’WPC4 ,LLŲ!@ HJÜæÜ WP_TV71411471WPTVWPC2WPTVWPC2®Aėń $ÜÜ !€!Ü------------------------------------------------------------------ÜÜ---- Ü!!8{WP}01Ķ•ĆĆĆĆĆĆĆĆĆĆĆĆĆĆĆĆĆĆĆ ĆĆ ĆĆ ĆĆ ĆĆ ĆĆĆĆĆĆĆĆĆĆĆŠ ŠŠHHŠŠ  ŠŠ2ŠŠ’’øę$’ŠŠHHŠŠŠŠŠŠŠŠ  ŠŠ ŠŠ ŠŃŃŃGenevaŃŃ ŃŃ € ŃŃ<Px–ČŃŅŅŅ443!#4ŅŅ$ *$ŅŅ$*$ŅŅ ’KK ŅŅ  Geneva ŅŅ  Geneva ŅÓ.,ÓÓÓÓ6ÓÓ ÓßßßßßßßßßßßßßßßßßßŃGenevaŃŃ ŃŅ  h HD: Geneva ŅŅ  h HD: Geneva Ņ)P€3.53.5, © 1989, 1995 Novell, Inc.(P€3.5Created with WordPerfectŖ 3.5.b¢^®Aėń WordPerfectxHHŲ(’į’āłFG(üHHŲ(d'`LLŲøšōŹTStlBStylNPSetZPJobfversrWDatŠSTR –PtPt¢7‚ <’’ “’’#“’’ 5³D’’ bµ8’’Ž·œæō’’ Ąŗp’’Š{WP}10{WP}01Ü' 'ÜŪ®Aėń{WP}10ŪŅ ’<<’KK ŅÓ.’.,’,..,,ÓŪŪĶ®AėńĶĶ®AėńĶŁ   ŁĶ®AėńĶĶ®AėńĶŁ   ŁĶ®AėńĶĶ®AėńĶŁ  ŁĶ®AėńĶĶ®AėńĶŁ  ŁĶ®AėńĶĶ®AėńĶŁ  ŁĶ®AėńĶĶ®AėńĶŁ  ŁĶ®AėńĶĶ®AėńĶŁ  ŁĶ®AėńĶĶ®AėńĶŁ  ŁĶ®AėńĶĶ®AėńĶŁ  ŁĶ®AėńĶĶ®AėńĶŁ  ŁĶ®AėńĶĶ®AėńĶŁ  ŁĶ®AėńĶÜ' 'Ü===========================================================Ü ÜDate: Mon, 6 May 1996 00:23:28 -0400Ü 0ÜFrom: KellySt@aol.comÜ0@ÜTo: T.L.G.vanderLinden@student.utwente.nl, kgstar@most.fw.hac.com, Ü@PÜstevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,ÜÜhous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,ÜÜDavid@interworld.com, lparker@destin.gulfnet.com, DotarSojat@aol.comÜÜCc: MLEN3097@mercury.gc.peachnet.eduÜ PÜSubject: Re: A new old memeberÜ  ÜÜ  °ÜMust be something in the air. Mike Leneski, whos been off line and out ofÜÜcontact for a while, just asked to be added to the list. Ü °ŠÜÜ ŠąÜOdd timing?Ü ąšÜÜ šÜMLEN3097@Mercury.GC.PeachNet.EDUÜ ÜÜ  ÜÜ 0ÜKellyÜ0@ÜÜ@PÜÜP`Ü===========================================================Ü`pÜDate: Mon, 6 May 1996 03:02:30 -0500 (CDT) From: Kevin C HoustonÜÜ To: Timothy van der LindenÜÜ Cc: KellySt@aol.com,ÜÜkgstar@most.fw.hac.com, stevev@EFN.ORG, Üp°Üjim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,ÜÜrddesign@wolfenet.com, David@InterWorld.com,ÜÜlparker@destin.gulfnet.com, DotarSojat@aol.com, neill@foda.math.usu.edu,ÜÜ101765.2200@CompuServe.COM Subject: Re: A new old memeber! :)ܰšÜMime-Version: 1.0Ü šÜÜ!ÜÜ" ÜÜ# 0ÜOn Sun, 5 May 1996, Timothy van der Linden wrote: Ü$0@ÜÜ%@PÜ>Hi all,Ü&P`ÜÜ'`pÜ>A few weeks ago, an old web-friend Emailed me. Some of the older LIT ÜJ(p€ pp((€ HJÜmembers > may still remember his name: Nick Tosh. æÜ) ÜÜ* °Ü>101765.2200@compuserve.com (Nick Tosh)Ü+°ĄÜÜ,ĄŠÜ>I assume he will introduce himself to the group when possible. Ü-ŠąÜÜ.ąšÜHeya Nick!Ü/šÜÜ0ÜHow are you doing. Things sure got busy around here when after you left. IÜÜgiven some serious thought to your idea of a lighter return craft. I thinkÜÜit's a good idea. Tim's worked out the math for a MARS-type mission.ÜÜalthough the only way so far to build the return trip's solar collectors andÜÜmaser array inside of ten years, is to assume some sort of self-ÜÜreplicating robot. Anyway, Glad to see you're re-connected. Ü 6pÜÜ 7p€ÜTake careÜ 8€ÜÜ 9 ÜKevin HoustonÜ : °ÜÜ;°ĄÜ.X-SenderĢ®AėņĢ: kgstar@pophost.fw.hac.comÜ<ĄŠÜMime-Version: 1.0Ü=ŠąÜĶ®AėńĶŁ   ŁŁ   ŁŁ  ŁŁ  ŁŁ  ŁŁ  ŁŁ  ŁŁ  ŁŁ  ŁŁ  ŁŁ  ŁĶ®AėńĶĶ®AėńĶÜ>ąšÜÜ?šÜ===========================================================Ü@ÜDate: Mon, 6 May 1996 08:40:22 –0500ÜA ÜTo: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) From:ÜB 0Ükgstar@most.fw.hac.com (Kelly Starks x7066 MS 10–39) Subject: Re:ÜC0@ÜPhysic helpÜD@PÜCc: KellySt@aol.com, kgstar@most.fw.hac.com, stevev@efn.org, ÜEP`Üjim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl,ÜF`pÜhous0042@maroon.tc.umn.edu, rddesign@wolfenet.com,ÜGp€ÜDavid@InterWorld.com, lparker@destin.gulfnet.com, DotarSojat@aol.com,ÜH€Üneill@foda.math.usu.edu, 101765.2200@compuserve.com,ÜI ÜMLEN3097@Mercury.GC.PeachNet.EDUÜJ °ÜÜK°ĄÜAt 11:29 AM 5/4/96, Timothy van der Linden wrote: ÜLĄŠÜ>I yesterday posted this by accident only to Kelly: Ü MŠąÜÜ!NąšÜ>– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – Ü"OšÜÜJ#P p(P HJÜ>To Kelly,æÜQ 0ÜÜR0@Ü>>Hi all you physics types. On my web page describing my explorer class'ÜS@PÜfusion reactors.ÜTP`ÜÜU`pÜ>>http://165.254.130.90:80/LIT/InterStellar/project/Explorer_ClassÜÜ/Bussard_Fus ion_systems.HTMLÜWpÜÜX Ü>>I have a table where I list various fusion fuel cycles. It lists theÜY °Üresulting energy in Mev. For those of us who arn't familure withÜ Z°ĄÜtranslating Mev into anything, could someone tell me what the speed ofÜ [ĄŠÜthe resulting particals is? Given that all the power in those reactions isÜ \ŠąÜcontained in the resulting kinetic energy of the particals. This should be aÜ ]ąšÜsimple calculation, and would certainly be a nice addition to the table.