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Magnetic Ring Could Launch Satellites, Weapons

MattSparkes writes, "A new study funded by the US Air Force has suggested a cheaper method of sending satellites (possibly missile weapons) into orbit. A 2-km-wide ring of superconducting magnets would contain and propel a payload, accelerating it over a period of hours, before suddenly flinging the satellite into space at 23 times the speed of sound. The satellites would be engineered to withstand the g-forces encountered (2,000 g), and be cased in an aerodynamic shell. A two-year study has been commisioned and will begin within a few weeks at LaunchPoint Technologies in Goleta, California." New Scientist points out that if such a launch ring were built, it would instantly become "one of the most important targets on the planet."

28 of 612 comments (clear)

  1. Re:"Moon is a Harsh Mistress" anybody?? by kayser_soze · · Score: 2, Informative

    No. I thought of the same thing too. Or perhaps the ribbons described in Frederik Pohl's Hechee novels.
    Good stuff. Glad to see someone else who enjoys old-school sci-fi. :-)

    [C]

  2. Here: by jbeaupre · · Score: 4, Informative
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  3. Re:"Moon is a Harsh Mistress" anybody?? by TubeSteak · · Score: 3, Informative
    They mention in TFA that "[M]ost have focused on straight tracks, which have to gather speed in one quick burst. Supplying the huge spike of energy needed for this method has proven difficult."

    But this quick burst seems to assume that the track is relatively short. Why not a longer track?
    Take a short track, connect the beginning to the end, and you now have a track of infinite length.

    So they are making a longer track.
    Which would then obviate the need for payloads or containers that could withstand such high gees (at least the angular ones).


    The reason the payload has to be built to withstand X,000 G's is because at some point or another, it is going to go off the track and run into a wall of air at very high speed.
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  4. Not a rail gun. by MoralHazard · · Score: 4, Informative

    Your lapse is forgivable, but only because the proliferation of terms like "Gauss gun", "rail gun", and "mass driver" in SF has overwhelmed their usage as technical terminology. But the point is, THIS IS NOT A RAIL GUN.

    A rail gun is a parallel, non-touching pair of conductive rails, joined at the back-end by a partial circuit capable of generating an extremely high current flow (amps) of electicity in a very, very short time. A conductive projectile is injected into the gap between the rails (so that it touches both rails at once), which completes the circuit. As current flows from one rail to the other, through the projectile, it generates a powerful magnetic field. The Lorentz force causes the projectile to be pushed toward the far end of the rails--the magnitude of the force depends on the current flow.

    Rail guns can achieve extremely high velocities, far higher than conventional explosive-charge guns. The velocity of a firearm projectile is limited by the velocity of the expanding explosive gasses that propel it out of the barrel; the gas velocity is in turn limited by the speed of sound in the gas medium, which has a physical upper limit for any type of explosive. Rail guns don't suffer from this limitation.

    I have seen references to a 'Gauss gun' which consists of a series of solenoids stationed along a tube barrel, timed to trigger so that a ferrous metal projectile will be pulled faster and faster down the barrel by each of the solenoids in turn. I don't know how valid this terminology is, though.

    1. Re:Not a rail gun. by cachorro · · Score: 2, Informative

      ...Rail guns don't suffer from this limitation...

      Actually, railguns have velocity limitations based on friction between the rails and a conducting solid armature, or drag in a conducting plasma armature. AFAIK, no railgun has ever exceeded the performance of a multistage light gas gun. The best shot I know of was 6 km/s which is certainly a sub-orbital velocity.

      Coilguns will not have this same problem.

  5. Re:Lost in space by megaditto · · Score: 5, Informative

    You are confusing pressure with acceleration. These are not the same.

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  6. Re:"Moon is a Harsh Mistress" anybody?? by mrchaotica · · Score: 2, Informative

    Who says they can't? It could just be that a straight-line version would be prohibitively expensive because instead of needing C magnets to span the circumference of the ring, they'd need N * C magnets to span the distance covered by the circumference times the number of revolutions.

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  7. Re:Anyone confirm this? by multiplexo · · Score: 2, Informative
    We've had laser guided artillery rounds since the 1970's. The 155mm Copperhead rounds have a target sensor and you had an forward observer with a laser designator to light up the targets. Some calculation is necessary, you have to make the calculations to get the round close to the target, but once you've done that the FO can illuminate the target and the round will home in on it, making it possible to use artillery to take out tanks.

