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Counter-rotary feature of quantum tunnelingCounte


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In 2000

https://m.youtube.com/watch?v=Hjw4BCmOxWQ

This landmark

https://i.ibb.co/qkjJCXM/screens-shot-842021116.jpg

Acts as my PhD because I'm explaining the source of the object that leveled this

https://a57.foxnews.com/static.foxnews.com/foxnews.com/content/uploads/2021/06/640/320/what-an-e-state-698074.jpg?ve=1&tl=1

And the lasers in the 2021 film Voyagers.

 

I need a negasonic teenage warhead

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Members must be able to participate without watching videos or going offsite. Please give a summary of the information you want people to have, and please also be clear about what you want to talk about

 
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I'm referring to an event that has been well documented as evidence of what secrets I hold I am tired of playing this game.

"

The historical codex, theory of everything, cracks QM, unlike relativity and newton's laws of universal motion it's a function f(t)=

Part 1; an 324 columb and 2 row [x y] coordinate matrix (sphere worlds) plus another 324 columb 2 row coordinate matrix (quantum gravity).

[n n]

[n n]

[n n]

[n n]

[n n]

[n n]

[n n]

[n n]

[n n]

[n n]

[n n]

[n n]

[n n]

[n n]

[n n]

[n n]

[n n]

[n n]

Etc...

Part 2; You take every integer in the first two matrices to the power of the product of those two matrices. Like Russian matryoshka dolls you have to do this n number times, n=radius of observable universe over the planck length. & we must root every exponent in the -z direction to superimpose the smaller and smaller x and y values in the +z direction. && everytime we do this we need to take out the top exponent from power towers and redo it for counting down to the x y coordinates of the z vector which is the center of the codex, let's call this operation V for vector. Problem with operation V, you will lose entire matrices of coordinates when you remove an exponent from the power tower, so now we must rectify that by putting op V through the power tower and plugging those values back into the base matrix, and then redoing the full power-tower with these adjusted values, but this time rooting every other iteration of exponential matrices in the power tower so that the values remain the same. 

Ex)

Vector +3

[3.14^[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]
[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]4.13^[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]
[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]]]
[3.14^[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]
[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]] 4.13^[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]
[√4.13^[√3.14 √4.13]
[√4.13 √3.14]
√3.14^[√3.14 √4.13]
[√3.14 √4.13]]]
 
&
 
Vector -3
 
[3.14^[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]
[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]4.13^[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]
[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]]]
[3.14^[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]
[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]] 4.13^[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]
[4.13^[3.14 4.13]
[4.13 3.14]
3.14^[3.14 4.13]
[3.14 4.13]]]
 
&&
Vector -2 part 1
 
[3.14^[4.13 3.14]
[4.13 3.144.13^[4.13 3.14]
[3.14 4.13]]
[3.14^[4.13 3.14]
[4.13 3.14]] 
4.13^[4.13 3.14]
[4.13 3.14]]
etc...
 
&&&
Vector -2 part 2
 
[36.34^[1^[1 1]
[1 1]
Etc...Etc..
 
(Note! + and negative vectors shown in example are reverse from what they should be for the Historical Codex because the values for the matrices in this example are >1 but the actual values you want will all be <1)
 
EDIT: And there's still one final grueling layer to part 2 I'm not accounting for. Of course the original 324 coordinates of three z matrices containing 9 spheres is not I'm 3.14 4.13 it's less than one meters! So that means in the code where the -z side of the power tower are all rooted exponents, eventually it's going to yield a value greater than 1! Eventually up to 13.8/2 billion light years over one meter! So wherever that happens, every exponent above it is not going to be rooted, whatever iteration exponential matrices that is, will say the nth exponent.
 
Part 3; Then you put your answer into a fibonacci sequence. The beginning of the sequence is 1 (first matrix) + 1 (second matrix how the gravity moves xyz up down forward back) = another matrix of sphere worlds which is multiplied by a different quantum gravity matrix which is the second iteration of this special fibonacci sequence, which is a potential universe at time equals 2 planck seconds, with 4 matrices. Etc..(note: you must know where the vectors talk to one another is equal to the generation of the gravitational matrices & when incorporating one gravitational matrix into another vector's matrices one must add or subtract it's transformational values by the difference of values between the matrices in those vectors)

Ex)

[n1 n1]____[n2 n2]____[n3 n3]

________+__________=_____________

[n1 n1]____[n2 n2]____[n3 n3]

&

[n1 n1]___[n3 n3]____[x1 y1]

_______+_________=2*______

[n1 n1]___[n3 n3]____[x1 y1]

&&

[n2 n2]___[n3 n3]__[x2 y2]

________+________=2*______

[n2 n2]___[n3 n3]__[x2 y2]

&&&

___[x1 y1]____[x2 ny2]__[n4 n4]

+/-_______+/-________=__________

___[x1 y1]____[x2 y2]__[n4 n4]

&&&&

[n3 n3]___[n4/2 n4/2]__[nx ny]___[n5 n5]

________+_____________+________=________

[n3 n3]___[n4/2 n4/2]__[nx ny]___[n5 n5]

where, nx & ny = geometric equations applied to the [n3 n3] columns' gravitational tug and [n4 n4] matrix gets divided by 2 because of the inverse square law for [n2 n2]'s gravitational tug applied to [n3 n3]'s position from when it was [n1 n1]

The practical use of the codex is evolving the observable universe from the moment of the big bang to now. 

You will acquire a printout of all of the atoms of the periodic table, and how their quantum particles behave, with a magnification of up to 26 orders of magnitude greater than that offered by an electron microscope, and how micro lasers will effect a quantum system every planck second after your instruments perturb a system.

Unlike normal quantum teleportation, which involves changes over lateral time, the codex can instead be used to teleport quantum objects using particle confusion, which is a result of analog quantum gravity, as gravity propagates in two directions, moving n-number of sphere worlds that collide to generate one higgs function to the exact quantum phonetic frequency of another higgs function, will duplicate that higgs function precisely where the other higgs function is located.

This works because all potential universes are so similar at the quantum scale."

https://www.scienceforums.com/topic/38131-historical-codex/?do=findComment&comment=394227

&&

"

The device in the experiment on harnessing controlled ballistic as a potential over unity device will involve hundreds of imperfect carbon nanotubes (more for conducting electricity and absorbing heat than durability), with 32 simultaneous and continuous rotary features powered by electric impulses (that's the energy input in the denominator of the over unity calculation because the sunlight is free), & 16 polarizing light filters. Whereas the numerator would be the overall energy output of the continuous laser pulse it generates if successful.

If successful as an over unity device, it would be installed into one of the satellites over earth that have static orbits in order to receive continuous sunlight along with and integrated into a miniaturized version of the tokamak fusion reactor."

 

&&&

https://www.scienceforums.com/topic/38229-counter-rotary-feature-of-quantum-tunneling/

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Moderator Note

There's just too much garbage in here to unpack. Overunity, numerology, and enough word salad to choke any herbivore. Trash, trash, trash. This is a science discussion forum. Don't do this here again, ever.

 
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