Ü ^šÜAssuming of course your half of and undergrate physics degree wasn't overÜ_Ü15 years stale!Ü` ÜÜa 0Ü>I'm a bit amazed you ask this because 2 paragraphs later you explain whatÜb0@Üan electron Volt is.Üc@PÜ>So I assume the problem is in the calculation of the velocities. ÜdP`ÜÜe`pÜA definition is one thing, knowing how to plug real numbers into anÜfp€Üequation is another.Üg€ÜÜh Ü>I'll assume the velocities are relativistic, to make the formulas usableÜi °Üfor all energies.Üj°ĄÜÜkĄŠÜ>Kinetic energy (in Joules) of a particle with mass m (in kg) moving withÜlŠąÜvelocity v:ÜmąšÜÜnšÜÜoÜ>1Üp Ü>–––––––––––––––Ü q 0Ü>2Ü!r0@Ü>gamma =Į ĮvĮĮgamma >= 1Ü"s@PÜ>SQRT(1 – ––––)Ü#tP`Ü>2Ü$u`pÜ>cÜ%vp€ÜÜ&w€Ü>2ÜJ'x  p(x  HJÜ>K = M c (gamma – 1) K=kinetic energyæÜy°Ą܀€>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. €€>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.€€€>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? €€€>Oh to make the thing complete here the translation from eV to joule: €€>1 eV = 1.6E–19 Joule€€>Note that eV is a measure for energy and that in particle physics the€mass of a particle is often given in eV also (this according to E=mc^2). €€€>Timothy€€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. 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!! €€Yeah I roughed out a calculation (I'm still not comfortable that I'm€following your math.) and get a specific impluse of about 1,600,000.€Which would demand hundreds of times the ships weight in fuel!!!! €€AAAAAAHHHHHHHHHHHH!!!!!€€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) €€––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––X–Sender:€€€€===========================================================€Date: Tue, 07 May 1996 00:02:03 +0100€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€From: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)€Subject: 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)X–Sender:€S9421793@mail.student.utwente.nl Mime–Version: 1.0€€€€===========================================================€Date: Tue, 7 May 1996 02:50:39 –0500 (CDT) From: Kevin C Houston€ To: interstellar drive group€<101765.2200@compuserve.com>, €David@InterWorld.com, DotarSojat@aol.com, jim@bogie2.bio.purdue.edu,€KellySt@aol.com, kgstar@most.fw.hac.com, lparker@destin.gulfnet.com,€mlen3097@mercury.gc.peachnet.edu, rddesign@wolfenet.com,€sl0c8@CC.USU.EDU, Steve VanDevender ,€T.L.G.vanderLinden@student.utwente.nl, zkulpa@zmit1.ippt.gov.pl Subject:€Insult to Injury€Mime–Version: 1.0€€First, a possible solution to Kelly's dilemma: €€Why not launch a fuel tank ahead of the explorer. This would be an€automated drone that would carry the decell fuel. This could be launched€using an ion beam/mag sail combination. Since there would be no crew to€worry about, acceleration could be several G's without difficulty.€assuming a .3C cruise speed, the drone gets pushed to .29C, and rendevous€takes place at the halfway point. The tanker is launched only a few months€ahead of the explorer, so the ship should be able to "see" their tanker at€all times. since we are adding an ion cannon (instead of a fuel launcher or€a maser array) the explorer could use that system to get up to some speed,€further saving on accel fuel. €€Also, maybe some form of charged–wire magnetic brake would help here, I€know we dismissed these for near light–speed ships, but perhaps at .3C€the explorer could benefit from it.€€€€Not to kick a man when he's down, but I heard a funny joke that I can twist€to fit the occasion:€€Three sub–warp ships pull into Deep–space Nine at the same time. An€anti–matter Rocket, a MARS, and a Fusion rocket. €€all three captains go into Quark's bar and order drinks. €€The captain of the Anti–matter Rocket says "give me a R.A." Quark smiles€and says "One Romulan Ale, coming up" €€The captain of the MARS says "give me a S.B." Again Quark smiles and says€"One Saurian Brandy coming up" €€The captain of the Fusion Rocket says "give me a Fifteen" Quark frowns€and says "fifteen? Whats a fifteen?" the captain says "Seven and Seven,€What else?" €€€Kevin€(who knows what it feels like to be slapped in the face by the cold hand of€reality)€€PS Don't give up so easily Kelly, There may yet be a way around your€problem.X–Sender: kgstar@pophost.fw.hac.com€Mime–Version: 1.0€€€€===========================================================€Date: Tue, 7 May 1996 07:47:50 –0500€To: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) From:€kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10–39) Subject: Re:€Physic help€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 12:02 AM 5/7/96, Timothy van der Linden wrote: €>>>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.