================================================================== X-Authentication-Warning: darkwing.uoregon.edu: majordom set sender to owner-starship-design using -f From: DotarSojat@aol.com Date: Tue, 20 Aug 1996 01:53:50 -0400 To: starship-design@lists.uoregon.edu Subject: (Re:)^4 starship-design: The Size of the Problem Sender: owner- starship-design@darkwing.uoregon.edu Precedence: bulk Reply-To: DotarSojat@aol.com At 3:08 PM 8/4/96, I sent the memo "The Size of the Problem." At 2:47 PM 8/8/96, I wrote: >Note that 1 cubic mile of (sea) water contains enough deuterium to provide about 5,400 USEs from the fusion reaction(s)-- >D + D + D --> He4 + n + p + 21.6 MeV . At 9:23 AM 8/12/96, Kelly Starks wrote: >The problem you listed was the difficulty in manufacturing (not to mention storing) the "fuel". Given that mining and using more conventional fuels (like Lithium, duterium, etc..) don't have the heavy power costs and system complexity problems. Actually it wasn't the "difficulty in manufacturing" the antimat- ter I listed, it was merely the equivalent energy content in USEs (units equal to the total production of U.S. Electricity in 1987 = 51.47 kg antimatter). I assumed the manufacturing method (unspec- ified) of antimatter was 100 percent efficient. My 8/8 note gave the volume of water (1/5400 of a cu mi) that would have to be pro- cessed to extract the deuterium to give 1 USE via the specified fusion reaction. >Hell yes I'ld rather have hundreds of times the fuel weight! (Obviously due to the weight of the fuel you'ld need more than 250 times as much, but it would be a lot easier to carry!) Using antimatter fuel, the example mission to 8 lt-yr at 1-g con- tinuous acceleration/deceleration is calculated (see my 8/12 note) to require a mass ratio of at least 15.11 for the acceleration phase alone. Using deuterium fuel in the above fusion reaction (with a Timothy "f" factor of 261) for the same mission, the cal- culated required mass ratio for the acceleration phase is at least 3.84E12 kg D for each kg delivered to the peak velocity half way to the destination. (Note to Timothy: This mass ratio is for an "optimum" exhaust velocity of 0.09580 c. One of the reaction products is a neutron, which can't be used as reaction mass.) With 1.44E8 kg D per cubic mile of ocean, this mass ratio says you have to extract all the deuterium from more than 26,000 cubic miles of ocean to provide the fusion fuel to accelerate 1 kg at 1 g over 4 lt-yr, half way to the destination. But there is a dramatic reduction in required mass ratio as the g-level is reduced (and the trip time is increased). The follow- ing table shows that the required mass ratio is reduced by about 9 orders of magnitude while the trip time is increased by about a factor of five: accel/decel Uend trip time mass ratio(accel) (g) (lt-yr/yr) (yr) 1.0 5.000 4.480 3.842E12 0.5 2.881 6.897 4.869E09 0.2 1.520 11.692 3.659E06 0.1 0.994 16.991 5.929E04 0.05 0.672 24.401 2.675E03 0.0132 0.333 48.064 6.057E01 For the last line, the mass ratio for the full trip would only be about 3,600 kg D for each kg of final mass, so one would only have to process about 2.5E-5 cubic miles of water for each kg delivered to 8 lt-yr in a continuous-g trip time of about 48 yr (24 yr with 1-g accel/coast/1-g decel; see below). To de- liver Kelly's 500,000-ton-dry-weight Explorer- class starship on this mission would require extracting all the deuterium from about 12,500 cubic miles of water. This may be a more "down-to-Earth" measure of the size of the problem. Now we're addressing the "difficulty in manufacturing." >Thought. What is the relative weight of an Anti-matter tank to the weight of the anti-matter? Would its weight added to the ship start outweighing the advantages of the lighter fuel? If the deuterium for the fusion engine is carried in a tank made of an alloy of lithium and aluminum, the anti-hydrogen could be carried in a tank made of an alloy of anti-lithium and anti- aluminum, with a mass fraction similar to that of the deuterium tank. ;) ["mass fraction" = (mass of contents)/(sum of masses of tank and contents)] >Also you would need to carry (and store) a full round trip worth of anti- matter since you couldn't refuel in the target system. Kelly beware! I'm on Zenon's side regarding "kamikaze" missions. ;) >You'ld probably get shorter trip times if you had used the same power in higher boosts at the start and end of the trips, with