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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Chris Bradley
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Converting drive voltage into collision energy.

Post by Chris Bradley »

Hi.

I've been picked up twice now in other threads for my conversion of drive voltage into collision energy (I say the latter is a half of the former, for unit charges like deuteron ions). If I understand them both, one says I am over by a factor of two, the other says I am under by a factor of 4.

Room for a specific discussion on this topic then!?

The actual collision energy experienced by particles driven through given voltages seems, to me, to be the most basic and important issue in trying to quantitatively enumerate the performance of IEC devices. So I'd like to nail this down well. At the moment, if I'm wrong I'm going to have to correct an awful lot of calculations.

Some of this may be actual scientific understanding, some may be 'just' nomenclatures. So I've broken it down into statements and you, dear reader, can let me know which you think are correct and which are wrong:

1. Let us say we have accelerated a deuteron ion through 100kV in the lab-frame and it approaches along a trajectory that brings it into a sticky collision (viz, fusion) with another, but lab-stationary, deuteron.

2. In actual numbers, this means that the relative velocity between them (as seen in any non-relativistic frame) is SQRT(2x100kVx1.6E19Jx1.6E-27kg) =~ 4,472,000m/s

3. The question is, what is the collision energy between these deuterons. I understand this term to mean the amount of kinetic energy converted into other [excitation] energy in that impact, i.e. it is the resultant kinetic energy loss.

4. If I am an observer at the centre of mass of these two deuterons, I will see them approaching at 2,236,000m/s, that is, each with a KE of 25keV.

5. They come into fused contact at their CofM (viz, where I am sitting) and thereafter neither have KE in my frame.

6. A total of 2 x 25keV has therefore been lost as kinetic energy.

7. The amount of energy released in the impact is 50keV to me as an observer at the CofM.

8. Try another frame; I am now an observer 'along for the ride' with one of the particles, so I see one particle as stationary. The other approaches at 4,472,000m/s with a KE of 100keV in my frame.

9. As they collide in front of me, they career off with a new velocity of 2,236,000m/s away from me (conservation of momentum). The particle that had been coming towards me has now lost 75keV, but the one I was with has just gained 25keV. Total loss = 50keV.

10. or another way, the resultant fused particle now has a mass twice that of a deuteron but is moving at 2,236,000, which again equals 50keV and therefore represents a loss of 100keV-50keV=50keV.

11. I could do this for any frame of reference, and still end up with a total KE loss of 50keV when a deuteron accelerated through 100kV comes into sticky collision with another deuteron not so accelerated.

12. You can repeat this for any drive voltage (V) with like particles and the KE loss is always ½ x V.e

If anyone doesn't agree, please identify which statement is in error. Statements of agreement also welcome!

Best regards,

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

Post by Chris Bradley »

Sorry, a quick self-correction: The absolute velocities were for protons, not deuterons. (Doesn't affect the final energy figures, though.)
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Re: Converting drive voltage into collision energy.

Post by Carl Willis »

Chris,

This looks like a correct treatment.

The trick with this kind of discussion is to be very clear about what quantities are being considered. It's easy to lose track and calculate a wrong number or inaccurately read someone else's numbers. If one is talking about the kinetic energy of individual particles, the oncoming particle in the lab frame has 4 times the energy of a particle in the CM frame. If one is talking about system kinetic energy, then the system in the target frame has 2 times the energy of the system in the CM frame. When one comes to the crucial consideration of cross-sections and yields, it's important to understand that by default they are reported in the target frame. The energy of relevance to a cross-section is the "collision energy" as you put it, a function of the relative velocities between the particles. In fact, as you noticed, calculations are much more intuitive if you use relative velocity as the variable: find relative velocity from the acceleration voltage, then determine the target-frame energy corresponding to this relative velocity, and look up CS or TTY at that energy. On the other hand, the convention is to report differential angular cross sections with angle in the CM frame and energy in the lab frame.

Confusion abounds!

-Carl
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Re: Converting drive voltage into collision energy.

Post by Chris Bradley »

So...if that's right so far, I understood reported fusion cross-section coefficients to be derived equivalent to CofM energy - having been converted from the target energy.

So in the case of DD's peak fusion cross section, at 1250keV and cross-section = 96millibarn, this is the collision/cross-section you'd get if you accelerate an initially lab-stationary deuteron through 2500kV of potential into another lab-stationary deuteron.

Y/N??
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Re: Converting drive voltage into collision energy.