€€I guess it would even out in the plasma. Collisions and such. €€>>>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)€€Thank you.€€>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. €€Supposedly not. Thats why the reactions have virtually no radiation, and€virtually all the energy can be converted to electricity. €€>>>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.€€Sorry, I should have caught that one.€€>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)€€Having no idea how to tie F to acctual fuels. This didn't help much. €€>Our best reaction is 2H + 3He ––> 4He +1H + 18.3 MeV with 3.51E14 J/kg€so that f=257€€??? That would give you a fuel mass ration of under 40 acording to your €table.€€>>AAAAAAHHHHHHHHHHHH!!!!!€€>That's why I tried pushing anti–matter as a fuel. €€Good power to weight ratio, but unproducable or handelable in the€quantities and weigh ratios we need.€€>Tim€€€>P.S. Don't remove the explorer pages, it makes a good example why not to€use fusion fuel. (just kidding)€€€eeeeEEEERRRRR!!!!€€€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) €€––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––X–Sender: kgstar@pophost.fw.hac.com€€€===========================================================€Date: Tue, 7 May 1996 08:03:29 –0500€To: Kevin C Houston From:€kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10–39) Subject: Re:€Insult to Injury€Cc: interstellar drive group <101765.2200@compuserve.com>, €David@InterWorld.com, DotarSojat@aol.com, jim@bogie2.bio.purdue.edu,€KellySt@aol.com, kgstar@most.fw.hac.com, lparker@destin.gulfnet.com,€mlen3097@mercury.gc.peachnet.edu, rddesign@wolfenet.com,€sl0c8@CC.USU.EDU, Steve VanDevender ,€T.L.G.vanderLinden@student.utwente.nl, zkulpa@zmit1.ippt.gov.pl €€At 2:50 AM 5/7/96, Kevin C Houston wrote: €>First, a possible solution to Kelly's dilemma: €€>Why not launch a fuel tank ahead of the explorer. This would be an€automated drone that would carry the decell fuel. This could be launched€using an ion beam/mag sail combination. Since there would be no crew to€worry about, acceleration could be several G's without difficulty.€assuming a .3C cruise speed, the drone gets pushed to .29C, and rendevous€takes place at the halfway point. The tanker is launched only a few months€ahead of the explorer, so the ship should be able to "see" their tanker at€all times. since we are adding an ion cannon (instead of a fuel launcher or€a maser array) the explorer could use that system to get up to some speed,€further saving on accel fuel. €€With my fuel launcher idea I was accelerating the ship without using any€onboard fuel. I need the 2,500,00 spec impuse so the ship could carry€enough braking fuel to stop from .3c.€€>Also, maybe some form of charged–wire magnetic brake would help here,€I know we dismissed these for near light–speed ships, but perhaps at .3C€the explorer could benefit from it.€€Some kind of drag chut would be nice, if their was anything to drag€against. €€AAAAHHHHHHHHHH!!!!€€>Kevin€>(who knows what it feels like to be slapped in the face by the cold hand€of reality)€€Yeah, we've all managed to chew up and spit out most all the ideas we've€come up with, and yes I rtemember you realizing that the MARS drag motor€could overcome the microwave thrust from the rectenna. €€>PS Don't give up so easily Kelly, There may yet be a way around your€problem.€€I suppose on the bright side were not doing much worse than others.€Maklov (sp) who wrote the starflight handbook was in a artical in Final€Frounteir magazine where he figured we launch thousand year solar sail€ships to the local starsystem in the next century. (Only if we all go€retarded!) €€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) €€––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––€€€€From: David Levine€Date: Wed, 8 May 1996 06:44:30 –0400€Mime–Version: 1.0€€Okay, slight problem here.€€Yesterday at around 7 or 8 PM, my system crashed. Hard. I have been here€since then trying to fix it. It is now 6:30 AM. My system now sort of€works, but EVERYTHING is gone. This means all of Kelly's Explorer€documents. This means the other LIT pages I've been editing when I get a€chance. This also means a hell of a lot of my actual work, too. Luckily, all€my source code is backed up with Visual Source Safe. But a lot of little€minor things aren't backed up... and, well, they add up.