Talk-Polywell.org

Okay, I think I see something. The Valencia paper seems to be saying that the fractional loss area is the key to mitigating line cusp losses. This makes intuitive sense given the picture of the trapping effect as a "wiffleball" (and one can easily realize that the same is probably true for point cus...

I think yes: running the vacuum pumps and feed-through ports at ground is easier than at a high positive voltage. In other words, after the first BFRs are built for power plants, newer designs might be able to put the collector and the outer wall together and save space. Okay, but the collector/inn...

The center of the reactor will have more ions than electrons, so it's positive, but just outside that center there will be more electrons than ions. Net field there is outwards for ions, in for electrons. As you get close to the MaGrid, the potential rises. It keeps rising until you get to the 2 Me...

I think the line cusp issue has been adequately addressed, at least considering that there is currently an experiment running. The Valencia paper , page 8 (which I believe was quoted on cosmiclog), mentions the existence of line cusps, and gives some criteria for mitigation in order to keep the trap...

Of course the inverse square law applies. I didn't say the field was going to be uniform; I said it was going to be uniformly inwards. It will be much stronger near the magrid. But without any charged structure in between the magrid and the wall, there's nothing to cause the field to reverse. That i...

[ASCII art] Why is the direct conversion collector separate from the outer wall? Practical considerations? Also, why is there current going through the direct conversion grid? The collector, not the decelerator, is what provides the electrons to neutralize the fusion products. Yes you will lose ele...

The ions will not be attracted to a point in the center, but will wonder aimlessly in a large volume. So much for annealing and head on collisions. Actually, that image looks just about right for a parabolic potential well. Remember that the electric field is the (negative) gradient of the potentia...

What do you mean by a "rigid" collision? There are "elastic" collisions, in which kinetic energy is conserved, and there are "plastic" collisions, in which kinetic energy is converted to heat. Collisions between macroscopic objects generally fall somewhere in between the two extremes of perfectly el...

No. Otherwise a Dirac delta function could never relax to a Maxwellian. Consider a frictionless elastic hard-sphere collision in which two particles moving at identical speeds at right angles to one another come into contact such that the line between their centres is along the axis of motion of par...

Even with identical ions, a collision that isn't perfectly head-on (ie: a particle's velocity vector doesn't go through the exact centre of its counterpart) will result in a change in momentum for each one. I'm just hoping that since the mass difference between the protons and borons is much greater...

I didn't say there could be no potential gradient inside a conducting closed surface. I said there could be no potential gradient inside an EMPTY conducting closed surface - ie: the charge on the conducting shell itself has no effect within that volume.

Doesn't the surface of the magrid have to be conductive?

I hadn't really thought about it.

I think your #3 is most likely, since that is where the electrons from ECR ionization would go. Also remember that it doesn't have to be 100 MW - it's just 50 amps, and at 20 kV, 50 A is just 1 MW - ie: 1% of reactor output.

I would add that there does seem to be an additional problem with alphas, due to the fact that they don't penetrate very far. Most of their energy is deposited near the surface, which can cause liberation of surface atoms, possibly in large numbers. This is referred to as the sputtering problem. At ...

So, we have: 1) A spherical (conducting) decelerator shell, which produces no E-field inside the device. 2) A quasi-spherical (conducting) magrid, which also should produce no E-field at radii less than the magrid itself, and which should act like a point-charge for radii greater than the magrid. 3...