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| Launching vehicles removes momentum from the rotor. Because the mass flow rate must be constant, la | Launching vehicles removes energy from the rotor. Assuming 98% energy efficiency, 400 tonnes per hour, and an average launch velocity of 10.3 km/s, that is 21.65 TJ/hr or 6 GW removed from the rotor. Rotor mass flow rate is conserved; it is 4.32 kg/m × 12500 m/s or 54 tonnes per second coming in and going out. Launch removes an average of 111111 Joules per kilogram of rotor; the rotor speed (in the fixed frame) drops from 12971 m/s to 12962.4 m/s, and the rotor density increases by 0.066%, presumably by reducing the distance between 10 meter rotor bolts by 6.6 millimeters. |
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| MoreLater | The rotor will stretch much more during the descent to the ground. Ignoring J₂, the gravitational energy difference is μ ( 1/R0 - 1/R80 ) = 398600.44 ( 1/6378 - 1/6458 ) = 0.774186 MJ/Kg = 774186 J/kg. The rotor speed in the fixed frame increases from 12962.4 m/s to 13022.0 m/s, so the rotor density must decrease, stretching by 0.46%, or 4.6 centimeters for a 10 meter bolt. |
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| the slower rotor gets slightly denser towards east station, complicating the computation of speed change. Before and after payload acceleration, the mass flow rate ( rotor density times rotor velocity ) remains constant, 4.15 * 12600 = 52300 kg/s at the start (and also at the end). The power in the rotor at the start of the launch track is $ { 1 \over 2 } m_r v_r^3 $ = 4.1516 TW (!). Since 24 GW of power has been removed at the end of the launch path, the power flow at that point (before the power of decelerating sleds is added back) is $ { 1 \over 2 } m_r v_r^3 $ = 4.1276 TW. Since $ m_r v_r $ remains constant, we can compute $ v_r $ as 12563.5 m/s, 99.71% of the starting speed. Rotor density must increase 0.289% to compensate, though that reduces to 0.241% after sled power is added back. | '''The stretch is helpful!''' The rotor must be disassembled into individual bolts for deflection and sorting on the surface. The rotor bolts will be held together longitudinally by long magnets connected to compliant springs; the 4.6 centimeter stretch will pull the bolts apart so they can be separated. The specific mechanism hasn't been defined, but we can exploit the stretch during normal system operation. |
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| . Again, these numbers are for a non-rotating Earth, and are a few percent pessimistic. Rotation helps launch payloads and adds centrifugal support to the track. | The uneven mass stream plus a resonator may help separate the bolts radially. The bolts must be separated and rotated for proper orientation into the deflection magnets; this may occur during the last few hundred meters of descent. The reverse will occur after the bolts pass through the motors and turnaround and the inspect/sort/replace switchyard. |
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| At east station, after the east end curve downwards from 100 km altitude to 50 km altitude, the sleds slow to a stop and are lowered back down to the surface with elevators, and shipped back to west surface station (as air or ocean cargo) for refurbishment and reuse. The method chosen will depend on sled cost (mostly magnet cost) and interest rates; hopefully cheap iron-nitride supermagnets will make sleds cheap and fast container ship transit cost-effective. Either way, the sleds must disassemble into a standard 40 foot shipping container, constraining width and length. Details of sled design are a different topic. | === Sled Processing === |
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| At the end of the launch path, the rotor drops 100 km, gains kinetic energy,increasing speed significantly. The gravitational energy difference is 398600.44 ( 1/6378 - 1/6478 ) km²/s² or 974.7 kJ/kg . If the rotor is NOT launching payload, and starts its descent at 12600 m/s, it speeds up to 12,676 m/s when it reaches the ground; half the velocity difference squared (kinetic energy per kilogram) is increased by the gravitational energy difference. The rotor is stretched by as much as 0.61%. Initially, when launch rates are slow and bursty, the generators at the surface will be small and unable to restore all the power extracted by a burst launch. The total mass of a burst will be limited by the compliance range of the rotor. If the rotor can slow as much as 0.6% overall, it will need 1.2% total stretch compliance between slowest (after many loops of burst launch) and fastest (rotor speed after descent