[ExI] lockheed's fusion video

[spike]

From: extropy-chat [mailto:extropy-chat-bounces at lists.extropy.org] On Behalf Of Adrian Tymes



>…For instance, if volume X has surface area Z, then 8*X will have 4*Z surface area...but still 8*Y neutron production, so the rate of neutrons crossing the surface went from Y/Z to 8*Y/4*Z = 2*(Y/Z).

Does the rate of neutron production not scale that way?

 

Oh OK, the way I was looking at it is to assume some physics magic where they contain the fusion reaction by some means.  We treat the reaction site as a point for the purposes of this thought experiment.  All of the fusions happen right at that point, not at all like a chemical reaction where the reactions happen everywhere inside the vessel.  With fusion, they focus a bunch of lasers at a point, and all the action happens right there.  That’s the way they did it back in the old days anyway.  Then you have a thermal neutron created for each fusion of a 3H with a 2H.  I haven’t heard that they figured out any way to fuse any other two isotopes besides those two.  These thermal neutrons fly outward radially from that reaction point at the focus of the lasers.  Neutrons hit the walls of the chamber and cause it to erode.

 

Here’s a sample calculation that you don’t even need to waste an envelope or look up anything.  As I understand it, the radius of a nucleus is about 4 orders of magnitude below the radius of an atom for the heavier eliments, so its area ratio is about 8 orders of magnitude, ja?  So if a neutron is to be captured by a nucleus, it must physically contact that nucleus, since the strong force is very short range.  So a neutron passing thru an atom or ion has about 1e-8 probability of being absorbed.  So if you have a metal such as lead, you need a layer around the reaction that is 1e8 atoms to have about a 1-1/e probability or about 60% chance of absorbing that neutron, and the heavy metals have atomic radii of about 200 pm if I recall, so diameter maybe 4e-10 meters times 1e8 is .04 meters of lead required to have about a 60% chance of stopping a neutron.

 

4 cm of lead gives 60% chance of absorption, 8 cm takes us to about 86%, 12 cm to 95%, 16 cm gets us 98% chance of absorption, and that’s where I can’t do these without an envelope or a spreadsheet anymore.  The only reason I knew how to do those three is that I memorized a long time ago 1-1/e, 1-e^-2 and 1-e-3 and 1-e-4: 0.63, 0.86, 0.95 and 0.98, and everything past that is over 0.99.  Those numbers have a lot of real-world applications.  Knowing just those four numbers gives you lots of opportunities to impress your geek friends.

 

 

Adrian …  Where should we go to learn about this?

 

>…Quite true.  I thought you might have already found some resources on this.

 

No.  I am currently distracted by another project, which I want to ask you guys under a different subject line.

 

spike

 

 

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