Talk-Polywell.org

I don't think we heat the ions with the electrons, we simply are using then to maintain a potential well to trap the ions. The ions are “heated” by simply falling down the potential well. By falling down the ions take energy from the electrons. Either they lower the well depth, thus taking potentia...

Is it really a good thing to get rid of the upscattered ions? Afterall the mechanism we use to give energy to the system (and thus the ions) is injection of hot electrons. Wouldn't an increase in electron energy cause a deeper potential well, thereby giving energy back to the ions (assuming most of ...

In the "Should Google Go Nuclear" pdf at askmar.com there's a picture of some single-turn devices emc2 used to test the polyhedron concept, look on page 14.
I think it was also shown in the video.

They are also mentioned in the IAC paper on page 10, but there isn't a picture.

If the electrons are left behind in the center they will be at high potential vs the grids. Just as they would be if shot in from an external electron gun. If the well depth isn't increasing the electrons in the center won't gain more potential energy than they had in the first place. Increasing we...

In an operating reactor - electron losses to the grid are replaced by "fusion" electrons. Electrons left behind by alphas. True, in an operating reactor with sufficient Q the alpha current could compensate all the electron losses. However in a working reactor the electrons lost are high energy, unl...

The main purpose of the electron guns isn't ionizing the gas, it's replacing the electrons lost to the grid (so the net charge stays negative) and heating the electrons coming from the fuel so they can produce a deep well.

Interesting idea, the question is how thick thick enough would be.

Could B10 be added into the water?
That way the energy would be absorbed directly where we want it.

1/2 the time we get a neutron and 3He, half the time we get T. The T presumably fuses (it's a lower energy reaction), giving another neutron per above table. In WB6 only a very small fraction of the fuel ions fused. A similarily small fraction of the T ions might have fused, but the amount of T ion...

Assuming you want to use a method running in O(n log n) or better computing the differences of the potential probably is the easiest way. Integrating over the charges to calculate the potential is the same as solving the poisson equation over the system. There are good algorithms to do that. Also yo...

I wrote 30 bit precision because MSimon estimated 33 bit as the precision double would allow us to get. This is about 10 decimal places. Apart from the memory tracking more particles requires more CPU power too and then there's also the issue of memory bandwith. The errors from having 'few' particle...

If we are talking 3D particle simulations machine precision really isn't an issue. It would take an extremly ridiculous number of particles to calculate field values with 30 bit precision. The problem is that it also takes a pretty big number of particles for a merely acceptable precision, due to th...

Ah ok, I was refering to electrons in the wiffle ball in general, not just to those in the center. Of course cold electrons in the center we do want. However cooling other electrons isn't a good idea (unless maybe if it could be limited to the transversal component). Specifically cooling those in th...

I'm with Indrek on this, the speed of the electrons at a given potential determine the maximal theoretically possible well depth.
If we wanted to cool the electrons, all we'd had to do is drop some heavy atoms in the fuel, bremsstrahlung would do the rest.

As I understand it, that has always been the main problem in simulating a polywell. I doubt this is a question of floating point precision though. The problem (for particle in cell approaches) is, if in a electron only simulation we need i.e. 1E5 macro particles to get an acceptable electrical field...

Indrek, I'm no expert, but since I wrote a similar (but not-quiet-working-yet) program here is one way to do it. The magnetic field is the curl of the magnetic vectorpotential, to calculate this potential you need to solve the poisson equation on the volume you want. Laplace( A(r) ) = -j(r) / (epsil...