Thin Film Electromagnetic Tokomak (TFET) Fusion: a peer review

On the Technological and Market Feasibility of Introducing Martian ores of Elements and Minerals into the General Marketplace: Thin Film Electromagnetic Tokomak Fusion (TFET)

Part A: Primary Mover Propulsion Systems

— Executive Summary of Reference Design Twenty-One Castle Mike

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Sunset at Gusev crater, Mars.

Work performed for a private consortium of the petroleum industry exploring the other side of peak oil.

Abstract: [redacted] Marine Construction, LLC dba [redacted], commissioned this study to determine if an investment of less than 25 billion USD would suffice to establish a trading infrastructure under the controlling interest of [redacted] in valuable ore material with the following characteristics:

1.) A return on investment of at least 3x in 10 years.

2.) A probability of technical success credibly estimated of greater than 90%

3.) A trading infrastructure of indefinite duration with extraction occurring within this solar system and with an intended international market.

4.) To position [redacted] as the sole commercial entity with the capacity to provide transportation within the solar system of both low and high mass payloads at a market tolerable price and risk.

5.) To position [redacted] as the sole commercial entity with the capacity to end-run peak oil

Note:

This is a summary of confidential papers and studies in the subject matter presented. Not all information may be publicly available. This document is intended for an audience of business executives and scientists and assumes cursory knowledge of spaceflight and general mathematical literacy. All units used in this document are Systeme Internationale (SI) and wherever units are not specified, SI units should be assumed.

Introduction

[redacted] is a privately held corporation incorporated in the State of DE with its combined, interested conglomerate assets exceeding 50 billion USD. Its primary mode of business is marine [redacted] for the petroleum industry. As of 2011, [redacted] has earned over 12 billion USD in the sale of [redacted]. An innovative paradigm shifter, [redacted] continually seeks heterodox ventures with the potential for large reward. This consortium is known internally as DynaLite AstroMarine, LLC.

This is one of several documents whose purpose is to provide the Business and Technical Requirements for a Reference Design model. In this document, the Reference Design discussed regards a direct kinetic fusion primary mover for spaceflight. The Reference Design specifies an operational (reusable) fusion reactor that is vacuum transit, VT, certified (known traditionally as space flight certified) and has a maximum rated power of 11 – 21 Terra watts continuous. The design was code-named Twenty-One Castle Mike to denote its power parameters; to wit, that this Reference Design shall specify a maximum power output of 21 Tw continuous (a Case Model), possibly slightly less depending on the design chosen. Due to considerable limitations in the current understanding of fusion power, we’ve opted to employ a novel design based on nuclear weapons research. Specifically, this technology involves the invention of several new composite technologies and is referred to overall as Thin Film Electric-Magnetic Tokomak fusion, or TFET, fusion (also Thin Film Electromagnetic Tokomak fusion, or TFET, fusion). Patents on this technology are now pending.

Thin Film Electric-Magnetic Tokomak Fusion (TFET)

The distance between Earth and Mars, for the purposes of illustration, can be estimated at 3.1 x 10⁹ meters (about 2 A.U.). We set Δv = 10000 m / s, the upper limit velocity before ablative shielding will be required (see later discussion).

3600 * 24 = 86400 seconds are in a day

to satisfy the fuel mass limitation, we shall keep the fuel expense well below the allocated 15000 kg with a value of 11140 kg mass one way (2 burns) and 22280 kg round trip (4 burns). Each burn is 5570 kg. The exact amount of deceleration will be mission dependent, but we allow for a maximum deceleration in any case.

ð Δv = (3.1 * 10⁹) / 432000 ≈ 10000 m / s

ð 10000 * 432000 = 4.32 * 10⁹ meters transit path length

The well-known Tsiolkovsky rocket equation is:

m_f / m_i = e ^{-( Δv / ve)}	[eq. 1.01]
Tsiolkovsky Rocket Equation

ð m_f = m_i* e ^{-( Δv / ve)}

ð m_f = (2500000) * e ^{-( 10000 / 0.015c)}

ð m_i =2505570 kg

ð 2505570 – 2500000 ≈ 5570 kg mass fuel.

We assume a total package mass of 2500000 kg. For mission purposes we’ve set the total allowed cargo and hull mass at 1000000 kg (10000 mT Earth weight) and the propulsion system at 1500000 kg (15000 mT Earth weight). then we can set our values, after some re-arranging, as:

Δv = v_e [ ln (m_i / m_f) ]	[eq. 1.02]
Mars transit baseline minimum delta v required
Reference Design => (0.015c) * (ln (2505570 / 2510000))≈ 10000 m / s

and

α = T_k / (m_i – f_m – c_m – h_m)	[eq. 1.03]
Mars transit baseline minimum specific power required
Reference Design=> ((710^12)/(1.5 10⁶)) ≈ 6 Mw / kg

and

T_k = α(m_i – f_m – c_m – h_m)	[eq. 1.04]
Mars transit baseline minimum kinetic thrust required
Reference Design => (3.5 * 10⁶)( 1500000) ≈ 5.25 Tw
Usable maximum power rating 7 Tw

where

f_m is the fuel mass, p_i is the propulsion system tare mass, m_i is the total mass of the entire assembly at mission start, c_m is the total cargo mass, h_m is the remaining assembly weight (superstructure – heat rejection panels are part of propulsion system mass) and T_k is the total kinetic thrust power. The three equations supra are the bunny trail that leads one inexorably to fusion kinetic propulsion for spaceflight. In the case supra we have not considered the burn time and we will address that directly.

These numbers will change dramatically when we consider the Gliese 581 mission profile, which will include 4 of the primary movers of the Reference Design Twenty-One Castle Mike in a more massive ship traveling at a maximum velocity of 0.9 c. This is addressed in a separate Reference Design. The trip time in that case will be around 7 years (crew) and 26 years (Sol-based observer).

The Confinement Problem: a Power Conversion interdependency

Current fusion research has had the opportunity to expose some fundamental voids of understanding of nature that represent extreme project risk. One of these is known as magnetic reconnection, which we suspect may be due to a lack of understanding of nature at a deep level. While that may not be the case, it is of sufficient concern to warrant discussion. Magnetic reconnection is the result of multiple magnetic field lines “bleeding” off into other nearby field lines when magnetic field strengths are very strong and confined. Fusion research has encountered, at least in some cases, stumbling blocks due to the fact that this magnetic reconnection phenomenon occurrs several orders of magnitude faster than the current state of the art predicts. This is a red flag. There are many reasons for saying this, but one very important one is the academic denial currently swirling around traditional Tokamak research (magnetic-only confinement). Tokamak research is generating billions in research grants and construction projects and the dirty little secret made so to keep the funds flowing is that they won’t work and researchers have no clue why. The reason is that this phenomenon known as “magnetic reconnection” occurs in magnetically confined plasmas (of moderate energy levels and beyond):

1.) With limited or no predictability [the second derivative of the reconnection changes without any understood cause]

2.) Faster than MagnetoHydroDynamic Theory (MHD) algebraically posits (ominous)

3.) In such a manner as to foul any attempts at confining a plasma for a useful length of time.

4.) In a space so small that experimental observation of the event is virtually impossible (which suggests serious issues if one tries to resolve this issue by gaining more insight into it).

5.) Is likely exponentially proportionate to energy output scaling (chasing your tail).

In magnetic-only confinement schemes a high value for the ratio of the plasma density to the magnetic field strength (called beta) is desirable. However, it is precisely in these regimes where magnetic confinement instabilities become severe. To summarize the salient points:

The conductivity of plasma drops as density increases; but this is the driving force of the entire reactor
Magnets lose their capacity to dampen inertial movements when density increases.
Much higher plasma densities are absolutely necessary for a practical power density regardless of the fuel used (a point to be proven in what follows)

The three most common instabilities in traditional Tokamak designs are known as ideal kink modes, resistive wall modes, and neoclassical tearing modes. This makes sense when one considers what beta means: if plasma density is considerably higher, relatively speaking, than the magnetic field strength, the fields usefulness in controlling the orbits of particles in a torus lessens since the higher density serves to accentuate small perturbations and instabilities such that the magnets cannot dampen them back out. The magnetic field is overwhelmed by the inertia of the higher density plasma motion. Electrostatic assistance (at about 10⁶ bar) would dampen these instabilities out handily. Large density gradients become very hard to control in a magnetic field of limited strength over the same region of space (the gradients of field strength and density do not match in such cases). When the density of plasma varies considerably as one moves outward from the central torus orbit (called the bootstrap value) instabilities occur if a so-called conducting wall is not present. In other words, the plasma literally needs to conduct current through to the external environment (not good for confinement). Again, this arises because variable density through the torus “beam” of particle flow translates, in neutral two-species plasma, to a higher density current which, when stronger at the edges, begs for a nearby conductor. This issue does not present with positively charged, electrostatically stabilized single species plasma. Finally, a so-called “Neo-Classical Tearing Mode” instability occurs when a plasma density which varies as position changes outward from the central orbit has a region in which this variance is non-monotonic. Again, 10⁶ bar of electrostatic force will readily eliminate this (most magnetic-only confinement schemes operate at just a few atmospheres). Operating at six orders of magnitude greater density while keeping the same field strength will cause beta to skyrocket, precisely what is desirable to dampen instabilities. Beta values are a telling clue that supports our contention that the magnetic field is itself causing these instabilities due to the fact that the field is not applied symmetrically throughout the plasma and because Magnetic fields are insufficient, by themselves, to attain stability in fusion reaction regimes. This is because the magnetic field strength on the outer orbits is necessarily weaker than the magnetic field on the inner orbits (due to the intrinsic geometric properties of a torus which bunch the magnets closer together on the inside of the torus).

Thus, increasing magnetic field strength won’t eliminate these problems, but increasing density by the application of a non-magnetic force will. This is so obvious it is frustrating to observe the endless attempts to eliminate instabilities using the very mechanism that is creating it and the apparent unwillingness to acknowledge this simple no-win strategy. Frustrating examples of dancing around this subject without addressing it are numerous, but one is illustrative. Many in the field of research have noted that flattening out the torus to a smaller diameter will help eliminate many of these instabilities. Of course it will. It reduces the asymmetric nature of the toroidal magnetic arrangement. But clearly it is not the solution. Symmetry must be enforced with virtual perfect precision and magnetism alone cannot do this. A symmetric force at least equal to the magnetic field strength must be applied to ensure perfect symmetry. In the Neo-Classical Tearing Mode the instability has been noted as being associated with the helical motion of the particle in its orbit. A symmetric dampening force that flattens out these helices eliminates this issue. Indeed, a symmetrically applied electrostatic force would have the effect of not just flattening out the helical travel, but shaping it symmetrically so that its helical orbit path radius is shorter on the higher charge side. The result is a stable, parabolic helix orbit about the main particle orbit path. This is exactly what an electrostatic force will do when applied symmetrically.

Due to the lengthy and well developed research already conducted in plasma fusion schemes we will propose a design that leans on this long history of research and utilizes a toroidal plasma as typically found in traditional Tokamak research reactors. However, we will also propose a new method for eliminating the difficulties currently encountered in MCF by developing a method that is complementary to MFC rather than divergent.

More about Thin Film Tokamak Fusion (TFET)

The solution to introducing sufficient power to the reaction chamber without direct use of materials is to use a device that delivers considerable levels of power via electromagnetic radiation directly into the plasma without spoiling it. This suggests a fission reaction in the power regimes between a standard fission reactor and an uncontrolled critical mass explosion. There are quite a few clever things one can do with Uranium to achieve this. And it gets even more interesting with an element known as Americium-242m. Ultra fast reactors have been constructed and experiments show them to be viable. Of course, we seek something a little hotter. What we envision is a very hot, very short burn of less than 30 minutes (probably just seconds – we will calculate this much more precisely infra). Fission explosions have positive power output over ranges in the 10^-9 second range. This is obviously too fast to control with any foreseeable technology. Doing the math, for a fission reactor with a 10 year refueling cycle (a little on the long side) suggests a median burn time of 3600 * 24 * 365 * 10 * 10^-9 / 2 ≈ 0.158 seconds. So, we’re a little on the slow side of an uncontrolled reaction. This can be achieved using an ultra fast reactor design using Americium 242 with a thin film deposition designed to preferentially vent critically excess fast nuclei outside the reaction mass. Much experimental work (much of it classified for a long time) shows this to be achievable. Recent research in Israel has proven it. The reaction mass is shaped concavely facing the target. This introduces alpha radiation (positively charged Helium nuclei), gamma radiation and neutrons directly into the plasma. Better still, it can be used to electrostatically accelerate fusion fuel product whose kinetic energy will heat the remaining reaction plasma. Boron and Li-Si shutters can be employed to control plasma destabilization and provide an “annealing” feature for the plasma. The heating effect would be staggering and would follow closely on the well-established principles developed in nuclear weapons research. Americium 242m provides approximately 2 Mw power per square meter of film. Given 1 such “shaped charge” with a surface area of 1000 m², this yields a total power output of about 7 Gw, placing the energy deliverable to the plasma in the thermonuclear explosion range after a 30 second burn. This energy, mostly in the form of fast ⁴He nuclei, can be contained via the electrostatic fields discussed infra. Be reflection material inside the torus can trap neutrons to aid in heating. A further refinement will be discussed in what follows.

The overall program here is to come up with a way to send fusion fuel product into the plasma without involving any contact between materials, and to do so in a way that forces of confinement and pressures are transmitted via fields only, ultimately “resting” against a material of sufficient strength to contain them (which we address infra).

The Fissile Fuel Injector; a “slow” shaped charge

We recommend the introduction of a purpose built fission reactor whose product is used to electrically accelerate ²H-¹¹B directly into the plasma at energies proportional to its β radiation power output. This can be achieved by using the fissile β radiation to accelerate the fuel by electrostatic attraction in a cylinder within a cylinder design. Magnets around the cylinders further shape the fissile product as it exits the fission reaction chamber. The fissile product can be vented to vacuum and thus keep the plasma pure. This follows closely the design principles behind so-called “Hydrogen bombs” which use fission “kickers”, called “first stages”, to induce a fusion reaction. Diagrams of the Fissile Kicker invention are included in the Twenty-One Castle Mike Reference Design diagrams.

We have proposed the use of Americium-243m as the First Stage, fissile material. It can heat a small region of fusion fuel in the Gw power regime for periods up to several days (thus, several restarts on one kicker “pack”).

Isotope	Half-Life	Specific Activity (Ci/g)	Decay Mode	alpha (α)	Beta (β)	Gamma (γ)
Am-241	430 yr	3.5	Α	5.5	0.052	0.033
Am-242m	150 yr	9.8	IT	0.025	0.044	0.0051
Am-242 (product of Am-242m decay)	16 hr	820,000	β, EC	0	0.18	0.018
Am-243	7,400 yr	0.2	Α	5.3	0.022	0.055
Np-239 (product of Am-243 decay)	2.4 days	230,000	Β	0	0.26	0.17

Key: IT = isomeric transition, EC = electron capture, Ci = curie, g = gram, and

MeV = million electron volts; a dash means that the entry is not applicable.

Americium-242 decays by two means:

by emitting a beta particle (83%) and by electron capture (17%). Certain

properties of americium-242 and neptunium-239 are included here because

these radionuclides accompany the americium decays. Values are given to

two significant figures.

Formally speaking, so-called “hybrid” fusion-fission systems are technically different in that they are inherently safer than what we propose (which is actually a fission-fusion hybrid, not a fusion-fission hybrid). In these systems a fusion reaction is the “kicker” that provides the neutrons for a fission reaction. Because a sub-critical fission reaction is driven by a fusion process, it is inherently safe (at least it can be made so). However, in our case, we are in fact using a fission reaction (which certainly can runaway) to “kick off” a fusion reaction. In the formal hybrid scheme positive feedbacks are not possible. In our scheme they very much are. Having said that, we are merely pointing out the dangers that have always been inherent in fission reactors and we do not feel that this risk is sufficient to cause concern for this project. In the end, we are simply throwing the most powerful source of heat and pressure we know of at the problem, just as was done when developing uncontrolled fusion reactions for bombs (which do in fact work).

We submit that this approach has not been contemplated amongst academicians, much less attempted, partly for cultural and political reasons. For civil power production the only advantage fusion offers over fission in a “kicker” design is raw power, albeit with still considerably less radioactivity. The fission reaction need only occur for startup. Once initiated, the fusion reaction can provide all necessary power. In any case, and though not widely acknowledged publicly, this method is the shortest and most certain path to controlled fusion power.

The fuel

We specify in this Reference Design a reaction involving the fusion of ²H (protons thereof) with ¹¹B pure ions. All the reasons for choosing this fuel will become evident in what follows, but the reader should keep in mind that in order to create the power levels we seek, dramatic increases in plasma pressure are required to bring up the reaction rates to truly useful levels. This makes the lower reaction rate of ²H – ¹¹B numerically insignificant in the final calculation, differences in commonly proffered fuel sources being orders of magnitude smaller than the gains we are about to demonstrate in reaction rates through increased plasma pressure. This represents yet another of the many “voids” of understanding in the popular literature regarding nuclear power: use of other fuels might be easier propositions short term, but in the long term a true, high powered fusion reactor cannot rely on these fuels for reasons we shall prove shortly.

¹¹B is prepared for tankage as isotopically pure (to prevent unwanted neutron flux). ²H is also isotopically pure. We specify its storage to be in liquid form and the ¹¹B storage to be in solid form. Though we have discussed scenarios involving fuel mixtures of equal parts ²H and ¹¹B for order of magnitude validations we here specifically indicate the requirement as being one part ²H for every eight parts ¹¹B. The reason for this is not at once obvious until we consider nuclear cross sections, which we will take up later. In summary, these proportions ensure the highest reaction rate possible. Since the fuel is loaded to a minimum of 30,000 kg mass over what is to be used for reaction, this does not require additional adaptation of the design.

The fuel injection system: the ignition sequence

The fuel required for the reference round-trip transit is 22,480 kg of ²H and ¹¹B. This fuel will be injected into the plasma at ignition sequence start. Once it is exhausted, a restart will be required. Therefore, we now detail the mechanism for fuel storage, plasma creation and fuel handling.

