This article originally appeared in the IEEE Spectrum back in March, 2009, but is only now available online:
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.
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 – (2×47=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
MAY 27,2010 BOWLING GREEN KENTUCKY
For a nuclesr fusion FOrmuls:
A(tritium) * B(Deuterium) * C(protium) =
(at,no,At mass integer) + constant of equality of energy.
For one gram moleculasr weight of 78 Platinum 195
A B C
Trit. deut Prot
39 39 0
40 37 1
41 35 2
42 33 3
43 31 4
44 29 5
45 27 6
46 25 7
47 23 8
48 21 9
49 19 10
50 17 11
51 15 12
52 13 13
53 11 14
54 9 15
55 7 16
56 5 17
57 3 18
58 1 19
Spot check arithmetic
A B C = 52,13, 13
at no check at mass integer check
52 x 1 = 52 52 x 3 = 156
13 x 1 = 13 13 x 2= 26
13 x 1 = 13 13 x 1 13
————– —————————-
at no 78 at mass integer 195
QUOD ERAT DEMONSTRANDUM
NOw for the Stoichiometry, the operastor hass less than 1 nanosecond to weight 52 tritium and get them fused nuclearly with nhe two other ingredients..suns temperature = very high pressure
likely required.
if youcant standtheheat,stay out of the kitchen.
So whats missing? I am looking for the coolest way to spend some bux that i inherited lately. Tempted on travelling to europe and try together ?