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 »

Hi Chris,

Various folks (Jon Rosenstiel, Wilfried Heil, me) do have voltage vs. counts data posted here. In the professional world, there are papers with this data. It's not too hard to find. I don't know if any of it covers the 10-20 kV range specifically. You have to look at the experiments and see what variables are controlled or not controlled, because it can be hard (if not impossible) to isolate the effect of potential from an experiment where current or pressure changes.

Whether or not most fusion comes from beam-on-beam or beam-on-neutral collisions depends on pressure and beam alignment. University of Wisconsin research has addressed this issue, and you're right, much fusion comes from beam-on-neutral collisions. The range of energies available in beam-on-beam collisions depends on whether the fusor is pulsed, continuous, or bunched as in the "oscillating plasma sphere" concept proposed by Park and Nebel at LANL. Bottom line is that to realistically estimate the fusion rate in a fusor, you can't just use the cross section but you have to compute the thick-target yield particular to your circumstances: the integral of energy-dependent cross-section with position- and energy-dependent target density, electronic stopping power, and beam current factored in, over the entire path of a particle. Furthermore, there are not only deuterons reacting, but also molecular ions like D2+ with their own cross-sections and stopping powers. Nobody that I know of has a reliable model of the discharge process in a Hirsch-type fusor to support such a computation of yield at different voltages.

-Carl
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Re: Converting drive voltage into collision energy.

Post by Chris Bradley »

Thanks for bearing with my line, Carl, I realise that picking over basic fundamentals isn't everyone's interest. I also recognise that there have been many previous threads that capture many of these details. I do do the searches before posting and I'm actually writing on the basis of the things I find there.

But to my mind there are few good practical conclusions to relate practice to theory, and I'm having a go at that. I recognise all that you are saying about the grand expanse of other mechanisms going on at the same time. Of course this is so. But surely accepting it all as impossible to rationalise will not advance fusor [or any other] technology as a science.

Sometimes modern science has taught us so much that we fear to simplify anything anymore because we know there will be all these other detailed complexities. But could we at least try to make some simplifications and just see which one gets us closer to a practical understanding? The one that gets us closest is then the starting point, then we work on the next one that gets us a bit closer. We may then begin to see the exceptions and the variations to these simplifications, but that's fine - it's just a 'straw man' to knock down and then make better. This is, after all, how modern science has got to where it's got to.

There is now so much material in so many disciplines that if one really were to throughly look through everyone else's research material that had any bearing on something new, I reckon in many cases they'd be dead before they read it all! Better to have a decent look proportionate to the amount of time you've got, then have a go and expect to get picked up by people directing you to the prior work.

best regards,

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

Post by MSimon »

No.

If the particles hit head on you require 1/4 the drive voltage (assuming a charge of +1) vs the final KeV of collision. Assuming of course that you are at .1 C or less. (that gives about a 10% error or less).

Let me add that I have checked my numbers vs Dr. Bussard's for p-B11 and had the equations worked over by a physics PhD.

All that matters is the relative velocities of the particles.
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Re: Converting drive voltage into collision energy.

Post by MSimon »

Actually if the B11 and the proton are accelerated from opposite directions by the same field 50 KV more or less will get you fusion at the resonance peak. If you have a well type device (virtual center electrode) about 65KV will be required to make up for losses in producing the well.
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Re: Converting drive voltage into collision energy.

Post by MSimon »

The calculations for any particle mass and charge operating with fixed drive voltage (no well)

is here:

http://iecfusiontech.blogspot.com/2007/ ... r-b11.html

with the calculations done more rigorously (same result) here:

http://iecfusiontech.blogspot.com/2007/ ... r-iec.html

They show a drive voltage of about 200 KV for p-B11 to get a 1.2 MeV collision energy. Exactly what Dr. Bussard says.
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Re: Converting drive voltage into collision energy.

Post by Carl Willis »

M.,

>To accelerate a proton into a lab-stationary, fixed, 11B target

That's the system Chris asked about, but not the one you're working in.

You're describing a CM system (possibly not, though....read the next bit).

>If the particles hit head on you require 1/4 the drive voltage (assuming a charge of +1) vs the final KeV of collision. Assuming of course that you are at .1 C or less. (that gives about a 10% error or less).

If the particles hit "head on?" This could be in the CM frame, the target frame, or some other frame, to which the phrase "head on" is completely indifferent. All "head on" means to me is that the momenta are oppositely directed. This is what I mean about confusion on this issue. To denote the center of momentum frame, you have to say "center of momentum frame" or CM for short. The 1/4 drive voltage thing is applicable in my example ONLY to DD collisions. If you are doing p-B fusion in a fusor, the lab frame is not the CM frame like it is in DD fusion. Each reactant particle has the same ENERGY, but NOT THE SAME MOMENTUM in the lab frame. So the CM frame is moving in a p-B fusor.