We specify an integrated and self-contained unit which, in its overall construction as a propulsion system, constitutes Twenty-One Castle Mike. Included in this system are 20 large fission reactor whose design is specified in Reference Design Low power services self contained, mobile fission reactor (MFR), a design derived of the United States SP-100 reactor design. It consists of a fully shielded system (to background) with 6, separate reactor cores individually controllable and providing a combined electrical power of 1900 kWe. Included in each system is a Brayton cycle that is VT certified and capable of operating in varying G environments. All MFRs brought online brings the system auxiliary maximum rated power to 38 MWe. Heat rejection is by liquid Li circulation.

Ignition sequence begins when, if not already online, all MFRs are brought to maximum rated power and diagnostics are performed to validate the requirement that at least 60 reactor cores remain functioning. This can be accomplished by the built in Li-Ion battery packs each MFR contains which, by remote command, can retract their control rods to start each of the six reactor cores.

We specify a startup plasma Earth weight of 70,000 kg and a final Earth weight at restart of 60,000 kg. This yields a total fuel load of ²H (the density of L²H is 68 kg / m³ at 20K.) and ¹¹B of approximately 10,000 kg. We further specify the use of a fission kicker as described supra and infra which, when the coolant/moderator is removed from the reaction chamber the shaped charge “fires”. This moderator removal is performed by auxiliary power discussed supra.

Plasma creation

One of the key reasons for using a thin film, apart from the fact that this is required to take advantage of electrostatic forces, is that we can reduce overall system weight significantly and dramatically improve the performance of “inducing” plasma current by the application of alternating charge in laminate layers of the vacuum vessel. This obviates the need for a central solenoid. But most importantly it means that current can be run much higher and therefore the forces created by the fields are consequently much greater.

The kicker is designed with Gd shielding of approximately 10 cm thickness. Inside that shield is a W liner which is honeycombed with coolant conduit passages. This kind of engineering is well established and we won’t dwell on the calculations here. Filling the voids of the W shielding is Nb which, upon kicker firing becomes liquid. Auxiliary power is used to circulate the liquid Nb through the W reactor wall and out to an ¹¹B fuel storage bin. A thermal energy transfer scheme is once again applied here and the ¹¹B, inside a bin set to 1 Pascal pressure (giving the ¹¹B a boiling point of 2075 C), is heated to its vapor pressure and just beyond. Once the desired thermal energy transfer from the Kicker to the ¹¹B fuel is complete the liquid Nb is rerouted to transfer thermal energy to the ships heat rejection panels. This is accomplished through the thermal transfer first to the liquid Li coolant used by the fusion reactor (which is circulating at 100 m / s – this requires new development in turbo pump technology).

Simultaneous to the events described supra, auxiliary power is used to depressurize a stream of L²H and run high voltage through it. The specification is that the electrical power applied shall be sufficient to ionize the ²H. ²H has a total ionization energy of 1312 kJ / mole. Therefore,

P_aux = E_i * M_kg * m_H	[eq. 3.01]
Auxiliary power required to start kicker power up sequence

Where P_aux is the auxiliary power required to initiate the kicker power startup procedure, E_i is the total ionization energy of the fuel used for startup, M_kg moles of said fuel per kg and m_H is the total mass of the fuel to be used in the fueling cycle being initiated.

The ²H is then pumped into a cylinder which runs the along the axis of the magnetic thruster unit at the systems extreme after end. This cylinder contains within it another cylinder, inside which β radiation from the kicker will flow. Magnets outside the cylinders deflect particles to maintain a trajectory along the axis of the cylinders. As the β radiation moves through the inner cylinder it “drags” the ²H with it and transfers momentum to the fusion fuel component. This ²H is then magnetically routed into a torus consisting of numerous electromagnets that provide resistance to the momentum of the each ²H particle flowing into the torus. This transfers kinetic energy to electrical energy. This process of decelerating ions is a well-established technology and we won’t dwell on it here.

As the ¹¹B is converted fully to its gaseous state it is pumped “downrange” of the kicker and aligned with the cylinders described supra. The β stream is directed to accelerate another stream of electrons by “pushing” electrons ahead of it (see diagrams). The electrons being “pushed” are outside the Gd shielding surrounding the outer cylinder. This secondary β stream we shall designate β₂. An ionization based on the technique used in Electron Beam Ion Traps is performed by colliding the β₂ stream with the ¹¹B fuel gas. The outer cylinder is divided into at least two sections to allow the flow of both ²H and ¹¹B at the same time. We can now utilize eq. 3.01 again by substituting ¹¹B. The result is the power transferred from the Kickers β stream to ionize the ¹¹B. Existing designs show that at least 5000 A / cm² charge density is required of the β stream to ionize the ¹¹B. The ionization energy for ¹¹B is 64740 kJ / mole.

Fuel injection

Once both fuel components are fully ionized the plasma valves for both fuels are opened and the plasma is directed back into the cylinder. As the Kicker emits β radiation through the cylinder and out toward the exhaust, it “drags” the fuel back up the cylinder. At a terminus point, when the fuel has acquired an energy greater than the potential energy thus presented by the electrostatic forces of confinement in the fusion torus, it is magnetically directed into the fusion reaction plasma. As the torus fills with fuel plasma a thin film develops inside the torus due to the geometry of the electrostatic forces of confinement. In addition, the high kinetic energy of the ²H-¹¹B fuel increases the temperature in the fusion torus. Plasma heating is performed by this process alone, that is, the fueling of the fusion reactor chamber by the forceful emission of β radiation from the Kicker, operating at approximately 5 GW power. Once all fuel has been thus injected, the liquid moderator is forced back into the Kicker reaction chamber and the Kicker shuts down. The circulating Nb is stilled and freezes in place. Once the Kicker shuts down, electrical power from the deceleration of ²H ceases. Auxiliary power is decoupled and the fusion reactor sends product to the same magnetic torus used previously and electrical power is generated by decelerating that product. Available electrical power, depending on the settings applied, can be measured in Tera Watts.

The entire Kicker runtime is about 30 minutes. It can also be run for the duration of a transit burn if necessary; a point we will explore later.

For reference, we note that the vapor pressure of ¹¹B is as follows:

Temperature/Vapor pressure:

2348K – 273.15 = 2075 C @ 1 Pa
2562K @ 10 Pa
2822K @ 100 Pa
3141K @ 1 k Pa
3545K @ 10 k Pa
4072K @ 100 k Pa

Density and pressure as a function of electrostatic forces

The ITER project proposes the construction of a monstrosity that weighs more than the Eiffel tower yet is orders of magnitude too weak to contain the pressures necessary for practical fusion. This is yet another aspect of fusion research that gets little coverage. Also related to the power conversion problem generally, plasma pressures will have to be extraordinary if a reasonable power density is to be expected, regardless of the fuel chosen. Indeed, herein lies yet another materials science impediment for which adequate materials for this purpose were available only in the last couple of years. Any viable fusion reactor design in the near future will be totally dependent on nanotube materials in order to contain these pressures. As we shall see, this Reference Design specifies a plasma pressure at maximum rated power of about 1.5 million bar. A blanket approximately 2 meters thick made of nanotube materials will be required for the construction of the reactor vessel.

1 Pascal is 1 N / m² means that the confinement forces must apply x amount of Newtons force over an area of 1 m^2, that is, it is the sum of all confinement forces acting over that area.

1 * 10⁶ bar = 1 * 10¹¹ Pascal. Therefore, the sum of all force acting over any given square meter surface must equal 10¹¹ Newtons force.

Let us begin our inquiry into this problem by tackling some fundamental physics first. The strength of a magnetic field can be expressed in the limit law form as:

dB = (κ_m / 4 * π ) / ( qv ds X r / \|r\|³ )	[eq. 3.02]
Biot-Savart Law

Which can be rewritten as

F_m12 = κ_m q₁ q₂ * (v₁ X (v₂ X r₁₂)) / r²₁₂	[eq. 3.03]
Magnetic Force acting on q₁ due to q₂

κ_m= 4π * 10^-7 T * m / A and is known as the “permeability of free space”, q₁ and q₂ are the charges in Coulombs of each observer 1 and 2, v₁ and v₂ are the velocities of the respective observers and r₁₂ is the right line distance between observers 1 and 2. Notice there is no mention of a magnetic field in that equation. This follows naturally from the fact that magnetic fields exist only in the presence of moving charge; that is, only in consequence of an already existing E field. This will become significant in MCF discussions infra.

Likewise, the electrostatic force acting on the same two observers can be found with a similar relation:

F_e12 = κ_e q₁ q₂ / r²₁₂	[eq. 3.04]
Electrostatic force acting on q₁ due to q₂

where κ_e = 8.987 * 10⁹ N * m² / C²and is the electrostatic proportionality constant and q denotes the charge of a particle in Coulombs. We can arbitrarily choose a value for |F_e12| and |F_m12| such that these two forces acting on two particles differ in magnitude by a known constant, C_p:

C_p [κ_e q₁ q₂ / r²₁₂] = κ_m q₁ q₂ * (v₁ X (v₂ X r₁₂)) / r²₁₂	[eq. 3.05]
Electrostatic and magnetic force compared for q₁ and q₂

which is just (4π * 10^-7 * |v₁| * |v₂|) / (8.987 * 10⁹)

But the values for |v₁| and |v₂| cannot exceed c. If we assume the maximum possible boundary value, then we can say that we have [(4π * 10^-7 * |v₁| * |v₂|) / (8.987 * 10⁹)] * c².

What this shows is that:

[(4π * 10^-7 * |v₁| * |v₂|) / (8.987 * 10⁹)] * c² = 12.567;

The problem, however, is that this assumes velocities far exceeding what even the most ambitious fusion reactor concepts have proposed. And these velocities would require advances beyond the current state of the art in materials science. Tokomak plasma rotational velocities (the highest velocity medium for this purpose) do not exceed 0.1 % the speed of c (e.g. 0.035% c path speed at TEXT facility, Texas). Nor is this needed for fusion. Any significantly higher velocity would simply be an energy loss. Thus, when we take the more reasonable figure the problem becomes evident. The figure above plummets to:

s_v = (4π * 10^-7 * (c * 0.01)²) / (8.987 * 10⁹) = 1.2567 * 10^-5	[eq. 3.06]
Ratio of strength of Magnetic field to Electric field at velocity of merit

Clearly, the obvious candidate for a realistic fusion reactor pressure providing force is the electric force. This follows immediately from the findings supra and from the fact that we cannot harness either the Weak or Strong nuclear forces because our materials science (especially in terms of tolerances) cannot operate at the scales over which those forces operate. And Gravity would require far too much mass. The remarkably high pressure we seek, 1.5 million bar, has yet to be justified, but a full treatment is promised in what follows.

In terms of traditional Tokomak design, this can be understood as a significant inefficiency in power conversion into the reactor fuel. It also dramatically undermines the specific power due to the fantastic mass requirements of Induction Magnets.

Charge density, the density of charge in Coulombs, is expressed:

Q =	[eq. 3.08]
Charge density of a charged volume

where x, y and z are the coordinates locating the total charge and ρ is the density function for charge as a function of r; a position vector within the total charge volume.

To find the total count of charge carriers in a volume we use:

N =	[eq. 3.09]
Number of charge carriers in a charged volume

Where N is the number of charge carriers in the volume and n ( r ) is the position dependent charge carrier count density.

In attempting to find a more efficient means of plasma confinement we examined the current state of the art in materials science in the area of conductor materials. In particular, we sought an ion superconductor (for reasons that will soon become clear) because of the power conversion issues that will arise if a conductor were to exhibit any Ohmic resistance greater than 0. As a template for the set of compounds already discovered we chose to examine the Ag₂Se superionic conductor and Nickel based ion superconductors that transport Ag ions. Some possible states for the Ag₂Se superionic conductor material are:

At -4 eV excitation, the peak region of charge density for Ag₂Se, the states for Ag and Se are 4d and 4 p respectively. This is two states for that material.

A charge density of 4 is known experimentally for each state => charge density of 8 per atom per eV. The time average of radial separation for Ag₂Se atoms at peak pair values in the superionic state is about 3 Angstrom. This determines the density of ions but does not speak to the density of electrons contained in the lattice atoms of Ag2Se, which are composed of different ions (non-flowing ions). Therefore, one meter of material will yield

1 / (6 * 10^-10) = 1.67 * 10⁹ [1 Angstrom = 10^-10 m]

positive ionic nuclei, equally divided between Ag and Se ions. Ag ions consist of 47 protons and Se ions consist of 34 protons. This gives an average of about 40 protons per ion (assuming the ionic charge of conduction is the same as the lattice atomic elements, which it apparently is). Therefore, the integral of charge from x₀ to x is:

= 40 * 1.67 * 10⁹ / m = 6.68 * 10¹⁰ ions per meter or,

N = = (6.68 * 10¹⁰)³ = 2.981 * 10³² ions / m³

which is (2.981 * 10³²) * 1.602176565 * 10⁻¹⁹ = 4.78 * 10¹³ C.

using eq. 3.03 and assuming plasma charge (which we will justify infra), then, at 10 cm distance this generates a force of:

F_i = [ 8.987 * 10⁹ * (4.78 * 10¹³) * (1.602176565 * 10⁻¹⁹ * 2.598 * 10²⁷)] / (0.1)² = 1.79 * 10³⁴ Newtons. We reduce by one order of magnitude to adjust for the vessel geometry:

1.79 * 10³³ Newtons force applied to the plasma.

This figure assumes a drift velocity of 1 m / s, which is conceivably achievable, especially with the large headroom this value gives us.

This Reference Design specifies a minimum force applied to the plasma of ~10¹¹ Newtons / m², the approximate force required to generate 1.5 million bar. Therefore, current technology in superionic conductor material is sufficient for our purpose.

This calculation was for making an order of magnitude estimate of charge density. This is the charge that passes a given point in the conductor per second per unit area. Because this figure exceeds the figure required for plasma compression, it readily exceeds the figure required for creation of magnetic fields for control of the plasma (this topic will be addressed in a following section). Reducing the Reference Design thickness of the conductor and modifying our drift velocity assumption in order to satisfy the pressure containment requirement while minimizing thickness, we have:

1 cm thickness at 1 mm / s drift velocity

ð 1.79 * 10²⁹ Pascal

This still leaves considerable headroom for materials science adjustments (operating temperature being one possible limitation). We shall examine this in more detail in what follows.

Given the findings supra we specify a reactor vessel of a generally toroidal shape and made primarily of nanotube materials. In order to make extraction of fusion product as technically easy and achievable as possible, we specify a reaction plasma generally located in the center of the torus that is a thin film, no more than 3 cm thick. Calculations we shall perform later will show mathematically the superior fusion product capture that will occur as a result of this geometry. We shall also find later that this provides other notable advantages.

A quick inquiry into the current state of the art in nanotube technology will show that the tensile strengths required of this material are currently sufficient to satisfy our requirements for a pressure of 1.5 million bar. However, limitations and restrictions abound. First, the thermal durability of these materials is still poor. This will require active and aggressive cooling. Second, and as alluded to supra, the tensile strength of the material is not the only concern in assessing its overall strength. Third, the plasma pressure to nanotube normal force interface is problematic. Even the hardest nanotube materials cannot withstand the pressure exerted on the inner surface of the material in the geometry we will require. Intrusion of charged particles into the nanotube material will erode (destroy) it. Fourth, there is as yet no manufacturing process to consistently manufacture the volumes of the material we will require. We specify a design that addresses all four concerns thusly:

A nanotube, laminate blanket is constructed in a toroidal shape and thus has a hollow center. Its inner surface shall be made of a neutron super mirror material (alternating layers of Ni and Ti) to be described later. It shall have a thickness of about 2 cm. It shall also have a negative electric charge applied during operation [layer 1]. Embedded in Layer 3 is a honeycomb of conduits equivalent to about 2 cm thickness into which liquid Li is pumped at high velocity (about 100 m / s). For diagrammatic purposes here, we treat it as a separate layer [layer 2]. Backing that layer is a layer of super pressed Gd about 4 cm thick and weighing about 225 mT [layer 3a]. Backing that layer is a layer of super pressed Os about 2 cm thick and weighing about 240 mT [layer 3b]. Backing that layer is a vacuum about 2 cm thick into which liquid N is pumped at modest velocity (about 10 m / s) [layer 4]. Backing that layer is a layer about 1 cm thick of Boron Nitride Wurtzite Nanotube, an electrical insulator and very hard material that resists N penetration. It also provides normal forces for containing plasma pressure [layer 5]. Backing that layer is an electron superconducting layer. The overall charge in this level is significantly less than that of the ion superconducting layers. It is used to moderate the electrostatic forces and will be described in greater detail in the text [layer 6]. Backing that layer is a repeat of layer 5 [layer 7]. Backing this layer is a layer of ion superconducting material about 10 cm thick to be described later. Ions in this conductor experience an electrostatic force exerted from ions within the plasma approximately 20 cm toward the center of the torus [layer 8]. Backing this layer is a layer identical in all respects to layer 5 [layer 9]. Backing this layer is a layer about 4 cm thick made of carbon nanotube material which acts as the primary stiffener and normal force to the electrostatic forces of the plasma [layer 10]. This lamination cycle repeats starting with a layer identical in all respects to layer 4. The total number of layers is approximately 90. Layers consisting of liquid coolants are bridged by carbon nanotube perpendicular “walls” assembled in a helical pattern about the torus. Superconductors have similar shaped insulators running the same helical course (the mathematical description of this helix will be provided later). Each “wall” is a V-shaped hollow vane to support electrostatic symmetry. Layers 2 and 4 receive several hundred very narrow return lines through the V-shaped walls. Return lines for layer 2 are lined with thermal insulators. Due to limitations in manufacturing tolerances, parallel processors are employed to anticipate out of symmetry charge densities and issue commands to correct these anomalies in real time. We can illustrate the layers thusly:

Internal layers (not repeated)

[layer 1]: alternating layers of Ni and Ti: 1 cm

[layer 2]: liquid Li: 2 cm (active Neutron and thermal absorber – flow rate: 100 m / s)

[layer 3a]: Gd: 3 cm (primary Neutron Transport backstop) flux is to background far side

[layer 3b]: Os: 1 cm (primary gamma backstop); is to background far side

Repeating layers

[layer 4]: liquid N: 1 cm

[layer 5]: Boron Nitride Wurtzite Nanotube: 1 cm

[layer 6]: electron superconducting material:1 cm

[layer 7]: Boron Nitride Wurtzite Nanotube: 1 cm

[layer 8]: ion superconducting material: 10 cm

[layer 9]: Boron Nitride Wurtzite Nanotube: 1 cm

[layer 10]: carbon nanotube material: 4 cm:

[layer 11]: Repeat pattern starting with layer 4.

