Image: Hawkins' theorem 1.  Apologies if link has expired.(Melchizedek Communique, MC120809) Gerald S. Hawkins, a retired astronomer, had studied the geometry of the crop circles. (Background: "Methodology of the Crop Circles", http://www.shout.net/~bigred/mc120709.html)

In one of the strange shapes in the wheat fields, Hawkins noticed that he could draw three straight lines, or tangents, so that each touched all three circles inscribed in the overall crop circle. Next, he drew an equilateral triangle using as points the centers of the three circles. By adding a circle centered on this triangle, Hawkins proved an original theorem. And he found the ratio of the diameter of the triangle's circumscribed circle to the diameter of the circles at each corner to be 4:3. (Source: "Theorems in Wheat Fields", by Ivars Peterson. June 30, 2003. http://www.maa.org/mathland/mathtrek_06_30_03.html)

But this relates to the Problem of the Three Revolving Bodies. "The problem of three gravitating bodies revolving together and about each other is one which, like the quadrature, has hitherto baffled all attempts of mathematicians to solve." In the 19th century, John A. Parker, engaged in studying the centuries-old puzzle known as "The Quadrature of the Circle" (a.k.a. "Squaring the Circle"), diverted himself by pondering the Problem of the Three Revolving Bodies. And Parker, over 100 years before Hawkins, also discovered this 4:3 ratio. (Source: Key to the Hebrew-Egyptian Mystery in the Source of Measures, by J. Ralston Skinner. 1894. Republished by www.kessinger.net)

Parker realized that each revolving body passed over an area equal to one and one-third. "In other words, their relative motion is as four to three." (One and one-third times 3 = 4.) (Skinner, op. cit.)

The revolving bodies, Parker discovered, each describes "a circumference equal to one and one-third the circumference of one diameter." (ibid.)

J. Ralston Skinner (op. cit.) concludes, "there is but one numerical form for the expression of these relations... and that is the Parker forms of 6561: 5153 x 4 = 20612..."

In a previous Melchizedek Communique was briefly described Parker's findings after exploring the problem of the Quadrature of the Circle. ("Quadrature is 'Source of Measures'", http://www.shout.net/~bigred/mc120609.html). Of course, it is now generally conceded the problem of squaring the circle is impossible to solve. Nonetheless, persons such as Peter Metius and John A. Parker came quite close to a solution. Parker's "6561: 5153 x 4 = 20612" findings had begun with an equilateral triangle with circumference of 3. From there he eventually arrived at a square (two equilateral triangles joined together) having area 6561. The 5153 was abstractly the area of a circle inscribed in a square having an area of 6561. The circle, having diameter 6561, is rectified into the circumference (4 x 5153) 20612.

According to Skinner (op. cit.), Parker proceeded further, with his 4:3 ratio and previous Quadrature of the Circle results, to finding a numerical value of a sidereal lunation.

Some preliminary checking gives this editor's own result for the circumference of the abovementioned circle with given area 5153 and diameter 6561. (The radius, half the diameter, would be 3280.5) The formula for circumference of a circle (if memory serves) is 2(pi)(radius). Plugging in the numbers, 2(pi)(3280.5) = 20611.9894... Not exactly a perfect Quadrature of the Circle, but close, since it is quite near the given 20612 number.

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