Polywell FOIA

[KitemanSA]

I cannot do this faith-based debate. Present your case, don't present the opinions of others because I cannot debate the reasons that other people have this opinion if neither you nor I know what those reasons are.

Chirismb,
Do you deny that in every case ever known of Farnsworth or ETW type fusors, that before net power can be achieved the grid will melt?

Is this an established FACT to you?

[chrismb]

I cannot do this faith-based debate. Present your case, don't present the opinions of others because I cannot debate the reasons that other people have this opinion if neither you nor I know what those reasons are.

Chirismb,
Do you deny that in every case ever known of Farnsworth or ETW type fusors, that before net power can be achieved the grid will melt?

Is this an established FACT to you?

I don't accept this, because I don't see any reason why, *if it works*, you couldn't make one to run at 1uW, and 1uW won't make a grid melt.

You're only imagining net power to mean megawatts. My approach is simple, in the first instance: while everyone is ramping up the power to get more neutrons, why not try to understand things by aiming for the same number of neutrons, but with less input power?

[Art Carlson]

I cannot do this faith-based debate. Present your case, don't present the opinions of others because I cannot debate the reasons that other people have this opinion if neither you nor I know what those reasons are.

Chirismb,
Do you deny that in every case ever known of Farnsworth or ETW type fusors, that before net power can be achieved the grid will melt?

Is this an established FACT to you?

I don't accept this, because I don't see any reason why, *if it works*, you couldn't make one to run at 1uW, and 1uW won't make a grid melt.

You're only imagining net power to mean megawatts. My approach is simple, in the first instance: while everyone is ramping up the power to get more neutrons, why not try to understand things by aiming for the same number of neutrons, but with less input power?

I once proposed to IPP that they do that with a tokamak. Since not only fusion power density but most loss terms scale with n^2, I imagined building an ITER-sized machine that could reach high Q, but with a low magnetic field and therefore cheap (since all the forces scale with B^2). Nobody took me seriously, but I never understood just why. It may be that at low densities the neutrals penetrate to the core and cause headaches there.

I think what KitemanSA should be saying (and maybe thinks he is saying) is that the power loss to the grid will always exceed the fusion power. At power levels interesting for commercial use, this will melt the grid, but it will prevent you from getting breakeven any way you cut it.

[MSimon]

I cannot do this faith-based debate. Present your case, don't present the opinions of others because I cannot debate the reasons that other people have this opinion if neither you nor I know what those reasons are.

Chirismb,
Do you deny that in every case ever known of Farnsworth or ETW type fusors, that before net power can be achieved the grid will melt?

Is this an established FACT to you?

I don't accept this, because I don't see any reason why, *if it works*, you couldn't make one to run at 1uW, and 1uW won't make a grid melt.

You're only imagining net power to mean megawatts. My approach is simple, in the first instance: while everyone is ramping up the power to get more neutrons, why not try to understand things by aiming for the same number of neutrons, but with less input power?

They do run them at 1 uW power out.

That is not net power though. It takes 500 to 1,000 W to get that 1 uW.

From what I have read Art has got it right.

In a fusor you need 10,000 transits to get reasonable Q. In a fusor you get 20 to 50.

[MSimon]

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".

Grid losses are the accepted reason fusors can not do net power. There may be additional reasons. Why look into them if the grid kills you before you start?

God is the accepted reason for why the Universe exists. That doesn't mean its true.

A right to vote is the accepted reason why democracy is considered "free and fair". That doesn't make it true, does it!?

I cannot do this faith-based debate. Present your case, don't present the opinions of others because I cannot debate the reasons that other people have this opinion if neither you nor I know what those reasons are.

OK. Present your reasons why you think grid losses are not an issue.

Here is mine:

You need 10,000 transits or more to get net power. i.e. loss ratio (per transit) of .9999 or better (1.0000 = no loss) (see Bussard) Grid losses per transit run from 2% to 5% or more. This is the premise of Polywell.

Now a two grid machine seems to do some better in reducing losses (by channeling the oscillations). Still not good enough. See the MIT Paper (by McDougal? - some Irish name I forget the exact name) on the sidebar at IEC Fusion Technology.

http://ssl.mit.edu/publications/theses/ ... Thomas.pdf

http://ssl.mit.edu/publications/theses/ ... chCarl.pdf

http://iecfusiontech.blogspot.com/

[chrismb]

They do run them at 1 uW power out.

That is not net power though. It takes 500 to 1,000 W to get that 1 uW.

And where does that 500 to 1000W go? If it all went into the grid then it *would* burn out!!

Clearly, most is lost by thermal transfer into the outer shell - and most of that is electron conduction, with a small amount due to the thermalisation losses with the background.

What is my argument for the grid not being the loss?? I already put it above, in the form of a question: Is the efficiency of a laser due to the size of the aperture the laser beam came through??

What is that you say? Fusor ions aren't like photons in a laser???!! I mean, bloomin' look at the thing!!!>>>>

image too big


Answer me two questions;

1) Do the beams in the above image look like they are intersecting the grid?