The overall layer pattern is a spiral that works its way outward from the center, providing a continuous, helical conductor surface from the interior of the reactor vessel to the outermost layer.

The key to this arrangement is subtle, so we will explicate it here. Charge density and possibly current in layer 6 is adjusted so that electrostatic forces between the plasma and layer 6 are virtually equal to the total force exerted by the plasma pressure (in other words, the electrostatic forces are adjusted so that they are almost, but not entirely sufficient to support the plasma pressure). If the force exerted on a square meter of surface in layer 6 consequent to the interaction of the charges located in the reaction plasma and layer 6 is denoted F₆ then the Reference Design requires that F_p- F₆= W_f ≪ F_p where F_p is the totalforce exerted (as an electrostatic potential energy) on the same square meter by the total plasma pressure. And W_f is the normal force exerted over that same square meter by the nanotube layer 7; which is in turn supported by layer 8. Thus, equilibrium is given by:

F_{n – 1}= F_n+ W_f@n ∀ n in blanket [ W_f@n ≪ F_{n – 1}]	[eq. 3.10]
Reactor Vessel Wall Equilibrium Condition (simplified)

where W_f ≪ F_p

And now; the denoument: the force F₆ experiences a normal force of equal magnitude from the next analogous layer backing it; that is, layer 12. And at that layer the same equilibrium condition is repeated. The difference however, is that the electrostatic normal forces decrease with each pattern repetition. The effect is to distribute the plasma pressure force evenly throughout the nanotube material such that the total force created by the pressure is not exerted on a single surface. In each case of electrostatic normal forces, the forces are acting on charged particle to charged particle and no material is involved. The pressure of the plasma, therefore, is transmitted via fields only. After a repetition of some 90 cycles over approximately 2 meters of blanket, the final material layer absorbs the remaining force delivered by plasma pressure by presenting a normal force to it. We will pause here with this overview and describe this pressure confinement system in much more detail in what follows. For now it is only important to understand that the strategy here is to complement and reinforce magnetic field lines that will be created as a consequence of this arrangement. This allows the Reference Design to follow the well-established experimental data already amassed in traditional Tokomak fusion.

It has been stated supra that “The choice of transit propulsion represents the greatest risk factor for this project”, and we can now reduce that statement to this:

The ability to create specific classes of carbon and Boron-Nitride nanotube material in quantity represents the greatest risk factor for this project. Once this problem is solved, the remainder is relatively short term development work.

Mass of an atom of ¹¹B is about 11 times that of a proton.

To determine speed of ²H -¹¹B product, we take 2 of the 3 helium nuclei produced by this reaction (because the 3^rd is a “slow” product) and calculate their velocity:

⁴He fast nuclei ²H-¹¹B fusion product v = 3.28 % c or 0.0328 * c; e.g.

(((4 * 10^6[energy of fast ⁴He in MeV]) * (1.60217646 * 10^-19 [conversion to Joules]) / (5 * 10^-27[mass of ⁴He]))^(1/2)) / 299792458 [speed of light].

A quantity of 2 of these fast ⁴He are generated with velocity vectors 155 degrees from each other. Coordinate system orientation is random.

Therefore, plasma velocity is best limited to about 0.001 c maximum. This yields:

0.001 * c ≈ 299,792 m/s

The ideal geometry for capturing product for kinetic thrust now appears to be a torus reduced to a cylinder in which fusion occurs in a thin film at the cylinder wall. It is possible to use cylinders with curvature in their walls along the plane of their radius but the math is made more complex. This configuration works best with the geometry of the ²H-¹¹B reaction due to the 155 degree separation between product vectors. An examination of this geometry shows that reaction product can be recovered from a little less than one-half of all reaction product kinetic energy.

We begin by first noting the percentage of total output lost due to re-absorption into the plasma and annealing. This we can estimate to be about 10% based on our “critical mass” calculations and a thin film of no more than six inches thick. In addition, neutron flux will account for as much as 0.1 % of total power output. At Tera Watt power levels this is substantial. Finally, we will need to take into account the loss of the slow ⁴He nuclei which is assumed in the ideal geometry. The total energy release from a ²H -¹¹Bo reaction is 8.7 MeV, of which 8 are accounted for by the fast nuclei. The remainder, 0.7 MeV is the kinetic energy of the slow nucleus. This energy is lost to the aggregate reaction rate power output by re-absorption and annealing.

The strategy is to find a way to capture the useable product mentioned by capturing it in a neutral E orbit outside the plasma. Once there, it can be redirected into accumulators.

An examination of the ideal geometry shows that probabilistically only about 32.5 percent of the reaction product can be recovered for kinetic thrust (this assumes a 10% loss of product by nuclei quantity as a result of product being captured by plasma and remaining therein). These calculations assume a successful deflection of product nuclei whenever incident upon the reactor vessel wall with an angle in the interval (π / 4, 0). Deflections at steeper angles will result in an electrical potential being produced in the electrostatic confinement field; thereby releasing approximately 32.5 % of the power output in the form of electric current. The remaining 35% will be power in the form of significantly reduced velocity of alpha particles and in the re-heating (and annealing) of the plasma; thereby providing the heating necessary to sustain the fusion reaction once initiated. These proportions are estimates that mathematically should be close to the real experimental values one would likely obtain. Some of the power recovered as current can be used to power the electromagnets and other reactor power devices. In addition, some of this power can be tapped for direct electrical within the spacecraft.

Distance between reaction plasma surface and reactor wall must be adjusted, with other related dependencies, to accommodate this optimum geometry. A curved cylinder wall, instead of a straight one, will improve the spatial tightness of the deflection. The final adjustment to make is to note that neutron flux may steal as much as 0.1 % of the total energy output available, therefore, adjusting for that, we can estimate recoverable power at 32.5 – (32.5 * 0.1) = 29.25 % as a very close figure for characterizing reactor efficiency. For our purposes, this alone will require a total reaction rate yielding a power production of 65 TW, enough to ensure the required 21 TW kinetic thrust we need and also take into account the power required to run the magnets and compensate for system energy losses (which, relative to these figures, will be tiny).

Heating of the plasma should be described to preclude any confusion. Heating is the process whereby the fuel is “vibrated” about its orbital axis as it spins in a helical orbit within the torus. The very strong electrostatic confinement coupled with the fine control of orbits given by the electromagnets ensures that even at very high temperatures the vibrations will result only in motions about the orbital path (a circular helix), but vibrations insufficient in strength to break free of this orbital path.

One can view this as a key requirement that the density and orbital confinement be tight enough to overcome the vibrations of heating.

Thus, when temperature is expressed in KeV what we are referring to is the kinetic energy, and hence velocity, of the fuel particles in vibration, not their velocity in their orbit, which is an entirely different energy measure. Thus, the total energy of a given particle is the sum of its kinetic energy due to heating and its kinetic energy imparted by the electromagnets and electrostatic confinement.

Given our previous considerations, we can calculate that our maximum plasma temperature will necessarily be about 122 KeV because we restricted the upper velocity of our fuel to a fraction of the product velocities, as described supra.

The nuclear binding energy of ¹¹Boron is 6.9277 MeV. This value corresponds to the average kinetic energy of collision particles that optimizes fusion reaction rates.

[1] But again, this is largely because scientists are not genuinely concerned with these issues. They are doing pure research and a single-mode is the better approach for their purpose.

If we assume a design pressure requirement of 1.5 * 10¹¹ Pascal then we need to calculate the particle count in a thin film whose density decreases as we move outward in accordance with the electrostatic equation for force:

F_p = φ Q_P / (ℓ₀- ℓ₄)²

Where we let φ = κ_e q_m1 where κ_e = (8.987 * 10⁹ N * m² / C²) is the electric constant and q_m1 is the charge of the particle placed under operating pressure (a “test” charge). Q_P is the sum of the charges of the rest of the plasma.

∴ as an order of magnitude estimate we know that the force will decrease as 1 / ℓ₀- ℓ₄)² and the potential energy is:

F_p ⨀ (ℓ₀- ℓ₄)= – φ Q_P / (ℓ₀- ℓ₄)²

and F and ℓ are perpendicular to the plane tangent to the plasma surface so that cosine ((π / 4) – ϕ) = 1 ∴

F_p = – φ Q_P / (ℓ₀- ℓ₄)

=> F = – φ Q_P ℓ (ℓ)^-1

=> F = – φ Q_P ln (|ℓ₀- ℓ₄|)

and if the magnitudes of ℓ and F are known constants (design requirements) then, letting ω = ln (|ℓ₀- ℓ₄|):

F / ω= – φ Q_P

We next treat our test particle as a sheet of charge with A = 1. Then q_m1 and Q_P are related by:

q_m1 = Q_P/ (ℓ₀- ℓ₄). Casting the problem as a test charge a distance (ℓ₀- ℓ₄) from a charged disk (r < 10 in this approximation since curvature will reduce fidelity):

e = F / q_m1 = { Q_P (1 – ([ln(ℓ₀- ℓ₄)] / √ [ [ln(ℓ₀- ℓ₄)]² + (ri)²])) } / [ 8ϵ₀π(ri)² ]

ð Q_P= F * [ 8ϵ₀π(ri)² ] / { q_m1 (1 – ([ln(ℓ₀- ℓ₄)] / √ [[ln(ℓ₀- ℓ₄)]² + (ri)²])) }

And the particle count, N₁, in a “charge” area q_m1 = Q_P / (|ℓ₀- ℓ₄|) is:

q_m1/ (ϵ)

where ϵ = 4.806 * 10^-19, the averaged charge per particle (at 50/50 mix). In a later section we will calculate the required reactor vessel wall charge.

The number of particles present must obey the relation:

[Q_P/ (4.806 * 10^-19)] = (2.791 * 10²⁷)^2/3 * (|ℓ₄- ℓ₅|)

ð [Q_P/ (4.806 * 10^-19)] = (2.791 * 10²⁷) * (0.1)

ð Q_P= (4.806 * 10^-19) * (2.791 * 10²⁷) * (0.1) = 1.34 * 10⁸ Coulomb per plasma 2D sheet.

This yields 1.34 * 10⁸ Coulomb per square meter of plasma and ((1.34 * 10⁸)^1/2)³

= (1.55 * 10¹²) Coulombs / m³. ∴

An explicit machine calculation gives:

Q_P = {|F| * (|ℓ₀- ℓ₄|) * ( 8ϵ₀π(ri)² ) / q_m1(1 – (|[ln(ℓ₀- ℓ₄)]| / √ [|[ln(ℓ₀- ℓ₄)]|² + (ri)²]))}

|F| = (1.34 * 10^8) / ((0.1) * ( 8*(8.85 * 10^(-12))*π(5)² ) / ( ((4.806 * 10^(-19)) * (2.791 * 10²⁷) / (0.1) ) * (1 – (|[ln(0.1)]| / √ [|[ln(0.1)]|² + (5)²]))})

|F| = (1.34 * 10^8) / ((0.1) * ( 8*(8.85 * 10^(-12))*pi *(5)^2 ) / ( ((4.806 * 10^(-19)) * (2.791 * 10^27)^(2/3) / (0.1) ) * (1 – ((ln(0.1)) / sqrt ((ln(0.1))^2 + (5)^2)))))

= 3.26 * 10¹⁸ Pascal

This is seven orders of magnitude above our target pressure of 10¹¹. We set that value infra because it represents the current technological limit of materials science.

The plasma total volume per film layer of 5,500 m³ cannot be increased without putting the specific power too low and creating an overall reactor vessel weight > 15000 mT. If we add 10 thin film layers to the design, we can recover an order of magnitude (+) thusly:

Q_P= (4.806 * 10^-19) * (2.791 * 10²⁶) * (0.1) = 1.34 * 10⁷ Coulomb per plasma 2D sheet.

|F| = (1.34 * 10⁷) / ((0.1) * ( 8*(8.85 * 10^(-12))*pi *(5)^2 ) / ( ((4.806 * 10^(-19)) * ((2.791 * 10²⁶))^(2/3) / (0.1) ) * (1 – ((ln(0.1)) / sqrt ((ln(0.1))^2 + (5)^2)))))

= 7.02 * 10¹⁶ Pascal

But we need 3.9 * 10¹¹ Pascal

while maintaining the same number of fuel ions in the plasma. We can increase those layers now by one more order of magnitude, to 100 layers, and get:

7.02 * 10¹⁵ Pascal.

The solution to this dilemma is to pass a negative charge carrier current through the innermost conducting vessel wall to “dope” the charge in the vessel wall, thereby reducing the effective electrostatic force between the wall and the plasma whilst retaining the same number of ions in the plasma. Notice that the plasma still remains a single-species, ionic plasma.

Having said this, we’ve completed this calculation totally neglecting (deliberately in order to demonstrate the value of electrostatic confinement) the magnetic field thus far, and the total amount of doping will be something less than indicated here, the exact amount determined by the strength of the magnetic field. We note that the limiting factor on doping will be the hardness of the Boron Nitride Wurtzite Nanotubes as each charge carrier will be “pushed” against it “harder” the fewer of these ions available in the vessel wall are.

If we make a reasonable estimate of two orders of magnitude for magnetic confinement (consistent with current research achievements) then the doping will need to be sufficient to reduce the pressure by two orders of magnitude, from 7.02 * 10¹⁵ to 7.02 * 10¹³ Pascal. And magnetic confinement can reduce it further to 7.02 * 10¹¹ Pascal. Additionally, the number of layers could be reduced to 10 with an order of magnitude increase in the “doping effect” or the magnetic confinement contribution (the upper limit of current research success).

With a correct proportion of “pressure” and particle count, we can continue

The strongest nanotubes to date have a specific strength of about 48000000 N·m·kg⁻¹. Using the maximum feasible weight for the reactor of 1000 kg / m³ gives:

1000 * 48000000 = 4.8 * 10¹⁰

which is ≈ 1.5 * 10¹¹ Newtons. This represents a current materials science limit. Thus we can now construct an specification reaction rate:

μ= n₁n₂ σ_R(E_p) v_Tai √ (N₂(ai))

where n₁ is the reactant number density of the projectile/s and n₂ is the reactant number density of the target. n₁₊₂ is one if no non-fuel species are present. σ_R(E_p) is the “cross section” as a function of E_p, E_p is the energy of the incident particle, and v_T is the relative particle velocity, all of which are due to relative motion only. E_r is the energy released from each reaction (8.7 MeV). N₂ is the particle density function of the “incident” particle stream, and N₁ is the total number of particles available for reaction.

μ= (0.2)*(0.8) * (ln(0.1)) * (1.2 * 10^-28) * (1.12 * 10^7)* ((2.791 * 10^27) ^(1/2))

= 2.6 * 10^-8

and (2.791 * 10²⁶) * (2.6 * 10^-8) = 7.26 * 10¹⁸ reactions / m³ / s (we justify 10²⁶ supra and infra)

And as discussed previously, the particle count, like power density, will decrease as the square of the distance from the center of the plasma. Thus “a” becomes ln a (which is the solution to the integral above).

2* (0.2)*(0.8) *(8.6605021 * 10^26) * [|ln(0.1)|] * (1.2 * 10^-28) * (1.12 * 10^7)* ((8.6605021 * 10^26) ^(1/2)) * (1.37 * 10^-12) ~ 3.46 * 10⁷ watts / m³

Using the velocity calculated in our cross section discussion, we can write an explicit machine calculation for a distribution of notable N₁ values as:

2* (0.2)*(0.8) *(2.791 * 10^27) * ((-1)*ln(0.1)) * (1.2 * 10^-28) * (1.12 * 10^7)* ((2.791 * 10^27) ^(1/2)) * (1.37 * 10^-12) = 2 * 10⁸ watts / m³

The reaction vessel contains 550 * x = y m³ plasma (where x is the number of thin-film layers used). ∴

(2 * 10^8)*(550 * 100) = 11 Tw total power output

Which is what we sought to specify.

So, this requirement works out to be 11 Tw maximum total output, 7 Tw usable maximum total output and 5.25 Tw guaranteed continuous; which is the requirement we sought to satisfy. The difference of 7 – 5.25 is a function of how the reactor is fueled and at what pressure range it operates since as it proceeds through a full burn it will consume some of that fuel and the internal pressure would be variable. At burn initiation there would be 7 Tw power slowly decreasing monotonically to a minimum power of 5.25 Tw before the burn ends.

Without meaning to beat a dead horse, we will seek to make it clear now that the addition of electrostatic forces to magnetic confinement tokomak schemes is the most productive path.

We can now see that as we increase the number of atoms in each m³ we increase the cross section. That in turn increases the reaction rate. The values for n₁ and n₂ are the concentration percentages for projectile specie and target specie in the plasma (for the ²H-¹¹B reaction it is recommended to be something like 8 to 1, however, by this equation alone, the obvious optimization is half and half).

Next, we consider the boundary limitations required for the plasma geometry and its magnetic and electric fields. The first thing we want to understand is whether or not thermal perturbations of toroidal orbits will be annealed within the plasma without destabilizing it. For magnetic only confinement this is a problem that cannot be easily calculated. However, in our scheme, the presence of an E field makes this much easier.

