Re: Physic help

[kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39)]

At 6:21 PM 5/10/96, Timothy van der Linden wrote:
>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?

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.

>>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.

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.

>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).

Not nessisarily.  If you don't slow down the partical (excluding losses in
forceing their way past the compression fields) you can direct that as a
plasma thrust.  Thats why I needed to find out what velocity they would
have, so I could work out the specific impulse and then the fuel mass
ratios.


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.

>- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
>
>>???  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
>>drive.
>>
>>I assume we both still agree that converting the fuel energy to
>>electricity, and electricly accelerating a reaction mass will not help us
>>any?
>
>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.

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

>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.

Pity we don't know of a way to make it work for us.

>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?

>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).
>
>
>Timothy

Thanks for the walk threw.  It was helpful.

Kelly


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Kelly Starks                       Internet: kgstar@most.fw.hac.com
Sr. Systems Engineer
Magnavox Electronic Systems Company
(Magnavox URL: http://www.fw.hac.com/external.html)

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