Weapon Mounts

As a general rule, a space warship is basically a "weapons platform." It is just a way to move some weapons that you control into a strategic position.

Single weapons and multi-weapon turrets are mounted on "hardpoints" or "weapon stations." These are positions on the spacecraft's hull that are designed to carry the mass of the weapon. One only hangs a heavy picture frame on a nail in a wall stud, not just the wall board. For the same reason only mount a heavy turret on a hardpoint, not on a flimsy stretch of hull. Some hulls are about as strong as the skin on a beer can.

Turrets pivot to allow aiming the weapon(s). Homing missiles are often mounted in "vertical launch systems" or "missile cells", because they do not have to be aimed. Fire and forget, they'll automatically find the target.


Naturally some people who are into hyper-optimization and min-maxing will quickly switch from mounting weapons on a ship to building the ship around a weapon. A monstrously huge weapon, with a fixed forward facing.

Of course you probably have to turn the entire spacecraft in order to aim the weapon, but the ship is going to smite the target with the most bang for your buck. It certainly will be the sort of ship that will blast the snot out of you if you are stupid enough to turn around and try running away. The ship will also have a similar outline as the weapon, probably long and skinny. Popular spinal mount weapons are coil guns, rail guns, and particle beam weapons, since those weapons inflict more damage the longer the weapon is.

The weapon can be mounted on the ship's nose, along the ship's side ("dorsally" or "ventrally", but RocketCat will rip your lips off if you use those terms), or along the ship's spine. In extreme cases the weapon is the ship's spine, this is what the Traveller RPG calls a "spinal mount". A good example is the "Wave motion gun" that forms the spine of Space Battleship Yamato.


Isaac Kuo has some interesting observations on the placement of laser turrets:

There's an interesting question of what the ideal number of turrets is. One thing that's counterintuitive is that the number of turrets has little effect on total firepower. Your laser engine(s) can fire the beam down a central corridor, with mirrors to select a branch toward any of the laser turrets. No matter how many turrets you have, you can concentrate all laser firepower through one turret.

I tend to favor two turrets on opposite sides. Besides providing all around coverage and some redundancy, it also allows use of a "hunter-killer" tactic. While one turret fires the laser to kill a target, the other turret can be scanning to "hunt" for the next target. This allows a near instantaneous switch from one target to the next, minimizing down time for the laser engine.

More importantly, this has a big tactical effect on the enemy's options. Suppose each of your ships only had one laser turret, and the enemy knows this. Then the enemy knows it takes some time for you to switch from the current targets to new targets. If the enemy notices that all of your ships are firing on particular targets, he can take advantage of this to open up sensitive sensors or radiators onboard the non-targeted ships. He knows that if you want to fire on a different target, he's got enough time to close protective "shutters". In contrast, with two turrets per ship nowhere is safe from being targeted.

Isaac Kuo

Nukes In Space

As you should know, there are two types of nuclear weapons. An "atomic bomb" is a weapon with a war-head powered by nuclear fission. An "H-bomb" or "hydrogen bomb" is a weapon with more powerful warhead powered by nuclear fusion.

You can read all about the (unclassified) details of their internal construction and mechanism here.

Occasionally you will find a fusion weapon referred to as a "Solar-Phoenix" or a "Bethe-cycle" weapon. This is a reference to the nuclear scientist Hans Bethe and the Bethe-Weizsäcker or carbon-nitrogen cycle which powers the fusion reaction in the heart of stars heavier than Sol.

A "neutron bomb" is what you call an "enhanced radiation bomb". They are specially constructed so more of the bomb's energy is emitted as neutrons instead of x-rays. This means there is far less blast to damage the buildings, but far more lethal neutron radiation to kill the enemy troops.

You will also occasionally find references to a nasty weapon called a "cobalt bomb". This is technically termed a "salted bomb". It is not used for spacecraft to spacecraft combat, it is only used for planetary bombardment. They are enhanced-fallout weapons, with blankets of cobalt or zinc to make large quantities of deadly radioactive dust.

Warhead

As far as warhead mass goes, Anthony Jackson says the theoretical limit on mass for a fusion warhead is about 1 kilogram per megaton. No real-world system will come anywhere close to that, The US W87 thermonuclear warhead has a density of about 500 kilograms per megaton. Presumably a futuristic warhead would have a density between 500 and 1 kg/Mt. Calculating the explosive yield of a weapon is a little tricky.

For missiles, consider the US Trident missile. Approximately a cylinder 13.41 m in length by 1.055 m in radius, which makes it about 47 cubic meters. Mass of 58,500 kg, giving it a density of 1250 kg/m3. The mass includes eight warheads of approximately 160 kg each.

Wildly extrapolating far beyond the available data, one could naively divide the missile mass by the number of warheads, and divide the result by the mass of an individual warhead. The bottom line would be that a warhead of mass X kilograms would require a missile of mass 45 * X kilograms, and a volume of 0.036 * X cubic meters (0.036 = 45 / 1250). Again futuristic technology would reduce this somewhat.

Nuclear weapons will destroy a ship if they detonate exceedingly close to it. But if it is further away than about a kilometer, it won't do much more than singe the paint job and blind a few sensors. And in space a kilometer is pretty close range.