a coast phase in the middle. Same power consumption, but higher average speed. Kelly, go to the blackboard and write 100 times "Power is the rate of use of energy." Whap, whap, whap. ;) If you replace the word "power" with "energy" you're right on. The table below shows the comparative trip times (to alpha Cen- tauri: distance = 4.35 lt-yr) for continuous accel/decel at g- levels below 1, with trip times for 1-g accel/coast/1-g decel with the same peak velocity (same energy requirement). The values of peak velocity, Energy/Mbo and mass ratio (accel alone) for each entry are given in the table in my note of 8/12 (and Kelly's response of 8/12). Accel/decel trip time | 1-g accel/decel w/coast (g) (yr) | trip time (yr) coast time (yr) 1.0 3.576 | 3.576 0.000 0.9 3.810 | 3.582 0.152 0.8 4.087 | 3.603 0.334 0.7 4.421 | 3.646 0.551 0.6 4.835 | 3.720 0.819 0.5 5.366 | 3.845 1.162 0.4 6.083 | 4.051 1.619 0.3 7.128 | 4.407 2.269 0.2 8.867 | 5.091 3.317 0.1 12.751 | 6.813 5.537 For acceleration periods that become a smaller fraction of the trip time, the average speed tends toward twice the average speed for continuous accel/decel, leading to cutting the trip time just in half. Note: For a reduction in energy requirement by a factor of 10 from the 1- g-all-the-way trip (an accel/decel g of about 0.17, from the 8/12 table), the trip time for 1 g with coast for the same energy is increased by less than a factor of 2. A factor of 10 reduction in energy requirement (for antimatter fuel) for less than a factor of 2 increase in time says the coast phase is worth considering. On the down side, however, are the requirements for heavier thruster (and power-system) weight for 1 g vs. reduced g, and increased complexity for artificial-gravity provisions during the coast time (the last column). >?? You still use goto's in your code! Bad Rex, Bad. Whap, whap, whap. ;) I can't imagine how to avoid them. Also, my Fortran compiler is dated May 1988 and doesn't allow ENDDO's, so it may not allow whatever gets around GOTO's. BTW, the code I provided in my 8/12 note had to be modified to cover fusion energy. But the size of the numbers above indicates to me that possibly nobody would be interested in using the modi- fied code (except to show I'm wrong). ;) RexX-Authentication-Warning: darkwing.uoregon.edu: majordom set sender to owner-starship-design using -f Date: Mon, 19 Aug 1996 23:37:54 -0700 From: Steve VanDevender To: DotarSojat@aol.com Cc: starship-design@lists.uoregon.edu Subject: (Re:)^4 starship-design: The Size of the Problem Sender: owner- starship-design@darkwing.uoregon.edu Precedence: bulk Reply-To: Steve VanDevender While I haven't had time to do a more complete write-up, I thought I would also mention an interesting corollary to Rex's analysis of the energy requirements of beaming power to accelerate a relativistic spacecraft. Not only is a large amount of power required, but the beaming equipment must be capable of (typically) output at a rate that can be over two orders of magnitude larger than is needed to accelerate the spacecraft at the start of the trip. I'm going to present some of the math without proof or demonstration at this time, but I'm sure it will be interesting fodder for discussion (either because Timothy or Rex will find any mistakes I might have made or because it shows another facet of difficulty to the problem of beaming power). I've recently been working on the physics of light signals between a "stationary" object and an object undergoing relativistic acceleration relative to it. Consider an object undergoing uniform accleration relative to itself; its frame position at its proper time t1 is: [ t x ] = [ 1/a * sinh(a * t1) 1/a * cosh(a * t1) ] At time t1 = 0, its position is [ 0 1/a ] (note again that for simplicity I am using geometrized units where c = 1 and acceleration has units 1/s (acceleration is fraction of c per unit time)). Consider an object beaming power to the object to accelerate it that also starts at that position, so that at time t=0 it coincides with the accelerated object at its proper time t1=0. If energy (light) from the beamer is emitted at time t, then the time t1 at which the accelerated object receives the energy is: t1 = -(ln(1 - a*t))/a Note that this is an asymptotic relationship -- as the frame time t of the beamer approaches 1/a, the object proper time t1 approaches infinity. This consequently means that the beamer must send