Diff for "LaunchLoopFAQ"

Launch Loop Frequently Asked Questions

Q1: The rotor is made of magnetic, at least somewhat conductive metal which passes repeatedly through a very large magnetic field which lifts it and curves it, and it does this at extremely high speeds (14 km/s). This can be expected to generate at least some eddy currents. Would the energy loss and rotor heating due to this effect be prohibitive to a launch loop? How hot does it get?

A1: Eddy currents occur when there are changes in the magnetic field that pass through a conductor. Provided the lift magnetic field is sufficiently uniform parallel to the motion of the rotor and changes slowly over distance, the currents will be small. However, vehicle acceleration puts very high eddy currents into the rotor - the 150,000 Newton force from a 5 ton payload accelerating at 3 gees heats the rotor by 85K as it passes. If 80 such payloads are launched per hour (averaging one every 45 seconds), the rotor heats up to 900 Kelvin, about 620 Celsius or 1150 degrees Fahrenheit. It cools by black body radiation to the sheath and from there to the surroundings.

Q2: If the rotor is constructed so that eddy currents can't occur, how can acceleration of the vehicle be achieved?

A2: Eddy currents are minimized when field changes are small and spread out over a very large distance. The turnarounds are almost 50 kilometers in radius, for example. The vehicle magnets generate rapidly varying magnetic fields at the 10s of centimeters scale, which induces much larger eddy currents and lots of drag. The vehicle magnets have an attractive "DC" component working against the repulsive "AC" component, increasing the repulsive AC forces and drag. The drag may be further increased by making a more complex rotor cross section that resonates electrically in synchronization with the passage of vehicle magnet sections - this may be used to control acceleration versus speed.

Q3: Do the catenary-like tensile cables give sufficient lateral stability to the structure?

The tensile cables give a static pull on the structure sufficient to curve it down to the horizontal, as well as help keep it pointed in the right direction. The cables allow lateral forces to be transmitted from spooling motors on the surface. These may take many seconds to arrive, but they can be used to correct long-wavelength low-frequency motions that exceed the capabilities of moving counterweights near the track.

Q4: Is there an issue with the vehicle balancing on top of the rotor like that?

The centre of mass of the vehicle is maintained near the cable, and the vehicle attitude can be controlled with attitude jets and/or momentum wheels.

Q5: If the rotor was to escape containment, can the structure survive the fall from 80km?

Some portions of the structure will be destroyed. Much of it can be connected to parachutes. There will need to be re-entry vehicles for station crew.

Q6: How much energy is stored in the rotor at full speed? How big an explosion would containment loss be?

A6: The rotor is designed to not be lost. The magnetic levitation is massively redundant, sections can fail completely without any problems as neighbouring levitation sections just work slightly harder to compensate.

The total energy in the rotor is huge, equivalent to a several hundred kiloton bomb. However, the round trip circuit for the rotor is nearly 7 minutes; it will not be released all at once. Perhaps the best way to release the energy is to use the turnarounds to fan the rotor outwards and downwards into the ocean, vaporizing the rotor and a lot of sea water.

The rotor that escape horizontally are moving faster than Earth escape velocity - they will go into long orbits around the sun. The Earth is a pretty small target relative to the radius of its orbit - the fragments will eventually impact the Earth, but it may be hundreds of millions of years before they get close enough to do so. They will be radar trackable, so there is plenty of time to find and collect them.

However, the same complex structure that amplifies vehicle drag may make the rotor fragment into small pieces when it encounters high gee forces and lateral stresses, which will be much higher during a reentry event than during normal operation. The small pieces will have a high surface-to-mass ratio, and will either burn up or fall relatively gently.

Q7: During startup in particular, as well as shutdown, will the cable go through resonances that may potentially damage it?

The rotor is actively stabilised, so will not go through resonances, as it is damped at all times.

Q8: Why is there a loading dock at 80km altitude, rather than at the ground?

Rrom the 1985 paper: "The long elevators to the stations are supported by pulleys from the anchor cables. The vehicles are brought up these cables rather than up the west incline to simplify the spacing controllers on the incline. Other benefits of this approach are minimized incline weight, shorter upward transit times, and less likelihood of sheath damage."

Q9: What's the difference between the $10 billion version and the $30 billion version? Could there be an upgrade?

They're mostly the same. The cheaper version has a quicker payback on investment, so the launch costs are set at $300/kg, and a much smaller power generation capacity, which limits the launch rate. (After the first year, the costs should go down significantly, once the hardware is paid off.) It also needs less powerful linear motors due to the lack of available power for them. The $10 billion version might be upgradeable, but probably there would need to be shut down for a few months to add sections to the motors.

Q10: Would the rotor fry due to runaway erosion due to the extremely high temperatures reached in the near vacuum of the sheath?

The rotor would be coated with a coating that prevents significant erosion, probably carbon nanotube or graphene. With an atomic weight of 12, a carbon atom moving at 14km per second has an energy of 14 electron volts. A glancing blow from this atom on the sheath will deposit some energy, perhaps kicking loose two carbon atoms, which could kick loose 4 when they hit the rotor, etc. This can lead to a "hypervelocity spalling cascade".

If the coating on the rotor and sheath is atomically smooth, the energy lost per collision will be small. If the coating is strong, it will absorb the energy without spalling off more atoms. And if the sheath has "diverter traps" along the bottom, loose atoms driven by the rotor velocity will be driven downwards and outwards away from the sheath, perhaps to be absorbed or pumped out by turbomolecular pumps below the sheath. However, the threat of spalling cascades does limit the maximum speed of the rotor; at some point, the energy per particle is too high.

Q11: Is the rotor suspension mechanism extraordinarily unstable?

Electromagnet suspension requires electronic systems to stabilize it. With an electronic system active, provided the system doesn't break, it is completely stable. In case of failures the rotor suspension mechanism has at least 10X redundancy; it needs at least 10 consecutive units to fail for the rotor to crash. Barring common mode failures (which have to be designed out of any system), such a failure is essentially impossible. Most normal systems have a redundancy ratio of 2 or 3. If the system did fail there would be a big explosion, but most of the expensive parts of the loop would be expected to survive; as would anybody launching at the time.

Q12: The rotor goes faster than orbital velocity, does this give a net antigravity effect and lift the structure?

No. The loop is pulled downwards by gravity in more or less the same direction along its whole length, and this net force must be resisted by the magnetic bearings. The east end and west end turnarounds are absorbing thousands of tons of force, and are spaced by many degrees of longitude so they point upwards relative to each other. The vector sum of these very large forces is upwards, and supports the rest of the launch loop in between them.

Q13: Does it take a lot of force to bend the rotor against its own stiffness?

The rotor is about the same weight and diameter as U.S. Schedule 10 "1.5 inch" (actually 1.93 inch, 4.9cm) pipe. If it was as stiff, then the bending force is about 100,000 N-m² divided by the square of the diameter of the turn in meters. To turn it into a 1 meter turn would require 100,000 Newtons, about a 10 ton weight (and it would break). To turn it in a 100 meter turn would require 10 Newtons, about the weight of 1 kilogram. The launch loop uses 28 kilometer diameter turns, so the bending forces are the same as a 13 milligram weight - far too small to notice, compared to the 42000 Newton/meter force needed to deflect the rotor into a 28 kilometer diameter turn.

For small scale, lower speed experimental launch loops, the bending forces will be significant. If a "back yard" 260m/s circular loop has a diameter of 10 meters, the bending forces will be 1000 Newtons, a noticable fraction of the deflection force, and perhaps enough to fracture welds.