We can frame the problem thusly. We seek a geometry and electrostatic force such that, whenever a particle of ¹¹B or ²H is perturbed from its orbital path, it’s deviation does not place the particle outside the electrostatic potential well. We can diagram it thusly:

Figure 1 (a) and (b) below

Let us begin by drawing a line segment in Figure 1 (b) in order to evaluate the electrostatic forces involved. We will mark each point with the letter ℓ_n∈ ck. U(p₀) is located at the point charge provided by the charge contained in the nanotube blanket’s outer layers (the total charge in the outer superconducting layers), call it ℓ₀. U(p₁) is located at the reactor vessel’s outer (meaning outer toroidal) interior wall. This is the ultimate orbital boundary which fusion product should never reach. We denote that point ℓ₁. Continuing to work our way inward from the outside of the torus toward the plasma, the next point is E_nout, call it ℓ₂. This is the point at which a locally, electrically neutral orbit can be found. This orbit exists because a modest negative charge exists on the solid wall surface at ℓ₁. It is considerably weaker than the primary charges we will consider, but is strong enough at ℓ₂ to have a local effect whereby any particle in the immediate vicinity (i.e. between any demarcated points in ℓ) will “feel” a force significant to the particles perturbations in this area which we will examine shortly. Continuing through the vacuum, we next consider point ℓ₃ (h_out) which is the outer horizon, or outer surface of the plasma. The next point, ℓ₄, is the component location at ck where the particle being acted upon is located. Here we will consider bj = 0 and ai > 0. The next point, ℓ₅, is the electrically neutral orbit at the center of the plasma. This is the orbit whereby the forces applied from Q_o and Q_i on a particle at this orbit cancel. It is a surface and has a toroidal shape. In the diagram of Figure 1 (b) it is given that ℓ₄ = ℓ₅ but this need not be so. However, the upper boundary we wish to test is for a reaction that occurs at the plasma surface, that is, ℓ₃, and we seek the condition ℓ₃ = ℓ₄. At this orbit the electrostatic potential energy to acquire in order to reach ℓ₁ is minimized. Continuing along the line segment we next arrive at ℓ₆which is the inner horizon and is analogous to ℓ₃. Points ℓ_6-9have like analogies to the outer points. Therefore, the condition we must enforce is:

Assuming ℓ₀ > ℓ₉the boundary condition is met at:

-U(ℓ₁)_m1 = [( F_ℓ0 - F_ℓ4) ⨀ (ℓ₀ - ℓ₄ )] – [(F_ℓ4 - F_ℓ9)⨀ (ℓ₄ - ℓ₉ )]	[eq. 3.19]
Electrostatic potential energy

where F denotes a force vector.

And it is required that:

|-650 KeV| = [( F_ℓ0 - F_ℓ4) ⨀ (ℓ₀ - ℓ₄ )] – [(F_ℓ4 - F_ℓ9)⨀ (ℓ₄ - ℓ₉ )]

ð 1.0414 * 10^-13 = [( F_ℓ0 - F_ℓ4) ⨀ (ℓ₀ - ℓ₄ )] – [(F_ℓ4 - F_ℓ9)⨀ (ℓ₄ - ℓ₉ )]

ð 1.0414 * 10^-13 / cos( θ ) = [( F_ℓ0 - F_ℓ4) * (ℓ₀ - ℓ₄ )] – [(F_ℓ4 - F_ℓ9)* (ℓ₄ - ℓ₉ )]

If we set s = 0.01 (1 cm) and cos( θ ) = 1 ( a good choice as this is the upper bound condition) then:

1.0414 * 10^-13 = [( F_ℓ0 - F_ℓ4) * (ℓ₀ - ℓ₄ )] – [(F_ℓ4 - F_ℓ9)* (ℓ₄ - ℓ₉ )]

ð 1.0414 * 10^-13 + [(F_ℓ4 - F_ℓ9)* (ℓ₄ - ℓ₉ )] = [( F_ℓ0 - F_ℓ4) * (ℓ₀ - ℓ₄ )]

Let φ = κ_e q_m1 where κ_e is the electric constant. Then the equations for potential energy, starting with forces, are:

F_o = φ Q_o / (ℓ₀- ℓ₁)² – φ Q_o / (ℓ₀- ℓ₄)²

F_i = φ Q_i / (ℓ₄- ℓ₉)² – φ Q_i / (ℓ₁- ℓ₉)²

F_p = φ Q_p / (ℓ₄- ℓ₅)² – φ Q_p / (ℓ₁- ℓ₅)²

F_R = F_o – F_i – F_p

ð U( Δℓ )_m1 = F_R ⨀ (ℓ₁- ℓ₄) ∴

We can expand thusly:

U( Δℓ )_m1 =

[φ Q_o / (ℓ₀- ℓ₁)² - φ Q_o / (ℓ₀- ℓ₄)²- φ Q_i / (ℓ₄- ℓ₉)² - φ Q_i / (ℓ₁- ℓ₉)²- φ Q_p / (ℓ₄- ℓ₅)² - φ Q_p / (ℓ₁- ℓ₅)²]

⨀ (ℓ₁- ℓ₄)

ð U( Δℓ )_m1 =

φ [Q_o / (ℓ₀- ℓ₁)² - Q_o / (ℓ₀- ℓ₄)²- Q_i / (ℓ₄- ℓ₉)² - Q_i / (ℓ₁- ℓ₉)²- Q_p / (ℓ₄- ℓ₅)² - Q_p / (ℓ₁- ℓ₅)²] ⨀

(ℓ₁- ℓ₄)

And since equilibrium requires that Q_i = Q_o we can simplify with

Q = Q_i = Q_o

ð U( Δℓ )_m1 =

φ Q[1 / (ℓ₀- ℓ₁)² - 1 / (ℓ₀- ℓ₄)²- 1 / (ℓ₄- ℓ₉)² - 1 / (ℓ₁- ℓ₉)²] – φ Q_p [1 / (ℓ₄- ℓ₅)² - 1 / (ℓ₁- ℓ₅)²] ⨀

(ℓ₁- ℓ₄)

And cosine returns 1 when examining the upper bounds so that,

U( Δℓ )_m1 =

φ Q(ℓ₁- ℓ₄)[1 / (ℓ₀- ℓ₁)² - 1 / (ℓ₀- ℓ₄)²- 1 / (ℓ₄- ℓ₉)² - 1 / (ℓ₁- ℓ₉)²] – φ Q_p (ℓ₁- ℓ₄) [1 / (ℓ₄- ℓ₅)² - 1 / (ℓ₁- ℓ₅)²]

And the boundary condition of interest is solving for Q or Q_p for given values in ℓ and the constraint that the electrostatic potential well must be greater than or equal to the highest thermal energy of any fuel particle (we examine fusion product separately). For the ²H-¹¹B reaction this value will be constant, namely;

T_e = 1.0414 * 10^-13

ð [ T_e / φ(ℓ₁- ℓ₄) ]+ Q_p [1 / (ℓ₄- ℓ₅)² - 1 / (ℓ₁- ℓ₅)²] =

Q [1 / (ℓ₀- ℓ₁)² - 1 / (ℓ₀- ℓ₄)²- 1 / (ℓ₄- ℓ₉)² - 1 / (ℓ₁- ℓ₉)²]

So that, solving for Q_p we have:

Q_p = Q[1 / (ℓ₀- ℓ₁)² - 1 / (ℓ₀- ℓ₄)²- 1 / (ℓ₄- ℓ₉)² - 1 / (ℓ₁- ℓ₉)²] – [T_e / φ(ℓ₁- ℓ₄) ] / [1 / (ℓ₄- ℓ₅)² - 1 / (ℓ₁- ℓ₅)²]

Or for Q:

Q = [T_e / φ(ℓ₁- ℓ₄) ]+ Q_p[1 / (ℓ₄- ℓ₅)² - 1 / (ℓ₁- ℓ₅)²] / [1 / (ℓ₀- ℓ₁)² - 1 / (ℓ₀- ℓ₄)²- 1 / (ℓ₄- ℓ₉)² - 1 / (ℓ₁- ℓ₉)²]

where κ_e = 8.987 * 10⁹ N * m² / C². We can state this equation as a function whose domain includes; T_e, ℓ₀, ℓ₁, ℓ₄, ℓ₅, ℓ₉, and q_m1. However, the general form would be:

Q_p (T_e, ℓ₀, ℓ₁, ℓ₄, ℓ₅, ℓ₉, φ(q_m1) )

Q (T_e, ℓ₀, ℓ₁, ℓ₄, ℓ₅, ℓ₉, φ(q_m1))

Assuming ℓ₀ > ℓ₉the boundary condition is met at:

Q_p =Q[1 / (ℓ₀- ℓ₁)² - 1 / (ℓ₀- ℓ₄)²- 1 / (ℓ₄- ℓ₉)² - 1 / (ℓ₁- ℓ₉)²] –[T_e / φ(ℓ₁- ℓ₄) ] / [1 / (ℓ₄- ℓ₅)² - 1 / (ℓ₁- ℓ₅)²]	[eq. 3.20]
Minimum total plasma charge required for thermal stability	{E_p ≥ 650 KeV}

And

Q =[T_e / φ(ℓ₁- ℓ₄) ]+ Q_p[1 / (ℓ₄- ℓ₅)² - 1 / (ℓ₁- ℓ₅)²] /[1 / (ℓ₀- ℓ₁)² - 1 / (ℓ₀- ℓ₄)²- 1 / (ℓ₄- ℓ₉)² – 1 / (ℓ₁- ℓ₉)²]	[eq. 3.21]
Minimum total wall charge required for thermal stability	{E_p ≥ 650 KeV}

The pressure of the plasma is just the sum of the forces acting on a test particle located at the point where pressure is measured; i.e.

The force acting at a given point can be solved by just choosing a point:

f_o = φ Q_o / (ℓ₀- ℓ₁)²

f_i = φ Q_i / (ℓ₄- ℓ₉)²

f_p = φ Q_p / (ℓ₄- ℓ₅)²

^OR

f_p = φ Q_p / (ℓ₀- ℓ₄)²

f_R = f_o + f_i + f_p

|f_R| = 1.5 * 10⁶ Bar

ð |f_R| = |f_o| – |f_i| – |f_p| = 1.5 * 10⁶ Bar = 1.5 * 10¹¹ Newtons

And the field is:

e_o = κ_e Q_o / (ℓ₀- ℓ₁)²

e_i = κ_e Q_i / (ℓ₄- ℓ₉)²

e_p = κ_e Q_p / (ℓ₄- ℓ₅)²

The rate change in these forces can be given by taking the ratio of any F to a unit distance lying along a boundary condition vector, say ℓ_bc. That is,

∇F_o = φ Q_o { ∂² / ∂ } [1 / (m₀- m₁)² - φ Q_o / (m₀- m₄)²] / (m₀- m₄)

∇F_i = φ Q_i { ∂² / ∂ } [1 / (m₄– m₉)² - φ Q_i / (m₁- m₉)²] / (m₀- m₄)

∇F_p = φ Q_p { ∂² / ∂ } [1 / (m₄- m₅)² - φ Q_p / (m₁– m₅)²] / (m₀- m₄)

∇F_R = ∇F_o + ∇F_i + ∇F_p

ð ∇U( Δℓ )_m1 = ∇F_R ⨀ (m₁- m₄)

where we use m to denote position vectors (all components).

Let:

κ_e = 8.987 * 10⁹

q_m1 = 4.806 * 10^-19

And (1.5 * 10¹¹) is the force required over one square meter, so q_m1 must be the charge present on a square meter surface of thin film; thus q_m1 becomes q_p.

Let:

φ = (8.987 * 10⁹) (4.806 * 10^-19).

ℓ₀= 0.55

ℓ₁= 0.50

ℓ₄= 0.45

ℓ₅= 0.40

ℓ₉= 0.25

We found supra that the number of particles per cubic meter in the plasma is about (6.66 * 10²⁷). Since the average charge for each particle was estimated at 4.806 * 10^-19 we will assume that value. And the electric field we denote e:

e_o = κ_e Q_o / (ℓ₀- ℓ₁)²

e_i = κ_e Q_i / (ℓ₄- ℓ₉)²

e_p = κ_e Q_p / (ℓ₄- ℓ₅)²

We will cast the problem by expressing Q as a charge density using the “force of a charged disk on a test charge” form:

e = ρ_Qo / 2ϵ₀ (1 – Δℓk / √ [Δℓk² + (ri)²])

ð e = F / q_p = { ρ_Qo(1 – (Δℓk / √ [Δℓk² + (ri)²])) } / 2ϵ₀

where ϵ₀ = 8.85 * 10^-12and ρ = Q / 4π(ri)² so,

e = F / q_p = { Q₀ (1 – (Δℓk / √ [Δℓk² + (ri)²])) } / [ 8ϵ₀π(ri)² ]

ð Q₀= F * [ 8ϵ₀π(ri)² ] / { q_p (1 – (Δℓk / √ [Δℓk² + (ri)²])) }

ð F = { q_p (1 – (Δℓk / √ [Δℓk² + (ri)²])) } / [ Q₀8ϵ₀π(ri)² ]

We can set |F| to |F_dr|, the Reference Design required pressure by constraining our evaluation to a square meter surface and casting F as the force required per particle:

Q₀= { (3.26 * 10¹⁸) / ((2.791 * 10^27)^(2/3)) } * [ 8ϵ₀π(ri)² ] / { q_p (1 – (Δℓk / √ [Δℓk² + (ri)²])) }

ð Q_i= { (3.26 * 10¹⁸) / ((2.791 * 10^27)^(2/3)) } * [ 8ϵ₀π(ri)² ] / { q_p (1 – (Δℓk / √ [Δℓk² + (ri)²])) }

ð Q_p= { (3.26 * 10¹⁸) / ((2.791 * 10^27)^(2/3)) } * [ 8ϵ₀π(ri)² ] / { q_p (1 – (Δℓk / √ [Δℓk² + (ri)²])) }

Q₀= { (3.26 * 10¹⁸) / ((2.791 * 10^27)^(2/3)) } * [ 8ϵ₀π(ri)² ] / { q_p (1 – (Δℓk / √ [Δℓk² + (ri)²])) }	[eq. 3.22]
Required charge at reactor vessel wall

q_p is a sheet of charge tangent to the plasma surface and located within it. Previously, when we calculated the required plasma charge, we used the symbol Q_P to denote the same thing. It corresponds to the horizon (h_out) in Figure 1.

Explicitly, we can set:

ℓ₀- ℓ₁ = 0.19 m

ℓ₄- ℓ₉ = 0.24 m

ℓ₄- ℓ₅ = 0.1 m

(ri) = 5

q_p = plasma surface = ((s)*(1.34 * 10⁷)) [see calculations infra]

s = plasma spin speed = 100000 m / s

Q₀= ( (7.02 * 10¹⁶) / ( (2.791 * 10^26)^(2/3) ) ) * ( 8 * ( 8.85 * 10^-12) * pi * (5)^2 ) / ((((1)*( (1.34 * 10⁷)) * ((4.806 * 10^-19)))) * (1 – ((0.01) / sqrt((0.019)^2 + (5)^2))) )

= 142.2 Coulomb

142.2 / (4.806 * 10^-19) = 2.96 * 10²⁰; the number of charge carriers / m³ in the wall and

(7.02 * 10¹⁶) / (2.96 * 10²⁰) = 2.37 * 10^-4 Newtons / particle / m²

100000 * ((2.791 * 10^26)) * (4.806 * 10^-19) * 2*(ℓ₄ – ℓ₅) = q_m1 = 2.68 * 10¹²

=> (ℓ₄ – ℓ₅) = q_m1 / 100000 * ((2.791 * 10^26)) * (4.806 * 10^-19) * 2

=> (ℓ₄ – ℓ₅) = (2.68 * 10¹²) / (100000 * ((2.791 * 10^26)) * (4.806 * 10^-19) * 2)

=> (ℓ₄ – ℓ₅) = 0.1 m

Is the thickness required of the plasma sheet and is partly the basis for characterizing this approach as thin film fusion.

The key Reference Design question is “what is the lowest wall charge we can allow and still be able to transfer the pressure, or force, of 1.5 * 10¹¹ Newtons to the torus structure”?

We can address this numerically by solving for the force exerted by a charge in the superconductor layer onto the Boron Nitride insulator layer. The key question is, does the particle have enough force to penetrate the very hard Boron Nitride surface? This now exposes the wisdom in choosing Boron Nitride Wurtzite nanotube material:

BNW begins to show noticeable deformation at around 1 GPa. But this is just beyond the pressures we have specified. While this cannot constitute an exact statement of the forces involved and considerable experimentation will be required, we can see that BNW has an order of magnitude sufficient hardness to resist penetration by conducting ions from the adjacent conductor layer. However, these hardness ratings are based on cross sections of about 1 mm. Taking the cross section of AgSe as an example, we know that one such ion occupies a space of very roughly:

(3 * 10^-10)²

And hardness tests are conducted by applying pressure over a region of about:

(1 * 10^-3)²

ð (1 * 10^-3)² / (3 * 10^-10)² = 1.1 * 10¹³

142 Coulomb wall charge:

(142) / 2π (ri)² = 0.9 Coulombs m²

0.9 / (4.806 * 10^-19) = 1.87 * 10¹⁸ charge carriers / m² ∴

(1.87 * 10¹⁸) / (10⁶) = (1.87 * 10¹²) and if we specify ~100 nanotube layers in the torus:

(1.87 * 10¹²) * (10²) / (1.1 * 10¹³) = 17.34 that is;

The BNW material is just within an order of magnitude, assuming 100 nanotube layers, of being able to maintain structural integrity under the force of a charge of only 142Coulombs. The reason for this is that the fewer charge carriers present the more force each has to impart to the surrounding material to obtain the normal force they require. Thus, the lowest wall charge that can likely be fabricated with current materials and development is about:

1 * 10^-3 Coulomb

Obviously, as a development risk, and given the very broad approximations we are making regarding this untested material, we specify an effective lower limit of:

1 * 10^-1 Coulomb

f_p = φ Q_p / (ℓ₀- ℓ₄)²

Q_o = [(8.987 * 10⁹) (4.806 * 10^-19)] / [{(1.5 * 10^11) / ((2.791 * 10^26)^(2/3))} * (.2)²]

The force acting to push plasma apart (destabilize it) is:

ð F = { q_p (1 – (Δℓk / √ [Δℓk² + (ri)²])) } / [ Q₀8ϵ₀π(ri)² ]

Which we find to be negligible under our thin film specification, but we must include it to accommodate changes.

