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Re: more physics (short)
- To: KellySt@aol.com, firstname.lastname@example.org, email@example.com, firstname.lastname@example.org, email@example.com, firstname.lastname@example.org, email@example.com, David@InterWorld.com, firstname.lastname@example.org, DotarSojat@aol.com, email@example.com, firstname.lastname@example.org, MLEN3097@Mercury.GC.PeachNet.EDU
- Subject: Re: more physics (short)
- From: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)
- Date: Tue, 04 Jun 1996 00:59:00 +0100
>>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
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:
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...