...

addendum to last post

One can either subtract the Khufu Pyramid corner angle,

from the tetrahedral "electron spin" angle 70.528~ {arctan sqrt 8},

or,

one can add tetrahedral 19.47122063 degrees to the Khufu Pyramid corner angle.

The only difference in the result,

is that the resultant angle tangent is flipped or reversed.

ie

first angle tangent result = sqrt {100 / 338}

second angle tangent result = sqrt {338 / 100}. ---> calculate fractions first, then take sqrt <---

I often talk of "transfer of code" in tangents, sines and cosines.

Such as fractions that end up in square roots,

as seen in the corner angles of these pyramid models.

We will use the same math operations to show how tangible "transfer of code"

will occur from the Khufu Pyramid,

into the resultant angle when the true tetrahedral angles are employed.

first the original quote:

Two things to clarify.

One is to view the Khufu Pyramid as shown below,

by setting the tangent {14 / 11},

with

assigned height as -- 14 units,

and assigend half base length as -- 11 units.

This conforms to the 280 cubit height and 440 cubit base geometry.

The surprise,

is that the corner to peak diagonal length -- from the base corner to the pyramid peak,

in the Khufu Pyramid geometry,

is sqrt 438,

where the number 438

aligns with the multiples of the prime 73,

like this:

{365 = 5 x 73 --- 438 = 6 x 73 --- and --- and 584 = 8 x 73}

thus aligns with:

the ancient calendar counts 365 and 584 {Venus synod and Earth year}.

In the math operation where the Khufu Pyramid Corner Angle,

is subtracted

from the "electron spin angle" 70.528~,

the resultant tangent,

sqrt {100 / 338},

produces a maxiumum convergence by virtue of the fraction {100 / 338},

coinciding with the Teotihuacan grid angle tangent fraction {see earlier posts}.

What we did not look at !

was the sine

and cosine

of that resultant angle.

Note here:

that Khufu Pyramid diagonal corner length from base corner to the pyramid peak:

= sqrt 438

Our resultant angle tangent is sqrt {100 / 338}, for 28.54302034 degrees.

sine 28.54302034 = 10 / sqrt 438 = sqrt {100 / 438},

cosine 28.54302034 = sqrt {338 / 438} = {sqrt 338 / sqrt 438}.

Now look at the sine of the angle, and it can be rewritten as such:

sine 28.54302034 = square root {100 x Khafre Pyramid slope tangent / 584 Venus synod}.

--- Khafre pyramid side face angle slope tangent = {4 / 3} --- {for the above math}

calculate the fraction first into decimal form,

then take the square root.

So the math works like this:

100 x {4 /3} ---> divided by ---> 584 = 0.228310502 <----> now take square root,

and once the square root is obtained,

that equals

sine 28.54302034

Point being that the diagonal corner length of the Khufu Pyramid,

{base corner to pyramid peak}

transfered the -- sqrt 438 -- length value,

from the geometry test with tetrahedral 70.528~,

into:

the sine and cosine of the resultant angle.

...

addendum to last post

One can either subtract the Khufu Pyramid corner angle,

from the tetrahedral "electron spin" angle 70.528~ {arctan sqrt 8},

or,

one can add tetrahedral 19.47122063 degrees to the Khufu Pyramid corner angle.

The only difference in the result,

is that the resultant angle tangent is flipped or reversed.

ie

first angle tangent result = sqrt {100 / 338}

second angle tangent result = sqrt {338 / 100}. ---> calculate fractions first, then take sqrt <---

I often talk of "transfer of code" in tangents, sines and cosines.

Such as fractions that end up in square roots,

as seen in the corner angles of these pyramid models.

We will use the same math operations to show how tangible "transfer of code"

will occur from the Khufu Pyramid,

into the resultant angle when the true tetrahedral angles are employed.

first the original quote:

Quote:When the Khufu Pyramid has the Side Face slope tangent as the fraction {14 / 11},

due to the ancient formula:

440 cubit base and 280 cubit height,

then,

the Corner Angles tangent are a function of square root 2 dynamics.

aka

corner angle tangent = {14 / 11 sqrt2} <---> = 0.899954085 --- 41.98575902 degrees.

Remember the earlier "electron spin angle" --- 70.528~ degrees = arctangent sqrt 8 <---

Let's test the Khufu Pyramid Corner Angle with the "electron Hi spin angle"

arctan sqrt 8 minus arctan{14 / 11 sqrt2}

70.52877937 minus 41.98575902 degrees = 28.54302034 degrees <---

tangent

28.54302034 degrees <---

equals

square root of fraction {100 / 338}

Now where did we see that fraction before ???

Go back,

to my Teotihuacan city grid angles --- the "16.5" degree angle tangent --- {100 / 338}.

tangent

28.54302034 degrees <---

also equals:

{100 x sqrt2} divided by 260 Tzolkin

Two things to clarify.

One is to view the Khufu Pyramid as shown below,

by setting the tangent {14 / 11},

with

assigned height as -- 14 units,

and assigend half base length as -- 11 units.

This conforms to the 280 cubit height and 440 cubit base geometry.

The surprise,

is that the corner to peak diagonal length -- from the base corner to the pyramid peak,

in the Khufu Pyramid geometry,

is sqrt 438,

where the number 438

aligns with the multiples of the prime 73,

like this:

{365 = 5 x 73 --- 438 = 6 x 73 --- and --- and 584 = 8 x 73}

thus aligns with:

the ancient calendar counts 365 and 584 {Venus synod and Earth year}.

In the math operation where the Khufu Pyramid Corner Angle,

is subtracted

from the "electron spin angle" 70.528~,

the resultant tangent,

sqrt {100 / 338},

produces a maxiumum convergence by virtue of the fraction {100 / 338},

coinciding with the Teotihuacan grid angle tangent fraction {see earlier posts}.

What we did not look at !

was the sine

and cosine

of that resultant angle.

Note here:

that Khufu Pyramid diagonal corner length from base corner to the pyramid peak:

= sqrt 438

Our resultant angle tangent is sqrt {100 / 338}, for 28.54302034 degrees.

sine 28.54302034 = 10 / sqrt 438 = sqrt {100 / 438},

cosine 28.54302034 = sqrt {338 / 438} = {sqrt 338 / sqrt 438}.

Now look at the sine of the angle, and it can be rewritten as such:

sine 28.54302034 = square root {100 x Khafre Pyramid slope tangent / 584 Venus synod}.

--- Khafre pyramid side face angle slope tangent = {4 / 3} --- {for the above math}

calculate the fraction first into decimal form,

then take the square root.

So the math works like this:

100 x {4 /3} ---> divided by ---> 584 = 0.228310502 <----> now take square root,

and once the square root is obtained,

that equals

sine 28.54302034

Point being that the diagonal corner length of the Khufu Pyramid,

{base corner to pyramid peak}

transfered the -- sqrt 438 -- length value,

from the geometry test with tetrahedral 70.528~,

into:

the sine and cosine of the resultant angle.

...