PAS means
everywhere (in a plain), here, south and north of me, east and west of me,
in all five places, wherefrom Greek pas pan 'all, every' and pente penta- 'five'.
TYR PAS named the supreme sky and weather god who overcame everybody
everywhere in weather and time that rule our lives but are also given to us
so that we make the best of them. TYR PAS would have been visualized by
the ringcross of the Bronze Age. A strongly polished version became French
temps 'weather, time', TYR PAS English time and PAS TYR English weather.
Greek theos 'god' begins with a Thaeta and derived from DhAG meaning able,
good in the sense of able. Zeus was a most able one, supreme god in the
Greek pantheon.
June 21, evening, 20 o'clock, end of hour 16 or hour Pi. The sun disappeared
below the western horizon. One version of the early Pi that survived in the
classical Greek alphabet can be seen as a high narrow gate, so the setting
sun passed an imaginary gate in disappearing. Pi from periphaeres 'circular,
round' wherefrom English periphery named the circumference and then also
the number of the circle, here the circle of the 24 hours that were mirrored
in the Ionic alphabet of 24 letters.
June 22, early morning, end of hour 24 or hour Omega. In the archaic alphabet
from Thera, 7th century BC, the long o was given as a circle with a central
dot, hieroglyph of Ra in Egypt, whereupon the Omega, apparently invented
in Milet, evokes the solar disc rising from the horizon, and the lower case
the solar disc on the horns of Hathor, Egyptian goddess in her emanation
of the heavenly cow (especially on an Egyptian standard). The wife of Zeus
had been cow-eyed Hera, descendant of the Divine Hind from Altamira who
called life into existence, also moon bulls, thus creating time, lunations
or synodic months, periods of 30 29 30 29 30 29 30 29 30 29 30 ... days.
As Divine Hind Woman she appeared in Orion below Aldebaran where the
moon bulls waited to go on their heavenly mission, while the sun passes
the Golden Gate framed by Aldebaran and the Pleiads.
From Alpha to Omega, from dawn to dawn, from sunrise to sunrise,
while the letters of the Ionic alphabet allowed to tell what happens in a day,
and from day to day ... Is it a coincidence that the Iliad and Odyssey have
each 24 books?
Now let us consider the numbers 1 8 16 24. As beginning of hour 1 (early
morning on the summer solstice) and end of hour 8 (midday) and end of
hour 16 (later evening) and end of hour 24 (early morning again) they
divide a cycle of 24 hours into 8 plus 8 plus 8 hours. Another arrangement
of the same numbers makes them grow beyond a day in a pattern that
can be prolonged
1 = 1 x 1
1 + 8 = 3 x 3
1 + 8 + 16 = 5 x 5
1 + 8 + 16 + 24 = 7 x 7
We have then a circular movement in the hours of a day mirrored in the
Ionic alphabet, and a linear one in a sequence of days and a written text.
As for the early morning of the summer solstice, also the glorious rotunda
of Lascaux represents that moment of the year, the red mare rising above
the ledge symbolizing the midsummer sun rising above the horizon, and
the proud white bull by her side a full moon occurring at the same time,
ideal start of an eight-year period in the lunisolar calendar of Lascaux.
Why was the King's chamber in the Great Pyramid left undecorated? Well,
the beauty and transformating power lies in the numbers. Jean-Philippe Lauer
(whom nobody can call a 'pyramidiot') discovered the Sacred Triangle 3-4-5
in the form of 15-20-25 royal cubits in the measurements of that chamber

width by length 10 by 20 royal cubits

diagonal of the short wall 15 royal cubits
length of the chamber 20 royal cubits
diagonal of the volume 25 royal cubits

Now the Sacred Triangle was the key to the Egyptian systematic method of
calculating the circle httw://www.seshat.ch/home/aim.pdf

The basis of the Great Pyramid measured originally 440 royal cubits.
Transform the cross-section into a circle of the same area. Using 22/7
for pi (first value gained by the above method when considering that the
arcs of the circle are slightly longer than the sides of the inscribed
polygons) we obtain 280 royal cubits for the diameter, which was the
original height of the Great Pyramid. And this is of symbolic meaning.
The pyramid was considered the body of the deified king. Transforming
the defining triangle into a circle of the same area was to make the king
accompany the supreme god Ra who manifested himself in the solar disc ...
Franz Gnaedinger
2017-06-12 06:35:40 UTC
Reply
Permalink
Post by Franz Gnaedinger
Why was the King's chamber in the Great Pyramid left undecorated? Well,
the beauty and transformating power lies in the numbers. Jean-Philippe Lauer
(whom nobody can call a 'pyramidiot') discovered the Sacred Triangle 3-4-5
in the form of 15-20-25 royal cubits in the measurements of that chamber
width by length 10 by 20 royal cubits
diagonal of the short wall 15 royal cubits
length of the chamber 20 royal cubits
diagonal of the volume 25 royal cubits
Now the Sacred Triangle was the key to the Egyptian systematic method of
calculating the circle httw://www.seshat.ch/home/aim.pdf
The basis of the Great Pyramid measured originally 440 royal cubits.
Transform the cross-section into a circle of the same area. Using 22/7
for pi (first value gained by the above method when considering that the
arcs of the circle are slightly longer than the sides of the inscribed
polygons) we obtain 280 royal cubits for the diameter, which was the
original height of the Great Pyramid. And this is of symbolic meaning.
The pyramid was considered the body of the deified king. Transforming
the defining triangle into a circle of the same area was to make the king
accompany the supreme god Ra who manifested himself in the solar disc ...
royal cubit and Horus cubit (measuring out the base of the Great Pyramid)

The base of the Great Pyramid measured 440 royal cubits or 230.336 meters,
yielding 52.34909... centimeters for one royal cubit. For the sake of
simplicity I use 52.36 cm (a value confirmed to me by Rainer Stadelmann
in a letter from 1992).

Combine the royal cubit with a Horus cubit so that

7 royal cubits are 11 Horus cubits

and you get a Horus cubit of 33.32 cm, corresponding to the length of
a kestrel from beak to tail.

Picture a model of the Great Pyramid: base 1 by 1 royal cubit, height
1 Horus cubit.

If the diameter of a circle measures 1 royal cubit, the circumference
2 royal cubits; and if the radius measures 1 Horus cubit, the area
1 royal cubit by 2 Horus cubits.

If the side of a square measures 10 Horus cubits, the diagonal 9 royal cubits;
and if the side measures 9 royal cubits, the diagonal 20 Horus cubits.

The latter formula would have been used for measuring out the square of
the base of the Great Pyramid around a large limestone hill that according
to Rainer Stadelmann might account for one third of the pyramd's volume.

Make a wooden frame in the shape of a square with the cross of the diagonals
and fix four protruding nails to the corners in the lateral distances of
9 royal cubits and diagonal distances of 20 Horus cubits. Place four wooden
blocks in the same distances from each other on the ground and mark their
tops of a softer wood with the nails of the wooden frame. Turn the frame by
90 degrees and check whether the nails match the nail impressions of the
former step. If not, correct the nails.

Then lay out a grid of leveled wooden blocks in the same distances around
the limestone hill, and mark the grid with nail impressions, always turning
the wooden frame and checking whether the nails fit in the previous nail
impressions. Thus you can achieve the amazing exactness observed in the
base: mistake of the northern side 1.2 cm and of the southern side 3.2 cm;
mistakes of the angles in the northwestern corner 1" (one single second!),
northeastern corner 58", southeastern corner 29", southwestern corner 16".

No mathe-free method could have achieved such a precision.
Franz Gnaedinger
2017-08-07 07:03:43 UTC
Reply
Permalink
Post by Franz Gnaedinger
royal cubit and Horus cubit (measuring out the base of the Great Pyramid)
The base of the Great Pyramid measured 440 royal cubits or 230.336 meters,
yielding 52.34909... centimeters for one royal cubit. For the sake of
simplicity I use 52.36 cm (a value confirmed to me by Rainer Stadelmann
in a letter from 1992).
Combine the royal cubit with a Horus cubit so that
7 royal cubits are 11 Horus cubits
and you get a Horus cubit of 33.32 cm, corresponding to the length of
a kestrel from beak to tail.
Picture a model of the Great Pyramid: base 1 by 1 royal cubit, height
1 Horus cubit.
If the diameter of a circle measures 1 royal cubit, the circumference
2 royal cubits; and if the radius measures 1 Horus cubit, the area
1 royal cubit by 2 Horus cubits.
If the side of a square measures 10 Horus cubits, the diagonal 9 royal cubits;
and if the side measures 9 royal cubits, the diagonal 20 Horus cubits.
The latter formula would have been used for measuring out the square of
the base of the Great Pyramid around a large limestone hill that according
to Rainer Stadelmann might account for one third of the pyramd's volume.
Make a wooden frame in the shape of a square with the cross of the diagonals
and fix four protruding nails to the corners in the lateral distances of
9 royal cubits and diagonal distances of 20 Horus cubits. Place four wooden
blocks in the same distances from each other on the ground and mark their
tops of a softer wood with the nails of the wooden frame. Turn the frame by
90 degrees and check whether the nails match the nail impressions of the
former step. If not, correct the nails.
Then lay out a grid of leveled wooden blocks in the same distances around
the limestone hill, and mark the grid with nail impressions, always turning
the wooden frame and checking whether the nails fit in the previous nail
impressions. Thus you can achieve the amazing exactness observed in the
base: mistake of the northern side 1.2 cm and of the southern side 3.2 cm;
mistakes of the angles in the northwestern corner 1" (one single second!),
northeastern corner 58", southeastern corner 29", southwestern corner 16".
No mathe-free method could have achieved such a precision.
Why was the sarcophagus in the Great Pyramid left undecorated? Its beauty
and transformational powers lie again in the numbers.

The royal cubit of the Great Pyramid was combined with seven complementary
measures I call Horus cubits, for they correspond to the length of a kestrel
from beak to tail and vary around some 32 centimeters, most important Horus
cubit A measuring 33.32 cm, defined as 7/11 of a royal cubit. All seven
Horus cubits are defined by relatively small-numbered ratios and help solve
demanding geometrical problems.

The lid of the sarcophagus of rose granite is missing, while the surviving
tub is not measured in royal cubits, which made Rainer Stadelmann wonder.
He gives the exact measurements of the tub. They are amazingly precise
multiples of four of the seven hypothetical Horus cubits (outer and inner
widths and lengths, while the outer and inner heights and the diagonal of
the inner space depend on the lost lid)

http://www.seshat.ch/home/egypt4.htm

For the tub covered by the lid I propose a double (outer and inner) right parallelepiped of these numbers, the seven Horus cubits A B C D E F G
drawn together into an ideal Horus cubit Hc

outer width 3 Hc
outer height 4 Hc
diagonal 5 Hc
according to the triple 3-4-5
outer length 7 Hc
thickness of stone 1/2 Hc
inner width 2 Hc
inner height 3 Hc
inner length 6 Hc
diagonal inner space 7 Hc
according to the quadruple 2-3-6-7
Franz Gnaedinger
2017-10-31 07:20:56 UTC
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Post by Franz Gnaedinger
Why was the sarcophagus in the Great Pyramid left undecorated? Its beauty
and transformational powers lie again in the numbers.
The royal cubit of the Great Pyramid was combined with seven complementary
measures I call Horus cubits, for they correspond to the length of a kestrel
from beak to tail and vary around some 32 centimeters, most important Horus
cubit A measuring 33.32 cm, defined as 7/11 of a royal cubit. All seven
Horus cubits are defined by relatively small-numbered ratios and help solve
demanding geometrical problems.
The lid of the sarcophagus of rose granite is missing, while the surviving
tub is not measured in royal cubits, which made Rainer Stadelmann wonder.
He gives the exact measurements of the tub. They are amazingly precise
multiples of four of the seven hypothetical Horus cubits (outer and inner
widths and lengths, while the outer and inner heights and the diagonal of
the inner space depend on the lost lid)
http://www.seshat.ch/home/egypt4.htm
For the tub covered by the lid I propose a double (outer and inner) right parallelepiped of these numbers, the seven Horus cubits A B C D E F G
drawn together into an ideal Horus cubit Hc
outer width 3 Hc
outer height 4 Hc
diagonal 5 Hc
according to the triple 3-4-5
outer length 7 Hc
thickness of stone 1/2 Hc
inner width 2 Hc
inner height 3 Hc
inner length 6 Hc
diagonal inner space 7 Hc
according to the quadruple 2-3-6-7
Rainer Stadelmann wonders why the sarcophagus tub of rose granite in the
King's Chamber of the Great Pyramid is not measured in royal cubits.

He gives this outer length of the tub: 227.6 centimeters. In a letter from
1992 he confirmed my value of 52.36 centimeters for the royal cubit of the
Great Pyramid. So we have for the length of the tub 4.346... royal cubits.
No integer.

I propose a set of complementary Horus cubits A B C D E F G. They vary about
the length of a kestrel (from the tip of the beak to the end of the tail).
Horus cubit D is defined as follows

18 royal cubits are 29 Horus cubits D

Horus cubit D = 32.499... centimeters

7 Horus cubits D = 227.495 centimeters
outer length of tub 227.6 cm (Stadelmann)
difference about one millimeter

Now we have an integer: the outer length of the tub measures 7 Horus cubits D.

Each of the seven Horus cubits allows at least one geometrical application,
in the given case a fine definition of the circle

radius 10 Horus cubits D
circumference 39 royal cubits

implicit pi value 377/120 or 3 '8 '60 or 3 '10 '24

This excellent value can be found via the Egyptian method of calculating
the circle, and, simpler, via the most important pi sequence

3/1 (plus 22/7) 25/8 47/15 69/22 ... 377/120

The sarcophagus remained undecorated. Its magic and transformational power
lie in the numbers.
Franz Gnaedinger
2017-11-02 08:21:55 UTC
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Post by Franz Gnaedinger
Rainer Stadelmann wonders why the sarcophagus tub of rose granite in the
King's Chamber of the Great Pyramid is not measured in royal cubits.
He gives this outer length of the tub: 227.6 centimeters. In a letter from
1992 he confirmed my value of 52.36 centimeters for the royal cubit of the
Great Pyramid. So we have for the length of the tub 4.346... royal cubits.
No integer.
I propose a set of complementary Horus cubits A B C D E F G. They vary about
the length of a kestrel (from the tip of the beak to the end of the tail).
Horus cubit D is defined as follows
18 royal cubits are 29 Horus cubits D
Horus cubit D = 32.499... centimeters
7 Horus cubits D = 227.495 centimeters
outer length of tub 227.6 cm (Stadelmann)
difference about one millimeter
Now we have an integer: the outer length of the tub measures 7 Horus cubits D.
Each of the seven Horus cubits allows at least one geometrical application,
in the given case a fine definition of the circle
radius 10 Horus cubits D
circumference 39 royal cubits
implicit pi value 377/120 or 3 '8 '60 or 3 '10 '24
This excellent value can be found via the Egyptian method of calculating
the circle, and, simpler, via the most important pi sequence
3/1 (plus 22/7) 25/8 47/15 69/22 ... 377/120
The sarcophagus remained undecorated. Its magic and transformational power
lie in the numbers.
Now for the outer breadth of the sarcophagus tub in the King's Chamber
of the Great Pyramid. Rainer Stadelmann gives 98.7 centimeters which
I transform into 3 Horus cubits E

22 royal cubits are 35 Horus cubits E

Horus cubit E = 32.912 centimeters

3 Horus cubits E = 98.736 centimeters
outer breadth of sarcophagus tub 98.7 cm
difference less than half a millimeter

diameter of a circle 1 royal cubit
circumference 5 Horus cubits E
implicit pi value 22/7 or 3 '7
4/1 (plus 3/1) 7/2 10£/3 13/4 16/5 19/6 22/7

side of a square 4 royal cubits
diagonal 9 Horus cubits E
implicit value for the square root of 2
99/70 or 1 '3 '15 '70

side of a square 4 royal cubits
diagonal 9 Horus cubits E
implicit value for the square root of 2
140/99 or 1 '3 '22 '33 '198

99/70 and 140/99 are mirror values from the basic number column approximating
the square root of 2, first lines

1 1 2, 2 3 4, 5 7 10, 12 17 24, 29 41 58, 70 99 140
Franz Gnaedinger
2017-11-08 08:15:21 UTC
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Post by Franz Gnaedinger
Now for the outer breadth of the sarcophagus tub in the King's Chamber
of the Great Pyramid. Rainer Stadelmann gives 98.7 centimeters which
I transform into 3 Horus cubits E
22 royal cubits are 35 Horus cubits E
Horus cubit E = 32.912 centimeters
3 Horus cubits E = 98.736 centimeters
outer breadth of sarcophagus tub 98.7 cm
difference less than half a millimeter
diameter of a circle 1 royal cubit
circumference 5 Horus cubits E
implicit pi value 22/7 or 3 '7
4/1 (plus 3/1) 7/2 10£/3 13/4 16/5 19/6 22/7
side of a square 4 royal cubits
diagonal 9 Horus cubits E
implicit value for the square root of 2
99/70 or 1 '3 '15 '70
side of a square 4 royal cubits
diagonal 9 Horus cubits E
implicit value for the square root of 2
140/99 or 1 '3 '22 '33 '198
99/70 and 140/99 are mirror values from the basic number column approximating
the square root of 2, first lines
1 1 2, 2 3 4, 5 7 10, 12 17 24, 29 41 58, 70 99 140
The inner length of the sarcophagus tub in the King's Chamber of the Great
Pyramid measures 198.3 centimeters (Rainer Stadelmann) or 6 Horus cubits B

12 royal cubits are 19 Horus cubits B

1 Horus cubit B = 33.069... centimeters

6 Horus cubits B are 198.416... cm
inner length of tub 198.3 cm (Stadelmann)
difference little more than one millimeter

a rectangle measures 1 by 3 Horus cubits B
diagonal 2 royal cubits
pseudo-triple 6-18-19
19/6 for the square root of 10

