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Re: Physic help



>>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?

>>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