and system energy restoration). For a 6000 km long rotor (500 km at each end of a 2000 km launch path, forward and back, total loop energy is around 2e15J, 0.6% of that is 12e12J, and total burst launch mass is 240 tonnes, about 10 minutes of burst launch, limiting the size of a "one pass mission assembly", about 5 times the Apollo mission stack. * A 12.6 km/s rotor completes one 6000 km circuit around the loop in 8 minutes; the "burst time" may be limited to that. The "disturbance distance" will propagate ahead of the point of first payload engagement at the "speed of sound" in the semi-stiff rotor, and the acceleration time to 10 km/s at 30 m/s² is 5.5 minutes, so the actual length and payload spacing of a burst will be more complicated than this estimate. Pinning down the details is what engineering teams are for, and verifying them requires detailed computer simulation and years of experimentation. The rate at which these 240 tonne missions can be launched depends on available power plant. If 20 GW (plus loss power) is available, this loop design can maintain 400 kg/s rates indefinitely, 1440 tonnes per hour, 34,000 tonnes per day. As the Earth rotates, this mass will be launched towards all orbital longitudes, but with patience and transfer orbits, it can collect at those rates in one place in one orbit. If only 140 MW power surplus is available after system losses, 240 tonne bursts will be limited to once per day. Much depends on available power, which grows rapidly after the launch loop launches its own source of space solar power. Remember that there should be three or more launch loops operating in parallel; if one or two fail, the remaining one can launch essential traffic. |
At east station, after the east end curve downwards from 100 km altitude to 50 km altitude, the launch sleds slow to a stop and are lowered back down to the surface with elevators, and shipped back to west surface station (as air or ocean cargo) for refurbishment and reuse. The method chosen will depend on sled cost (mostly magnet cost) and interest rates; hopefully cheap iron-nitride supermagnets will make sleds cheap, and fast container ship transit cost-effective. Either way, the sleds must disassemble and fit into a [[ https://en.wikipedia.org/wiki/Intermodal_container | standard 40 foot shipping container ]], constraining (folded?) length to 12 meters. Details of sled design are a different topic.. |
Rotor Speed
A 12.5 km/s, 4.32 kg/m rotor
Expanded from the old 14km/s argument below. A velocity-transformer track increases efficiency and launch rate, and a somewhat slower rotor allows for more mass in the rotor. This also reduces maximum launch speed, but the loop can still launch smaller vehicles beyond escape velocity.
The Earth's standard gravitational parameter (ignoring J_2 oblateness effects) is 398600.44 km³/s², so gravity at 80 km altitude (6458 km radius) is 9.55744 m/s² plus the J₂ oblateness term of 0.01514 m/s (proportional to 1/r⁴), for a total of 9.5728 m/s². The track is rotating at 72.9211509 microradians per second with the Earth, so that adds 471 m/s to the velocity and a centrifugal acceleration of -0.03434 m/s at altitude to the gravity, or 9.5834 m/s² acting on the fixed track
The rotor moves 12.5 km/s relative to the rotating-with-the-Earth track at altitude, so it moves at 12,971 m/s in the fixed frame. The centrifugal force on the rotor is ρV²/R = 112.55 N/m outwards. The gravitational force on the rotor is 4.32 kg/m × 9.5728 m/s² or 41.35 N/m, with a net 71.2 N/m to support the ( 71.2 N/m / 9.5834 ) or 7.43 kg track and suspension cables. Total track and cable mass for the 2000 km acceleration path between east and west stations is about 15,000 tonnes.
The new design transformer track converts rotor energy to payload velocity with nearly 100% efficiency (system efficiency less because of power dissipated in power conversion, residual gas drag, and coil resistance losses). Assuming 400 kg/s maximum payload rate, 10 km/s launch velocity, 5 tonne payloads, and 1 tonne launch sleds, then the loop emits 20 GW of launch power into the payloads, and temporarily puts an additional 4 GW into the sleds (mostly recovered at the east end).
Escape velocity from 6458 km altitude is √( 2 μ / r ) = 11.11 km/s, or 10.64 km/s relative to the rotating track. Launches to Earth orbits will be slower; for example, vehicles to a high, slow ConstructionPort cargo orbit will launch at (10.51-0.47) or 10.04 km/s relative to the rotating track. Large interplanetary missions will assemble (and test) in a Construction orbit, then add extra delta V at apogee with high Isp electric-thrust rockets for trips to Mars and the asteroids.