€€Anyway, I have a conference to go to next week where we're demoing this€software package I've been working on to a lot of people. I was behind€already. Now I think I have to be here around the clock until we leave. €€Anyway, the upshot of all of this is that I may not have time over the next€week or two to reset up all the accounts again. I'll try, but I can't promise€anything.€€And Kelly... (and anyone else who happened to upload)... I sure hope you have€a backup of the Explorer stuff. I wish I had more backups for my stuff.€€–David€€(Oh, BTW, this refers to the LIT staging–prototype server–type thingy...not€the SunSITE address...)€€€===========================================================€€€===========================================================€Date: Thu, 09 May 1996 17:36:37 +0100€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€From: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)€Subject: Re: Physic help€€Kelly wrote:€€>The info was taken from a paper by Bussard on reactors using these fuels.€The reactors converted virtuall all (99+%) of the fueles energy directly to€electricity. No shielding or cooling system included. €€>I was impressed.€€>Probably the random collisions will cause heat and light. But not until its€clear of the system (or the energy is converted to electricity). €€Well like I wrote you before, I still haven't the faintest idea how the€Bussard engine works. Is there anyone who can tell me the principle (ie.€tell me more than that it uses fusion as initial energy source)? (are there€any good electronic references?) €€>>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)€€>I was only calculating the ratio in the explorer craft for deceleration€(acceleration fuel isn't carried by the ship). With the velocities/specific€impulse that was giving the fuel ration was far higher. ???€€But in your own calculations you didn't use a fusion fuel with f=257. You€used one with f=439, I did show that number just under the table, maybe€you have not seen it? When looking in the table for f=439, you'll see that€the number is somewhere near your own calculated ratio. €€TimothyX–Sender: kgstar@pophost.fw.hac.com€Mime–Version: 1.0€€€€===========================================================€Date: Thu, 9 May 1996 11:25:37 –0500€To: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) From:€kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10–39) Subject: Re:€Physic help€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 5:36 PM 5/9/96, Timothy van der Linden wrote: €>Kelly wrote:€€>>The info was taken from a paper by Bussard on reactors using these€fuels. The reactors converted virtuall all (99+%) of the fueles energy€directly to electricity. No shielding or cooling system included. €€>>I was impressed.€€>>Probably the random collisions will cause heat and light. But not until€its clear of the system (or the energy is converted to electricity). €€>Well like I wrote you before, I still haven't the faintest idea how the€Bussard engine works. Is there anyone who can tell me the principle (ie.€tell me more than that it uses fusion as initial energy source)? (are there€any good electronic references?) €€As near as I can figure it, it seems like a simple system. An electron cloud€is compressed in the center of a hollow steel sphere by a voltage charge€on the sphere. The fuel ions are feed into the center of the electron cloud,€which forms the fuels containment 'chanber' (i.e they are repeled by the€electrons and compresed into the center.). The fuels chosen release the€energy of the fusion reaction only in the kinetic energy of the released€waste particals. Those ions are going fast enough to blow threw the€electron cloud, and slam into voltage gradiant. They are decelerated by the€voltage. I.E. their kinetic energy is converted into a electrical current in€the reactor systems.€€€>>>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)€€>>I was only calculating the ratio in the explorer craft for deceleration€(acceleration fuel isn't carried by the ship). With the velocities/specific€impulse that was giving the fuel ration was far higher. ???