Re: Physic help

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

At 5:36 PM 5/9/96, Timothy van der Linden wrote:
>Kelly wrote:
>
>>The info was taken from a paper by Bussard on reactors using these fuels.
>>The reactors converted virtuall all (99+%) of the fueles energy directly to
>>electricity.  No shielding or cooling system included.
>>
>>I was impressed.
>>
>>Probably the random collisions will cause heat and light.  But not until
>>its clear of the system (or the energy is converted to electricity).
>
>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.).  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.


>>>                          End velocity -->
>>>          +------+-------+-------+------+-------+--------+
>>>          | 0.20 |  0.30 |  0.40 | 0.50 |  0.60 |  0.70  |
>>>    +-----+------+-------+-------+------+-------+--------+
>>>  f | 200 |  7.6 |  22.2 |  69.5 |  244 |  1032 |   5906 |
>>>    | 250 |  9.7 |  31.9 | 114.6 |  467 |  2338 |  16422 |
>>> || | 300 | 12.0 |  44.4 | 180.0 |  839 |  4896 |  41401 |
>>> || | 350 | 14.6 |  60.2 | 272.7 | 1439 |  9662 |  96907 |
>>> || | 400 | 17.6 |  79.8 | 401.4 | 2376 | 18191 | 213876 |
>>> \/ | 450 | 21.0 | 104.1 | 577.2 | 3805 | 32958 | 449882 |
>>>    | 500 | 24.7 | 133.8 | 813.9 | 5941 | 57820 | 908988 |
>>>    +-----+------+-------+-------+------+-------+--------+
>>>
>>>   Where f is  c^2 / 2.058E14 = 439        (c=3E8 m/s)
>
>>I was only calculating the ratio in the explorer craft for deceleration
>>(acceleration fuel isn't carried by the ship).  With the
>>velocities/specific impulse that was giving the fuel ration was far higher.
>>???
>
>But in your own calculations you didn't use a fusion fuel with f=257. You
>used one with f=439, I did show that number just under the table, maybe you
>have not seen it? When looking in the table for f=439, you'll see that the
>number is somewhere near your own calculated ratio.
>
>Timothy


???  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?


REX!  Can you shine some light on this?!!

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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