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Re: Physic help

To Kelly,

>>Well like I wrote you before, I still haven't the faintest idea how the
>>Bussard engine works. Is there anyone who can tell me the principle (ie.
>>tell me more than that it uses fusion as initial energy source)?
>>(are there any good electronic references?)
>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?

>The fuels chosen
>release the energy of the fusion reaction only in the kinetic energy of the
>released waste particals.  Those ions are going fast enough to blow threw
>the electron cloud, and slam into voltage gradiant.  They are decelerated
>by the voltage.  I.E. their kinetic energy is converted into a electrical
>current in the reactor systems.

To keep the electrons in the center the voltage at the wall of the steel
sphere should be negative, the ions from the reaction are positive and will
thus be accelerated instead of decelerated.

It isn't necessary to decelerate the ions from the reaction, they will also
give an electrical current when they move fast.

But now that I have some fague idea how a Bussard engine works, I see that
all calulations you asked for are meaningless because you transfer the
fusion energy first to electric energy by decelerating particles. Then you
use the electric energy to accelerate other (or the same) particles by a
lineac (I suppose).

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

>???  In you first responce to my request for exaust velocity info you said:
>>>For 2.058E14 Joule/kg this would give 2E7 m/s or 0.067c
>I responded that a 2e7m/s exaust velocity would translate to a specific
>impulse of about 2,000,000.  Which would mean the explorer class would need
>carry 148 times its own weight in fuel to get down from .3 c.
>So what is f in your equation?  How do you get a lower F for a fuel?  Given
>the 2.058E14 in:
>>>>   Where f is  c^2 / 2.058E14 = 439        (c=3E8 m/s)
>It would imply you were using the same fuel, He3 as listed in my table.
>But your geting wildly differnt fuel ratios with it.
>In short I am totally confused.  Worse I'm completly confused in a critical
>peace of info about the system!!  I.E. the specific impluse of the fusion
>I assume we both still agree that converting the fuel energy to
>electricity, and electricly accelerating a reaction mass will not help us

I don't see why you need to be confused, I get a ratio of about 100 and you
get a ratio of about 150. I'm not sure what exactly causes this difference
but I assume it is caused somewhere in your calculation.

About "converting the fuel energy to electricity, and electricly
accelerating a reaction mass", it depends on several factors whether this is
true, but generaly it does not help us much.

f = The fuel factor = (Total mass of the fuel) divided by (mass of the fuel
that can be converted to energy).

So say that you have a fusion fuel and are able to "squeeze" 2E14 joules per
kilogram out of it. Now all you have to do is determine the mass equivalence
of that amount of energy with E=m*c^2 (m=E/c^2).
In this case that makes m=2E14/9E16=0.00222 kg
The total mass of the fuel was 1 kg
So f=1/0.00222=450
Then look in the table for a final velocity of 0.3c and we see a fuel to
ship ratio of 104.
This table doesn't show anything about exhaust velocities but assumes the
best possible (which in this case is 0.06333c assuming 100% efficiency).