energy for any possible trip within a time 1/a, no matter how far the acclerated object goes, and that the rate at which power is sent increases asymptotically to infinity as t approaches 1/a. The relative rate of time passage between the beamer and the accelerated object at frame time t has the relationship dt1 / dt = 1/(1 - a*t) In the case where a = 9.8 m/s^2 (or in geometrized units, 3.267e-8 c/s), the asymptote is reached within about year of beamer time (3.06e8 s). The good news is that to boost an object at 1 g up to its turnaround point and then provide deceleration power to its destination, you beam power for no more than two years, no matter how far away you send the object. The bad news is that at the turnaround point you are beaming some large multiple of the power needed to keep the object accelerating at 1 g at the beginning and end of the trip, because of the relative rate of time lapse between the beamer and the accelerated object. In fact, given the relationship between t1 and t, we can characterize just what this multiple is based on halfway trip time of the object. Solving t1 = -(ln(1 - a * t))/a for t, we get: t = (1 - e^(-a*t1))/a Substituting into 1/(1 - a*t), we get: dt1 / dt = e^(a*t1) In other words, the maximum power output at turnaround is exponentially related to trip time for the object. Perhaps the worse news is that because the relative rate of time lapse is asymptotic, even the schemes proposed for exponentially self-reproducing power generation equipment ultimately run up against the asymptotic limit; the asymptotic relationship always reaches a point where it is growing faster than the exponential function. Hopefully this makes sense to at least some of you. I hope to have time to cover more of the background soon, as I'm sure this is confusing without it. Timothy knows I've been working on analysis of accelerated objects in ================================================================== X-Authentication-Warning: darkwing.uoregon.edu: majordom set sender to owner-starship-design using -f X-Sender: kgstar@pophost.fw.hac.com Mime-Version: 1.0 Date: Tue, 20 Aug 1996 09:05:13 -0500 To: DotarSojat@aol.com From: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) Subject: (Re:)^4 starship-design: The Size of the Problem Cc: starship- design@lists.uoregon.edu Sender: owner-starship-design@darkwing.uoregon.edu Precedence: bulk Reply-To: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) At 1:53 AM 8/20/96, DotarSojat@aol.com wrote: >At 3:08 PM 8/4/96, I sent the memo "The Size of the Problem." >At 2:47 PM 8/8/96, I wrote: >>Note that 1 cubic mile of (sea) water contains enough deuterium to provide about 5,400 USEs from the fusion reaction(s)-- >>D + D + D --> He4 + n + p + 21.6 MeV . >At 9:23 AM 8/12/96, Kelly Starks wrote: >>The problem you listed was the difficulty in manufacturing (not to mention storing) the "fuel". Given that mining and using more conventional fuels (like Lithium, duterium, etc..) don't have the heavy power costs and system complexity problems. >Actually it wasn't the "difficulty in manufacturing" the antimat- ter I listed, it was merely the equivalent energy content in USEs (units equal to the total production of U.S. Electricity in 1987 = 51.47 kg antimatter). I assumed the manufacturing method (unspec- ified) of antimatter was 100 percent efficient. My 8/8 note gave the volume of water (1/5400 of a cu mi) that would have to be pro- cessed to extract the deuterium to give 1 USE via the specified fusion reaction. May I inquire what unspecified technology you expected to be able to generate anti-mater with 100% efficency? I wasn't particularly interested in the energy costs. Assuming high efficency systems that wouldn't melt the nieghborhood, power generation in a fixed installation is mainly a cost issue. We could build power plant hundreds thousands of times the size and capacity of the current ones, but what would be the point? MOre importantly to this project, what would be the cost!? >>Hell yes I'ld rather have hundreds of times the fuel weight! (Obviously due to the weight of the fuel you'ld need more than 250 times as much, but it would be a lot easier to carry!) >Using antimatter fuel, the example mission to 8 lt-yr at 1-g con- tinuous acceleration/deceleration is calculated (see my 8/12 note) to require a mass ratio of at least 15.11 for the acceleration phase alone. Using deuterium fuel in the above fusion