The inner width of the tub measures 2 Horus cubits C. If it measured 2 Horus
cubits B instead, the diagonal measured 4 royal cubits.

rectangle 1 by 3 royal cubits
diagonal 5 Horus cubits B
pseudo-triple 19-57-60
60/19 for the square root of 10
mirror value of 19/6 above

The inner width of the tub measures 68.1 centimeters (Rainer Stadelmann)
or 2 Horus cubits C

13 royal cubits are 20 Horus cubits C

1 Horus cubit C = 34.034 centimeters

2 Horus cubits C are 68.068 cm
inner width of tub 68.1 cm (Stadelmann)
difference one third of a millimeter

radius of a circle 3 royal cubits
circumference 29 Horus cubits C
implicit pi value 377/120

3/1 (plus 22/7) 25/8 47/15 69/22 ... 311/99 ... 377/120
Franz Gnaedinger
2017-11-09 08:26:07 UTC
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Post by Franz Gnaedinger
The inner length of the sarcophagus tub in the King's Chamber of the Great
Pyramid measures 198.3 centimeters (Rainer Stadelmann) or 6 Horus cubits B
12 royal cubits are 19 Horus cubits B
1 Horus cubit B = 33.069... centimeters
6 Horus cubits B are 198.416... cm
inner length of tub 198.3 cm (Stadelmann)
difference little more than one millimeter
a rectangle measures 1 by 3 Horus cubits B
diagonal 2 royal cubits
pseudo-triple 6-18-19
19/6 for the square root of 10
The inner width of the tub measures 2 Horus cubits C. If it measured 2 Horus
cubits B instead, the diagonal measured 4 royal cubits.
rectangle 1 by 3 royal cubits
diagonal 5 Horus cubits B
pseudo-triple 19-57-60
60/19 for the square root of 10
mirror value of 19/6 above
The inner width of the tub measures 68.1 centimeters (Rainer Stadelmann)
or 2 Horus cubits C
13 royal cubits are 20 Horus cubits C
1 Horus cubit C = 34.034 centimeters
2 Horus cubits C are 68.068 cm
inner width of tub 68.1 cm (Stadelmann)
difference one third of a millimeter
radius of a circle 3 royal cubits
circumference 29 Horus cubits C
implicit pi value 377/120
3/1 (plus 22/7) 25/8 47/15 69/22 ... 311/99 ... 377/120
(follow two more messages on the sarcophagus, the one below and the next one;
then we shall have a look at the Grand Gallery and the recently discovered
'great void')

The inner height of the sarcophagus tub measures 87.4 centimeters (Rainer
Stadelmann) or 21/8 Horus cubits A

7 royal cubits are 11 Horus cubits A

1 Horus cubit A = 33.32 cm

21/8 Horus cubits A = 87.465 centimeters
inner height of tub 87.4 cm (Stadelmann)
difference less than one millimeter

radius of a circle 1 Horus cubit A
circumference 2 royal cubits
area 1 royal cubit by 2 Horus cubits A
implicit pi value 22/7

side of a square 10 Horus cubits A or 9 royal cubits
diagonal 9 royal cubits or 20 Horus cubits A respectively
99/70 and 140/99 for the square root of 2 respectively

1 1 2, 2 3 4, 5 7 10, 12 17 24, 29 41 58, 70 99 140

model of a pyramid
base 1 by 1 royal cubit
height 1 Horus cubit A

The inner height of the closed sarcophagus - tub and lost lid - would have
measured 3 Horus cubits A or 99.96 centimeters.

The outer height of the tub measures 105.1 centimeters (Rainer Stadelmann)
or 22/7 Horus cubits F

23 royal cubits are 36 Horus cubits F

1 Horus cubit F = 33.452... centimeters

22/7 Horus cubits F = 105.135... cm
outer height of tub 105.1 cm (Stadelmann)
difference less than half a millimeter

rectangle 2 by 4 royal cubits
diagonal 7 Horus cubits F
pseudo-triple 72-144-168

rectangle 7 by 14 Horus cubits F
diagonal 10 royal cubits
pseudo-triple 161-322-360

mirror values 161/72 and 360/161 for the square root of 5

1 1 5, 2 6 10, 1 3 5, 4 8 20, 2 4 10, 1 2 5, 3 7 15 ... 72 161 360

For the outer height of the closed sarcophagus - tub and lid - we may
assume 4 Horus cubits F or 133.808... centimeters.
Franz Gnaedinger
2017-11-10 07:46:24 UTC
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Post by Franz Gnaedinger
(follow two more messages on the sarcophagus, the one below and the next one;
then we shall have a look at the Grand Gallery and the recently discovered
'great void')
The inner height of the sarcophagus tub measures 87.4 centimeters (Rainer
Stadelmann) or 21/8 Horus cubits A
7 royal cubits are 11 Horus cubits A
1 Horus cubit A = 33.32 cm
21/8 Horus cubits A = 87.465 centimeters
inner height of tub 87.4 cm (Stadelmann)
difference less than one millimeter
radius of a circle 1 Horus cubit A
circumference 2 royal cubits
area 1 royal cubit by 2 Horus cubits A
implicit pi value 22/7
side of a square 10 Horus cubits A or 9 royal cubits
diagonal 9 royal cubits or 20 Horus cubits A respectively
99/70 and 140/99 for the square root of 2 respectively
1 1 2, 2 3 4, 5 7 10, 12 17 24, 29 41 58, 70 99 140
model of a pyramid
base 1 by 1 royal cubit
height 1 Horus cubit A
The inner height of the closed sarcophagus - tub and lost lid - would have
measured 3 Horus cubits A or 99.96 centimeters.
The outer height of the tub measures 105.1 centimeters (Rainer Stadelmann)
or 22/7 Horus cubits F
23 royal cubits are 36 Horus cubits F
1 Horus cubit F = 33.452... centimeters
22/7 Horus cubits F = 105.135... cm
outer height of tub 105.1 cm (Stadelmann)
difference less than half a millimeter
rectangle 2 by 4 royal cubits
diagonal 7 Horus cubits F
pseudo-triple 72-144-168
rectangle 7 by 14 Horus cubits F
diagonal 10 royal cubits
pseudo-triple 161-322-360
mirror values 161/72 and 360/161 for the square root of 5
1 1 5, 2 6 10, 1 3 5, 4 8 20, 2 4 10, 1 2 5, 3 7 15 ... 72 161 360
For the outer height of the closed sarcophagus - tub and lid - we may
assume 4 Horus cubits F or 133.808... centimeters.
(sarcophagus and King's Chamber)

For the outer and inner measurements of the sarcophagus in the King's Chamber
we found these numbers

3 Horus cubits E x 4 Horus cubits F x 7 Horus cubits G
2 Horus cubits C x 3 Horus cubits A x 6 Horus cubits B

The Horus cubits vary around the length of a kestrel (beak to tail). If we
ignore the relatively small differences and simply speak of a Horus cubit
we obtain this definition

3 by 4 by 7 Horus cubits (outside)
2 by 3 by 6 Horus cubits (inside)
inner diagonal 7 Horus cubits
quadruple 2-3-6-7

The small faces of the closed sarcophagus (tub and lid) would have a height
of 4 and a breadth of 3 Horus cubits, diagonal 5 Horus cubits, triple 3-4-5.

Jean-Philippe Lauer discovered the Sacred Triangle 15-20-25 royal cubits
in the King's Chamber of the Great Pyramid

diagonal of a small wall 15 rc
length of chamber 20 rc
diagonal volume 25 rc

The Sacred Triangle 3-4-5 or 15-20-25 is the key to the Egyptian systematic
method of calculating the circle.

The length of the sarcophagus measures 7 Horus cubits G, while the height
of the King's Chamber measures 18 Horus cubits G

41 royal cubits are 66 Horus cubits G

1 Horus cubit G = 32.526... centimeters

7 Horus cubits G = 227.686... centimeters
length of sarcophagus tub 227.6 cm (RS)

18 Horus cubits G = 585.48 centimeters
theoretical chamber height 585.4 centimeters
123/55 for the square root of 5

1 1 5, 2 6 10, 1 3 5, 4 8 20, 2 4 10, 1 2 5 ... 55 123 275

The combination of royal cubit (52.36 cm) and Horus cubits, most important
Horus cubit A (33.32 cm), works amazingly well and finds a confirmation
in the seemingly odd measurements of the sarcophagus tub in the King's
Chamber of the Great Pyramid.

Next time: Grand Gallery and 'great void'
Franz Gnaedinger
2017-11-11 08:43:16 UTC
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Post by Franz Gnaedinger
(sarcophagus and King's Chamber)
For the outer and inner measurements of the sarcophagus in the King's Chamber
we found these numbers
3 Horus cubits E x 4 Horus cubits F x 7 Horus cubits G
2 Horus cubits C x 3 Horus cubits A x 6 Horus cubits B
The Horus cubits vary around the length of a kestrel (beak to tail). If we
ignore the relatively small differences and simply speak of a Horus cubit
we obtain this definition
3 by 4 by 7 Horus cubits (outside)
2 by 3 by 6 Horus cubits (inside)
inner diagonal 7 Horus cubits
quadruple 2-3-6-7
The small faces of the closed sarcophagus (tub and lid) would have a height
of 4 and a breadth of 3 Horus cubits, diagonal 5 Horus cubits, triple 3-4-5.
Jean-Philippe Lauer discovered the Sacred Triangle 15-20-25 royal cubits
in the King's Chamber of the Great Pyramid
diagonal of a small wall 15 rc
length of chamber 20 rc
diagonal volume 25 rc
The Sacred Triangle 3-4-5 or 15-20-25 is the key to the Egyptian systematic
method of calculating the circle.
The length of the sarcophagus measures 7 Horus cubits G, while the height
of the King's Chamber measures 18 Horus cubits G
41 royal cubits are 66 Horus cubits G
1 Horus cubit G = 32.526... centimeters
7 Horus cubits G = 227.686... centimeters
length of sarcophagus tub 227.6 cm (RS)
18 Horus cubits G = 585.48 centimeters
theoretical chamber height 585.4 centimeters
123/55 for the square root of 5
1 1 5, 2 6 10, 1 3 5, 4 8 20, 2 4 10, 1 2 5 ... 55 123 275
The combination of royal cubit (52.36 cm) and Horus cubits, most important
Horus cubit A (33.32 cm), works amazingly well and finds a confirmation
in the seemingly odd measurements of the sarcophagus tub in the King's
Chamber of the Great Pyramid.
Next time: Grand Gallery and 'great void'
(Grand Gallery and 'great void')

The Grand Gallery leading to the King's Chamber measures 46.71 meters = 89
royal cubits and the angle 26 degrees 2 minuts 30 seconds according to Rainer
Stadelmann. From these numbers I deduced a pair of ideal models

slope 89 royal cubits or 46.6004 meters
run 80 royal cubits
rise 39 royal cubits
triple 39-80-89

slope 140 Horus cubits A or 46.648 meters
run 80 royal cubits
rise practically 39 1/5 royal cubits

44 x 44 oblique height (in rc/55)
44 x 49 rise
44 x 100 run
49 x 100 slope

average slope 46.624... meters
actual slope 46.71 meters (RS)

average angle 26 degrees 2 minuts 49.357... seconds
actual angle 26 degrees 2 minuts 30 seconds (RS)

Half the pyramid's base measured 220 and its height 280 royal cubits,
while the slope measured practically 356 royal cubits, according to
the golden sequence

4 4 8 12 20 32 52 84 136 220 356

or about 560 Horus cubits A - four times the length of the gallery.

If the recently discovered 'great void' above the gallery, 30 meters long,
ascends at about the same angle (as indicated by one of the two versions
provided by the Scan Pyramids team), then there is a mathematical solution
parallel to the one of the gallery

slope 56 royal cubits or 88 Horus cubits A or 29.3216 meters
tangent 1/2, angle 26 degrees 33 minutes 54.1784... seconds
rise 25.043... or practically 25 royal cubits
run 50.087... or practically 50 royal cubits

The pseudo-triple 25-50-56 goes along with a variant of the number column
approximating the square root of 5, here the first lines

1 4 5, 5 9 25, 14 34 70, 7 17 35 ... 25 56 125

(remember, a number column can start with any pair of numbers).

The height of the pyramid measured 280 royal cubits or 440 Horus cubits A
- five times the length of the 'void' in the above numbers.

One might call the Great Pyramid a mathematical poem.

Next time: symbolical aspect of the 'great void'
Franz Gnaedinger
2017-11-13 08:30:23 UTC
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Post by Franz Gnaedinger
(Grand Gallery and 'great void')
The Grand Gallery leading to the King's Chamber measures 46.71 meters = 89
royal cubits and the angle 26 degrees 2 minuts 30 seconds according to Rainer
Stadelmann. From these numbers I deduced a pair of ideal models
slope 89 royal cubits or 46.6004 meters
run 80 royal cubits
rise 39 royal cubits
triple 39-80-89
slope 140 Horus cubits A or 46.648 meters
run 80 royal cubits
rise practically 39 1/5 royal cubits
44 x 44 oblique height (in rc/55)
44 x 49 rise
44 x 100 run
49 x 100 slope
average slope 46.624... meters
actual slope 46.71 meters (RS)
average angle 26 degrees 2 minuts 49.357... seconds
actual angle 26 degrees 2 minuts 30 seconds (RS)
Half the pyramid's base measured 220 and its height 280 royal cubits,
while the slope measured practically 356 royal cubits, according to
the golden sequence
4 4 8 12 20 32 52 84 136 220 356
or about 560 Horus cubits A - four times the length of the gallery.
If the recently discovered 'great void' above the gallery, 30 meters long,
ascends at about the same angle (as indicated by one of the two versions
provided by the Scan Pyramids team), then there is a mathematical solution
parallel to the one of the gallery
slope 56 royal cubits or 88 Horus cubits A or 29.3216 meters
tangent 1/2, angle 26 degrees 33 minutes 54.1784... seconds
rise 25.043... or practically 25 royal cubits
run 50.087... or practically 50 royal cubits
The pseudo-triple 25-50-56 goes along with a variant of the number column
approximating the square root of 5, here the first lines
1 4 5, 5 9 25, 14 34 70, 7 17 35 ... 25 56 125
(remember, a number column can start with any pair of numbers).
The height of the pyramid measured 280 royal cubits or 440 Horus cubits A
- five times the length of the 'void' in the above numbers.
One might call the Great Pyramid a mathematical poem.
Next time: symbolical aspect of the 'great void'
(marriage of Nut and Geb)

The sky goddess Nut arched herself over the Earth and had an imaginary
presence in the Great Pyramid in form of the big inscribed hemisphere

Loading Image...

The radius equals the golden major of the pyramid's height, 173 of 280
royal cubits, or 272 of 440 Horus cubits A, according to the golden sequences

2 7 9 16 25 41 66 107 173 280

8 8 16 24 40 64 104 168 272 440

Nut carried the sun child in her womb

Loading Image...

The home of the sun child - young Ra, also the king reborn - might be the
hypothethical sun chamber on top of the imaginary hemisphere, 90.6 meters
above the base. Measurments of the sun chamber in Horus cubits A

length 20 (E-W)
width 16 (S-N)
lateral heights 12
middle height 15

Geb was the Earth god, husband of Nut, lying under her, sometimes with erected
phallus. Now if the 'great void' has a symbolical meaning beyond a mere static
function, it might represent Geb's phallus, or, less graphic, the marriage of
Nut and Geb. While the Grand Gallery leads to the King's chamber (sarcophagus)
the 'great void' might be a part of a hidden upway to the hypothetical sun
chamber (new life); length of the gallery one fourth of the pyramid's slope,
length of the 'void' one fifth of the pyramid's height.

Ra would emerge from the sun chamber in form of a big imaginary circle,
vertical diameter given by the pyramid's height, area by the one of the
defining triangle (base 440 royal cubits, height 440 Horus cubits A,
area triangle and circle each 220 royal cubits by 440 Horus cubits A).