Please understand: I am NOT saying that nuclear warheads are ineffective. I am saying that the amount of damage they inflict falls off very rapidly with increasing range. At least much more rapidly than with the same sized warhead detonated in an atmosphere.

But if the nuke goes off one meter from your ship, your ship will probably be vaporized. Atmosphere or no.

George William Herbert says a nuke going off on Terra has most of the x-ray emission is absorbed by the atmosphere, and is transformed into the first fireball and the blast wave. There ain't no atmosphere in space so the nuclear explosion there is light on blast and heavy on x-rays. In fact, almost 90% of the bomb energy will appear as x-rays behaving as if they are from a point source (specifically 80% soft X-rays and 10% gamma), and subject to the good old inverse square law (i.e., the intensity will fall off very quickly with range). The remaining 10% will be neutrons.

For an enhanced radiation weapon (AKA "Neutron Bomb") figures are harder to come by. The best guess figure I've managed to find was up to a maximum of 80% neutrons and 20% x-rays.

The fireball and blast wave is why nuclear warheads detonating in the atmosphere will flatten buildings for tens of kilometers, but detonations in space have a damage range under one kilometer.

If you want to get more bang for your buck, there is a possibility of making nuclear shaped charges. Instead of wasting their blast on a spherical surface, it can be directed at the target spacecraft. This will reduce the surface area of the blast, thus increasing the value for kiloJoules per square meter.

According to John Schilling, with current technology, the smallest nuclear warhead would probably be under a kiloton, and mass about twenty kilograms. A one-megaton warhead would be about a metric ton, though that could be reduced by about half with advanced technology.

Eric Rozier has an on-line calculator for nuclear weapons. Eric Henry has a spreadsheet that does nuclear blast calculations, including shaped charges, on his website. For bomb blasts on the surface of the Earth or other planet with an atmosphere, you can use the handy-dandy Nuclear Bomb Effects Computer. But if you really want to do it in 1950's Atomic Rocket Retro style, make your own do-it-yourself Nuclear Bomb Slide Rule!

EMP

When it comes to the dreaded EMP created by nuclear detonations, matters become somewhat complicated. Please, do NOT confuse EMP (electromagnetic Pulse) with EM (electromagnetic Radiation).

Most SF fans have a somewhat superficial understanding of EMP: an evil foreign nation launches an ICBM at the United States, the nuke detonates in the upper atmosphere over the Midwest, an EMP is generated, the EMP causes all stateside computers to explode, all the TVs melt, all the automobile electrical systems short out, all the cell phones catch fire, basically anything that uses electricity is destroyed.

This is true as far as it goes, but when you start talking about deep space warfare, certain things change. Thanks to Andrew Presby for setting me straight on this matter.

First off, the EMP I just described is High Altitude EMP (HEMP). This EMP can only be generated if there is a Terra strength magnetic field and a tenuous atmosphere present. A nuke going off in deep space will not generate HEMP. Please be aware, however, if a nuke over Iowa generates a HEMP event, the EMP will travel through the airless vacuum of space just fine and fry any spacecraft that are too close.

Secondly, EMP can also be generated in airless space by an e-Bomb, which uses chemical explosives and an armature. No magnetic field nor atmosphere required. This is called a Non-nuclear electromagnetic pulse (NNEMP). As with all EMPs, once generated they will travel through space and kill spacecraft.

Thirdly, there is System Generated EMP (SGEMP) to consider. HEMP is created when the gamma rays from the nuclear detonation produce Compton electrons in air molecules, and the electrons interact with a magnetic field to produce EMP. But with SGEMP, gamma rays penetrating the body of the spacecraft accelerated electrons, creating electromagnetic transients.

SGEMP impacts space system electronics in three ways. First, x-rays arriving at the spacecraft skin cause an accumulation of electrons there. The electron charge, which is not uniformly distributed on the skin, causes current to flow on the outside of the system. These currents can penetrate into the interior through various apertures, as well as into and through the solar cell power transmission system. Secondly, x-rays can also penetrate the skin to produce electrons on the interior walls of the various compartments. The resulting interior electron currents generate cavity electromagnetic fields that induce voltages on the associated electronics which produce spurious currents that can cause upset or burnout of these systems. Finally, x-rays can produce electrons that find their way directly into signal and power cables to cause extraneous cable currents. These currents are also propagated through the satellite wiring harness.

Dr. George W. Ullrich

Impulsive Shock

A one kiloton nuclear detonation produces 4.19e12 joules of energy. One kilometer away from the detonation point defines a sphere with a surface area of about 12,600,000 square meters (the increase in surface area with the radius of the sphere is another way of stating the Inverse Square law). Dividing reveals that at this range the energy density is approximately 300 kilojoules per square meter. Under ideal conditions this would be enough energy to vaporize 25 grams or 10 cubic centimeters of aluminum (in reality it won't be this much due to conduction and other factors).

