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## Re: How to Derive the Cubit from the Meter

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• ... You are welcome Chris. ... The original idea was to divide the arc of the meridian into 10 million equal parts or the polar circumference into 40 million
Message 1 of 10 , Jan 31, 2007

--- In sacredlandscapelist Chris wrote:

>
> Hi John, that was interesting. Thanks for sending it along.

You are welcome Chris.

> For the benefit of others on the list, I would like to illustrate that
> you aren't putting the cart before the horse. The meter was originally
> formed from a conceptual belt or perimeter around the earth made up of
> 10 million units.

The original idea was to divide the arc of the meridian into 10 million equal parts or the polar circumference into 40 million equal parts. The French mismeasure of the earth fixed the length of the meter imperfectly.  If it measured the polar circumference of the earth correctly there would be 40 million meters in it - not 39940671.42790.   One 1/40000000 of the Polar Cicrumference equals .99851678 meters not 1 meter.  The mile measures the polar circumference more accurately than the meter.  When the edge of a cube equals 1 mile = 1609.344 meters, the length of the radius of the sphere equal in volume to the cube is 998.35734 meters which equals 1/40,000 of the polar circumference 998.51678 meters accurate to 1/2 foot

1609.344³ = 4/3Pi(998.35734)³ =

Volume of Cube of Edge Length 1 Mile = Volume of Sphere of Radius equal to Polar Circumference / 40000

1 kilometer equals 1/40,000 of the polar circumference accurate to 5 feet

If one designs the meter as 1/40000000 of the geometric circumference of the earth -40007770 meters - ( the square root of the product of the equatorial and polar circumferences ) - the kilometer is more accurate than the mile.

> When we go back to the cubit, do you believe the ancients had
> sophisticated knowledge of the earth's size?

I do not know.  It is common knowledge that there are

360 * 365,221 feet in the equatorial circumference of the earth

Does it indicate the length of the year and the length of the equator integrated into the length of the foot?

Clearly, there is a very simple geometrical relationship between the kilometer and the mile, and the cubit and the meter, the kilometer being the radius of the sphere equal in volume to the cube of edge length one mile, and the cubit the ratio of the volume of the sphere of diameter one meter and the area of a square meter.  Since there are 1760 yards in one mile, the number of cubits in the perimeter of the Great Pyramid, it is possible to relate the cubit to the yard, although it is a bit complex.  The occurance of 1760 yards in one mile, and the relationship between the kilometer and the mile, and the meter and the cubit, suggests the possibility of the memory of a tradition that there are 1760 cubits in the perimeter of the Great Pyramid.   Is there a reason why there are 5280 feet or 1760 yards in a mile?   Although I have researched possible relations between the mile and kilometer and the cubit and meter, I have not explored in depth the possible relations between the different feet of antiquity.

> In Greek units, for
> instance, the earth's circumference measures 216,000 stades (60 x 60 x
> 60 stades). In Berriman's opinion, he thinks this is an accident: "no
> such traditional size was   >known

What length does Berriman give for the stadium in modern measure, and what length does he give for the Greek foot, the Greek cubit, the Babylonian foot and cubit, and the Egyptian foot and cubit?  How accurately does 216,000 stades actually measure the earth's circumference?  If it is exact then 1 stadium equals 1/10 of a minute or six arc seconds of whatever circumference circle it measures.

To know that the sun does not go around the earth all you have to do is head true north at the height of the summer and you will see the sun circle the horizon without setting!   Up there in the land of Inuit and Tornasuk, the Greek constellation Ursa Major circles directly overhead.

216,000 is of course related to one of the most famous numbers in antiquity 432,000 - the number of years in the Kali Yuga of the Vedic tradition; also the total number of Warriors at the 540 Gates of the Rooms of Vahalla in the Scandinavian Edda.

22. Valgrind is the gate    that wards the gods,
holy, nigh holy doors;
old is that wicket,    nor wot many
with what bolt that gate is barred.

23. Five hundred rooms    and forty withal
I ween that in Bilskirnir be;
of all the halls    which on high are reared
the greatest I see is my son's.

24. Five hundred doors    and forty withal
I ween that in Valholl be:
eight hundred warriors    through one door hie them
when they fare forth to fight the Wolf.

Geometrically, a regular pentagon has 540 degrees.

432000 is the number of English miles in the radius of the sun; and the number of years from the beginning of creation to the flood and years of the reign of the 10 antediluvian god-kings of Babylon listed by Berossis - while 43200 = seconds in one half day as well as 1/2 the number of Sabbaths from the dawn of creation to the Flood of the 10th generation of the line of the covenant of Adam, the Great Flood of Noah.  The pyramid scale of the earth 1/43200 allows the perimeter cubit of the Great Pyramid to be calculated to the accuracy of 1 micron from a circle of equatorial latitude:

Polar Circumference² / Equatorial Circumference / 43200 = Circle of Equatorial Latitude / 43200 =

( 12713511 * Pi )² / 40075035 / 43200 = 921.453 meters ~ 921.455 meters =

Cole's measure of the Perimeter of the Great Pyramid = 230.253 + 230.454 + 230.391 + 230.357 accurate to 2 millimeters

So dividing the Circle of Equatorial Latitude by 43200 and then by 1760 yields the cubit of the perimeter of the Great Pyramid accurate to 1 micron.  At least in its relation to the equatorial latitude circle the cubit divides that circle of the earth into 43200 * 1760 parts.