The 'great void' may be the enabler setting in motion the mythological
mechanism of the Great Pyramid, leading from the big imaginary hemisphere
to the big imaginary circle, releasing sun and sky from the primeval hill,
and sending the reborn king on his heavenly mission in a sun bark of Ra.
Franz Gnaedinger
2018-05-08 06:48:32 UTC
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Post by Franz Gnaedinger
(marriage of Nut and Geb)
The sky goddess Nut arched herself over the Earth and had an imaginary
presence in the Great Pyramid in form of the big inscribed hemisphere
http://www.seshat.ch/home/heaven.GIF
The radius equals the golden major of the pyramid's height, 173 of 280
royal cubits, or 272 of 440 Horus cubits A, according to the golden sequences
2 7 9 16 25 41 66 107 173 280
8 8 16 24 40 64 104 168 272 440
Nut carried the sun child in her womb
http://www.seshat.ch/home/chamber1.GIF
The home of the sun child - young Ra, also the king reborn - might be the
hypothethical sun chamber on top of the imaginary hemisphere, 90.6 meters
above the base. Measurments of the sun chamber in Horus cubits A
length 20 (E-W)
width 16 (S-N)
lateral heights 12
middle height 15
Geb was the Earth god, husband of Nut, lying under her, sometimes with erected
phallus. Now if the 'great void' has a symbolical meaning beyond a mere static
function, it might represent Geb's phallus, or, less graphic, the marriage of
Nut and Geb. While the Grand Gallery leads to the King's chamber (sarcophagus)
the 'great void' might be a part of a hidden upway to the hypothetical sun
chamber (new life); length of the gallery one fourth of the pyramid's slope,
length of the 'void' one fifth of the pyramid's height.
Ra would emerge from the sun chamber in form of a big imaginary circle,
vertical diameter given by the pyramid's height, area by the one of the
defining triangle (base 440 royal cubits, height 440 Horus cubits A,
area triangle and circle each 220 royal cubits by 440 Horus cubits A).
The 'great void' may be the enabler setting in motion the mythological
mechanism of the Great Pyramid, leading from the big imaginary hemisphere
to the big imaginary circle, releasing sun and sky from the primeval hill,
and sending the reborn king on his heavenly mission in a sun bark of Ra.
No further chambers in the tomb of Tutenkhamun. But interesting news from
the pyramid El Castillo, Chichén Itzá, Mexico. It has been scanned with
the Electrical Resistance Tomography method and revealed a smaller and
a still smaller pyramid within. One day this method will also be applied
to the Great Pyramid, we can hope.
Franz Gnaedinger
2018-08-27 06:40:42 UTC
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Post by Franz Gnaedinger
No further chambers in the tomb of Tutenkhamun. But interesting news from
the pyramid El Castillo, Chichén Itzá, Mexico. It has been scanned with
the Electrical Resistance Tomography method and revealed a smaller and
a still smaller pyramid within. One day this method will also be applied
to the Great Pyramid, we can hope.
Having seen a tv documentary on the Pyramid Scan project I am happy to say
that the region of the hypothetical sun chamber has not (yet) been scanned:
floor 173 royal cubits or 90.6 meters above the center of the geometrical
base, on the zenith of the imaginary hemisphere that symbolizes the sky
once enclosed in the primeval hill, personified by the sky goddess Nut
arching herself over the earth, her husband Geb, the hypothetical sun chamber
symbolizing her womb wherein she carries the mythical sun child. The radius
of the imaginary hemisphere and floor height of the sun chamber, 173 rc,
is the golden major of the pyramid's height, 280 rc, and found by means
of a golden number sequence

9 16 25 41 66 107 173 280

A team of the Pyramid Scan project got a special permission to explore
the cavity in an edge of the pyramid, height 112 meters above the base.
112 meters correspond to 214 royal cubits, 41 rc above the floor of the
hypothetical sun chamber and 66 rc below the top of the former pyramidion.
The numbers 41 and 66 belong to the above golden sequence.

Original base 440 by 440 rc, original height 280 rc. Royal cubit of the
Great Pyramid 52.36 centimeters (a value confirmed to me by Rainer Stadelmann
in a letter from the early 1990s). The plateau (horizontal cross-section)
at the height of 214 rc measured originally 84 by 84 rc. 84 rc or 44 meters
is the floor height of the King's chamber.

The Great Pyramid is a mathematical cosmos enfolded in the symbolical
primeval hill.
Franz Gnaedinger
2018-09-01 07:16:04 UTC
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Post by Franz Gnaedinger
Having seen a tv documentary on the Pyramid Scan project I am happy to say
floor 173 royal cubits or 90.6 meters above the center of the geometrical
base, on the zenith of the imaginary hemisphere that symbolizes the sky
once enclosed in the primeval hill, personified by the sky goddess Nut
arching herself over the earth, her husband Geb, the hypothetical sun chamber
symbolizing her womb wherein she carries the mythical sun child. The radius
of the imaginary hemisphere and floor height of the sun chamber, 173 rc,
is the golden major of the pyramid's height, 280 rc, and found by means
of a golden number sequence
9 16 25 41 66 107 173 280
A team of the Pyramid Scan project got a special permission to explore
the cavity in an edge of the pyramid, height 112 meters above the base.
112 meters correspond to 214 royal cubits, 41 rc above the floor of the
hypothetical sun chamber and 66 rc below the top of the former pyramidion.
The numbers 41 and 66 belong to the above golden sequence.
Original base 440 by 440 rc, original height 280 rc. Royal cubit of the
Great Pyramid 52.36 centimeters (a value confirmed to me by Rainer Stadelmann
in a letter from the early 1990s). The plateau (horizontal cross-section)
at the height of 214 rc measured originally 84 by 84 rc. 84 rc or 44 meters
is the floor height of the King's chamber.
The Great Pyramid is a mathematical cosmos enfolded in the symbolical
primeval hill.
Sorry for a silly mistake of mine. The plateau (horizontal cross-section)
at height 214 rc measured originally 103c 5p by 103c 5p, not a telling
number as far as I can see, while the floor height 84 rc of the King's Chamber
would connect Khufu's pyramid with the Red Pyramid of his father Sneferu,
and this in a most significant manner

base 420 by 420 or 5x84 by 5x84 rc

original height 200 rc

slope exactly 290 rc, triple 20-21-29

Remember the basic number column for the calculation of the square

1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
and so on

The lines of this number column hide infinite sequences of alternating triples
and quadruples, for example

5 7 10 / 5 3+4 / 3-4-5 // 29 41 58 / 29 20+21 / 20-21-29

Imagine a sphere packed into the ideal geometrical pyramid. How long is the
radius? Exactly 84 rc, the key number again. The imaginary sphere would
symbolize the sun once enclosed in the primeval hill, and if there is a hidden
chamber in the Red Pyramid, then in the center of the virtual solar sphere,
84 royal cubits or 44 meters above the center of the base.
Franz Gnaedinger
2019-02-05 08:41:00 UTC
Reply
Permalink
Post by Franz Gnaedinger
Sorry for a silly mistake of mine. The plateau (horizontal cross-section)
at height 214 rc measured originally 103c 5p by 103c 5p, not a telling
number as far as I can see, while the floor height 84 rc of the King's Chamber
would connect Khufu's pyramid with the Red Pyramid of his father Sneferu,
and this in a most significant manner
base 420 by 420 or 5x84 by 5x84 rc
original height 200 rc
slope exactly 290 rc, triple 20-21-29
Remember the basic number column for the calculation of the square
1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
and so on
The lines of this number column hide infinite sequences of alternating triples
and quadruples, for example
5 7 10 / 5 3+4 / 3-4-5 // 29 41 58 / 29 20+21 / 20-21-29
Imagine a sphere packed into the ideal geometrical pyramid. How long is the
radius? Exactly 84 rc, the key number again. The imaginary sphere would
symbolize the sun once enclosed in the primeval hill, and if there is a hidden
chamber in the Red Pyramid, then in the center of the virtual solar sphere,
84 royal cubits or 44 meters above the center of the base.
out of an actual reason: workings of the mind, hyperkooks, Garry Denke et al.

(part 1)

Philosophy and neurology can't explain consciousness. However, neurologists
gained a lot of insights over the past twenty years and propose a model of
how the mind works.

The human brain contains one hundred billion neurons, each connected with
a thousand other neurons - a high complexity corresponding to a mathematical
space of a dozen dimensions.

All neurons together form a large neural network. Within the large network
are partial networks that communicate with each other: an old network that
is concerned with the outer world via the senses, followed by a younger
network that is concerned with our inner life, and a variety of smaller
partial networks, all of them connected by 'white lines' that can be seen
as information highways.

Now this approach to the workings of the mind raises a question: what if
the partial networks are out of sync?

(part 2)

No grammar center has been found in the human brain. The forming of a sentence
is now understood as 'recursive sequencing' that can be compared to the way
we move an arm: a first muscle makes a beginning, a second muscle adds a
contribution, a third muscle a correction, and so on. We may then assume that
speaking involves the various partial networks of the brain.

Consider also that vision occupies more than thirty areas of the brain, in all
half of the brain.

Gothic saiwala English soul inspired a Magdalenian compound: SAI POL saiwala
sawol soul, SAI meaning life, existence, and POL naming a fortified settlement
- a well organized (early) town as metaphor for the soul ... Freud compared
the soul to Rome, an imaginary Rome where all the buildings of all times
are still standing, while quickly built modern towns and quarters are sometimes
called soulless, German seelenlos.

We have then a multiple analogy connecting soul and brain and language and
vision and a well organized town.

What if the complex communication and cooperation between the partial networks
in the brain fail?

(part 3)

Neurology made progress in understanding the mind by identifying partial neural
networks in the brain and studying the communication and cooperation between
them, however, this won't be the final chinup, the mountain grows while we
are climbing ...

Each neuron is a cell and contains the entire genome. How do genetics and
epigenetics influence our mind? (an aspect ignored by the Blue Brain or
Human Brain Project).

Biological stasis accounts for the coherence of a species, in my opinion
with a genetic 'blueprint' or a specific set of solutions to the genetic
'life equations' around which the individual members of the species oscillate.
How does that inner 'map' condition our mind?

Embodiment is a further issue. A flatworm's memory is expanded over tis whole
body. An octopus has additional brains in all eight arms. And we humans have
a second brain of one hundred million neurons along the intestines (a thousand
times less than in than the actual brain, but still).

Microbiot' is the name for the sum of all microbes living on our skin and
in our body - ten times more than cells (the eukariotic cell itself a lucky
symbiosis of bacteria, Lynn Margulis), between one-and-a-half and three
kilograms in an adult -, while 'microbiom' is the name of their gene pool.
By far the most microbes are our helpers. They defend us against harmful
bacteria, communicate with our brain, and keep us in a healthy balance.
How far their influence reaches becomes evident in a strange case. A certain
bacterium can enslave the brain of a mouse and make it seek the proximity
of a cat, which ends the life of the rodent but starts a new cycle in the
reproduction of a parasite (hope I remember this correctly). Bacteria are
powerful, and very luckily mostly on our side.

A living organism is highly complex - endowed with a complexity of the
highest order, coping with inevitable failures. Considering what can go wrong
it is a miracle how much goes well most of the time.

(part 4)

Freud believed in a neurological basis of the mind, its workings, achievements,
and failures.

Hypothesis: the hyperkooks that ruin sci.logic, sci.math, humanities.lit.
authors.shakespeare, sci.archaeology and sci.lang suffer from problems of
communication and cooperation between partial neural networks in the brain,
try to heal themselvess by their frantic posting that should enhance the
internal communication, but fail, and instead ruin one or several fora,
in the case of Garry Denke sci.archaeology and sci.lang.

Remember the multiple analogy of soul and brain and language and vision and
a once well organized town - meanwhile devastated, a Syrian town laid in ruins,
IS fighters hiding between heaps of rubble.

Garry Denke must invoke powers from heaven and hell and use dirty swearing
when it comes to the Heel Stone of Stonehenge, center of his phantasm under
which all forces of the universe gather in his name, from Garry G God to
Denke D devil, imploring help from the Heel Stone read as Heal Stone,
a solid menhir but wavering promise of healing.

Conclusion (valid not only for Garry Denke alias alias alias): the massive
posting of a hyperkook, desperate and shameless at the same time, compensates
for a blocked cerebral communication in a failed attempt of self-healing.
Franz Gnaedinger
2019-02-14 09:20:01 UTC
Reply
Permalink
reconstructing the Nebra calendar in numbers, e-mail to Professor Harald
Meller from this morning, an English translation will follow


Zürich, den 14. Februar 2019

Sehr geehrter Herr Professor Meller,
im Schweizer Radio SRF2 lief eine spannende Sendung über Ihr Nebra-Buch
worauf ich es erwarb und als Experte in Sachen früher Mathematik
( http://www.seshat.ch/home/mathe.pdf ) gerne einen Vorschlag für
Ihren Kalender einbringen möchte.

Antike Zählweise von Lunationen oder synodischen Monaten

30 29 30 29 30 ... Tage für 1 2 3 4 5 ... Lunationen

354 Tage für 12 Lunationen
354 Tage für 12 weitere Lunationen
384 weitere Tage für 13 Lunationen

insgesamt 37 Lunationen oder 1'092 Tage (exakt 1'092.64... Tage)

In anderer Form

12 Lunationen plus 12 Lunationen plus 12 Perioden von 32 Tagen
gemäss den 32 Sternen von Phase eins

Ein Horizontbogen der Phase zwei verdeckt zwei Sterne, so bleiben
30 Sterne sichtbar, gemäss der Anfangszahl des Algorithmus 30 29 30 ...

Hat eine Woiche 7 Tage und ein Monat vier Wochen oder 28 Tage
so sind 1'092 Tage 156 Wochen oder 39 Monate, gemäss den 39 Löchern
länges dem Rand der Scheibe in Phase vier. Diese Löcher könnten
ein astronomisches Observatorium auf dem Hügel von Nebra suggerieren,
ein Kreis von 39 Pfosten, jeder Pfosten-Abstand ein Monat von 28 Tagen,
ein voller Umgang 37 Lunationen oder 1' 0'92 Tage, 9 Abstände ein
Winkel von 83 Grad (theoretisch 83.07...Grad), Winkel der Horizont-
Bögen ab Phase zwei.

1'092 Tage sind praktisch 3 Jahre. Den kleinen Fehler kann man beheben
indem man alle 6 Jahre eine Schaltwoche einfügt. So bekommt man
6 Jahre oder 313 Wochen oder 2'191 Tage.

30 Jahre (Zahl der Sterne ab Phase zwei) wären dann 1'565 Wochen
oder 10'955 Tage (anstelle der genauen 10'957.26... Tage).

Wievîele Lunationen gehen in 30 Jahre? Die Antwort gibt ein 'Ariadne-
Faden', eine additive Reihe von Verhältnissen welche Lunationen
und Jahre verbindet, und damit sozusagen durch das astronomische
Labyrinth führt

L/J 37/3 99/8 136/11 235/19 371/30

12 plus 12 plus 13 Summe 37 Lunationen für 3 Jahre finden sich im
keltischen Kalender; 99 Lunationen für 8 Jahre im Kalender von Lascaux
(mathe.pdf); 235 Lunationen für 19 Jahre im minoischen Kalender von
Knossos, codiert im der Sage vom Minotaurus (mathe.pdf); und 371
Lunationen für 30 Jahre im minoischen Kalender von Mallia. Während
37 Lunationen für fast drei Jahre (ein Umgang des Pfosten-Kreises)
und 371 Lunationen für 30 Jahre (zehn Umgänge plus eine Lunation)
den Kalender von Nebra bestimmen dürften. 371 Lunationen sind
10'955.84... Tage, was sehr gut mit dem obigen Wert von 10'995
Tagen einhergeht.

Die wichtige Zahl 7 erscheint in der Form 6 plus 1 in den Plejaden
(6 Jahre plus 1 Schaltwoche), ebenso im Sternbogen um den Vollmond
(7 Sterne für die Tage einer Woche). Wiederkehrende Zahlen sind
ein typisches Merkmal früher Mathematik, erlauben Quasi-Formeln,
einfache Visualisierungen komplexer Verhältnisse. Meine Formel für
das Verstehen früher Kulturen: einfach aber komplex (inspiriert von
Goethe).

Vielleicht finden sich noch Reste der hypothetischen 39 Pfosten
auf der Anhöhe von Nebra? Diese wäre dann eine Kalender-Sternwarte
gewesen, die Scheibe ihr Emblem ein Kalender von 3 bis 30 Jahren.

Plejaden und Sichel des jungen Mondes hätten den Beginn eines
Jahres angezeigt, auch den Beginn der Feldarbeit, wie in Ihrem
Buch erklärt.

Mit freundlichen Grüssen

Franz Gnaedinger, Zürich
Franz Gnaedinger
2019-02-15 08:20:23 UTC
Reply
Permalink
Post by Franz Gnaedinger
reconstructing the Nebra calendar in numbers, e-mail to Professor Harald
Meller from this morning, an English translation will follow
Zürich, den 14. Februar 2019
Sehr geehrter Herr Professor Meller,
im Schweizer Radio SRF2 lief eine spannende Sendung über Ihr Nebra-Buch
worauf ich es erwarb und als Experte in Sachen früher Mathematik
( http://www.seshat.ch/home/mathe.pdf ) gerne einen Vorschlag für
Ihren Kalender einbringen möchte.
Antike Zählweise von Lunationen oder synodischen Monaten
30 29 30 29 30 ... Tage für 1 2 3 4 5 ... Lunationen
354 Tage für 12 Lunationen
354 Tage für 12 weitere Lunationen
384 weitere Tage für 13 Lunationen
insgesamt 37 Lunationen oder 1'092 Tage (exakt 1'092.64... Tage)
In anderer Form
12 Lunationen plus 12 Lunationen plus 12 Perioden von 32 Tagen
gemäss den 32 Sternen von Phase eins
Ein Horizontbogen der Phase zwei verdeckt zwei Sterne, so bleiben
30 Sterne sichtbar, gemäss der Anfangszahl des Algorithmus 30 29 30 ...
Hat eine Woiche 7 Tage und ein Monat vier Wochen oder 28 Tage
so sind 1'092 Tage 156 Wochen oder 39 Monate, gemäss den 39 Löchern
länges dem Rand der Scheibe in Phase vier. Diese Löcher könnten
ein astronomisches Observatorium auf dem Hügel von Nebra suggerieren,
ein Kreis von 39 Pfosten, jeder Pfosten-Abstand ein Monat von 28 Tagen,
ein voller Umgang 37 Lunationen oder 1' 0'92 Tage, 9 Abstände ein
Winkel von 83 Grad (theoretisch 83.07...Grad), Winkel der Horizont-
Bögen ab Phase zwei.
1'092 Tage sind praktisch 3 Jahre. Den kleinen Fehler kann man beheben
indem man alle 6 Jahre eine Schaltwoche einfügt. So bekommt man
6 Jahre oder 313 Wochen oder 2'191 Tage.
30 Jahre (Zahl der Sterne ab Phase zwei) wären dann 1'565 Wochen
oder 10'955 Tage (anstelle der genauen 10'957.26... Tage).
Wievîele Lunationen gehen in 30 Jahre? Die Antwort gibt ein 'Ariadne-
Faden', eine additive Reihe von Verhältnissen welche Lunationen
und Jahre verbindet, und damit sozusagen durch das astronomische
Labyrinth führt
L/J 37/3 99/8 136/11 235/19 371/30
12 plus 12 plus 13 Summe 37 Lunationen für 3 Jahre finden sich im
keltischen Kalender; 99 Lunationen für 8 Jahre im Kalender von Lascaux
(mathe.pdf); 235 Lunationen für 19 Jahre im minoischen Kalender von
Knossos, codiert im der Sage vom Minotaurus (mathe.pdf); und 371
Lunationen für 30 Jahre im minoischen Kalender von Mallia. Während
37 Lunationen für fast drei Jahre (ein Umgang des Pfosten-Kreises)
und 371 Lunationen für 30 Jahre (zehn Umgänge plus eine Lunation)
den Kalender von Nebra bestimmen dürften. 371 Lunationen sind
10'955.84... Tage, was sehr gut mit dem obigen Wert von 10'995
Tagen einhergeht.
Die wichtige Zahl 7 erscheint in der Form 6 plus 1 in den Plejaden
(6 Jahre plus 1 Schaltwoche), ebenso im Sternbogen um den Vollmond
(7 Sterne für die Tage einer Woche). Wiederkehrende Zahlen sind
ein typisches Merkmal früher Mathematik, erlauben Quasi-Formeln,
einfache Visualisierungen komplexer Verhältnisse. Meine Formel für
das Verstehen früher Kulturen: einfach aber komplex (inspiriert von
Goethe).
Vielleicht finden sich noch Reste der hypothetischen 39 Pfosten
auf der Anhöhe von Nebra? Diese wäre dann eine Kalender-Sternwarte
gewesen, die Scheibe ihr Emblem ein Kalender von 3 bis 30 Jahren.
Plejaden und Sichel des jungen Mondes hätten den Beginn eines
Jahres angezeigt, auch den Beginn der Feldarbeit, wie in Ihrem
Buch erklärt.
Mit freundlichen Grüssen
Franz Gnaedinger, Zürich
Before I can go on with my work (for example the Nebra calendar) I have to
cope with the massive attack on sci.lang and sci.archaeology by Garry Denke.
He is getting ever more angry. One can tell by the increasing number of
exclamation marks that should enforce his insain slandering of the Queen.
What is happening? I never opened one of his threads, yet his titles alone
combined with the range of his aliases reveal a soul-splitting drama.