We have validated that this plasma can possess the charge necessary to provide a plasma pressure of 1.5 million Bar. We now turn our attention to the question of whether or not the current materials state of the art can provide a material that can hold sufficient charge for the requirements derived infra.

Average charge of fuel particle in plasma:

(5+1/2 = 3) * 1.602 * 10^-19 = 4.806 * 10^-19 and

Earlier we noted that a m³ of plasma at 1.5 * 10⁶ Bar contained 2.791 * 10²⁶ of fuel particles mixed half and half. To get the charge counts, we can average over that value thusly:

5 +1=6

ð 6 / 2 = 3 is the average charge per particle and

ð 3 * 8.6605021 * 10²⁶ is the charge carrier count present at a material density of 9.4227 kg / m³

But since we are looking for a charge contained in the laminate solid layers (proton superconductors) we must modify this density to be more realistic. Let us pick a density for an order of magnitude approximation by assuming the density of copper. This is around 9000 kg / m³. And 9000 / 9.4 ~ 10³. Therefore, for a laminate layer we could realistically expect to squeeze in as much as

3 * (2.791 * 10²⁶) * (4.806 * 10^-19) * 10³ * 5500= 2.2 * 10¹⁵

yielding results similar to what we obtained for the plasma. At this point we are now in a much better position to begin describing the magnetic fields and the field interactions.

As we can see, thermal instabilities are easily excluded from this plasma. Ironically, the challenge we will face here won’t be instabilities in plasma. This plasma will follow B lines like solid rock. However, the intense pressure of the plasma is what makes this challenging technologically. We will next calculate whether or not this many charge carriers will prevent the fuel product from escaping to the neutral E orbit.

A related question however, is will the product overshoot the neutral surface and impinge on the vessel wall? To prevent this the interesting relation must hold:

Fs > Fp + 4 MeV / ℓ	[eq. 3.23]
Boundary for separation of product from reactor vessel wall

The product plasma is also sufficient by itself to deflect all particles incident upon it due to heating. This means that should the reaction plasma and electrostatic forces from solid materials be insufficient to fully correct and stabilize a particle, the product plasma can force any particle of any heat energy back into the reaction plasma where the electrostatic forces acting therein are sufficient to stabilize its orbit.

To explicitly verify that our Reference Design lies within an order of magnitude from what is achievable we can use a simple construction. In Figure 1, h_in denotes the plasma inner surface and U(p₁) denotes the potential energy of a particle reaching the outer reactor vessel wall (by “outer” we mean the interior surface of the reactor vessel furthest from the center of the torus). To establish the boundary condition a little more precisely, we seek the condition such that a nucleus of ⁴He product at 4MeV discharging from h_in with a velocity vector, v_{0 ,}pointed directly at U(p₁); that is, perpendicular to the tangent plane of the reactor vessel wall. This is the most energy efficient path a reaction product can take to reach the reactor vessel wall. In this mode the particle has an initial component magnitude of velocity in this direction that is equal to its speed. In order to prevent it from striking the wall and to possibly contain it, we need a force of deflection perpendicular to v_i with a force sufficient to compel an acceleration which in turn is sufficient to change the vector v₀ by π / 2 while | v₀ | remains constant. Furthermore, this angular change must place v₀parallel to the orbital path defined by magnetic field lines for a particle with its initial conditions.

The line lying along the points denoted at h_out – U(p₁) in Figure 1 is the vacuum gap through which the particle in question will take a parabolic path around the torus. After a deflection of π / 2 the particle will promote to orbit.

Applying some trigonometry to the problem, we first note that deflections do not occur in one basis axis only, but occur on the i’th and j’th bases (see Figure 1). Therefore, there are two magnetic fields in play here (we will look at this more closely later) and the deflection of fusion product will occur across a plane, not an axis. Therefore, the acceleration required of each deflected particle is determined by calculating the trigonometric component values for v₀. If s is the arc length of the parabolic path traveled and | v₀ | remains constant, the i’th acceleration component required across the i-j plane is given by:

\| a_i \| = ½ (\| v₀\| cos (α))² / s	[eq. 3.24]
Reactor vessel geometric dependency; α = π / 4
\|a_i\|= ½ (1.3836 * 10⁷) * cos (π / 4) / (1/4 * π * 75) = 8.3 * 10⁴ m / s / s

Where α is the angle between the i’th and j’th components of v₀. Note that the circumference is divided by 4 because the interval, s, extends only for ¼ rotation. But what we would like to know is how much force must act on the particle to impart this acceleration to it. That will finally tell us how strong of a B field must be present (we will describe the B fields in detail infra). Of course, we know that F = dp / dt, where p is a classic momentum vector, which just works out to be F = ma. ∴

|F| / m = ½ | v₀ |² / s [because F ⨀ s = ½ mv²]

where m is the mass of the deflected particle. Recall the mass of the proton and Neutron, respectively,

1.672621777 × 10⁻²⁷ kg,

1.674927351 × 10⁻²⁷ kg

In this Reference Design the mass of the ⁴He product is

m = 2(1.672621777 × 10⁻²⁷) + 2(1.674927351 × 10⁻²⁷)

and at 4 MeV we have:

| v₀ | = √ [2 * (4000000 * (1.60217646 * 10^-19)) / (6.695 * 10^-27)]

ð | v₀ | = 1.3836 * 10⁷m / s and

ð | v₀ | / c = 0.04615

With a Reference Design specification of 75 meters diameter for the circle marking the thin film plasma, we set s = 75*π meters. This is a conservative figure as it only represents circumference, not a helical path. Finally, since F = ma we get:

F = (6.695 * 10^-27) * (8.3 * 10⁴) = 5.56 * 10^-22 Newtons per reaction.

The amperage of the plasma is calculated by solving for the amount of charge passing a cross section of the torus in one second. For now we will naively assume a velocity for the plasma which we may refine later, but one based on typical plasma speeds in today’s tokomaks; about 100,000 m / s.

The total number of charge carriers in the Reference Design plasma is, as indicated above, 2.791 * 10²⁶ * 53000 = 1.48 * 10³¹. For a plasma constant pressure gradient (this gradient is only constant for vectors lying in the tangent plane to the plasma surface. The vector v₀ must by definition lie in this plane) the charge is (4.806 * 10^-19) * (1.48 * 10³¹) = 7.11 * 10¹² Coulombs.

With a circumference of about 75*π m the torus is traversed 100,000 / 75*π = 4.19 * 10³times per second. ∴, the total plasma current is:

i = (4.806 * 10^-19) * (1.48 * 10³¹) * (4.19 * 10³) = 2.98 * 10¹⁶ Amperes

Since μ = (2.6 * 10^-8), the number of reactions occurring per unit length is:

((2.6 * 10^-8) * (2.791 * 10^26))^(1/3) = 1.94 * 10⁶

=> (5.56 * 10^-22) * (1.94 * 10⁶) = 1.08 * 10^-15 Newtons

And to get the required Tesla for this deflection we multiply:

(2.98 * 10¹⁶)*(1.08 * 10^-15) = 32.18 Tesla

We recall from eq. 3.02 (magnetic force acting on q₁ due to q₂) that,

F_m12 = κ_m q₁ q₂ * (v₁ X (v₂ X r₁₂)) / r²₁₂ ∴

7.315 * 10^-22 = (1.2567 * 10^-6) q₁ q₂ * (v₁ X (v₂ X r₁₂)) / r²₁₂

As an order of magnitude estimate we will choose a point charge representing one cubic meter of “local” reaction plasma. This results in a charge, q₂ = (2.791 * 10²⁶) * (1.60217646 * 10^-19) = 4.47 * 10⁷ Coulombs. We can set q₁ to the charge of ⁴He product; i.e. (2 * 1.60217646 * 10^-19), however, the figure we seek is the product of this value and the total number of product nuclei in the cubic meter we are considering which undergo deflection. Therefore, q₁ = (μ = 7.26 * 10¹⁸) * (2 * 1.60217646 * 10^-19) = 2.33 Coulombs. r is a bit trickier for estimation purposes. Since the electrostatic force drops with the square of the distance, we need an “average” distance in a 1 m³ volume, assuming q₁ lies on that averaged surface within the cubic meter evaluated and q₂ is at the center. Without delving into the mathematical theorems that justify this claim, for now we will note that a distance of ½ * 1/3 = 1/6 meter from q₂ is a sufficient approximation. ∴

F_m12 = 1.4 * 10^-24 = (1.2567 * 10^-6) * (2.33) * (4.806 * 10^-19) * (v₁ X (v₂ X r₁₂)) / (1/6)²

which just reduces to:

(1.4 * 10^-24) / (2.1709 * 10^-23) = 0.0645 = (v₁ X (v₂ X r₁₂))

0.0645 = (|v₁| sin (ω) (|v₂| sin (δ) / 6))

However, the upper boundary condition we seek to validate consists of a geometry ∃

sin (ω) = sin (δ) = 1 so,

6 * 0.0645 = |v₁| * |v₂|

ð 0.388 = |v₁| * |v₂|

There is little need to go further as we can quickly see that the deflection forces are orders of magnitude more than sufficient to deflect product from the reactor vessel wall. Please note that this likewise means that the confinement forces which force fuel particles into their orbits upon thermal heating to 650 KeV are yet stronger; by several orders of magnitude, than the thermal energies. And this is just the magnetic field. But there is a related question that must be addressed. How much energy does it require to deflect these particles? This is a pertinent question since the product possesses the very energy we seek to harness and we would not want deflection to reduce its energy by any appreciable amount.

Recall that |v₀| = 1.3836 * 10⁷m / s. Thus the energy of the particle at creation is,

k = ½ m|v₀|² = 4 MeV and the energy expended to deflect the particle is,

U = – F ∙ s

The problem, of course, is that the generous magnetic field created allows for a wide range of values for s. Therefore, we will pick a reasonable value for what it is worth and try to validate this better once a smaller range for s can be determined (in what follows). Let s = 0.001 m. This is a very steep deflection but appears to be consistent with the forces we are seeing (we shall see that our choice was modest). And again, cosine (φ) = 1 in the geometry of the boundary conditions we are validating. Therefore,

-U = (7.315 * 10^-22) * (0.001) = 7.315 * 10^-25 Joules

and k = 4 MeV or 4000000 * (1.60217646 * 10^-19) = 6.408 * 10^-13 Joules and

even at a steep deflection (which will give us plenty of margin to adjust electrostatic forces for other purposes) the ratio of energy expended to energy gained is:

(7.315 * 10^-25) / (6.408 * 10^-13) = 1.142 * 10^-12; that is,

the free body system is gaining 10¹² times the energy it is losing for this needed deflection and the energy loss is virtually zero.

e_d = ( F ∙ s ) / 6.408 * 10^-13 Joules (or 4 MeV)	[eq. 3.25]
Ratio of deflection energy expended to product kinetic energy
Reference Design Twenty-One Castle Mike => (7.315 * 10^-25) / (6.408 * 10^-13) = 1.142 * 10^-12

Derivation of Mean Free Path in thin film plasma

A simple form of mean free path can be estimated by the relation:

ℓ = 1 / (σ * n)	[eq. 3.26]
Mean Free Path (parallel to average density surface)
Reference Design Twenty-One Castle Mike=>1/((1.2 * 10^-28)(2.79110^26)) = 29.86 m

ℓ = 1/ (σ * n)

where ℓ is the mean free path, n is the number of particles present per cubic meter in the volume considered and σ is the effective cross-section of the particle whose mean free path is sought.

From previous calculations we know that σ = (1.2 * 10^-28)and that n = (2.791*10^26) so,

ℓ = 1/((1.2 * 10^-28)*(2.791*10^26)) = 29.86 m

Of course, this won’t suffice since the plasma pressure is variable as we pass from the outer surface to the inner surface of the plasma. ∴ we need a term that describes the mean free path, ℓ, as a function of height “above” or “below” E_n(see Figure 1). Refining our approach, we get:

ℓ = 1/ ( (1.47 * 10^-11) * N(ℓ_k) = 29.86 m,

where N(ℓ_k) is a function that describes the particle count as a function of ℓ_k, that is, the k’th component of ℓ ∀ ℓ ∈ k. This function is proportional to the electrostatic force as the square of ℓ_k; i.e. N_average = c / ℓ_k², where c is some constant. In order to solve for the mean free path one must integrate over that path to count the number of particles present. ∴

N(ℓ_k) = cℓ_k^-1

ð N(ℓ_k) = c ℓ_k^-1

ð N(ℓ_k) = c [ ln (ℓ_k)_f - ln (ℓ_k)_i ]

ℓ = 1 / (σ c [ ln (ℓ_k)_f - ln (ℓ_k)_i ] / 10¹⁵)	[eq. 3.27]
Mean Free Path (general form)

Reverting again to Figure 1, let the mean free path be denoted m₁, and let us solve for ϕ,

|m₁|² = δ²i + ε²k

ð ϕ = tan^-1(ε / √ [|m₁|² - ε²])

ð ϕ = tan^-1(ε / √ [ [ 1 / (σ c [ ln (ℓ_k)_f - ln (ℓ_k)_i ] / 10¹⁵) ]² – ε²])

Which suggests that, on average, for any ϕ < tan^-1(ε / √ [ [ 1 / (σ c [ ln (ℓ_k)_f - ln (ℓ_k)_i ] / 10¹⁵) ]² – ε²]) the fusion product is re-absorbed into the plasma and:

T_e = P_t {ϕ < tan^-1(ε / √ [ [ 1 / (σ c [ ln (ℓ_k)_f - ln (ℓ_k)_i ] / 10¹⁵) ]² – ε²])}	[eq. 3.28]
Thermal plasma input due to fusion

where T_e is the thermal energy introduced to the plasma from fusion reactions and P_t is the total power output of the reactor. The above equation is the reason for choosing a thin film plasma since the ratio of mean free path to plasma thickness determines the proportion of fusion energy output discharged as thermal heating of the plasma to power output remainder. And as we have seen, the mean free path is on the order of 30 m.

A color rgraphic showing the helical layering scheme. Working outward from the plasma this pattern is repeated about 100 times.

This verifies the efficacy of the geometry chosen for thin film plasma.

The next step is to verify the efficacy of propellant capture and exhaust. In the case of fusion product we consider a significantly higher energy than that experienced from thermal instabilities. Maximum fusion product will be about 4 MeV. Recalling that 1 electron volt = 1.60217646 × 10^-19 joules we have:

(4 * 10⁶) * 1.60217646 × 10^-19= 6.408 * 10¹³ Joules. Based on our previous calculations we can see that the reaction plasma will have no chance of confining this and virtually 100% of the fusion product (including the “slow” product) will easily reach either the α – ρ neutral E orbit surface or the β – ρ neutral E orbit surface, depending on its velocity components. But once it arrives, unless its angle of incidence is rather steep, it will also readily become circularized into this E neutral orbit. Note that the coordinate system in which the product emanates is already in relative motion in consequence of the orbits induced by the magnetic fields around the torus. This means that regardless of the products vector, a tendency to circularize its orbit and find its way to an accumulator at the top or bottom is inevitable. Manipulation of the force dot product as done supra will show that cos θ will indeed have to be less than 1 in order to capture product, however, the lost product will be mostly re-circulated back into the reaction plasma or slowed and captured by the vessel wall as mentioned previously.

Vessel capture is not as serious of an issue as it may sound like at first. Looking at the equations already worked for thermal heating, we can see that product reaching the wall from steep incident angles will have a reduced energy and be captured by an Osmium-Beryllium-¹⁰Boron vessel sleeve. This sleeve can be replaced on regular maintenance cycles and due to the type of fusion reaction involved, neutron fluxing of the sleeve will not render it too “hot”. Because product moving within the vessel is in contact with a strong electrostatic field, unstable movements therein will cause voltage potentials to oscillate within the torus and this can be used to recapture much of that energy. At first blush this would appear to result in a statistical cancelling effect. However, if the torus is shaped … like a torus, as opposed to a straight cylinder, the differences in electrostatic forces can be separated by computer control.

Confinement of the Fuel Source Undergoing Fusion Reactions

Bremsstrahlung radiation losses are a major concern for the ²H-¹¹B reaction we shall specify. However, these losses only occur in the presence of electrons. The lower the proportion of electrons to plasma is the smaller the energy loss due to “Brems”. Thus, it might be useful to perform a sanity check on the very idea of single species plasma in fusion applications. Those in the magnetic confinement Kool-Aid camp will quickly advise that the Brillouin density limit renders single species, positively charged ion plasmas unworkable in fusion applications. This is false. The Brillouin density limit is a well-known limit on the achievable density of plasma as radially confined by magnets. It is not a fundamental density limitation any more than a black hole is subject to a density limitation. This density limitation is an artifact of the interaction of centrifugal forces, the electric force and the magnetic force. Therefore, compressing single-species ionized plasma up to the Brillouin density limit can be readily achieved with magnetic fields and densities beyond that can be readily achieved by other, non-magnetic means.

The Brillouin limit is irrelevant in a torus in which an electrostatic charge is presented both outside and inside the torus, symmetrically compressing the plasma as it rotates. Given sufficient power the magnets are free to compress the plasma to arbitrary densities beyond the Brillouin limit because the centrifugal forces causing it are stabilized by electrostatic forces. The scheme we propose addresses and solves both the Bremsstrahlung radiation issue and the Brilouin density limit that plagues magnetic confinement-only schemes.

1 * 10⁶ bar = 1 * 10¹¹ Pascals. Therefore, the sum of all force acting over any given square meter surface must equal 10¹¹ Newtons force.

Charge density, the density of charge in Coulombs, is expressed:

Q =	[eq. 3.08]
Charge density of a charged volume

where x, y and z are the coordinates locating the total charge and ρ is the density function for charge as a function of r; a position vector within the total charge volume.

To find the total count of charge carriers in a volume we use:

N =	[eq. 3.09]
Number of charge carriers in a charged volume

Where N is the number of charge carriers in the volume and n ( r ) is the position dependent charge carrier count density.

At -4 eV excitation, the peak region of charge density for Ag₂Se, the states for Ag and Se are 4d and 4 p respectively. This is two states for that material.

