DRAWING THE “JESUS TOMB” SYMBOL, PART 16

The tomb symbol angle rotated 90 degrees becomes the red line.

I am continuing here with my series about the large symbol found on the Talpiot tomb in Israel (aka “The Lost Tomb of Jesus”).  The symbol is a stone relief sculpture of a small circle within an upside-down Y-shape.  (More on the tomb here.)  

I have been making designs derived from the tomb symbol.  Generally, I am going by what I observe in my drawings; I am not attempting mathematical proofs, except where I am obviously doing math as in this post.  As I go along I define whatever terms I need to, but might not repeat definitions in each post.  So this post should be read in conjunction with the previous ones.  (Index.)

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Here I have drawn the tomb symbol in the context of a hexagram (six-pointed star).  The tomb symbol angle (lime green) has its apex at the midpoint of the upper horizontal line of the hexagram, and its ends at the lower lateral peripheral points of the hexagram.  The tomb symbol circle (black-green) is one-fourth the circle enclosing the hexagram, and its circumference intersects the center of the design and the midpoint of the lower vertical radius.  The red line (red) passes through the center of the tomb symbol circle and also intersects the upper lateral peripheral point of the hexagram, and a point where the interlocking equilateral triangles of the hexagram intersect.  The image above illustrates how the tomb symbol angle rotated 90 degrees, and re-anchored on the upper lateral peripheral point of the hexagram, becomes the red line.

In the following image, a design based on the hexagram, there are numerous equilateral triangles.  Each angle of an equilateral triangle is 60 degrees; half that is 30 degrees.  The sine of 30 degrees, very conveniently happens to be 1/2.  Thus if I say segment DJ equals 1, then side JO of equilateral triangle JOS equals 2 (sin 30 degrees = opposite / hypotenuse = DJ / JO = 1 / 2).

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Using the Pythagorean theorem, the height DO of triangle JOS equals the square root of 3 (written here as √3).  (1² + x² = 2²;  1 + x² = 4; x² = 3; x = √3.)

For tomb symbol angle CED, find the tan (opposite / adjacent). 

Tan CED =  CD / DE

Where CD = 3 x DJ = 3 x 1 = 3

Where DE = 2 x DO = 2 x √3 = 2 √3

CD / DE = 3 / 2 √3= √3 / 2

Show that the red line (segment FG) forms an angle EFG with the same tan ratio:

Tan angle EFG = EG / EF

Where EG = 3/4 DE = 3/4 x 2 √3 = 1.5 √3

Where EF = CD = 3

EG / EF = 1.5 √3 /3 = √3 / 2

Thus, the tan ratios are the same, both √3 / 2.  Hooray!

That the red line does indeed pass through point J and the center of the tomb symbol circle at G, is shown by tan angle DJG = DG / DJ = 1/4 DE / 1 = 1/4 x 2 √3 / 1 =  1/2 √3 / 1= √3 / 2.  Again, same ratio.  (Parallel lines create equal angles.)

This ratio √3 / 2, happens to also be the ratio of the height to the base in an equilateral triangle.  (For example, DO = √3 (height); JO = JS = 2 (base).)  Really neat!

Does the fact that the red line, alias the rotated tomb symbol angle, passes through the center of the tomb symbol circle (where I’d placed it based on my observation of the actual sculpture), mean that I’ve anchored the circle in the right place?  I don’t know.  There could be other relationships that are even more astounding.  But maybe I got it right.

I should mention here that some of my images are very wide and when shrunk down to fit in this column, they lose a lot of detail.  To view a full-size image in this series, click on “Full-size image” under the image, then click on the image to open or click “open.”  Then hover your cursor over the image to get a pop-up button in the lower right corner (or other feature depending on your browser – this will take a few moments), and click on that to get the actual full-size.

Continued here.

-2009-

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 This post was posted on May 28, 2009.