In Annie Dillard's Pilgrim at Tinker Creek she notes that the Solar System is sailing in the direction of Hercules at a velocity of 43200 miles per hour

43200 can be derived easily from the Great Pyramid independent of knowing the dimensions of the earth.  The sum of its cubit height 280, and the width of the square it rises from, 440 cubits, equals 720 cubits = 6! = 360 * 2 cubits = the number of degrees in two circles = 12 * 60 = the number of minutes in 1/2 day.  Multiplied by the 60 cubits of the circuit of the base of the walls of the King chamber it equals 720 * 60 = 43200 the number of seconds in a half day.

The area of the celestial sphere divided by 1760 the perimeter of the pyramid equals the angle of the intersection of the plane of the solar system and the celestial equator - accurate to less than a third of a second of arc.

The root base number 432 also corresponds to the number of Buddhas sitting in the Lotus position at the quadrangular Borobudur Temple in Indonesia :

http://www.pbs.org/treasuresoftheworld/a_nav/boro_nav/bnav_level_1/2revealed_borofrm.html

> In the case of Egypt , if you subscribe to the school of thought that
> posits a relationship between Egyptian linear measures and the earth,
> then the connection between the Egyptian cubit and the meter would, at
> some point, become moot and be highly commensurate. So I take your
> example as a very refined instance of this.

Even if the cubit is earth commensurate, the fact that the meter is such a terrible measure of the polar circumference will not make a great case for the accuracy of the cubit's measure of the earth.  The cubit works best when it measures the volumetric circumference of the earth http://www.jqjacobs.net/astro/cosmo.html of

40030005.6967 meters

It seems more logical to base an earth commensurate measure on either the volumetric circumference or radius, or the geometric circumference or radius, so as to take account of the ellipsoidal shape of the earth, rather than just use one of its axes or circumferences.

The cubit which works best with the volumetric circumference is Petrie's value for the mean cubit of ancient Egypt 20.63 inches, which is also Newton's value for the length of the cubit determined from Greaves' measure of the base of the walls of the King chamber.   If the meter perfectly measured the volumetric circumference of the earth so that it equals exactly 40 million meters, the actual length of the meter would be

40030005.6967 / 40000000 = 1.000750 meters

So let this equal the length of the edge of  a cube, then the volume of the sphere inscribed in that cube divided by the area of the face of the cube, or the volume of a sphere of diameter 1.000750 meters, divided by the square of that diameter, equals

4/3Pi(1.000750/2)³ meter³ / 1.000750² meter² = .523991474... meter

which equals the length of the cubit of Newton and the mean cubit of ancient Egypt accurate to 10 microns

20.63 inches * .0254 meter per inch = .524002 meter = .523991474... meter = earth commensurate cubit

How accurate is it?  It measures the volumetric circumference of the earth 800 meters greater than it is.   Can anyone say with certainty that the cubit intentionally incorporates the measure of the earth?

It sure is easy to measure the earth through it.  All one need do is take oldest cubit of Egypt by the reckoning of Newton and Petrie and derive the length of the volumetric meter of the earth through it by setting it equal to the equation

cubit = volume of sphere of diameter meter / meter²

and solve for the unknown variable meter

meter = 6 / Pi cubit =

the product of one Newton cubit and the ratio of the Volume of a Cube and the Volume of the Sphere inscribed it.   The perfect ancient Egyptian measure of the earth equals

6/Pi cubit * 40,000,000 = volumetric circumference of the earth =

6/Pi * 20.63 inch cubit of Newton * .0254 meters per inch * 40,000,000 =

40030804 meters ~ 40030006 meters = Actual Volumetric Circumference of Earth

The cubit measures one volumetric meter to the accuracy of 20 microns.

- John

• Hi John, I ll get back to you a bit later on your other points. Gonna celebrate with my son momentarily. But thought I d better seed the net with Berriman s
Message 2 of 10 , Feb 3, 2007
Hi John,

I'll get back to you a bit later on your other points. Gonna celebrate with my son
momentarily. But thought I'd better seed the net with Berriman's measures. Anything to
counter the confusion of metrum.org:

1 Remen = 370 mm = 20 digits
1 Sumerian ft. = 335 mm
1 Sumerian cb. = 502 mm = 24/25 Royal cb. = 1/10 pole
1 Egyptian Royal ft. = 349 mm = 2/3 Royal cb.
1 Egyptian Royal cb. = 524 mm = sq.rt. Remen = pi/6 metre
1 "version of the Royal cb." = 519 mm = 28 digits
6/7 Royal cb. = 449 mm = pi/7 metre
1 Assyrian ft. = 329 mm = 8/9 Remen
1 Assyrian cb. = 494 mm = 4/3 Remen
1 Greek ft. = 308 mm = 5/6 Remen
1 Greek cb. = 463 mm = 5/4 Remen = 25 digits
1 Talmudist cb. = 555 mm = 3/2 Remen
1 Roman ft. = 296 mm = 16 digits
1 Roman cb. = 444 mm = 4/9 metre = 24 digits
1 Palestinian cb. = 641 mm = cb.rt. Remen

For myself, I've been using the following values:

1 Sumerian cb. = 506 mm
1 Royal cb. = 524 mm
1 Greek cb. (stylobate) = 464 mm
1 Greek cb. (apothem) = 462 mm

Hope it helps,

-Chris

How
> accurately does 216,000 stades actually measure the earth's
> circumference? If it is exact then 1 stadium equals 1/10 of a minute or
> six arc seconds of whatever circumference circle it measures.

