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Monday, September 7, 2015

nlc part neuf=how many turns? *inserted*

It never ceases to amaze me how much damage we do to our lives for things unseen and, probably, not even there.  Gods and religions, and people we love.
To find the right algorithm for the universe is the first step to understanding it better.  In this section I have gone farther back to the origins, the pre-nlc, g-space that I have ever gone before, explaining how a spontaneous universe made up of information could arise from a primordial dimensionless space to give rise to an exponential universe of pluses and minuses.
The relationship of decompressing non-linearity was revived by having ct1 be non-linearity and the remaining states being gradual compression.
We have the intersection of the two states giving rise to higher compression states but also having, at least up the 50% mark, there being less intersection in the less compressed states than would be visible later.
It we have the two contacting spirals and the idea of the ct1 spiral "playing" the information in the higher clock time states as it passes through them like the needle on a phonograph or the laser against a disc we can see how the compression equation is affected by both the turning of the compressing spiral and the sampling by the decompression spiral.
If we assume that the ct1 grows in dimensions, picking up states as it compresses the remainder of the universe by contacting the compressing spirals, eventually we would have the to equal sides necessary for the final compression, one the accumulated ct1 and the other the worn down ct2-ctx.
If we can count the states produced so far by the interaction, and we know how much information there is in the universe,  we would know how many of the loops are in place and can calculated the number remaining.  We could also calculate (since at 2pir (or possibly at 180 degrees or pi(r)) each loop is compressed to the next higher state or clock time) with only two of the cycles what the moment of the compression cycle is, assuming the cycle is constant, using the appropriate equation for the type of spiral defined.  Likewise, with a single rotation, the same estimate can be made if we can determine the equation for the curve.
This leaves quite a bit left to do, however, figuring why different compression states are expressed when compressed into existence, into forces or dimensional coordinates.
For now, if we accept the spiral equation and assume that a stage of compression occurs at 2pi (each revolution of the spiral) then we come up with a very simple solution for the amount of "time" in the universe.
First we can get the number of "turns" by solving for n in the very simple equation:
I(tot)=2^n. N is the total number of turns of the spiral and the maximum compression (or this number N divided by 2 if we assume 180 degree (pi(r)) intersecting points).
where I(total) is the total number of quantum bits of information (at any one conserved one point or with the past, present and future being a little unclear to me in this case, but a one point being suggested by fact that the total information at each turn is the same)
In some of the writing I intimate, incorrectly, that the number of turns was much higher, as much as half the amount of information at one point in time, but the amount of compression at these states is so high that it would take a huge amount of separate high density information states to combine to make this final conversion which explains why, in such a diffuse seeming universe we can be 5% to the total compression.
The total information is merely the solution to the total gravity of the universe divided by the quantum amount of gravity (minimum amount) or, alternatively, by taking how far the universe expanded at ct1 and dividing that by the very small quantum length of the universe.


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