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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Carl Willis
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Re: Converting drive voltage into collision energy.

Post by Carl Willis » Fri May 09, 2008 9:20 pm

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 » Fri May 09, 2008 11:39 pm

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 » Sat May 10, 2008 12:34 am

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 » Sat May 10, 2008 12:42 am

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 » Sat May 10, 2008 12:46 am

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 » Sat May 10, 2008 4:39 am

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 » Sat May 10, 2008 4:55 am

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 » Sat May 10, 2008 6:37 am

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 » Sat May 10, 2008 7:09 am

>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 » Sat May 10, 2008 10:09 am

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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