IEEE Spectrum: Fusion on a Budget

This article originally appeared in the IEEE Spectrum back in March, 2009, but is only now available online:

.... it’s pretty simple, according to Paul Schatzkin, who runs Fusor.net, a Web site where amateur ”fusioneers” congregate to swap equipment and advice: ”Find two stainless steel half-spheres, seal them together around a wire grid, suck the air out of it, apply some high voltage to the grid, inject a bit of deuterium into the chamber, and sit back and count the neutrons.” Don’t expect to reach energy breakeven, Schatzkin says, but at least you’ll be failing to achieve practical fusion at only a millionth the cost of a tokamak....

Building a fusor is simple enough for amateurs to contemplate because of the enormous global inventory of used lab equipment, says veteran fusioneer Richard Hull. For example, fusion containers need only achieve pressures of about one millitorr; vacuum pumps with that capability make their way to eBay for US $10 to $100. High-voltage feedthroughs are in the same range—or free, if you build your own from microwave-oven salvage. The most expensive items, according to Schatzkin, are neutron detectors, which have to be purchased new. But a cheap workaround is available, Hull says: You can prove your fusor is working by irradiating a piece of silver and watching the decay of the Ag-108 and Ag-110 isotopes with a simple Geiger counter.

Yes, ITER and NIF may be getting all the press (I still haven't heard from NYT's Friedman...), but the real action is taking place in basements and garages.  Follow those two links and look at those two facilities.  They can't be serious. 

Comments

[john kirk gray]

june 17,2009 oling green kENTUCKY usa
jnokgray@yahoo.com = john kirk gray
310 j. bragg rd., edmonton,ky 42129 USA.
For mathemastical expression of three variables to describe any natural isotope and many lab- made nuclear FUSION products
!. let A1 +B1 + C1 = at no. "A"
for all chemical species 1 to 118,this year,sofar...
2/ Let 3A + 2B + 1C = Atomic mass integer
"Z" within limit: 1 At no "A" is lees than or equal to Z which is lees thanor equal to 3 atomic numbers" 3A" (most nuclides and isotopes lie between two and three Atomic numbers in the Atomic mass integer
e.g. For 79 Au 197, 197 lies between 2 x 79(158) and 3 x 79 (237)
in association with the two equations bove
combined: A (1,3) + B(1,2 ) + C(1,1) = (79,197) = {A,Z)

To solve for AB And C:
Set C = 0
from Z at 197 subtract 2 A or 158
197 - 158 = 39 This is A of (tritium(1H3))
For B of Deuterium
From At no 79 subtrat A (t) just obtained
79-39 = 40 -= B(deuterium)
Plug into formula
239 Tritium + 40 deuterium + 0 Protium = 79,197 = energy k

A confirmation of the validity of the expression may be easier with Mercury...
40 tritium = 40 deuterium + 0 protium = 80(Mercury: Hg) 200 Inthat it is liquid at room temperature,shiny,silvery color and subject to gravity pull...If not lost somewhere inthe home built reactor chamber,shoul be very easy to spot visually...confirming the efficacious thesis,
AND with one less deuterium
39 tritium + 39 deuterium + 0 protium = 78Platinum 195, transmutation of hydrogen nuclides may prove less es xpensive if there is some Platinum to SELL on EBAy or at the money getting places...
for silver 47 108
set c= 0,find a then find B and plug in formula..

108 - (2x47=94) =14 Tritium
from 47 subtract 14 for B deutreium
47 -14 = 33 = B deuterium?
14 Tritium + 33 Deuterium + 0 protium = 47,Silver 108 (orBUST!!) = energy constant K.

I got NO Kilowatthours saved up at my bank or in the checking account there. no elwctron volts in the SAVINGS AND LOAN either... My credit sucks...

from 108 subtract 2 x 47 = 94