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*To*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Subject*: Re: more physics (short)*From*: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39)*Date*: Tue, 4 Jun 1996 07:59:26 -0500*Cc*: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl, hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com, David@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com, neill@foda.math.usu.edu, 101765.2200@compuserve.com, MLEN3097@Mercury.GC.PeachNet.EDU

At 12:59 AM 6/4/96, Timothy van der Linden wrote: >>>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. Well eiather way I can't remember where I got my atomic mass constant from. Eiather one seems to get similar numbers, so its probably not a biggy. >>>(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 >electricity. Why? The velocities seem to be similar to the 'optimum' velocities in your table. I thought you figured that by the time you converted the particals momentum to elec, and then convert that elec back to the exaust velocity of a mass it would all come out even? (Give or take a lot of tonage of power equipment.) Please CC me if you have anything. >>>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) Thanks for the equation. As to my question about why are you geting larger numbers for 75E6m/s, 125 E6m/s, and 150 E6m/s, but smaller for 100 E6m/s? My guess would be you calculated for .3c not 1/3rd c again. >>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. Best is of course a relative term, and so obviously is the term simple. ;) >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 Thank you. Thats a table that would have made your other table useful. Kelly ---------------------------------------------------------------------- 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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