Ok just found some very sloppy old math to get you started on the proper formula to build the pyramid.. The formula turns a solid sphere inside out. A
true pyramid is literally a doorway between static and geometric universes where one can be used to effect the other.. So here is what I have on the
math: The math is to 1/1,000th scale of the Great Pyramid of Giza..
A = Height = 5.772 inches
B = width.. Height multiplied by 1/2 Pi [1.570796327] = Width 9.066636398 inches. [B = width..]
C = Length of faces = Height multiplied by 1/4 pi [.785398163] = Length of faces, "if they were not indented" 7.349138656 inches. [C = Length of
B & C - files.abovetopsecret.com...
D = Face, "surface" area at bottom of, "indented" faces = Length of faces divided by 1/4 pi [.785398163] = 9.357213961 inches; surface area at
bottom of the, "indented" faces.. [D = surface" area at bottom of, indented faces.]
and more specifically:
[Time Lifes' "Mystic Places" for references on indentation of faces..]:
Math varies a little cause I have no need for, "passageways" etc..
E = Area to be added to width through indentation at the bottom of the faces = Face area at bottom of faces [9.357213961] minus actual width
[9.066636398] equals area to be added at the bottom of the faces by making an indentation: .290577563 inches.. [E = Area to be added to width through
indentation at the bottom of the faces; ".290577563 inches"..]
F = E [ .290577563 ] inches multiplied by 6 = 1.743465378 inches for outside compass point from center. [F = outside compass setting from
G = E [ .290577563 ] inches multiplied by 6.75 = 1.96139855025 inches for compass setting away from curving face. [ G = compass setting away from
curving face ]
H = [ .290577563 ] inches multiplied by 2.25 = 0.65379951675 inches compass setting for curve in center of face, " non-indented".. [H = compass
setting for curve in center of face non-indented..]
Compass settings here: files.abovetopsecret.com...
I = B [ Width] squared multiplied by A [height] = 474.480885250463727939888
J = I [474.480885250463727939888] divided by 2 = 237.240442625231863969944 cubic inches of volume in pyramid, "without" indentations.. J = cubic
inches of volume in pyramid, without indentations..
K = The indentation takes exactly 1.1544 square inches of ground area," off of the base area on each face; 1.1544 multiplied by 4 faces = 4.6176
square inches.. [K = total square inches taken off base area of pyramid.]
L = B [Width] squared = 82.203895573538414404 square inches of ground area; without indentations.. 82.203895573538414404 subtracting K [4.6176 ] =
leaving a, [ 77.58629557 square inch base area].. [ K = square inch base area with indentations..]
Now here's where I get lost now days because my mind can no longer, "go there": There are exactly 2.886 cubic inches taken out of each face
resulting from the indentations.. I just can't remember the equations I used in order to determine the radiuses.. They were very complicated, "to
That is where one of you fine mathematicians might just be very welcome/invited; to come mathematically save the day for those of us who are not so
mathematically well endowed..
Indentations and flat areas on pyramid: files.abovetopsecret.com...
M = The 4 faces, 1.1544 square inches each, multiplied by 2.886 = 11.544 cubic inches of volume taken out of the pyramid volume; being the result of
the indentations.. [M = cubic inches of volume taken out of the pyramid volume; being the result of the indentations.. ]
N = K [77.58629557] multiplied by A [5.772] = 223.91404901502 cubic inches volume of pyramid with indentations. [N = cubic inches volume of pyramid
O = 237.240442625231863969944 cubic inches of volume in pyramid, "without" indentations minus K-cubic inches volume of pyramid with indentations
[223.91404901502] = 13.326393610211863969944 inches total volume to be removed from pyramid to equal a 223.91404901502 cubic inch ball inverted.. [O =
13.326393610211863969944 cubic inches of volume to be removed from pyramid..]
P = O [13.326393610211863969944] minus the 11.544 inches of volume taken out of the faces through indentations equals 1.782393610211863969944 cubic
inches to be removed from the center of mass with the pyramid.. [P = cubic inches to be removed from the center of mass within the pyramid..]
For the, "kings chamber" I used the unit Phi [1.618034]in the formula to remove the remaining volume.
Q = The cube root of P [1.782393610211863969944] is 1.212461263.. Q [1.212461263] inches; width to landmass.. [Q = Kings chamber width: 1.212461263
inches towards greater land mass..]
R = Q [1.212461263" multiplied by phi = 1.961803547216942 inches.. [R = kings chamber height 1.961803547216942 inches]
S = Q [1.212461263] inches; divided be PHI [1.618034] = 0.749342265366 inches depth for the kings chamber towards greater sea mass.. [S =
0.749342265366 inches depth to greater sea mass..]
Ok so there's the math I do have for the proper construction of the, "Osmium" pyramid at the top of the geometric invitation.. There might be some
added info for it, if I choose to continue and divulge the fullness of the geometric invitation. Waiting to decide.. Not many responses. Oh-Well..
Edit to say the formula is a mathematicians dream in that it demonstrates bases for the American inch, demonstrates the pyramid inch being 1.1544
American inches, and demonstrates base purposed for measurements in ones, 5s, and tens; instead of 3 feet to a yard, for example..
[edit on 26-9-2009 by noconsequence]