    The laser designator for the Copperheads was quite large, the ones I saw were vehicle mounted. I would imagine that in the 20+ years since I saw them that they've gotten smaller and smarter.

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  8. a_c = - \omega^2 r by Kadin2048 · · Score: 5, Informative

    Except that the proposed design accelerates the payload around in a circle -- using magnets arranged inside a torus -- not a long straight runway. I doubt a linear runway would be practical; it would just be too long. The advantage of a torus is you can keep using the same magnets to accelerate the payload, over and over, until you've reached sufficient speed to let it fly.

    Unless the circle was ridiculously large (probably the size of a continent or better), you're not going to be able to get up to escape velocity before you'd (as a human being) would be crushed by the effects of the centripetal acceleration.

    I'm not going to do the math right now, but I'm pretty confident that of the 6,000 Gs they're quoting, most of them are in the radial direction and not in the tangential, so that even if you brought the payload up to speed slowly, you'd still be crushed. It would be just like being in a centrifuge.

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    1. Re:a_c = - \omega^2 r by timster · · Score: 1, Informative

      Well, all you need to do to fix that is to use a small black hole as your centripetal force. As all matter experiences gravitation equally, the body's structure wouldn't experience centrifugal stress.

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    2. Re:a_c = - \omega^2 r by radtea · · Score: 4, Informative

      I'm not going to do the math right now,

      The speed of sound at sea level is 330 m/s, and a = v*2/r, so at 23*330 = 7590 m/s you would need r ~ 600 km to get a under 10 g.

      Of course, there's going to be a bit of bump when the capsule hits the atmosphere, and there's also the bit of a trick about getting the thing oriented so the capsule if flung upward...

      As a satelite launcher this sounds like a great technology, although I'm not sure who would be "targeting" it or for what purpose...advertisers, maybe? Painting thier logos on it or something? Or some guy hiding in a cave someplace that we're supposed to be all afear'd of?

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    3. Re:a_c = - \omega^2 r by radtea · · Score: 4, Informative

      The advantages of being high up are... ...negligable. Realistically, you can only get a few kilometers up, unless you're proposing to build it in the Himalayas. It is well known from other mass-driver studies that the aerodynamic advantage of hitting 80 bar at Mach 23 are no big improvement over hitting 100 bar at Mach 23.

      The reason why I mentioned pointing it up is that there is a big advantage to passing through the atmosphere as quickly as possible. Firing a capsule out normal to the local vertical will result in minutes being spent in getting to the top of the atmosphere, by which time you will have lost most of the initial velocity, to say nothing of broken all the windows for kilometers around. If you do the math, it takes about 13 seconds to travel 100 km at Mach 23 (just under 8 km/s). So a 30 degree incline nearly doubles that (you get some benefit from the curvature of the Earth) and things get rapidly worse from there on.

      As the whole point of my calculation was to show how big the thing would have to be to keep the acceleration below 10 g there is no way a 30 degree incline is going to happen--you've have to have a curve so long that the top of it really would be above a significant fraction of the atmosphere.

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  9. Centripetal Acceleration by LeDopore · · Score: 2, Informative

    I think a lot of folks here are confused about the "2000 gs" part of this device. This acceleration is from the centripetal acceleration needed to keep the payload moving in a circular path.

    Here's the math:
    The acceleration A needed to keep something moving at speed V in a circle of radius R is V^2/R.

    A = (8 000 m/s)^2 / (1000 m) = (64 000 m/s/s) = 6 400 gs.

    TFA says "More than 2000 gs" - my guess is that this is a mixture of sloppy journalism, and maybe confusion over the minimum acceleration needed to get to escape velocity (about 5.5 km/s). If they did get their wires crossed and report the 8 km/s figure but the g force of getting to escape velocity, the needed A is:

    A = (5 500 m/s)^2 / (1000 m) = about (30 000 m/s/s) = 3 000 gs, so they're still wrong.