--- In sacredlandscapelist@yahoogroups.com, "jdepompeo" <john.depompeo@...> wrote:
>
>
> --- In sacredlandscapelist Chris wrote:
> >
> > Hi John, that was interesting. Thanks for sending it along.
>
> You are welcome Chris.
>
> > For the benefit of others on the list, I would like to illustrate that
> > you aren't putting the cart before the horse. The meter was originally
> > formed from a conceptual belt or perimeter around the earth made up of
> > 10 million units.
>
> The original idea was to divide the arc of the meridian into 10 million
> equal parts or the polar circumference into 40 million equal parts. The
> French mismeasure of the earth fixed the length of the meter
> imperfectly. If it measured the polar circumference of the earth
> correctly there would be 40 million meters in it - not 39940671.42790.
> One 1/40000000 of the Polar Cicrumference equals .99851678 meters not 1
> meter. The mile measures the polar circumference more accurately than
> the meter. When the edge of a cube equals 1 mile = 1609.344 meters, the
> length of the radius of the sphere equal in volume to the cube is
> 998.35734 meters which equals 1/40,000 of the polar circumference
> 998.51678 meters accurate to 1/2 foot
>
> 1609.344³ = 4/3Pi(998.35734)³ =
>
> Volume of Cube of Edge Length 1 Mile = Volume of Sphere of Radius equal
> to Polar Circumference / 40000
>
> 1 kilometer equals 1/40,000 of the polar circumference accurate to 5
> feet
>
> If one designs the meter as 1/40000000 of the geometric circumference of
> the earth -40007770 meters - ( the square root of the product of the
> equatorial and polar circumferences ) - the kilometer is more accurate
> than the mile.
>
>
> > When we go back to the cubit, do you believe the ancients had
> > sophisticated knowledge of the earth's size?
>
> I do not know. It is common knowledge that there are
>
> 360 * 365,221 feet in the equatorial circumference of the earth
>
> Does it indicate the length of the year and the length of the equator
> integrated into the length of the foot?
>
> Clearly, there is a very simple geometrical relationship between the
> kilometer and the mile, and the cubit and the meter, the kilometer being
> the radius of the sphere equal in volume to the cube of edge length one
> mile, and the cubit the ratio of the volume of the sphere of diameter
> one meter and the area of a square meter. Since there are 1760 yards in
> one mile, the number of cubits in the perimeter of the Great Pyramid, it
> is possible to relate the cubit to the yard, although it is a bit
> complex. The occurance of 1760 yards in one mile, and the relationship
> between the kilometer and the mile, and the meter and the cubit,
> suggests the possibility of the memory of a tradition that there are
> 1760 cubits in the perimeter of the Great Pyramid. Is there a reason
> why there are 5280 feet or 1760 yards in a mile? Although I have
> researched possible relations between the mile and kilometer and the
> cubit and meter, I have not explored in depth the possible relations
> between the different feet of antiquity.
>
> > In Greek units, for
> > instance, the earth's circumference measures 216,000 stades (60 x 60 x
> > 60 stades). In Berriman's opinion, he thinks this is an accident: "no
> > such traditional size was >known
>
> What length does Berriman give for the stadium in modern measure, and
> what length does he give for the Greek foot, the Greek cubit, the
> Babylonian foot and cubit, and the Egyptian foot and cubit? How
> accurately does 216,000 stades actually measure the earth's
> circumference? If it is exact then 1 stadium equals 1/10 of a minute or
> six arc seconds of whatever circumference circle it measures.
>
> To know that the sun does not go around the earth all you have to do is
> head true north at the height of the summer and you will see the sun
> circle the horizon without setting! Up there in the land of Inuit and
> Tornasuk, the Greek constellation Ursa Major circles directly overhead.
>
> 216,000 is of course related to one of the most famous numbers in
> antiquity 432,000 - the number of years in the Kali Yuga of the Vedic
> tradition; also the total number of Warriors at the 540 Gates of the
> Rooms of Vahalla in the Scandinavian Edda.
>
> 22. Valgrind is the gate that wards the gods,
> holy, nigh holy doors;
> old is that wicket, nor wot many
> with what bolt that gate is barred.
>
> 23. Five hundred rooms and forty withal
> I ween that in Bilskirnir be;
> of all the halls which on high are reared
> the greatest I see is my son's.
>
> 24. Five hundred doors and forty withal
> I ween that in Valholl be:
> eight hundred warriors through one door hie them
> when they fare forth to fight the Wolf.
>
> Geometrically, a regular pentagon has 540 degrees.
>
> 432000 is the number of English miles in the radius of the sun; and the
> number of years from the beginning of creation to the flood and years of
> the reign of the 10 antediluvian god-kings of Babylon listed by Berossis
> - while 43200 = seconds in one half day as well as 1/2 the number of
> Sabbaths from the dawn of creation to the Flood of the 10th generation
> of the line of the covenant of Adam, the Great Flood of Noah. The
> pyramid scale of the earth 1/43200 allows the perimeter cubit of the
