Single circuits generate nuclear reactions

[UncleFester]

Hi Uncle Fester,

Understand the numbers above but slightly confused as to some values.  If  2000VDC @ .68uf will excite a rod, (I have everything to duplicate this level right now except for carbon)  why have we been so hung up on the high value 108J per pulse except for high power output. 

This level of energy will not see a effect. It's only very small amounts of current and only half the input voltage. Output sine wave is still whatever the input is. In my case it was 35Khz from the neon driver HV circuit. The current was just enough to self run and go into a runaway mode, but not enough to light a bulb or do anything useful. Thus the required 108-110 joules is the requirement to see the full effect where you have 2/3 the input voltage and many times the current needed to self run. I believe if you talked to JLN and Valee they would say the 110 joules was found from experimentation from Valee and probably where Naudin got his info to run his experiments, hence the reason he used that amount of energy as input.

We may find later on as with many other schemes: I.E. Meyer, Bedini, Gray, Johnson, Newman that high voltage in very short pulses increases the effect many times over. Using a triggered spark gap or some other form running in the 2KV to 25KV range. Although if larger amounts of current are present at those voltages it will be much more difficult to make the power usable I.E. possibly a pole transformer or something similar would be required to drop the voltage levels down to 480-120VAC. Obviously a few amps at 2KV would be highly lethal, let alone 25KV @ 5 ampere! Windings on the toroid would of course also be special wound with HV magnet wire.

A physicist worked on the following equations. According to him there should also be Gamma rays through this process.......

The way I evaluated the dose rate of Boron beta decay is as follows:
First, 1 joule is equal to 6.24 * 10^12 MeV, and the decay energies of both Boron 12 & 13 round off to 13.4 MeV.

And since the natural aboundance of Carbon 12 is ~99% and Carbon 13 is ~1% I divided the 6.24 * 10^12 MeV of total decay energy for 1 joule into 6.18 * 10^12 MeV for Boron 12 decay and 6.24 * 10^10 MeV for Boron 13 decay.

As per the usual convention I omitted decay modes that comprise less than 1% of the total energy release.

Percent & amount ofenergy   Betas      Gammas         Count each Particle


[ Boron 12 -- Total energy 6.18 * 10^12 MeV ]


92.2%   6.01 * 10^12 MeV   13.4 MeV            4.48 * 10^11

1.50%   9.27 * 10^10 MeV   5.71 MeV   3.21 MeV & 4.43 MeV   6.92 * 10^9

1.23%   7.60 * 10^10 MeV   8.93 MeV   4.43 MeV      5.67 * 10^9


[ Boron 13 -- Total energy 6.24 * 10^10 MeV ]


91.2%   5.75 * 10^10 MeV   13.4 MeV            4.29 * 10^9

7.60%   4.74 * 10^9  MeV   9.75 MeV   3.68 MeV      3.54 * 10^8


Derived Gamma Averages:

Gamma      Total Count   Total Energy

4.43 MeV   1.26 * 10^10   5.58 * 10^10 MeV

3.21 MeV   6.92 * 10^9   2.22 * 10^10 MeV

3.68 MeV   3.54 * 10^8   1.30 * 10^9  MeV

--------   ------------   ----------------

3.99(Avg) MeV   1.99 * 10^10   7.93 * 10^10

From this I derived the dose rate using the unshielded dose rate equation number two in the "Shielding for Gamma Radation.pdf".

kSEUen/P

-------

4 Pi r^2

I used:

k = 1.60 * 10^(-10)   Value for grays per second

S = 1.99 * 10^10    Gammas per second

E = 3.99 MeV

Uen/P = 2.045 * 10^(-2) Soft tissue & 4 MeV

r = 100cm

1.60*10^(-10) * 1.99*10^10 * 3.99 * 2.045*10^(-2)
------------------------------------------------- = 2.07*10^(-6) grays per joule
      4*Pi*100^2

The result gives actually gives the dose rate per joule of raw emitted decay energy.  Multipy this figure by the watts output by the Boron to get grays per second.

[k4zep]

This level of energy will not see a effect. It's only very small amounts of current and only half the input voltage. Output sine wave is still whatever the input is. In my case it was 35Khz from the neon driver HV circuit. The current was just enough to self run and go into a runaway mode, but not enough to light a bulb or do anything useful. Thus the required 108-110 joules is the requirement to see the full effect where you have 2/3 the input voltage and many times the current needed to self run. I believe if you talked to JLN and Valee they would say the 110 joules was found from experimentation from Valee and probably where Naudin got his info to run his experiments, hence the reason he used that amount of energy as input.

We may find later on as with many other schemes: I.E. Meyer, Bedini, Gray, Johnson, Newman that high voltage in very short pulses increases the effect many times over. Using a triggered spark gap or some other form running in the 2KV to 25KV range. Although if larger amounts of current are present at those voltages it will be much more difficult to make the power usable I.E. possibly a pole transformer or something similar would be required to drop the voltage levels down to 480-120VAC. Obviously a few amps at 2KV would be highly lethal, let alone 25KV @ 5 ampere! Windings on the toroid would of course also be special wound with HV magnet wire.

A physicist worked on the following equations. According to him there should also be Gamma rays through this process.......