    Incidentally, I love the ring idea, but it could only ever launch pretty specialized cargo due to the g forces needed. What I'd love to see would be a linear accelerator which got a rocket up to about 3-4 km/s, then the rocket would take over. EM launching systems with reasonable length can be built for low speeds, and rockets have high efficiency only when they're already moving fast (otherwise, most kinetic energy goes into making the exhaust, and not the payload, go fast), so a switchover plan seems pretty natural (except that it demands all the infrastructure of a small EM launcher as well as all the problems of a chemical fuel rocket - although some of these problems are less of an issue if you can accelerate the rocket to faster than the fuel's exhaust velocity before it reaches the muzzle of the EM launcher - then your shiny equipmetn doesn't get burned.)

    My 2. Enjoy!

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    1. Re:Centripetal Acceleration by joto · · Score: 2, Informative

      Incidentally, I love the ring idea, but it could only ever launch pretty specialized cargo due to the g forces needed.

      That's ok. After all, it's not like we have a tendency to send ordinary items into space today either.

      and rockets have high efficiency only when they're already moving fast (otherwise, most kinetic energy goes into making the exhaust, and not the payload, go fast

      So, with your ideas of physics, newton's third law is no longer valid? At low speed the exhaust will receive higher force than the rocket, and at high speed the rocket will receive higher force than the exhaust. Please explain.

      although some of these problems are less of an issue if you can accelerate the rocket to faster than the fuel's exhaust velocity before it reaches the muzzle of the EM launcher - then your shiny equipmetn doesn't get burned.)

      Still going on about this? Sorry, but can you please go back and review your high-school physics? Using an EM launching system to assist a rocket is a good idea, for at least two obvious reasons (1: less fuel needed to be carried with the rocket, 2: rocket can be hurled past the lower atmosphere). But not because Newtons third law doesn't apply for rockets!

  10. Re:"Moon is a Harsh Mistress" anybody?? by saider · · Score: 3, Informative

    It is not about hypoxia, but rather the fact that you would have your soft tissue torn away from your stronger skeletal tissue. Most of your organs remain in place as long as the connecting tissue is intact.

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  11. Bad math? by Bender0x7D1 · · Score: 5, Informative

    Am I crazy, or did they get the math wrong in the article?

    The acceleration equation for circular motion is: a = v^2 / r

    We are given:

    Velocity: 10 kilometers/s

    Width of ring = 2 kilometers, so radius = 1 kilometer

    So:
    v = 10,000 m/s
    r = 1,000 m

    a = (10,000 m/s * 10,000 m/s) / (1,000 meters) = 100,000 m/s^2

    The acceleration due to gravity is about 10 m/s^2

    This gives: (100,000 m/s^2) / (10 m/s^2) = 10,000 g

    So it seems that their 2,000 g is way off. Even if we use 2 km for the radius it is still 5,000 g.

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    1. Re:Bad math? by doctor_nation · · Score: 5, Informative

      Your math is correct. I have an abstract from a presentation these guys gave last week and it lists the radial force at 20 MN (that's mega-Newtons) for a 200 kg projectile = 10,000 G. They don't list the acceleration in G anywhere so it's probably a New Scientist math error.

  12. Re:Sounds Good, except by NoData · · Score: 5, Informative
    I know there's a relationship between bird migration and magnetic fields, too, as a lot of them blindly smack into the brick walls at a local MRI center.


    Cute, but you gotta be kidding. I work with a 3T research MRI magnetic. Both the machine and the facility are heavily shielded, and the field drop-off is very steep. While the isocenter of the bore is at 3 Tesla (30,000 Gauss), the 5 Gauss line is only a few meters (about 5 in the axial direction, 3 in the radial direction) from the isocenter. By comparison, a kitchen magnet is maybe 100-250 Gauss.
  13. Re:"Moon is a Harsh Mistress" anybody?? by roystgnr · · Score: 2, Informative

    How long do you think that straight track would have to be to obviate the need for high-g payloads?

    Shorter than a ring with the same g requirements. If you have a ring of radius r and you want to launch a payload at velocity v, then the "g force" on the payload will be at least centrifugal acceleration, v^2/r. If you stretch that out into a straight track of length 2*pi*r and you want to launch a payload at velocity v, then you need acceleration a such that v = a*t and 2*pi*r = a*t^2/2 = v^2/2a, so you need a = v^2 / (4*pi*r), an order of magnitude less force. You can try to cheat by making the straight track length 2*r instead, so it wouldn't just be as long as the ring it would fit inside the ring entirely, and the straight track would still have lower acceleration requirements.