He externalizes inner conflicts he can't resolve by projecting them under
the Heel Stone of Stonehenge - may it be his Heal Stone, may it provide
healing! The forces of the universe gather in his name under this menhir,
from Garry G God to Denke D devil, the latter hailed as O LUCIFER the
devil Satan. Confined in Garrry Denke's Ark of the Covenant they are kept
in a fragile balance. But apparently the royal court of England allowed
an archaeologist to stabilize that big menhir with a block of cement,
and now that block blocks the exchange between the competing powers,
and when their equilibrium collapses, hell must break loose. Already the
Lucifer alias prevails. What will happen next? ever more exclamation marks?
Lucifer escaping and making a knot into spacetime and letting the world end?

By the way, the Halle museum directed by professor Harald Meller is getting
bombarded with mail, among them a lot of kreisy stuff that reminds of the
world's end hysteria in sci.archaeology in the final years of the last
millennium. A group of 39 people told Meller that the Nebra disc announces
the end of the world, only 39 people shall survive and be transported to
the Pleiads. If he publishes their theory, he will be saved as number 40.
Others interpret the Nebra disc as a time machine. An astronomer told one
of them: We must absolutely discuss your time machine. Call me again
yesterday.
Franz Gnaedinger
2019-02-18 09:20:29 UTC
Reply
Permalink
Post by Franz Gnaedinger
reconstructing the Nebra calendar in numbers, e-mail to Professor Harald
Meller from this morning, an English translation will follow
Zürich, den 14. Februar 2019
Sehr geehrter Herr Professor Meller,
im Schweizer Radio SRF2 lief eine spannende Sendung über Ihr Nebra-Buch
worauf ich es erwarb und als Experte in Sachen früher Mathematik
( http://www.seshat.ch/home/mathe.pdf ) gerne einen Vorschlag für
Ihren Kalender einbringen möchte.
Antike Zählweise von Lunationen oder synodischen Monaten
30 29 30 29 30 ... Tage für 1 2 3 4 5 ... Lunationen
354 Tage für 12 Lunationen
354 Tage für 12 weitere Lunationen
384 weitere Tage für 13 Lunationen
insgesamt 37 Lunationen oder 1'092 Tage (exakt 1'092.64... Tage)
In anderer Form
12 Lunationen plus 12 Lunationen plus 12 Perioden von 32 Tagen
gemäss den 32 Sternen von Phase eins
Ein Horizontbogen der Phase zwei verdeckt zwei Sterne, so bleiben
30 Sterne sichtbar, gemäss der Anfangszahl des Algorithmus 30 29 30 ...
Hat eine Woiche 7 Tage und ein Monat vier Wochen oder 28 Tage
so sind 1'092 Tage 156 Wochen oder 39 Monate, gemäss den 39 Löchern
länges dem Rand der Scheibe in Phase vier. Diese Löcher könnten
ein astronomisches Observatorium auf dem Hügel von Nebra suggerieren,
ein Kreis von 39 Pfosten, jeder Pfosten-Abstand ein Monat von 28 Tagen,
ein voller Umgang 37 Lunationen oder 1' 0'92 Tage, 9 Abstände ein
Winkel von 83 Grad (theoretisch 83.07...Grad), Winkel der Horizont-
Bögen ab Phase zwei.
1'092 Tage sind praktisch 3 Jahre. Den kleinen Fehler kann man beheben
indem man alle 6 Jahre eine Schaltwoche einfügt. So bekommt man
6 Jahre oder 313 Wochen oder 2'191 Tage.
30 Jahre (Zahl der Sterne ab Phase zwei) wären dann 1'565 Wochen
oder 10'955 Tage (anstelle der genauen 10'957.26... Tage).
Wievîele Lunationen gehen in 30 Jahre? Die Antwort gibt ein 'Ariadne-
Faden', eine additive Reihe von Verhältnissen welche Lunationen
und Jahre verbindet, und damit sozusagen durch das astronomische
Labyrinth führt
L/J 37/3 99/8 136/11 235/19 371/30
12 plus 12 plus 13 Summe 37 Lunationen für 3 Jahre finden sich im
keltischen Kalender; 99 Lunationen für 8 Jahre im Kalender von Lascaux
(mathe.pdf); 235 Lunationen für 19 Jahre im minoischen Kalender von
Knossos, codiert im der Sage vom Minotaurus (mathe.pdf); und 371
Lunationen für 30 Jahre im minoischen Kalender von Mallia. Während
37 Lunationen für fast drei Jahre (ein Umgang des Pfosten-Kreises)
und 371 Lunationen für 30 Jahre (zehn Umgänge plus eine Lunation)
den Kalender von Nebra bestimmen dürften. 371 Lunationen sind
10'955.84... Tage, was sehr gut mit dem obigen Wert von 10'995
Tagen einhergeht.
Die wichtige Zahl 7 erscheint in der Form 6 plus 1 in den Plejaden
(6 Jahre plus 1 Schaltwoche), ebenso im Sternbogen um den Vollmond
(7 Sterne für die Tage einer Woche). Wiederkehrende Zahlen sind
ein typisches Merkmal früher Mathematik, erlauben Quasi-Formeln,
einfache Visualisierungen komplexer Verhältnisse. Meine Formel für
das Verstehen früher Kulturen: einfach aber komplex (inspiriert von
Goethe).
Vielleicht finden sich noch Reste der hypothetischen 39 Pfosten
auf der Anhöhe von Nebra? Diese wäre dann eine Kalender-Sternwarte
gewesen, die Scheibe ihr Emblem ein Kalender von 3 bis 30 Jahren.
Plejaden und Sichel des jungen Mondes hätten den Beginn eines
Jahres angezeigt, auch den Beginn der Feldarbeit, wie in Ihrem
Buch erklärt.
Mit freundlichen Grüssen
Franz Gnaedinger, Zürich
Schneewittchen, eine Erinnerung an den Beginn der Bronzezeit in Mitteleuropa

Zürich, den 18. Februar 2019

Sehr geehrte Damen und Herren,
haben Sie mein e-mail für Herrn Professor Meller vom 14. Februar
bekommen?

Die Hypothese oder mittlerweile Theorie eines Reiches von Nebra
scheint mir sehr plausibel.

Besonders interessiert mich auch der Zinnhandel mit Cornwall,
da ich einen temporären Zinnweg von Cornwall über die
Schweizer Alpen - Falera Greina Olivone - in die Aegäis vermute.
Eine der Goldrauten des Fürsten vom Bush Barrow und die ovale
Bronzescheibe der grossen 'Nadel' von Falera codieren denselben
Kalender vom Stonehenge-Typus
http://www.seshat.ch/home/stonehen.htm

Dann gibt es eine sprachliche Erinnerung an den Beginn der Bronzezeit
in Mitteleuropa: die historisch/mythologische Ebene des Märchens
von Schneewittchen oder Schneeweisschen, verschlüsselt in Symbolen

schöne Königin - Verkörperung von Kupfer, für Jahrtausende
das wertvollste Material, ihr Spiegel eine polierte Kupferscheibe

Schneewittchen hinter den sieben Bergen - Zinn von weither

vergifteter Apfel - Arsen im Kupfer

die sieben Zwerge - Bergleute, nördlich des Erzgebirges?
Knaben von vierzehn Jahren an, klein geblieben wegen
ungenügender Ernährung und mangelnder Sonne,
grantig aber warmherzig, eine anrührende Hommage
an die Bergleute denen wir einen grossen Teil
unseres Wohlstandes verdanken

Dagegen berichten die Homerischen Epen vom Ende der Bronzezeit
in der Aegäis, auch wieder mit Symbolen: die schöne Helena eine
Verkörperung von Zinn, ihre weissen Arme runde weisse Zinnbarren,
ihre langen glitzernden Roben die sie selber anfertigte Halden des
glitzerndedn Zinnerzes Kassitterit, ihr Faden Zinndraht, einst aus
gehämmertem Zinnblech geschnitten; ihr Gatte xanthos Menelaos
Kupfer, gelbes braunes rötliches Erz; ihre gemeinsame Tochter,
die liebliche Hermione welche der goldenen Aphrodite glich
die frisch gegossen goldglänzende Bronze ... In Griechenland
gibt es kein Zinn, das mykenische Zinn kam aus Zentralasien
und passierte den Hellespont wo die Trojaner ihre Hand auf die
kostbare Fracht legten, sozusagen die schöne Helena entführten ...

Wieder zum Beginn der Bronzezeit in Mitteleuropa, vielleicht sogar
des Reiches von Nebra. War der Apfel ein Zeichen der Herrschaft,
eine goldgelbe Szepter-Kugel aus Bronze von der Grösse eines Apfels
die im späteren goldenen Reichsapfel überdauerte? die Hütte der
Zwerge der Beginn des Reiches von Nebra? die Zwerge Untertanen?
was auf ihren Tellerchen fehlte Abgaben, Steuern? Schneewittchen
die Prinzessin? wachgeküsst von einem wagemutigen Prinzen
wie demjenigen des Märchens von Dornröschen?

Um der Logik des Märchens willen müsste der Prinz der Kupfer-
familie der Königin angehört haben. Vielleicht war die Königin
einst eine Göttin des Kupfers, der Prinz (ihr Sohn?) ein Kupfergott,
und Schneewittchen die Zinngöttin? auf deren Abstammung
sich die Könige von Nebra beriefen?

Mit freundlichen Grüssen

Franz Gnaedinger, Zürich

Ps Wäre schön wenn ich eine Bestätigung für den Empfang
meiner e-mails bekäme
Franz Gnaedinger
2019-02-19 09:28:24 UTC
Reply
Permalink
Post by Franz Gnaedinger
Schneewittchen, eine Erinnerung an den Beginn der Bronzezeit in Mitteleuropa
Zürich, den 18. Februar 2019
Sehr geehrte Damen und Herren,
haben Sie mein e-mail für Herrn Professor Meller vom 14. Februar
bekommen?
Die Hypothese oder mittlerweile Theorie eines Reiches von Nebra
scheint mir sehr plausibel.
Besonders interessiert mich auch der Zinnhandel mit Cornwall,
da ich einen temporären Zinnweg von Cornwall über die
Schweizer Alpen - Falera Greina Olivone - in die Aegäis vermute.
Eine der Goldrauten des Fürsten vom Bush Barrow und die ovale
Bronzescheibe der grossen 'Nadel' von Falera codieren denselben
Kalender vom Stonehenge-Typus
http://www.seshat.ch/home/stonehen.htm
Dann gibt es eine sprachliche Erinnerung an den Beginn der Bronzezeit
in Mitteleuropa: die historisch/mythologische Ebene des Märchens
von Schneewittchen oder Schneeweisschen, verschlüsselt in Symbolen
schöne Königin - Verkörperung von Kupfer, für Jahrtausende
das wertvollste Material, ihr Spiegel eine polierte Kupferscheibe
Schneewittchen hinter den sieben Bergen - Zinn von weither
vergifteter Apfel - Arsen im Kupfer
die sieben Zwerge - Bergleute, nördlich des Erzgebirges?
Knaben von vierzehn Jahren an, klein geblieben wegen
ungenügender Ernährung und mangelnder Sonne,
grantig aber warmherzig, eine anrührende Hommage
an die Bergleute denen wir einen grossen Teil
unseres Wohlstandes verdanken
Dagegen berichten die Homerischen Epen vom Ende der Bronzezeit
in der Aegäis, auch wieder mit Symbolen: die schöne Helena eine
Verkörperung von Zinn, ihre weissen Arme runde weisse Zinnbarren,
ihre langen glitzernden Roben die sie selber anfertigte Halden des
glitzerndedn Zinnerzes Kassitterit, ihr Faden Zinndraht, einst aus
gehämmertem Zinnblech geschnitten; ihr Gatte xanthos Menelaos
Kupfer, gelbes braunes rötliches Erz; ihre gemeinsame Tochter,
die liebliche Hermione welche der goldenen Aphrodite glich
die frisch gegossen goldglänzende Bronze ... In Griechenland
gibt es kein Zinn, das mykenische Zinn kam aus Zentralasien
und passierte den Hellespont wo die Trojaner ihre Hand auf die
kostbare Fracht legten, sozusagen die schöne Helena entführten ...
Wieder zum Beginn der Bronzezeit in Mitteleuropa, vielleicht sogar
des Reiches von Nebra. War der Apfel ein Zeichen der Herrschaft,
eine goldgelbe Szepter-Kugel aus Bronze von der Grösse eines Apfels
die im späteren goldenen Reichsapfel überdauerte? die Hütte der
Zwerge der Beginn des Reiches von Nebra? die Zwerge Untertanen?
was auf ihren Tellerchen fehlte Abgaben, Steuern? Schneewittchen
die Prinzessin? wachgeküsst von einem wagemutigen Prinzen
wie demjenigen des Märchens von Dornröschen?
Um der Logik des Märchens willen müsste der Prinz der Kupfer-
familie der Königin angehört haben. Vielleicht war die Königin
einst eine Göttin des Kupfers, der Prinz (ihr Sohn?) ein Kupfergott,
und Schneewittchen die Zinngöttin? auf deren Abstammung
sich die Könige von Nebra beriefen?
Mit freundlichen Grüssen
Franz Gnaedinger, Zürich
Ps Wäre schön wenn ich eine Bestätigung für den Empfang
meiner e-mails bekäme
Nebra, calendar and mythology (English summary)

Imagine an observatory on top of a hill with a wide horizon,
a circle of 39 poles, one of them marking due South. 10 and 10
lateral poles (radial angle 83 degrees) cover the horizons of the
rising and setting sun. The distance from pole to pole represents
a month of 28 days, four weeks of 7 days, the whole circle 39
months or 156 weeks or 1,092 days, nearly 3 years, or 12 12 13
sum 37 lunations or synodic months, or 12 lunations plus 12
lunations plus 12 periods of 32 days

30 29 30 29 30 ... days for 1 2 3 4 5 ... lunations

354 354 384 sum 1,092 days for nearly 3 years

2lun 32d 2lun 32d 2lun 32d 2lun 32d sum 364 days

Add one leap week for 6 years and you get 313 weeks or 2,191
days (exactly 2,191.45... rounded 2,191 days). Multiply these
numbers by a factor of 5 and you obtain 30 years or 1,565 weeks
or 10,955 days (exactly 10'957.26... days). How many lunations
are in 30 years? The answer is given by an 'Ariadne thread',
an additive sequence of ratios that combine lunations (l) and
years (y) and guided a Minoan astronomer through the 'labyrinth'
of his demanding calculations

l/y 37/3 99/8 136/11 235/19 371/30

12 12 13 sum 37 lunations for 3 years are found in the Celtic calendar;
99 lunations for a cylce of 8 years in the calendar of Lascaux; 235
lunations for 19 years in the Minoan calendar of Knossos, encoded
in the legend of the Minotaur, and 371 lunations for 30 years in an
alternative Minoan calendar from Mallia.

The ratios 37/3 and 371/30 were used for the Nebra calendar,
combined with a rule for the beginning and end of the farming year
adopted from Babylon according to Rahlf Hansen.

Harald Meller postulates a kingdom (Reich) of Nebra. He misses
a written source like Homer. But maybe there is a written memory
of that lost kingdom - the historical and mythical level of a famous
fairy tale, Snow White behind the seven mountains, here told as
a story of supernatural beings, maybe of a half divine rank.

The copper queen was most beautiful and loved to look at herself
in a polished copper disc. One day her mirror told her that there
is an even more beautiful woman, Snow White behind the seven
mountains, tin princess. Hereupon the queen got mighty jealous
and tried to eliminate her rival with a poisened apple (arsenic
used for hardening copper). But she had a son, the copper prince.
This one saved Snow White, married her, and together they had
a son who founded a royal dynasty which brought forth the later
founder of the kingdom of Nebra, while the seven dwarfs were
miners who came from the Ore Mountains, and their hut was
the humble origin of the kingdom of Nebra ...

