# Re: Physic help

```>>       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.

>>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)

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.

>>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.

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)

Our best reaction is 2H + 3He --> 4He +1H + 18.3 MeV
with 3.51E14 J/kg so that f=257

>AAAAAAHHHHHHHHHHHH!!!!!

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

Tim

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

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