Only assumption 1b for the buildplan of the pyramids in Giza results in a special value for the

surface measure over all 5 surfaces, which is: phi² a²

(phi = golden section, a = length of edge at basis)

As shown below, the assumption, that the surface measure of each side equals the square over

the heigt of the pyramid (1b), leads to a height of the pyramid of

h = sqrt(phi)/2 a (h² = phi/4 a²)

and a height of a face-triangle of

s = phi/2 a, surface A(tri) = 1/2 a s = phi/4 a² (=h²)

The sum of the surface measure over the 4 triangles and the basis-square is then:

(phi/4+phi/4+phi/4+phi/4+1) a² = (phi +1) a² = phi² a² (because of D5)

(phi is the value of the Golden Cut/Ratio/Section and

the result (limes) of the fibonacci-sequence)

But why phi ?

Is it a hint to the relations between each distance planet-Sun ?

(a set of circles with radii increasing by factor phi are located nearly always between min. and

max. radius of the planet orbits around the sun, see table at

http://home.t-online.de/home/astro-uni/Forschung/gold.htm#gs und son )

Corrections for mail below

1: D2) 1.61803398875 = phi

D3) 0.61803398875 = 1/phi

2.61803398875 = phi²

2: ansatz 1b h = a/2 sqrt(phi) h/a=0.6360098247 h = 147.22 m

3: Why isn't it twice as big (because of 2:sqrt(5):3:4:5 = 34:38:51:68:85)

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Sophisticated Pyramid

What have you accepted as facts of the Great Pyramid ?

Here is your ultimate list of answers !

(hope you'll be convinced)

Statements:

1. there are three relations for the heigt of the pyramid to the edgde of

the basis, which cannot all be true at same time, but differ only a very

small amount from each other:

a) the height of the pyramid is related to each edge in the ratio 7 / 11

b) the surface measure of each side equals the square over the heigt of

the pyramid

c) the sum over the four edges of the basis is related to the heigt of the

pyramid

in the same ratio as the circumfence of a circle to its radius

2. each edge of the basis equals the distance which a point on the equator

moves in exactly half a second

3. the height of each side of the pyramid is approximately one classic

greek stadion of 600 feet

4. measurements were taken in multiple integers of an royal ell of 52,4 cm

and a foot of 30,8 cm (calibration)

5. the "king's" chamber represents the pythagorean triangles 3 : 4 : 5 and

2 : sqrt(5) : 3

Assumptions:

A : pyramid

A1) 231 m ~ length of each edge at basis

A2) 147 m ~ height

A3) 187 m ~ height of each side

A4) 51° 50' ~ angle between each side and basis

B : "king's" chamber

B1) 5,25 m ~ witdh

B2) 5,80 m ~ height

B3) 10,50 m ~ length

C : equator

C1) 40,000,000 m = length of equator per definitionem (old)

D : math

D1) 11/7 - pi/2 = 0.0006324....

D2) 1.68103398875 = phi

D3) 0.68103398875 = 1/phi

D4) 1 + 1/phi = phi

D5) 1 + phi = phi²

D6) phi = 0.5 * (1+sqrt(5))

Conclusions

For 1a: the heigt of the pyramid is related to each edge in the ratio 7 /

11

tan 51° 50' 33" = 14/11 = 7/(0.5 * 11) = h/a2 with a2 = 1/2 a means h/a =

7/11

For 1b: the surface measure of each side equals the square over the heigt

of the pyramid

a = edge of basis

h = height of pyramid

s = heigth of side

pythagorean triangle: s (height of side) from base to top, h from top to

ground, "a2" = a/2 from the center of the base square to the middle of the

edge of the basis

area of each side

A = h² = 1/2 a s, s² = 4 (h² /a)²

pythagoras

s²=(a/2)² + h²

gives

1/4 a² + h² = 4 (h²/a)² times 4/a²

1 + 4 h² / a² = (4 h² / a²)² which is the same as 1 + phi = phi²

4 h² / a² = phi

h = 1/2 sqrt(phi) a

a2 = 1/2 a

arc tan (h/a2) = arc tan ( sqrt(phi)) = 51° 49' 38"

For 1c: the sum over the four edges of the basis is related to the heigt

of the pyramid

in the same ratio as the circumfence of a circle to its radius

4 a = 2 pi h

h/a = 2/pi

h/a2 = 4/pi

arc tan (4/pi) = 51° 51' 14"

Result for statement 1 complete (a = 6250/27 m = 231.48 m) :

ansatz 1a h = 7/11 a h/a=0.636363636 h = 147.31 m

s=a sqrt(317)/22 s/a=0.809295173 s = 187.34 m

ansatz 1b h = a/2 sqrt(phi) h/a=0.630098247 h = 147.22 m

s = a/2 phi h/a=0.809016994 s = 187.27 m

ansatz 1c h = 2/pi a h/a=0.636619772 h = 147,37 m

s=a/2pi*sqrt(16+pi²) s/a=0.809496593 h = 187.38 m

For 2 : each edge of the basis equals the distance which a point on the

equator moves in exactly half a second

40,000,000m / (24*60*60*2) = 5*5*5*5*10/(3*3*3) m = 6250/27 m = 231.48 m

For 3 : the height of each side of the pyramid is approximately one

classic stadion of 600 feet

1 stadion = 40,000,000m / (60*60*60) = 5000/27 m = 185.18 m

(=600*30,864cm) after 1 the height of each side is ~187.3 m which is 606,8

feet

For 4: measurements were taken in multiple integers of an royal ell of

52,4 cm and a foot of 30,8 cm

440 royal ell = length of edge at basis = 6250/27 m

1 royal ell = 625/(27*44) m = 52.6094 cm

750 pyramid feet = length of edge = 6250/27 m

1 pyramid foot = 25/81 m = 30.8614 cm

after 1a heigth of pyramid h = 440 ells * 7/11 = 280 ells

pythagoras: 280² + 220² = 356,09²

600 pyramid feet a 25/81 m = 1 stadion = 352 royal ells

(440=11*40, 280=7*40, 352=11*32 352/440=4/5=0.8)

1 pyramid inch (ounce=1/12) = 2.54 cm ! (no, only a joke, 2.57cm is

correct)

For 5 : the "king's" chamber represents the pythagorean triangles 3 : 4 :

5 and 2 : sqrt(5) : 3

royal ell / pyramid foot = 625 / (27*44) * 81/25 = 75/44 = 1.70454545...

which means 17 pyramid feet are nearly 10 royal ells and give 17*25/61m =

5,25 m

now we take b = 8.5 feet, 17 feet = b * 2 (=5.24m, width)

8.5 feet * sqrt(5) = 19,0066 feet = b * sqrt(5) (=5.87m, height)

8.5 feet * 3 = 25.5 feet = b * 3 (=7.87m, diagonal)

34 feet = b * 4 (=10.5m, length)

8.5 feet * 4 = 25.5 feet = b * 3 (=7.87m, diagonal)

8.5 feet * 5 = 42.5 feet = b * 5 (=13.1m, room diagonal)

Why isn't it twice as big ? (because of 2:sqrt(5):3:4:5 = 34:38:51:86)

This corresponds to that one edge is only HALF a (time) second times the

velocity at the equator Perhaps symbols for 1/2, time and space.

Now there are three such pyramids, arranged as the stars of the orion belt...