2) How much input power do you think that grid can take, in vacuo, before it'd start glowing like a light bulb?

[chrismb]

In a fusor you need 10,000 transits to get reasonable Q. In a fusor you get 20 to 50.

Oh, JEEEZERS.... How many times do I have to go through this!!!

It needs MILLIONS of passes just to summon up a 'reaction rate', let alone before you even begin to measure efficiency!

[MSimon]

In a fusor you need 10,000 transits to get reasonable Q. In a fusor you get 20 to 50.

Oh, JEEEZERS.... How many times do I have to go through this!!!

It needs MILLIONS of passes just to summon up a 'reaction rate', let alone before you even begin to measure efficiency!

According to Dr. B (and possibly the Navy) that is incorrect.

Some of the amateur fusors are going to liquid cooled grids to get more powerful fusors.

BTW much of that 500 to 1,000 watts is in pumping power and eqpt losses. A low power fusor runs at 10KV @ 10 mA. Not counting ballast resistor losses that is 100 W. You get to 500 W and the grids melt. And you still don't get net power.

And actually it takes a chris constant (10^10) to get to net power ;-)

[MSimon]

Measurable reaction rate is dependent on number of collisions. Not the ratio of power out to power in.

[chrismb]

I've gone through the calculation of fusion cross-section before, to show you how you need, on average, millions of passes for a fusion event, even at the supposedly enormous pressures that will be found in a Polywell, let alone at the micron range of fusors. I'm not going to bother repeating myself if some folks can be bothered to educate themselves about calculating fusion reaction rates.

(Art, can you see some other way that I can explain fusion-rate dependency on cross-section? What can I say to improve my intelligibility?)

[MSimon]

I've gone through the calculation of fusion cross-section before, to show you how you need, on average, millions of passes for a fusion event, even at the supposedly enormous pressures that will be found in a Polywell, let alone at the micron range of fusors.

That is a pretty good proof that fusors have never generated a single neutron ever. None of them gets millions of passes.

[TallDave]

It needs MILLIONS of passes just to summon up a 'reaction rate', let alone before you even begin to measure efficiency!

A fusing ion needs on average a million or so passes to fuse, but the system doesn't need anything near that kind of confinement to get some small measurable fusion (because this is roughly the same as saying one out of a million ions will fuse on a given pass). That's why fusors produce neutrons.

This paper has a more detailed look at confinement measured in passes. They also give some interesting fractions of ion population needed to fuse for net power (p46). For D-D at Q=5, you need your fusion rate to equal your loss rate -- so if we say millions of passes are needed for a ion to fuse, ion lifetime must also be millions of passes. OTOH for D-T at Q=1 it's in the tens of thousands -- so Simon is roughly correct to say

In a fusor you need 10,000 transits to get reasonable Q.

http://ssl.mit.edu/publications/theses/ ... Thomas.pdf

[chrismb]

I've gone through the calculation of fusion cross-section before, to show you how you need, on average, millions of passes for a fusion event, even at the supposedly enormous pressures that will be found in a Polywell, let alone at the micron range of fusors.

That is a pretty good proof that fusors have never generated a single neutron ever. None of them gets millions of passes.

Not at all. It just shows that you are out into the multiple-sigma end of the distribution when generating neutrons, hence why it'll never get to break even.

Look at it this way - to get break even with a, say, 2.5MeV neutron energy capture from a cadre of 100keV deuterons, you'd need at least 1 in 25 to fuse to get Q=1.

The mean free path to a fusion event of a 100keV deuteron in 1 micron is 1/(1E19.1E-29)=1E10m. So if you need at least 1 in 25, so that they each then need to cover 1E10m/25=400,000km so that the total 'mileage' they've covered is the 1E10m total.

OK, now let's think of statistics and the distribution: If a fusor is sized to be 1m total flight per pass, we need 1E10 particles to set off to get the 1E10m total flight required, for a fusion event, in the first pass. In the meantime, the mfp for those deuterons to hit a background is around 1/(1E19.1E-20), or 10m. So we can estimate that this first 1m generates one single fusion event and knocks out 10% of the deuterons. 1E9 deuterons @ 100keV = 1E14eV. The fusion event liberates 2.5E6eV. So the efficiency would be (2.5E6/1E14)=2.5E-8.

Now.... let me guess.... what'd'ya think the efficiency of a fusor is????....

[MSimon]

Now.... let me guess.... what'd'ya think the efficiency of a fusor is????....

1/10^10

[TallDave]

chrismb,

The link suggests a mfp of 1e8M.

Now calculate for D-T with a confinement of 10,000 passes (assuming 1e8 passes per fusion) and you'll see why it's wrong to say you need confinement to be in the millions of passes to even get a reaction rate.

From the paper:

Fusion mean free paths can be estimated with a density of 1020 m-3 and a peak 35 cross-section of 1 barn (10-28 m-2), yielding 108 meters or 200 million passes through our characteristic device. Current state-of-the-art confinement is well below these levels, in the 10 pass area and this thesis deals with the effect of increasing that to 1000 - 10000 passes.