€€>But in your own calculations you didn't use a fusion fuel with f=257. You€used one with f=439, I did show that number just under the table, maybe€you have not seen it? When looking in the table for f=439, you'll see that€the number is somewhere near your own calculated ratio. €€>Timothy€€€??? In you first responce to my request for exaust velocity info you said: €€>>For 2.058E14 Joule/kg this would give 2E7 m/s or 0.067c €€I responded that a 2e7m/s exaust velocity would translate to a specific€impulse of about 2,000,000. Which would mean the explorer class would€need carry 148 times its own weight in fuel to get down from .3 c. €€So what is f in your equation? How do you get a lower F for a fuel? Given€the 2.058E14 in:€€>>>Where f is c^2 / 2.058E14 = 439‚(c=3E8 m/s)€€It would imply you were using the same fuel, He3 as listed in my table.€But your geting wildly differnt fuel ratios with it. €€In short I am totally confused. Worse I'm completly confused in a critical€peace of info about the system!! I.E. the specific impluse of the fusion€drive.€€I assume we both still agree that converting the fuel energy to€electricity, and electricly accelerating a reaction mass will not help us€any?€€€REX! Can you shine some light on this?!! €€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) €€––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––X–Sender: S9421793@mail.student.utwente.nl Mime–Version: 1.0€€€€===========================================================€Date: Fri, 10 May 1996 18:21:59 +0100€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€From: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)€Subject: Re: Physic help€€To Kelly,€€>>Well like I wrote you before, I still haven't the faintest idea how the€Bussard engine works. Is there anyone who can tell me the principle (ie.€tell me more than that it uses fusion as initial energy source)? (are there€any good electronic references?) €€>As near as I can figure it, it seems like a simple system. An electron€cloud is compressed in the center of a hollow steel sphere by a voltage€charge on the sphere. The fuel ions are feed into the center of the electron€cloud, which forms the fuels containment 'chanber' (i.e they are repeled by€the electrons and compresed into the center.). €€I assume the ions positive, are you sure they are repelled? €€>The fuels chosen€>release the energy of the fusion reaction only in the kinetic energy of the€released waste particals. Those ions are going fast enough to blow threw€the electron cloud, and slam into voltage gradiant. They are decelerated by€the voltage. I.E. their kinetic energy is converted into a electrical current€in the reactor systems.€€To keep the electrons in the center the voltage at the wall of the steel€sphere should be negative, the ions from the reaction are positive and will€thus be accelerated instead of decelerated. €€It isn't necessary to decelerate the ions from the reaction, they will also€give an electrical current when they move fast. €€But now that I have some fague idea how a Bussard engine works, I see€that all calulations you asked for are meaningless because you transfer€the fusion energy first to electric energy by decelerating particles. Then€you use the electric energy to accelerate other (or the same) particles by€a lineac (I suppose).€€– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – €€>??? In you first responce to my request for exaust velocity info you said: €€>>>For 2.058E14 Joule/kg this would give 2E7 m/s or 0.067c €€>I responded that a 2e7m/s exaust velocity would translate to a specific€impulse of about 2,000,000. Which would mean the explorer class would€need carry 148 times its own weight in fuel to get down from .3 c. €€>So what is f in your equation? How do you get a lower F for a fuel? Given€the 2.058E14 in:€€>>>>Where f is c^2 / 2.058E14 = 439‚(c=3E8 m/s)€€>It would imply you were using the same fuel, He3 as listed in my table.€But your geting wildly differnt fuel ratios with it. €€>In short I am totally confused. Worse I'm completly confused in a critical€peace of info about the system!! I.E. the specific impluse of the fusion€drive.€€>I assume we both still agree that converting the fuel energy to€electricity, and electricly accelerating a reaction mass will not help us€any?€€I don't see why you need to be confused, I get a ratio of about 100 and you€get a ratio of about 150. I'm not sure what exactly causes this difference€but I assume it is caused somewhere in your calculation. €€About "converting the fuel energy to electricity, and electricly€accelerating a reaction mass", it depends on several factors whether this€is true, but generaly it does not help us much. €€f = The fuel factor = (Total mass of the fuel) divided by (mass of the fuel€that can be converted to energy).