1e8 watts per square centimeter for about a microsecond will melt part of the surface of a sheet of aluminum. 1e9 W/cm2 for a microsecond will vaporize the surface, and 1e11 W/cm2 for a microsecond will cause enough vaporization to create impulsive shock damage (i.e., the surface layer of the material is vaporized at a rate exceeding the speed of sound). The one kiloton bomb at one kilometer only does about 3.3e7 W/cm2 for a microsecond.

One megaton at one kilometer will do 3.3e10 W/cm2, enough to vaporize but not quite enough for impulsive shock. At 100 meters our one meg bomb will do 3.3e12 W/cm2, or about 33 times more energy than is required for impulsive shock. The maximum range for impulsive shock is about 570 meters.

Luke Campbell wonders if 1e11 W/cm2 is a bit high as the minimum irradiation to create impulsive shock damage. With lasers in the visible light and infrared range, 1e9 W/cm2 to 1e10 W/cm2 is enough. But he allows that matters might be different for x-rays and gamma rays due to their extra penetration.

As to the effects of impulsive damage, Luke Campbell had this to say:

First, consider a uniform slab of material subject to uniform irradiation sufficient to cause an impulsive shock. A thin layer will be vaporized and a planar shock will propagate into the material. Assuming that the shock is not too intense (i.e., not enough heat is dumped into the slab to vaporize or melt it) there will be no material damage because of the planar symmetry. However, as the shock reaches the back side of the slab, it will be reflected. This will set up stresses on the rear surface, which tends to cause pieces of the rear surface to break off and fly away at velocities close to the shock wave velocity (somewhat reduced, of course, due to the binding energy of all those chemical bonds you need to break in order to spall off that piece). This spallation can cause significant problems to objects that don't have anything separating them from the hull. Modern combat vehicles take pains to protect against spallation for just this reason (using an inner layer of Kevlar or some such).

Now, if the material or irradiance is non-uniform, there will be stresses set up inside the hull material. If these exceed the strength of the material, the hull will deform or crack. This can cause crumpling, rupturing, denting (really big dents), or shattering depending on the material and the shock intensity.

For a sufficiently intense shock, shock heating will melt or vaporize the hull material, with obvious catastrophic results. At higher intensities, the speed of radiation diffusion of the nuke x-rays can exceed the shock speed, and the x-rays will vaporize the hull before the shock can even start. Roughly speaking, any parts of the hull within the diameter of an atmospheric fireball will be subject to this effect.

In any event, visually you would see a bright flash from the surface material that is heated to incandescence. The flash would be sudden, only if the shock is so intense as to cause significant heating would you see any extra light for more than one frame of the animation (if the hull material is heated, you can show it glowing cherry red or yellow hot or what have you). The nuke itself would create a similar instant flash. There would probably be something of an afterglow from the vaporized remains of the nuke and delivery system, but it will be expanding in a spherical cloud so quickly I doubt you would be able to see it. Shocks in rigid materials tend to travel at something like 10 km/s, shock induced damage would likewise be immediate. Slower effects could occur as the air pressure inside blasts apart the weakened hull or blows out the shattered chunks, or as transient waves propagate through the ship's structure, or when structural elements are loaded so as to shatter normally rather than through the shock. Escaping air could cause faintly visible jets as moisture condenses/freezes out - these would form streamers shooting away from the spacecraft at close to the speed of sound in air - NO billowing clouds.

Luke Campbell

Nuke vs. Spacecraft

Dr. John Schilling describes the visual appearance of a nuclear strike on a spacecraft.

First off, the weapon itself. A nuclear explosion in space, will look pretty much like a Very Very Bright flashbulb going off. The effects are instantaneous or nearly so. There is no fireball. The gaseous remains of the weapon may be incandescent, but they are also expanding at about a thousand kilometers per second, so one frame after detonation they will have dissipated to the point of invisibility. Just a flash.

The effects on the ship itself, those are a bit more visible. If you're getting impulsive shock damage, you will by definition see hot gas boiling off from the surface. Again, the effect is instantaneous, but this time the vapor will expand at maybe one kilometer per second, so depending on the scale you might be able to see some of this action. But don't blink; it will be quick.

Next is spallation - shocks will bounce back and forth through the skin of the target, probably tearing chunks off both sides. Some of these may come off at mere hundreds of meters per second. And they will be hot, red- or maybe even white-hot depending on the material.

To envision the appearance of this part, a thought experiment. Or, heck, go ahead and actually perform it. Start with a big piece of sheet metal, covered in a fine layer of flour and glitter. Shine a spotlight on it, in an otherwise-dark room. Then whack the thing with a sledgehammer, hard enough for the recoil to knock the flour and glitter into the air.

The haze of brightly-lit flour is your vaporized hull material, and the bits of glitter are the spallation. Scale up the velocities as needed, and ignore the bit where air resistance and gravity brings everything to a halt.

Next, the exposed hull is going to be quite hot, probably close to the melting point. So, dull red even for aluminum, brilliant white for steel or titanium or most ceramics or composites. The seriously hot layer will only be a millimeter or so thick, so it can cool fairly quickly - a second or two for a thick metallic hull that can cool by internal conduction, possibly as long as a minute for something thin and/or insulating that has to cool by radiation.

