## HOW CAN I PROVE THE ORIGINAL HEIGHT OF THE GREAT PYRAMID OF GIZA ? (Including Capstone) by Michael Lawrence Morton(c) 2000

First ... we define the Polar circumference of Earth as 21,600 nautical miles. This is a fact.

Second ... use the Tetrahedral Grid LAT of the Tetrahedral Latitude. I took Richard C.Hoagland's latitude, north and south of planetary equators ... 19.47122061 arc-degrees ... exactly. That's the correct, precise angle. (It's Sine is 1/3rd, or 0.333333333).

Simply translate the degrees ... into degrees, minutes, and seconds. And then ... MULTIPLY THOSE NUMBERS !!!

19 (deg) X 28 (min) X 16.46707131 (sec) north/south ... = 8760.48194 North/South. {Morton, 1999}

Third ... Use the Polar Radius of Earth .... simply divide the 21,600 by exactly 2Pi ... 21600 / 6.283185307 = 3437.746771 nautical miles.

Fourth ... 8760.48194 / 3437.746771 = 2.548320898 ... the Grid POINT Value of SILBURY HILL in England (Munck, 1992).

Here are the Silbury Hill numbers ... precise latitude and longitude placement ... from which that Grid POINT Value is calculated :

Grid LAT 51 (deg) X 24 (min) X 55.43987036 (sec) North ... = 67858.40132 North ... which is exactly (Pi X 21600) itself !!!

Grid LONG 32 (deg) X 59 (min) X 14.10416949 (sec) W.Giza ... = 26628.672 W.Giza. [ W.Greenwich 01 deg 51 min 13.30416949 sec ].

Grid POINT Value SILBURY HILL ... 67858.40132 / 26628.672 ... = 2.548320898

Fifth ... multiply times 60 ... then times Pi ... 2.548320898 X 60 X 3.141592654 = 480.3471728 ... regular British Feet; original height including capstone, of The Great Pyramid.

NOT SATISFIED ? THEN TRY THIS ...

Take that Tetrahedral Grid LAT ... 8760.48194 .. (see above) ...

1) Multiply by Pi ... 3.141592654 X 8760.48194 = 27521.8657 ...

2) Divide by The Radian (deg)*... assuming we are using 360 arc-degrees on one circumference ... 27521.8657 / 57.29577951 = 480.3471728 regular British Feet; original height, including capstone, of The Great Pyramid.

* 57.29577951 (deg) = 360 / (6.283185307).

This corroborates and effectively proves Munck's figure of 480.3471728 regular British Feet ... as precisely correct.

This also directly supports Hoagland's "Tetrahedral Latitude" as true and precisely correct at 19.47122061 arc-degrees, north/south of planetary equators. And this is integrated into the "ancient matrix" by Morton's "Tetrahedral Grid LAT" of 8760.48194 north/south of planetary equators.

Further .. this supports the "Grid Matrix" calculation methodology, involving the multiplication of degrees X minutes X seconds, of latitude and longitude.

-- Michael Lawrence Morton
http://www.farshore.force9.co.uk/mlmindex.htm
http://www.greatdreams.com/gem1.htm
http://mission-ignition.tripod.com