Polywell FOIA

Point out news stories, on the net or in mainstream media, related to polywell fusion.

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MSimon
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Post by MSimon »

chrismb wrote:
MSimon wrote: And what is this 10^10 obsession? Why not 1E8 or 1E17?
Typical total collision cross-section for fusible H or D = ~10^-16 cm^2

Typical fusion cross-section for fusible H or D = ~10mbarns

ratio = ~10^10

Sure, that can be modified (a little bit) but is around that value, particularly for the 'lower energy' 10s of keV experiments.
Dude,

Where are you getting your info?

D-D fusion has a cross section of .1 barn depending on energy.

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chrismb
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Post by chrismb »

I don't understand what you are saying. I have said it is around 10mbarn for a few 10's of keV and you show me a plot showing a cross-section of 10mbarn at 25keV.

errr.....?

Please clarify your point.

KitemanSA
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Post by KitemanSA »

Getting back to

POLYWELL FOIA...


eleven.

Hmmph!

TallDave
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Post by TallDave »

Sure, OK maybe they have spent their time thinking about it. I spent my time calculating.
Other people have spent orders of magnitude more time calculating.
People long to see such a simple device as a fusor working. As the grid is the only thing people can physically see in a fusor, so they blame that for it "not working".
Or because it's blindingly obvious you can't have a giant hunk of metal sitting in the middle absorbing ion impacts, rather than because of their "longings."
You are right that people have thought the issue is the grid.
Again, I'm not sure why you have so much trouble with the concept of "not the only issue." Everyone who has studied IEC agrees the grid is not the only issue. It is generally accepted the grid losses are a showstopper in addition to whatever other problems exist. You are the only one who thinks it is "minor." So, either everyone in the field is an idiot or you're wrong (perhaps you are longing to see yourself as some sort of superstar lone genius). Which way is the safer bet?
This is human nature - to imagine whatever is the nature of a thing and then hold on to that belief tenaciously whilst rejecting objective counter arguments.
Oh, the irony.
Last edited by TallDave on Sun Mar 07, 2010 5:05 pm, edited 1 time in total.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

chrismb
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Post by chrismb »

TallDave wrote: Or, because it's blindingly obvious.
Just show me these calculations!

We've had this debate before, and nothing objective came out of the "the-grid-is-the-problem" camp then, and I presume it'll be the same now.

Fusor beams are self-generating along the beam lines. As the ion current builds up along a particular path, so the majority of the ionisation happens there. Beams self-support themselves, and they self-organise to miss the grid.

Ions that hit the grid 'die out' quickly. Ions that happen to be on the right path then 'breed' more ions along that path.

It's a bit like saying it is blindingly obvious that the limiting efficiency in a laser is the tube that contains it.

If it is blindingly obvious that the grids are the problem, then why don't fusor beams intersect the grid!!! THEN it would be blindingly obvious and I would agree with you. They don't, so I don't.

TallDave
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Post by TallDave »

Just show me these calculations!
There's about a million papers on IEC fusors out there, chock full of calculations telling us at what power a gridded IEC system will overheat and fail. Let me know when you've read them.
We've had this debate before, and nothing objective came out of the "the-grid-is-the-problem" camp then, and I presume it'll be the same now.
Because you can't seem to grasp the obvious. The grids MELT. Not enough ions miss them, self-organized or not. You can't accelerate anything with a melted grid. Not a minor problem.

A gridless IEC fusor doesn't instantly become a net power machine, but it does solve a major problem. Hopefully the FOIA will give us some clue how difficult the remaining problems will be.
Last edited by TallDave on Sun Mar 07, 2010 5:19 pm, edited 2 times in total.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

MSimon
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Post by MSimon »

chrismb wrote:I don't understand what you are saying. I have said it is around 10mbarn for a few 10's of keV and you show me a plot showing a cross-section of 10mbarn at 25keV.

errr.....?

Please clarify your point.
I don't know how this can be done but if you could possibly raise the drive voltage (do we have the technology for that?) you could get it up to 100 mbarn. And possibly beyond.

But I'll grant you 10^10. For something. What was it again?

Well maybe we will find out in the FOIA.
Engineering is the art of making what you want from what you can get at a profit.

chrismb
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Post by chrismb »

TallDave wrote: Because you can't seem to grasp the obvious. The grids MELT. Not enough ions miss them, self-organized or not. You can't accelerate anything with a melted grid. Not a minor problem.
You are right that a conventional discharge fusor will have some limiting power at which the grids cannot function.