After this, if the shock is strong enough, the hull is going to be materially deformed. For this, take the sledgehammer from your last thought experiment and give a whack to some tin cans. Depending on how hard you hit them, and whether they are full or empty, you can get effects ranging from mild denting at weak points, crushing and tearing, all the way to complete obliteration with bits of tin-can remnant and tin-can contents splattered across the landscape.

Again, this will be much faster in reality than in the thought experiment. And note that a spacecraft will have many weak points to be dented, fragile bits to be torn off, and they all get hit at once. If the hull is of isogrid construction, which is pretty common, you might see an intact triangular lattice with shallow dents in between. Bits of antenna and whatnot, tumbling away.

Finally, secondary effects. Part of your ship is likely to be pressurized, either habitat space or propellant tank. Coolant and drinking water and whatnot, as well. With serious damage, that stuff is going to vent to space. You can probably see this happening (air and water and some propellants will freeze into snow as they escape, BTW). You'll also see the reaction force try to tumble the spacecraft, and if the spacecraft's attitude control systems are working you'll see them try to fight back.

You might see fires, if reactive materials are escaping. But not convection flames, of course. Diffuse jets of flame, or possibly surface reactions. Maybe secondary explosions if concentrations of reactive gasses are building up in enclosed (more or less) spaces.

Dr. John Schilling

Radiation Flux

Crew members are not as durable as spacecraft, since they are vulnerable to neutron radiation. A one megaton Enhanced-Radiation warhead (AKA "neutron bomb") will deliver a threshold fatal neutron dose to an unshielded human at 300 kilometers. There are also reports that ER warheads can transmute the structure of the spacecraft into deadly radioactive isotopes by the toxic magic of neutron activation. Details are hard to come by, but it was mentioned that a main battle tank irradiated by an ER weapon would be transmuted into isotopes that would inflict lethal radiation doses for up to 48 hours after the irradiation. So if you want to re-crew a spacecraft depopulated by a neutron bomb, better let it cool off for a week or so.

For a conventional nuclear weapon (i.e., NOT a neutron bomb), the x-ray and neutron flux is approximately:

Fx = 2.6 x 1027 * (Y/R2)

Fn = 1.8 x 1023 * (Y/R2)

where:

  • Fx = X-ray fluence (x-rays/m2)
  • Fn = Neutron fluence (neutrons/m2)
  • Y = weapon yield (kilotons TNT)
  • R = range from ground zero (meters)

There are notes on the effects of radiation on crew and electronics here.

Nuclear Shaped Charges

Back in the 1960's, rocket scientist came up with the infamous "Orion Drive." This was basically a firecracker under a tin can. Except the tin can is a spacecraft, and the firecracker is a nuclear warhead.

Anyway, they realized that about 90% of the nuclear energy of an unmodified nuclear device would be wasted. The blast is radiated isotropically, only a small amount actually hits the pusher-plate and does useful work. So they tried to figure out how to channel all the blast in the desired direction. A nuclear shaped charge.


Remember that in the vacuum of space, most of the energy of a nuclear warhead is in the form of x-rays. The nuclear device is encased in a radiation case of x-ray opaque material (uranium) with a hole in the top. This forces the x-rays to to exit only from the hole. Whereupon they run full tilt into a large mass of beryllium oxide (channel filler).

The beryllium transforms the nuclear fury of x-rays into a nuclear fury of heat. Perched on top of the beryllium is the propellant: a thick plate of tungsten. The nuclear fury of heat turns the tungsten plate into a star-core-hot spindle-shaped-plume of ionized tungsten plasma. The x-ray opaque material and the beryllium oxide also vaporize a few microseconds later, but that's OK, their job is done.

The tungsten plasma jet hits square on the Orion drive pusher plate, said plate is designed to be large enough to catch all of the plasma. With the reference design of nuclear pulse unit, the plume is confined to a cone of about 22.5 degrees. About 85% of the nuclear device's energy is directed into the desired direction, which I think you'd agree is a vast improvement over 10%.


About this time the representatives of the military (who were funding this project) noticed that if you could make the plume a little faster and with a narrower cone, it would no longer be a propulsion system component. It would be a nuclear directed energy weapon. Thus was born project Casaba-Howitzer.

Details are scarce since the project is still classified after all these years. Tungsten has an atomic number (Z) of 74. When the tungsten plate is vaporized, the resulting plasma jet has a relatively low velocity and diverges at a wide angle (22.5 degrees). Now, if you replace the tungsten with a material with a low Z, the plasma jet will instead have a high velocity at a narrow angle ("high velocity" meaning "a recognizable fraction of the speed of light"). The jet angle also grows narrower as the thickness of the plate is reduced. This is undesirable for a propulsion system component (because it will destroy the pusher plate), but just perfect for a weapon (because it will destroy the enemy ship).

The report below suggests that the practical minimum half angle the jet can be focused to is 5.7° (0.1 radians).

They would also be perfect as an anti-ballistic missile defence. One hit by a Casaba Howitzer and a Soviet ICBM would be instantly vaporized. Which is why project Casaba-Howitzer's name came up a few times in the 1983 Strategic Defense Initiative.