Launching vehicles removes energy from the rotor. Assuming 98% energy efficiency, 400 tonnes per hour, and an average launch velocity of 10.3 km/s, that is 21.65 TJ/hr or 6 GW removed from the rotor. Rotor mass flow rate is conserved; it is 4.32 kg/m × 12500 m/s or 54 tonnes per second coming in and going out. Launch removes an average of 111111 Joules per kilogram of rotor; the rotor speed (in the fixed frame) drops from 12971 m/s to 12962.4 m/s, and the rotor density increases by 0.066%, presumably by reducing the distance between 10 meter rotor bolts by 6.6 millimeters.
The rotor will stretch much more during the descent to the ground. Ignoring J₂, the gravitational energy difference is μ ( 1/R0 - 1/R80 ) = 398600.44 ( 1/6378 - 1/6458 ) = 0.774186 MJ/Kg = 774186 J/kg. The rotor speed in the fixed frame increases from 12962.4 m/s to 13022.0 m/s, so the rotor density must decrease, stretching by 0.46%, or 4.6 centimeters for a 10 meter bolt.
The stretch is helpful! The rotor must be disassembled into individual bolts for deflection and sorting on the surface. The rotor bolts will be held together longitudinally by long magnets connected to compliant springs; the 4.6 centimeter stretch will pull the bolts apart so they can be separated. The specific mechanism hasn't been defined, but we can exploit the stretch during normal system operation.
The uneven mass stream plus a resonator may help separate the bolts radially. The bolts must be separated and rotated for proper orientation into the deflection magnets; this may occur during the last few hundred meters of descent. The reverse will occur after the bolts pass through the motors and turnaround and the inspect/sort/replace switchyard.
Sled Processing
At east station, after the east end curve downwards from 100 km altitude to 50 km altitude, the launch sleds slow to a stop and are lowered back down to the surface with elevators, and shipped back to west surface station (as air or ocean cargo) for refurbishment and reuse. The method chosen will depend on sled cost (mostly magnet cost) and interest rates; hopefully cheap iron-nitride supermagnets will make sleds cheap, and fast container ship transit cost-effective. Either way, the sleds must disassemble and fit into a standard 40 foot shipping container, constraining (folded?) length to 12 meters. Details of sled design are a different topic..
From 0 to 12.6 km/s: Starting a Launch Loop
A pre-deployment launch loop stretched out near the ocean's surface will need external power for the levitation magnets and to start moving the rotor. Without rotor movement past generator coils, this power must be provided externally, at frequent intervals along the 2000 km length of the launch and return tracks.
Older stuff: At 14 km/s, lighter rotor
As the rotor circulates in the launch loop, it ascends from ground level to the 100 kilometer altitude of the launch path, then descends back down to the surface for the ground anchored ambits. If the 3 kilogram-per-meter rotor moves at 14 km/second at altitude, it will gain energy, speed up, and stretch out as it descends. The stretch is because the rotor "flux" passing any given point must be 3 kg/m * 14 km/s = 42 tonnes per second.
How much does it speed up? If the surface is at 6378 km, and the launch path is at 6478 km (100 km altitude), then the energy difference per kilogram is ( 398600.44 km³/s² × ( 1/6378 km - 1/6478 km ) = 0.96474 km²/s² , which is added to the ½ × (14 km/s)² = 98 km²/s² (per kg) of the rotor at altitude, resulting in a ground level kinetic energy of 98.96474 km²/s² (per kg), corresponding to a rotor speed of 14.069 km/s, 69 m/s faster
- Note: A second order wrinkle is the rotation velocity of the Earth, 0.4724 km/s at 100 km and 0.4651 km/s at the surface (a speed difference of , which is added to the westbound rotor speed and subtracted from the eastbound rotor speed. That adds another 3 m/s to the westbound rotor at the surface.
The eastbound rotor has the same kinetic energy at altitude as the westbound rotor, and the same velocity with respect to a nonrotating inertial frame, but the rotating earth ...
MoreLater . If the easttbound rotor moves at a ground relative speed of 14 km/s, it is moving at 14.474 km/s in the inertial frame, and this is what provides track support. The westbound rotor will also be moving at 14.474 km/s in the inertial frame, it is moving at 14.948 km/s in the ground-referenced rotating frame. I MUST THINK THIS THROUGH.