The Homeric epics tell a comparable story from the end of the
Bronze Age in the Aegaean. Beautiful Helen symbolizes tin,
her white arms round white ingots, her long glittering robes
she made herself heaps of the glittering tin ore cassitterite,
her thread tin wire, by then cut out from hammered tin sheet;
her husband xanthos Menelaos symbolizes copper, the color
xanthos covering all hues of copper ores, yellow brown reddish;
and their daughter, lovely Hermione who resembles golden
Aphrodite symbolizes bronze, alloy of copper and tin, of a golden
shine when freshly cast ... Now there is no tin in Greece, the
Mycanaean tin came from Central Asia and passed the Hellespont
where the Troians laid hands on the precious cargo, abducting
Helen, as it were ...
Franz Gnaedinger
2019-02-20 10:19:36 UTC
Reply
Permalink
Post by Franz Gnaedinger
Nebra, calendar and mythology (English summary)
Imagine an observatory on top of a hill with a wide horizon,
a circle of 39 poles, one of them marking due South. 10 and 10
lateral poles (radial angle 83 degrees) cover the horizons of the
rising and setting sun. The distance from pole to pole represents
a month of 28 days, four weeks of 7 days, the whole circle 39
months or 156 weeks or 1,092 days, nearly 3 years, or 12 12 13
sum 37 lunations or synodic months, or 12 lunations plus 12
lunations plus 12 periods of 32 days
30 29 30 29 30 ... days for 1 2 3 4 5 ... lunations
354 354 384 sum 1,092 days for nearly 3 years
2lun 32d 2lun 32d 2lun 32d 2lun 32d sum 364 days
Add one leap week for 6 years and you get 313 weeks or 2,191
days (exactly 2,191.45... rounded 2,191 days). Multiply these
numbers by a factor of 5 and you obtain 30 years or 1,565 weeks
or 10,955 days (exactly 10'957.26... days). How many lunations
are in 30 years? The answer is given by an 'Ariadne thread',
an additive sequence of ratios that combine lunations (l) and
years (y) and guided a Minoan astronomer through the 'labyrinth'
of his demanding calculations
l/y 37/3 99/8 136/11 235/19 371/30
12 12 13 sum 37 lunations for 3 years are found in the Celtic calendar;
99 lunations for a cylce of 8 years in the calendar of Lascaux; 235
lunations for 19 years in the Minoan calendar of Knossos, encoded
in the legend of the Minotaur, and 371 lunations for 30 years in an
alternative Minoan calendar from Mallia.
The ratios 37/3 and 371/30 were used for the Nebra calendar,
combined with a rule for the beginning and end of the farming year
adopted from Babylon according to Rahlf Hansen.
Harald Meller postulates a kingdom (Reich) of Nebra. He misses
a written source like Homer. But maybe there is a written memory
of that lost kingdom - the historical and mythical level of a famous
fairy tale, Snow White behind the seven mountains, here told as
a story of supernatural beings, maybe of a half divine rank.
The copper queen was most beautiful and loved to look at herself
in a polished copper disc. One day her mirror told her that there
is an even more beautiful woman, Snow White behind the seven
mountains, tin princess. Hereupon the queen got mighty jealous
and tried to eliminate her rival with a poisened apple (arsenic
used for hardening copper). But she had a son, the copper prince.
This one saved Snow White, married her, and together they had
a son who founded a royal dynasty which brought forth the later
founder of the kingdom of Nebra, while the seven dwarfs were
miners who came from the Ore Mountains, and their hut was
the humble origin of the kingdom of Nebra ...
The Homeric epics tell a comparable story from the end of the
Bronze Age in the Aegaean. Beautiful Helen symbolizes tin,
her white arms round white ingots, her long glittering robes
she made herself heaps of the glittering tin ore cassitterite,
her thread tin wire, by then cut out from hammered tin sheet;
her husband xanthos Menelaos symbolizes copper, the color
xanthos covering all hues of copper ores, yellow brown reddish;
and their daughter, lovely Hermione who resembles golden
Aphrodite symbolizes bronze, alloy of copper and tin, of a golden
shine when freshly cast ... Now there is no tin in Greece, the
Mycanaean tin came from Central Asia and passed the Hellespont
where the Troians laid hands on the precious cargo, abducting
Helen, as it were ...
Schiff der Zeit / ship of time (English summary below)

Esrt eine Korrektur. 10'955 Tage wären 10 Umläufe im Kreis der
39 Pfosten, plus 1 Monat von 28 Tagen, ein weiterer Pfosten,
plus eine Woche von 7 Tagen.

Ein alternativer 'Ariadne-Faden' liefert eine genauere Umrechnung
von 30 Jaren in Tage

365/1 (plus 1461/4) 1826/5 3287/9 ... 9131/25 ....

5 Jahre sind 1'826 Tage, 5 reguläre Jahre von 365 Tagen plus
1 Schalttag. 25 Jahre sind 9'131 Tage, 25 reguläre Jahre plus
6 Schalttage. Zusammengezählt bekommen wir für 30 Jahre
10,957 Tage (exakt 10'957.26... Tage), 30 reguläre Jahre
plus 7 Schalttage. Schon wieder die Zahl 7.

Dagegen sind 371 Lunationen für 30 Jahre etwas kürzer,
10'955.84... Tage. Kalendarische Formeln sind immer nur
Näherungen. Am besten fährt man wenn man mehrere von
ihnen kombiniert.

Möglicherweiser kannten die Astronomen von Nebra die Formel
vom Göbekli Tepe (63 kontinuierliche Perioden von 30 Tagen
sind 1'890 Tage und entsprechen 64 Lunationen, Fehler weniger
als eine Minute per Lunation, oder ein halber Tag auf ein Leben),
und zwar in der Minoisch/Helladischen Version (ein Jahr symbolisiert
als Rosette von acht Blütenblättern, je ein langer Monat von 45
Tagen oder 5 Homerischen Wochen, der kleine Kreis in der Mitte
zusätzliche 5 und manchmal 6 Tage). 15 und 17 Lunationen
gezählt im 30 29 30 Modus ergeben 443 und 502 Tage, zusammen
945 Tage für 32 Lunationen. Mal vier bekommen wir für Nebra
taugliche Formeln

a) 128 Luntionen oder 135 Monate von 28 Tagen

b) 2 x 2 x 2 x 2 x 2 x 2 x 2 Lunationen oder 135 Pfosten

c) 8 x 16 Lunationen oder 9 x 15 Pfosten (Monate)

Man achte auf die sieben Faktoren 2 in Formel b. Formel c kann
sehr einfach notiert werden, ohne Schrift: als Felder von 8 x 16
und 9 x 15 Punktuationen auf einer Holz- oder Knochenplatte
oder einer Keramikfläche.

135 Pfosten im Kreis der 39 Pfosten wären 3 Umläufe plus 18
Pfosten für das Boot auf seiner Himmelsbahn, welches daher
als ein 'Schiff der Zeit' angesehen werden kann. Der südliche
Pfosten - auf der Scheibe das untereste Loch - markiert sowohl
die Sommersonnwende als auch (ziemlich genau) die Mitte
des Rumpfes. Wieviele Sterne sind auf dem Schiff zu sehen?
7. Immer wieder die Zahl Sieben.

ship of time (brief English summary)

The astronomers of Nebra might have known the calendar formula
of the Goebekli Tepe in the Minoan/Helladic version. 15 and 17
lunations counted in the 30 29 30 mode yield 443 and 502 days
respectively, together 945 days for 32 lunations. Multiply these
numbers by a factor of 4 and you get a formula that can be used
for Nebra: 135 months of 28 days, represented by 135 poles,
correspond to 128 or 2x2x2x2x2x2x2 (seven times the same factor)
lunations - when the 'ship of time' starting from the midsummer
pole in the South completed 3 full circles plus 18 posts, 128 lunations
have passed. How many stars are on the ship? 7. Always the number
seven.
Franz Gnaedinger
2019-02-23 08:55:05 UTC
Reply
Permalink
Post by Franz Gnaedinger
Schiff der Zeit / ship of time (English summary below)
Esrt eine Korrektur. 10'955 Tage wären 10 Umläufe im Kreis der
39 Pfosten, plus 1 Monat von 28 Tagen, ein weiterer Pfosten,
plus eine Woche von 7 Tagen.
Ein alternativer 'Ariadne-Faden' liefert eine genauere Umrechnung
von 30 Jaren in Tage
365/1 (plus 1461/4) 1826/5 3287/9 ... 9131/25 ....
5 Jahre sind 1'826 Tage, 5 reguläre Jahre von 365 Tagen plus
1 Schalttag. 25 Jahre sind 9'131 Tage, 25 reguläre Jahre plus
6 Schalttage. Zusammengezählt bekommen wir für 30 Jahre
10,957 Tage (exakt 10'957.26... Tage), 30 reguläre Jahre
plus 7 Schalttage. Schon wieder die Zahl 7.
Dagegen sind 371 Lunationen für 30 Jahre etwas kürzer,
10'955.84... Tage. Kalendarische Formeln sind immer nur
Näherungen. Am besten fährt man wenn man mehrere von
ihnen kombiniert.
Möglicherweiser kannten die Astronomen von Nebra die Formel
vom Göbekli Tepe (63 kontinuierliche Perioden von 30 Tagen
sind 1'890 Tage und entsprechen 64 Lunationen, Fehler weniger
als eine Minute per Lunation, oder ein halber Tag auf ein Leben),
und zwar in der Minoisch/Helladischen Version (ein Jahr symbolisiert
als Rosette von acht Blütenblättern, je ein langer Monat von 45
Tagen oder 5 Homerischen Wochen, der kleine Kreis in der Mitte
zusätzliche 5 und manchmal 6 Tage). 15 und 17 Lunationen
gezählt im 30 29 30 Modus ergeben 443 und 502 Tage, zusammen
945 Tage für 32 Lunationen. Mal vier bekommen wir für Nebra
taugliche Formeln
a) 128 Luntionen oder 135 Monate von 28 Tagen
b) 2 x 2 x 2 x 2 x 2 x 2 x 2 Lunationen oder 135 Pfosten
c) 8 x 16 Lunationen oder 9 x 15 Pfosten (Monate)
Man achte auf die sieben Faktoren 2 in Formel b. Formel c kann
sehr einfach notiert werden, ohne Schrift: als Felder von 8 x 16
und 9 x 15 Punktuationen auf einer Holz- oder Knochenplatte
oder einer Keramikfläche.
135 Pfosten im Kreis der 39 Pfosten wären 3 Umläufe plus 18
Pfosten für das Boot auf seiner Himmelsbahn, welches daher
als ein 'Schiff der Zeit' angesehen werden kann. Der südliche
Pfosten - auf der Scheibe das untereste Loch - markiert sowohl
die Sommersonnwende als auch (ziemlich genau) die Mitte
des Rumpfes. Wieviele Sterne sind auf dem Schiff zu sehen?
7. Immer wieder die Zahl Sieben.
ship of time (brief English summary)
The astronomers of Nebra might have known the calendar formula
of the Goebekli Tepe in the Minoan/Helladic version. 15 and 17
lunations counted in the 30 29 30 mode yield 443 and 502 days
respectively, together 945 days for 32 lunations. Multiply these
numbers by a factor of 4 and you get a formula that can be used
for Nebra: 135 months of 28 days, represented by 135 poles,
correspond to 128 or 2x2x2x2x2x2x2 (seven times the same factor)
lunations - when the 'ship of time' starting from the midsummer
pole in the South completed 3 full circles plus 18 posts, 128 lunations
have passed. How many stars are on the ship? 7. Always the number
seven.
checking a set of higher astronomical numbers with Mesopotamian /
Minoan / Helladic methods

334 years are practically 121,992 (exactly 121,990.89... days) and
4,131 (exactly 4,131.0010...) lunations.

15 and 17 lunations are 443 and 502 days respectively, together
945 days for 32 lunations. 17 15 17 15 17 sum 81 lunations correspond
to 502 443 502 443 502 sum 2,392 days. Multiply these numbers by
3 x 17 = 51 and you obtain 121,922 days for 4,131 lunations,

What about years?

l/y 37/3 99/8 136/11 235/19 371/30

17 times 19 years are 323 years, plus 11 years are 334 years.
17 times 235 plus 136 are 4,131 lunations.

Working with 'Ariadne' series becomes rather easy when you gathered
some experience and use a calculating bord subdivided in fields that
give a pebble the shifting value of 1 or 10 or 100 or 1,000 ..

An interesting case occurs when you consider a pole 1 day and the circle
of 39 poles a round of 39 days. 184 rounds are then 243 lunations,
3,128 rounds 121,922 days or 334 years or 4,131 lunations.

Methods that work well for centuries also work fine for decades and
a lifetime.

The astronomers of Nebra might have had contact with colleagues
from Mycene and Knossos, the Minoans probably having come from Ebla
in Syria, where they were acquainted with Babylonian astronomy.
Franz Gnaedinger
2019-02-25 08:21:02 UTC
Reply
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Post by Franz Gnaedinger
checking a set of higher astronomical numbers with Mesopotamian /
Minoan / Helladic methods
334 years are practically 121,992 (exactly 121,990.89... days) and
4,131 (exactly 4,131.0010...) lunations.
15 and 17 lunations are 443 and 502 days respectively, together
945 days for 32 lunations. 17 15 17 15 17 sum 81 lunations correspond
to 502 443 502 443 502 sum 2,392 days. Multiply these numbers by
3 x 17 = 51 and you obtain 121,922 days for 4,131 lunations,
What about years?
l/y 37/3 99/8 136/11 235/19 371/30
17 times 19 years are 323 years, plus 11 years are 334 years.
17 times 235 plus 136 are 4,131 lunations.
Working with 'Ariadne' series becomes rather easy when you gathered
some experience and use a calculating bord subdivided in fields that
give a pebble the shifting value of 1 or 10 or 100 or 1,000 ..
An interesting case occurs when you consider a pole 1 day and the circle
of 39 poles a round of 39 days. 184 rounds are then 243 lunations,
3,128 rounds 121,922 days or 334 years or 4,131 lunations.
Methods that work well for centuries also work fine for decades and
a lifetime.
The astronomers of Nebra might have had contact with colleagues
from Mycene and Knossos, the Minoans probably having come from Ebla
in Syria, where they were acquainted with Babylonian astronomy.
from the Nebra calendar to Bronze Age geometry

Picture a calendar rope measuring 39 'months' or 156 'weeks' or 1092 'days'.
Make a loop of it by connecting the ends. Now you can lay out five
elementary forms and solve related problems with Bronze Age methods.

Lay out a rectangular triangle. Find a solution of integer numbers.
How long are the sides? 52 and 39 and 65 'w'.

Lay out an equilateral triangle. The sides are 52 'w'. What is the height?
45 'w' according to the following number column

1 1 3
2 4 6
1 2 3
3 5 9
8 14 24
4 7 12
11 19 33
30 52 90 * 52
15 26 45 * 45

Lay out a square. The side measures 39 'w' or 273 'd''. How long is
the diagonal?

1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140

side 99 99 70 5 sum 273 'd'
diagonal 140 140 99 7 sum 386 'd'

Lay out a double square. It measures 26 by 52 'w' or 182 by 364 'd'.
How long is the diagonal?

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 20
13 29 65
42 94 210
21 47 105
68 152 340
34 76 170
17 38 85
55 123 275
178 398 890
89 199 445
288 644 1440
144 322 720
72 161 360

short side 72 55 55 sum 182 'd'
diagonal 161 123 123 sum 407 'd'

Lay out a circle, circumference 39 'm'. Transform the area into
a square

3/1 (plus 22/7) 25/8 47/15...311/99...377/120...1521/484

1521/484 equals 39x39/22x22. The square of the same area
measures 22/2 by 22/2 or 11 by 11 'm'.

Bronze Age mathematics was taught by telling examples. A calendar
rope measuring 39 'm' or 156 'w' or 1'092 'd' could have been such
an example.

Who would have thought that the odd number 39 enfolds not only
an elaborate calendar but also a 'book' of Bronze Age geometry?
Franz Gnaedinger
2019-02-28 08:31:52 UTC
Reply
Permalink
Post by Franz Gnaedinger
from the Nebra calendar to Bronze Age geometry
Picture a calendar rope measuring 39 'months' or 156 'weeks' or 1092 'days'.
Make a loop of it by connecting the ends. Now you can lay out five
elementary forms and solve related problems with Bronze Age methods.
Lay out a rectangular triangle. Find a solution of integer numbers.
How long are the sides? 52 and 39 and 65 'w'.
Lay out an equilateral triangle. The sides are 52 'w'. What is the height?
45 'w' according to the following number column
1 1 3
2 4 6
1 2 3
3 5 9
8 14 24
4 7 12
11 19 33
30 52 90 * 52
15 26 45 * 45
Lay out a square. The side measures 39 'w' or 273 'd''. How long is
the diagonal?
1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
side 99 99 70 5 sum 273 'd'
diagonal 140 140 99 7 sum 386 'd'
Lay out a double square. It measures 26 by 52 'w' or 182 by 364 'd'.
How long is the diagonal?
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 20
13 29 65
42 94 210
21 47 105
68 152 340
34 76 170
17 38 85
55 123 275
178 398 890
89 199 445
288 644 1440
144 322 720
72 161 360
short side 72 55 55 sum 182 'd'
diagonal 161 123 123 sum 407 'd'
Lay out a circle, circumference 39 'm'. Transform the area into
a square
3/1 (plus 22/7) 25/8 47/15...311/99...377/120...1521/484
1521/484 equals 39x39/22x22. The square of the same area
measures 22/2 by 22/2 or 11 by 11 'm'.
Bronze Age mathematics was taught by telling examples. A calendar
rope measuring 39 'm' or 156 'w' or 1'092 'd' could have been such
an example.
Who would have thought that the odd number 39 enfolds not only
an elaborate calendar but also a 'book' of Bronze Age geometry?
Stonehenge 1, circle of Aubrey holes

Harald Meller believes in a connection between Stonehenge and Nebra.
The concentric ditch and ringwall and 56 Aubrey holes are considerably
older than Nebra and may encode a similar calendar as the one of the
hypothetical 39 poles indicated by the 39 holes along the rim of the
Nebra disc.

