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Re: Physic help



>>>As near as I can figure it, it seems like a simple system.  An electron
>>>cloud is compressed in the center of a hollow steel sphere by a voltage
>>>charge on the sphere.  The fuel ions are feed into the center of the
>>>electron cloud, which forms the fuels containment 'chanber' (i.e they are
>>>repeled by the electrons and compresed into the center.).
>>
>>I assume the ions positive, are you sure they are repelled?
>
>Hum, good question.  Maybe it had an ion cloud of fuel in the center not
>electrons.  I'll have to dig that paper up again and check.

I indeed think that makes more sense, that way they positive ions are also
decelerated when they make it out of the inner ion-cloud.

>>It isn't necessary to decelerate the ions from the reaction, they will also
>>give an electrical current when they move fast.
>
>I think that if you run them threw a voltage gradiant they  lose their
>energy to it.  If they don't lose kinetic energy somewhere, you don't get
>electricity.

Well in theory they create a charge difference, the inside of the core gets
more negative while the outside gets more positive. The idea is a bit like
charging a capacitor.
Of course while creating this potential difference, the particle
decelerates, and when decharging it, the particles decharge even more. But
anyway, I don't care, as long as we get energy from it.

>The liniac would have certain advantages.  But also extra complexity.  So
>unless it would give us some performance advantage I'll assume the drive
>systems are using direct plasma thrust.

But then we are back to the photons inside the engine, since we are talking
about a dense plasma, a lot of collisions will be going on and then you
readely get photons.

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>I was runing it through the programs on the LIT servers computer center.
>Do you know if their is an error in that software?

I'm not sure, I only used it once. I think Steve did check several programs
once and found one to be not correct.
One thing I'm quite sure of, the program at the LIT server is probably made
for chemical fuels, in fusion fuels the energy density is much greater (and
thus a significant amount of mass is transferred to "mass-less" energy).

>>f = The fuel factor = (Total mass of the fuel) divided by (mass of the fuel
>>that can be converted to energy).
>
>Oh yeah.  Thats why I never used that table.  Strange number.  How would I
>find out what the fuel factor number for any of my fuels is?  (Yeah ok you
>added the equation below, but thats not a big help for someone tring to use
>the table.)  Or, why would you use the fuel factor in a table?

I personally find it not a stange number at all, it shows very clear what
part of the initial mass can be converted to energy (for anti-matter&matter
mixture f=1, or said differently, all mass can be converted to energy).
I assumed that everyone who would read my document did know about E=mc^2 and
thus could calculate it (as you could see the calculation was rather
straight forward.
I already had added a table with fusion fuels and their f-ratios, I think it
isn't visible on the web at the moment since all my web-sites seem to have
collapsed lately.
Also originally I had a little (but unclear) example of how to calculate f,
maybe I should add it again.

Tim