    The only reason I can see for using a circular track is to cut the power requirements - that centrifugal acceleration is all perpendicular to your velocity, so it doesn't directly cost you any energy. With a linear track every bit of acceleration costs power, and trying to add 20,000 m/s^2 to an 8,000 m/s payload should cost you at least 160 megawatts per kilogram. It might be nice to add that kinetic energy more gradually.

    Of course, maybe I'm just doing my math wrong. v^2/r at v = 7,800 m/s and r = 1,000 m gives you over 6,000 g's, not 2,000. Did I get something wrong or did the article?

  14. Re:"Moon is a Harsh Mistress" anybody?? by dgatwood · · Score: 4, Informative

    Peak for shuttle launch is 3Gs, and for Apollo reentry, exceeded 7Gs (source paper with cited sources). For a launch abort on the Apollo design, stress would have exceeded 16Gs, and this was deemed uncomfortable, but survivable (albeit with an assumed inability to operate controls during the process). (source LBJ Space Center.)

    So limiting it to 2Gs of total stress is very arbitrary and unnecessary.

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  15. Fast dead mass is still REALLY useful if its cheap by Big_Breaker · · Score: 4, Informative

    This ring could fling mass up to a skyhook to recharge its orbit. Imagine a LEO skyhook that catches dozens of dead weight shots from this gun and uses that momentum to promote its orbit to a highly eccentric one. Then the satellite can exchange this orbit potential with a target at its low altitude point through a tether or skyhook style method. The target could be a large satellite in LEO or even a suborbital payload. Once the potential is transfered the target can have its orbit promoted to GEO or other significant altitude.

    This method saves a lot of reaction mass in a heavy lifter because you can aim for a high alitutde but a suborbital trajectory. IE it's easier to shoot straight up than curve towards an orbital path at sufficient speed. For instance the X prize is all about sub-orbital. LEO is much harder and GEO is even harder still.

  16. Re:"Moon is a Harsh Mistress" anybody?? by superflyguy · · Score: 2, Informative

    drag resistance in fluid varies as a cube of the velocity, so twice the velocity is 8 times the air resistance: 2.3 times the velocity is 12.167 times the air resistance. It's more than an order of magnitude more air resistance, and building missiles to travel 10 times the speedo of sound is not an easy task.

  17. Re:I've already covered that. by Anonymous Coward · · Score: 1, Informative

    A Copperhead is fired using the same powder charge as a standard artillery round.

    I'm an ex-Artillery NCO.

  18. Re:"Moon is a Harsh Mistress" anybody?? by wishmechaos · · Score: 2, Informative

    Consider you're accelerating horizontally until you reach sufficient speed. You'd just have 1g pulling you downwards (or upwards, if you're upside-down) and 2g pulling you sideways (centripetal or lineal acceleration). That adds up to perceived 3g.

  19. Re:"Moon is a Harsh Mistress" anybody?? by Rei · · Score: 4, Informative

    I was reminded of Gerald Bull, one of the great "mad scientists" of our day, and Project HARP. :) Check out the plume leaving the barrel of their research gun. That had to be quite something to see in person.

    Of modern ballistic launch mechanisms, there are lots of neat options ranging from light gas guns to ram accelerators. I also find the concept of ballistically-launched scramjets to be pretty nifty. :)

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  20. Re:"Moon is a Harsh Mistress" anybody?? by 246o1 · · Score: 2, Informative

    I don't think that's how it works. Gravity should add like vectors, so 1g down and 2g horizontal should provide sqrt(1^2+2^2) or (sqrt5)g in a down-ish horizontal fashion.

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  21. Re:More math... by doctor_nation · · Score: 2, Informative

    This was a point brought up at the presentation. One of the linear accelerator guys was pretty sure that the struts holding the tracks in place would be transmitting huge amounts of energy, thereby heating the super-conducting magnets and possibly causing the struts to fail. The Launchpoint guy was sure that they had looked at the problem thoroughly though and that there wouldn't be an issue. Time will tell on that one...

  22. Re:"Moon is a Harsh Mistress" anybody?? by Unique2 · · Score: 2, Informative

    Wikipedia has a snippet about an accidental human exposure to near vacuum. "his last conscious memory was of the water on his tongue beginning to boil" ... gnarly.

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