Converting drive voltage into collision energy.

It may be difficult to separate "theory" from "application," but let''s see if this helps facilitate the discussion.
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Re: Converting drive voltage into collision energy.

Post by Carl Willis »

The little resonance peak is at 162 keV, meaning, by the math I just went through, that to hit this little bump (and it is tiny, peaking out at only ~60 millibarn) the fusor would need to operate at 96 kV.

>Think of it. You could be the first in the world to do continuous p-B11 fusion.

Well, it looks like this is a task only you might be sold on, so I'd say get to work and that settles it. Pragmatists around here have a good comprehension of the challenges with p-B and have neglected it in favor of other callings (personally, I'm into fusors for the neutrons). There are always people who come on the forum and tout the great virtues of aneutronic fusion, share their grandiose plans for doing it, talk smack about how nobody else is doing it, and basically just make noise. If someone does an experiment with p-B and THEN opens his mouth, he'll earn my respect and maybe my interest. But to date the p-B cheerleading here has consisted of much sound and fury signifying nothing: a poorly-informed racket of the quality one might expect to find among Trekkies on a Sci-Fi Channel discussion board at 3:00 AM.

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Re: Converting drive voltage into collision energy.

Post by MSimon »

This is the cross section graph I used.

By careful measurement I come up with about 140 KeV of collision energy.

However even assuming 170 KeV collisions that is still only 62KV of drive according to the included spread sheet. Well the spread sheet wouldn't load. I may park it some where and leave a link in a bit.

The spread sheet gives the correct number according to hand calculation at a couple of points. One of those points is Dr. Bussards 200 KV drive for the pB11 peak at 580 KeV collision energy.

In any case there is a link above to IEC Fusion Technology where I give the equations. You can make your own spread sheet and check my work.

Here is a link to the spreadsheet:

http://www.mediafire.com/?0xvmsrumy8x
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Re: Converting drive voltage into collision energy.

Post by Frank Sanns »

It matters not if it is 4x or 2x or 1x or 0.5x, the efficiency is still 1.00000_1 for what we do. It is about like taking a deep breath and blowing out the back window of an SUV and trying to use it a efficient propulsion. Sure, there is thrust just like there is fusion but the net outcome is not very practical.

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We have to stop looking at the world through our physical eyes. The universe is NOT what we see. It is the quantum world that is real. The rest is just an electron illusion. ---FS
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Re: Converting drive voltage into collision energy.

Post by Chris Bradley »

Much as I tend to pick points of a graph when I'm being lazy, this isn't a good approach!!

The actual experimental data is a bit patchy and your graph is just a smoothed out representation of the data.

I don't think it's very important - I can't see anyone fine-tuning their p-11B reactor anytime soon, but for the record my texts say that the resonance is a peak in the astrophysical factor of 3500MeV.barn, which occurs at 148keV. That would be a proton through 162kV into a 11B target. So I guess that's Carl's figure, that's the 'EN' figure (right notation now, Carl??)

However, after a quick calc. I can also agree with M Simon's figure of about 50keV drive to both p and 11B to get to the 148keV resonance. To be more accurate with the arithmetic, 57.6kV, in fact.

It's not just about the lab frame being the CofM frame, it can be a bit more efficient that that, as follows:

Accelerate a p through 57.6kV and you get it up to 3,394,100m/s in the lab frame. Accelerate a (5+)11B through 57.6kV and you get it up to 2,288,300m/s. Total relative velocity = 5,682,400m/s.

Now, if a p and an 11B come up towards each other at this 5,682,400m/s rate, then the p approaches their CofM at 5,208,900m/s (135,662keV worth) and the 11B approaches it at 473,500m/s (12,333keV worth). Total CofM energy = 148keV.

The issue with respect to your concern about this drive voltage figure, Carl, is that after both p and 11B are accelerated through a lab voltage of 57.6keV, the lab is no longer in the CofM frame. It, the CofM, is now moving at 1,814,800m/s relative to the lab. This is because of the differential charge/mass ratios of the two particles accelerated by that lab voltage. The lab stays in the CofM frame only if the particles have the same charge/mass ratio.

(Not that I subscribe to the idea you can get these things doing that collision electrostatically, nor that the figures are exact, but that's the maths.)

best regards,

Chris MB.
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Re: Converting drive voltage into collision energy.

Post by Carl Willis »

M.,

I haven't read through all the spreadsheet because it contains this problem buried amid a lot of other undocumented complexity (what the heck is DV?). However, the most obvious issue with the spreadsheet as it pertains to the p-B example I wrote up is that it treats yet a different problem...a fully stripped B-11 nucleus (+5 charge). A spreadsheet can only do so much for you if you don't give it the right input.