The way I evaluated the dose rate of Boron beta decay is as follows:
First, 1 joule is equal to 6.24 * 10^12 MeV, and the decay energies of both Boron 12 & 13 round off to 13.4 MeV.

And since the natural aboundance of Carbon 12 is ~99% and Carbon 13 is ~1% I divided the 6.24 * 10^12 MeV of total decay energy for 1 joule into 6.18 * 10^12 MeV for Boron 12 decay and 6.24 * 10^10 MeV for Boron 13 decay.

As per the usual convention I omitted decay modes that comprise less than 1% of the total energy release.

Percent & amount ofenergy   Betas      Gammas         Count each Particle


[ Boron 12 -- Total energy 6.18 * 10^12 MeV ]


92.2%   6.01 * 10^12 MeV   13.4 MeV            4.48 * 10^11

1.50%   9.27 * 10^10 MeV   5.71 MeV   3.21 MeV & 4.43 MeV   6.92 * 10^9

1.23%   7.60 * 10^10 MeV   8.93 MeV   4.43 MeV      5.67 * 10^9


[ Boron 13 -- Total energy 6.24 * 10^10 MeV ]


91.2%   5.75 * 10^10 MeV   13.4 MeV            4.29 * 10^9

7.60%   4.74 * 10^9  MeV   9.75 MeV   3.68 MeV      3.54 * 10^8


Derived Gamma Averages:

Gamma      Total Count   Total Energy

4.43 MeV   1.26 * 10^10   5.58 * 10^10 MeV

3.21 MeV   6.92 * 10^9   2.22 * 10^10 MeV

3.68 MeV   3.54 * 10^8   1.30 * 10^9  MeV

--------   ------------   ----------------

3.99(Avg) MeV   1.99 * 10^10   7.93 * 10^10

From this I derived the dose rate using the unshielded dose rate equation number two in the "Shielding for Gamma Radiation.pdf".

kSEUen/P

-------

4 Pi r^2

I used:

k = 1.60 * 10^(-10)   Value for grays per second

S = 1.99 * 10^10    Gammas per second

E = 3.99 MeV

Uen/P = 2.045 * 10^(-2) Soft tissue & 4 MeV

r = 100cm

1.60*10^(-10) * 1.99*10^10 * 3.99 * 2.045*10^(-2)
------------------------------------------------- = 2.07*10^(-6) grays per joule
      4*Pi*100^2

The result gives actually gives the dose rate per joule of raw emitted decay energy.  Multipy this figure by the watts output by the Boron to get grays per second.

Whoooooow Nellie,

Now you have went over my experimenters head...I never went beyond basic Calculus and it kicked my ass, I guess that's why I'm not a mathematician.  .....I'll have to take your word for it......Puttering around still, As I just don't understand the math involved, I can not visualize what is going on there.  I have seen that the output pulse is approximately the same as the input pulse so I guess the effect rides on that waveform if and when it can integrate itself on top of the induced pulse via the 1 turn to X number of turns in the Toroid.  With just a 50/1 turn ratio, I am seeing a very high pulse voltage and current out of a toroid with my simple circuit but the duty cycle is 1% or less so output is zilch really. It becomes obvious quickly that we have to have at least a 50% duty cycle and maybe more to make a sine wave output or we have to store the pulses via a bridge into a large cap and use downstream.  So much work to get viable information!!!!!
THIS IS NOT SIMPLE! 

Ben

Ben

[UncleFester]

Looks like you are not seeing much of an effect. Decay takes place over 20mS timespan and falls off sharply. Sharp rise and fall as the reaction takes place and sharp fall off as it ends. This forms a sine wave which should take place over about 12 to 15mS time period. Collapse of the field follows and the toroids field should collapse and form the negative side of the sine wave.

The previous post basically say's you should also be getting Gamma rays from the process, which to me was not good news since it would be much harder to deal with Gamma rather than alpha or beta rays.

[AbbaRue]

Keep in mind that the Th in the tungsten rod that Naudin used is a Gamma source.
Welders work with this all day long without any serious effects.
So the small amount of gamma this gives off shouldn't be a problem.

[k4zep]

Looks like you are not seeing much of an effect. Decay takes place over 20mS timespan and falls off sharply. Sharp rise and fall as the reaction takes place and sharp fall off as it ends. This forms a sine wave which should take place over about 12 to 15mS time period. Collapse of the field follows and the toroids field should collapse and form the negative side of the sine wave.

The previous post basically say's you should also be getting Gamma rays from the process, which to me was not good news since it would be much harder to deal with Gamma rather than alpha or beta rays.

3:00 a.m.  Have done a lot of experimenting tonight, have much to digest.............have a lot of thinking and things to recheck before I come to any conclusions with any certainty.  Have seen a lot I can not explain, have a much better understanding of workings, basically when small cap used, very short pulse in and out, would have to be a high frequency device.  This causes all sorts of new electronic design problems. I wonder what the freq. of  the 6 watt unit was?   have seen ideas and statements I have to digest, I have to recheck this integration into the toroid and reaction that causes full sine wave production effect from 1/2 wave burst effect.......Arrrrgggggghhhhhhhhhh Charley Brown.

Ben