Each Aubrey hole - or space between two holes - may have represented
a month of 30 days, one round of 56 holes 1,680 days, 5 rounds 8,400
days or 23 years, 9 rounds 15,120 days or 512 lunations.

How is the number of 8'400 days in 23 years found? by means of
a number sequence (which I ascribe to the astronomers of Tell Halaf
in Syria, 9,000 years ago)

365/1 (plus 1461/4) 1826/5 3287/9 4748/13 6205/17

7670/21 9131/25

21 years are 7,670 days (exactly 7,670.08... days). Add 365 days
and again 365 days for two more years and you get 8,400 days for
23 years (8400.57... days).

Now for the lunations. 15 and 17 lunations counted in the 30 29 30
mode yield 443 and 502 days respectively, 17 15 17 15 sum 81 lunations
1,890 days or 63 continuous periods of 30 days (Goebekli Tepe definition,
12,000 years ago, mistake less than one minute per lunation, or half
a day in a lifetime). Multiply the numbers by a factor of 8 and you get
512 lunations or 15,120 days (15,119.66... days).

A week may have been 7 days, a round of 56 months 240 weeks, 5 rounds
or 23 years 1,200 weeks, 9 rounds or 512 lunations 2,160 weeks, 8 times
64 lunations 270 weeks.

A basic year had 12 months or 360 days, a regular year 365 days, an
occasional leap year 366 days. For example 21 years are 7,670 days
(number sequence above), 16 regular years and 5 leap years.

Mike Parker Pearson discovered that Stonehenge 1 of around 3000 BC
was a cemetary. The ashes of the the defunct were buried in the plain,
and the ashes of high ranking nobles (we may assume) in the Aubrey
holes, whereupon the circle of 56 bluestones from the remote Presely
Mountains (hills) had been erected in the Aubrey holes.

The circle of bluestones may then have been seen as a 'generator'
of time for the living and for the worthy souls in a heavenly beyond ...

Maybe there had been much earlier connections to the Aegaean,
or a knowledge of basic number sequences already existed in Neolithic
Europe, wherever it came from.

O LUCIFER the Devil Satan
2019-02-15 15:52:32 UTC
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Permalink
[snipped nonsense and ineffective rant]
The Creator, Garry Denke, has been called worse.
You can do better than that! Surely you can.

John the Baptist
O LUCIFER the Devil Satan
2019-02-15 15:58:32 UTC
Reply
Permalink
[snipped nonsense and ineffective rant]
The Creator, Garry Denke, has been called much worse.
You can do much better than that! Surely you can.

James the Christ
Franz Gnaedinger
2015-12-16 07:41:05 UTC
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Post by Franz Gnaedinger
Post by Franz Gnaedinger
http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
Egyptian method of calculating the circle (illustrations)
Key figure, grid 10 by 10, square 10 by 10 royal cubits or 70 by 70 palms
or 280 by 280 fingerbreadths, diagonal practically 99 palms (1 1 2, 2 3 4,
5 7 10, 12 17 24, 29 41 58, 70 99 ...), twelve rational points of the
virtual circle, four points given by the ends of tha axes and eight points
defined by the Sacred Triangle 3-4-5 doubled to the rectangle 4 by 3 or
3 by 4 diagonal 5
http://www.seshat.ch/home/key1.GIF
http://www.seshat.ch/home/key2.GIF
The shorter arcs of the circle indicated by the dozen points measure
practically 40 fingerbreadths each, and the longer ones 90 fingerbreadths,
yielding a circumference of 880 fingerbreadths or 220 palms, and if we
divide the 220 palms by the diameter 70 palms we obtain 22/7 for pi.
Drawings of the first four polygons that have 12 20 28 36 shorter and
shorter sides
http://www.seshat.ch/home/poly1.GIF
http://www.seshat.ch/home/poly2.GIF
http://www.seshat.ch/home/poly3.GIF
http://www.seshat.ch/home/poly4.GIF
12 20 28 36 points of the grid evoke the circle
http://www.seshat.ch/home/polyg1a.GIF
http://www.seshat.ch/home/polyg1b.GIF
http://www.seshat.ch/home/polyg1c.GIF
http://www.seshat.ch/home/polyg1d.GIF
The corners of the rounder and rounder polygon are generated by a turning
angle of arctan3/4 (36.86989... degrees), mirror each point around the
axes (y = 0, x = 0) and the oblique cross (y = +-x)
http://www.seshat.ch/home/poly5.GIF
Another way of generating the triples is to draw up a number column
that uses a factor of minus four
1 1 -4
2 -3 -8 3 8 3-4-5
-1 -11 4
-12 -7 48 7 48 7-24-25
-19 41 76
22 117 -88 117 88 44-117-125
139 29 -556
168 -527 -672 527 672 336-527-625
-359 -1199 1436
-1558 237 6232 237 6232 237-3116-3125
and so on
Many more drawings can be found via egypt2 and egypt3 (links above),
here for example the hemisphere of Nut (below) and circle of Ra (above)
http://www.seshat.ch/home/pyramid2.GIF
how to calculate the square (a lesson in experimental math-history)

The Babylonian mathematicians had two values for the square root of 2,
the simple value 17/12, or 1;25 in their sexagesimal number system,
and the fabulous value 1;24,51,10 (mentioned on the clay tablet YBC 7289
from around 1600 BC). How did they find the latter value? Draw up the
number column for the calculation of the square

1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
169 239 338
408 577 816
985 1393 1970
and so on

Divide 1393 by 1985 and you obtain 1;24,51,10,3,2... Let go the small
numbers ...3,2... and keep 1;24,51,10 (1.414219... for 1.4142135...).

Back to the Old Kingdom of Egypt and calculating with integers. How long
is the diagonal of a square whose side measures 83 royal cubits or 581
palms or 2324 fingerbreadths? Using the above number column you may proceed
as follows

side 83 or 70 5 5 3 rc
diagonal 99 7 7 4 sum 117 rc
(exactly 117.379... rounded 117 rc)

side 581 or 577 2 2 p
diagonal 9816 3 3 sum 822 p
(exactly 821.658... rounded 822 p)

side 2324 or 1393 577 239 99 12 2 2 f
diagonal 1970 816 338 140 17 3 3 sum 3287 f
(exactly 3286.632... rounded 3287 f)

Operate with finer and finer units and the result will be better and better.

Important is the algorithm

a b 2a
a+b b+2a 2(a+b)

You can begin with any number pair a and b, for example 4 and 11, and even
make mistakes

4 11 8
15 19 30
34 49 68
83 117 166 (note the numbers 83 and 117)
200 283 400
481 683 964 (mistake, 481 instead of 483)
1164 1647 2328
and so on

The number column starting with 4 11 and containing a mistake approximates
the square root of 2 more slowly, but still.

By the way, the regular number column starting with 1 1 and adding all
numbers correctly has an exact equivalent in the continued fraction
(1;2,2,2,2,2,2...).

And what was the origin of the above number column? I have reasons to
assume a Stone Age formula

If the side of a square measures 5 paces
or a multiple thereof,
the diagonal measures 7 paces
or a multiple thereof,
and if the side measures 7 paces
or a multiple thereof,
the diagonal measures 2x5 = 10 paces
or a multiple thereof

Then, still a long time ago, a clever early mathematician realized that
if the side measures a multiple of 5 + 7 = 12 paces or ropes, the diagonal
measures a multiple of 7 + 12 = 17 paces or ropes ...
Franz Gnaedinger
2015-12-18 09:13:41 UTC
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Post by Franz Gnaedinger
Post by Franz Gnaedinger
http://www.seshat.ch/home/egypt2.htm
http://www.seshat.ch/home/egypt3.htm
how to calculate the square (a lesson in experimental math-history)
The Babylonian mathematicians had two values for the square root of 2,
the simple value 17/12, or 1;25 in their sexagesimal number system,
and the fabulous value 1;24,51,10 (mentioned on the clay tablet YBC 7289
from around 1600 BC). How did they find the latter value? Draw up the
number column for the calculation of the square
1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
169 239 338
408 577 816
985 1393 1970
and so on
Divide 1393 by 1985 and you obtain 1;24,51,10,3,2... Let go the small
numbers ...3,2... and keep 1;24,51,10 (1.414219... for 1.4142135...).
Back to the Old Kingdom of Egypt and calculating with integers. How long
is the diagonal of a square whose side measures 83 royal cubits or 581
palms or 2324 fingerbreadths? Using the above number column you may proceed
as follows
side 83 or 70 5 5 3 rc
diagonal 99 7 7 4 sum 117 rc
(exactly 117.379... rounded 117 rc)
side 581 or 577 2 2 p
diagonal 9816 3 3 sum 822 p
(exactly 821.658... rounded 822 p)
side 2324 or 1393 577 239 99 12 2 2 f
diagonal 1970 816 338 140 17 3 3 sum 3287 f
(exactly 3286.632... rounded 3287 f)
Operate with finer and finer units and the result will be better and better.
Important is the algorithm
a b 2a
a+b b+2a 2(a+b)
You can begin with any number pair a and b, for example 4 and 11, and even
make mistakes
4 11 8
15 19 30
34 49 68
83 117 166 (note the numbers 83 and 117)
200 283 400
481 683 964 (mistake, 481 instead of 483)
1164 1647 2328
and so on
The number column starting with 4 11 and containing a mistake approximates
the square root of 2 more slowly, but still.
By the way, the regular number column starting with 1 1 and adding all
numbers correctly has an exact equivalent in the continued fraction
(1;2,2,2,2,2,2...).
And what was the origin of the above number column? I have reasons to
assume a Stone Age formula
If the side of a square measures 5 paces
or a multiple thereof,
the diagonal measures 7 paces
or a multiple thereof,
and if the side measures 7 paces
or a multiple thereof,
the diagonal measures 2x5 = 10 paces
or a multiple thereof
Then, still a long time ago, a clever early mathematician realized that
if the side measures a multiple of 5 + 7 = 12 paces or ropes, the diagonal
measures a multiple of 7 + 12 = 17 paces or ropes ...
Hesy and Apollo (calculating cube roots)

A comment on the double set of 14 and 14 continually increasing cylindrical
barrels in the Third Dynasty Tomb of Hesy at Saqqara says that their
calculation required the cube roots of 2 and 5 and 10, moreover the
knowledge of pi.

We already discussed the number of the circle. Now for the cube root of 2.
An elegant way to approximate it is by drawing up a further number column

1 1 1 2
2 2 3 4
4 5 7 8
9 12 15 18
3 4 5 6

3 4 5 6
7 9 11 14
16 20 25 32
36 45 57 72
12 15 19 24

12 15 19 24
27 34 43 54
61 77 97 122
138 174 219 276
46 58 73 92

46 58 73 92
104 131 165 208
235 276 373 470
531 669 843 1062
177 223 281 354

400 504 635 800
200 252
100 126
50 63

The surviving Greek version of an outer-European legend tells us how
the oracle at Delos helped the Athenians avert a plague in 430 BC
- by doubling the volume of Apollo's altar. If his altar should have
measured 50 by 100 by 200 units, the new one of the same proportions
would have measured 63 by 126 by 252 units. Or if the old one was a cube
measuring 50 by 50 by 50 units, the doubled volume would have measured
63 by 63 by 63 units. Moral of the story? the gods want us to do science
in order to cope with the hardships of life.

(My Paleo-linguistic studies reveal Apollo as god from Asia Minor
who guided shamans to a place where to build a new town and protect it
with a wall, AD POL LOP Apollo - toward AD fortified settlement Greek
polis POL enveloping hedge or fence or palisade or wall LOP - and did it
with omina, for example an arrow of light, a sun beam breaking through
the clouds and illuminating a favorable place.)

Number sequences (like the pi sequences explained previoulsy) provide
good values from poor and mediocre ones, and excellent values form
mediocre and good ones. Let us use them for calculating the cube roots
of 5 and 10.

17/10 is a little less than the cube root of 5, 12/7 slightly more

5 x 10x10x10 = 5000 17x17x17 = 4913

5 x 7x7x7 = 1715 12x12x12 = 1728

Draw up a number sequence

17/10 (plus 12/7) 29/17 41/24 53/31 65/38

65/38 is a fine value for the cube root of 5.

13/6 is a little more than the cube root of 10, 28/13 slightly less

10 x 6x6x6 = 2160 13x13x13 = 2197

10 x 13x13x13 = 21970 28x28x28 = 21952

13/6 (plus 28/13) 41/19 69/32 97/45 125/58 153/71 181/84
209/97 237/110 265/123

265/123 is a very fine value for the cube root of 10. But you can also
pick up another value that comes handy. If the edge of a cube measures
3 royal cubits or 21 palms or 84 fingerbreadths and you have to multiply
the volume by a factor of 10 you choose the value 181/84 - new edge 181
fingerbreadths, or 6 cubits 3 palms 1 finger.
Franz Gnaedinger
2015-12-21 07:14:03 UTC
Reply
Permalink
sci.archaeology has become nice and quiet. A peaceful forum where one can
focus on a topic. Today I work over my previous message on cube roots.
Next time will follow a message on mirror values and a method of drawing
any square root.

Hesy and Apollo (calculating cube roots)

A comment on the double set of 14 and 14 continually increasing cylindrical
barrels depicted on a wall in the Third Dynasty Tomb of Hesy at Saqqara
says that their calculation required the cube roots of 2 and 5 and 10,
moreover the knowledge of pi.

We already discussed the number of the circle. Now for the cube root of 2.
An elegant way to approximate it is by drawing up a further number column

1 1 1 2
2 2 3 4
4 5 7 8
9 12 15 18
3 4 5 6

3 4 5 6
7 9 11 14
16 20 25 32
36 45 57 72
12 15 19 24

12 15 19 24
27 34 43 54
61 77 97 122
138 174 219 276
46 58 73 92

46 58 73 92
104 131 165 208
235 276 373 470
531 669 843 1062
177 223 281 354

400 504 635 800
200 252
100 126
50 63

The surviving Greek version of an outer-European legend tells us how
the oracle at Delos helped the Athenians avert a plague in 430 BC
- by asking them to double the volume of Apollo's altar. If his altar
should have measured 50 by 100 by 200 units, the new one of the same
proportions would have measured 63 by 126 by 252 units. Or if the old one
was a cube measuring 50 by 50 by 50 units, the doubled volume would have
measured 63 by 63 by 63 units. Moral of the story? the gods want us to do
science in order to cope with the hardships of life.

(My Paleo-linguistic studies reveal Apollo as god from Asia Minor
who guided shamans to a place where to build a new town and protect it
with a wall, AD POL LOP Apollo - toward AD fortified settlement Greek
polis POL enveloping hedge or fence or palisade or wall LOP -, and did it
with omina, for example an arrow of light, a sun beam breaking through
the clouds and illuminating a favorable place.)

Number sequences (like the pi sequences explained previously) provide
good values from poor and mediocre ones, and excellent values from
mediocre and good ones. Let us use them for calculating the cube roots
of 5 and 10.

17/10 is a little less than the cube root of 5, 12/7 slightly more

17/10 x 17/10 x 17/10 = 4913/1000 = 4.913

12/7 x 12/7 x 12/7 = 1728/343 = 5.0379...

Draw up a number sequence

17/10 (plus 12/7) 29/17 41/24 53/31 65/38

65/38 is a fine value for the cube root of 5.

13/6 is a little more than the cube root of 10, 28/13 slightly less

13/6 x 13/6 x 13/6 = 2197/216 = 10.1712...

28/13 x 28/13 x 28/13 = 21952/2197 = 9.9918...