> Great Pyramid to be calculated to the accuracy of 1 micron from a circle
> of equatorial latitude:
>
> Polar Circumference² / Equatorial Circumference / 43200 = Circle of
> Equatorial Latitude / 43200 =
>
> ( 12713511 * Pi )² / 40075035 / 43200 = 921.453 meters ~ 921.455
> meters =
>
> Cole's measure of the Perimeter of the Great Pyramid = 230.253 + 230.454
> + 230.391 + 230.357 accurate to 2 millimeters
>
> So dividing the Circle of Equatorial Latitude by 43200 and then by 1760
> yields the cubit of the perimeter of the Great Pyramid accurate to 1
> micron. At least in its relation to the equatorial latitude circle the
> cubit divides that circle of the earth into 43200 * 1760 parts.
>
> In Annie Dillard's Pilgrim at Tinker Creek she notes that the Solar
> System is sailing in the direction of Hercules at a velocity of 43200
> miles per hour
>
> 43200 can be derived easily from the Great Pyramid independent of
> knowing the dimensions of the earth. The sum of its cubit height 280,
> and the width of the square it rises from, 440 cubits, equals 720 cubits
> = 6! = 360 * 2 cubits = the number of degrees in two circles = 12 * 60 =
> the number of minutes in 1/2 day. Multiplied by the 60 cubits of the
> circuit of the base of the walls of the King chamber it equals 720 * 60
> = 43200 the number of seconds in a half day.
>
> The area of the celestial sphere divided by 1760 the perimeter of the
> pyramid equals the angle of the intersection of the plane of the solar
> system and the celestial equator - accurate to less than a third of a
> second of arc.
>
> The root base number 432 also corresponds to the number of Buddhas
> sitting in the Lotus position at the quadrangular Borobudur Temple in
> Indonesia:
>
> http://www.pbs.org/treasuresoftheworld/a_nav/boro_nav/bnav_level_1/2reve\
> aled_borofrm.html
> <http://www.pbs.org/treasuresoftheworld/a_nav/boro_nav/bnav_level_1/2rev\
> ealed_borofrm.html>
>
> > In the case of Egypt, if you subscribe to the school of thought that
> > posits a relationship between Egyptian linear measures and the earth,
> > then the connection between the Egyptian cubit and the meter would, at
> > some point, become moot and be highly commensurate. So I take your
> > example as a very refined instance of this.
>
> Even if the cubit is earth commensurate, the fact that the meter is such
> a terrible measure of the polar circumference will not make a great case
> for the accuracy of the cubit's measure of the earth. The cubit works
> best when it measures the volumetric circumference of the earth
> http://www.jqjacobs.net/astro/cosmo.html
> <http://www.jqjacobs.net/astro/cosmo.html> of
>
> 40030005.6967 meters
>
> It seems more logical to base an earth commensurate measure on either
> the volumetric circumference or radius, or the geometric circumference
> or radius, so as to take account of the ellipsoidal shape of the earth,
> rather than just use one of its axes or circumferences.
>
> The cubit which works best with the volumetric circumference is Petrie's
> value for the mean cubit of ancient Egypt 20.63 inches, which is also
> Newton's value for the length of the cubit determined from Greaves'
> measure of the base of the walls of the King chamber. If the meter
> perfectly measured the volumetric circumference of the earth so that it
> equals exactly 40 million meters, the actual length of the meter would
> be
>
> 40030005.6967 / 40000000 = 1.000750 meters
>
> So let this equal the length of the edge of a cube, then the volume of
> the sphere inscribed in that cube divided by the area of the face of the
> cube, or the volume of a sphere of diameter 1.000750 meters, divided by
> the square of that diameter, equals
>
> 4/3Pi(1.000750/2)³ meter³ / 1.000750² meter² = .523991474...
> meter
>
> which equals the length of the cubit of Newton and the mean cubit of
> ancient Egypt accurate to 10 microns
>
> 20.63 inches * .0254 meter per inch = .524002 meter = .523991474...
> meter = earth commensurate cubit
>
> How accurate is it? It measures the volumetric circumference of the
> earth 800 meters greater than it is. Can anyone say with certainty
> that the cubit intentionally incorporates the measure of the earth?
>
> It sure is easy to measure the earth through it. All one need do is
> take oldest cubit of Egypt by the reckoning of Newton and Petrie and
> derive the length of the volumetric meter of the earth through it by
> setting it equal to the equation
>
> cubit = volume of sphere of diameter meter / meter²
>
> and solve for the unknown variable meter
>
> meter = 6 / Pi cubit =
>
> the product of one Newton cubit and the ratio of the Volume of a Cube
> and the Volume of the Sphere inscribed it. The perfect ancient
> Egyptian measure of the earth equals
>
> 6/Pi cubit * 40,000,000 = volumetric circumference of the earth =
>
> 6/Pi * 20.63 inch cubit of Newton * .0254 meters per inch * 40,000,000 =
>
> 40030804 meters ~ 40030006 meters = Actual Volumetric Circumference of
> Earth
>
> The cubit measures one volumetric meter to the accuracy of 20 microns.
>
> - John
>
• ... That should be: 1 Egyptian Royal cb. = 524 mm = (sq.rt. 2) (1 Remen) = pi/6 metre 1 Palestinian cb. = 641 mm = (sq.rt. 3) (1 Remen)
Message 3 of 10 , Feb 3, 2007
Correction:

> 1 Egyptian Royal cb. = 524 mm = sq.rt. Remen = pi/6 metre
> 1 Palestinian cb. = 641 mm = cb.rt. Remen

That should be:

1 Egyptian Royal cb. = 524 mm = (sq.rt. 2) (1 Remen) = pi/6 metre
1 Palestinian cb. = 641 mm = (sq.rt. 3) (1 Remen)
• ... Doh, I conveniently forgot that part. Guess I got confused over the radius of a spherical earth being 10 million meters. My memory clearly prefers unity to
Message 4 of 10 , Feb 3, 2007
> The original idea was to divide the arc of the meridian into 10 million
> equal parts or the polar circumference into 40 million equal parts.

Doh, I conveniently forgot that part. Guess I got confused over the radius of a spherical
earth being 10 million meters. My memory clearly prefers unity to quaternity. :)

> Clearly, there is a very simple geometrical relationship between the
> kilometer and the mile, and the cubit and the meter, the kilometer being
> the radius of the sphere equal in volume to the cube of edge length one
> mile, and the cubit the ratio of the volume of the sphere of diameter
> one meter and the area of a square meter.

That is a very tidy way to correlate the kilometer and the mile. Nice.

>Since there are 1760 yards in
> one mile, the number of cubits in the perimeter of the Great Pyramid, it
> is possible to relate the cubit to the yard, although it is a bit
> complex. The occurance of 1760 yards in one mile, and the relationship
> between the kilometer and the mile, and the meter and the cubit,
> suggests the possibility of the memory of a tradition that there are
> 1760 cubits in the perimeter of the Great Pyramid. Is there a reason
> why there are 5280 feet or 1760 yards in a mile? Although I have
> researched possible relations between the mile and kilometer and the
> cubit and meter, I have not explored in depth the possible relations
> between the different feet of antiquity.

I've been researching many different ancient sites in order to compare their boundaries
and axial dimensions. Contemplating meaning in measurement, I suppose. "Man is the
measure of all things" as always.

You ask why there are 5280 feet in a mile. This is a familiar question but in a different
context. As Joseph Campbell asks, "should we marvel the more at the sexagesimal system
or at the Sumerians who invented it." Their calendrical festival year was reckoned in
mathematical--not natural--terms of 72 five-day weeks plus 5 intercalated festival days.
5 x 72 = 360. But 360 x 72 = 25, 920 thus yielding "...a mathematically found "great year"
whose coincidence with the observable astronomical "great year" might indeed have been
the reesult only of a sheer (but then how really wonderful!) accident."

The background to all it is, of course, that 60 is the mathematical key to a time and space
based system of reckoning, which we still use today in the familiar face of the clock and
degrees in a circle.

> What length does Berriman give for the stadium in modern measure, and
> what length does he give for the Greek foot, the Greek cubit, the
> Babylonian foot and cubit, and the Egyptian foot and cubit? How
> accurately does 216,000 stades actually measure the earth's
> circumference? If it is exact then 1 stadium equals 1/10 of a minute or
> six arc seconds of whatever circumference circle it measures.

This is what Berriman says:

"In Egyptian metrology there was a linear unit, of 20 digits, called remen and a distance of
500 remens became the stade used for itinerary measurements by the Greeks and
Romans.

Therefore 1 geodetic mile = 5000 Egyptian remens = 10 stades (Greek and Roman)
Therefore circumference of the spherical earth = 60^3 = 216,000 stades

No such traditional size was known to the Greeks. Eratosthenes of Cyrene, a younger
contemporary of Archimedes, observing that at Syene at the summer solstice the sun cast
no shadow from an upright pole, while at Alexandria (assumed to be on the same
meridian) the legnth of the shadow implied that the rays of the sun were inclined to the
vertical bgy (1/50)th of the circle, calculated the circumference to be 50 x 5000 =
250,000 stades on the assumption that the distance between these two places was 5000

> The root base number 432 also corresponds to the number of Buddhas
> sitting in the Lotus position at the quadrangular Borobudur Temple in
> Indonesia:

I've been there. Plausible but many of the Buddhas were stolen. Search through our SL
archives for a good discussion of this between myself and Barry Carrol (just search for 108
and my name, Chris). Good ol' Barry--what is he up to these days?

> It seems more logical to base an earth commensurate measure on either
> the volumetric circumference or radius, or the geometric circumference
> or radius, so as to take account of the ellipsoidal shape of the earth,
> rather than just use one of its axes or circumferences.

Why is it more logical to do this?

Fascinating. So what is your background? Do you do math professionally?

