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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*: Correction*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Thu, 21 Mar 1996 01:56:03 +0100

Hello Rex, I wrote your letter during, recalculating my formulas, when I though to have finished the calculations, I forgot to remove some drag from the letter. Everything between the --- and === is still the same, the rest after the === is different at several points. Sorry, for the inconvenience, here the right text: ----------------------------------------------------------------------------- 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 and CORRECT 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. dumping unused fuel L[t]). However, this is only valid for bigger fuel-factors f>6.295 where the point of intersection between 11.1 and 11.2 is the minimum. But for smaller fuel-factors f<6.295 (when 11.2 starts much later than 11.1) this minimum doesn't exist anymore and is replaced by one with a different origin. This minimum originates from 11.1 only. (11.1 has a minimum from itself, 11.2 doesn't) The reason for this minimum I cannot explain yet (but hope to do tomorrow) >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. I can't comment right now but will do tomorrow (I hope) >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 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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