Post by Carl Willis »

>reported fusion cross-section coefficients to be derived equivalent to CofM energy - having been converted from the target energy.

It depends whose work you read. It may be that in astrophysics or fusion energy fields (not my fields, by the way), data gets reported in CM by default. I do know that if you look up cross-sections in CSISRS or another nuclear data service, and see energy "EN" without "CM" next to it, you're dealing with a lab-frame energy. If you see "EN-CM," then you have the center-of-momentum collisional energy being used. It's too bad researchers are sometimes not explicitly clear on this point and rely on what may be dubious convention.

>So in the case of DD's peak fusion cross section, at 1250keV and cross-section = 96millibarn, this is the collision/cross-section you'd get if you accelerate an initially lab-stationary deuteron through 2500kV of potential into another lab-stationary deuteron.

>Y/N??

N.

This cross-section is obtained from a nuclear data service like CSISRS, right? There's a nice evaluated cross-section courtesy of Liskien available there which shows the peak in the vicinity of 1.6 MeV(that would be in the target frame) at 106 mb. This 1.6 MeV implies that the deuteron beam is accelerated through the equivalent of a 1.6 MV potential difference.

The deuteron is moving relative to the other one at Vrelative = SQRT(2*1.6 MeV / Mdeuteron). In the CM frame, this deuteron has a velocity relative to the observer of Vrelative / 2. Its kinetic energy is 1/4 that of 1.6 MeV, but the system ("collision") kinetic energy is 1/2 of 1.6 MeV because there are two deuterons with equal energy in this system.

With a fusor, the operator is in the CM frame, a departure from the usual conditions for reported cross-sections. The particles of charge q are accelerated by a cathode voltage of U volts and the system energy for a head-on collision is 2*U*q. But this is not where you look up the cross-section. You'd look that up at 4*U*q. Example would be my fusor "Carl's Jr." running at U = 70 kV. Each deuteron ideally ends up in the center at 70 keV. The system energy in a head-on collision is 2*70 keV = 140 keV, but I look up the cross-section for this in CSISRS or similar at 4*70 keV = 280 keV. This is one of the advantages of arranging head-on, center-of-momentum collisions in the lab--you can get by with lower voltages (and smaller feedthroughs, transformers, etc). As I think you pointed out in another post though, energy is conserved regardless of what frame you do the calculation in.

This horse has been near beat to death in previous discussions on this board, dating way back. I suspect if you search "center of momentum" or similar, you'll dredge up the goods.

-Carl
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Re: Converting drive voltage into collision energy.

Post by Chris Bradley »

It is relevant of you to draw out that astrophysical texts differ - the sources I've read are, indeed, generally such texts.

I have also paid particular regard to a freely down-loadable chapter from a book on nuclear fuel reactions, which is also explicitly CM oriented; http://fds.oup.com/www.oup.co.uk/pdf/0-19-856264-0.pdf

Further, you have been talking about head-to-head collisions, both particles moving, whereas I have been talking about a beam into a stationary particle. I suspect we've been discussing convention of notations, then, and nothing more.

May I please test this, then, by setting one final agreed 'datum' of data before closing the thread. By way of example (using differential masses and a specific energy) there is a peak fusion resonance for p-11B at a lower energy than its main peak, at 148keV (CM). To accelerate a proton into a lab-stationary, fixed, 11B target and to get to this resonant cross-section, the proton needs to be moving with a velocity of 5,682,500m/s towards a target 11B nucleus and thus the proton needs to be accelerated though a lab potential of 161.5kV to get it to that velocity in the boron's stationary lab-frame.

Is this right?

best regards,

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Re: Converting drive voltage into collision energy.

Post by Carl Willis »

Chris,

Without going through the paces of doing the whole problem myself, I'm inclined to agree that this sounds right. The boron is a heavy nucleus compared to the proton, and of course in the limit that the total system mass is no different than the target mass, the CM and lab frames are the same. The lab energy you found for the proton in the lab frame is only slightly higher than the CM frame, so it looks right.

-Carl
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Re: Converting drive voltage into collision energy.

Post by Chris Bradley »

Thanks, Carl. That one's done now. I think it was just the conventions of notation/terminology that were the issue, not the science.