-Chris
• oh John, I thought you might like to take a look at this page if you haven t already: http://www.odeion.org/atlantis/chapter-2.html -Chris ... system ... days.
Message 5 of 10 , Feb 9, 2007
oh John, I thought you might like to take a look at this page if you haven't already:

http://www.odeion.org/atlantis/chapter-2.html

-Chris

--- In sacredlandscapelist@yahoogroups.com, "Chris" <groups@...> wrote:
>
>
> > The original idea was to divide the arc of the meridian into 10 million
> > equal parts or the polar circumference into 40 million equal parts.
>
> Doh, I conveniently forgot that part. Guess I got confused over the radius of a spherical
> earth being 10 million meters. My memory clearly prefers unity to quaternity. :)
>
> > Clearly, there is a very simple geometrical relationship between the
> > kilometer and the mile, and the cubit and the meter, the kilometer being
> > the radius of the sphere equal in volume to the cube of edge length one
> > mile, and the cubit the ratio of the volume of the sphere of diameter
> > one meter and the area of a square meter.
>
> That is a very tidy way to correlate the kilometer and the mile. Nice.
>
> >Since there are 1760 yards in
> > one mile, the number of cubits in the perimeter of the Great Pyramid, it
> > is possible to relate the cubit to the yard, although it is a bit
> > complex. The occurance of 1760 yards in one mile, and the relationship
> > between the kilometer and the mile, and the meter and the cubit,
> > suggests the possibility of the memory of a tradition that there are
> > 1760 cubits in the perimeter of the Great Pyramid. Is there a reason
> > why there are 5280 feet or 1760 yards in a mile? Although I have
> > researched possible relations between the mile and kilometer and the
> > cubit and meter, I have not explored in depth the possible relations
> > between the different feet of antiquity.
>
> I've been researching many different ancient sites in order to compare their boundaries
> and axial dimensions. Contemplating meaning in measurement, I suppose. "Man is the
> measure of all things" as always.
>
> You ask why there are 5280 feet in a mile. This is a familiar question but in a different
> context. As Joseph Campbell asks, "should we marvel the more at the sexagesimal
system
> or at the Sumerians who invented it." Their calendrical festival year was reckoned in
> mathematical--not natural--terms of 72 five-day weeks plus 5 intercalated festival
days.
> 5 x 72 = 360. But 360 x 72 = 25, 920 thus yielding "...a mathematically found "great
year"
> whose coincidence with the observable astronomical "great year" might indeed have
been
> the reesult only of a sheer (but then how really wonderful!) accident."
>
> The background to all it is, of course, that 60 is the mathematical key to a time and
space
> based system of reckoning, which we still use today in the familiar face of the clock and
> degrees in a circle.
>
> > What length does Berriman give for the stadium in modern measure, and
> > what length does he give for the Greek foot, the Greek cubit, the
> > Babylonian foot and cubit, and the Egyptian foot and cubit? How
> > accurately does 216,000 stades actually measure the earth's
> > circumference? If it is exact then 1 stadium equals 1/10 of a minute or
> > six arc seconds of whatever circumference circle it measures.
>
> This is what Berriman says:
>
> "In Egyptian metrology there was a linear unit, of 20 digits, called remen and a distance
of
> 500 remens became the stade used for itinerary measurements by the Greeks and
> Romans.
>
> Therefore 1 geodetic mile = 5000 Egyptian remens = 10 stades (Greek and Roman)
> Therefore circumference of the spherical earth = 60^3 = 216,000 stades
>
> No such traditional size was known to the Greeks. Eratosthenes of Cyrene, a younger
> contemporary of Archimedes, observing that at Syene at the summer solstice the sun
cast
> no shadow from an upright pole, while at Alexandria (assumed to be on the same
> meridian) the legnth of the shadow implied that the rays of the sun were inclined to the
> vertical bgy (1/50)th of the circle, calculated the circumference to be 50 x 5000 =
> 250,000 stades on the assumption that the distance between these two places was
5000
>
> > The root base number 432 also corresponds to the number of Buddhas
> > sitting in the Lotus position at the quadrangular Borobudur Temple in
> > Indonesia:
>
> I've been there. Plausible but many of the Buddhas were stolen. Search through our SL
> archives for a good discussion of this between myself and Barry Carrol (just search for
108
> and my name, Chris). Good ol' Barry--what is he up to these days?
>
>
> > It seems more logical to base an earth commensurate measure on either
> > the volumetric circumference or radius, or the geometric circumference
> > or radius, so as to take account of the ellipsoidal shape of the earth,
> > rather than just use one of its axes or circumferences.
>
> Why is it more logical to do this?
>
> Fascinating. So what is your background? Do you do math professionally?
>
> -Chris
>
• ... Anything to counter the confusion of metrum.org: Stecchini mentions Berriman: I have traced the first pyramidite book back to 1704; ever since the flood
Message 6 of 10 , Mar 1, 2007

--- In sacredlandscapelist Chris wrote:

> But thought I'd better seed the net with Berriman's measures.  Anything to counter the confusion of metrum.org:

Stecchini mentions Berriman:

"I have traced the first pyramidite book back to 1704; ever since the flood of pyramidite literature has continued unabated up to the present day. The last major effort appeared in 1953 as the Historical Metrology of A. E. Berriman."