Casaba Howitzers fired from orbit at ground targets on Terra would be inefficient, which is not the same as "does no damage." A nuclear warhead fired at a ground target would do far more damage, but the Casaba Howitzer bolt is instantaneous, non-interceptable, and would still do massive damage to an aircraft carrier.

Scott Lowther has done some research into a 1960's design for an Orion-drive battleship. It was to be armed with naval gun turrets, minuteman missiles with city-killing 20 megatons warheads, and Casaba-Howitzer weapons. It appears that the Casaba-Howitzer charges would be from subkiloton to several kilotons in yield, be launched on pancake booster rockets until they were far enough from the battleship to prevent damage (several hundred yards), whereupon they would explode and skewer the hapless target with a spear of nuclear flame. The battleship would probably carry a stockpile of Casaba-Howitzer weapons in the low hundreds.

Mr. Lowther estimates that each Casaba-Howitzer round would have a yield "up to a few kilotons" and could deliver close to 50% of that energy in the spear of nuclear flame. Three kiltons is 1.256 × 1013 joules, 50% of that is 6.276 × 1012 joules per bolt.

This is thirty-five times as powerful as a GBU-43/B Massive Ordnance Air Blast bomb, the second most powerful non-nuclear weapon ever designed. Per bolt.

Get a copy of the report for more details, including a reconstruction of a Casaba-Howitzer charge.

What is the mass and volume of a Casaba-Howitzer charge? Apparently this also is still classified. An Orion Drive nuclear pulse unit would be about 1,150 kg, have a blast yield of about 29 kilotons, and be a cylinder with a radius of 0.4 meters and a height of 0.87 meters. The volume would therefore be about 0.4 cubic meters. As previously mentioned a Casaba-Howitzer charge would have a yield ranging from sub-kiloton to a few kilotons, so presumably it would be smaller and of lower mass than a pulse unit.

DIRECTED THERMONUCLEAR EXPLOSIVES

Another device being investigated by both SDI architects and weapon designers is "a kind of nuclear shotgun with little pellets" named Prometheus. According to a Congressional report that was otherwise quite pessimistic about SDI, Prometheus "may have nearer-term applications for picking out warheads from decoys" (in the midcourse phase of ballistic-missile flight) than the Neutral Particle Beam (NPB), a leading contender for that role. Encouraged by experiments already conducted, SDI officials in 1987 ordered an acceleration of the Prometheus project for "concept verification," using funds from that year's $500 million supplemental SDI request.

One research engineer familiar with the project described the device as operating much like a rifle, using a polystyrene-filled barrel to help couple a plate to the "gunpowder-like" blast of a directed nuclear charge. After the impulse from the explosion generates an intense shock wave, the plate "fractionates" into millions of tiny particles. Of course, these would vaporize if in direct contact with the bomb, but as configured, the pellets have reportedly achieved speeds of 100 kilometers per second without vaporization.

Thermonuclear shaped charges, one of the better understood third-generation concepts§, have much in common with conventional shaped-charge explosives already used extensively in military and commercial applications. Both conventional and thermonuclear shaped charges tailor an explosive burn-wave using a detonation front that releases energy along a prescribed path. Both can produce jets of molten metal having velocities greatly in excess of the detonation velocity.*

For thermonuclear fuels such as deuterium plus tritium, the burn-wave can be directed by placing hollow bubbles or inert solids in the path of the detonation front in order to alter its velocity. Of course, ignition of a thermonuclear burn in a warhead requires a fission trigger to achieve the necessary compression and temperature (about 100 million K), but even with such a (nondirected) trigger, the overall directivity of a thermonuclear shaped charge can still be significant.

Velocities achievable with thermonuclear shaped charges are impressive. Unlike molten jets produced by conventional shaped charges, which are limited to about 10 kilometers per second (about four times the velocities of the gases resulting from chemical explosions), thermonuclear shaped charges can in principle propel matter more than two orders of magnitude faster. Since fusion temperatures reach 100 million K, the detonation front of a thermonuclear explosive travels at speeds in excess of 1,000 kilometers per second. Using a convergent conical thermonuclear bum-wave with a suitable liner, one could theoretically create a jet traveling at 10,000 kilometers per second, or 3 percent of the speed of light.

Up to 5 percent of the energy of a small nuclear device reportedly can be converted into kinetic energy of a plate, presumably by employing some combination of explosive wave-shaping and "gun-barrel" design, and produce velocities of 100 kilometers per second and beam angles of 10-3 radians*. (The Chamita test of 17 August 1985, reportedly accelerated a 1-kilogram tungsten/molybdenum plate to 70 kilometers per second. ) If one chooses to power 10 beams by a single explosion, engaging targets at a range of 2,000 kilometers with a kill energy of 40 kilojoules per pellet (one pellet per square meter), then such a device would require an 8-kiloton explosive and could tolerate random accelerations in the target, such as a maneuvering RV or satellite, of up to 0.5 g (5 m/s2).

