Is Mill's Hydrino the c-Hydrogen ?

Discussion concerning the first major re-evaluation of Dewey B. Larson's Reciprocal System of theory, updated to include counterspace (Etheric spaces), projective geometry, and the non-local aspects of time/space.
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Horace
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Is Mill's Hydrino the c-Hydrogen ?

Post by Horace » Sat Apr 08, 2017 6:18 pm

Is Randall Mill's Hydrino the c-Hydrogen ?

Addition of what motion can convert the m-Hydrogen to c-Hydrogen ?
In Mill's apparatus he clams that it is a collision of m-Hydrogen with an Acceptor: a Lithium atom or a Potassium atom or a monomolecular isolated H2O molecule. Does this make sense?

See these videos:
https://www.youtube.com/watch?v=CtA4FdRrCkY
https://vimeo.com/194678559

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bperet
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Re: Is Mill's Hydrino the c-Hydrogen ?

Post by bperet » Sun Apr 09, 2017 10:34 am

Horace wrote:
Sat Apr 08, 2017 6:18 pm
Is Randall Mill's Hydrino the c-Hydrogen ?
Yes. Well, almost...
Horace wrote:
Sat Apr 08, 2017 6:18 pm
Addition of what motion can convert the m-Hydrogen to c-Hydrogen ?
In Mill's apparatus he clams that it is a collision of m-Hydrogen with an Acceptor: a Lithium atom or a Potassium atom or a monomolecular isolated H2O molecule. Does this make sense?
A big mistake made by conventional science that Larson cleared up is that hydrogen is NOT an atom--it is a compound particle (proton + neutrino). Atomic number 1 is actually deuterium, with two double-rotating systems.

So don't look as hydrogen as atomic, but simply as a proton. All that is needed is to accelerate the proton beyond the unit speed boundary and it inverts to the cosmic form of the same particle (as do all particles--but not atoms).

The "rotational datum" (where rotation rotates all the way back to the start) for particles is 2 units, since particles can never have a displacement higher than 1. (It is 4 for atomic magnetic and 8 for atomic electric), so:

In order to convert displacements for a motion that crosses the unit speed boundary, you just subtract the rotational datum from the particle:

m-proton - datum = c-proton
1-1-(1) - 2-2-(2) = (1)-(1)-1

(Remember this is "displacement math" form, so (1) - (2) = 1; the parenthesis are symbols of spatial displacement, not negation, so a negative result means you flip aspects.)

The production of these c-protons probably have nothing to do with chemistry--the accelerations used to create those "collisions" are causing protons to cross over the unit speed boundary and switch to a cosmic structure.

NOTE: this is NOTHING NEW. Nature does it all the time and was documented in the early 1900s by Viktor Schauberger under the term "living water." This SAME concept was later promoted by "Joe" of Australia as the "Joe Cell," a functional device that used living water as a source of c-protons to run a combustion engine. (Joe's unique contribution was that his "cells" were self-running at the "seeding" stage.) This was later rediscovered by Yule Brown as "Brown's Gas," then by a Japanese researcher labeling it as "Ohmasa's Gas," and most recently is being billed as "HHO."

So this company is just re-branding old ideas as something "new," most likely to secure patents for the process--without understanding it, but with sufficient data to block any future research into c-proton based technology via patent infringement claims. They may actually get to a device, but as soon as they do, Big Energy will buy them out and lock everything up, and send hordes of drooling lawyers at anyone that attempts to develop similar tech. After all, that is the modus operandi these days.
Every dogma has its day...

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Horace
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Re: Is Mill's Hydrino the c-Hydrogen ?

Post by Horace » Sun Apr 09, 2017 3:35 pm

bperet wrote:
Sun Apr 09, 2017 10:34 am
In order to convert displacements for a motion that crosses the unit speed boundary, you just subtract the rotational datum from the particle:
m-proton - datum = c-proton
1-1-(1) - 2-2-(2) = (1)-(1)-1
The production of these c-protons probably have nothing to do with chemistry--the accelerations used to create those "collisions" are causing protons to cross over the unit speed boundary and switch to a cosmic structure.
But how? If temperature alone is enough then any flame around 3000º should be converting m-Hydrogen to c-Hydrogen ubiquitously. This means that any OxyAcetylene welding torch should be making c-Hydrogen and kilowatts of bright light when free m-Hydrogen remains in the flame after imperfect combustion or deliberate addition.

Mills' newest apparatus demonstrates what appears to be a self-sustaining conversion without any electric current input, just temperature ( see this video )
The only things in the apparatus are: silver, argon, hydrogen and OH group + temperature.
Is that enough to subtract 2-2-(2) ?


Anyway, what happens after the m-proton gets converted to the c-proton? AFAIR c-proton does not gravitate in space. What happens to it when it encounters regular matter?
Also, why the continuous hard UV radiation with a 10.1nm cutoff wavelength occurs?

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Horace
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Re: Is Mill's Hydrino the c-Hydrogen ?

Post by Horace » Sun Apr 23, 2017 5:14 am

Horace wrote:
Sun Apr 09, 2017 3:35 pm
This means that any OxyAcetylene welding torch should be making c-Hydrogen and kilowatts of bright light when free m-Hydrogen remains in the flame after imperfect combustion or deliberate addition.
...also when any water is present, because it thermolyses to hydrogen and oxygen above 2000º.
 
 
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duane
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Re: Is Mill's Hydrino the c-Hydrogen ?

Post by duane » Wed Apr 26, 2017 6:30 pm

in his book, "Occult Ether Physics: Tesla's hidden space propulsion system and the conspiracy to conceal it"
by William Lyne Second Revised Edition. 1997,1998 published by CREATOPIA PRODUCTIONS (TM)

Mr. Lyne talks about the extra energy obtained during OxyAcetylene welding
over and above the "calculated" amount determined by combustion.
"How does the atomic hydrogen obtain its energy, if not from the "ether"? No wonder establishment science
doesn't want you to know there is an ether. If we are to believe the "law of conservation of energy",
as interpreted by establishment (relativistic, ether excluding) 'science',
this process is impossible, yet using data available from 'standard' texts.
I have shown that the input energy of 103 cal./gram molecule is somehow
either 'magnified' to 109,000 cal./gram molecule of hydrogen----a multiplication
of over 1,058 times---or that, by use of the hydrogen as a "medium",
that the 103 calories is 'seed' energy (called the "activation energy"),
triggering the atomic hydrogen's apprehension of a net 108,897 cal./gram molecule, from the "ether"
page 96

the important thing is the mono or atomic condition of the hydrogen (H) rather than the molecular (H2) form
could we be seeing the same thing just "looked at" by different theories?

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Horace
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Re: Is Mill's Hydrino the c-Hydrogen ?

Post by Horace » Thu Apr 27, 2017 5:01 pm

duane wrote:
Wed Apr 26, 2017 6:30 pm
the important thing is the mono or atomic condition of the hydrogen (H) rather than the molecular (H2) form
could we be seeing the same thing just "looked at" by different theories?
AFAIR the OxyAcetylene flame has a temperature of 3400º and if you look at the thermolysis graph in the previous post then you'll see that half of the molecular H2 cannot survive at this temperature and is broken down to monoatomic hydrogen.

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