5 ms·
What is the possibility of using these for space launches? edit: obviously would have to be slower then these for the sake of any fragile cargo (including peop
by zoomzoom 16y ago
What is the possibility of using these for space launches?
edit: obviously would have to be slower then these for the sake of any fragile cargo (including people)...
- dasil003 16y agoHard to imagine anything useful withstanding that acceleration. You could make a longer launch tower, but how tall would it have to be?
- cperciva 16y agoFor humans who can survive an acceleration of about 3 g, you'd need a 1000 km long railgun in order to deliver enough energy. For cargo a railgun a few miles long would probably be enough, but designing a missile which can survive surface air densities while travelling at orbital velocities would be very difficult.
- stcredzero 16y agoLaunching bulk cargo would still be a very worthwhile thing to do. Solar power satellites, anyone? The main obstacle is the launch cost. If we could reduce the shipment of bulk items to orbit to freight rates, then solar power satellites would become a reality. The US military already has specs for electronic components that can withstand 100,000 gravities. (Yes, the number of zeroes is correct!) I bet we could develop lightweight thin-film solar panels that could be rolled up and packaged to survive 1000.
- SerpentJoe 16y agoWhy not aim for the horizon? That way your track can be on the ground, and even better your impulse is spent approaching orbital speed rather than struggling against gravity drag. Even traditional space rockets turn east as quickly as possible, and a (semi) ballistic launch can do it from the start. That said for human payloads you'd need a really long track, tens or hundreds of kilometers. I wouldn't want to live near the launch end. (edit: to get a useful amount of speed; still assuming rockets to get us the rest of the way)
- bpd1069 16y agoBetter yet - tunnel it underground.
- tsotha 16y agoNah. You'd be better off building it up the side of a mountain (or mountain range). You want the air to be as thin as possible when the projectile hits it going mach 25 (or whatever, depending on where it's going).
- bpd1069 16y agotunnel can be environmentally controlled, which is good to reduce the possibility of combustion (argon gas). The force against the rails can be mitigated by the constraints of the earth it self at tremendous pressures. The exit can be out of the side of the mountain, its somewhat arbitrary. More importantly the tunnel/launch tube should be at/near the equator. And yes, since the first half of the tunnel is sloping downwards you are using a large fraction of g to accelerate the slug/projectile, thus reducing the power requirements. Its an engineering problem, not a physics problem. Unfortunately a project of this scale is beyond the largest organizations that exist today, nation-states. Maybe later? When the day comes for the need of a super-structure to be built in orbit, this technology will likely be used. This or a space elevator. Getting far enough away from our own gravity well will free us from this rock.
- stcredzero 16y agoIt's feasible to build a structure about 10 miles high using even conventional construction techniques. (Which would allow for orbital velocities at under 500 gravities.) Some folks at JPL have determined that aircraft building materials make 60 mile high structures possible. I think our government should build such a launch accelerator in the desert somewhere. (Tilted at 45 degrees or so, to get a cargo drone out of the atmosphere and give it a big boost towards orbital velocity.) The ability to place bulk cargo into orbit would give the US a tremendous economic advantage. (Selling solar power satellites, colonization of Mars...)
- kgermino 16y agoWell lets see: Assuming: Launch Speed: 17,180 mph (Speed of Space Station) [1] Launch Acceleration: 25g (Based on my understanding of what humans can survive) EDIT [2] Just using that as a rough estimate and Wolfram Alpha we get 242,250 meters, or about 150 miles. http://www.wolframalpha.com/input/?i=17,239.2+mph+/25+gs http://www.wolframalpha.com/input/?i=17,239.2+mph+/25+gs http://www.wolframalpha.com/input/?i=31.434+seconds+at+17,239.2+mph+to+meters http://www.wolframalpha.com/input/?i=31.434+seconds+at+17,23... Please feel free to check my math and yell at me. [1] http://en.wikipedia.org/wiki/International_Space_Station http://en.wikipedia.org/wiki/International_Space_Station [2] cperciva [http://news.ycombinator.com/item?id=1993399 http://news.ycombinator.com/item?id=1993399] used 3g. Based on that you would have to multiply the launch distance by (25/3) and get 2,018,750 meters, about 1,250 miles. Again please correct me if I got something wrong.