13/6 (plus 28/13) 41/19 69/32 97/45 125/58 153/71 181/84
209/97 237/110 265/123

265/123 is a very fine value for the cube root of 10. But you can also
pick up another value that comes handy. If the edge of a cube measures
3 royal cubits or 21 palms or 84 fingerbreadths and you have to multiply
the volume by a factor of 10 you choose the value 181/84 - new edge 181
fingerbreadths, or 6 cubits 3 palms 1 finger.
Franz Gnaedinger
2015-12-28 09:27:02 UTC
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Permalink
Post by Franz Gnaedinger
Hesy and Apollo (calculating cube roots)
A comment on the double set of 14 and 14 continually increasing cylindrical
barrels depicted on a wall in the Third Dynasty Tomb of Hesy at Saqqara
says that their calculation required the cube roots of 2 and 5 and 10,
moreover the knowledge of pi.
We already discussed the number of the circle. Now for the cube root of 2.
An elegant way to approximate it is by drawing up a further number column
1 1 1 2
2 2 3 4
4 5 7 8
9 12 15 18
3 4 5 6
3 4 5 6
7 9 11 14
16 20 25 32
36 45 57 72
12 15 19 24
12 15 19 24
27 34 43 54
61 77 97 122
138 174 219 276
46 58 73 92
46 58 73 92
104 131 165 208
235 276 373 470
531 669 843 1062
177 223 281 354
400 504 635 800
200 252
100 126
50 63
The surviving Greek version of an outer-European legend tells us how
the oracle at Delos helped the Athenians avert a plague in 430 BC
- by asking them to double the volume of Apollo's altar. If his altar
should have measured 50 by 100 by 200 units, the new one of the same
proportions would have measured 63 by 126 by 252 units. Or if the old one
was a cube measuring 50 by 50 by 50 units, the doubled volume would have
measured 63 by 63 by 63 units. Moral of the story? the gods want us to do
science in order to cope with the hardships of life.
(My Paleo-linguistic studies reveal Apollo as god from Asia Minor
who guided shamans to a place where to build a new town and protect it
with a wall, AD POL LOP Apollo - toward AD fortified settlement Greek
polis POL enveloping hedge or fence or palisade or wall LOP -, and did it
with omina, for example an arrow of light, a sun beam breaking through
the clouds and illuminating a favorable place.)
Number sequences (like the pi sequences explained previously) provide
good values from poor and mediocre ones, and excellent values from
mediocre and good ones. Let us use them for calculating the cube roots
of 5 and 10.
17/10 is a little less than the cube root of 5, 12/7 slightly more
17/10 x 17/10 x 17/10 = 4913/1000 = 4.913
12/7 x 12/7 x 12/7 = 1728/343 = 5.0379...
Draw up a number sequence
17/10 (plus 12/7) 29/17 41/24 53/31 65/38
65/38 is a fine value for the cube root of 5.
13/6 is a little more than the cube root of 10, 28/13 slightly less
13/6 x 13/6 x 13/6 = 2197/216 = 10.1712...
28/13 x 28/13 x 28/13 = 21952/2197 = 9.9918...
13/6 (plus 28/13) 41/19 69/32 97/45 125/58 153/71 181/84
209/97 237/110 265/123
265/123 is a very fine value for the cube root of 10. But you can also
pick up another value that comes handy. If the edge of a cube measures
3 royal cubits or 21 palms or 84 fingerbreadths and you have to multiply
the volume by a factor of 10 you choose the value 181/84 - new edge 181
fingerbreadths, or 6 cubits 3 palms 1 finger.
mirror values (from number columns to Heron)

First eight lines of the basic number column approximating the square
root of 2

1 1 2 line 1*
2 3 4 line 2*
5 7 10 line 3
12 17 24 line 4*
29 41 58 line 5
70 99 140 line 6
169 239 338 line 7
408 577 816 line 8*

Each line provides a pair of what I call mirror values

1/1 and 2/1
3/2 and 4/3
7/5 and 10/7
17/12 and 24/17
41/29 and 58/41
99/70 and 140/99
239/169 and 338/239
577/408 and 816/577

The product of a pair of mirror values equals 2, while the exact value
of the square root of 2 lies in between them, suggesting that we take
their average in get a better value

1/1 plus 2/1 equals 3/1 halved 3/2 mirror value 4/3
3/2 plus 4/3 equals 17/6 halved 17/12 mirror value 24/17
17/12 plus 24/17 equals 577/204 halved 577/408 mirror value 816/577

The new values belong to the same number column, but instead of proceeding
slowly, from line to line, 1 2 3 4 5 6 7 8 ..., we are now leaping forward
in ever bigger steps, above marked with asterisks, 1 2 4 8 16 32 64 128 ...

The method of adding a pair of mirror values and halving their sum
was mentioned by Heron in Metrica, and therefore named Heron's method,
but I am certain that it was already known to the Egyptians, because
it springs directly from the number columns, and allows to draw any
square root, for the example the one of 23

1 ? 23

1 x 23 = 25 5 x 5 = 25 (close to 23)

1 5 23

5/1 plus 23/5 equals 48/5 halved 24/5 mirror value 115/24

24/5 plus 115/24 equals 1151/120 halved 1151/240

1151/240 = 4.7958333...
sqrt 23 = 4.7958315...

1151/240 is already a very fine value for the square root of 23.
Franz Gnaedinger
2016-01-15 08:43:06 UTC
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Post by Franz Gnaedinger
mirror values (from number columns to Heron)
First eight lines of the basic number column approximating the square
root of 2
1 1 2 line 1*
2 3 4 line 2*
5 7 10 line 3
12 17 24 line 4*
29 41 58 line 5
70 99 140 line 6
169 239 338 line 7
408 577 816 line 8*
Each line provides a pair of what I call mirror values
1/1 and 2/1
3/2 and 4/3
7/5 and 10/7
17/12 and 24/17
41/29 and 58/41
99/70 and 140/99
239/169 and 338/239
577/408 and 816/577
The product of a pair of mirror values equals 2, while the exact value
of the square root of 2 lies in between them, suggesting that we take
their average in get a better value
1/1 plus 2/1 equals 3/1 halved 3/2 mirror value 4/3
3/2 plus 4/3 equals 17/6 halved 17/12 mirror value 24/17
17/12 plus 24/17 equals 577/204 halved 577/408 mirror value 816/577
The new values belong to the same number column, but instead of proceeding
slowly, from line to line, 1 2 3 4 5 6 7 8 ..., we are now leaping forward
in ever bigger steps, above marked with asterisks, 1 2 4 8 16 32 64 128 ...
The method of adding a pair of mirror values and halving their sum
was mentioned by Heron in Metrica, and therefore named Heron's method,
but I am certain that it was already known to the Egyptians, because
it springs directly from the number columns, and allows to draw any
square root, for the example the one of 23
1 ? 23
1 x 23 = 25 5 x 5 = 25 (close to 23)
1 5 23
5/1 plus 23/5 equals 48/5 halved 24/5 mirror value 115/24
24/5 plus 115/24 equals 1151/120 halved 1151/240
1151/240 = 4.7958333...
sqrt 23 = 4.7958315...
1151/240 is already a very fine value for the square root of 23.
square root of 4 (how number columns work)

The square root of 4 is 2. Let us calculate it by drawing up a further
number column

1 1 4
2 5 8
7 13 28
20 41 80
61 121 244
182 365 728

The mirror values approach 2 from below (1/1) and above (4/1). We obtain
the exact mirror values by alternatingly adding and subtracting one to or
from the middle number

1 2 4
2 4 8
7 14 28
20 40 80
61 122 244
182 364 728

The absolute mistake of the middle number in the initial number column
1 1 4 is always one, while the numbers get larger and larger, so the
relative mistake is getting ever smaller, meaning the number column
1 1 4 really approaches 2. Studying this number column may have allowed
the Ancient Egyptians and the Mesopotamians to understand how the other
number columns work - 1 1 2, 1 1 3, 1 1 5, 1 1 1 2.
Franz Gnaedinger
2016-01-20 09:15:27 UTC
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Permalink
Post by Franz Gnaedinger
square root of 4 (how number columns work)
The square root of 4 is 2. Let us calculate it by drawing up a further
number column
1 1 4
2 5 8
7 13 28
20 41 80
61 121 244
182 365 728
The mirror values approach 2 from below (1/1) and above (4/1). We obtain
the exact mirror values by alternatingly adding and subtracting one to or
from the middle number
1 2 4
2 4 8
7 14 28
20 40 80
61 122 244
182 364 728
The absolute mistake of the middle number in the initial number column
1 1 4 is always one, while the numbers get larger and larger, so the
relative mistake is getting ever smaller, meaning the number column
1 1 4 really approaches 2. Studying this number column may have allowed
the Ancient Egyptians and the Mesopotamians to understand how the other
number columns work - 1 1 2, 1 1 3, 1 1 5, 1 1 1 2.
humor in Egyptian mathematics (RMP 32)

The Rhind Mathematical Papyrus RMP was written by one Ahmose or Ahmes
in around 1 650 BC, as copy of a lost scroll from around 1 850 BC.

In problem 32 (modern counting) Ahmes divides 2 by 1 1/3 1/4 and obtains
1 1/6 1/12 1/114 1/228

2 divided by 1 '3 '4 equals 1 '6 '12 '114 '228

Beginners learn how to handle unit fraction series. Advanced learners would
have been given a more demanding task. Imagine a right parallel-epiped
measuring 2 by 1 '3 '4 by 1 '6 '12 '114 '228 units. How long is the
diagonal of the volume?

Impossible to calculate?

No, quite simple. The diagonal of the volume measures exactly

1 '3 '4 plus 1 '6 '12 '114 '228

1 1 plus '3 '6 plus '4 '12 plus '114 '228

2 '3 '3 '76 units

There is an underlying theorem. Divide 2 by any number A and you obtain B.
Let a right parallel-epiped measure 2 by A by B units and the diagonal of
the volume measures exactly A plus B units.

Back to the basic number column approximating the square root of 2

1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140

Splitting the middle number of each line

1 0+1 2
2 1+2 4
5 3+4 10
12 8+9 24
29 20+21 58
70 49+50 140

These lines contain alternatingly a triple and a quadruple

0-1-1 quasi-triple

2-1-2-3 quadruple

3-4-5 triple (Sacred Triangle)

12-8-9-17 quadruple

20-21-29 triple

70-49-50-99 quadruple

Let us consider the line 12 17 24 or 12 8+9 24 and the quadruple 12-8-9-17.
If a square measures 12 by 12 plams, the diagonal measures practically 17
palms. If a right parallel-epiped measures 12 by 8 by 9 palms, the diagonal
of the volume measures exactly 17 palms.

Now let us divide the numbers 12 and 8 and 9 by 6

12 divided by 6 equals 2
8 divided by 6 equals 1 '3
9 divided by 6 equals 1 '2

These numbers satisfy the equation of the 'magic parallel-epiped'

2 divided by 1 '3 equals 1 '2

The joke of RMP 32 ('immpossible' calculation solved easily) might well
go back to the study of the simplest number column that was already known
to the mathematicians of the Old Kingdom, the line 12 17 24 having been
used by Imhotep at Saqqara, and the line 70 99 140 by Hemon at Giza.
Franz Gnaedinger
2016-02-01 08:08:32 UTC
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Post by Franz Gnaedinger
humor in Egyptian mathematics (RMP 32)
The Rhind Mathematical Papyrus RMP was written by one Ahmose or Ahmes
in around 1 650 BC, as copy of a lost scroll from around 1 850 BC.
In problem 32 (modern counting) Ahmes divides 2 by 1 1/3 1/4 and obtains
1 1/6 1/12 1/114 1/228
2 divided by 1 '3 '4 equals 1 '6 '12 '114 '228
Beginners learn how to handle unit fraction series. Advanced learners would
have been given a more demanding task. Imagine a right parallel-epiped
measuring 2 by 1 '3 '4 by 1 '6 '12 '114 '228 units. How long is the
diagonal of the volume?
Impossible to calculate?
No, quite simple. The diagonal of the volume measures exactly
1 '3 '4 plus 1 '6 '12 '114 '228
1 1 plus '3 '6 plus '4 '12 plus '114 '228
2 '3 '3 '76 units
There is an underlying theorem. Divide 2 by any number A and you obtain B.
Let a right parallel-epiped measure 2 by A by B units and the diagonal of
the volume measures exactly A plus B units.
Back to the basic number column approximating the square root of 2
1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
Splitting the middle number of each line
1 0+1 2
2 1+2 4
5 3+4 10
12 8+9 24
29 20+21 58
70 49+50 140
These lines contain alternatingly a triple and a quadruple
0-1-1 quasi-triple
2-1-2-3 quadruple
3-4-5 triple (Sacred Triangle)
12-8-9-17 quadruple
20-21-29 triple
70-49-50-99 quadruple
Let us consider the line 12 17 24 or 12 8+9 24 and the quadruple 12-8-9-17.
If a square measures 12 by 12 plams, the diagonal measures practically 17
palms. If a right parallel-epiped measures 12 by 8 by 9 palms, the diagonal
of the volume measures exactly 17 palms.
Now let us divide the numbers 12 and 8 and 9 by 6
12 divided by 6 equals 2
8 divided by 6 equals 1 '3
9 divided by 6 equals 1 '2
These numbers satisfy the equation of the 'magic parallel-epiped'
2 divided by 1 '3 equals 1 '2
The joke of RMP 32 ('immpossible' calculation solved easily) might well
go back to the study of the simplest number column that was already known
to the mathematicians of the Old Kingdom, the line 12 17 24 having been
used by Imhotep at Saqqara, and the line 70 99 140 by Hemon at Giza.
A Japanese myon scan of a glacier in the Swiss Alps revealed
the underside of the huge ice mass. If there is a method that could
possibly detect the hypothetical sun chambers in the Red Pyramid
(44 meters above the base) and in the Great Pyramid (90.6 m)
then this one.

I will go on with my thread until we have myon scans of those pyramids.
Elijahovah
2016-08-09 22:08:16 UTC
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Permalink
INSANITY AMONG PYRAMIDOLOGY

People who porclaim to be scholars (schooled) and perhaps were schooled by other idiots like themselves are worshippers of magic numerology in mathematics. Am i judging? Not at all. Their conclusions do the judging of themselves.

The Genesis of the Bible holds opposite views of the pyramids based on longevity of all organics making a crashing plummet. Simply put the short genesis claimed as truth by Jews states that 4 dynasties were kings who outlived dynasty kings of 5 thru 8. This means they die dyring the 12th dynasty of which fact is they are buried with 12th dynasty objects, and OUR scholars like to claim tombs were opened to put this stuff there. The double dating such as very old mummies with very young 12th dynasty carbon does not carry weight because it is the lesser C-14 in those people and animals that caused them to live so much longer and die with very little C-14 in them. Thus a man who lives 400 years from 2300-1900bc appears to be 3000bc because he has the extremely low C-14 of immediate post-Flood years. Your main mass of C-14 in DNA is not lost to breathing out burned carbons. You do not digest and burn your DNA carbon. It was put there the first as your main mass bulk during the first 20-30 years of your life. Thus the C-14 of a Man (2300-1900bc) will be far less and matching the 30 years 2300-2270bc than the man who lives 200 as 2150-1950bc and has the C-14 of 2150-2120bc. The man dying at 400 in 1900bc will appear 3000bc while his son dying before him in 1950bc will appear as 2000bc.

THE POINT then is long Genesis implies 400-year old ancestral parents died centuries before later dynasties and so pyramids are tombs. BUT Hebrew genesis says pyramids were built before longevity deaths started from the 8 people on the ark. Specifically the tombs of Ur in 2030bc. Egypt claims Peleg of Ur is 2321bc dying when UNAS Sakkara died.

NEXT you people keep Americanizing things which corrupt real Egyptian. The British spelling of the sun god Re has its reason. It is pronounced as Ray not as Rah. We have sun rays not sun rahs. The rays of the sun, Ray was born on day 40 of the Flood. There is no Shamash before the Flood, no distinct shadow line to measure geometry and trigonometry with. Thus nor is theri any INSIDE ray of the pyramid unless a shaft lets Ray down inside. There will be no chmaber with no shaft because Ray shines on a pyramid,. not inside one, unless a shaft to the east or south or west lets him in. Thus the year, the date, the simple calendar is what would locate any entrance you people cannot find yet. I have a device, a globe, that shows these points. BUT assholes are greedy people who seek their fame, their wealth, their ego of what THEY think. So i rather die pushing what i know of the near-asteroid Armageddon and the only one way to survive it.

People go back to vomit. 95% of books claim the reflective Great Pyramid shaft from descent to ascend is an exact reflection. 5% list the true precession change for 92 years so as to fail the 100-year Seth cycle the pyramid was built to observe as difference between 100 of 360-day and 100 of 365-day and its 25 leap days.

Those years are 2170-2078bc, ascending shaft 2078-1986bc, last chamber (kings) being the holy and most holy for sarcophagus records (writings in the coffin of Noah's ark thru the Flood) is 1986-1894bc which is why THE TOWER is confused with first year of Babylon (Babel 2240bc, but its first king 1894bc). The sarcophagus is thus not for the body of Kay-Ef (KF which is british Che-Off which people are saying as Cheeee-ops like chimpanzee, while others say KhuFu when i never heard the letters K and F called Ku and Fu). THE POINT is the coffin in the Great Pyramid is for text records, as in presuming NOah's ark still has text records left behind on it, from the preFlood world whose writing had no global system yet until Ur's king Reu in 2207bc.

The idea then is that the coffin sarcophagus is the same exact purpose as ark of the covenant by Moses. And the kings chamber is actually a Holy and Most Holy just as Moses had been in and so likewise designed a duplicate as that Great Pyramid.