Chris makes a good point about your data. Where is it from? Can you get a table? What is the reference frame? With my data from CSISRS, you get the graph and a table of data, and you know that the reference frame is that of the boron atom.

Chris, your astrophysical factor may be reported in the CM frame. You'll have to check carefully on that. The CM frame always has the lowest system energy for given relative velocity.

-Carl
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Re: Converting drive voltage into collision energy.

Post by MSimon »

*

http://fds.oup.com/www.oup.co.uk/pdf/0-19-856264-0.pdf

*

The proton–boron reaction
p + 11B → 3α + 8.6 MeV, 1.43
is particularly interesting, because it does not involve any radioactive
fuel, and only releases charged particles. Its cross section exhibits
a very narrow resonance at = 148 keV, where the S factor peak
at 3500 MeV·barn and a broader resonance at = 580 keV, where
S ≈ 380 MeV·barn.

From page 13 of the pdf for all the Greek letters etc.
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Re: Converting drive voltage into collision energy.

Post by Carl Willis »



http://fds.oup.com/www.oup.co.uk/pdf/0-19-856264-0.pdf


This document handles the theory here clearly and concisely. The relevant section is accurate. It doesn't say anything that hasn't been said already, though. It does give preference to the CM reference frame.

-Carl
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Re: Converting drive voltage into collision energy.

Post by Chris Bradley »

I'm happy with the figures. If you check the lab-framed energies, they would otherwise be (1.e*57.6kV)+(5.e*57.6kV)=345.6keV for the two particles, as those are the respective e-field accelerations. If you follow my calc. above you get 148keV for the CofM frame that I reckon is moving relative to the lab. So, yes, this CofM which I describe as being in motion relative to the lab frame does look like it's got the lowest system energy.

best regards,

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Re: Converting drive voltage into collision energy.

Post by Carl Willis »

Good deal.

-Carl
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Re: Converting drive voltage into collision energy.

Post by Richard Hull »

I found it amusing regarding the glib nature of the B at +5 comment. I also found it wise of Carl to note what it takes to get a B (+5). It is quite a special thing to strip out a B atom to +5 and hold a wad of them in that state to warrant a 5+ ionization at collision.

Like Carl, I await the first fusioneer to actually succeed at, and report on, P-B11 fusion. Been waitin' a long time for that report, too.

To listen to the pundits and boosters, it shouldn't be all that tough. Odd how the real-life millions are being spent on the easiest fusion and not the "best" or eco-friendly fusion.

If you can't seem to get it to click doing it the easiest possible way, why struggle at a more difficult or invovled process.

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Re: Converting drive voltage into collision energy.

Post by MSimon »

I remember the early days of amateur computing. A lot of things were going on in parallel.

I was still struggling with I/O and memory while others were adding disk drives to their systems. Later folks just bought built systems and started doing software experiments. I do have the honor of designing and manufacturing the I/O board that went into the world's first BBS. I didn't even have a modem at the time.

Other folks (Ted Nelson for example) were on the side lines dreaming of what could be and encouraging the more technically minded to have at it.

Of course you have been waiting a long time to see p-B11 fusion. There are not enough people working on the problem. The early days of amateur computing were like that. With no standard systems and everything custom roll your own advances were slow. Things didn't start popping until Jan '75 when Popular Electronics came out with the Altair on their front cover. (That is when I joined up). Then we got inflation of the ranks. The second major jump came in the early 80s with the IBM PC.

Right now we are in the pre-'75 era. That will change when some one comes out with a $5,000 fusion kit. (BTW I gave away a $5,000 S-100 dual 8" disk drive computer to a friend when I moved on to IBM PCs so a move to commercial products will advance the abilities of even the most broke hobbiests - in time).
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Re: Converting drive voltage into collision energy.

Post by Chris Bradley »

Just to clarify - I'm happy to do the pure theoretical calculation, but that doesn't mean I'd suggest it is viable (I think I added that caveat!).

I presume that any real serious look at p11B should come to the conclusion that the effort of getting 11B into motion seems considerably outweighed when compared with the relatively simple task of accelerating a proton, notwithstanding the x 3 voltage required. It's the relative velocity, not lab energy, so no real benefits to accelerating the heavier particle. I never quite understood that part of Bussard's argument, that 11B would only need 1/5th of the voltage drive.

In fact, given the calcs above, if at all possible you would have to put in 345keV of lab energy to get a 'head-to-head' p11B/148keV collision going, whereas just accelerating a proton into a 11B lab-target takes just 162keV of energy. You'd instantly double your efficiency by a beam-target approach!

best regards,

Chris MB.
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Re: Converting drive voltage into collision energy.

Post by Dustinit »

I noticed that the velocities are exceeding that of 1% of the speed of light,
and was curious whether that reduces the velocities (ev) required as the apparent mass increases and cross sectional area would begin to increase due to relativistic effects.

mmm
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Re: Converting drive voltage into collision energy.