€€So say that you have a fusion fuel and are able to "squeeze" 2E14 joules€per kilogram out of it. Now all you have to do is determine the mass€equivalence of that amount of energy with E=m*c^2 (m=E/c^2). In this case€that makes m=2E14/9E16=0.00222 kg The total mass of the fuel was 1 kg€So f=1/0.00222=450€Then look in the table for a final velocity of 0.3c and we see a fuel to ship€ratio of 104.€This table doesn't show anything about exhaust velocities but assumes the€best possible (which in this case is 0.06333c assuming 100% efficiency). €€€TimothyX–Sender: kgstar@pophost.fw.hac.com€Mime–Version: 1.0€€€€===========================================================€Date: Fri, 10 May 1996 12:07:36 –0500€To: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) From:€kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10–39) Subject: Re:€Physic help€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 6:21 PM 5/10/96, Timothy van der Linden wrote: €>To Kelly,€€>>>Well like I wrote you before, I still haven't the faintest idea how the€Bussard engine works. Is there anyone who can tell me the principle (ie.€tell me more than that it uses fusion as initial energy source)? (are there€any good electronic references?) €€>>As near as I can figure it, it seems like a simple system. An electron€cloud is compressed in the center of a hollow steel sphere by a voltage€charge on the sphere. The fuel ions are feed into the center of the electron€cloud, which forms the fuels containment 'chanber' (i.e they are repeled by€the electrons and compresed into the center.). €€>I assume the ions positive, are you sure they are repelled? €€Hum, good question. Maybe it had an ion cloud of fuel in the center not€electrons. I'll have to dig that paper up again and check. €€>>The fuels chosen€>>release the energy of the fusion reaction only in the kinetic energy of€the released waste particals. Those ions are going fast enough to blow€threw the electron cloud, and slam into voltage gradiant. They are€decelerated by the voltage. I.E. their kinetic energy is converted into a€electrical current in the reactor systems.€€>To keep the electrons in the center the voltage at the wall of the steel€sphere should be negative, the ions from the reaction are positive and will€thus be accelerated instead of decelerated. €€>It isn't necessary to decelerate the ions from the reaction, they will also€give an electrical current when they move fast. €€I think that if you run them threw a voltage gradiant they lose their€energy to it. If they don't lose kinetic energy somewhere, you don't get€electricity.€€>But now that I have some fague idea how a Bussard engine works, I see€that all calulations you asked for are meaningless because you transfer€the fusion energy first to electric energy by decelerating particles. Then€you use the electric energy to accelerate other (or the same) particles by€a lineac (I suppose).€€Not nessisarily. If you don't slow down the partical (excluding losses in€forceing their way past the compression fields) you can direct that as a€plasma thrust. Thats why I needed to find out what velocity they would€have, so I could work out the specific impulse and then the fuel mass€ratios.€€€The liniac would have certain advantages. But also extra complexity. So€unless it would give us some performance advantage I'll assume the drive€systems are using direct plasma thrust.€€>– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – €€>>??? In you first responce to my request for exaust velocity info you€said: €€>>>>For 2.058E14 Joule/kg this would give 2E7 m/s or 0.067c €€>>I responded that a 2e7m/s exaust velocity would translate to a specific€impulse of about 2,000,000. Which would mean the explorer class would€need carry 148 times its own weight in fuel to get down from .3 c. €€>>So what is f in your equation? How do you get a lower F for a fuel? Given€the 2.058E14 in:€€>>>>>Where f is c^2 / 2.058E14 = 439‚(c=3E8 m/s)€€>>It would imply you were using the same fuel, He3 as listed in my table.€But your geting wildly differnt fuel ratios with it. €€>>In short I am totally confused. Worse I'm completly confused in a€critical peace of info about the system!! I.E. the specific impluse of the€fusion drive.€€>>I assume we both still agree that converting the fuel energy to€electricity, and electricly accelerating a reaction mass will not help us€any?