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| The rotor and track are collinear and follow a very straight path, with control sections on the track spaced about one meter. The control section spacing is a fraction of the shortest structurally undamped vibration modes in the rotor (assuming the stiff rotor has some structural damping). The rotor can be deflected sideways by perturbations (particle impacts, jostling by payloads, etc) but these are very small compared to the kinetic "forces" in the rotor. | The rotor and track are collinear and follow a very straight path, with control sections on the track spaced about one meter. The control section spacing is a fraction of the shortest structurally undamped vibration modes in the rotor (assuming the stiff rotor has some structural damping). The rotor can be deflected sideways by perturbations (particle impacts, jostling by payloads, etc) but these are very small compared to the kinetic "forces" in the rotor. |
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| We must deflect the track to match the rotor - slowing the 3kg/m rotor lateral velocity, adding to the 7kg/m track velocity, until the lateral velocity of the two mass streams match. We need to do that before the spacing changes too much. The nominal spacing is 1cm, and if we use up half that, D=0.005m, then we must accelerate the track and rotor relative to each other by V^2^/2D or 144 meters/second^2^, about 14 gees. The acceleration is applied for 2D/V seconds, or 8.3 milliseconds. The 3kg/m rotor gets pushed one way, the 7kg/m track gets pushed the other way. If the track is accelerated vertically by 43.2 m/s^2^, and the rotor by 100.8 m/s^2^ the other way, the relative accelerations add up. That is a force of approximately 300 N/m between them. The unperturbed track force is approximately 7kg*10m/s^2^ or 70N/m . | We must deflect the track to match the rotor - slowing the 3kg/m rotor lateral velocity, adding to the 7kg/m track velocity, until the lateral velocity of the two mass streams match. We need to do that before the spacing changes too much. The nominal spacing is 1cm, and if we use up half that, D=0.005m, then we must accelerate the track and rotor relative to each other by V^2^/2D or 144 meters/second^2^, about 14 gees. The acceleration is applied for 2D/V seconds, or 8.3 milliseconds. The rotor moves 116 meters in that time. If some of the controller sections fail and are inert, or in the worst case "stuck the wrong way", then the other controller sections can compensate. The 3kg/m rotor gets pushed one way, the 7kg/m track gets pushed the other way. If the track is accelerated vertically by 43.2 m/s^2^, and the rotor by 100.8 m/s^2^ the other way, the relative accelerations add up. That is a force of approximately 300 N/m between them. The unperturbed track force is approximately 7kg*10m/s^2^ or 70N/m . |
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| The main trick in the launch loop is to let Newton move the power, while the control electronics only deflect it. | The main trick in the launch loop is to let Newton move the power, while the control electronics only deflect it. The control electronics communicate with each other at a sizeable fraction of the speed of light, and can make millions of control calculations per second. The rotor moves very slowly compared to controller compute speed, and the lateral forces much more slowly. The control system adds and subtracts lateral forces only, between rotor and track, counterweights and ground cables. "Packets" of lateral momentum can be added to the rotor as it passes by, and subtracted farther down the system. Packets can be moved from the eastbound to the westbound rotor. The most likely failure mode will not be from uncontrollable large forces, but from bad control algorithms. These must be completely debugged in power storage systems before applying them to launch systems. |
PSU Physics Seminar
January 11, 2010
Questions and Answers
Question: Positional control of the iron within the tube will, by itself, require an enormous amount of energy. Assume for the loop control the position at x depends on forces applied 14,000 meters previous. This introduces an inherent delay instability (a gain of 2) of the system that fundamentally limits the ability of the control loop to work, that is, this system will be oscillatory unstable unless the structures are totally rigid within this characteristic length. Shortening this length requires increasing the control loop power.
The rotor and track are collinear and follow a very straight path, with control sections on the track spaced about one meter. The control section spacing is a fraction of the shortest structurally undamped vibration modes in the rotor (assuming the stiff rotor has some structural damping). The rotor can be deflected sideways by perturbations (particle impacts, jostling by payloads, etc) but these are very small compared to the kinetic "forces" in the rotor.
A 5 ton payload deflects the rotor with a force of 50kN - the deflection angle is 50kN divided by 3*14000*14000 Joules/meter (1 Joule/meter == 1 Newton), or 85 microradians. Assume the payload puts all the force only on the rotor, and the control sections must transmit that force to the track and deflect that by a corresponding 85 microradians. The vertical velocity change is 1.2 meters/second . That is the key concept - the rotor is ballistic, and it takes a huge force to perturb it from its ballistic path, so large forces cause small lateral velocity changes.
We must deflect the track to match the rotor - slowing the 3kg/m rotor lateral velocity, adding to the 7kg/m track velocity, until the lateral velocity of the two mass streams match. We need to do that before the spacing changes too much. The nominal spacing is 1cm, and if we use up half that, D=0.005m, then we must accelerate the track and rotor relative to each other by V2/2D or 144 meters/second2, about 14 gees. The acceleration is applied for 2D/V seconds, or 8.3 milliseconds. The rotor moves 116 meters in that time. If some of the controller sections fail and are inert, or in the worst case "stuck the wrong way", then the other controller sections can compensate.
The 3kg/m rotor gets pushed one way, the 7kg/m track gets pushed the other way. If the track is accelerated vertically by 43.2 m/s2, and the rotor by 100.8 m/s2 the other way, the relative accelerations add up. That is a force of approximately 300 N/m between them. The unperturbed track force is approximately 7kg*10m/s2 or 70N/m .
Magnet forces are adjusted by adding or subtracting energy to the field (increasing or decreasing currents and voltages). The energy stored in the magnetic field that is doing useful work is the force times the gap. 300N/m times a 1cm gap is 3 Joules per meter. There is also leakage flux that is not doing useful work, so assume 5 Joules/meters total. That is a tiny fraction of the rotor energy of 290 Megajoules per meter. Getting that energy in and out of the magnetic field will require a hefty power pulse, but tiny compared to the rotor power.
That is worst case - you can couple forces from the payload into the track as well as the rotor, and subtract some of the payload energy to add and subtract current to the control coils. You can do half the velocity adjustment before payload passage, and half after. This reduces the peak power handled by the electronics, at the expense of some complexity.
The power put into the payload, meanwhile, is 5000Kg * 30m/s * Velocity. Payload exit velocities can approach 11000 meters per second, or 1.7GW for our 5 ton payload. /THAT/ is enormous power, but it is not being directly controlled by electronics, only indirectly via spacing adjustments. The main trick in the launch loop is to let Newton move the power, while the control electronics only deflect it.
The control electronics communicate with each other at a sizeable fraction of the speed of light, and can make millions of control calculations per second. The rotor moves very slowly compared to controller compute speed, and the lateral forces much more slowly. The control system adds and subtracts lateral forces only, between rotor and track, counterweights and ground cables. "Packets" of lateral momentum can be added to the rotor as it passes by, and subtracted farther down the system. Packets can be moved from the eastbound to the westbound rotor.
The most likely failure mode will not be from uncontrollable large forces, but from bad control algorithms. These must be completely debugged in power storage systems before applying them to launch systems.