- nkurz 16y agoIt's been a long time since I've thought about rail guns, but I think you might be off in your second equation. But seeing as my other comment today is about how I may not have fully recovered mentally from a debilitating disease, I have to doubt myself. Anyway, aren't you forgetting that most of the time is spent at a lower speed, so that you want to integrate the acceleration rather than simply multiplying by the escape velocity? d = 1/2 at^2, where t = escape velocity / 3 G? http://www.wolframalpha.com/input/?i=escape+velocity+%2F+3gs http://www.wolframalpha.com/input/?i=escape+velocity+%2F+3gs t = 6.3 min http://www.wolframalpha.com/input/?i=distance+travelled+in+6.3+min+at+3+gs+of+acceleration http://www.wolframalpha.com/input/?i=distance+travelled+in+6... d = 2100 km (1300 miles) While we're at it, I'm not sure what we really should be using for escape velocity. We should probably account for the deceleration after we leave the muzzle while still in the atmosphere. And if we presume a surface mounted rail gun, we'd probably want the escape velocity for a launch aimed just above the horizon, which involves more of this.[1] Or if we're somehow presuming vertical, we'd want the escape from that height. But if vertical, we'd have to account for the extra gravitational force on the passengers, unless this is already accounted for in the human limits. In any case, I'm not sure that the speed of the Space Station (or the surface velocity I used above) really makes sense here. [1] I vaguely recall that ignoring air resistance, the velocity is the same regardless of direction, but I don't have confidence in this, and my quick searching hasn't turned up anything definitive. Is this possibly right?
- tsotha 16y agoRail guns wouldn't be very useful for delicate things, but you could, in theory, use one to launch bulk supplies like fuel and water. The problem is the projectile comes out of the gun so fast most of the weight is going to be taken up by shielding. If you have to put up with a sucky mass factor you may as well build a rocket instead. A ram accelerator is probably more practical in any case.
- stcredzero 16y agoThe problem is the projectile comes out of the gun so fast most of the weight is going to be taken up by shielding. Just how long does the shielding have to withstand high temps? We already solve a much harder problem with reentry of orbital vehicles. The problem here is much simpler, as we don't actually want to shed most of the velocity. In fact, we want to lose as little of our kinetic energy as possible. If ablative shielding can handle reentry, it can handle launch. AFAIK, the mass ratio for reentry shielding is quite good.
- tsotha 16y agoThe difference is where you start. Coming back from orbit you're going to start your run at orbital speed in no air and gradually increase air density. You'll dissipate most of your energy by the time you reach the thick air of the lower atmosphere. The rail gun cargo is going in the opposite direction - you start in the thick air at a speed greater than orbital velocity. It's been years since I could grind through the numbers, but it makes a big difference
- stcredzero 16y agoYou'll dissipate most of your energy by the time you reach the thick air of the lower atmosphere. This speaks to my point. In launch accelerator mode, you want to dissipate as little energy as possible. In reentry mode, you want to dump the majority of a big honking chunk of kinetic energy. I'm sure there are all sorts of optimizations to consider. How about a mostly vertical trajectory, with most of the horizontal velocity provided by lasers heating a heat-exchanger? By keeping the trajectory mostly vertical, one can avoid most of the lower atmosphere by stationing the opening to the atmosphere miles above the surface. This also limits the downrange danger-zone for falling malfunctioning cargo pods. Additionally, it makes the cargo pods easier to target with the lasers, since the powered flight would be entirely in the upper atmosphere. It would also make the rocket nozzles easier to design.
- edj 16y agoThere appears to be potential. A quick google search turned up 'Launch to Space With an Electromagnetic Railgun' in IEEE Transactions on Magnetics. [1] "The extension of this technology [rail guns] to the muzzle velocities ( 7500 m/s) and energies ( 10 GJ) needed for the direct launch of payloads into orbit is very challenging, but may not be impossible." [1] http://research.lifeboat.com/ieee.em.pdf http://research.lifeboat.com/ieee.em.pdf
- hartror 16y agoNot much use on earth given the length of the rail required to launch at a reasonable g force but very practical on the moon where there is a lot more space and a lot less gravity (and no atmosphere to speak of).