1 / (6 * 10^-10) = 1.67 * 10⁹ [1 Angstrom = 10^-10 m]

= 40 * 1.67 * 10⁹ / m = 6.68 * 10¹⁰ ions per meter or,

N = = (6.68 * 10¹⁰)³ = 2.981 * 10³² ions / m³

which is (2.981 * 10³²) * 1.602176565 * 10⁻¹⁹ = 4.78 * 10¹³ C.

using eq. 3.03 and assuming plasma charge (which we will justify infra), then, at 10 cm distance this generates a force of:

F_i = [ 8.987 * 10⁹ * (4.78 * 10¹³) * (1.602176565 * 10⁻¹⁹ * 2.598 * 10²⁷)] / (0.1)² = 1.79 * 10³⁴ Newtons. We reduce by one order of magnitude to adjust for the vessel geometry:

1.79 * 10³³ Newtons force applied to the plasma.

This figure assumes a drift velocity of 1 m / s, which is conceivably achievable, especially with the large headroom this value gives us.

1 cm thickness at 1 mm / s drift velocity

ð 1.79 * 10²⁹ Pascal

This still leaves considerable headroom for materials science adjustments (operating temperature being one possible limitation). We shall examine this in more detail in what follows.

Internal layers (not repeated)

[layer 1]: alternating layers of Ni and Ti: 1 cm

[layer 2]: liquid Li: 2 cm (active Neutron and thermal absorber – flow rate: 100 m / s)

[layer 3a]: Gd: 3 cm (primary Neutron Transport backstop) flux is to background far side

[layer 3b]: Os: 1 cm (primary gamma backstop); is to background far side

Repeating layers

[layer 4]: liquid N: 1 cm

[layer 5]: Boron Nitride Wurtzite Nanotube: 1 cm

[layer 6]: electron superconducting material:1 cm

[layer 7]: Boron Nitride Wurtzite Nanotube: 1 cm

[layer 8]: ion superconducting material: 10 cm

[layer 9]: Boron Nitride Wurtzite Nanotube: 1 cm

[layer 10]: carbon nanotube material: 4 cm:

[layer 11]: Repeat pattern starting with layer 4.

The overall layer pattern is a spiral that works its way outward from the center, providing a continuous, helical conductor surface from the interior of the reactor vessel to the outermost layer.

F_{n – 1}= F_n+ W_f@n ∀ n in blanket [ W_f@n ≪ F_{n – 1}]	[eq. 3.10]
Reactor Vessel Wall Equilibrium Condition (simplified)

where W_f ≪ F_p

And now; the denoument: the force F₆ experiences a normal force of equal magnitude from the next analogous layer backing it; that is, layer 12. And at that layer the same equilibrium condition is repeated. The difference however, is that the electrostatic normal forces decrease with each pattern repetition. The effect is to distribute the plasma pressure force evenly throughout the nanotube material such that the total force created by the pressure is not exerted on a single surface. In each case of electrostatic normal forces, the forces are acting on charged particle to charged particle and no material is involved. The pressure of the plasma, therefore, is transmitted via fields only. After a repetition of some 90 cycles over approximately 2 meters of blanket, the final material layer absorbs the remaining force delivered by plasma pressure by presenting a normal force to it. We will pause here with this overview and describe this pressure confinement system in much more detail in what follows. For now it is only important to understand that the strategy here is to complement and reinforce magnetic field lines that will be created as a consequence of this arrangement. This allows the Reference Design to follow the well established experimental data already amassed in traditional Tokamak fusion.

It has been stated supra that “The choice of transit propulsion represents the greatest risk factor for this project”, and we can now reduce that statement to this:

Mass of an atom of ¹¹B is about 11 times that of a proton.

To determine speed of ²H -¹¹B product, we take 2 of the 3 helium nuclei produced by this reaction (because the 3^rd is a “slow” product) and calculate their velocity:

⁴He fast nuclei ²H-¹¹B fusion product v = 3.28 % c or 0.0328 * c; e.g.

(((4 * 10^6[energy of fast ⁴He in MeV]) * (1.60217646 * 10^-19 [conversion to Joules]) / (5 * 10^-27[mass of ⁴He]))^(1/2)) / 299792458[speed of light].

A quantity of 2 of these fast ⁴He are generated with velocity vectors 155 degrees from each other. Coordinate system orientation is random.

Therefore, plasma velocity is best limited to about 0.001 c maximum. This yields:

0.001 * c ≈ 299,792 m/s

The strategy is to find a way to capture the useable product mentioned by capturing it in a neutral E orbit outside the plasma. Once there, it can be redirected into accumulators.

One can view this as a key requirement that the density and orbital confinement be tight enough to overcome the vibrations of heating.

The nuclear binding energy of ¹¹Boron is 6.9277 MeV. This value corresponds to the average kinetic energy of collision particles that optimizes fusion reaction rates.

[1] But again, this is largely because scientists are not genuinely concerned with these issues. They are doing pure research and a single-mode is the better approach for their purpose.

Density and pressure as a function of electrostatic forces

F_p = φ Q_P / (ℓ₀- ℓ₄)²

∴ as an order of magnitude estimate we know that the force will decrease as 1 / ℓ₀- ℓ₄)² and the potential energy is:

F_p ⨀ (ℓ₀- ℓ₄)= – φ Q_P / (ℓ₀- ℓ₄)²

and F and ℓ are perpendicular to the plane tangent to the plasma surface so that cosine ((π / 4) – ϕ) = 1 ∴

F_p = – φ Q_P / (ℓ₀- ℓ₄)

=> F = – φ Q_P ℓ (ℓ)^-1

=> F = – φ Q_P ln (|ℓ₀- ℓ₄|)

and if the magnitudes of ℓ and F are known constants (design requirements) then, letting ω = ln (|ℓ₀- ℓ₄|):

F / ω= – φ Q_P

We next treat our test particle as a sheet of charge with A = 1. Then q_m1 and Q_P are related by:

q_m1 = Q_P/ (ℓ₀- ℓ₄). Casting the problem as a test charge a distance (ℓ₀- ℓ₄) from a charged disk (r < 10 in this approximation since curvature will reduce fidelity):

e = F / q_m1 = { Q_P (1 – ([ln(ℓ₀- ℓ₄)] / √ [ [ln(ℓ₀- ℓ₄)]² + (ri)²])) } / [ 8ϵ₀π(ri)² ]

ð Q_P= F * [ 8ϵ₀π(ri)² ] / { q_m1 (1 – ([ln(ℓ₀- ℓ₄)] / √ [[ln(ℓ₀- ℓ₄)]² + (ri)²])) }

And the particle count, N₁, in a “charge” area q_m1 = Q_P / (|ℓ₀- ℓ₄|) is:

q_m1/ (ϵ)

where ϵ = 4.806 * 10^-19, the averaged charge per particle (at 50/50 mix). In a later section we will calculate the required reactor vessel wall charge.

Figure 1

The number of particles present must obey the relation:

[Q_P/ (4.806 * 10^-19)] = (2.791 * 10²⁷)^2/3 * (|ℓ₄- ℓ₅|)

ð [Q_P/ (4.806 * 10^-19)] = (2.791 * 10²⁷) * (0.1)

ð Q_P= (4.806 * 10^-19) * (2.791 * 10²⁷) * (0.1) = 1.34 * 10⁸ Coulomb per plasma 2D sheet.

This yields 1.34 * 10⁸ Coulomb per square meter of plasma and ((1.34 * 10⁸)^1/2)³

= (1.55 * 10¹²) Coulombs / m³. ∴

An explicit machine calculation gives:

Q_P = {|F| * (|ℓ₀- ℓ₄|) * ( 8ϵ₀π(ri)² ) / q_m1(1 – (|[ln(ℓ₀- ℓ₄)]| / √ [|[ln(ℓ₀- ℓ₄)]|² + (ri)²]))}

|F| = (1.34 * 10^8) / ((0.1) * ( 8*(8.85 * 10^(-12))*π(5)² ) / ( ((4.806 * 10^(-19)) * (2.791 * 10²⁷) / (0.1) ) * (1 – (|[ln(0.1)]| / √ [|[ln(0.1)]|² + (5)²]))})

|F| = (1.34 * 10^8) / ((0.1) * ( 8*(8.85 * 10^(-12))*pi *(5)^2 ) / ( ((4.806 * 10^(-19)) * (2.791 * 10^27)^(2/3) / (0.1) ) * (1 – ((ln(0.1)) / sqrt ((ln(0.1))^2 + (5)^2)))))

= 3.26 * 10¹⁸ Pascal

This is seven orders of magnitude above our target pressure of 10¹¹. We set that value because it represents the current technological limit of materials science.

Q_P= (4.806 * 10^-19) * (2.791 * 10²⁶) * (0.1) = 1.34 * 10⁷ Coulomb per plasma 2D sheet.

|F| = (1.34 * 10⁷) / ((0.1) * ( 8*(8.85 * 10^(-12))*pi *(5)^2 ) / ( ((4.806 * 10^(-19)) * ((2.791 * 10²⁶))^(2/3) / (0.1) ) * (1 – ((ln(0.1)) / sqrt ((ln(0.1))^2 + (5)^2)))))

= 7.02 * 10¹⁶ Pascal

We will mitigate this high pressure using a negative charge carrier conducting layer mixed with the nanotube laminate layers. The resultant pressure will then be ~10¹¹. By doing this, we will be able to introduce ~10²⁷ particles / m³ / s for reaction. We shall see the implications of this choice in the continuing discussion.

Description of the multi-layered thin film

This is why polywell fusion and other electrostatic schemes like it cannot work. Using the Ideal Gas Law is extremely deceptive when dealing with a single species, or highly charged, plasma. Gas and plasma have completely different properties. If is astonishing in the extreme that such a proposal has gained so much attention when the basic physics behind it are not only unproven but haven’t even been correctly calculated on paper.

The strongest nanotubes to date have a specific strength of about 48000000 N·m·kg⁻¹. Using the maximum feasible weight for the reactor of 1000 kg / m³ gives:

1000 * 48000000 = 4.8 * 10¹⁰

which is ≈ 1.5 * 10¹¹ Newtons. This represents a current materials science limit.

as the design requirement.

Confinement materials science at pressures up to 10⁶ Bar

Another daunting challenge in this reference design was in what to do about pressure.

Straightforward calculations quickly revealed that any commercially and practical alternative to fission power would require enormous operating pressures.

This is because the reaction rate depends so heavily on it and plasmas are not normally very dense. This simple observation has escaped many fusion cheerleaders who still don’t seem to realize that there is little point in designing a fusion reactor that does not outperform fission reactors by orders of magnitude.

This represents yet another area of myth and fantasy in the public literature regarding fusion power. And this requires massive pressures. It is simply impractical to build a reaction vessel with the volume of stadium. And if you don’t have that much reaction going on you are better off with fission.

We will now continue the discussion we initiated supra by deriving the essential equations related to the nanotube blanket there described. In particular, we wish to characterize the proportional relationships between the E and B fields created and how they affect the plasma.

Equations of motion for this plasma are derived based on MHD in the Reference Design primary document.

magnetic force lines

[TBC]

Control of the Fuel Source Undergoing Fusion Reactions

Lampe and Manheimer [2] (L and M) have emphasized that the collisional rate νpB for

(p-B) momentum exchange scattering is 37 times faster than the fusion reaction rate νF ,

and that the collisional (p-B) drag rate although slower than νpB by a factor of twice the

mass ratio mp/mB is still faster than νF by a factor 7. They discussed a variety of collisional

processes which would destroy the colliding beam equilibrium on time scales much faster

than the fusion reaction time. They considered an explicit equation for the time evolution

of the transverse spread of the proton beam, based on the Fokker-Planck kinetic equation,

and their calculations clearly showed the heating of the proton beam due to p-B collisions

in the limit νpB > νpe, where νpe is the collisional rate for proton-electron (p-e) momentum

transfer. They concluded that the required colliding beam equilibrium “cannot be sustained

for long enough to provide fusion gain.”

(d_scF_R / dL)² = (_cF_E )² + (dF_s / dL)²

dα = tan^-1((d_scF_r / (dF_s / dL))²)

Wherever

∫ tan-1 ((_cF_E)_i / (_cF_E))_f dL = ∫ tan^-1(d_scF_r / (dF_s) dL

is taken over the line representing the inner lining of the torus shell the plasma is confined at any arbitrary pressure and/or temperature provided sufficient power is present. This is a field equation and there is no dependency on mass-bearing material in this relation. This relation requires two magnetic fields, one primary created by a solenoid and the other the result of moving positive charge in a proton superconductor induced by the electrostatic repulsion of positive ions in the plasma. An ultimate materials science limitation will present if a sufficiently large current in the proton superconductor forces the material into a non-superconducting regime. At that point, increases in current will be limited to the capacity of active cryogenic cooling and the thermal behavior of the superconductor. This is a materials science power conversion limitation and not a direct pressure/temperature relation.

Thermal management of the plasma

Obviously, if all of the ⁴He product is being prejudicially scooped out of the plasma the instant it is created there will be no intrinsic heating mechanism to maintain operating temperature. However, not all ⁴He can be removed. Therefore, the key question is whether or not that portion that cannot be removed would possess sufficient summed energy to heat the plasma. This can be estimated by examining the geometry of the plasma. First, the plasma is only about 3.3 cm thick. By taking the neutral orbit within the plasma as the average reaction locus some simple trigonometry will give us an idea of how much ⁴He is captured by the plasma, as previously shown in the derivation of mean free path.

Power Conversion out of a Fusion Fuel Source: why alternative fuels for fusion are just more of the all-too common digression we see into Con-Fusion Wonderland.

One of the previously noted problems with Tokamak reactors is that no matter how we may design them, and especially when using super light fuels which are technically easier for other reasons, tokamaks invariably create too many necessary interfaces with physical matter. This means that thermal losses will skyrocket when any attempt is made to actually harness the power from fusion reactions. Use of thermal power transfer, as is so common with fission reactors, is simply unrealistic with fusion as it will invariably run up power transfer losses. Put another way, the materials science needed to control power levels when tapping them thermally is decades in the future. The current ITER plan is to use Li-Si blankets to “capture” energetic product from fusion reactions which is embedded round the torus. The very plasma destabilizing events that magnets are good at controlling will skyrocket when fusion product is allowed to physically interact, by physical contact, with the chamber. This unworkably high proportion of power transfer out of the system by Neutron Transport will quickly become a heat sink sucking all the heat out of the plasma. And there is the more basic question of, what will happen to fusion plasma in a “dirty” deuterium or tritium fuel scenario? Ash from these reactions will easily foul the plasma because, as in the case of the Li-Si blankets, the ash will form a heat sink by which all heat will, with high preference, conduct out of the plasma. To understand the magnitude of the heat sink problem, note that for the reactions being considered at ITER, temperature regimes of something like 150 million K are involved (at these thermal variances between system and external, does an exact number really matter for a heat sink?). A small reliance on common sense would tell you that any such system must be almost completely isolated from potential heat sinks (when we speak of a “heat sink” we are loosely referring to the fact that the outer materials are super frigid compared to the plasma and act as a heat sink whenever any power transfer path presents). Tokamaks, as practical reactors and by current design, have no way of achieving this.

Whatever power output scheme is used, it must be one that bypasses materials altogether in order achieve power output rates in the TeraWatt range. The only way to do that is by clever use of the forces of nature, the only two to which we have practical access being the Electric and Magnetic forces of nature. For some background on this subject, Neutron Transport is the study of how Neutrons from nuclear reactions interact with matter. It is important because this is the only known way in which Neutrons can interact with the universe. They do not “respond” to the Electric or Magnetic forces. But this necessarily makes any power transfer from or to Neutrons a materials science issue. And it is precisely the “easy” fusion fuels in consideration by most current research efforts that produce their power output in such large proportions as Neutron energy; also called Neutron flux. Any undergraduate physics student should quickly see this problem and ignore talk of any use of such fuels; unless, of course, some truly eye-popping development in materials science is just around the corner. But ITER isn’t about solving the fusion riddle. It is about funding. This should be an encouragement to the reader, because it suggests that fusion power is not as far off as it may seem. We shall see in what follows that, even using a supposedly aneutronic fuel source will result in problems of considerable difficulty in thermal management. And we shall provide its solution.

The ²H-¹¹B reaction

The ²H-¹¹B reaction involves several individual reactions which need to be examined. We will first look at neutron production:

Neutrons come primarily from the reaction

¹¹B + α → ¹⁴N + n + 157 keV

The reaction itself produces only 157 keV, but the neutron will carry a large fraction of the alpha energy, which will be close to E_fusion/ 3 = 2.9 MeV. Another significant source of neutrons is the reaction

¹¹B + p → ¹¹C + n − 2.8 MeV

These neutrons will be less energetic, with energy comparable to the fuel temperature. In addition, ¹¹C itself is radioactive, but will decay to negligible levels within several hours as its half life is only 20 minutes.

Since these reactions involve the reactants and products of the primary fusion reaction, it would be difficult to further lower the neutron production by a significant fraction. A clever magnetic confinement scheme could in principle suppress the first reaction by extracting the alphas as soon as they are created, but then their energy would not be available to keep the plasma hot. The second reaction could in principle be suppressed relative to the desired fusion by removing the high energy tail of the ion distribution, but this would probably be prohibited by the power required to prevent the distribution from thermalizing.

Before we proceed, we need to explain what we mean by “branching probability”. One can think of a branching process as a random walk. Let S_i denote the state in period i, and let X_i be a random variable that is iid over all i. Then the recurrence equation is

S_{i + 1} = S_i + X_i+1 – 1 =	[eq. 3.29]
Recurrence Equation

with S₀ = 1. To gain some intuition for this formulation, one can imagine a walk where the goal is to visit every node, but every time a previously unvisited node is visited, additional nodes are revealed that must also be visited. Let S_i represent the number of revealed but unvisited nodes in period i, and let X_i represent the number of new nodes that are revealed when node i is visited. Then in each period, the number of revealed but unvisited nodes equals the number of such nodes in the previous period, plus the new nodes that are revealed when visiting a node, minus the node that is visited. The process ends once all revealed nodes have been visited.