>1 Remen = 370 mm = 20 digits

>1 Sumerian ft. = 335 mm
>1 Sumerian cb. = 502 mm = 24/25 Royal cb. = 1/10 pole
>1 Egyptian Royal ft. = 349 mm = 2/3 Royal cb.
>1 Egyptian Royal cb. = 524 mm = sq rt 2 * Remen = pi/6 metre
>1 "version of the Royal cb." = 519 mm = 28 digits
>6/7 Royal cb. = 449 mm = pi/7 metre
>1 Assyrian ft. = 329 mm = 8/9 Remen
>1 Assyrian cb. = 494 mm = 4/3 Remen
>1 Greek ft. = 308 mm = 5/6 Remen
>1 Greek cb. = 463 mm = 5/4 Remen = 25 digits
>1 Talmudist cb. = 555 mm = 3/2 Remen
>1 Roman ft. = 296 mm = 16 digits
>1 Roman cb. = 444 mm = 4/9 metre = 24 digits
>1 Palestinian cb. = 641 mm = sq rt 3 * Remen

Thanks for posting the info on cubits and remens Chris - if I get a chance I will respond in more detail in a later post.   Wikipedia puts the Greek pous or foot at 316 mm a mean of lengths which range from Ionic 296 mm to Doric 326 mm.  http://en.wikipedia.org/wiki/Ancient_Greek_units_of_measurement .  If we use Eratosthenes stadia of 750 pous, and the 308 mm pous of Berriman, one stadia equals 231 meters - close to the 440 * 523.6 mm mean side of the Great Pyramid - 230 1/2 meters.  It makes one wonder whether Eratosthenes not only measured the distance from Alexandria to Syrene, and the angle of the elevation of the sun at each location to determine the length of the circumference of the earth, but also took the measure of the Great Pyramid.  In fact, when the length of the side of the second great Pyramid equals 410 cubits, its cubit and the number of perimeter cubits in one side of the Great Pyramid allows one stadia to be almost exactly derived ( 750 * 308 mm ~  440 * 525.03 mm )   Hence we can see how to derive the volumetric circumference of the earth from the

Mean length of the side of pyramid of Khafre 8475.9 inches / 410 cubits * (366 1/4) /(365 1/4)(43200) * 1760 = 40029 kilometers =

Cubit of Khafre * rotations in a year / days in a year * 43200 seconds in a night * number of perimeter cubits in perimeter of Great Pyramid =

Volumetric circumference of the Earth 40030 kilometers, accurate to a kilometer.    The length of 2 Eratosthenes stadia times the number of seconds in one rotation of the earth equals nearly the length of the volumetric circumference of the earth.

2 Stadia * Seconds in Rotation = 1500 pous * 308 mm  * 86636.55 = 40026 kilometers  ~ 40030 kilometers

The cubit which divides 1 stadia into 440 equal parts is 525 mm long and the cubit which divides the side of Khafre into 410 equals parts is 525.03 mm long.  This latter cubit in conjunction with the ratio

solar / sidereal day = 86400 / 86164 = rotations in a year / days in a year = 366.256 / 365.256 = 1.0027378

is interesting in light of the international meter, and the ratio of the volume of a sphere and the area of the square of its diameter, since the volume of a sphere inscribed in a cube of edge length 1.0027378 meters, divided by the area of the face of the cube, identically the volume of a sphere of diameter 1.0027378 meters divided by the area of the square of its diameter equals

4/3Pi(1.0027378/2)³ meter³ / 1.0027378² meter² = .525032... meter = .525030... meter = side of Khafre / 410 = cubit of Khafre ? accurate to 2 microns

The ratio of Volume of a sphere inscribed in a cube / Area of one face of the cube, derives the perimeter cubit of Khufu accurate to a couple of microns when the edge of the cube has a length of one meter, and derives the perimeter cubit of Khafre accurate to a couple of microns when the length of the edge of the cube in meters equals the number of orbits of the earth in a year and a day - 1.0027378

This is true when the side of Khafre equals 410 cubits.  When the side of Khafre is divided by 411 the cubit equals the mean cubit of Khufu accurate to 5 microns, and using that mean cubit and the 3-4-5 triangle, the basic dimensions of Khafre are all integral multiples of 137.

Thus where A = 1.0027378 = number of days per rotation of the earth, 43200 equals number of seconds in a night, 1760 equals number of cubits in perimeter of the Great Pyramid,

4/3Pi( A/2 )³ meter³ / A² meter² * A * 43200 * 1760 = 40030 kilometers = volumetric circumference of the earth = 2 Eratosthenes stadia  * A * 86409

There are 86409 sabbaths in 1656 sidereal years ( the year of the flood of Genesis )

The foot of length 308 mm is called the artabic foot by Stecchini, and the Attic or common foot by a number of modern scholars, although its existence is debated.  See the article by Mark Wilson Jones:  Doric Measure and Architectural Design 1:  The Evidence of the Relief from Salamis.   Concerning the foot of 308 mm he writes:  "Already in the 16th century the foremost architects in Rome knew of excellent approximations to the length of the ancient Roman foot.   Baldassare Peruzzi, for example, left record on one of his drawings of a conversion factor that yields a value of 295.9 mm, one that comes very close to the peak of the distribution curve associated with actual ancient measuring instruments ( fig. 1 ).  The contrast is painfully clear in Robert Hussey's 1836 publication on Greek and Roman metrology; he was able to synthesize various Roman evidence in favor of the same foot value as Peruzzi's, but for Greek evidence he was left groping in the dark.  All he could do was rely on literary sources such as Pliny ( Natural History 2.21 ), who put the ratio between the Greek and Roman feet as 25:24, and observe that the value of 308.2 mm for the former thereby obtained yielded 100 feet for the stylobate of the Parthenon, thus explaining the appellation hekatompedon.  This unit, sometimes called the "common" foot, continues to be uncritically cited as a statement of fact, although surviving standards of about this length are known to date only from Egypt and the Levant, and William Bell Dinsmoor insisted that this was a unit "which no Greek ever employed."  p 73-74

http://www.ajaonline.org/index.php?ptype=content&aid=232

In footnote 10 he cites a "small Ptolemaic foot" of 308.6 mm, and a Syrian foot of 304.8 mm - the exact length of the English foot.  The measurements available online of the front of the Parthenon tend to cluster at 30.89 meters, yielding 100 "attic" feet of 308.9 mm.   The earlier Parthenon survey of Stuart and Revett yields a Greek foot of 308.3 mm, and the latter survey of Penrose yields a foot of 308.9 mm.   See http://www.metrum.org/key/athens/early.htm

> >accurately does 216,000 stades actually measure the earth's circumference?

