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 co-linear and follow a straight path, with control sections on the track spaced about one meter that provide only transverse (sideways vertical and horizontal, not longitudinal). The control section spacing is a fraction of the shortest mechanically undamped transverse vibration modes in the rotor (assuming the stiff rotor has some mechanical damping). The rotor can be deflected sideways by perturbation forces (particle impacts, jostling by payloads, etc) but these forces are very small compared to the kinetic "forces" in the rotor.

The key concept is that the rotor is ballistic, and it takes a huge force to perturb it from its ballistic path. Large transverse forces cause only small transverse velocity changes. While that means that correcting a large transverse velocity perturbation requires huge forces, large velocity perturbations do not come out of nowhere. The system is Newtonian, and momentum is conserved.

An example: Assume 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. To put it another way, the 50 kiloNewtons is deflecting 3*14000 kilograms per secont to the side, yielding 1.2 meters per second downwards velocity, which forms an 85 microradian deflection angle compared to 14000 meters per second longitudinal velocity. That deflection would soon cause trouble - the rotor is now dropping 8.5 centimeters per kilometer, and would run into the track - but we use the controllers to keep the spring constant.

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 MJ/m stored in the rotor.

Another example: Things are more interesting for a moving payload. Both the rotor and the track will have vertical velocity changes. Let's say the 5000kg payload is moving 4000 meters per second, about half of orbital velocity. Because it is "partly in orbit", the payload's 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 this paper. We will solve it numerically for this one 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 reduce that change 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 the same distances, 0.18 m/s approaching the payload, 0.18 m/s leaving. 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 1/2 of the velocity change at front or back ( half of the total velocity change ) multiplied by the time in front or back ( half the total time) . A similar formula describe the track, deflecting in the other direction. The total deviation is D = 0.25*0.36m/s*L/(4000m/s) + 0.25*(0.36m/s)*L/(10000m/s). L solves to approximately 160 meters if D = 0.005m . So the force change between rotor and track is about 68 N/m, about the same as 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 1 joule 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 1 joule 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 400 watts, 40 watts per meter. 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.

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.

Significant forces can also occur if stability cables break. Under some circumstances, the impulsive forces from a ruptured main cable may be too great for the control system to handle, and the loop will fail. Small cable breakages will be less significant than a payload passage.

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.