The following reaction is the primary source of the neutron flux we seek to manage. It is caused by ⁴He product being re-absorbed back into the plasma. Since this is the product, it is desirable that most of it is removed:

¹¹B + ⁴He → ¹⁴N + n + 157 keV

Where the neutron produced will carry about 2.9 MeV. The rest of the neutrons come from the reaction:

¹¹B + p → ¹¹C + n − 2.8 MeV

Where the neutron produced will carry about 650 KeV (or, the thermal energy of the plasma).

¹¹B + p → ¹²C + γ + 16.0 MeV

with a branching probability relative to the primary fusion reaction of about 10⁻⁴. This just means that there are, probabilistically averaged, about 0.04 of the above reactions for every one primary reaction. This introduces another issue that will be addressed infra.

Finally, isotopically pure fuel will have to be used and the influx of impurities into the plasma will have to be controlled to prevent neutron-producing side reactions like these:

¹¹B + d → ¹²C + n + 13.7 MeV

d + d → ³He + n + 3.27 MeV

Other reactions include:

¹¹B + p→_0 + ⁸Be + 8.586 MeV,

¹¹B + p→_1 + ⁸Be* + 5.65 MeV,

⁸Be*→_12 + _12 + 3.028 MeV.

The kinetic energy interval for this reaction product is [3.4 MeV, 10 MeV]; though 90% of the total power produced is from particles with kinetic energy < 5 MeV.

Experimentally determined reactions and values (4 channels total):

¹¹B + p → ¹²C→ α₀ + ⁸Be→ α₀ + α₀₁ + α₀₂ μ = 4.4 mb/s E = 8.583 MeV ϕ = 154 deg

¹¹B + p → ¹²C→ α₁ + ⁸Be→ α₁ + α₁₁ + α₁₂ μ = 6.8 mb/s E = 5.683 MeV ϕ = 165 deg

¹¹B + p → ¹²C→α₂ + α₃+ α₄ μ = ? barns E = [2.0 MeV, 2.7 MeV]

¹¹B + p → ¹²C→¹²C + γ μ = 10 barns E = [2.0 MeV, 2.7 MeV]

To understand thermal losses we begin by an examination of Neutron reflectors. These are surfaces, also known as Neutron super mirrors, that reflect Neutron flux rather than absorbing it. This, obviously, is the preferred method of managing thermal losses due to Neutron flux. Whatever incident radiation is reflected is not only managed but fed back into the reaction plasma.

Ion sputtered (polished) neutron super mirrors made of pure isotope (with ⁵⁸Ni and ⁶²Ni) NiC/Ti laminates have very high reflectivities (nearly 90% incident). However, for an 11 Tw Reference Design we would need a reflectivity of something like this since:

11 Tw @ 1% neutron flux => 0.11 Tw to the wall => 90% reflectivity is therefore 11 Gw. This is still a monstrous amount of thermal energy to manage. The good news is that neutrons from the presenting fusion reaction are weak; with energies insufficient to produce fission reactions. Therefore, a reflector normally used for fast neutrons would likely reach or exceed a 99% reflectivity.

Use of ion sputtering and laminate structures of pure isotopes ⁵⁸Ni and ⁶²Ni, as well as surface oils and other “tricks”, could raise the reflectivity yet higher.

This results in a highly likely achievable specification of 99% reflectivity which yields 11 Tw * 0.01 ≈ 110 Gw and 110 Gw * 0.01 = 1.1 Gw. This is the power level that must be thermally managed in our reference design. This is still considerable and will require new innovations in thermal management never before used in fission reactors.

But we must point out one important fact: our assumptions here are worse case in that we are assuming that ⁴He product is not removed from the plasma. Of course, we know it must be removed to make use of the power output. However, product vectors lying parallel or near parallel to the tangent plane of the thin film will be re-absorbed. The safe bet therefore, is to specify the thermal management herein until experimental data can settle the issue more precisely.

Additionally, we noted supra that gamma radiation will be emitted (at 4, 12 and 16 MeV) which we can average as having an energy of 12 MeV and which occurs 0.04 times each time the primary reaction, which releases 8.7 MeV, occurs. Thermal (and radiation) management of this must also be considered. This results in an effective contribution to total power on the order of 0.04 / 12000000 ≈ 3.33 * 10^-9 eV for every 8.7 MeV primary reaction. Taking a ratio of the two shows that:

(3.33 * 10^-9) / 8700000 ≈ 3.83 * 10^-16

which delivers (31 * 10¹²) * (3.83 * 10^-16) = 0.0118 watts to the reactor wall. While this may seem like very little radiation, and it is in some respects, the effect over time could be harmful to persons or equipment. Therefore, a thin super pressed Os shell outside the Gd shell is indicated to absorb the gamma radiation thus emitted. This shell is described infra.

Energy reaching the wall through slow neutron flux:

Let

r = ratio of neutron flux reflected to total incident flux

p = full power output of reactor

n = ratio of neutron flux power to total reaction product power for the ²H ¹¹B fusion reaction.

and

f_i = incident flux power = p*n

f_w = neutron flux power that is absorbed by wall

Then,

f_w = f_i (1 – r)	[eq. 3.30]
Power absorbed by reactor wall via Neutron Transport, ²H ¹¹B

Now we see that an enormous amount of thermal energy will be generated as neutron flux by what is a presumably a-neutronic fuel. Imagine the energy involved if we were talking about a neutronic fuel? Wonderland. We simply lack the materials science to utilize any fuel other than Boron and Hydrogen, which is what we alluded to earlier.

We recommend that a thin neutron super mirror made of pure isotopes of Ni, C and Ti form the inner reactor vessel liner. We specify a required reflectivity of slow neutrons be 99% or better. Additionally, we specify a proportion of Carbon to Nickel (to be an alloy in one layer of the mirror laminate) adequate to raise the melting point of the Ni alloy up to about 2000 deg C; the approximate melting point of Ti. We suspect this will result in a mix of about 80% Ni and 20% C. Directly behind this we recommend a 2 cm layer of liquid pure isotopes of Li coolant. The maximum coolant temperature we specify at or about 1450 deg C (the boiling point for Li is 1615 deg C and its melting point is about 450 deg C). The upper bound is low enough that the Nickel-Carbon and Titanium can tolerate the temperatures while still providing a large enough temperature gradient to make use of heat rejection panels highly efficient. For the reflector to maintain structural integrity, this leaves a temperature gradient of approximately 1000 deg Celsius. This thermal energy would be dissipated over heat rejection panels with a minimal differential temperature between near absolute zero and 450 deg Celsius. This leverages Wien’s displacement law to the extent feasible with current material science and is essential for tackling the problem of waste heat in a spacecraft application.

Finally, behind the Li coolant a 4 cm solid layer of super pressed Gadolinium is employed. We will justify this thickness infra. It is pressed to a density > 8,000 kg / m³. The higher this value the longer its service life; and it should be designed to be replaceable on a regular maintenance schedule, as should the super mirror. For this Reference Design we assume 10000 kg / m³. This yields a total vessel weight due to the Gadolinium shell of about 2,200,000 kg, or 2200 mT. This (and the Os layer) is one of the heaviest objects requiring orbital delta v. Based on these calculations, the Reference Design can expect a Gadolinium shell useful service life of about 15 months at reactor full power. At that point excessive embrittlement will require a full replacement. This is the reactor’s primary neutron flux absorber and safety backstop for neutron flux. And it absorbs only what remains of the flux that was not reflected by the super mirror.

Backing the Gd shell we specify a super pressed Os shell of 4 cm thickness. We specify a pressing of the Os elemental solid to create a density not less than 25,000 kg / m³. This is the reactor’s primary gamma radiation absorber. This is one of the heaviest components to be lifted to Earth orbit; weighing approximately 5,500,000 kg.

Let ρ_Os = the density of the pressed Os liner. Then the mass is,

M_Os = (ρAT)_Os / 9.8 m / s / s ≈ 56 mT	[eq. 3.31]
Mass of the Osmium reactor vessel liner

Where A and T are the liner Area and Thickness, respectively.

Neutron transport through circulating liquid Li and Gd liner

Let ρ_Gd = the density of the pressed gadolinium liner. Then the mass is,

M_Gd = (ρAT)_Gd / 9.8 m / s / s ≈ 22.45 mT	[eq. 3.32]
Mass of the Gadolinium reactor vessel liner

Where A and T are the liner Area and Thickness, respectively.

The thermal values are achieved at thermal equilibrium. The actual energy carried away from the reactor vessel walls depends on the rate flow of the fluids. Behind the Gadolinium layer we specify a liquid Nitrogen layer followed finally by the first, Boron Nitride Wurtzite nanotube layer of the reactor vessel.

We specify a coolant flow rate of at least 100 m / s for the Li coolant. This results in the ability to remove the equivalent energy of 100,000 deg C / m² / s. Unfortunately, this requires, at minimum, an equal drop of energy over the heat rejection panels. As per reference design, this suggests the rejection by electromagnetic transfer to vacuum of about 100,000 deg C for each of some 700 m² of reactor wall surface. And that

ð 700 * 100,000 = 70 million deg C / s

This places the reference design loosely in the right order of magnitude with a drop of 0.01% of total reactor power output. That is;

70 * 10⁶ / 0.0001 = 700 billion deg C / s and, as per reference design, if the heat rejection panels have a total area of;

A = 100 meters * 33 meters * 48 then

A = 100 * 33 * 48 = 158000 m²

which yields a minimum required rejection rate at the surface of each panel of:

70 * 10⁶ / 158000 = 443 deg C / m² / s

If rejecting equally from both sides it is 443 / 2 ≈ 220 deg C / m² / s per side

This satisfies the minimal temperature differential indicated supra of 450 deg C.

let b = 2.8977685 * 10⁻³ m·K (Wien’s proportionality constant) then the optimum wavelength for thermal rejection is:

λ_max = b / T	[eq. 3.33]
Wien’s Displacement Law

where T is the absolute temperature in Kelvin and λ_max is the wavelength at which emissivity is maximum. At thermal equilibrium the electromagnetic emission temperature is the maximum coolant operating temperature; e.g. about 1500 deg C. Therefore if [K] = [°C] + 273.15

=> 1773.15 K

=> λ_max = (2.8977685 * 10⁻³) / 1773.15 ≈ 1.63424893 * 10^-6 m

Equations of motion for the following are found in the primary Reference Design document:

finish neutron flux thermal management
Determine required thickness of Gd liner

Revisiting the fuel injection system and heat rejection panels

We are now in a position to specify the requirements for heat rejection.

Fusion Product Capture for Outbound Power Conversion

Even with the most advanced, clever design, MFC or not, these realizations suggest that the super light fuels are simply not workable with first generation fusion reactors. It is so far beyond materials science to convert 15 TeraWatts of thermal power from Neutron Transport to useable power that discussion of it is borderline pathological. But this is exactly the “path” ITER is following. But the practicably nearest supposedly “aneutronic” fuel alternative, ²H-¹¹B, which can afford power transfer without physical contact, has other conditions that make it a tough proposition. First and foremost, the triple product for this reaction is higher. But once that is solved, one has to consider that such a plasma of ²H-¹¹B fuel will have to be put in a state that is not, technically speaking, plasma. Known as single species plasma, ²H-¹¹B would have to be stripped clean of all or most electrons so that all its fusion power isn’t produced in a form for which no conceivable power conversion exists: Bremsstrahlung radiation. This problem is similar to the Neutron Transport issue, but not as challenging. The simplest answer is not to use electrons in the reaction plasma, which are the source of it. As positively charged, single species plasma, the ²H-¹¹B fuel would orbit in helical paths around a torus in one direction only.

However, another problem is encountered in that, for magnetic confinements of this type, a density limit is realized (known as the Brillouin density limit). Density is one of the factors of the triple product. We have proposed stabilizing the magnetic confinement using electrostatic positive charge to collapse the plasma into its magnetically defined orbits (relying on a multi-confinement scheme). This allows the Brillouin density limit to be exceeded because this limit is an artifact of how magnets force orbits.

Let us assume then, for the sake of discussion, that a setup in which power transfer can occur without mass contact (no thermal conduction) is employed by using the ²H-¹¹B reaction. Then we know that electrostatic, symmetric forces will have to be applied to the plasma to “assist” the magnetic forces. Otherwise, the exponentially growing instabilities of the plasma will cause exponentially increasing power transfer losses as we try to pass the triple product (due to the limitations of magnetic reconnection in this regime and, most generally, due to the fact that the locating of magnets in a torus is inherently asymmetric). Once a fusion reaction is reached (beyond ignition) the fusion energetic product will be two fissile products that immediately decay into three ⁴He nuclei and spread at very high velocities with positive charge. So, the power form this reaction gives us is moving electric charge, something we can tap without thermal contact. Having said this, we are still not in the clear. Mass to mass contact is unavoidable. When the fusion reaction ramps up to full power density a nasty quantity of power will present. The highly energetic, ⁴He ions (the primary ²H-¹¹B reaction product) cannot all be subjected to negative accelerations to tap off their energy into a useable current. There simply isn’t enough space and, statistically speaking and perhaps more significantly, not all particles will have velocity vectors that can be exploited. Let us take a walk inside the torus to explain what we mean. Remember, we have to keep the plasma stable, so trying to perform this deceleration outside the tokamak is not a winning strategy. When a fusion reaction occurs a more or less random direction of flight for the Helium product is “decided”. Some of these particles will travel with the plasma rotation, more or less. Some will travel in the opposite direction. But some will be ejected at right angles to the orbits. Once again, electrostatic confinement helps out. Let us set an electrostatic neutral zone in the center of the plasma, a special, neutral orbit path. As we move outward from that path a massive (and it will need to be about 10⁶ Bar at between stable and neutral E) electrostatic positive charge will push us back toward the neutral orbit. This helps. But the Helium nuclei can easily overcome this electrostatic force. But, the pressure from the electrostatic force will tend to influence the flight vector of the particles into a torus orbit in which, given a sufficient component at right angles to the neutral orbit, will cause the Helium product to arc either out toward the torus shell or in toward its center. The Helium nucleus will then depart the plasma and head for the wall. It is in this wall that the Helium product, with carefully designed symmetry, can be captured, but not fully stopped (not yet), into a stream of Helium nuclei collected at the torus edge and central axis. A helical collector can then force these nuclei into a deceleration using inductive stators to generate current. At this point, we are outside the plasma so no destabilization occurs as a result. A selected remainder can be vented outside the apparatus for thrust. But these are the happy paths.

Some statistical fraction of ⁴He will invariably impinge on the walls due to unfavorable initial kinetic vectors, straight on collisions with the vessel wall or by simply being too energetic to capture. Though we can minimize these “losses”, they will generate heat when the power levels skyrocket. Minimizing this kind of loss will be crucial but is far, far easier than the challenges faced by today’s tokamaks (like ITER) as regards the same issue. And as the fusion reaction heats up, this tiny power loss still incurred even in the “best” design here suggested will be enormous. In other words, a surplus of heat energy will be needed; one far beyond the surplus capable of being generated by neutral particle injection or RF heating (with foreseeable materials science technology). Given a limited power loss as described, cooling can probably be practically achieved, but the loss will still force us to reconsider the existing heating schemes now in use. What is needed is a heating mechanism that allows us to extend the confinement time indefinitely. Even at the theoretical level there is only one known, certain power transfer device that can do this: A fission reaction.

Based on our considerations supra we now specify the performance and characteristics of Reference Design Twenty-One Castle Mike:

Torus Height (inner vessel wall shell): 25 meters

Torus Diameter (inner vessel wall): 7 meters

Run: ((((pi *14) * 25) * 0.01) * 100) – ((((pi *7) * 25) * 0.01) * 100) = 550 as an approximate volume [100 thin-film layers each 1 cm thick]

Total structural weight (locates all components): 100mT

Reactor vessel weight: 12000mT

Magnet weight: 1000mT

Total system Earth weight: 15000 mT

Total Fusion output power: 11 Tw

Total thermal power recirculated to sustain reaction: 3 Tw

Total electric power:

electromagnetic systems power requirement to sustain reaction: 0.25 Tw

coolant and heat rejection system electric power requirement: 0.001 Tw

total power available to Brayton cycle (onboard ship services): 0.01 Tw

[for each of 3 units, all redundant]

Total useable kinetic power (at maximum rated power): 7 Tw

Total thermal power loss to reactor vessel wall via fast ⁴He: 0..5 Tw

Total thermal power loss due to inefficiency of recycling thermal losses due to Neutron Transport: 100 Gw

Total power loss to Brensstrahlung: ~1 Mw

Specific Power: 3.5 Mw / kg

Am-242m film power: 2Mw / m²

Total Kicker power output: 7 Gw

Kicker runtime at rated maximum power per fuel emplacement: 10 days

Gd reactor vessel liner lifetime at maximum rated power: 15 months

Hours of operation at maximum rated power between overhauls: 160 hours

Neutron flux at Gd liner exterior surface at maximum rated power: ~0

Plasma reaction mass volume per layer: 5525 m³

Number of plasma layers: 25

Plasma operating density at maximum rated power: 9.4427 kg / m³

Plasma Reaction mass pressure at maximum rated power: 1.5*10⁶ Bar

Mars expedition proposal total mass ²H-¹¹B, fast burn + 1/2 slow burn: 30,000 kg

These numbers were derived from the reaction rates and masses discussed infra. The Space Shuttle Main Engine turbo pumps, by comparison, had a specific power of about 153 kW / kg. It is often stated that the vast distances of space make interplanetary travel a challenge. This is not entirely correct. Movement in space is demonstrably more efficient than in or on just about any medium. The issue isn’t distance, its mass. It takes precious little force to impart a noticeable velocity to several thousand tons of bulk mass in LEO. There are professional drivers who drive a million miles in just a few years. And missions to Jupiter using chemical propulsion are measured in similar time frames. But since Jupiter is some 930 million miles distant the added efficiency of propulsion in space is made up for in distance. But getting a truly powerful power plant, something far more powerful than, say, a diesel engine, into orbit, is costly. For example, an 800 horsepower diesel engine (very large) is about 600 Mw power. Multiplying by a factor of ten (6 Gw) with a rocket, travel times to Jupiter make sense if light weight components and virtually no mass to speak of is carried as cargo. The problem, however, is when we contemplate moving large, worthwhile masses across the solar system. So, we can easily see why power outputs can climb into the Tw range. And that is the issue facing the Mars proposition.