Using the standard Greek stadion of 600 pous, 216000 stadia divided into the polar circumference of the earth equals 308.18 millimeters ~ the length of Hussey's 1836 estimate of the Greek Attic foot - 308.2 mm

Polar Circumference / 360 / 360,000 = 308.18... millimeters = Attic foot ?

> > It seems more logical to base an earth commensurate measure on either
> > the volumetric circumference or radius, or the geometric circumference
> > or radius, so as to take account of the ellipsoidal shape of the earth,
> > rather than just use one of its axes or circumferences.
>
> Why is it more logical to do this?

Maybe it is not more logical.

Ice cap melting and equatorial sea rise will balance out any change in the length of a volumetric cubit better than a cubit length determined by just the major or the minor axis.  Gravity causes a greater mass of water to mass at the equator.  Also 6/Pi is a cube/sphere volumetric ratio.  Aside from ensuring a meter length fixed to what is in reality axes lengths changing with climate, a geometric designer might intend a measure to reflect the rectified unity of the sphere being measured, and in the case of ancient Egypt, if a priestess truely knew the ellipsoidal surface and dimensions of the sphere of the earth, she may have thought it wise to measure both the height and girth of Geb, the Egyptian god of the Earth, and enshrine it in the length used to measure the earth.  If the goddess and god of Heaven and Earth, Nut and Geb, are identified in the cosmogonic act of their union, the etymology of Nut - Oval and Ball - is suggestive, as the ellipses of the oval ball of Nut and Geb will more perfectly determine a measure of that ball.   It has the added virtue of being even more accurate when the ratio 5/GoldenRatio² is used.

5/GoldenRatio² * 20.63 inch cubit * .0254 meter an inch * 40000000 =

5/GoldenRatio² Newton Cubits of Memphis * 40000000 = 40030191 meters ~ 40030174 meters =

NASA's measure of the volumetric circumference of the earth, and Newton's Egypt cubit accurate to 1/5 a micron in length

http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html

Or since 20,000 cubits equals one Egyptian Itrw - River Measure, using the cubit of Newton, the length of the volumetric circumference of the earth equals

( 100 / GoldenRatio )² Itrw

The Golden Ratio equals 1.618034 compared to its actual value 1.61803398...

BTW the length of the cubit in meters also happens to be within an arc minute or two of the latitude of the Great Pyramid in radians:  Pi/6 = 30 degrees in radians.

> oh John, I thought you might like to take a look at this page if you haven't already:

An interesting and well designed site.  I have not had the time to verify his math - although when I do have some time I will follow his lead and calculate a great circle for Gizeh and compare it to the cubit and the pyramid of Hemiunu, as he does for the latitude of Athens.

>
> Fascinating.

Glad you like it.  It is not a popular topic in the modern world!  :)

- John

• ... kilometers ~ 40030 kilometers ... into ... foot ? To end this little aside on the Greek foot we can examine which period of time and which circumference of
Message 7 of 10 , Mar 2, 2007

> The length of 2 Eratosthenes stadia times the number of

> seconds in one rotation of the earth equals nearly the length of the
> volumetric circumference of the earth.

> 2 Stadia * Seconds in Rotation = 1500 pous * 308 mm * 86636.55 = 40026 kilometers ~ 40030 kilometers
>
> Using the standard Greek stadion of 600 pous, 216000 stadia divided into
> the polar circumference of the earth equals 308.18 millimeters ~ the
> length of Hussey's 1836 estimate of the Greek Attic foot - 308.2 mm
>
> Polar Circumference / 360 / 360,000 = 308.18... millimeters = Attic foot ?

To end this little aside on the Greek foot we can examine which period of time and which circumference of the earth works best with the Berriman Greek foot of 308 mm.   One may divide it into the length of the polar, equatorial, volumetric, or equatorial latitude circle of the earth, and further divide by seconds in rotation 86164, seconds in day 86400, or seconds in orbit rotation 86636.55.    Using 2 stade of Eratosthenes ( 750 Greek feet a stade )

Polar Circumference / 2 Stade / 86400 = 308.184... mm

Volumetric Circumference / 2 Stade / 86636.55 = 308.03 mm

Equatorial Circumference / 2 Stade / 86636.55 = 308.376... mm

Circle of Equatorial Latitude / 2 Stade / 86164 = 307.992... mm =

Polar Circumference² / Equatorial Circumference / 2 Stade / 86164

So  the circle of equatorial latitude and rotation of the earth in seconds work best, calculating the Berriman Greek foot to about 8 microns accuracy.

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