Please keep a friendly look out for any similar mis-interpretations of mine in other threads - I am working from first principles here and it is quite probable that there are many other commonly held conventions of which I am not remotely aware and need guidance/correction on.

best regards,

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

Post by Richard Hull »

Thanks are due to Carl for belaboring this point to the bitter end. I was originally very confused about this myself back on songs and actually went to the old Ask a Scientist at the princeton fusion website. I was given careful tutelage on this very issue and reported it in Songs.

It comes back to what Carl noted. Who is doing the talking? Where you are picking up your cross sections from? What the particles under discussion are doing. Even after all this, many remain confused. (including some pro's trying to cross over to other disciplines.)

For us, it is simple. Using the fusion cross section charts mentioned by Carl that were derived for fusion research, the defaualt in our fusors demands for the most ideal and fortuitous collisonal scenario that the cross section be read at 4*V applied*q.

To amplify on Carl. This is the sole great glory of the fusor, as we use it. Most every other aspect of the fusor is a loser or rife with issues and problems due to its extreme simplicity. So, extreme simplicity and a super cross section achieved per unit voltage applied are the attractors for the amateur community.

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Re: Converting drive voltage into collision energy.

Post by Chris Bradley »

I am sad that you find this boring, but as you say it is about misunderstanding the context.

I was originally picked up for dividing the fusor drive voltage by 2 to get the collision energy. I was unequivocal with the scenario I was calculating for. It was a lab-fast particle into a lab-stationary particle. I was therefore right to divide the drive voltage by two.

The subsequent issue was then discussed because the presumption was that collisions in the fusor are lab-fast into lab-fast, with the lab at rest with the CofM. This would clearly be a <drive voltage x 4> factor, as was then raised as a correction to my submission.

In point of fact, I cannot see where we actually disagreed with the calculation after that, it was just this misunderstanding of setting, not of science.

A factor of x8 between two results may be here-nor-there with a fusor because it is a case of bunging in a good load of 10's of kV to get the best performance, and it always will be.

Though this may satisfy the practical, experimental interests of fusioneers, it does not satiate my desire to understand exactly what is happening.

I believe it is quite wrong to be so sure that fusions in the fusor are lab-fast into lab-fast (the x4 drive voltage factor). Some may happen this way, but I believe that the majority are lab-fast into lab-stationary (which is therefore the x0.5 factor).

I can provide the basis of a proof either way by means of the very calculation which has been boringly discussed here:

Let us compare the neutron outputs of two otherwise identically operated fusors, one operating at 20kV, the other at 10kV ('10kV!!' you may say... exactly my point, please read on..):

A) If the fusor operates lab-fast into lab-fast particles, the collision energy is 80keV for the higher drive voltage and 40keV for the lower voltage. Using an astrophysical factor of S(0)=56 and Gamow energy of 986kV to calculate cross sections for DD (as the experimental data is a bit sparse in the lower energy ranges), we end up with cross-sections of 10millibarn for the fusor collisions operating at 10kV and 20millibarn for that operating at 20kV. Recall that reaction rate is also related to velocity (which therefore differs by SQRT(2)). This means that whatever the neutron rate you get out of a fusor operating at 20kV, you should also be able to achieve 35% of that rate at 10kV. (I presume this means that 10kV rates should be within the possibility of being measured? - Just run it for (less than) 3 times as long.)

B) If the fusor operates lab-fast into lab-stationary (i.e. only collisions with background particles) then the collision energies are then 10keV and 5keV which gives cross sections of 0.27millibarn and 9microbarn respectively. This means that comparing neutron rates at 10kV drive voltage should be (including velocity factor) about 3% what you would get at 20kV drive voltage, which I suspect would be generally below measurement capability, or a very long run.

Summary: I do not have the practical data to show one way or the other, but if neutron rates at 20kV for a given time period are detectable but neutron rates at 10kV, by the same kit, are not measurable with a run of just 3 times the duration of the first, then this is at least prima facie evidence that the collisions are not lab-fast into lab-fast, and that the lab is not in the CofM frame for the fusion collisions.

I have been digging around to find any one's reported neutron output rates below 10kV but to no avail. I therefore have to suspect that the fusor does, indeed, function by lab-fast into lab-stationary particles as otherwise I'd expect measurable rates in the 5-10kV range because they would be fairly comparable and within the same order of magnitude as that with, say, a 20kV drive voltage. I will tend to think this is correct, unless there is experimental evidence to the contrary.

best regards,

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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,

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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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Frank Sanns
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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.

Frank S.
Achiever's madness; when enough is still not enough. ---FS
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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Chris Bradley
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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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Carl Willis
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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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