The initial plate for each beam in this Casaba-like device would weigh only 32 kilograms but would have to fractionate into tiny particles to be an effective weapon—4 million evenly spaced pellets to produce one per square meter at 2,000 kilometers range. If such pellets could be created uniformly, which is highly questionable, then, at a velocity of 100 kilometers per second, they would each weigh 8 milligrams, carry 40 kilojoules of energy (the amount of energy in 10 grams of high explosive), and travel 2,000 kilometers in 20 seconds. Such hypervelocity fragments could easily punch through and vaporize a thin metal plate and could cause structural damage in large soft targets such as satellites and space-based sensors, but they would have little probability of striking a smaller RV, or even disabling it if a collision did occur.§

10-kiloton ASAT
Nuclear yield10 kilotons
Number of beams10
Mass per plate32 kg
Mechanism50 kilojoules per pellet impact kill
Assumptions4 × 106 particles per beam
uniformly spaced 1 per m2
at 2,000 kilometers
Range2,000 kilometers

‡ SPARTA, Inc., Workshop on Interactive Discrimination, 1986, unclassified. The velocity of 100 kilometers per second falls between the goal of 50 kilometers per second in the 1960s, only a fraction of which was achieved, and the 1,000 kilometers per second velocities possible with the plasma howitzer concept. The latter allegedly operates at 10 percent efficiency up to about 1 megaton, although with only about 10-2 radian beam directivity. Speeds of 1,000 kilometers per second are inevitably accompanied by ionization, and because charged particles curve in the earth's magnetic field, they would not be useful for long-range applications. Velocities up to 200 kilometers per second, however, are believed possible without vaporization.

§ See, for example, the detailed analysis of nuclear shaped-charges by R. Schall, "Detonation Physics," in P. Caldirola and H. Knoepfel, eds., Physics of High Energy Density, (New York: Academic Press, 1971), pp.230-244.

* Friedwardt Winterberg, The Physical Principles of Thermonuclear Explosive Devices, (New York: Fusion Energy Foundation, 1981), p.117. Conventional shaped charges have been applied to demolition, antisubmarine weapons, and advanced ordnance antitank munitions—all being further developed at Livermore—as well as for igniting the fission triggers in thermonuclear warheads. Cf. Energy & Technology Review, Lawrence Livermore National Lab, (June-July 1986), pp.I4-15.

† Devices based on this principle were pursued in the 1960s. Project Orion examined their potential for space propulsion. Casaba and "nuclear howitzer" were names for weapon applications.

‡ The detonation front shock-wave velocity is (32 kT/3M)½, where M is the average mass per ion of the thermonuclear fuel. Suitable geometries can propel matter at many times the detonation front velocity. Using cone geometry, the jet speed is v/sinθ, where v is the detonation-front velocity and θ is the cone's half-angle. A practical minimum for θ has reportedly been found to be θ ≈ 0.1. See Winterberg, Thermonuclear Physics, p.41,122

* SPARTA Workshop, 1986. This scaling presumably holds up to about 50 kilotons but, due to blackbody x-ray emission, decreases to about 1 percent for larger yields.

† Robert S. Norris, Thomas B. Cochran, and William M. Arkin, "Known U.S. Nuclear Tests July 1945 to 31 December 1987," Nuclear Weapons Databook Working Paper NWD 86-2, Natural Resources Defense Council, September 1988.

‡ The energy fluence per beam, E in J/m2, is approximately ηY/(NbR2θ2), where η is the fraction of overall yield transferred to the pellets, Y is the bomb yield (1 kiloton is equivalent to 4.2 × 1012 joules), Nb is the number of individual beams being driven by one bomb, R is the distance to the target, and θ is the individual full-beam divergence angle. A maneuvering target could accelerate out of the path of the beam if amR/vf2 > θ, where am is the magnitude of the target's average acceleration, vf is the particle velocity, and τ = R/vf is the particle fly-out time. (For comparison, the average acceleration of ICBMs is about 40 m/s2.) To deliver this energy requires a total mass per beam of Mb = 2E(Rθ)2/vf2.

§ For instance, even if an RV were coated with aluminum, a more volatile material than might be expected, the resulting vapor blow-off would only push a 350-kilogram RV off course by about 15 meters in 20 minutes of flight (about five times the amount if there were no ablation), thus failing to degrade significantly the ≈150 meter accuracy of a modern ICBM. Of course, if the collision caused the RV to tumble upon re-entry, the results would be less predictable

From The Effects of Nuclear Test-ban Regimes on Third-generation-weapon Innovation by Dan L. Fenstermacher. Science & Global Security 1990, Volume 1, pp. 187-223

There are a few more crumbs of information in the report Fourth Generation Nuclear Weapons: Military effectiveness and collateral effects. They note that harnessing the x-rays from a nuclear blast is not only good for making deadly jets of atomic fire, but can also be used to pump x-ray lasers and energize EMP weapons. Not to mention accelerating projectiles to very high velocities by means of x-ray ablation, or by means of neutrons from the nuclear explosion (see report for cites on this).

So the report points out that the x-rays and neutrons can be used to drive or self-forge several projectiles or fragments (a "nuclear gun" or "nuclear grenade"). X-rays and neutrons can also be used to heat a working fluid and form hot jets (the above-described "nuclear shaped charge").