reaction (with a Timothy "f" factor of 261) for the same mission, the cal- culated required mass ratio for the acceleration phase is at least 3.84E12 kg D for each kg delivered to the peak velocity half way to the destination. (Note to Timothy: This mass ratio is for an "optimum" exhaust velocity of 0.09580 c. One of the reaction products is a neutron, which can't be used as reaction mass.) With 1.44E8 kg D per cubic mile of ocean, this mass ratio says you have to extract all the deuterium from more than 26,000 cubic miles of ocean to provide the fusion fuel to accelerate 1 kg at 1 g over 4 lt-yr, half way to the destination. >But there is a dramatic reduction in required mass ratio as the g-level is reduced (and the trip time is increased). The follow- ing table shows that the required mass ratio is reduced by about 9 orders of magnitude while the trip time is increased by about a factor of five: >accel/decel Uend trip time mass ratio(accel) >(g) (lt-yr/yr) (yr) >1.0 5.000 4.480 3.842E12 >0.5 2.881 6.897 4.869E09 >0.2 1.520 11.692 3.659E06 >0.1 0.994 16.991 5.929E04 >0.05 0.672 24.401 2.675E03 >0.0132 0.333 48.064 6.057E01 >For the last line, the mass ratio for the full trip would only be about 3,600 kg D for each kg of final mass, so one would only have to process about 2.5E-5 cubic miles of water for each kg delivered to 8 lt-yr in a continuous-g trip time of about 48 yr (24 yr with 1-g accel/coast/1-g decel; see below). To de- liver Kelly's 500,000-ton-dry-weight Explorer- class starship on this mission would require extracting all the deuterium from about 12,500 cubic miles of water. >This may be a more "down-to-Earth" measure of the size of the problem. Now we're addressing the "difficulty in manufacturing." You and I are starting to get into and apples vrs oranges argument here. I've pretty well droped considering continuous G missions, or a Tau Ceti Mission, due to the extream power/fuel requirements. As your table above shows, you eiather wind up with unusably long flight times or rediculas fuel mass ratios. Note. I'm not clear if the fule ratios listed above are for acceleration/deceleration, or even if they are the fule weight on the ship vrs fuel needed to fuel the anti-mater generators. >>Thought. What is the relative weight of an Anti-matter tank to the weight of the anti-matter? Would its weight added to the ship start outweighing the advantages of the lighter fuel? >If the deuterium for the fusion engine is carried in a tank made of an alloy of lithium and aluminum, the anti-hydrogen could be carried in a tank made of an alloy of anti-lithium and anti- aluminum, with a mass fraction similar to that of the deuterium tank. ;) ["mass fraction" = (mass of contents)/(sum of masses of tank and contents)] I might have some technical issues with Lithium and aluminum tanks, I flat cant take seriously anti-Lithium and anti-aluminum! How does one attach and anti-matter tank to a ship? How do you manufacture it? And most importantly do you seriously expect to see any of this technology by 2050?! >>Also you would need to carry (and store) a full round trip worth of anti- matter since you couldn't refuel in the target system. >Kelly beware! I'm on Zenon's side regarding "kamikaze" missions. ;) A bizzar and politically unastute mindset to put it mildly! Oddly this issue may get public debate soon. I ran into someone on the sci.space.tech (or something board) that is working up a paper for presentation at the next Case for mars conference. He is going to propose that NASA consider one way missions. His analysis was that it was cheaper to send one exploration crew and yearly supply flights 30 years, than to cycle back crews every year or two. >>You'ld probably get shorter trip times if you had used the same power in higher boosts at the start and end of the trips, with a coast phase in the middle. Same power consumption, but higher average speed. >Kelly, go to the blackboard and write 100 times "Power is the rate of use of energy." Whap, whap, whap. ;) If you replace the word "power" with "energy" you're right on. The table below shows the comparative trip times (to alpha Cen- tauri: distance = 4.35 lt-yr) for continuous accel/decel at g- levels below 1, with trip times for 1-g accel/coast/1-g decel with the same peak velocity (same energy requirement). The values of peak velocity, Energy/Mbo and mass ratio (accel alone) for each entry are given in the table in my note of 8/12 (and Kelly's response of 8/12). >Accel/decel trip time | 1-g accel/decel w/coast >(g) (yr) | trip time (yr) coast time (yr) >1.0 3.576 | 3.576 0.000 >0.9 3.810 | 3.582 0.152 >0.8 4.087 | 3.603 0.334 >0.7 4.421 | 3.646 0.551 >0.6 4.835 | 3.720 0.819 >0.5 5.366 | 3.845 1.162 >0.4 6.083 | 4.051 1.619 >0.3 7.128 | 4.407 2.269 >0.2 8.867 | 5.091 3.317 >0.1 12.751 | 6.813 5.537 >For acceleration periods that become a smaller fraction of the trip time, the average speed tends toward twice the average speed for continuous accel/decel, leading to cutting the trip time just in half. >Note: For a reduction in energy requirement by a factor of 10 from the 1- g-all-the-way trip (an accel/decel g of about 0.17, from the 8/12 table), the trip time for 1 g with coast for the same energy is increased by less than a factor of 2. A factor of 10 reduction in energy requirement (for antimatter fuel) for less than a factor of 2 increase in time says the coast phase is worth considering. On the down side, however, are the requirements for heavier thruster (and power-system) weight for 1 g vs. reduced g, and increased complexity for artificial-gravity provisions during the coast time (the last column). Your right about the heavyier drive systems, but your stuck with the heavy artificial G systems regardless. As you've shown you cant sustain a continuous 1g during flight, and obviously need to stop once you reach the starsystem. So the only way to give the crew a 1 G environment is to make your own. Actually it doesn't really cost any extra mass to provide the 1G environment. The only exception might be if you shielding mass is spread out due to the larger centrafuge tourus. But give the fuel masses we'ld need anyway. Shielding may just be a repackaging of the fuel mass. >>?? You still use goto's in your code! Bad Rex, Bad. Whap, whap, whap. ;) >I can't imagine how to avoid them. Also, my Fortran compiler is dated May 1988 and doesn't allow ENDDO's, so it may not allow whatever gets around GOTO's. Strange. Structured extensions were added to FORTRAN by the late '70's and structured code was the standard I was using when I stoped programing profesionally in '86. (Did you get a really cheap compiler? ;) ) >BTW, the code I provided in my 8/12 note had to be modified to cover fusion energy. But the size of the numbers above indicates to me that possibly nobody would be interested in using the modi- fied code (except to show I'm wrong). ;) >Rex Kelly ------------------------------------------------------------------ ---- Kelly Starks Phone: (219) 429-7066 Fax: (219) 429-6859 Sr. Systems Engineer Mail Stop: 10-39 Hughes defense Communications 1010 Production Road, Fort Wayne, IN 46808-4106 Email: kgstar@most.fw.hac.com ------------------------------------------------------------------ ---- ================================================================== X-Authentication-Warning: darkwing.uoregon.edu: majordom set sender to owner-starship-design using -f X-Sender: kgstar@pophost.fw.hac.com Mime-Version: 1.0 Date: Wed, 21 Aug 1996 16:54:30 -0500 To: starship-design@lists.uoregon.edu From: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) Subject: starship-design: New draft of site Sender: owner-starship- design@darkwing.uoregon.edu Precedence: bulk Reply-To: kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39) I uploaded updates to the LIT beta site (Daves w/s). Marine now has a bunch of web links to interesting (I hope) sites. A few corrections here and there (like the correction to the Solar system Dev Banor. Argosey/tradewinds, M.A.R.S., and fuel sail now have token pages loaded for them. (Just what I could rattle off on the top of my head. I still didn't get a readable copy of Brains draft of his Argosey/tradwinds page, and since none of you forwarded one to me. I guess you didn't eiather. BRIAN!! Mail it to me on a floppy (I'll refund postage!!), ftp it to Daves, whatever. Needs some link fixes and I have to do the Newsletter function, but otherwise its about down from my end. Kelly ------------------------------------------------------------------ ---- Kelly Starks Phone: (219) 429-7066 Fax: (219) 429-6859 Sr. Systems Engineer Mail Stop: 10-39 Hughes defense Communications 1010 Production Road, Fort Wayne, IN 46808-4106 Email: kgstar@most.fw.hac.com ------------------------------------------------------------------ ---- ==================================================================