Rotor Speed

At 12.6 km/s, heavier 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 (though it reduces maximum launch speed). The "lift" of the 14km/s, 3.0 kg/m rotor was (for a non-rotating Earth at 100 km altitude, 6478 km radius) is \rho v^2/r or 90.77 N/m . Earth's standard gravitational parameter (ignoring J_2 oblateness effects) is 398600.44 km³/s², so gravity is 9.50 m/s², and the centrifugal acceleration will support 9.56 kg/m, 6.56 kg/m of track and stabilization cables in addition to the 3.0 kg rotor. Total track mass for the 2000 km acceleration path between east and west stations is 13,100 tonnes.

An important second order detail is the rotation of the Earth, which adds 472 m/s of rotational velocity for eastbound launches, makes the effective rotor speed faster, and adds 0.03 m/s² of centrifugal acceleration "lift" to the track, reducing the effective gravitational force. The effect is opposite on the westbound return rotor (which does not launch payloads and can have a simpler track). However, the westbound track can be used for at-speed rotor segment replacement, so lift considerations can apply to that as well.

Assume the same track mass m_s = 6.56 kg/m. For v_r = 12.6 km/s, the rotor must be more massive to lift the same track mass. For a rotor mass m_r , a track mass m_s = 6.56 kg/m, a circular radius r = 6478 km, and gravity g = 9.50 m/s², we can compute the unknown m_r from g ( m_s + m_r ) ~=~ ( {v_r}^2 / r ) m_r or m_r = m_s / ( ( {v_r}^2 / ( g r ) - 1 ) = 6.56 / 1.58 = 4.15 kg/m .

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).

The rotor slows while doing this, but because the mass flow rate must be constant, 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.

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.

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.

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.

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

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.

RotorSpeed (last edited 2017-10-10 21:53:48 by KeithLofstrom)