€€>I don't see why you need to be confused, I get a ratio of about 100 and you€get a ratio of about 150. I'm not sure what exactly causes this difference€but I assume it is caused somewhere in your calculation. €€I was runing it through the programs on the LIT servers computer center.€Do you know if their is an error in that software? €€>About "converting the fuel energy to electricity, and electricly€accelerating a reaction mass", it depends on several factors whether this€is true, but generaly it does not help us much. €€Pity we don't know of a way to make it work for us. €€>f = The fuel factor = (Total mass of the fuel) divided by (mass of the fuel€that can be converted to energy).€€Oh yeah. Thats why I never used that table. Strange number. How would I€find out what the fuel factor number for any of my fuels is? (Yeah ok you€added the equation below, but thats not a big help for someone tring to use€the table.) Or, why would you use the fuel factor in a table? €€>So say that you have a fusion fuel and are able to "squeeze" 2E14 joules€per kilogram out of it. Now all you have to do is determine the mass€equivalence of that amount of energy with E=m*c^2 (m=E/c^2). In this case€that makes m=2E14/9E16=0.00222 kg The total mass of the fuel was 1 kg€>So f=1/0.00222=450€>Then look in the table for a final velocity of 0.3c and we see a fuel to€ship ratio of 104.€>This table doesn't show anything about exhaust velocities but assumes€the best possible (which in this case is 0.06333c assuming 100%€efficiency). €€€>Timothy€€Thanks for the walk threw. It was helpful. €€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) €€––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––X–Sender: S9421793@mail.student.utwente.nl Mime–Version: 1.0€€€€===========================================================€Date: Sat, 11 May 1996 00:14:10 +0100€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€From: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)€Subject: Re: Physic help€€>>>As near as I can figure it, it seems like a simple system. An electron€cloud is compressed in the center of a hollow steel sphere by a voltage€charge on the sphere. The fuel ions are feed into the center of the electron€cloud, which forms the fuels containment 'chanber' (i.e they are repeled by€the electrons and compresed into the center.). €€>>I assume the ions positive, are you sure they are repelled? €€>Hum, good question. Maybe it had an ion cloud of fuel in the center not€electrons. I'll have to dig that paper up again and check. €€I indeed think that makes more sense, that way they positive ions are also€decelerated when they make it out of the inner ion–cloud. €€>>It isn't necessary to decelerate the ions from the reaction, they will€also give an electrical current when they move fast. €€>I think that if you run them threw a voltage gradiant they lose their€energy to it. If they don't lose kinetic energy somewhere, you don't get€electricity.€€Well in theory they create a charge difference, the inside of the core gets€more negative while the outside gets more positive. The idea is a bit like€charging a capacitor.€Of course while creating this potential difference, the particle€decelerates, and when decharging it, the particles decharge even more. But€anyway, I don't care, as long as we get energy from it. €€>The liniac would have certain advantages. But also extra complexity. So€unless it would give us some performance advantage I'll assume the drive€systems are using direct plasma thrust.€€But then we are back to the photons inside the engine, since we are talking€about a dense plasma, a lot of collisions will be going on and then you€readely get photons.€€– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – €€>I was runing it through the programs on the LIT servers computer center.€Do you know if their is an error in that software? €€I'm not sure, I only used it once. I think Steve did check several programs€once and found one to be not correct.€One thing I'm quite sure of, the program at the LIT server is probably made€for chemical fuels, in fusion fuels the energy density is much greater (and€thus a significant amount of mass is transferred to "mass–less" energy). €€>>f = The fuel factor = (Total mass of the fuel) divided by (mass of the€fuel that can be converted to energy).