16 thoughts on “High-Speed Train”

  1. One power outage and the whole thing comes crashing to the ground.

    I don’t think building a structure three times the height of mount Everest is the answer we’re looking for.

    But I like the out of the box thinking. I still think ballooons are more viable.

  4. A little math, based on “accelerate it for five straight minutes to speeds up to 9.1 kilometres per second” yields a tunnel 13650 km long, basically 1/3 of the way around the Earth. That is one very long superconducting cable in the tunnel carrying 20 million Amps, and another 13650 km long cable carrying 200 million Amps underground for the length of the tunnel. The first solar storm after this is put up will induce huge currents in the anchor cables, probably whipping them around the way an aurora moves.

    2. I think your math must be in error (perhaps by 9.806?)

      3 G’s gets you to 9.1 km/sec in 311 seconds over an 885 miles track.

      MfK suggested that California would start building it in the middle, and oddly that’s exactly the way to build it. First off, build 156 miles of track that ends by climbing a mountain to 15,000 feet, with a final angle of 20 degrees to vertical.

      Run the acceleration at 17 G’s for 56 seconds, which average people can tolerate horizontally for several minutes, and you exit the track at 9.4 km/sec (bumped from 9.1 km/sec to make up for drag losses) with a vertical velocity of 10,600 feet per second (exiting the bulk of the atmosphere is going to happen very quickly). Guessing at their drag numbers that produced a 3G decel at 65,000 feet, the ship would probably have an initial decel of -27Gs, which is survivable but not recommended.

      So first off, double the mass of the craft without changing its frontal cross section, or cut the frontal cross section in half to get the decel down to -14 Gs, very close to what ordinary people can take in the eyes-out direction (12 Gs). Of that added mass, much of it could be solid rocket motors with a linearly tapering thrust that falls to near zero in about 5 seconds, knocking as much off the exit decel as the engineers find feasible or cost effective.

      Ignoring that enhancement, you start out at 15,000 feet and -13.8 G’s.
      At T+1 second you’re at 25,500 feet facing 9.4 G’s.
      T+2 you’re at 36,000 feet and 6.3 G’s
      T+3 you’re at 46,200 feet and 4.1 G’s
      T+4 is 56,400 feet and 2.6 G’s
      T+5 is 66,000 feet and 1.5 G’s
      T+6 is 76,000 feet and 0.8 G’s

      So with just 166 miles of track and no outrageous vertical supports, you’ve got a workable (though unpleasant) launch system. If the price was linear with track length it should cost about $10 blllion instead of their $60, but since it doesn’t have the vertical supports it would probably be cheaper still.

      Once it’s operational for cargo and really ballsy crew, start extending the track in both directions. Along land it’s reducing the constant 17G acceleration, falling to 3 G’s with a track length of a thousand miles. Extending the launch end drops the initial deceleration numbers by altitude (my above table should give an approximate idea), reducing the need for compensating solid rocket motors. Making the vehicle more like a rod of depleted uranium (narrow and dense) would also decrease the drag problem, and since the vehicle is launched by a ground based system instead of a rocket, the cost penalties aren’t nearly as harsh.

      1. I’m not sure how much good putting a bend in the track is going to do. With an exit velocity of 9.4 km/sec and an angle of 20 degrees you have a horizontal component of 8.8 km/sec and a vertical component of 3.2 km/sec. If you assume you started your acceleration horizontally and added the vertical component at the end (ski jump, like Fireball XL5 or When Worlds Collide) you need about 20 seconds at 16 g’s. The ski jump ends up being 175 km long, 33 km high. That doesn’t seem doable–you’re going to have to build the track in a straight line.

      2. Good catch. Unless the launch starts dozens of miles underground, it couldn’t produce that vertical velocity while staying within G limits.

        The only solution that comes to mind is to reduce the launch angle by a large amount. If we up the muzzle velocity to 10.5 km/sec and exit at 5 degrees from the horizontal, the vertical velocity is only 916 m/sec (3,000 fps). Across the 15,000 foot climb the astronaut would pull 9.4 G’s down in their seat for 10.5 seconds. Going to 20,000 feet would reduce that to 7 G’s for 13.3 seconds.

        But it would take 30 seconds to get to 80,000 feet and the muzzle velocity had to go to 10.5 km/sec. But if the craft kept pulling vertical G’s after launch (a hypersonic waverider with a L/D ratio of perhaps 5) the situation might improve. If it could pull 4 G’s it would get to 80,000 feet in just 18 seconds with the vertical angle climbing from the intial 5 to 8 degrees. The penalty is about one extra G of decel due to the increased drag (depending on the L/D ratio and how many vertical G’s you go for).

        This suggest that if you can compensate for the high drag in the lower atmosphere for 10 or 20 seconds, the cheapest approach might be to let the craft gain the vertical velocity mostly on its own through lift. The waverider design problem is also simplified because the vehicle only goes through a vary narrow range of mach numbers because it’s not slowly accelerated in the atmosphere, it’s shot out of a giant maglev gun. And once you’ve got the maglev gun you can shoot aerodynamic waverider test models all day long. 🙂

        1. You can also extend the track from the other end. Dani Eder a few decades back was looking at a Space Bridge (I think that’s what he called it)–light gas gun on the earth’s surface, tether in orbit. Shoot packets out to the tether, grow the tether so that the lower end moves slower and slower, making it easier and easier to get to. If you reduce the velocity needed to dock with the tether by 10% you reduce the acceleration required over a fixed length track by 20% (well, nearly).

