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Re: more physics (short)

>>Thus the energy per kg is:
>>1.394E-12 / 1.993E-26 = 6.995E13 J/Kg
>So we used the same numbers (thou in a different order) and got about the
>same number.

I guess so.

>One suprize, I had that the mass of Protons and neutrons as about 1.673
>E-27 kg, where you have 1.661E-27.  Is that a typo?  We came up with the
>same velocity, which seems odd if we were using a different constant.

I used the definition for 'u', which finds it origin in the mass of a Carbon
12 atom. The weight of 12C is defined exactly 12*u. A single proton or
neutron is heavier than the protons and neutrons in a atom-core of multiple
nuclei (hence fusion frees energy).

But like I said before, the accuracy of this number only makes sense if you
know the (nearly) exact numbers of atom-masses.

>>(Don't use much more significant numbers, if you do use them you need to
>>define some things more accurate and cannot simply think that 11 H weighs as
>>much as 11B.)
>>Using E=0.5 m v^2 I still get v=11.8E6 m/s, so I really wonder how Bussard
>>can get a higher velocity, unless he dumps some of the reaction mass instead
>>of accelerating it.
>No the numbers were not based on an engineering output.  That was supposed
>to be the direct output from the reactions.  Could be I misread something
>and the 14E6 number was for something else (though I can't see what).

I don't know, but I've been discussing this with Rex lately and he thinks
that the efficiency of a DIRECT fusion engine may be low. With direct I mean
that one doesn't convert the kinetic energy of the reaction products to

>>I assume this time relativistics is the origin (for 0.5c gamma=1.15 which
>>means its not almost equal to 1 anymore) for our differences
>Why are you geting larger (often much larger) numbers for 75E6m/s, 125
>E6m/s, and 150 E6m/s, but smaller for 100 E6m/s?  Also the deltas don't
>seem even.  Hum.  I suppose I should rerun my numbers with relativistic
>equations.  With our luck I'll come up with a third set of numbers.

I think the deltas are exponential, this causes rapid deviations when one
neglects higher terms (ie. use classical instead of relativistic formulas)
I believe the relativistic rocket equation is:

M=Mo Exp[Vend/(g Vexh)]  where g=1/Sqrt[1-Vexh^2/c^2]

(Vexh is relative to the ship)

>Pity you gave no way to use you magic number.  (Like the value for given
>fuels?)  Also the F number table didn't give the numbers for the fuels,
>which naturally  didn't have even F numbers.

Well simply devide c^2 (9E16) by the J/kg numbers (eg. 9E16/6.995E13=1287)
Compared to 257 for the best fusion fuel we have come up with.

Somewhere in my document there are a few fusion reactions with f, Mev and J/kg.

Here is something that looks like it:

                                           J/kg     f
2H  +  3He  ->   4He +   1H  + 18.3 MeV  3.51E14   257
2H  +  3H   ->   4He +    n  + 17.6 MeV  3.38E14   267
3He +  3He  ->   4He + 2 1H  + 12.9 MeV  2.06E14   437
6Li +  6Li  -> 3 4He         + 20.0 Mev  1.60E14   564  (Combined)
3He +  6Li  -> 2 4He +   1H  + 16.0 MeV  1.28E14   704
1H  +  6Li  ->   4He +   3He +  4.0 MeV  5.47E13  1645
1H  + 11B   -> 3 4He         +  8.7 Mev  6.99E13  1287


P.S. To Kevin, I'm working on your latest letter...