So with these facts do we gather and go forward? NO. Because scholars are assholes. They bicker, they dispute, they have their better ideas. They seek their own funding. So i don't choose the doctrine that men are evil and Armageddon destroys them. I choose the words of Jesus that the idiots take no notice when it happens that they could have foreseen it and predicted just like Noah building an ark and getting on it. Did you know Mayan calendar proves that asteroid impact struck 7 days before Noah closed the door and the tsunami raised the ark while the tar was still wet on that door? Think about it, because you are all about to die. YOU don't know your pyramids.
Harmonica Harp
2019-02-05 13:15:56 UTC
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Post by Elijahovah
INSANITY AMONG PYRAMIDOLOGY
People who proclaim to be scholars (schooled) and perhaps were schooled by other idiots like themselves are worshipers of magic numerology in mathematics. Am I judging? Not at all. Their conclusions do the judging of themselves.
The Genesis of the Bible holds opposite views of the pyramids based on longevity of all organics making a crashing plummet. Simply put the short Genesis claimed as truth by Hebrews states that 4 dynasties were kings who outlived dynasty kings of 5 thru 8. This means they die during the 12th dynasty of which fact is they are buried with 12th dynasty objects, and OUR scholars like to claim tombs were opened to put this stuff there. The double dating such as very old mummies with very young 12th dynasty carbon does not carry weight because it is the lesser C-14 in those people and animals that caused them to live so much longer and die with very little C-14 in them. Thus a man who lives 400 years from 2300-1900 BC appears to be 3000 BC because he has the extremely low C-14 of immediate post-Flood years. Your main mass of C-14 in DNA is not lost to breathing out burned carbons. You do not digest and burn your DNA carbon. It was put there the first as your main mass bulk during the first 20-30 years of your life. Thus the C-14 of a Man (2300-1900 BC) will be far less and matching the 30 years 2300-2270 BC than the man who lives 200 as 2150-1950 BC and has the C-14 of 2150-2120 BC. The man dying at 400 in 1900 BC will appear 3000 BC while his son dying before him in 1950 BC will appear as 2000 BC.
THE POINT then is long Genesis implies 400-year old ancestral parents died centuries before later dynasties and so pyramids are tombs. BUT Hebrews Genesis says pyramids were built before longevity deaths started from the 8 people on the ark. Specifically the tombs of Ur in 2030 BC. Egypt claims Peleg of Ur is 2321 BC dying when Unas Saqqara died.
NEXT you people keep Americanizing things which corrupt real Egyptian. The British spelling of the Sun God Re has its reason. It is pronounced as Ray not as Rah. We have Sun rays not Sun rahs. The rays of the Sun, Ray was born on day 40 of the Flood. There is no Shamash before the Flood, no distinct shadow line to measure geometry and trigonometry with. Thus nor is there any INSIDE ray of the pyramid unless a shaft lets Ray down inside. There will be no chamber with no shaft because Ray shines on a pyramid, not inside one, unless a shaft to the east or south or west lets him in. Thus the year, the date, the simple calendar is what would locate any entrance you people cannot find yet. I have a device, a globe, that shows these points. BUT assholes are greedy people who seek their fame, their wealth, their ego of what THEY think. So I'd rather die pushing what I know of the near-asteroid Armageddon and the only one way to survive it.
People go back to vomit. 95% of books claim the reflective Great Pyramid shaft from descent to ascend is an exact reflection. 5% list the true precession change for 92 years so as to fail the 100-year Seth cycle the pyramid was built to observe as difference between 100 of 360-day and 100 of 365-day and its 25 leap days.
Those years are 2170-2078 BC, ascending shaft 2078-1986 BC, last chamber (kings) being the holy and most holy for sarcophagus records (writings in the coffin of Noah's ark thru the Flood) is 1986-1894 BC which is why THE TOWER is confused with first year of Babylon (Babel 2240 BC, but its first king 1894 BC). The sarcophagus is thus not for the body of Kay-Ef (KF which is british Che-Off which people are saying as Che-ops like chimpanzee, while others say KhuFu when I never heard the letters K and F called Ku and Fu). THE POINT is the coffin in the Great Pyramid is for text records, as in presuming Noah's ark still has text records left behind on it, from the pre-Flood world whose writing had no global system yet until Ur's king Reu in 2207 BC.
The idea then is that the coffin sarcophagus is the same exact purpose as the ark of the covenant by Moses. And the kings chamber is actually a Holy and Most Holy just as Moses had been in and so likewise designed a duplicate as that Great Pyramid.
So with these facts do we gather and go forward? NO. Because scholars are assholes. They bicker, they dispute, they have their better ideas. They seek their own funding. So I don't choose the doctrine that men are evil and Armageddon destroys them. I choose the words of Jesus that the idiots take no notice when it happens that they could have foreseen it and predicted just like Noah building an ark and getting on it. Did you know the Mayan calendar proves that asteroid impact struck 7 days before Noah closed the door and the tsunami raised the ark while the tar was still wet on that door? Think about it, because you are all about to die. YOU don't know your pyramids.
https://www.urbandictionary.com/define.php?term=Rah

There you go.

O John the Baptist
O LUCIFER the Devil Satan
2019-02-15 12:47:03 UTC
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♥ Pauline LaFon, a French (Brittany) woman, Solved Stonehenge
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MelchizedekKingofSalem
@MelchizedekKingofSalem
7 Comments ♥ 1 Like Received
LATEST ACTIVITY

♥ Privacy ♥
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MelchizedekKingofSalem
1h ♥ Edited

The Longest Pit Stop, Ever! by Pauline LaFon
Best rejected Super Bowl ad ($5.5M refund)

We like Tucker Corporation's Race Car AD 24 the best. The One where Moses, the Father of Texas, is pointing due West from atop Pisgah on Mount Nebo, then it flashes to Ark of the Covenant electro-magnetic Race Car driver Moses peeling out of there making a Pit Stop at the Heel Stone underground Garage of Stonehenge, next crossing the Atlantic Ocean atop the Spruce Goose landing in Nacogdoches the Oldest Town in Texas, where it drives off Spruce Goose's starboard wing across The Alamo finish line, and ends with Moses, the Father of Texas, smiling at his Ark of the Covenant electro-magnetic Race Car silhouetted by a San Antonio sunset in the Promised Land.

The Alamo: 29.4260° N, 98.4861° W
Mount Nebo: 31.7683° N, 35.7253° E

Happy Valentine's Day!
Mummy
_______________________________________________________

MelchizedekKingofSalem
1h ♥ Edited

When you're Right you're Right (okay will do!) - William Harry
_______________________________________________________

MelchizedekKingofSalem
3h ♥ Edited

The South will Rise Again (save our statues!) - Stonehenge Black-skin Cheddar Man is from The South (Gobekli Tepe, Anatolia). Stonehenge Olivine-rich Altar Stone is from The South (W. Pontides, Turkey). It's from Brittany? C'mon man!
Go South, young man. Go South.
_______________________________________________________

MelchizedekKingofSalem
11h ♥ Edited

Popeye the Sailor (a/k/a Recep Tayyip Erdoan) said that Gobekli Tepe's where Stonehenge Altar Stone's from. He loves Spinach, and Turkey too. Go figure?
_______________________________________________________

MelchizedekKingofSalem
12h ♥ Edited

President of Turkey (Anatolia), Recep Tayyip Erdoan, said that Turkey (Anatolia) will pay the shipping cost for the Return of Stonehenge Altar Stone to its capitol, Ankara, Turkey (Anatolia); so its shipping cost is covered.
Sole Proprietor of Stonehenge, A Texas Tycoon, Garry Denke, said that A Texas Tycoon will pay the shipping cost for the Return of Stonehenge Altar of Burnt Offering and its contents: 1. gold Mercy Seat, 2. gold Ark of the Testimony, 3. gold Table for the Shewbread, 4. gold Candlestick, 5. gold Ephod-Girdle, 6. gold Breastplate, 7. gold Altar of Incense (1-7 partial list of contents) to San Antonio, Texas (the Alamo); so its shipping cost is covered.
Prime Minister of United Kingdom, Theresa Mary Brasier, said that Elizabeth Alexandra Mary will pay the shipping costs for the Return of Stonehenge Bluestones, Sarsens, and High Tor Limestone to Preseli, Marlborough, and Gower, respectively; so their shipping costs are covered.

Ship the stolen Bluestones back to Preseli because its stolen property,
Ship the stolen Sarsens back to Marlborough because its stolen property,
Ship the stolen High Tor Limestone back to Gower because its stolen property,
Ship the stolen Altar Stone back to Anatolia (Turkey) because its stolen property,
and Ship the stolen Mishkan (1.2m below Heel Stone) back to G-D (Texas) because
its stolen property.

A tad on the expensive side You're saying?
Dadgum less than 2 Armadillo Tunnels!
_______________________________________________________

MelchizedekKingofSalem
1d ♥ Edited

Ancients survived Ice Age inside Coal Caves;

Stonehenge Car Park Postholes, Stonehenge Mesolithic Postholes
Westernmost post "A" represents "Pembrokeshire Coal Field"
The Centre post "B" represents "South Wales Coalfield"
Easternmost post "C" represents "Bristol Coalfield"

Avebury coal duster, Cursus coal duster, Durrington Walls coal duster, Long Barrow coal duster, Robin Hood's Ball coal duster, Stonehenge coal duster, Woodhenge coal duster, etc, all being originally surface coal hunting failures. Every one of them were coal exploration sites that did not yield any coal.

Take away all of the dressed up cemetery headstone rocks and what have you got? Nothing more than a bunch of coal exploratory ditches and holes, that is what. Afterwards, these ditches and holes were utilized as grave plots, for tired disappointed coal explorers, and their cold disheartened families.

Three Mesolithic Posts: The 40 Mile Markers;

1. Glaciers transport Bluestones to the Salisbury Plain.
2. Ice Age hunters destroy forests in Wales - England.
3. Coal miners from the West see their same Rocks.
4. Exploration for Surface Coal below chalk begins.
5. Pembrokeshire-South Wales-Bristol trend fails.
6. Wales, England coal miners build cemeteries.
7. Shoring a Coal Mine the Ancient's Message.

Melchizedek King of Salem
Coal Exploration Dept.
_______________________________________________________

MelchizedekKingofSalem
1d ♥ Edited

Prolly why I hid my Tabernacle at Hele Stone (after shipping).
_______________________________________________________

Stonehenge mystery solved? Prehistoric sailors may have been responsible for legendary str...♥ Stonehenge is perhaps the most famous rock structure...♥
https://www.foxnews.com/science/stonehenge-mystery-solved-prehistoric-sailors-may-have-been-responsible-for-legendary-structure
♥ foxnews.com

Conversation on Fox News ♥ 246 Comments

Jesus Christ! Space missed! You idiot!
Hey, I'm Deep UNDERCOVER here.
Cause NO MORE dadgum errors!

O LUCIFER
the Devil
Satan
O LUCIFER the Devil Satan
2019-02-15 16:09:48 UTC
Reply
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Post by O LUCIFER the Devil Satan
♥ Pauline LaFon, a French (Brittany) woman, Solved Stonehenge
https://media.spokesman.com/photos/2004/12/16/16_OBIT_PAULINE_GORE_12-16-2004_ML3AFDD_t400.jpg?fd5af0684d698ce74dd4392bafb4f89a6dc66ee3
MelchizedekKingofSalem
@MelchizedekKingofSalem
7 Comments ♥ 1 Like Received
LATEST ACTIVITY
♥ Privacy ♥
_______________________________________________________
MelchizedekKingofSalem
1h ♥ Edited
The Longest Pit Stop, Ever! by Pauline LaFon
Best rejected Super Bowl ad ($5.5M refund)
We like Tucker Corporation's Race Car AD 24 the best. The One where Moses, the Father of Texas, is pointing due West from atop Pisgah on Mount Nebo, then it flashes to Ark of the Covenant electro-magnetic Race Car driver Moses peeling out of there making a Pit Stop at the Heel Stone underground Garage of Stonehenge, next crossing the Atlantic Ocean atop the Spruce Goose landing in Nacogdoches the Oldest Town in Texas, where it drives off Spruce Goose's starboard wing across The Alamo finish line, and ends with Moses, the Father of Texas, smiling at his Ark of the Covenant electro-magnetic Race Car silhouetted by a San Antonio sunset in the Promised Land.
The Alamo: 29.4260° N, 98.4861° W
Mount Nebo: 31.7683° N, 35.7253° E
Happy Valentine's Day!
Mummy
_______________________________________________________
MelchizedekKingofSalem
1h ♥ Edited
When you're Right you're Right (okay will do!) - William Harry
_______________________________________________________
MelchizedekKingofSalem
3h ♥ Edited
The South will Rise Again (save our statues!) - Stonehenge Black-skin Cheddar Man is from The South (Gobekli Tepe, Anatolia). Stonehenge Olivine-rich Altar Stone is from The South (W. Pontides, Turkey). It's from Brittany? C'mon man!
Go South, young man. Go South.
_______________________________________________________
MelchizedekKingofSalem
11h ♥ Edited
Popeye the Sailor (a/k/a Recep Tayyip Erdoan) said that Gobekli Tepe's where Stonehenge Altar Stone's from. He loves Spinach, and Turkey too. Go figure?
_______________________________________________________
MelchizedekKingofSalem
12h ♥ Edited
President of Turkey (Anatolia), Recep Tayyip Erdoan, said that Turkey (Anatolia) will pay the shipping cost for the Return of Stonehenge Altar Stone to its capitol, Ankara, Turkey (Anatolia); so its shipping cost is covered.
Sole Proprietor of Stonehenge, A Texas Tycoon, Garry Denke, said that A Texas Tycoon will pay the shipping cost for the Return of Stonehenge Altar of Burnt Offering and its contents: 1. gold Mercy Seat, 2. gold Ark of the Testimony, 3. gold Table for the Shewbread, 4. gold Candlestick, 5. gold Ephod-Girdle, 6. gold Breastplate, 7. gold Altar of Incense (1-7 partial list of contents) to San Antonio, Texas (the Alamo); so its shipping cost is covered.
Prime Minister of United Kingdom, Theresa Mary Brasier, said that Elizabeth Alexandra Mary will pay the shipping costs for the Return of Stonehenge Bluestones, Sarsens, and High Tor Limestone to Preseli, Marlborough, and Gower, respectively; so their shipping costs are covered.
Ship the stolen Bluestones back to Preseli because its stolen property,
Ship the stolen Sarsens back to Marlborough because its stolen property,
Ship the stolen High Tor Limestone back to Gower because its stolen property,
Ship the stolen Altar Stone back to Anatolia (Turkey) because its stolen property,
and Ship the stolen Mishkan (1.2m below Heel Stone) back to G-D (Texas) because
its stolen property.
A tad on the expensive side You're saying?
Dadgum less than 2 Armadillo Tunnels!
_______________________________________________________
MelchizedekKingofSalem
1d ♥ Edited
Ancients survived Ice Age inside Coal Caves;
Stonehenge Car Park Postholes, Stonehenge Mesolithic Postholes
Westernmost post "A" represents "Pembrokeshire Coal Field"
The Centre post "B" represents "South Wales Coalfield"
Easternmost post "C" represents "Bristol Coalfield"
Avebury coal duster, Cursus coal duster, Durrington Walls coal duster, Long Barrow coal duster, Robin Hood's Ball coal duster, Stonehenge coal duster, Woodhenge coal duster, etc, all being originally surface coal hunting failures. Every one of them were coal exploration sites that did not yield any coal.
Take away all of the dressed up cemetery headstone rocks and what have you got? Nothing more than a bunch of coal exploratory ditches and holes, that is what. Afterwards, these ditches and holes were utilized as grave plots, for tired disappointed coal explorers, and their cold disheartened families.
Three Mesolithic Posts: The 40 Mile Markers;
1. Glaciers transport Bluestones to the Salisbury Plain.
2. Ice Age hunters destroy forests in Wales - England.
3. Coal miners from the West see their same Rocks.
4. Exploration for Surface Coal below chalk begins.
5. Pembrokeshire-South Wales-Bristol trend fails.
6. Wales, England coal miners build cemeteries.
7. Shoring a Coal Mine the Ancient's Message.
Melchizedek King of Salem
Coal Exploration Dept.
_______________________________________________________
MelchizedekKingofSalem
1d ♥ Edited
Prolly why I hid my Tabernacle at Hele Stone (after shipping).
_______________________________________________________
Stonehenge mystery solved? Prehistoric sailors may have been responsible for legendary str...♥ Stonehenge is perhaps the most famous rock structure...♥
https://www.foxnews.com/science/stonehenge-mystery-solved-prehistoric-sailors-may-have-been-responsible-for-legendary-structure
♥ foxnews.com
Conversation on Fox News ♥ 246 Comments
Jesus Christ! Space missed! You idiot!
Hey, I'm Deep UNDERCOVER here.
Cause NO MORE dadgum errors!
O LUCIFER
the Devil
Satan
https://groups.google.com/forum/#!topic/alt.security.terrorism/PnqXFvGSMZ0
https://groups.google.com/forum/#!topic/soc.history.ancient/LfucyJf3EXk
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https://groups.google.com/forum/#!topic/sci.archaeology/di5dU8hn4Cc
https://groups.google.com/forum/#!topic/alt.atheism/eTZVMAMvmWg
https://groups.google.com/forum/#!topic/sci.physics/rBLe4MOlVtI
https://groups.google.com/forum/#!topic/anti-theism/aNMN5nn3Lwc
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https://groups.google.com/forum/#!topic/antitheism/mpO14VKedIo
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Major Tom Threepersons
2019-02-05 11:37:25 UTC
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Oy, my Dad (Garry Denke) has been called worse.
You can do better than that! Surely.

O Lord Jesus Christ
O LUCIFER the Devil Satan
2019-02-14 10:00:38 UTC
Reply
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Post by Major Tom Threepersons
Oy, my Dad (Garry Denke) has been called worse.
You can do better than that! Surely.
O Lord Jesus Christ
President of Turkey (Anatolia), Recep Tayyip Erdoan, said that Turkey (Anatolia) will pay the shipping cost for the Return of Stonehenge Altar Stone to its capitol, Ankara, Turkey (Anatolia); so its shipping cost is covered.
Sole Proprietor of Stonehenge, A Texas Tycoon, Garry William Denke, said that A Texas Tycoon will pay the shipping cost for the Return of Stonehenge Altar of Burnt Offering and its contents: 1. gold Mercy Seat, 2. gold Ark of the Testimony, 3. gold Table for the Shewbread, 4. gold Candlestick, 5. gold Ephod-Girdle, 6. gold Breastplate, 7. gold Altar of Incense (1-7 partial list of contents) to San Antonio, Texas (the Alamo); so its shipping cost is covered.
Prime Minister of United Kingdom, Theresa Mary Brasier, said that Elizabeth Alexandra Mary will pay the shipping costs for the Return of Stonehenge Bluestones, Sarsens, and High Tor Limestone to Preseli, Marlborough, and Gower, respectively; so their shipping costs are covered.

Ship the stolen Bluestones back to Preseli because its stolen property,
Ship the stolen Sarsens back to Marlborough because its stolen property,
Ship the stolen High Tor Limestone back to Gower because its stolen property,
Ship the stolen Altar Stone back to Anatolia (Turkey) because its stolen property,
and Ship the stolen Mishkan (1.2m below Heel Stone) back to G-D (Texas) because
its stolen property.

A tad on the expensive side You're saying?
Dadgum less than 2 Armadillo Tunnels!

https://www.foxnews.com/science/stonehenge-mystery-solved-prehistoric-sailors-may-have-been-responsible-for-legendary-structure

Popeye the Sailor (a/k/a Recep Tayyip Erdoan) said that Gobekli Tepe's where
Stonehenge Altar Stone's from. He loves Spinach, and Turkey too. Go figure?

Melchizedek
King of Salem
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