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
faces.}
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.}
See,
pymd.com...
and more specifically:
{Time Lifes' "Mystic Places" for references on indentation of faces..}:
www.amazon.com...
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
center...}
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 } =
77.58629557
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
me"..
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..
{Please}
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
with indentations..}
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..
No consequencE..
Thank you.
[edit on 26-9-2009 by noconsequence]