Re: Physic help

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

At 12:20 AM 5/8/96, Timothy van der Linden wrote:
>>>Yes, but that still means that the momentum and energy are not necessary
>>>equally distributed. Rex has given a 2 particle calculation but I wonder if
>>>that makes much sense, since all energy/momentum is finally directed in one
>>>direction namely backwards.
>>
>>I guess it would even out in the plasma.  Collisions and such.
>
>Yes, something like this I had in mind.
>
>>>What isn't mentioned in your fusion reactions is that within the engine
>>>probably many photons are generated too. You should see these photons just
>>>as a form of energy and not as an ingredient of the reaction.
>>
>>Supposedly not.  Thats why the reactions have virtually no radiation, and
>>virtually all the energy can be converted to electricity.
>
>I'm a bit amazed by that, maybe not initially but as soon as the particles
>start colliding with each other I think photons may be created. I don't know
>when or why photons are created, but I assume that something like blackbody
>radiation will certainly create a bunch of them. Rex, do you know about that?

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).

>>>Didn't I show this a while ago? It shows the fuel to ship ratios.
>>>
>>>                          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)
>>
>>Having no idea how to tie F to acctual fuels.  This didn't help much.
>
>I'm sorry, I hope you have an idea by now: f=c^2/(J/kg)
>
>>>Our best reaction is 2H + 3He --> 4He +1H + 18.3 MeV
>>>with 3.51E14 J/kg so that f=257
>>
>>???  That would give you a fuel mass ration of under 40 acording to your
>> table.
>
>Yes that's right, but keep in mind that these ratios are just for
>acceleration or deceleration, if you want to accelerate and decelerate
>without refilling in between then square the numbers (40^2=200).
>
>Tim

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.
???

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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