PSU Physics Seminar

January 11, 2010

Questions and Answers

Question: The technique is not perfect yet, and sounds a little dangerous. Protect the loop from many kinds of destruction is a concern.

Indeed, the loop stores a lot of energy. Failures will happen, and emergency systems must dissipate that energy without harm. That will require good design, as well as testing to destruction of expensive test systems. Constructing the loop in mid-ocean, thousands of miles from shore, will help protect large populations from those energy releases, as well as help protect the loops from attack and sabotage. Not perfectly, though.

Much like current airline transportation, there will be a tradeoff between safety and privacy. A large enough bomb in a payload can destroy a launch loop. Preventing that will require intrusive inspection. Launch loops could be used to launch military payloads, or be the target of military weapons. The best way to minimize these problems (they will never go away) is to benefit as many people as possible, and to threaten as few as possible.

Humanity will be handling 50 Terawatts by the end of this century. There is no way to do this with perfect safety. There will be stupid accidents, and innocent people will be killed. However, with enough energy and enough experience, we can design systems that are much safer than the cars and airplanes and rockets, and the power plants and power lines we use now. If we can access space solar power, the launch loop can replace deadly power sources and transportation systems on the ground, improve living standards for everyone, and reverse the destruction of the environment. That will save millions or billions of lives.

Question: Will the efficiencies and costs be as low as presented?

It is very difficult to predict costs for a large system - I can only compare energy costs. Boeing cannot accurately budget time and expense for a new airliner like the 787 within a factor of two. The construction of the Panama Canal a century ago involved moving 30 times as much material as the first French team predicted, even after their successful canal at Suez. A simpler canal was completed by the U.S. using different methods.

Building a launch loop without prior experience and without an established market is impossible. Before it is practical, or the project cost and times are predictable, there must be a much higher demand for space launch, as well as years of experience for power storage loops, as well as large established factories to make the pieces. Fortunately, the intermediate steps can be small, and profitable. The companies and nations that take on these tasks can make huge profits, and will gain access to vast resources, if they proceed prudently but quickly.

If power storage loops can reduce the expense of generating electricity by load-sharing between continents, it will rapidly pay for itself. If the launch loop provides inexpensive access to space solar power, it will reduce the major cost of operating a launch loop - electrical grid power. While the cost-per-kilogram numbers presented at the seminar are surprisingly small, they are based on 12cent/kWhr electricity. If the construction costs are higher than expected, but electric energy costs drop, the launch loop costs will be cheaper than presented. It is difficult to know without much more experience.

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 at rest deflects the 3kg/m rotor with a force of 50 kiloNewtons - 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 of the rotor 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.

Since we are at rest, the 7kg/m track is not moving at all. Under the payload, it too makes a 85 microradian change, but there is no movement - it just hangs there on the magnetic field between track and rotor, with a force of 70N/meter. The energy stored in the field is approximately the force times 0.01 meter (1cm) gap. or about 0.7 Joules per meter. Some energy ends up in leakage flux, magnetic field not directly contributing to suspension. If the total energy is 1 Joule per meter in the deflection field, that is a tiny fraction of the 290 megajoules stored in the rotor.

Things are more interesting for a moving payload. Let's say the 5000kg payload is moving 4000 meters per second, half of orbital velocity. Because it is "partly in orbit", its gravitational weight is partly balanced by centrifugal acceleration, for a total downwards force of 37 kilonewtons. In the inertial frame of the payload, 3kg/m rotor is approaching from behind at 10kg/meter and getting deflected down, and 7kg/m track is approaching from the front and deflected down. If the track and rotor are instantaneously deflected by the same angle, the angle is 37kN/(7*(4km/s)^{2 + 3*(10km/s)}2), or 90 microradians. The rotor velocity changes downwards by 90 microrad*10 km/s, or 0.9 m/s. The track velocity changes downwards by 90 micronrad*4 km/s, or 0.36 m/s .

In a worst-case design, the force from the payload is coupled only to the rotor - the track must change the magnetic field and the relative accelerations to decelerate the vertical velocity of the rotor somewhat, coupling that velocity (through the magnetic field) to the track. The acceleration must be high enough to keep the track-to-rotor spacing within a few millimeters of the nominal 1cm spacing. The math is tricky. The full algebra is in the paper online. We can solve it numerically for this particular case.

The rotor in isolation deflects by a larger angle - 126 microradians, a vertical velocity change of 1.26 m/s . We need to change that to 0.9 m/s over some distance before and after the payload. The track needs to change velocity by 0.36 meters per second over that same distances. And the spacing between the track and rotor should not exceed, say, 1.5 centimeters. The deviation of 0.5cm is maximum under the payload, increasing parabolically on both sides from zero.

If the force per meter increases instantaneously by F at distance L to each side from the payload, then the force is acting on the track for a time of 2L/(4000m/s). That time, multiplied by the force F and divided by the track mass of 7kg/m, equals the 0.36m/s vertical velocity change of the track. F * L for the track solves to 10,800 newtons. The total force acting on the rotor to change its vertical speed by 0.36m/s (from 1.26m/s to 0.9m/s ) is also 10,800 newtons (the other way)over time 2L/(10000m/s). The rotor and track are pulled together by 10,800 newtons to maintain spacing.

The rotor's upward deviation from a sharp angle is 1/4 of the average velocity change multiplied by the time. The same formula is true for the track, in the other direction. The total deviation is D = 0.25*0.36m/s*2L/(4000m/s) + 0.25*(0.36m/s)*2L/(10000m/s). L solves to approximately 80 meters if D = 0.005m . So the force change between rotor and track is about 135 N/m, about twice the unperturbed track force. Again, this is the force that conforms the track to the rotor. With leakage flux, that is an additional energy of perhaps 2 joules per meter. Tiny compared to the rotor's 290MJ/m .

The main problem is power, not energy. If we try to add the energy of 2 joules over, say, 10 meters of track, moving at a relative velocity of 4000 meters per second, we have 2.5 milliseconds to do so. That is a power level of 800 watts. We also need to keep adding power as the spacing increases to 1.5cm, as we are filling more space with magnetic field energy. That power pulse requires significant power handling cost in the electronics.

That is worst case. We 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. 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/s² * 4000 m/s, or 600MW. 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 launch 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. Momentum packets can be moved between the eastbound to the westbound rotor, as needed for stability.

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.