Author Topic: Herman Anderson  (Read 24168 times)

0 Members and 2 Guests are viewing this topic.

Offline Login to see usernames

  • Hero member
  • ****
  • Posts: 609
Re: Herman Anderson
« Reply #24 on: March 28, 2016, 00:30:48 am »
[/quote from Steve]
Fabio, I am still struggling to find out where the neutrons are coming from in Hermans setup...
I See the hydrogen ions being accelerated by the hv and soft xrays.
Any ideas on that?
[/quote]

Oxygen... I don't really think it cares how its structured (hydroxyl or hydroxide ect.)
« Last Edit: March 28, 2016, 07:07:01 am by KS »

Online Login to see usernames

  • Administrator
  • Hero member
  • ****
  • Posts: 4443
    • water structure and science
Re: Herman Anderson photoneutrons
« Reply #25 on: March 28, 2016, 17:16:23 pm »
42
The contribution to the neutron energy spectrum due to absorption of a photon of given energy by any one of these processes will be proportional to the product of the partial cross section for that process and the number of photons at that particular energy. The energy of the released neutron is obtained from the kinematics of the process and the energy of the final state. The total spectrum is obtained by summing over the possible decay modes and integrating over the photon energy. For a compound the sum is also over the constituent elements with a weighting proportional to the relative abundances.
RESULTS
Neutron energy spectra in carbon, nitrogen and oxygen were calculated for various bremsstrahlung end-point energies and, from these, tissue spectra were calculated using the tissue equivalent molecular formula C5H._01ftN. As an example of the results, the photoneutron spectrum from tissue for a photon end-point energy of 28 MeV is shown in figure 1, together with the contributions from the constituent elements. The neutron spectra from tissue over the energy range from 12-30 MeV are summarised in table 1, where average neutron energies and kerma conversion factors are displayed.
Full details of this work can be found in the published report (Allen and Chaudhri 1982).
REFERENCES
1. Allen P D, and Chaudhri M A, (1982) Phys. Med. Biol. 27:553.
2. berman B L, Fultz F C, Caldwell J T, Kelly M A, and Dietrich S S, (1970) Phys. Rev. C2:2318.
3. Caldwell J T, Bramblett R L, Berman B L, Harvey R R, and Fultz S C, (1965) Phys. Rev. Lett. 15:976.
4. Caswell R S, Coyne J J, and Randolph M L, (1980) Rad. Res. 83:217.
5. Fultz S C, Caldwell J T, Berman B L, Bramblett R L, and Harvey R R, (1966) Phys. Rev. 143:790.
6. Schiff L I, (1951) Phys. Rev. 83:252.

Online Login to see usernames

  • Administrator
  • Hero member
  • ****
  • Posts: 4443
    • water structure and science
Re: Herman Anderson
« Reply #26 on: March 28, 2016, 17:22:50 pm »
X-rays are basically produced by high-energy electrons bombarding a target, especially targets that have a high proton number (Z). When bombarding electrons penetrate into the target, some electrons travel close to the nucleus due to the attraction of its positive charge and are subsequently influenced by its electric field. The course of these electrons would be deflected, and a portion or all of their kinetic energy would be lost. The principle of the conservation of energy states that in producing the X-ray photon, the electron has lost some of its kinetic energy (KE):

final KE of electron = initial KE of electron - energy of X-ray photon
The 'lost' energy is emitted as X-ray photons, specifically bremsstrahlung radiation (bremsstrahlung is German for 'braking radiation'). Bremsstrahlung can have any energy ranging from zero to the maximum KE of the bombarding electrons (i.e., 0 to Emax), depending on how much the electrons are influenced by the electric field, therefore forming a continuous spectrum. The 'peak' of the spectrum typically occurs at approximately one-third of Emax so for a bremsstrahlung spectra with an Emax value of say 120 keV, the peak of the spectrum would be at approximately 40 keV.

The intensity of bremsstrahlung radiation is proportional to the square of the atomic number of the target (Z), the number of unit charges of the bombarding particle (z) and inversely with the mass of the bombarding particle (m): Z² z / m. It follows that light particles such as electrons and positrons bombarding targets of high atomic number are more efficient producers of bremsstrahlung radiation than heavier particles such as alpha particles or neutrons (which can also cause X-rays to be produced through bremsstrahlung, though it's much more unlikely than with electrons).

Online Login to see usernames

  • Administrator
  • Hero member
  • ****
  • Posts: 4443
    • water structure and science
Re: Herman Anderson
« Reply #27 on: March 28, 2016, 17:38:13 pm »
Ok, I think I am getting a complete picture here.
Herman the ion king was real. For real, when he said that he just wanted to prove that water could work as a fuel.

