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*To*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Subject*: Re: New idea Laser launcher/scoop systems*From*: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39)*Date*: Mon, 11 Mar 1996 13:27:54 -0500*Cc*: 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

At 5:40 PM 3/11/96, Timothy van der Linden wrote: >>>Kelly, how did you calculate these numbers? >>> >>>Tim >> >>found the equation. >> >>A.1 The rocket equation >> >>The "rocket equation" tells you that >>the ratio of rocket mass when full of fuel to rocket mass after >>burning the fuel is: >> >> m(i) / m(f) = exp ( dV / v(e) ) > >Yups, I already found it when you pointed out where to look for it. Pity I didn't read that until after I sent the equation. Oh, well at least we are on the same equation. >=============================================================================== > >>>I already see what makes the difference (besides the fact that formula >>>doesn't take into account what the energy:mass ratio of the fuel is). For >>>slightly different exhaust speeds (around 0.08c) the ratios are optimal. I >>>must have overlooked that in Kellies table, where at the top this speed of >>>25M is used. I don't understand however why all the other numbers are shown >>>in the table. >> >>I wanted to show all mass ratio's with various specific impulse numbers. > >OK, unfortunately (as you stated below) the paper didn't show all there was, >your numbers only showed a decrease of mass for lower exhaust-velocities. >I found out that for high exhaust speed the mass goes up because of the >extra kinetic energy (could be overcome by beaming) and for lower speeds the >mass goes up too, because of the waste of energy (ie. using fusion fuel as >reaction mass). I'm not sure I follow this. By waste energy are you implying the same specific impulse ratio could be developed with lower fuel to reaction mass ratios? Given that the fusion reactions are listed as releasing all the fusion energy in the kinetic energy of there particals, I assumed that was about the best we could do. >So one needs to find the valley between the (ever rising) hills. >The formula to do this: > > BestG:=1+1/(f-1); (the relativistic gamma) > BestV:=c*Sqrt(Sqr(BestG)-1)/BestG (the accompanying velcity) > >The optimum exhaust-velocity is only dependent on the mass:energy ratio of >the fuel (thus not on the final velocity). > > f exhaust velocity (in c) >200 0.09987 >250 0.08935 >300 0.08158 >350 0.07554 >400 0.07067 >450 0.06663 >500 0.06321 > >(f is the mass:energy ratio (about 270 is the best for fusion untill now)) > >The next table shows the ship:fuel ratios needed to accelerate upto the >final velocities that vary horizontally. Vertically the energy:mass ratios >of the fuel are varied. >Every time the optimal exhaust speed is used (this can be looked up in the >previous table). > > End velocity --> > 0.10 0.20 0.30 0.40 0.50 0.60 0.70 >200 2.7 7.6 22.2 69.5 244 1032 5906 >250 3.1 9.7 31.9 114.6 467 2338 16422 >300 3.4 12.0 44.4 180.0 839 4896 41401 >350 3.8 14.6 60.2 272.7 1439 9662 96907 >400 4.1 17.6 79.8 401.4 2376 18191 213876 >450 4.5 21.0 104.1 577.2 3805 32958 449882 >500 4.9 24.7 133.8 813.9 5941 57820 908988 > ^ > +--- Energy:mass ratio of the (fusion) fuel > > >If anything is unclear, don't hesitate to ask. > >Timothy Mass energy ration of what? Kinetic energy to mass? Thermal energy to mass? etc... Your table apears to show that I could get lower ship to fuel mass ratios than I expected. Kelly ---------------------------------------------------------------------- Kelly Starks Internet: kgstar@most.fw.hac.com Sr. Systems Engineer Magnavox Electronic Systems Company (Magnavox URL: http://www.fw.hac.com/external.html) ----------------------------------------------------------------------

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