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does it yet stand to be disproved?

parameters: gas, anode diameter, length, fill pressure, in FoFu-1 differ from that of the paper;
i expect lee’s calculator to quantify a prediction for them, though.

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MTd2 wrote: The papers are just about the interval up to 50KV, and this is clearly not enough since the LPPX is 45Kv. 90KV is clearly the ideal, but I don`t know the lower value. I`d have to play with the values and I`d have to play with the simulation spreadsheet first. Let me make an estimative though. Saturation occurs when all wasted energy is due the dynamical resistance. So, that means an asymptotic value of V=RI. For an optimized configuration of 100KV, see table 3, ( https://focusfusion.org/index.php/forums/viewreply/7483/ ) is about 1.5MA, we achieve a pinch saturation at 1.5MA @ 35KV. Given that we want saturation to occur at 3MA, we must double the voltage, to 70KV. Give a bit of safety margin, to avoid the flatten curve before saturation, we could say 75KV. So yes, your guess is right.

by the way, the quoted articles are using a fill pressure from 2.7 to 3.5 Torr. there are several parameters that are different in FoFu-1.
eg: How does fill pressure affect Ipeak? FoFu-1, if i recall, will be running at 40 Torr.

“the pinch current limitation is not a simple effect, but is a combination of […] the various inductances involved in the plasma focus processes abetted by the increasing coupling of C0 to the inductive energetic processes, as L0 is reduced.”

“once the model parameters have been fitted to a machine for a given gas, these model parameters may be used with some degree of confidence when operating parameters such as the voltage are varied[26,27]” —

reference 26 turns out to be online at:
http://plasmafocus.net/IPFS/modelpackage/File1RADPF.htm

and there is a calculator

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on the contrary,
“Radioactive particles that slam into the gold push out a shower of high-energy electrons. They pass through carbon nanotubes and pass into the lithium hydride from where they move into electrodes, allowing current to flow.”

this shower of high-energy electrons process is like what we want to see happen in the onion.

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MTd2 wrote: …
Summing up, not even with Poseidon, which has banks of 750KJ and a voltage of 80KV, were capable of achieving more than 1.5MA of current in the pinch. This is real bad since it means a neutron yield one magnitude lower than what is desired for LPPX for an equipment with much higher capacity.

by the way,
the forum software, here, mangles URLs that embed spaces or punctuation, but you can use a shortener:

http://www.plasmafocus.net/IPFS/Papers/PP1Published APPLAB922021503_1witherratum.pdf
–> http://tinyurl.com/2uho7a9

http://www.intimal.edu.my/school/fas/UFLF/Papers/PP3PublishedPPCF 50 (2008) 065012 .pdf
–> http://tinyurl.com/38u9lfd

http://www.plasmafocus.net/IPFS/2010 Papers/Energies PP.pdf
–> http://tinyurl.com/346zl96

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Francisl wrote: As long as the DPF units are using deuterium and producing neutrons, can they be used to make radioisotopes for sale? Go to page 461 of the Congressional Budget Request

i thought most (all?) medical radioisotopes are fission products?

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MTd2 wrote:

Long distance transmission lines double the voltage to reduce the current by 50%. Using P=(I^2) R, doubling the transmission voltage results in 25% of the losses. Not sure how this relates to raising voltage to raise current…

There are 2 things here. The wasted energy in collisions and the free electrons. The wasted energy will go like P=(I^2)*R anyway, like in a wire. This is a fraction of the electrons that effectively behave like they were in a resistive medium.The increase of I of an input just makes things go even more wasted.

But note that “P” is not the total energy of the capacitor banks, just the part where it is wasted. So, by increasing the total voltage AND keeping BOTH total resistance (which would happen naturally anyway) and total current CONSTANT, is the same as diminishing the current that goes through the resistive process. So, not only R goes down, as also I (resistive) go down so that conservation of energy is respected. So, more current will be available to be absorbed by the pinch.

