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*To*: 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*Subject*: Re: more physics (short)*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Sun, 02 Jun 1996 18:37:46 +0100

>I went over the Bussard paper and found 14E6 as the upper exaust vel for p >+ 11B. Which obviously isn't what I'm getting. Eiather I miscalculated >the watts per Kg (I sent my equations for that in another letter) or I >misread Bussards paper. (I suppose I could write him.) I'ld hate to write >off about 15% of spec impulse if I didn't need to. (See table below.) >If you know how to get the exaust speed from Mev numbers let me know. I'm >sure those numbers from the papers are correct. Watch carefully: The reaction as found in my tables book: H + 11B --> 3 4He + 8.7 MeV Let's assume energy conversion is 100% and ALL reaction products will be accelerated in the 100% effective lineac. First turn MeV in to Joules: 8.7 MeV = 8.7E6 * 1.602E-19 = 1.394E-12 Joule Then determine the weight of the particles in kg: H + 11B is approx 12 u (u=atomic mass unit) 12 u = 12 * 1.661E-27 = 1.993E-26 Kg Thus the energy per kg is: 1.394E-12 / 1.993E-26 = 6.995E13 J/Kg (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. >I was working up the following table of the fuel mass ratios needed to get >to or from certain speeds given the exaust velocities. > >NEW numbers > >Fuel --> Exhaust >Vexh 75E6m/s 100 E6m/s 125 E6m/s 150 E6m/s > >p + 11B --> 3 4He > 11,800,000 m/s 576.0 4,790.0 39,900 332,000 > >6Li + 6Li --> 3 4He > (Combined) > 17,800,000 m/s 67.6 275.0 1,120 4,570 > >3He + 3He --> 4He + 2 p > 20,300,000 M/s 42.5 138.0 472 1,620 > >De + 3He --> 4He + p > 26,500,000 m/s 16.9 43.5 112 287 Of course ;) I get completly different numbers (at least for higher Vend) 674 2,680 82,000 1,200,000 73 181 1,728 10,000 44 97 708 3,377 18 34 154 509 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 >Amazing how touchy the fuel ratios are to changes in exaust vel/specific >impulse. Look at the difference between 6Li + 6Li and 3He + 3He! ;) Well yes, that is because of the energy per kilogram of fuel. Which is 3 times higher for 3He + 3He. (This is what my "magic" number f says) Timothy

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