New Fusion Method_What do you think?

[Joseph Chikva]

So how close do you think two nucleii need to get to before nuclear forces begin to have an action?

I do not think anything on that.
I only need a proper collision energy initially and also do not to loose that energy during some distance till near 100% of particles will not react.

[chrismb]

So how close do you think two nucleii need to get to before nuclear forces begin to have an action?

I do not think anything on that.
I only need a proper collision energy initially and also do not to loose that energy during some distance till near 100% of particles will not react.

This is not enough to think like this. When two nucleii come together, fusion is just one possible outcome, and not very possible at that. There are a host of other possible outcomes, mainly scattering and non-elastic collisions (e.g. O-P for high deuteron energy). Fusion is never, ever an outcome where 'the majority of particles' get to fuse. Start doing some reading, and some maths. This is no good. Your are not making any progress with your understanding of fusion.

[Joseph Chikva]

This is not enough to think like this. When two nucleii come together, fusion is just one possible outcome, and not very possible at that. There are a host of other possible outcomes, mainly scattering and non-elastic collisions (e.g. O-P for high deuteron energy). Fusion is never, ever an outcome where 'the majority of particles' get to fuse. Start doing some reading, and some maths. This is no good. Your are not making any progress with your understanding of fusion.

Really?
You did not read the description.
Particle scattered at a small angle will return to the axis magnetically. And this will be repeated and repeated before fusion event occure.
Non-elastic collisions will have near-zero probability at energies at which fusion cross section is maximum or close to maximum.
O-P for the reactions of interest we can exclude at all.

[chrismb]

So, how big an angle are you proposing to be able to pull back into the beam. 45 degrees? Tell me this, then; what is the scattering cross-section for a 45 degree scatter [or whatever your maximum is] of a deuteron off a triton at 64keV?

[Joseph Chikva]

So, how big an angle are you proposing to be able to pull back into the beam. 45 degrees? Tell me this, then; what is the scattering cross-section for a 45 degree scatter [or whatever your maximum is] of a deuteron off a triton at 64keV?

45 degrees is not a small angle. Usualy one scattering event declines particle on angles lower on two-three orders. Then particle returns to the axis. Then again scattering or fusion. And so on.

[chrismb]

So how big an angle of deuteron scatter will your system be able to recover. Worst case. Just some estimate will do.

[Joseph Chikva]

So how big an angle of deuteron scatter will your system be able to recover. Worst case. Just some estimate will do.

Particles scattered at any angles and also any charged particles produced as a result of reaction from any direction will be directed at the axis thanks to self-magnetic field.

[chrismb]

So the magnetic field you are proposing to generate will be able to prevent a 64keV deuteron from exiting the beam when it is initially kicked out on a 90 degree trajectory? Am I understanding your inference correctly?

[Joseph Chikva]

So the magnetic field you are proposing to generate will be able to prevent a 64keV deuteron from exiting the beam when it is initially kicked out on a 90 degree trajectory? Am I understanding your inference correctly?

Firstly deuteron will have much more energy. 64 keV is collision energy in center-of-mass frame. And yes, self-magnetic field will be able to return reaction products with energy not keV-s but MeV-s directed even oppositely.
So, the Method can be used even for space propulsion in aneutronic reaction case.

[chrismb]

So the beam has a 5 barn cross-section and the magnetic field will be sufficient to keep a [let's just say] 64keV deuteron within it, even if it is initially scattered at 90 deg?

Am I understanding your idea correctly?

[Joseph Chikva]

So the beam has a 5 barn cross-section and the magnetic field will be sufficient to keep a [let's just say] 64keV deuteron within it, even if it is initially scattered at 90 deg?

Am I understanding your idea correctly?

Not beam but deuterium+tritium reaction at center-of-mass collision energy case 64keV (according another source a little bit more) has cross section 5 barns.
Amount of nucleus scattered at 90 degrees by a single event will be very limited (near zero).
But beam's self-magnetic field will be able to turn back to the axis any charged particle even from 180 degrees (namely charged particles - reaction products). This is very convenient for direct energy conversion.
Or for space propulsion.

[chrismb]

So the beam has a 5 barn cross-section and the magnetic field will be sufficient to keep a [let's just say] 64keV deuteron within it, even if it is initially scattered at 90 deg?

Am I understanding your idea correctly?

Not beam but deuterium+tritium reaction at center-of-mass collision energy case 64keV (according another source a little bit more) has cross section 5 barns.

How could you get maximum fusion if the beam is wider than 5 barns?

[Joseph Chikva]

How could you get maximum fusion if the beam is wider than 5 barns?

I do not understand the question.

[chrismb]

How could you get maximum fusion if the beam is wider than 5 barns?

I do not understand the question.

If you do not understand the question, then I do not understand your proposal.

Let us know once you've filed the patent and want to discuss it further.

[Joseph Chikva]

If you do not understand the question, then I do not understand your proposal.

Let us know once you've filed the patent and want to discuss it further.

I did not understand the question because that was not put correctly.
Good luck.