Where did you first obtain the measurements, Michael? Did you obtain your own measurements, observing first hand the angles and deviations? Gordon
Michael responds:
I got the measurements from a book by Raymond E.Capt, of Scotland. The book is entitled, "The Great Pyramid Decoded". In this book, he meticulously gives many precise measurements involving The Great Pyramid, including the so-called "pyramid-inch" measurements of the length and width of the floor of The King's Chamber. He also notes that the "pyramid-inch" is approximately 1.001 longer than the regular inch.
Here's a "cut & paste" from my article at ... http://www.farshore.force9.co.uk/mat_11.htm
Michael Lawrence Morton ___ 1998 ___
Using Regular Inches Instead Of "Pyramid Inches" ...
"On page 54, in the book "The Great Pyramid Decoded", by E. Raymond Capt, the length of The King's Chamber is given in "pyramid inches" as 412.13186, the width as 206.06593, and the height as 230.38871.
We are told that a "pyramid inch" is slightly larger than a regular inch, a ratio of approximately 1.000965 or so.As soon as I saw the first 3 digits of the given length (above) ... "412" ...
I thought of the decimal harmonic of the Surface Area On A Sphere .... 4125296125 (to ten digits). This figure comes from a geometry 'formula' ; using the RADIAN (deg) *as the radius* of a 'generic sphere'. In other words,we are taught (hopefully) in school that the "formula for the Area Of A Circle" is .... "Pi x 'r' Squared". What I'm saying here is to actually substitute the figure 57.29577951 (degrees of arc) for the "r" in that generic formula we learn in school. The RADIAN (deg) is equal to (assuming 360 degrees on one complete circumference) 57.29577951 degrees-of-arc on any true circumference. This curved length ... the RADIAN (deg) ... is equal, as close as possible, to the straight-line length of the actual radius of that *same* circle or sphere.
The formula (generic) for Surface Area On A Sphere is :
4 x Pi x ('r' Squared) So ; if we substitute 57.29577951 (deg) for "r" .... 4 x 3.141592654 x (57.29577951 x 57.29577951) = 41252.96125 .... an actual number value we can use.
What would the ratio be, of "pyramid inch" to regular inch, if we look at the possibility of the length of The King's Chamber as "412.5296125" regular inches ?
412.5296125 / 412.13186 = 1.00096511
Yes ; this is ***very close*** to the estimate of 1.000965 given above.
I think we have probably found something very significant here. The work of Carl P. Munck (self-published) has proven the reality of a lost, deeply ancient ... and evidently suppressed ... advanced planetary mathematical matrix involved in precise latitude and longitude placements of pyramids, mounds, stone circles, monuments, and earthen effigies. This advanced mathematical matrix has also been found to be frequently *encoded* into the structural designs and into the internal geometry and measurements of the structures themselves.
As noted above, the width of The King's Chamber is exactly half its length. So, the width in terms of regular inches would be a decimal harmonic of the Surface Area On A Hemisphere ..... 2062648063 (to ten digits).
Area of The King's Chamber Floor
Taking the length times the width, in regular inches, will give us the area in regular square inches, of the floor :
412.5296125 x 206.2648063 =3D 85090.34062 Sq. in.
If we now divide this area by the Square of the RADIAN (deg) : 85090.34062 / 3282.80635 = 25.92
We find here a decimal harmonic of one complete Precession Cycle of the Equinoxes of Earth in terms of years. This would be :
25920 / 1000 =3D 25.92 (yrs)".
-------------------------------
So ... I think that shows, pretty well, the logic and methodology behind my conclusions on this topic. By the way ... this book, by Raymond E.Capt, is quite revered by many researchers ... for its meticulous attention to "accuracy of detail". So this is probably as accurate as we can get, in terms of getting the precise measurements I quoted.
The other alternative is to ask someone like Chris Dunn ... if he's measured the King's Chamber himself. I'll try to get in touch with him and ask him if he's measured it.
BUT ... it stands to reason that the figures quoted, above, indicate an obvious precision calculated on the basis of The Radian (deg). And THAT is, of course, my main point !!
-- Michael Lawrence Morton
http://www.farshore.force9.co.uk/mlmindex.htm
http://www.greatdreams.com/gem1.htm
http://mission-ignition.tripod.com