(It might be worth while to review the difference between a shaped charge and a self-forging projectile, they are similar enough to be confused together, but are quite different in end result.)

Thirdly, the forwards and backwards flux of x-rays and neutrons from a single nuclear device can be used to drive a multi-warhead weapon, e.g., a single weapon that fires a self-forging penetrator followed a few microseconds later by a jet of hot plasma. Talk about a one-two punch! The penetrator cracks the armor, allowing the hot jet to enter the target's interior and vaporize the soft chewy center.

The report also estimates, that for the use in military conflicts on the surface of the Earth, these weapons will probably be powered by nuclear devices in the 1 to 100 tons of TNT range (subkiloton range). Whether this will also hold true in the space environment is a question above my pay grade.

Freeman's analysis of nuclear explosions in a vacuum, resulting in a series of three short papers titled Free Expansion of a Gas, was central to the feasibility of Orion. It was also central to the feasibility of directed-energy nuclear weapons, and led directly from Orion to a project code-named "Casaba-Howitzer," described as "a one-shot version of Orion, like Orion except without any ship." Casaba-Howitzer, conceived by Moe Scharff while still at Livermore, would be resurrected many years later as the basis for the "Star Wars" space-weapons program, known as the Strategic Defense Initiative or SDI. "Whereas Orion directed a dense plasma at relatively low velocity at a wide angle, this was to direct a lower-density plasma at a higher velocity and a narrower angle," Scharff explains. "Orion was a space vehicle. Casaba-Howitzer could be consid­ered space weaponry. It could even have been things carried aboard an Orion, for example, if Orion was a battleship."

Casaba-Howitzer's descendants remain under active investigation and Scharff is unable to give any further details beyond the origins of the name. "They had been naming things after melons and the good ones were gone already. They were on a melon kick that year. The one con­nection was seeds—many of those melons have seeds, like the particles we were projecting." Casaba-Howitzer was derived directly from Orion, and later versions of Orion drew heavily on Casaba-Howitzer's experi­mental and theoretical results. Funding for Casaba-Howitzer kept the Orion team going after funding for Orion dwindled out. But there was a costly side to the bargain—a shroud of secrecy that has lingered long after any plans for battleship Orion were shelved. Conversely, if we ever decide to build something like Orion, it will be the continued work on directed-energy weapons—and how to protect surfaces against them— that will allow us to pick up where Project Orion left off.

Anything in the near vicinity of a nuclear explosion gets vaporized into a plasma—a cloud of material so hot that its atoms are stripped of their electrons—that cools as it expands. It was a simple mathematical problem to draw some conclusions relating the shape and density of the initial object that gets vaporized to the shape and density of the result­ing cloud of gas. "The model should be simple enough so that the hydro-dynamical equations can be integrated exactly," Freeman explained. "A real cloud of gas will not have precisely the density-distribution ot the model, but still one may expect the behavior of a real cloud to be quali­tatively similar to that of the model." Freeman set up the equations and the numbers were run on General Atomic's IBM 650 card-programmed calculator, one of the workhorse machines that had handled many of the early bomb and blast-wave calculations at Los Alamos and had not yet been superseded by the IBM 704 that General Atomic acquired in 1959.

According to Freeman's model, something originally in the shape of a cigar expands into the shape of a pancake, and something originally in the shape of a pancake expands into the shape of a cigar. This was "very directly relevant to the expansion of a bomb," he explains. "If you have something that starts in the form of a pancake and you heat it up to a very high temperature it will expand more sideways along the axis, and less at the edges. The pressure gradient is highest along the axis, so then after a while, since the velocity is highest along the axis, it becomes cigar-shaped. So you get inversion, something that begins like a pan­cake becomes like a cigar, and something that begins as a cigar becomes a pancake, if you just let it expand freely. It goes roughly with the square root, if you start with a pancake where the ratio of the diameter to thick­ness is ten, then it will end up as a cigar where the ratio of the length to the diameter is square root of ten, roughly speaking. That would be quite helpful, of course, if you had a real Orion, to start out with a pan­cake and it will produce then a jet that is collimated within 20 degrees or so quite nicely. The fact that it's so easy to make an asymmetrical explosion may still be classified, for all I know."

The right pancake in the right place can focus a significant fraction of the bomb's output into a narrow jet of kinetic energy, directed construc­tively at the pusher plate of a nearby spaceship—or destructively at something else. The thinner the pancake, the narrower the jet. In the early days of Orion, with a huge pusher plate as the target, the propellant was assumed to be a thick slab of something light and cheap like polyeth­ylene; later versions of Orion, with smaller pusher plates, required a thin­ner slab of higher-density material, such as tungsten, to focus the bomb's energy into a narrower cone. Exactly how narrow remains a secret, though a look at the later configurations of Orion permits a guess. This is one of the reasons that detailed design information about Orion, such as the exact standoff distance between the pulse unit and the pusher plate, remains classified, even after forty years have passed.