>Let me add that I have checked my numbers vs Dr. Bussard's for p-B11 and had the equations worked over by a physics PhD.

This is not Ph. D. material. As you said, all that matters for the energetics is the relative velocities. At least that's a more intuitive way to approach these problems. I'm willing to bet your Ph. D. is right, but you are mis-applying what s/he said.

-Carl
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Re: Converting drive voltage into collision energy.

Post by Carl Willis »

Hi Chris,

I didn't bring up the complexity of the fusor in order to say that it defies all analysis. I just wanted to point out that a rigorous analytical approach to the fusion yield using the tools of cross-sections, etc. doesn't make much sense right now.

If you have a pure D+ ion gun looking at a D or T target at some potential relative to the gun, the analytical approach to predicting thick-target yields is easy, even formulaic, and comports well with experiment. There's a nice paper by Jasmina Vujic and some others associated with UC Berkeley that lays this all out in formulas.

But the fusor as most of us know it is just too complicated for this approach, because it involves a self-sustaining hollow-cathode discharge and thus all kinds of complexities not present with high-vacuum ion beams. Until the discharge can be modeled so we can determine ion charge states, velocities, and densities, we can't meaningfully apply this kind of thick-target calculation to the fusor.

That's all I was saying.

-Carl
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Re: Converting drive voltage into collision energy.

Post by MSimon »

Nothing like misreading the problem.

In that case the energy (in KeV) is delivered with a drive voltage equal to the energy (assuming a charge of +1).
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Re: Converting drive voltage into collision energy.

Post by Carl Willis »

>Actually if the B11 and the proton are accelerated from opposite directions by the same field 50 KV more or less will get you fusion at the resonance peak.

Homework problem: is the above statement accurate (or even close)?

To restate the problem, we have a proton (mass Mp) colliding head-on with a singly-charged B-11 ion (mass Mb) in the center of a U = 50 kV Farnsworth fusor. An evaluated cross-section data set (Hale, 1979) retrieved from CSISRS is shown below. The peak in this cross-section, 0.8 b, occurs at 625 keV. The frame-of-reference for the kinetic energy in this cross-section data is the boron atom.

>Step 1. Compute the relative velocity between the proton and the boron in the fusor.

In the center of the fusor, the proton has kinetic energy E = U*q = 50 keV. The boron also has energy E = U*q = 50 keV. The speed of the proton is | vp | = SQRT(2*E / Mp), while the speed of the boron is | vb | = SQRT(2*E / Mb). Since this is a head-on collision, the relative velocity is just the sum vr = vb + vp. That is, from the point-of-view of the boron, the proton moves at vr = SQRT(2*E / Mb) + SQRT(2*E / Mp).

>Step 2. Compute the kinetic energy of the proton in the boron's frame of reference (==Eb).

Eb = Mp * vr^2 / 2. Algebra shows that Eb = U*q*Mp*[1/Mb + 1/Mp + 2/SQRT(Mp*Mb)]. You can see that if Mb = Mp, Eb = 4*U*q as I stated earlier for the DD case. In this case though, Mb is about 11*Mp, so the math reduces to Eb = 1.694*U*q = 84.7 keV. In simplest language, this 50 kV fusor creates central collisions at a maximum energy, in the boron frame, of 84.7 keV. Where is that big resonance peak in the cross section? Daggone it, why that's out at Eb = 625 keV! Nowhere close to 84.7 keV. B-11(p,a) is nothing short of a pipe dream for those of us without a monster feedthrough and a few drums of Shell Diala oil to our name.

This is not hard math, it's not a hard concept (but it can be hard to communicate clearly verbally). If you have another source of cross-sections that you trust more than my choice tonight, if you want to see the problem worked in the center of mass, if you want to see it with a fully-stripped B-11 nucleus rather than the +1 ion, if you want to look at a different reaction...we can do that. If you think I f*ed up, tell me where (it's possible, but exceedingly unlikely since I checked through what I wrote). It would be nice if we could all get this much right.

-Carl
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Re: Converting drive voltage into collision energy.

Post by MSimon »

I was thinking of the resonance peak of .1 barn that occurs at about 150KeV collision energy.

It looks twinky on a linear scale. It should work fine in a test reactor if you can hold the voltages close enough.

The energy gain is about 20X if you can use that peak vs a gain of about 8 at the higher voltage peak. It does mean a bigger reactor for a given output unless you can do some enhancing (like POPS). The higher gain may be worth it. It certainly makes the experimental supplies cheaper. Not to mention feed throughs etc.

An amateur fusor might accomplish this by coating the electrodes with elemental Boron 11 or if that is not available ordinary boron would do the job with some loss of efficiency. Feed in hydrogen and away you go. Of course detecting high energy alphas would be a trick.

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

Really. There is so much that hobbiests could contribute to understanding fusion and its possibilities.
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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.

-Carl
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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,

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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.
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