Post by Chris Bradley »

Numerically, the variation of mass is ~2 parts in 10,000. Experimentally, I would suggest that's unmeasurable at this velocity for this type of experiment.

I'm not sure the question makes sense with regards cross-section. Again, cs is an experimental measure that I understand to have much bigger uncertainties than this, but when measured it's referenced to the CofM collision energy anyway which, I presume, must work out to be frame invariate for any observer else you'd never know what reference frame to actually use for the calculation!
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Re: Converting drive voltage into collision energy.

Post by giorgos giorgakis »

Hi sorry for the necro post but what is the required energy for p-B11 fusion after all? I see that you use Boron 5+ for your calculations( which is hard to make ) and with what frame of reference? The papers you mentioned say 150keV plasma energy not ion energy which is different and you also have to take relativistic effects into account in such speeds. For 56.7 keV I get 3.32x10^6 m/s for the proton and 3.10x10^6m/s for the B5+ since you have to take the charge also into account, with the lab as a reference frame, so the relative speed for the two ions would be ~6.4x10^6 m/s given the reference frame of the lab, much higher than the combined 150keV you said in the first place. The speed for B+ is 1.07602*10^6 though which is more realistic. If 150keV absolute energy is needed then the speeds are different. When you take relative speeds with reference to the B5+ w/ lorentz transformations you get 4.31984*10^6 for the proton.
Last edited by giorgos giorgakis on Mon Nov 09, 2015 6:40 pm, edited 3 times in total.
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Re: Converting drive voltage into collision energy.

Post by John Futter »

giorgo
how about following the forum rules and introducing yourself first before posting elsewhere
and please change your logon to your full real name
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Re: Converting drive voltage into collision energy.

Post by giorgos giorgakis »

Ok then although there arent many things to say about me. The real question would be 150 keV on what reference frame?
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Fuel Cycles, Energy Levels

Post by Paul_Schatzkin »

what is the required energy for p-B11 fusion after all?
Giorgos,

Welcome and thanks for using your real, full name (and thanks John F for monitoring the case. Still amazes me how hard it is for people to follow simple instructions... <*sigh*>.

I have been reading up on fuel cycles a bit myself lately, maybe this has the answer to your question?

http://www.visionofearth.org/industry/f ... they-work/

This might be the money quote:
There is only one major disadvantage to this fuel cycle, but it is a big one. The reaction rate peaks at an energy of 123 keV! This is a very high energy, so high in fact that many people consider this fuel cycle to be a dead end in terms of research.
Does that answer your question?

--PS
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Re: Converting drive voltage into collision energy.

Post by Richard Hull »

Giorgos,

There is no "required energy" for p-B11 fusion. The fusion reaction picks up as the applied potential, (relative ion velocity increases). There is always a maximum or optimum production energy which if exceeded will actually result in less fusion due to an oppenhiemer-philips type reaction which is not fusion.

As most fusion attempts are thermal ones P-B11 is sort of a dead end as there are other reactions in the thermal range which would be far easier and yet, even they have never been successful in producing net energy.

In a collisional accelerator regime the P-B11 reaction is much easier to achieve, but the same ultimate limitation appears so far as net energy production is concerned. Far more input energy is required and wasted in the effort than fusion energy created by the effort.

In general, in a fusor type, collisional accelerator regime, you would need a collisional energy for the fuel element ions on the order of 125kev to be near an early optimum peak. Note: the real maximum peak is way out at 700kev

It is easy to talk of getting ions to that target energy, but another to actually herd them to that level to do advantageous fusion.

I am rather stunned you did not do a simple google search on the cross sectional data...... I did and all this has been covered in great detail here in prior postings

http://www.crossfirefusion.com/nuclear- ... actor.html



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Re: Converting drive voltage into collision energy.

Post by giorgos giorgakis »

Hi thanks for the replies and links , all sources I've found treat the ions as gases and they should for a large amount of ions but I want to calculate the energy needed for a head on collision of 2 ions p-B+ without taking into account the collision cross section and other gas theories.
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Re: Converting drive voltage into collision energy.

Post by Richard Hull »

Head on kinetic collision energy for P-B11 ions could easily be 5 eV or any actual kinetic energy imagined.. For fusion it is another matter and that is purely one of collision cross sectional energies only. A head on collision of any ion pair happens constantly in air! The result is some simply mechanical process of field interactions. Fully ionized B11 would demand an energy input to the molecule of perhaps a thousand eV or more just to strip all the electrons.

There is no hard surfaces for head on collisions of ions, only field interactions. To do something other than simply kinetic ricochett reactions in a "head on" to the ions, (a few eV to thosands of eV), you would have to do quantum tunneling, (fusion - collisonal cross section chart - hundreds of thousnads of eV), or stripping reactions, (millions of eV).

