[Steven Sesselmann]

Hi Guys,

F.I.C.S. fusion is a work in progress. I am now moving forward in incremental steps. The reactor seems to be working, and I make an off the record claim that I have measured a TIER of at least 200,000 n/s, but it is still early days, and I have not had the voltage over 25 kV. Initial runs highlighted the need for a few improvements.

http://www.fusor.net/board/view.php?bn=fusor_construction&key=1334382403&st=25

Steven

http://www.beeresearch.com.au

[Steven Sesselmann]

Without knowing more about the research than what I read on that page, I would go back and look at all my wiring. I am currently running my own neutron background experiment with an He3 detector, and these things are so sensitive you only need to sneeze to get a false count, so Imagine what a near by lightning strike would do. The signal to noise ratio on most sensitive neutron detectors is poor, which makes them hard to work with in areas where there are plasma discharges. I am sure you guys at LPP agree with that. One of the reasons we like to use bubble detectors, as these are insensitive to EMF noise.

I suspect this might turn out to be another “faster than light neutrino” embarrassment.

Steven

[Steven Sesselmann]

Ivy Matt wrote: I can’t help wondering what would happen if you aimed two of those at each other, with a dilithium crystal in between….

This is what happens… http://youtu.be/Y7bDMFsGisk

Well di-lithium was a little hard to come by, so I substituted it with di-deuterium 🙂

Steven

[Steven Sesselmann]

Ivy Matt wrote: Is there any particular reason deuterium is preferred, apart from the relative ease of fusing deuterium as opposed to heavier elements?

Thanks for your question, Apart from the availability and the known ease of fusing Deuterium, no, there is no particular reason, the other fuels have also been mentioned in the patent. Clearly a fuel having a positively charged fusion product is preferred, as those reactions contribute in the charge separation process.

I think my idea of using the kinetic energy of the fusion ash to separate and amplify a charge through an electrical potential is novel, and if the process works, it will offer another way to convert fusion energy into electricity directly.

The p + B11 reaction is troublesome, due to it’s small cross section, and so far as I am unaware, no amateur scientist has achieved p+B11 fusion in any measurable way or form, although I do have a friend who has been trying for the last 10 years. Who knows, one day he might succeed.

Steven

[Steven Sesselmann]

Hello all,

Some images from the construction of my F.I.C.S. fusion invention.

http://www.beejewel.com.au/research/Bee_Research/My_Albums/Pages/FICS_Construction.html

Before asking questions read the discussion of fusor.net here http://www.fusor.net/board/view.php?bn=fusor_theory&key=1326524523

Cheers…

Steven

[Steven Sesselmann]

James,
Not a simple 1:1 relationship, ionizing the atom at 10^-5 volts potential might lower the coulomb barrier by a factor as follows…

The mass energy potential of +1 hydrogen ion at ground potential, less the negative potential divided by the total mass energy at ground potential.

So I think the coulomb barrier might be lower by a factor of ((10^9) – (10^5)) / (10^9) = 0.9999

Not much in the scheme of things, but maybe enough to see an improvement in the fusion rate.

Steven

[Steven Sesselmann]

jamesr wrote: The only significant influence on one ion is the other ion.

I agree with most of your post, just this last part that I would like to single out…

In my hypothesis the above quoted statement is false, there is a third factor which can not be ignored, namely the observer. A particles preferential potential is a function of its mass, and as such, the preferred potential of a deuteron is above ground potential. We know this, because if you drop a deuteron it does not fall down….it floats up!

Therefore by dragging it below ground potential we are taking it further away from its preferential potential, the deuteron will in my opinion do anything it can, to get back to where it belongs, and as sure as water runs down hill, nature will find a way, and we know that the only way out is fusion.

This is the equivalent of putting the deuteron (king) in check, and the only way out for the deuteron (king) is fusing (towering).

Steven

[Steven Sesselmann]

jamesr wrote:

what I do believe is that ionizing particles below ground potential, lowers the relative Coulomb barrier.

In a word… No.

snip…

The electric field created by one ion at the position of the other at this distance is E=1/(4pi*eps0) *q/r^2 ~10^20 V/m

snip…

James, before you throw the baby out with the bathwater, stop for a minute and consider the observers potential.

The equation above is valid for particles generated at ground potential, ie, the experimenter and the particles origin is the same.

Not so when the ions are generated below ground potential, in the case of my FICS fusion reactor, the plasma holds a negative potential with respect to the observer, and I think one needs to multiply the equation above by 10^-5, when observing the event from ground potential.

Steven

[Steven Sesselmann]

asymmetric_implosion wrote: I’m a bit confused by the video and it seems like you are glossing over a few important points in the physics. You mention alpha particles being created in the fusor; are you using D-T fuel or a p-11B fuel? D-D fuel cannot produce 4He in a low pressure environment due to momentum conservations considerations. If you are referring to D-T or p-11B, why does the alpha particle only have 100 keV? Shouldn’t any particles created in the fusion event have their fraction of the change in binding energy (14.1 MeV in D-T) plus the potential difference of 100 keV. In that case, the 100 keV seems pretty small compared to the fusion yield. Could you clarify please?