The 70kv spark in the hole on the outside of his electrode had to be as thin as possible, so that the soft x rays could enter the cell and in that way to accelerate electrons and hydrogen ions and to add photons to kick neutrons off the oxygen , so the accelerated hydrogen ions could merge with the neutrons. All done in a pressurised cell.
Nickel plated iron electrodes.
70kv
Certain plate distance
Koh
20 amps 12volts
Etc
Bingo....


The last theory is done.
Now to figure out how to make it work in real life.

Offline Login to see usernames

  • Global Moderator
  • Hero member
  • ****
  • Posts: 3607
Re: Herman Anderson
« Reply #28 on: March 29, 2016, 04:58:31 am »
Horvath also talk about xrays right... for xrays to happen there also must be vaccum ... i dont know how this neutron stuf is ... what i think really meyer was doing was breaking apart some hydrogen atoms tranforming mass into energy.. he usually say that the oxygen missing electrons wouldnt allow for water to reform and so the reaction keep going until another structure is formed... i think he mean that thermoexplosive energy will result probably because of the decay of few atoms into particles with high energy in a container pressurized such as ice 

this is indeed a nice idea to investigate fusion or whatever it maybe happening..

so maybe you just need an xray tube than..   their target usualy are made of tungsten for being heat resistant,,,

Steve do you know how do you make a 100ev xray?

http://www.svpvril.com/WRadis.html

Offline Login to see usernames

  • Hero member
  • ****
  • Posts: 609
Re: Herman Anderson
« Reply #29 on: March 29, 2016, 05:08:20 am »
Steve posted and shared the most important document maybe on the first or second page of my replication section...a study done at Stanford that perfectly explains what Herman was doin in part of his radiolysis.
Herman says if you go higher than 70kV 10 mA in his apparatus it could create another isotope (tritium) ...something we do not want to do.

Online Login to see usernames

  • Administrator
  • Hero member
  • ****
  • Posts: 4443
    • water structure and science
Re: Herman Anderson
« Reply #30 on: April 09, 2016, 14:09:08 pm »
Hydrogen will appear at the cathode (the negative electrode, where electrons enter the water), and oxygen will appear at the anode (the positive electrode)

Herman spoke of the facts that the hydrogen will be release on the ANODE side and moves to the Kathode side.
As far as i understand electrolysis, on the kathode side, two hydrogen ions will merge into hydrogen molecule.

I also learned that ions of hydrogen are so tiny, that they move easy thru metal. Thats what for example a Joe cell is doing.
Now the question. What do ions do with that creapy hole in the anode?

Kevin, i assume that the electrode with that famous hidden chamber was the anode?

Offline Login to see usernames

  • Hero member
  • ****
  • Posts: 609
Re: Herman Anderson
« Reply #31 on: April 10, 2016, 04:20:44 am »
Best guess is yes being Herman said "that's where we split it at "
I like the other drawing with the air holes better tho...that Stanford study that you posted is by far the most revielling  study I have ever seen....I have tried several times short of calling Mr.Ricketts at MTSU to get a full length unedited copy of that that interview without success....I emailed him again last week with no reply yet.
You will notice at the end of this link he mentions using a hydrogen enrichment add on way back in 08....it would be interesting to at least learn the details of the generator he and his students used.Mr.Ricketts is the man that interviewed Herman on the portion we bought from James....I still get upset that after all these years we still haven't seen that whole interview,just like with Stans tapes we only get bits and pieces of...I really don't even like thinking about that...it aggravates me that people that have all of those tapes are so damn scared if the rest of us had a look at it we might get it finished first.....
All we can do is take what information we have ,let it take us to things not mentioned and figure it out ourselves.
There is nothing or nobody on these forums can do more or better than the universities and corporations can and have already done in terms of hydrogen....this is all suppose to be for fun and sharing,something to do rather than bar hop and nothing or what ever.....why is that we still haven't seen but bits and pieces of certain tapes and interviews is beyond me!!!!!!


http://cem.mtsu.edu/k-12/presenter/dr-cliff-ricketts

I just watched the webcast Mr.Rickets had with some students explaining his coast to coast trip....wasn't exactly what I was expecting ...the electrolysis unit appeared to just be a single pipe cell and he keeps calling the anode negative and the cathode positive so idnt know if he just misspoke or tries to fuel an argument but either way he hasn't answered none of my emails.
« Last Edit: April 10, 2016, 05:53:23 am by KS »