€€>Oh yeah. Thats why I never used that table. Strange number. How would I€find out what the fuel factor number for any of my fuels is? (Yeah ok you€added the equation below, but thats not a big help for someone tring to use€the table.) Or, why would you use the fuel factor in a table? €€I personally find it not a stange number at all, it shows very clear what€part of the initial mass can be converted to energy (for€anti–matter&matter mixture f=1, or said differently, all mass can be€converted to energy). I assumed that everyone who would read my€document did know about E=mc^2 and thus could calculate it (as you could€see the calculation was rather straight forward.€I already had added a table with fusion fuels and their f–ratios, I think it€isn't visible on the web at the moment since all my web–sites seem to€have collapsed lately.€Also originally I had a little (but unclear) example of how to calculate f,€maybe I should add it again.€€Tim€€€===========================================================€Date: Fri, 10 May 1996 21:12:23 –0700€From: Steve VanDevender To:€T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) 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€Subject: Re: Physic help€€Timothy van der Linden writes:€>>I was runing it through the programs on the LIT servers computer center.€Do you know if their is an error in that software? €€>I'm not sure, I only used it once. I think Steve did check several programs€once and found one to be not correct.€€I only recall checking the relativistic acceleration/time/distance€program, but it was incorrect.€€The correct equation to use is:€€t = c / a * acosh(1 + a * d / c^2)€€where€€a = acceleration (m/s^2)€d = distance (m)€t = time (s)€c = speed of light (m/s) (duh)€acosh = inverse hyperbolic cosine€€I recall the program in the LIT archives used alog (inverse common€logarithm) instead, apparently because it was translated from BASIC to C€by a rather poor automated translator.X–Sender:€S9421793@mail.student.utwente.nl Mime–Version: 1.0€€€€===========================================================€Date: Sat, 11 May 1996 16:18:33 +0100€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€From: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)€Subject: Re: Physic help€€>I recall the program in the LIT archives used alog (inverse common€logarithm) instead, apparently because it was translated from BASIC to C€by a rather poor automated translator.€€Steve, thank you for refreshing my mind. €€Kelly, maybe I'll take a look at the source code one of these days, what€was the name of the program you used?€€€Timothy€€€===========================================================€Date: Sun, 12 May 1996 23:25:49 –0400€From: KellySt@aol.com€To: T.L.G.vanderLinden@student.utwente.nl, 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€Subject: Re: Physic help€€The lit programs I used were the ones that calculated the specific impulse€when given a exaust velocity, and the ones for calculating the terminal€speed/fuel mass ration/specific impulse. Give 2 of the three and it€figures out the third. There were actually 2 of the latter programs (both€for staged and single stage craft) and the numbers were close. €€I don't see how energy content of the fuel would effect the calculations,€since that wasn't an imput. Only the exaust velocity or specific impulse. €€Kelly€€€===========================================================€Date: Sun, 12 May 1996 23:25:59 –0400€From: KellySt@aol.com€To: T.L.G.vanderLinden@student.utwente.nl, 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€Subject: Re: Physic help€€>>>f = The fuel factor = (Total mass of the fuel) divided by (mass of the €fuel€>>>that can be converted to energy).€€>>Oh yeah. Thats why I never used that table. Strange number. How would I€find out what the fuel factor number for any of my fuels is? (Yeah ok you€added the equation below, but thats not a big help for someone tring to €use€>>the table.) Or, why would you use the fuel factor in a table? €€>I personally find it not a stange number at all, it shows very clear what€part of the initial mass can be converted to energy (for€anti–matter&matter mixture f=1, or said differently, all mass can be€converted to energy). €€True, after you do the calculation, but initself not a very helpfull number. €In this case, its sort of an intermediate step in a calculation, rather than€a usefull result in itself. Also kind of confusing. €€>I assumed that everyone who would read my document did know about€E=mc^2 €and€>thus could calculate it (as you could see the calculation was rather€straight forward.€€Only after you remenber the units of measure that are to be used in the€equation. An exaple WITH UNITS does help a lot. €€Kelly€€€X–Sender: S9421793@mail.student.utwente.nl Mime–Version: 1.0€€