# Re: Physic help

```At 12:02 AM 5/7/96, Timothy van der Linden wrote:
>>>       2
>>>K = M c (gamma - 1)   K=kinetic energy
>>>
>>>
>>>In fact this is all you need, but there is a bit of a problem, namely that
>>>there is more than one particle in each reaction, and that sometimes not all
>>>particles have the same mass. This makes that not all particles have the
>>>same final speed, ofcourse one could figure out a mean velocity, but I'm not
>>>sure how to do that best.
>>
>>In the case of Li6 the resulting exaust is all 4He?  3 per 20 MeV.
>
>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.

>>>If I would do a very rough approximation, I would use the Watts/kg numbers
>>>in your table (by the way it should be Joule/kg).
>>>
>>>Since then the velocities wouldn't be relativistic at all I can simply use:
>>>
>>>E = 0.5 m v^2
>>>
>>>or
>>>
>>>v = SQRT[2 E/m]
>>
>>For example is E energy in Joules, watts, voltage...?  Is m in kilos,
>>proton mass, atomic weight?  Without this the equation isn't useable to me.
>>In physics as I remember only a few standard units are used.  In
>>Engineering anything is used.
>
>Yes, forgot that again, maybe subconsiously assuming that the formula looked
>familiar.
>
>E=energy (Joules)
>m=mass of all the particles left over after the reaction (kg)
>v=mean velocity of the particles (m/s)

Thank you.

>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.

>>>For 2.058E14 Joule/kg this would give 2E7 m/s or 0.067c
>>
>>What are you using for mass?  Are you assuming a 4He and 2 p, or assuming
>>that the energy of the p's are transfered to the next reaction that are
>>involved in?  Or assumed the energy is lost?
>
>Since I was talking about mean values, I wasn't thinking about particles
>anymore, it was enough to just talk about energy per amount of fuelmass.
>So for the mass I simply used 1 kg since it is 2.058E14J per 1 kg. I could
>also have used 2 kg because it is 4.116E14J per 2 kg. I hope If I'm childish
>here, forgive me.

Sorry, I should have caught that one.

>So when using v = SQRT[2 E/m], forget about electron volts, atomic mass etc.
>just use the energy-per-mass number in your table.
>
>If you like to know how those energy-per-mass (J/kg) numbers were calculated
>give a yell and I will show it to you. (Then I will need eV and amu again)
>
>>Damn, this is bad 2e7 would translate to a specific impulse of 2,000,000.
>>Which would mean the explorer class would need a fuel to weight ratio of
>>148 to 1 to get down from .3 c.
>
>I get something around 100:1 but I guess that doesn't matter much anymore.
>
>>And given that you used 2.058E14, which is
>>the result for He3 not Li6, and Li6 reactions divide 1.596 E14 over about
>>twice as much mass, then a Li6 fueled ship would have a much lower specific
>>impulse.  I.E. the Explorer class couldn't work!!
>
>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.

>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.

>>AAAAAAHHHHHHHHHHHH!!!!!
>
>That's why I tried pushing anti-matter as a fuel.

Good power to weight ratio, but unproducable or handelable in the
quantities and weigh ratios we need.

>Tim
>
>
>P.S. Don't remove the explorer pages, it makes a good example why not to use
>fusion fuel. (just kidding)

eeeeEEEERRRRR!!!!

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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