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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, bmansur@oc.edu, DotarSojat@aol.com*Subject*: Re: Review of Calculations*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Wed, 20 Mar 1996 22:34:24 +0100

To Rex, >I have studied your document that describes some calculations >for relativistic self-fueled rockets. I think that you are the first, thanks. >I believe I have verified your physics and mathematics using >two other independent approaches, but I'm not completely sure; >I get slightly different numerical results from yours. We need >to resolve the difference in numbers. Yes. >In my opinion, the last four sentences in the first paragraph >are unnecessary; the integration is straightforward. I added those lines to make people with less mathematical background a bit more confident, but then again the math is so ugly (compared to F=m*a) that they are scared away even before reading those lines. >Also, >I didn't find the expression for power in your write-up, so >the involvement of acceleration is not made clear. The power used is shown implicitly in the formula of k[t] (3) k[t] = j[t] (g - 1) Energy mass needed from there substitute j[t] and then on of the two M[t]. Once you have done that multiply by c squared (E=mc^2) I will add this in the new release (formula 12.1, 13.1 and 12.2, 13.2). >Your >other results don't depend on time (time cancels out in the >derivation of the rocket equation), so absent the expression >for power your specification of acceleration is distracting. Indeed after the integration all time dependence vanishes (I even stated somewhere that the apparent dependence on "a" was because of formula substitution) I must say that before I started the calculation I hadn't really thought about the influence of "a", so your comment is partly justified. I hope that you now see why I used the acceleration in my formulas. Mainly to calculate the power (in my little Pascal program) using formulas defined in the document. (If you feel my comment isn't justified, then please correct me) >I had trouble with the "Energy/Mo" heading in your table of >results. I finally figured that it must be "total exhaust >energy (in joules) per kg of empty mass of the ship." Yes, Kelly also mentioned this when I sent those tables to our group a week ago. This ratio is the same as that what I defined by f. I thought people would understand that, I will try to make this clearer in the new release, thanks. >(Your >assumption of 100% conversion of nuclear energy to exhaust >energy needs to be acknowledged and justified.) Indeed, in my whole calculation I assumed that there where no energy losses. With losses I mean that energy which cannot be controlled. (eg. heat) So I'm not talking about fuel efficiency, since that can be incorporate in that f-ratio. >Your observation of a minimum in the ratio of total exhaust >energy to final vehicle energy is astute, but I don't follow >your explanation. I haven't tried to explain the minimum >myself (even though I had calculated its existence non-relat- >ivistically in the past), In the next few lines I try to restate a bit what I wrote in my "calc.txt", I hope it's clearer. Could you write me what part you can't follow? I found out that for high exhaust speed the total mass of the fuel goes up because of the extra kinetic energy (Energy is squared dependent on Vexh) and for lower speeds the mass goes up too, because of the waste of energy (ie. using unused fusion fuel as reaction mass). >but I can't believe that the rela- >tions ever call for using anti-matter for exhaust mass. This >may not be a point of interest to anyone other than a systems >analyst, however. Yes, this seems a bit strange. But when f=1 the optimum exhaust velocity is infinite, so you don't need to use anti-matter as exhaust mass then (because you will need all the energy you can get). I should not that this optimum exhaust velocity does decreases fast when f increases (for f=2 -> Vexh=0.866c) , and in practice might even be impossible. But the whole idea is that if you don't do that, you will accelerate slower. The explanation is simple, the extra energy you would need when increasing the exhaust velocity would be more than >Following is a comparison of your and my results: > > M[0]/Mo "Energy/Mo"(xE-16) > Vexh/c Vend/c Yours Mine Yours Mine >f=1 > 0.4 0.76 12.18 12.068 8.40 9.06 > 0.54 0.76 6.37 6.327 7.65 9.01 > 0.9 0.76 3.75 3.025 10.3 23.6 > 0.99 0.76 2.75 2.735 13.5 94.9 > 0.99 0.99 14.49 14.489 104. 738. >f=257 > 0.4 0.76 1.9E23 1.61E23 (didn't try) About the M[0]/Mo numbers, I don't know why, but indeed mine were wrong. About the Energy/Mo numbers, mine are wrong but the new numbers are all close to the old one, but the differ much from yours. >If you check your arithmetic and still disagree, let me know. >I'll send you a description of my analysis. For the Energy/Mo we do not agree so for this I need a futher description. (Energy/Mo is the same as Mk/Mo (formula 11)) I have figured out something new about the minima. The formula to calculate the minimum for low f has a completely different origin than for high f. Unfortunately this new minimum cannot be calculated algebraically, I'm trying to work out a computer program to do some calculations. (I'm not sure this latter makes some sense, it's hard to visualize this from words, I my HTML-release that I'm working on, I will add some graphs that would clarify some things) I will sent you my newer release of the "calc.txt" document. For the others, I will put it on my web-site in a while. Timothy

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