Conclusion

Only this very year (2012) have all the component technologies reached the maturation assumed by this Reference Design (the last hurdle being the production rate for nanotubes reached in the summer of 2012), meaning that development of the primary mover system proposed is a developmental work which, having considerable research and development ahead of it, is based on what today are existing technologies. We recommend adoption.

Because of the enormous profit potential such a power source offers for ground based electrical power production, we recommend as an alternative funding option the development, production and sale of self-contained units that can be shipped by sea to coastal locations across the United States. It is unclear at this point how much profit might be realized from this approach but we recommend it be studied in any case. The same is true for the 41 GLASS Pack, which will likely open up the space launch market rather widely. We therefore have the potential to own that market.

At current U.S. consumption rates; if one Twenty-One Castle Mike were to tap into the U.S. electrical power grid, the plant would have a load of about ½ Tw on average year round and generate 43 billion USD per year in fees based on the average cost of electrical power in the U.S. At wholesale rates and after paying extortion fees to the distributors you would likely draw about 20 billion a year. To make it attractive, you could cut the cost in half, making 10 billion a year in the U.S. alone.

And yes, you read that correctly. One Twenty-One Castle Mike could supply all the electrical power used in the United States year-round about 20 times over. Hence the Peak Oil end run.

Finally, we recommend the production facility for Twenty-One Castle Mike be located on the mainland arm of the Aleutian chain which will provide considerable safety stand-off (even if just in the public’s mind), provide a low profile for operations and provide a sea-based delivery scheme for shipping to ground stations across the coastal U.S. as well as for shipping components out to the south pacific for launch. The State of Alaska is generally friendly to energy initiatives and would likely be so to this proposal as well. As of this writing a large tract of land has already been purchased for this purpose near Kings Cove, AK, under a corporate pseudonym. Cold Bay airport provides air service to the area and we recommend improvements by adding an additional runway on this tract as well as 800 foot piers. As mating of components to each 41 GLASS Pack will be the most complicated of all, we further recommend that this sizable tract be utilized as the launch operations home port. We recommend a sister home port near San Diego, CA to support high volume launching.

We also recommend a dedicated, two order of magnitude increased production facility for nanotubes be located in Anchorage, AK.

Appendix

A Cold Plasma Electric Tensor (CPET) Operator

Since 1918 it has been well-known (and was likely known by Einstein) that the fundamental metric tensor of General Relativity yields a set of equations that, in principle, allow for the artificial manipulation of gravitational fields. If we reduce the field equations to a linearized form (and we can check that this is valid by placing back into their original form to check our work) we see that since 1918 a massive elephant in the living room has been sitting right before our eyes. That this is so poorly known generally is a story by itself, but it will suffice here to simply relate those equations:

Derivation of equations of motion from GR – planar rotation	[A1.00]

(dp/dt)_i = (1 / 2c²) ui(Gω²∫_u1^u2du m_u/uR_u)

=> (dp/dt)_i = (1 / 2c²) ui[ ln (u₂R_u2) – ln (u₁R_u1) ]Gω² / 2c²R_u

(dp/dt)_j = (1 / 2c²) vj(Gω²∫_v1^v2 dv m_v/vR_v)

=> (dp/dt)_i = (1 / 2c²) vj[ ln (v₂R_v2) – ln (v₁R_v1) ]Gω² / 2c²R_v

- (dp/dt)_k = (1 / 2c²) wk(Gω²∫_w1^w2 dw m_w/wR_w)

=> – (dp/dt)_i = (1 / 2c²) wk[ ln (w₂R_w2) – ln (w₁R_w1) ]Gω²

Where R is the radius of a massive object spinning about the k’th basis, ω is the angular velocity with which this massive body spins about k, m and M are mass, G is the gravitational constant (= 6.67 * 10-11 m³/kg-sec²), c is the speed of light in vacuum and i, j, and k are the basis vectors for a Cartesian coordinate system whose origin lies at the center of the spin axis. Coordinates u,v and w specify the location of a test mass with respect to the i, j, and k basis. R is the radius of the massive body for each coordinate u, v and w.

Evidently, if we spin a ring of enough mass fast enough we will likely be able to measure a decrease in the ambient gravitational field along the k’th basis. Assume zero ambient gravitational or inertial force. Assume further that a confinement force in the i-j plane exists to limit motion as R increases. Then the resultant motion of the test mass is a helical path, spiraling through the center of the rotating mass along the k’th basis. In other words, and when fully confined to prevent any motion, the test mass is being “pulled” by the gravitational field into a cone shape.

When the spinning mass is a sphere rather than a “ring” or “ribbon” (planar) the following equations can be derived directly from GR (this has been known since 1918, btw):

Derivation of equations of motion from GR – spherical rotation	[A2.00]

(dp/dt)_i = [(i4uω² /5 – 8 ωs_j) / 3c²] m_uG∫_u1^u2du 1 / uR_u

=> (dp/dt)_i = [i 4uω² /5 – 8 ωs_j)] m_uG [ ln (u₂R_u2) – ln (u₁R_u1) ] / 3c²R_u

(dp/dt)_i = [(j4vω² /5 – 8 ωs_i) / 3c²] m_vG∫_u1^u2dv 1 / vR_v

=> (dp/dt)_j = [i 4uω² /5 – 8 ωs_i] m_vG [ ln (v₂R_v2) – ln (v₁R_v1) ] / 3c²R_v

(dp/dt)_k = – kw[(8Gω²∫_w1^w2 dw m_w/wR_w) / 15c³

(dp/dt)_k = - kw8m_wGω²[ln (w₂R_w2) – ln (w₁R_w1)] / 15c³

From this we can see that should the sphere (large mass) be reposed whilst the test mass moves inside it then a gravitational “force” acts on the test charge inside the sphere; acting as the product of and the linear velocity of the sphere (the cross product of the two). What the equation above (2.00) explicitly shows is that the velocity of the test mass, here denominated “s”, is analogous to the interaction of the electric and magnetic fields.

Unfortunately, the superficial understanding of General Relativity means that many who approach this problem harbor a decidedly narrow view of what it implies. Use as an “anti-gravity” scheme is just one such example. The value in these equations is not in reducing weight by reducing gravitational acceleration but, rather, the fact that this could entail modifying space-time itself.

The analogies one can readily observe in nature between this gravitational phenomenon and the electromagnetic force are not at first obvious because the materials science required to exploit the gravitational effect is only now being reached.

We can start with a toroidal shape much like our tokomak discussed infra. The value of the electric field at the center of a torus is given by:

Electric field at center of torus	[A3.00]
E = – B = + (d/dt) (µNIr²/4πR²)

Where R is the radius of the torus, r is the radius of one of the loops of wire wound around it, and N is the total number of turns.

The paucity of research in this area leaves some doubt as to whether or not limitations exist in the strength of the field created. Even more uncertainty enters when we take the analogy of the electromagnetic field as gospel by substituting quantities into the known electromagnetic relation:

An EM analogue	[A3.00]
G = -K = – (d / dt) [ηNTr² / 4πpiR²]

Where G is the gravitational field generated by the total mass current NT (T being the mass flow), and η = 3.73*1026 m/kg would be the gravitational stand-in for the magnetic permeability.

Thus we can see that overall, what is needed to generate a non-negligible gravitational field is high density and high angular velocities. The need for high density can be seen by noting that for smaller values of R we increase the induced gravitational field. Due to the limitations that centrifugal forces play on rotational speed, some clever work-arounds are needed. First, just as in the case with fusion we quickly see materials science limitations. We’ve already shown that with current nanotube technology pressures of about 1.5 * 10⁶ bar can be generated. This would indeed result in a very high value of ω. But therein lies an additional obstacle. The densest materials known, if spun up to such high values of ω, will internally deform even when supported by nanotube structures, jackets or blankets. But any “out of round” or asymmetrical deformation would be disastrous at such high values of ω.

Second to the issue of nanotube materials science, the chief innovation needed here is one that deals with symmetry issues.

The only way to achieve this; that is, to ensure that a uniform material density is maintained at all values of ω, is to use a superfluid, super-dense cold plasma. By using the heaviest ion that can be handled as a superfluid, we can maintain a uniform density in all regimes by the same confinement techniques discussed for fusion. And that brings us to another point. It would likely be necessary to use both magnetic and electric fields, as we proposed in the fusion discussion, in order to maintain full control over the plasma and to thus be able to produce magnetic lines upon which ions will complete their orbits in an orderly manner. By greatly reducing or eliminating fluid friction we allow the ions to spread uniformly (remain along their magnetic field lines) and not be heterogeneously thrown off their orbital path. By keeping the plasma cold we reduce the thermal agitation that tends to lead to non-uniform distributions of density. Of course, by using a plasma we pay a heavy price in density as even at 1.5 * 10⁶ bar we are not as dense even as Pb.

But once this problem is solved, one approach might be to circulate this dense ionic fluid through “conductors” analogous to the conductors used in electromagnets. By winding a helix with these “conductors” an interesting parallel can be established between the EM and the gravitational force.

From these equations we can now ask, does any of this have any relevance to this reference design inasmuch as it being a quality desirable for the mission enunciated?

The answer to this question is complex and uncertain because we lack solid information. While many have made elaborate arguments for how this can be used as an “anti-gravity” source the derivations above don’t support this view. The reasons are pragmatic in nature. This is because, in our research we’ve found no publicly available calculations to address the resultant mass / weight gains. In other words, how massive would such a device need to be to generate a given field? This is quite analogous to the same problem addressed in our fusion discussion when we considered specific power. What we really need to know is how much mass is needed to create a field with strength |F|? Simply claiming that a device can reduce rest mass by 89% is meaningless. If the vehicle has a “natural” weight of 5 million pounds reducing its weight to 500,000 lbs. isn’t going to make it fast or efficient by any stretch. Given the similar hints of the great mass required to build such an object as we got from our fusion discussion, we suspect that any such field, to be practical, would need to obey the relation:

m₀|a₀| > m₀|a_e|

where m₀ is the rest mass of a test mass, |a₀| is the acceleration of the artificially generated gravitational field and |a_e| is the acceleration of the gravitational field at Earth’s surface. A more general relation, however, might be more useful,

The actuarial limitation	[A4.00]
m₀\|a₀\| > m₀\|a_t\|

where we subsitute a_t for a_e to denote that the field of interest need not be limited to the surface of Earth but may be any test field we wish to evalute. We can almost assert this out of the gate because of the high likelihood that such a device would be very, very heavy. First, a rather stout primary mover is needed to power it (a fission reactor?) and the apparatus itself couldn’t possibly be of typical aeronautical weight. There are no performance gains without a total neutralization of the resultant field.

Having established this first criteria, we are now prepared to ask, assuming we can generate such a field, where and how would it be useful to us in the context of the over-arching reference design (Master Reference Design Exo-energy Infrastructure Level 1 AKA Reference Design Echo India One)? We are asking this question to fully vet all available technological options, even if it means examining the most exotic options conceivable (as we did for matter/antimatter primary movers). On the face of it, this technology seems mostly limited to “punt technology” (“punt” refers to aerospacecraft used solely for translation between the surfaces of gravitational bodies and to transit ships). Presently, we’ve recommended this role be taken by the 41 GLASS pack, a staged, kerosene powered rocket limited to use on Earth. Other punts have been proposed for Mars and other planets but they all rely on rocketry.

Is there a use beyond this? We will now propose a novel design which takes its lead from the novel laminate layer solution we discussed in our fusion conversation. This design would allow essentially infinite rotational speeds and densities using existing materials science. While we do not propose this as a punt or ship “drive” mechanism for Echo India One, it is so interesting we’ve included it in the appendix. It is as follows:

Consider a superfluid such as ⁸⁸Sr (cooled to about 2µK). Current research reveals that this is the heaviest element that can currently be cooled to a superfluid state. There may be more, and the temperature at which it attains this state might be higher if ionized. But for now, we’ll assume use of ⁸⁸Sr. Ionized, ⁸⁸Sr becomes a single-species plasma, just as in our fusion example. It is essentially a Bose-Einstein Consdensate (BEC). In accordance with equation A1.00, we now rotate this plasma at high velocity, confining it as we did in the fusion example by using an electrically charged outer shell reinforced by nanotube backing. The object of the exercise is to decrease R and increase ω as much as material science allows; that is, to an effective pressure of 1.5 * 10⁶ bar, as equation A1.00 suggests. Now, we employ the same magic as we did in the fusion discussion.

Hypothesis: we consider a layered laminate of several layers of ⁸⁸Sr of thin film plasma all rotating within an value of R and a value of ω limited only by the materials employed and sufficient (this is what we must test – can this be done) distortion of the space-time manifold containing the rotating layers within it. What we see right away is that the focus on “anti-gravity” is a decidedly Earth-based focus and that it misses the point. From the fundamental metric tensor of General Relativity what we see is that what is occurring is better characterized as an artificial manipulation of the space-time manifold in the same way that natural, gravitational bodies distort that field. In other words, these equations, when exploited technologically, can cause distortions of space-time, bending space as the design dictates. This “bending” is what the metric tensor characterizes: a tensor is a mathematical construct used to characterize, in some instances, in what directions and by what amounts a “fabric” is “pulled” or distorted. In this case the “fabric” is space itself, one of the key findings of GR that surprised most who saw it the first time. It seemed odd that space can be treated like a “fabric” and that it can be pulled and stretched. But we now know GR to be a largely intact and correct representation of nature.

Next, we denote each layer of ⁸⁸Sr cold plasma with integer multiples, beginning with 1 at the innermost layer and p as the final, outermost. Next, consider any arbitrarily selected layer, ℓ_n , where n is less than or equal to p, greater than 1. Let m be any integer multiple such p > q > m > 1; q being an arbitrarily chosen value that satisfies the aforementioned relation.

As a brief aside, there are other more exotic elemental alloys that can be used that might result in greater ionic mass and higher superfluid temperatures. I have heard of Mercury-Thallium-Barium-Calcium-Copper-Oxide (Hg12Tl3Ba30Ca30Cu45O125), but have no way to confirm this. It is supposedly a superfluid at about 150 Kelvin. This is remarkable considering the fact that Hg isn’t superfluid until it hits about 3 Kelvin. After some research I identified the element bearing the most favorable characteristics for this purpose. As it turns out, Hg is superfluid at 3 Kelvin and consists of 200 protons. It is very heavy. The problem, however, is how to achieve a temperature of only 3 Kelvin at 1.5 * 10⁶ bar. One way to greatly assist in this effort is to ionize the Hg. Whatever the case, much research into this question would be needed.

At sufficiently obtained values for R and ω for n_q the centrifugal force exerted by the spin in ℓ_m is reduced by a proportionate amount. So, the program is the same as the fusion program. We are reducing the forces acting on the materials. The values of R increase as we approach p and ω increases as we approach 1, precisely what equation A1.00 calls for. An explicit calculation will be needed to determine if the sum of pressure at p of 1.5 * 10⁶ bar is sufficient to achieve a adequate result in layer 1. We will do this next.

But before we do that a digression into the inner workings of General Relativity is needed. Unfortunately, this topic is not well understood or known at all to most. So, we’ll offer a brief primer. The fundamental metric tensor of General Relativity, often just called the metric, is an order 4, rank 2 tensor. In sloppy terms, this just means that it operates in a 4-dimensional space-time framework and each index is “operated on” by a function or functions in two successive steps (and if covariant, we mean to say that we can perform this operation in both directions and get the same results going back and forth – and we say it is invariant if we cannot). Typically, the “steps” taken are simply a reference to the number of derivatives taken over an expression, such a a lorentz transformation. So, if we take an expression and differentiate it twice, it is rank 2. If it is covariant, we should be able to take the result and integrate it twice (solve for its anti-derivative twice successively) to get our original expression back.

A common example of derivates would be position vectors. A position vector differentiated once with respect to time is velocity. Differentiated again is acceleration. That is:

d/dt [f(x)] = f’(x) = v

and

d²/dt [f(x)] = f”(x) = a

where v is velocity and a is acceleration. In this special case, the superscript 2 matches the “rank” of the tensor; that is, rank 2.

We can take the antiderivative of a and get v:

f’(x) =∫_a1^a2dt f”(x)

and over again:

f(x) =∫_v1^v2dt f’(x)

and if it is covariant we are guaranteed to get our function f(x) back, for any x evaluated.

The program of General Relativity is really quite simple. We just take the lorentz relations and differentiate and integrate as needed. Of course, we do this with partial derviatives and, being a derivative, we get a tangent to any curved basis 0,1,2 or 3 (the bases t, i, j and k; time and the three spatial axes). Of course, people that tout GR and tensor talk like to try to look smart so they make it all sound much more obscure than this (ho hum) but the result of doing this is to create a tangent space. Recalling that a derivative generates a tangent line (when understood geometrically); that is, it is the line tangent to the point on the curve being evaluated, we know we’ll be generating exactly “order” number of tangent lines, or 4 tangent lines for each basis in the case of the metric. So, we can think of all four of these tangent lines as creating a “tangent space” by treating each tangent line as a kind of basis vector of a new space. If we imagine the set of all vectors that can be defined in that space (which invokes notions of linear dependence for those linear algebra purists) we have what tensor-talk types like to call a “tangent bundle”. Whatever.

So, why do we need to calculate velocities and accelerations of these bases? Well, we don’t. Remember that this is just the physical analogy usually used when speaking of first and second derivatives. There is something yet more interesting derivatives can tell us. They are also a way to account for the curvature of a space-time at the point x evaluated. And that is why they appear as the “operators” of the metric.

to be continued

In the presence of a gravitational field the distance along x constricts and the time axis dilates. Motion is initiated along a curved path in x because we are “traveling” through time even when reposed in x. Thus the resultant vector induces motion in the x component. Where no central force is found (artificial gravity) the constriction in x is principally unbound. As the reader might begin to notice this isn’t about “anti-gravity” at all. It’s about artificially generating “warped” space-time. The plot thickens.

- kk

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