Talk-Polywell.org

but in all the photographic cases, you are doing tomography. That it true but it would tell a lot. I absolutely do not believe that the plasma will be a small sphere in the middle, like what is shown on this forum's logo. Indrek's simulations show electrons spending a lot of time "recirculating" ou...

What about simply photographing the plasma so one can know its shape?

2) The above would give us a discrete set of energy states called modes (100s of KeV hopefully) and their probabilities of existence inside the well for each discrete (i.e. quantum) states. The natural mode ideally should look like a spheical cosine (or Bessel) (with a big bulge at the center of th...

There is enough helium at current consumption for over 200 years, see http://minerals.usgs.gov/minerals/pubs/ ... -heliu.pdf

I'm not sure I agree with that. Bussard seemed quite confident that a p-B11 reactor is little more than a slightly bigger version of the D-D reactor. If so, I expect to see p-B11 be the first one to market, because (1) boron and hydrogen are a lot cheaper than deuterium, and (2) avoiding the vast m...

From http://en.wikipedia.org/wiki/Polywell Despite initial difficulties in spherical electron confinement, at the time of the 2005 research project's termination, Bussard had reported a neutron rate of 10e9 per second (based on detection of roughly three neutrons per test[6], giving a wide confidenc...

TallDave wrote:I get 1E+9 fusions x 17.6 MeV per fusion x 1.6E-19 eV per joule = .00282 joules/second = .00282 watts. Seems pretty dim.

He used D-D, so two fusions per neutron, and about 0.001 watts. With D-T he would probably get 10 to 30 times as much fusion, but tritium is too much trouble for an experiment.

Wouldn't the Faraday effect just mean no external impinging E fields? I thought it was the combination of the internal E field from the charge on the trapped electrons and the positive charge on the magrid that kept the positively charged ions confined. Yes, there are no external impinging E fields...

The positive charge on the ions keeps them in the center region due to the positive charge on the grids. Those ions that do get near the grid will be low energy. Really? Inside the grid there are almost no E fields because the grid forms a Faraday cage. You can not compress ions into a region of em...

There is one quite similar called the Z-pinch (Z machine) which is been tested in Sandia Labs and it is close to breakeven. Close to breakeven? The Sandia site uses it for an X-Ray generator, and hardly mentions energy break even. I did see a claim of 10% efficiency of conversion from electricity t...

From the paper: I have built an exceptionally crude NFG system and ACHIEVED POSITIVE EXCESS ENERGY. However, he has no photographs, graphs of data, tables of data, etc. I could find no evidence of any real measurement. If he really has made excess energy from hydrogen fusion, then his lab should be ...

So the opposing views used by most antagonists are based on a masters thesis? No. I studied plasma physics at Princeton in 1976 and 1977, and the infeasibility of aneutronic fusion because of bremsstrahlung was well known even then. The basic problem with fusion is that for every fusion interaction...

Nanos wrote:Would pyrolytic graphite be a better choice ?

I've ordered some of that myself to play with.

Why? Because it is diamagnetic? Diamagnetism is caused by the orbitals of the electron. When it is ionized, it will just be an ordinary plasma. See http://en.wikipedia.org/wiki/Diamagnetic

Sounds like nonsense to me. "Citation needed" has been on the page since june 21, so I removed it. It would take an enormous amount of energy to fully ionize Bismuth, and it has a nuclear charge of 83 vs. 5 for Boron and 1 for hydrogen. It would be very hard to get a hydrogen ion close to the nucleus.

The levitated dipole would have most of the electrons on the outside, like the Van Allen belts of the earth. With the Van Allen belts, there is zero density in the center, because the earth is there. See http://en.wikipedia.org/wiki/Van_Allen_radiation_belt.