Megalithic Yard Unearthed     |   home Megalithic Yard Surprises   |   Was Thom misled?   |   Woodhenge   |   Stonehenge and Sarsen circle   |   The Lintel Circle   |   Hawkins'  Photogrammetry   |   Trilithon ellipse   |   Megalithic Foot and Megalithic Inch   |   Arcs, chords & sagitta   |   Pi-Pyramid and the Megalithic Yard   |   Map Scales   |   Miles and the Equator   |   Additions, Vara etc.   |   Contact Me   |   CALCULATOR Pi-Pyramid and the Megalithic Yard       After finding 2.72... emerging from various geometrical figures, it is inevitable that the pyramid must be examined.      A 'Pi-Pyramid' is one whose vertical height to the perimeter of it's base, bears the same relationship as the Radius of a circle does to its Circumference. The slope angle of such a Pyramid is 51°51¢14×3²      Quite simply, the number of Ö3 units in the vertical height of such a pyramid, is the same number of 2.720699... units in one of the base sides. (i.e. If the height is 60Ö3 feet, the base is 60 MY).      A circle whose diameter is the vertical height, has a circumference of twice the number of units in the base side. (i.e. if the base is 60MY, then the circumference is 120MY).      Now in a 30/60 triangle, if the base side is 10,000 inches, the height will be 5773.5026... inches or 481.1252... feet. Which I do believe is as near as one can get to the actual height of the Gt. Pyramid. As such, the circumference of a circle on this measure (as a diameter) will be 555.555... MY, which means the base is 277.777... MY, or 755.749...feet, or 9068.99... inches. Flinders Petrie gives a mean measure of the four base sides as 9068.8 inches.     His height is 5776.0 +/- 7 inches, for his slope angle is steeper at 51°52¢      It also means that the base side is 1/360th of 100,000 MY, and thus the circumferenc must be 1/180th of100,000 MY., and that the height is 277.777...Ö3 feet.     Dividing the units in the base side of a Pi-type Pyramid by (2.720699... /2), produces the base side of an Equilateral triangle having  the same vertical height . In the case of Gt.Pyramid, this Equilateral triangle will have sides of 555.555...feet. So it looks as though our Imperial inch governs the dimensions of the Gt. Pyramid., and Piazzi Smyth's numerical coincidence he found between the pyramid's height and the distance from Earth to the Sun was perfectly correct.                                      93,000,000 / (5773.5... * Ö3) = 9300 Now repeat the exercise using the quoted dimensions in cubits: base 440 cubits., therefore height = circumference (880) / Pi = 280.11269... cubits. Transfer this to a 30/60 triangle, and the base will be 485.169... cubits.   (which makes a cubit equal to 5773.5026... inches / 280.11269...  = 20.61135... inches or  523.528... mm.)    This number , albeit 4851.69... , has been met before. It is the number of English feet in a Roman Mile. (Petrie also gives his  Roman Mile as 58220+/- 4 inches, whereas mine is 58220.33 inches.)    It is also 13200 ft / 2.72069...,, and 13200 ft. is 2.5 Statute Miles or 4851.69... Meg Yards. Back      Home        Next                                                                                            Ó Hugh Franklin, May 2007