===========================================================Date: Tue, 30 Apr 1996 23:06:41 -0400From: KellySt@aol.comTo: T.L.G.vanderLinden@student.utwente.nl Cc: kgstar@most.fw.hac.comSubject: Re: JPG or GIF>Hi Kelly,>I was looking at your pages on the Net once more. I noticed that some ofyour hand-drawn images are in JPEG format. I'm not sure if you know, butnormally simple line drawn images are best compressed (smaller size ANDbetter image quality) when using GIF format . Oh, thats right JPEG for images and GIF for line drawings. >Oh, the ray-traced image of the hab-ring is quite dark, maybe you canretrace it with an extra lightsource. Probably need to redo it with a lighter background too. The dark onedoesn't contrast well after its cut down to 256 colors. Thanks for the comments.Kelly===========================================================Date: Wed, 1 May 1996 12:11:05 -0500 (CDT) From: Kevin C Houston<hous0042@maroon.tc.umn.edu> To: interstellar drive group<David@interworld.com>, DotarSojat@aol.com, jim@bogie2.bio.purdue.edu, KellySt@aol.com, kgstar@most.fw.hac.com,lparker@destin.gulfnet.com, rddesign@wolfenet.com, sl0c8@cc.usu.edu,Steve VanDevender <stevev@efn.org>,T.L.G.vanderLinden@student.utwente.nl, zkulpa@zmit1.ippt.gov.pl Subject:Long range scans of T.C. I ;)Mime-Version: 1.0The first images from the fly-by recon mission are in planet I is definetlya water world, and may have life. check outhttp://www.tc.umn.edu/nlhome/m056/hous0042/taucet.html KevinPS you must have netscape 2.0 to see it spin. The animation sequence maybe useful to show mission plans.PPS I'm only sending this to show off my web skills. If we could havegotten this much info from a flyby recon mission, we wouldn't be sendinga humanFrom: David Levine <David@InterWorld.com> To:"'kgstar@most.fw.hac.com'" <kgstar@most.fw.hac.com> Subject: RE: MapOpps!===========================================================Date: Sat, 04 May 1996 11:29:42 +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 From:T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) Subject:Re: Physic helpI yesterday posted this by accident only to Kelly: - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - To Kelly,>Hi all you physics types. On my web page describing my explorer class'fusion reactors.>http://165.254.130.90:80/LIT/InterStellar/project/Explorer_Class/Bussard_Fus ion_systems.HTML>I have a table where I list various fusion fuel cycles. It lists theresulting energy in Mev. For those of us who arn't familure withtranslating Mev into anything, could someone tell me what the speed ofthe resulting particals is? Given that all the power in those reactions iscontained in the resulting kinetic energy of the particals. This should be asimple calculation, and would certainly be a nice addition to the table.Assuming of course your half of and undergrate physics degree wasn't over15 years stale!I'm a bit amazed you ask this because 2 paragraphs later you explain whatan electron Volt is.So I assume the problem is in the calculation of the velocities. I'll assume the velocities are relativistic, to make the formulas usable forall energies.Kinetic energy (in Joules) of a particle with mass m (in kg) moving withvelocity v:1---------------2gamma =	v	gamma >= 1SQRT(1 - ----)2c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.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 2.058E14 Joule/kg this would give 2E7 m/s or 0.067c 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===========================================================Date: Sun, 5 May 1996 16:28:42 -0400From: DotarSojat@aol.comTo: KellySt@aol.comCc: 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, sl0c8@cc.usu.edu,DotarSojat@aol.comSubject: Re: Physic helpOn 5/3/96 at 9:07 am EDT, Kelly Starks wrote- >I have a table where I list various fusion fuel cycles. It lists theresulting energy in Mev. For those of us who arn't familure withtranslating Mev into anything, could someone tell me what the speed ofthe resulting particals is? Yes, but I don't know what you can do with the answer. First of all, you have to coax two charged particles to penetrate eachother's Coulomb barriers with some amount of accelerator (bombardment)energy, a few MeV, before there can be a nuclear reaction (or use very hightemperature and pressure to cause a thermonuclear reaction). Thisbombardment energy conveys a motion to the center of mass of thereacting particles which must be added vectorially to the velocity (whichis in a random direc- tion) of each resulting particle with respect to theircenter of mass after the reaction. We ignore this center-of-mass motionin the following analysis.Let's define the following quantities-m = mass of lighter reaction product, in atomic mass units (e.g., mass of proton = 1.00813 amu)M = mass of heavier reaction product, in amu E = "resulting energy" in MeVof reaction products (this is with respect to their center of mass)= [(m1 + M1) - (m2 + M2)] 931 MeV(where 1 designates particles before the reaction and 2 designates particles after the reaction; for a mass of 1.00000 amu,mc^2 = 931 MeV)v = velocity of the lighter reaction product with respect to the center of massV = velocity of the heavier reaction product with respect to the center of massAfter the reaction, the momentums of the two particles are equal andopposite (and in a random direction), and the sum of the kin- etic energiesof the particles is equal to "the resulting energy," i.e.,m v = M V0.5 m v^2 + 0.5 M V^2 = E0.5 m v^2 + 0.5 M (m v/M)^2 = E0.5 m v^2 [1 + (m/M)] = E(v/c)^2 = 2[E/(931 MeV)]M/[m(M + m)]So, the velocity of the lighter particle, in units of c, is v/c = sqrt(2[E/(931 MeV)]M/[m(M + m)])and the velocity of the heavier particle, in units of c, is V/c = (m/M)v/c= sqrt(2[E/(931 MeV)]m/[M(M + m)])I hope this answers your question.(The velocity calculations can be made simpler without great loss ofaccuracy by considering only whole numbers of amu, e.g., 1 in- stead of1.00813 for a proton.)Regards, RexX-Sender: S9421793@mail.student.utwente.nl Mime-Version:1.0===========================================================Date: Sun, 05 May 1996 00:35:22 +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 From:T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden) Subject: Anew old memeberHi all,A few weeks ago, an old web-friend Emailed me. Some of the older LITmembers may still remember his name: Nick Tosh.He has been off-line for a long time (6 months) and likes to join theSD-group again. His connection is still not completely up and running, so Iwill do the initial correspondence for him and ask everybody to please addhis address to the Cc-list:101765.2200@compuserve.com (Nick Tosh)I assume he will introduce himself to the group when possible. Thanks, Timothy