Thank you for pointing this out, I was talking without a script, and could have used better terminology.

My experiment will be done with Deuterium fuel only so we are discussing D+D fusion, and the reactions as we know are D+D -> T + p and D+D -> 3He + n, in my talk I was referring to the positively charged fusion products, ie. the proton and the 3He nucleus.

If my predictions hold true, the fusing molecules will automatically orientate themselves so that the positive particle/nucleus (p or 3He) can go to ground via the flight tube, ie. up the potential energy slope. You can imagine the potential energy well is in the order of 100 Kev, while the fusion products have energies in the order of Mev, so they will have no trouble escaping.

A proton or an He3 nucleus with a charge of +1 climbing through a potential of 100 Kv liberates 100 Kev of power, so at 10^14 fusion reactions per second , you get !00,000 ev * 10^14 – 10^19 ev = 1.6 watt seconds. The fusion products are capable of climbing through a higher potential than that. The output is therefore limited only by our ability to insulate the cathode from ground.

I would go so far as to say that fusion is the only escape route for these trapped deuterons, and providing the temperature of the plasma can be maintained high enough, this escape route is the only way out, and nature will find it.

Steven

[Steven Sesselmann]

jamesr wrote: While I applaud your enthusiasm, I’m not sure you really get how fusion works. You mention half way through (08:00) that you don’t consider the ion collisions as the cause of fusion! It is all about this!

James, thanks for your reply and valuable comments. I probably did not make myself clear in the video (it was a single take, one shot unedited recording), I do not disagree that particle collision is necessary to initiate fusion, what I do believe is that ionizing particles below ground potential, lowers the relative Coulomb barrier.

You point to a weakness in my design that I am also aware of, the issue of keeping the plasma lit, and hot enough to fuse. The Lawson criterion still applies, and enough particles need to be confined for long enough at a high enough temperatures to fuse, with the only difference in my thinking being that the numerical value of the criterion is lower when the plasma has been ionized at a negative potential.

If my first experiment fails, it will be because the plasma is too cold, and I will need to look at other ways to keep the plasma lit. Clearly if enough fusion reactions can be initiated, the plasma will be self heating, the question is at what point.

If anyone wants to help me with the numerical modelling I would be more than appreciative.

Remember water runs down hill, all we can do is tilt the hill a bit …

Steven Sesselmann

[Steven Sesselmann]

ikanreed wrote:

…snip…

Up = Fdr = mc^2

…snip…

This is not how relativistic force works at all. Equation for total kintetic energy from wikipedia, which at low velocities approximates F*dr. Now, I don’t know if this is “backyard lab” verifiable, but it’s been experimentally tested since 1908. E = MC^2 was Einstein’s rest mass formula, not total energy formula. This is Physics 101 stuff.

You have not grasped my hypothesis, read it one more time. It’s all potential energy, you me and the rest of it.

When I talk about Fdr, I mean the creation of matter from absolutely nothing to matter at ground potential.

Please read it again…

https://focusfusion.org/index.php/forums/viewthread/1010/
http://www.fusor.net/board/view.php?bn=fusor_other&key=1307190918

Steven

[Steven Sesselmann]

Theoretically all fusion reactors produce net energy…

Since most of the reactor designs are intended for converting fusion energy into heat, one could theoretically say that all the input energy is also converted to heat, so if you input one 1MW to get one watt out, you could theoretically recover 1,000,001 watt, ergo a Q of 1.000001, however the conversion of thermal energy to useful electric power is at best 60%, so we go back to the drawing board.

Clearly a fusion scheme that takes advantage of direct charge to current conversion will have a huge advantage. Focus fusion promises to do this, as well as my soon to come FICS fusion.

Steven

[Steven Sesselmann]

James, I love it…, Feynman hits the nail on the head.

One good thing is that the more you know about the world and about physics, the better you get at guessing.

I am still working on my guess, and it’s slowly coming together im my head, in a nutshell, I am hoping to redefine F, where;

Up = Fdr = mc^2

Apart from being good at guessing, I have a GUT feeling about this one…

Steven 🙂

[Steven Sesselmann]

The list is not comprehensive, it just lists IEC reactors, but yes, for IEC that’s about the range.

Tokamak reactors such as JET in the UK, have demonstrated a Q of around 0.75, and ITER is expected to show a Q more than 1.

Steven

[Steven Sesselmann]

ikanreed wrote:
1. Give an example where the measurements are more accurate to your theory versus the standard model(Null hypothesis validation)
or
2. Give where the calculations get the same results for existing measurements, but simplify the formulas(Occam’s razor validation)

That’s it. Those are the only two ways you can have a decent improvement on existing science.

You are quite right, and what I have proposed so far is nothing more than a hypothesis, and now it is a matter as quoted above, to show mathematically that it is either more accurate or at least simpler than existing theory,

…and if it turns out to be rubbish, then it’s just another piece of paper for the basket, I have had plenty of those.

I believe it is better to try many ideas than to sit around waiting for a perfect one 🙂

Steven

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