As you can see "head-on" ion collisions can do three totally different things based on kinetic energy be it lab frame, center of mass or whatever.

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Re: Converting drive voltage into collision energy.

Post by Dan Tibbets »

The KE needed for good P-B11 fusion cross section optimization depends on several assumptions.
400 KeV is often given as the target. This is not the cross section peak but it is where the curve is leveling out and pushing further becomes counter productive (mostly due to Bremsstyruhlung considerations). this assumes beam - target or beam - background collisions. One particle is at 400KeV and the other is at 0 KeV in the lab reference frame. If there is beam - beam fusion collisions, each participant is traveling with 200 KeV energies relative to the lab reference and they collide head on. This requires minimization of any low energy background particles like neutral particles for the beam - beam collisions to dominate. The advantages are obvious, Only one half of the accelerating voltage is needed. Bremsstruhlung radiation will also be ~ 4 times less. Some graphs also show a resonant peak in the P-B11 fusion cross section at ~ 123 KeV. This comes close to the cross section value at 400 KeV, but it is narrow and non thermalized plasma is required to take advantage of this. In my imagination this is the solution. Only 123 KeV is needed, or rather ~ 60 KeV with beam - beam fusion. This would produce nearly as much fusion as the 400 KeV goal, and the Bremsstruhlung would be about 40 times less, or about 10 times less if the beam- beam 200 KeV comparison was used.
Rider's conclusions, which arguably discounted the usefulness of P-B11 fusion become much less significant with these perhaps fanciful assumptions. Note that even without these assumptions there are theoretical methods to make the Bremsstruhlung losses tolerable. This includes diluting the boron relative to the protons, and in the Polywell's case, having an energy distribution of the electrons with low energy in the central core where interacting ion concentrations are greatest.

D Tibbets
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Re: Converting drive voltage into collision energy.

Post by giorgos giorgakis »

Thanks for the replies guys , in a perfect world you would be able to put 2 ions on rails and launch them so that they hit each other without taking the cross section into consideration or the wave nature of particles hence no statistics would be needed. I found this link here http://www.thepolywellblog.com/2010/01/ ... spark.html it's a good read and says that D-T fusion only takes 10 keV.
This CF fusion device page http://www.crossfirefusion.com/nuclear- ... actor.html says that this CF machine can accelerate neutral particles with it's moving standing waves so why hasn't it worked then?

If there were a vector field k*r^n similar to the coulomb force without any point charge needed wouldn't a spherical cloud of ions hit each other at the center and would these ions fuse if they had enough kinetic energy?
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Re: Converting drive voltage into collision energy.

Post by Richard Hull »

It must be always kept in mind that........

1. Doing nuclear fusion is abysmally easy and inexpensive
2. Losses in doing all known fusion processes always predominate over energy production.
3. Doing fusion by any method currently under study to produce significantly distributable electrical energy for consumption at a price slightly above, near or below the current rates of distrubuted power is absolutely impossible.

We can talk and theorize and calculate forever, but as of this moment we cannot do over unity, economically feasible nuclear fusion, just like we always have and will be into the distant future.

In short, nothing works in fusion to produce even one millionth of one watt of fusion energy that is over unity which can or has been converted to 24-7-365 reliable electrical energy. No one millionth of one watt!

Fusion in reactors, to-date, have had terawatts of impulse energy input and done fusion. Others have used hundreds of thousands of watts continuously when on and done fusion. Not one of these monsters has actually produced one millionth of one watt of demonstrable electricity or lit so much as a single flashlight bulb due solely to nuclear fusion.

Fission based electricity was demo'd in the thousand watt delivered class only 4 years after the first atom bomb was dropped. 61 years after first H bomb, not one millionth of one watt of fusion based electricity has ever flowed in a closed electrical circuit.

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Re: Converting drive voltage into collision energy.

Post by giorgos giorgakis »

Still I think that it's not impossible, it just hasn't happened yet. If we find the weaknesses of all fusion devices then we can try to eliminate them, let's say the fusor has a physical electrode at the center which many ions collide with and even if they pass they will be decelerated for as much as the electrode radius, the polywell and many IEC devices loose most of their energy as radiation because of their magnetic confinement configurations that use the gyrating movement of charges(there are ways to eliminate this). If you could make a fusor but without the physical electrode wouldn't this be better? What would be the minimum speed needed for fusion to happen then? A way to get the energy out would be the same way as in the polywell , with a spherical electrode charged at megavolts surrounding the hypothetical fusor that is the positive electrode of a capacitor. http://polywellnuclearfusion.com/Polywe ... lectr.html




Have you guys used any 3d simulator? I plan on using blender for a rough simulation of a fusion device but I haven't got the grips of it yet.
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