In this space, visitors are invited to post any comments, questions, or skeptical observations about Philo T. Farnsworth's contributions to the field of Nuclear Fusion research.
Subject: Re: Cross sectional collision energy
Date: Mar 07, 8:35 pm
Poster: Jim Lux
On Mar 07, 8:35 pm, Jim Lux wrote:
>The fusor collides deuterons, therefore, every deuteron in the fusor is in motion!
>Under ideal situations the deuteron is accelerated through the full fusor applied potential to collide head on with a like deuteron accelerated from the opposite side of the fusor.
>Thus, if we have a 30kv accelearation potential, we have two 30kev deuterons collide. This is NOT a total of 60kev to be looked up on the collisonal cross section graph!
Except that in fusors at the typical pressures, and so forth, there are a couple of other factors that bite you in the a**...
1) There are a significant number of essentially unionized deuterons sitting around, and you only get the 30 keV whack for those.
2) Sure, for head to head collisions, you get the 120 keV whack, but in a typical fusor, many of the beams are NOT at 180 degrees, but are at 60 or 90 or 45 or some other angle, so the relative velocities get scaled by cos(theta).
I have a spreadsheet at http://home.earthlink.net/~jimlux/nuc/beam2.xls
that breaks all this out for a hypothetical grid that is a icosahedron (20 triangles), you could integrate that sheet with the ones that have the cross section data at
to figure out the actual reaction cross section.
Interestingly, if you average all the beams together, the average eV energy (expressed as eV, not velocity) is about 1.14 times the accelerating voltage (assuming all the ions are created at the outer grid/shell)..
So.. as a practical matter, I'd say you should just go into the tables with your accelerating voltage.
Richard, how do your measured rates compare with theoretical expectations?
- Re: Cross sectional collision energy - Richare Hull Mar 08, 10:32 am