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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*Subject*: Re: Physic help*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Sat, 04 May 1996 11:29:42 +0100

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

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