      3. I think your math must be in error (perhaps by 9.806?)

        3 G’s gets you to 9.1 km/sec in 311 seconds over an 885 miles track.

        My math was indeed in error. I hate that when that happens. It should have read 1365 km. (Slightly different from your result as I used 300 seconds). In any event, that is more than 1/30th of the way around the world, and the tethering cables are still varying from zero to 19km long at the end. Large varying electric currents would still be applied to the superconducting cables and the tethers in every solar storm, whipping the train tracks around like a flag in a breeze.

      4. Yes, I think the seriously underestimate what high winds do to cable stabilized structures. Even buildings sway in the wind, and they are vastly beefier and shorter.

        I think the lack of rigity might be a fatal flaw in the scheme without active compensation, such as the inner track aligning itself relative to the support frames by laser beam to ensure straightness.

        As an example of how bad the problem is, suppose you’re zipping down the end of the track at 9 km/sec. There’s a local wind that shifts the middle of a 1 kilometer section by 1 meter. You’ll cross the half-kilometer to the maximum deviation in 55.5 milliseconds, inducing a 66 G sideways acceleration that would be like a high-speed side impact in a car. I think it equates to hitting the pavement from six stories up. It would probably be fatal, and of course if the wind produced a 10 meter deviation it would be a 660 G acceleration.

        But the track is to be stabilized by long tethers. At 65,000 feet, assuming they formed an equilateral triangle, the one meter deviation would represent a 0.001% elongation. A 5 degree F temperature change in a steel cable of that length would produce the 1 meter deviation, as side winds would undoubtedly produce severe changes.

        So the system has to have vastly greater rigidity, or active controls.

  5. Some resemblance to a Lofstrom loop, but a different suspension method, and independent propulsion.

    …run a superconducting cable through the ground beneath the ascending maglev tunnel carrying 200 million amperes and cable in the launch tube itself bearing 20 million amperes…

    Doesn’t each of those cables run one way? I’m thinking 63.2MA going around a loop, one way along the ground, back through the launch cable.

  6. 19-kilometres high is too low for 32,000-kilometres-per-hour vehicle.
    Even if on top of mt Everest [8850 meters high] it’s still to low 8.8 km/sec or Mach 25.
    Didn’t the blackbird fly at around this elevation with max mach of 3.2 and required powerful engines and got hot from the drag. hmm:
    Maximum speed: Mach 3.3 (2,200+ mph, 3,530+ km/h, 1,900+ knots) at 80,000 ft (24,000 m).
    Service ceiling: 85,000 ft (25,900 m)
    http://en.wikipedia.org/wiki/SR-71_Blackbird
    “In addition, cruising at Mach 3.2 would heat the aircraft’s external surface well above 500 °F (260 °C) and the inside of the windshield to 250 °F (120 °C), so a robust coolant system was vital.”

    Or: “The initial version, the X-43A, was designed to operate at speeds greater than Mach 7 (4,700 mph; 7,600 km/h) at altitudes of 30,000 m or more.
    “The X-43A research vehicle was boosted to 95,000 feet for a brief preprogrammed engine burn at nearly Mach 7, or seven times the speed of sound. During its third and final flight – at nearly Mach 10 – the X-43A research vehicle flew at approximately 7,000 mph at 110,000 feet altitude, setting the current world speed record for an air-breathing vehicle. ”
    http://www.nasa.gov/centers/dryden/news/FactSheets/FS-040-DFRC.html

    So seems if want to go faster than mach 10, it should over 30 Km high.

  7. “I wonder how they came up with the sixty-billion number? It would be a lot better deal than SLS.”

    SLS gets you campaign contribution- it’s priceless.
    The train wouldn’t work, but suppose it did, would worth 60 billion.
    In others words, it’s built and it works, would anyone buy it for 60 billion.
    Lot’s question like how much does cost to maintain, how payload, etc.
    Let say maintenance cost $1 per year and you can lift 10 tons of payload. Included in 60 billion is one “train car” or whatever is used. But the land is not included say that is an addition cost.
    So question is how payloads would you need to sell per year and at what price per lb to make back your 60 billion?
    Suppose you had 100 tons per year. 200,000 lb at $1000 per lb is $200 million per year- requires 300 years. 1000 tons- 30 years. 10,000 tons per year it requires 3 years.
    Musk would bury you.

    Try something different. Suppose you were a rocket maker and you could lease a launch pad- say for 10 years [or longer lease if you want].
    The question is how would it be worth per launch using launch pad?
    Not included in price per launch is whatever infrastructural is needed to process payload and rocket integration- suppose typical prices. And one has good road access, rail, dock, and/or airport near site.
    What you buying is use of pad and range and tracking.

    Suppose with some rocket from a pad it takes 9.5 km/sec of delta-v to reach LEO. And we call that pad A.
    So Pad A is worth x amount dollars to use.
    Pad B requires instead of 9.5 km/sec, for same rocket require 9 km/sec.
    If for whatever reason Pad B require .5 km/sec less than Pad A how more is it worth per launch?
    Next suppose Pad C required 8 km/sec of delta-v to get to LEO- how much more would this be worth?
    And finally Pad D required 6 km/sec to reach LEO- how would a rocket maker pay per launch?

    1. Just don’t sign that maintenance agreement with Patentes Talgo S.A. The maintenance alone will come out to 6 billion/year, and you will need another 30 billion up front for the vehicle check-out building.

    2. At 1000 tons/year a “train to orbit” like this doesn’t make much sense. At 100,000 tons/year, making multiple launches an hour, well within the limits of the system if it has enough electrical feed, the economics shift.

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