Secret Bible Code

squaring_the_circle

It will, wrote J. Ralston Skinner in his book, Key to the Hebrew-Egyptian Mystery in the Source of Measures (1894), “afford much food for thought, as to whether so sublime a work as the Holy Record can be a refuge for that so much oppressed and bedeviled idea of ‘squaring the circle’…”

Does the Old Testament contain, via the Hebrew alphabet, secrets pertaining to the ancient problem of squaring the circle? Extrapolating from the investigations of John A. Parker as contained in his book, The Quadrature of the Circle, Skinner began to see connections between a particular ratio (6561: 20612) and the Biblical “cubit” measurement. (Background: “Finding The Cubit”, Ersjdamoo’s Blog entry of August 12, 2013.)

The ancient Hebrews, says Skinner, “had no numeral system, apart from their literal one – i.e., their alphabet held their numerals…” In his book, Skinner includes a reference table of the Hebrew alphabet showing the numerical value of each letter. From this, Skinner deduced that, for example Jared, father of Enoch, has something to do with the current “yard” measurement. Enoch himself, who lived for 365 years, is thought by Skinner to secretly symbolize the number of days in the year.

Also thought by Skinner to contain values based upon an ancient forgotten solution of the “squaring the circle” problem are the Great Pyramid of Giza and King Solomon’s Temple.

So if John A. Parker had at least come close to re-solving the ancient “squaring the circle” problem, this may have revolutionary implications for our understanding of portions of the Old Testament.

In yesterday’s blog entry (“In The Beginning Was The Point”), Parker’s fourth proposition from chapter two of his book (Proposition 2-4) was delved into. “The circumference of any circle being given, if that circumference be brought into any other shape formed of straight lines and of equal sides and angles, the area of that shape is equal to the area of another circle, which circle being circumscribed by another and similar shape, the area of such shape circumscribing the last named circle is equal to the area of the circle whose circumference is given.”

Plate_12

In the Proposition 2-4, we went from Circle E with circumference of 3, to Triangle F with perimeter of 3, to Circle G having the same area as Triangle F, and finally to Triangle H which contains Circle G. Parker shows how Triangle H has the same area as the original Circle E. In proposition 5 from chapter 2 (Proposition 2-5), Parker proceeds in an inverse direction to that of Proposition 2-4. As illustrated in Parker’s Plate 12 (above), a six-sided polygon H contains an inscribed circle G. The area of G is equal to that of another six-sided polygon F. The circumference of polygon F equals the circumference of another circle, E. And circle E has the same area as the polygon H. This is formally stated by Parker in his Proposition 2-5 as follows: “The circumference of a circle by the measure of which the circle and the square are made equal [i.e., squaring the circle], and by which the properties of straight lines and curved lines are made equal, is a line outside of the circle, wholly circumscribing it, and thoroughly enclosing the whole area of the circle, and hence, whether it shall have breadth or not, forms no part of the circle.”

Advertisements

About ersjdamoo

Editor of Conspiracy Nation, later renamed Melchizedek Communique. Close associate of the late Sherman H. Skolnick. Jack of all trades, master of none. Sagittarius, with Sagittarius rising. I'm not a bum, I'm a philosopher.
This entry was posted in Uncategorized. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s