Now... tell me why ramping up the power improves fusor efficiency?

Usually it is simply a linear function up to some current value, then neutron rate tails off (which we have come to presume is due to space charge effects).

In other words, lower power inputs (down to some minimal limit) give you better efficiency in a fusor.

Further, the Wisconsin group have also peer review publications to show the grid is a relatively minor thing, and others that show the numerical simulations support this conclusion. I have linked to those before, somewhere.

chrismb
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Post by chrismb »

MSimon wrote: I don't know how this can be done but if you could possibly raise the drive voltage (do we have the technology for that?) you could get it up to 100 mbarn. And possibly beyond.
Yes, that's right. And as you wind up the particle energy you also reduce the total collision cross-section. Win-win, as far as it goes, but still not enough to reach the prize - over unity. You have to find a 10 oom improvement for that, and you can't get it just by input/particle energy alone.

TallDave
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Post by TallDave »

chrismb,

Sure, that's all fine... if you don't care about ever getting to net power. If you do, the grid is a major problem. You cannot produce megawatts of power with a hunk of metal in the center.

I'm still not sure why you think Coulomb scattering is a bigger problem. Where are you saying the energy goes?
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

chrismb
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Post by chrismb »

TallDave wrote:chrismb,

Sure, that's all fine... if you don't care about ever getting to net power. If you do, the grid is a major problem. You cannot produce megawatts of power with a hunk of metal in the center.

I'm still not sure why you think Coulomb scattering is a bigger problem. Where are you saying the energy goes?
It goes into heating up the background matter. Plain, and simple. That's what all the 'glow' is about.

The question is whether you really ever could end up with a situation in which there is so little 'slow' stuff in a Polywell that all the bouncing around is all radial. Thing is, you've got ions moving in and out and if you get them colliding just nanometers aways from the dead centre then they will start scattering into a Maxwellian distribution, and that's presuming you've got the impossible 'perfect' vacuum, into which you never get any sputtered products off wall bombardment, &c., &c..

But that's to confuse the specific question of the grid.

You should still be able to get net power at mW (not counting support equipment, of course), if the device works as intended. The grid is a bar to very high input power - agreed, no challenge - but it is not the problem for getting net power, it's about 10% of that problem, according to my understanding of Wisconsin/simulations. I would've thought it was less than that myself, but will happily pay due respect to those who have doen the experiment.

Art Carlson
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Post by Art Carlson »

TallDave wrote:I'm sure there's some spread of velocities, but Chacon's paper suggests partially relaxed distributions could yield large Q values.
With D-T, if you neglect cusp losses. But the ion energy distributions with high Q are not "partially" relaxed, but "nearly fully" relaxed.

TallDave
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Post by TallDave »

It goes into heating up the background matter.
There isn't any background matter to heat in a Polywell. There are electrons and accelerated ions. The ions are all falling into the well, so they have high energy at the bottom.

If you're going to claim the velocity distribution will spread out, sure. If you want to claim they pass too much energy to electrons, ok maybe. But you can't reduce the energy of the whole system by positing a bunch of slow ions at the core without explaing where that energy went.

Does anyone remember the average life of a fuel ion at reactor conditions? I know I saw it somewhere. Maybe I'll dig around for it.

Art,

I'm prepared to neglect them pending WB-8 results (the hotspots haven't been on the wall, at least not yet). I agree p-B11 is a whole 'nother story.

At this point I really just want to know what the electron losses look like.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

chrismb
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Post by chrismb »

TallDave wrote:There isn't any background matter to heat in a Polywell.
You're living in a fantasy world thinking this kind of stuff.

So you mean it's gonna operate at 10^-22 torr?

These are the sorts of ideas that idealistic inventors dream up, then wonder why their invention didn't work.

Problem is, you won't need many loose neutrals to spoil the party, cos once just one gets shunted by a fast ion, then that fast ion is no longer very fast, then you've got two not-full-speed ions, and then 4, and then xxx.. This is thermalisation, and Polywell will not resist it. Polywell's only hope is that this, supposed, annealing process will somehow pull up those slowed ions back to full energy, a process that I am not holding my breath to see.

The only way you can avoid thermalisation is to ensure there really are no neutrals floating around at all, and I presume you do understand that this is just a fantasy.

TallDave
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Post by TallDave »

Are you even being serious anymore? You think all the energy is going into background neutrals??

You do understand this thing is full of electrons, right?
Last edited by TallDave on Sun Mar 07, 2010 10:46 pm, edited 1 time in total.
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...

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