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#21
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Question on the space elevator
So it would take about 16 kg of fuel+oxidizer, give or take, to send 1
kg up the elevator. But you also have to carry the fuel to lift the fuel, and the fuel to lift the fuel to lift the fuel, and so on. Mathematically, you would have to carry e^16= 9 million kg of fuel+ox. (Factors of 2 or 1/2 at that point become very important.) Anyway, beaming the power becomes economical at that point. That number of 9 million kg seems a little large... we've sent much larger things out of the gravity well on much less fuel, using less efficient processes. |
#22
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Question on the space elevator
JRS: In article , seen in
news:sci.space.science, Gordon D. Pusch g_d_pusch_remove_underscores@xn et.com posted at Tue, 6 Apr 2004 21:08:29 : Your proposal is not _totally_ implausible. The energy required to climb a beanstalk is only a small fraction of the energy required to accelerate a payload into Low Earth Orbit; the fuel and oxygen tankage required would be large, but not prohibitively so. Earth radius is 7 Mm, GSO radius is 42 Mm, so in units of the Earth's radius 1 and 6. Potential energy in an inverse square field is inverse linear, so with it being zero at infinity, it is 1/6 at GSO and 1 on the ground, difference 5/6. But the energy required for LEO is half that for escape, so climbing the stalk to GSO requires 5/3 of the energy to LEO. "Centrifugal force" will supply a significant part of that energy (by lengthening the day), but not enough to leave only a small fraction. Much of the fuel and oxygen will need to be lifted, though there is the advantage that the exhaust need only carry a little energy other than that of height. -- © John Stockton, Surrey, UK. / © Web URL:http://www.merlyn.demon.co.uk/ - FAQish topics, acronyms, & links. Correct = 4-line sig. separator as above, a line precisely "-- " (SoRFC1036) Do not Mail News to me. Before a reply, quote with "" or " " (SoRFC1036) |
#23
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Question on the space elevator
The current answer to the last question isn't magnets but .. lasers.
Free Electron lasers beam power to the climber, which converts the energy into mechanical energy (wheels or treads). IIRC, a FEL has been designed that can do the job. Please give me a link to this design. If you are talking about simply a beamer that could perhaps be used in an unknown design that incorporates aiming (you know, to hit that elevator), suitable transfer medium, and receiver system, which can operate on these lurdicous lenghts, then dont bother. |
#24
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Question on the space elevator
Gordon D. Pusch wrote:
For any reasonable strength of material, the amount of additional energy that can be stored by pressurizing the tanks is negligible compared to the chemical energy stored in the propellants themselves. About all you will do Consider tanks that contain solid hydrogen instead of liquid hydrogen - or more simply, contain say supercoold hydrogen at twice the pressure while weighting no more than present ones. by pressurizing the tanks is to allow you to eliminate the mass of the turbopumps and drive turbines, which is likely to be marginal compared to the additional mass of high-pressure propellant tanks. Pressure-fed rockets But as you used much better materials, the mass of the tanks went down or at least didn't increase much. _may_ be justifiable on the basis of lower cost or higher reliability, but are =VERY= unlikely to provide significantly better performance than pump-fed rockets. But it is not really a question of replacing the pumps with a pressure-fed system, its a question of using better materials whichmeans that either you can leave the mass the same - while getting higher performance *or* reducing the mass of all compoents while still getting same performance. -- Gordon D. Pusch perl -e '$_ = \n"; s/NO\.//; s/SPAM\.//; print;' -- Sander +++ Out of cheese error +++ |
#25
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Question on the space elevator
There's a lot of traffic in this thread about powering the climber. Why
can't it simply have a diesel/gasoline engine with its own oxygen supply? just possibly something to this... Or run electrical cables up the elevator to power an electric motor? But this was a bad idea. Why make things more complicated than they need to be? Your proposal is not _totally_ implausible. The energy required to climb a beanstalk is only a small fraction of the energy required to accelerate a payload into Low Earth Orbit; the fuel and oxygen tankage required would be large, but not prohibitively so. I did a quick calculation assume modern diesels assuming a 10 tonne payload starting with 10 tonnes of fuel. From my matlab script: G=6.67300e-11; %gravitional constant Mc=10e3; %Mass of cargo, 10tonnes Me=5.95e24; %mass of earth e=0.5; %efficiency of typical 4 stroke Esp=40e6; %fuel energy per kg (assume air breathing) r0=12760e3/2; %start point (radius of earth) Mf=10e3; %mass of fuel Equating the energy required to get out of the gravity well to the energy available in the fuel tank: GMe(Mf+Mc)(1/r0-1/r1) = e Esp Mf This makes three assumptions, One is that we're air-breathing all the way, which might not be too unreasonable because the hardest part (where g is highest and the engine will be working its hardest) has plenty of air. Even at high altititudes a super-charger or similar technology could be applied to allow the engine to breath the air, only switching to a liquid oxygen tank, or oxygen-less fuel when truely in space. The second is that the mass of the fuel in the tank remains constant - obviously it would be used up. The third is that the 10tonnes of payload would of course include the heavy engine. I get from this about 1000km of altitude - only about 1/30th of the altitude needed. So perhaps not entirely unreasonable, but difficult nonetheless. Rest of matlab script: r1=1/(1/r0 - (Esp*e*Mf)/(G*Me*(Mf+Mc))); %calculate height altitude=(r1-r0)/1e6 %million meters (need about 30e6) |
