Crossfire Fusor

[cosmos]

What is the strength required of the magnets in order to prevent 3 to 5 MeV alphas from hitting the walls?

for 3MeV is about 1.2595 Tesla,
I would choose 3 Tesla or higher.

[MSimon]

I think your calculations are incorrect.

Please show all your steps.

In addition you assume all reactions take place in a small core.

And how do you prevent divergence? And how do you keep your fuels in the core? How do you get your exhaust gases to go around the bends without hitting the walls ?

How do you handle cusp losses?

[cosmos]

Cosmos,

And how about heat load from undeflected 4He?

The plasma is ionized, there is low probability of it occurs.
If it occurs, there is the heat exchange system for cooling the superconducting magnets.

[cosmos]

I think your calculations are incorrect.

Please show all your steps.

In addition you assume all reactions take place in a small core.

And how do you prevent divergence? And how do you keep your fuels in the core? How do you get your exhaust gases to go around the bends without hitting the walls ?

How do you handle cusp losses?

My calculations are correct. I was starting from this point:
10 × (3MeV) / (207.60172×E-27 Kg) = 1.44507E32 eV/Kg
Sorry, I am very busy for doing calculations for you.

small core? divergence?
Penning Traps has no problem of this type and it confines efficiently the fuel.
Please read more about Penning Traps.
cusp losses? I use continuous injection of ions and there is no recirculation of electrons.

[MSimon]

I think your calculations are incorrect.

Please show all your steps.

In addition you assume all reactions take place in a small core.

And how do you prevent divergence? And how do you keep your fuels in the core? How do you get your exhaust gases to go around the bends without hitting the walls ?

How do you handle cusp losses?

My calculations are correct. I was starting from this point:
10 × (3MeV) / (207.60172×E-27 Kg) = 1.44507E32 eV/Kg
Sorry, I am very busy for doing calculations for you.

small core? divergence?
Penning Traps has no problem of this type and it confines efficiently the fuel.
Please read more about Penning Traps.
cusp losses? I use continuous injection of ions and there is no recirculation of electrons.

Where did the "10" come from?

[cosmos]

the "10" come from decaborane (B10H14).

[MSimon]

the "10" come from decaborane (B10H14).

That makes no sense.

[cosmos]

the "10" come from decaborane (B10H14).

That makes no sense.

The following calculations take decaborane (B10H14) as example:
1H + 11B + 123KeV → 3 4He + 8.68MeV (66 TJ/Kg ≈ 18 GWh/Kg)

decaborane has "10"(1H + 11B)
and a rest of "4"(H)

[MSimon]

http://www.crossfirefusor.com/

The travel between Earth and Alpha Centauri star, including acceleration and deceleration, will take 2 years and 11 months and a decaborane consumption of 275.2Kg, reaching a maximum velocity of 3.27131×10E9Km/h, about three times the speed of light, at the midway.

Heck of a rocket ship you have there Johnny.

[scareduck]

This site is getting overrun with cranks.

[chrismb]

Cosmos;

The Newtonian physics has no restrictions about the velocity. Only the Theory of Relativity restricts the velocity to speed of light...
The Theory of Relativity was not fully proven...

I cannot resist the question - who, exactly, fully proved Newtonian physics?

If the inference that the theory of relativity is unproven, then why are you implicity accepting that the 8.68MeV/c^2 rest-mass loss in the p+11B->3He reaction results in 8.68MeV of energy? Please do let me know if you need more explanation of this question.

best regards,

Chris MB.

[cosmos]

Cosmos;

The Newtonian physics has no restrictions about the velocity. Only the Theory of Relativity restricts the velocity to speed of light...
The Theory of Relativity was not fully proven...

I cannot resist the question - who, exactly, fully proved Newtonian physics?

If the inference that the theory of relativity is unproven, then why are you implicity accepting that the 8.68MeV/c^2 rest-mass loss in the p+11B->3He reaction results in 8.68MeV of energy? Please do let me know if you need more explanation of this question.

best regards,

Chris MB.

I only share same opinions and critiques with other people concerning some theories. It is only critiques and opinions.

Newtonian physics works well in a real world.
I believe in the definition of electronvolt:

In physics, the electron volt (eV) is a unit of energy. By definition, it is equal to the amount of energy gained by a single unbound electron when it accelerates through an electrostatic potential difference of one volt.

Maybe the polemic is from this:

The formula E=½mv² refers to the action and reaction (Newton's third law) between the spacecraft and the exhausting products and not between the spacecraft and the origin point or launching site.
In the outer space, half of the energy goes to the exhausting products:
½E=½mv² → E=mv² → 200e6=500000× v²→ v=20 m/s → a=∆v/∆t → a=20m/s² → g-force= 2.0 g

the Tsiolkovsky rocket equation diverges a lot from E=½mv²
taking in mind the NASA's Space Shuttle of 2000tons at 12GWatts as reference.

[chrismb]

I only share same opinions and critiques with other people concerning some theories. It is only critiques and opinions.

Science is not based on opinions, it is based on repeatable observations that support some predicate logic.

The formula you have for kinetic energy is wrong. Kinetic energy is the difference between the rest mass of an object and its relativistic mass. KE=m.c^2-m(o).c^2.

You can fiddle with this to get v=SQRT(c^2-[c^3/({E/m}+c^2)]^2), if that helps you out. The formula v=SQRT(2E/m) is Newtonian and is only approximate for low velocities.

You do not seem to understand that the fusion energy from p+11B is not measured with some sort of 'alpha energy' meter. Instead, 3 alphas are observed and then, in some completely different experiment, the rest masses of the proton, alpha and 11B are measured and *then* these relativistic formulas are applied to get a value for the fusion energy.

Those people that do not 'believe' in relativity therefore also cannot 'believe' in fusion energy (and there are a few). But the two things go together. I regret to point out that your lack of understanding that you actually lack the understanding means you have a mountain of auto-didactic information to spend your time studying.

If you progress down this route, you will come to understand that fusion occurs so seldomly that to achieve net fusion energy output you have to overcome Coulomb, inelastic and radiative scattering losses. Fusion is the easy bit, overcoming the scattering losses is the 'big idea' that a net-energy fusion idea MUST explicitly disclose for it to be of any merit.

I hope your interest will take you down a route of self-learning that you come to understand these things, and you will be enriched for it.

best regards,

Chris MB.

[cosmos]

You are right.
E/m=66.03921e12 J/Kg
v=SQRT(2E/m) = 11.49254e6 m/s
v=SQRT(c^2-[c^3/({E/m}+c^2)]^2) = 11.48622e6 m/s

I am not an expert in Relativity.
I rely on acceleration (g-force) for generating a gravitational effect in the spacecraft.
When the matter is acceleration, the Relativity sounds very strange for me.
At speed of light, will the g-force disappear?

[MSimon]

You are right.
E/m=66.03921e12 J/Kg
v=SQRT(2E/m) = 11.49254e6 m/s
v=SQRT(c^2-[c^3/({E/m}+c^2)]^2) = 11.48622e6 m/s

I am not an expert in Relativity.
I rely on acceleration (g-force) for generating a gravitational effect in the spacecraft.
When the matter is acceleration, the Relativity sounds very strange for me.
At speed of light, will the g-force disappear?

No worries mate. You can't reach the speed of light if you have any rest mass. I suggest starting with a kilo or two. That should be enough to prevent the problem from arising.