So, what are the quantitative predictions around this?
do you think the voltage will have to be raised to, say, 75kV?
lower? higher?

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the feed conductors in the FoFu are flat sheets, placed some distance apart. the capacitor discharge propagates along these sheets, from outside to inside.
for total current of 2.5MA, there are four sections of 3 each, (total of 12 switches).

the problems with your notion, are two:
1) there is no practical way to wrap a coil around, to test this; but,
2) i am unconvinced that there is any physical effect, regardless.

i don’t believe your description is unclear. it’s just wrong.

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mchargue wrote:

the field generated in the plasmoid will quickly dominate the process. i doubt there is any benefit

As I understood the process, the axial field is used to start the field rotating in order to assist the process, and that it is an important part of the process. So, I think that tuning the action of the axial field has merit

*tuning* has merit. but, once the pinch has started, there isn’t much you can do. what use would it be to affect .00000001% of the field, if you have only a few nanoseconds to sense and correct?

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the field generated in the plasmoid will quickly dominate the process. i doubt there is any benefit

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unfortunately, this means the public is convinced that fusion is too hard

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Francisl wrote:
The idea is that an auxiliary low voltage, high current power source would be connected to the conductors for a few nanoseconds or however long it takes to establish a stable magnetic field. Then the high voltage capacitor discharges before the magnetic field collapses in the conductors. That way the high voltage current isn’t slowed down by having to establish the magnetic field that is the cause of self-inductance.

i understand what you’re saying, but i don’t believe the physics works that way. the magnetic field isn’t the cause of self-inductance, but a consequence of the current flow. there will be no effect on the high-voltage current.

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Francisl wrote: Is self-inductance in the cables and distribution system the main problem preventing faster discharges? I suggest making inductance a useful thing. That would require using a second circuit to pre-charge the conduction system with a large current to create a strong stable magnetic field around the conductors just before the high voltage system discharges.

I don’t see how that can work.

But i dont see induction as the problem. It’s a matter of synchronization. You have up to 2.4 MA current, fed by 12 switches, so each one rises from nothing to 200 kA, in as few nanoseconds as possible.

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MTd2 wrote: 1.So, it is possible that right now you guys achieved 100J but no one besides you or the investors know, right?
2.If this is the case, will you publish on Nature?

I think the above commentary covers it. at this time, it might be better for those of us awaiting patiently,
to kick back with a big bowl of popcorn, and continue to wait, patiently.

P-)

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JimmyT wrote: this is getting pretty close to the defination of theoretical break even.

Although not continuous, the principle of dynamic equilibrium easily translates to plasma focus, when you consider it to be a sequence of repeating cycles. Q shouldn’t really be that hard to understand.

but you also have: gain

let’s not redefine/muddle the terms. they’re supposed to be mathematically concise.

i understand gain more in terms of each individual fusion reaction, where it should be the simple ratio of
(energy out) / (energy in).

if you think about it, if ANY fusion occurs, at all, this should be >1. which is why, i think its more useful to have
Q = (energy out – energy in) / (energy in), per cycle.

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Evan Carew wrote: Using the below formula, if you consume 100Kg of fuel on a 2000 Metric Ton vehicle, you end up traveling
.049 kps. This doesn’t seem to make sense. What am I doing wrong, or what is missing here? Is the exaust velocity an order of magnitude (or so) slow?

the mechanics of your calculations are correct. with 100 kg of reaction mass and effective exhaust velocity of 1×10^6 m/s,

Vf = Ve * ln(Mi/Mf)
= 1×10^6 * ln( (2 x10^6 +100 ) / 2×10^6)
= 49.99875 m/s

what’s wrong is conceptual: rockets expel a huge amount of reaction mass. as much as 99% of their total mass may be consumed.

as you can surmise, interstellar travel is infeasible unless/until an alternative to rocket propulsion is invented. since that appears to require new physics, it wont be happening any time soon on this planet.

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