#26
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Question on the space elevator
Makhno wrote:
This makes three assumptions, One is that we're air-breathing all the way, which might not be too unreasonable because the hardest part (where g is highest and the engine will be working its hardest) has plenty of air. Even at high altititudes a super-charger or similar technology could be applied to allow the engine to breath the air, only switching to a liquid oxygen tank, or oxygen-less fuel when truely in space. Could you be better off stopping part way up to fill the O2 tanks using those superchargers/whatever instead of lugging them up filled from the surface? After all, you could just hang there on the tether right? rick jones -- firebug n, the idiot who tosses a lit cigarette out his car window these opinions are mine, all mine; HP might not want them anyway... feel free to post, OR email to raj in cup.hp.com but NOT BOTH... |
#27
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Question on the space elevator
In article , Bob
Martin wrote: That number of 9 million kg seems a little large... we've sent much larger things out of the gravity well on much less fuel, using less efficient processes. The energetics are different. If you are halfway up the beanstalk, then burning a kg of fuel will give you the same amount of energy as it would at the base. think of the energy wasted as you drop the ash from your fuel over the side and let it just fall back to Earth. (If you could use it as a counterweight to pull you up, then you'd be better off.) In a rocket, the amount of kinetic energy you get from burning a certain amount of fuel (thrusting along the velocity vector) is proportional to your speed. Intuitively, this is because the fuel in your tanks has kinetic energy that you partially recover--if your exhaust velocity is equal to your speed (in some frame of reference) then the burned fuel ends up with zero kinetic energy and the rocket gets it all. -- David M. Palmer (formerly @clark.net, @ematic.com) |
#28
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Question on the space elevator
Dr John Stockton writes:
JRS: In article , seen in news:sci.space.science, Gordon D. Pusch g_d_pusch_remove_underscores@xn et.com posted at Tue, 6 Apr 2004 21:08:29 : Your proposal is not _totally_ implausible. The energy required to climb a beanstalk is only a small fraction of the energy required to accelerate a payload into Low Earth Orbit; the fuel and oxygen tankage required would be large, but not prohibitively so. Earth radius is 7 Mm, GSO radius is 42 Mm, so in units of the Earth's radius 1 and 6. Potential energy in an inverse square field is inverse linear, so with it being zero at infinity, it is 1/6 at GSO and 1 on the ground, difference 5/6. But the energy required for LEO is half that for escape, so climbing the stalk to GSO requires 5/3 of the energy to LEO. "Centrifugal force" will supply a significant part of that energy (by lengthening the day), but not enough to leave only a small fraction. Much of the fuel and oxygen will need to be lifted, though there is the advantage that the exhaust need only carry a little energy other than that of height. The significant difference is that, when climbing a beanstalk, one only has to supply the change in _energy_, not the change in _momentum_, which is is instead provided by the lateral force exerted by the beanstalk, which very efficiently extracts momentum from the angular momentum of the Earth. By contrast, much of the energy stored in the propellant of a rocket is uselessly wasted, because it goes into accelerating the _rocket exhaust_, rather than the _payload_... -- Gordon D. Pusch perl -e '$_ = \n"; s/NO\.//; s/SPAM\.//; print;' |
#29
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Question on the space elevator
Rick Jones writes:
Makhno wrote: This makes three assumptions, One is that we're air-breathing all the way, which might not be too unreasonable because the hardest part (where g is highest and the engine will be working its hardest) has plenty of air. Even at high altititudes a super-charger or similar technology could be applied to allow the engine to breath the air, only switching to a liquid oxygen tank, or oxygen-less fuel when truely in space. Could you be better off stopping part way up to fill the O2 tanks using those superchargers/whatever instead of lugging them up filled from the surface? After all, you could just hang there on the tether right? Please note that the density of the atmosphere is already negligible at a mere 150 km of altitude, less than 0.4% of the way to GSO, whereas the gravitational force at that altitude is still essentially the same as that at "sea level." It will =NOT= be possible to use "superchargers" to get oxygen from the air --- LOX tanks will be needed. -- Gordon D. Pusch perl -e '$_ = \n"; s/NO\.//; s/SPAM\.//; print;' |
#30
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Question on the space elevator
Gordon D. Pusch wrote:
Dr John Stockton writes: The significant difference is that, when climbing a beanstalk, one only has to supply the change in _energy_, not the change in _momentum_, which is is instead provided by the lateral force exerted by the beanstalk, which very efficiently extracts momentum from the angular momentum of the Earth. Does the beanstalk have to be straight up and down? If canted so as to lag the Earth's rotation (at least during a climb maneuver)how would that make a difference? |
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