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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: Energy:momentum ratio*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Tue, 11 Jun 1996 00:06:42 +0100

To Steve, >If you want to use kinetic energy as a quantity in your relativistic >dynamics problems, be prepared for your analysis to get _a lot_ more >complicated. Yes, I assume you know that I and Rex independently have done such analysis's. We both have made documents that include some algebra that isn't always fun to look at. We also came to the conclusion that every fuel (combined with a Vend of the starship) has its own optimal Vexhaust. (More about this optimum in a few sentences) >The most unlikely result of your derivation is your claim that lower >exhaust velocities are better. Does that mean that zero exhaust >velocity is best, or is there instead an optimal velocity that is >non-zero? I think you misunderstood the meaning of these formulas. In essence these formulas are just showing p/E ratios, they tell little about how they should be use, indeed for self-powered spaceships their use it limited (will come back on that in a minute), but for beaming stations on Earth they may be of direct use (that comes in another couple of minutes). To answer you question about the optimum, "Yes there is an optimum bigger than zero (and less than c)". Let me quote a few lines from my HTML-document: (Keep in mind that this assumes a certain constant Vend of the starship) The minimum exists because on the one hand, the higher Vexh the worse the energy/momentum ratio of the exhaust mass or because of the dumping of excess mass. And on the other hand, for low Vexh more mass is needed to get the same momentum, this means that unused fuel is used as propellant mass (this could easely be changed by increasing the fuel-factor). The complete document (large) should be here (with Dave): http://165.254.130.92/LIT/calc/calc.html To add some extra confusion ;) or to make things more complete I like to make clear why for beaming mass would be cheaper (in terms of energy) than beaming photons. For self-powered starships low exhaust velocities mean enormous amounts of propulsion mass to get the same total momentum. As one could imagine, part of that mass needs to move with the starship during the acceleration phase. This isn't handy because it makes everything harder to acclerate. But when one beames from Earth (or another source), it doesn't matter much how much mass is beamed, since one doesn't have the accelerate-the-mass-with you-problem (any other terminology?) that I described above for the starship. For constant momentum (the momentum that the starship needs for a specific Vend), and decreasing Vbeam the energy levels decrease: (Vbeam=v=the velocity of the beamed mass) E = p c {(c-Sqrt[c^2-v^2])/v} where {(c-Sqrt[c^2-v^2])/v} decreases for decreasing v Of course decreasing v means increasing M (gamma*m*v=fixed) but as long as the total energy decreases we don't care. (Mass is cheaper than energy isn't it?) OK, hope that your not confused by all the increases and decreases (and that I made no mistake in it). Now a more practical problem, the beam velocity cannot stay low at all times, since the starship accelerates and the beamed mass needs to catch up with the ship. Well actually this is not really a problem, only the p/E ratio gets worse, but it will always be better than when one uses photons. (It is only the same when mass moving with Vbeam=c) Timothy

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