As the jet of propellant is targeted more narrowly in space, its impact against the pusher plate is spread out more widely in time. The result is more effective horsepower and a softer ride. "In the end we did come up with some designs that were very tight in their angular distribution of momentum," says Bud Pyatt, without mentioning specific numbers, but revealing that "you had to have it pointing at the center of the pusher plate, it couldn't even be five degrees off without stressing the shock absorber too much."

Project Orion: The True Story of the Atomic Spaceship by George Dyson

A propellant plate in the form of a pancake expands into a plume shaped like a cigar. And the reverse is true: a propellant plate in the form of a cigar/cylinder would expand into a plume shaped like a pancake. Specifically:

(Dplume / Lplume) = 1 / sqrt(Dplate / Lplate)

where:

  • Dplume = plume diameter (perpendicular to direction of travel)
  • Lplume = plume length (in direction of travel)
  • Dplate = plate diameter (perpendicular to direction of travel)
  • Lplate = plate length (in direction of travel)

So if the plate had a diameter of 4 and a length of 1 (diameter to length ratio of 4/1 or 4), the plume would have a diameter to length ratio of 1/2, or a diameter of 1 and a length of 2. Equation is from Nuclear and Plasma Space Propulsion by M. Ragheb.

Boom Table

The Boom Table has been moved here.

Laser Cannon

There is a great summary of the various issues of directed-energy weapons. Luke Campbell has an in depth analysis of laser weapons for science fiction on his website, don't miss the on-line calculator for laser weapon pulse parameters. Eric Rozier has another on-line calculator for laser weapons. Rick Robinson's analysis Space Warfare V: Laser Weapons is also quite good. You also might want to look over this 1979 NASA report on using nuclear reactions to directly power a laser beam. (Thanks to Andrew for suggesting this link.)

Before we get to all the boring equations, lets have some juicy details. Say that the habitat module of your combat starship gets penetrated by an enemy laser beam. What happens? Luke Campbell and Anthony Jackson have the straight dope:

That depends on the parameters of the beam.

A single pulse with a total energy of 100 MJ would have the effect of the detonation of 25 kg of TNT. Everyone in the compartment who is not shredded by the shrapnel will have their lungs pulverized by the blast.

That same 100 MJ delivered as 1,000,000 pulses of 100 J each could very well drill a hole. The crew see a dazzling flash and flying sparks. Some may be blinded by the beam-flash. Anyone in the path of the beam has a hole through them (and the shock from the drilling of that personal hole could scatter the rest of them around the crew compartment). Everyone else would still be alive and would now be worrying about patching the hole.

Although it occurs to me that the jet of supersonic plasma escaping from the hole being drilled could have the combined effect of a blowtorch and grenade on anyone standing too close to the point of incidence, even if they are not directly in the beam. The effect would probably be similar to the arc flash you can get in high power, high voltage electrical systems, where jets of superheated plasma can cause severe burns from contact with the plasma, blast damage from the shock waves, blindness from the intense light produced, and flash burns from the radiated heat.

A continuous beam could have enough scattered and radiant heat to cause flash burns to those near the point of incidence, along with blinding those who are looking at the point of incidence when the beam burns through. If it burns a wide hole, people die quickly when the compartment explosively decompresses, throwing everyone into deep space. If it burns a narrow hole, the survivors who can see can just slap a patch over the hole to prevent the escape of their air.

Luke Campbell

Luke Campbell said: "Although it occurs to me that the jet of supersonic plasma escaping from the hole being drilled could have the combined effect of a blowtorch and grenade..."

Well, it really depends on what you're standing next to, and on how wide the beam is. The energy release at any point along the beam path will be equal to the energy required to drill through the object (so you'll get pulses of heat from each object hit), and it won't really be explosive. Flash burns is the most likely consequence.

Flash burns start at about 5 J/cm2 on exposed skin, and can go above 100 J/cm2 with reasonable protection. At a range of 1 meter, that requires an energy release of 0.63MJ, and once the beam is substantially inside the object, most of the flash will be deposited on the rest of the inside of the object, so it's really only object shells we need to worry about.

If the beam has an area of 50 square centimeters ( AV:T scale) to emit a total of 630 kJ it must be emitting 12.6 kJ/cm2. About the same amount is probably consumed drilling through the object. 1mm of steel requires about 6 kJ/cm2, so anything with a casing of at least 2mm steel, or anything comparable, will cause flash burns within 1 meter.

This is not particularly terrifying, unless of course the beam drills through something like a high pressure steam line, at which point it's suddenly very exciting, though not because of the laser per se.

Anthony Jackson

Anthony Jackson said: "so you'll get pulses of heat from each object hit, and it won't really be explosive"

My thought was that the shocks could coalesce. All shocks are supersonic to the material they have not gone through, and subsonic to the material they have traveled through. As a consequence, a second shock will catch up to a previous shock until they merge into a single, stronger shock. If the beam is pulsed at a high rate (say, a MHz or so) a good number of the individual blasts could coalesce within a short distance to create a more potent blast that might cause significant problems.

The physics of shocks is tricky, and for spherically expanding shocks you get into issues of rarefaction and backflow, which should limit the number of shocks that can coalesce. While I have a highly recommended text on shock physics, I've not had the time to look through it yet, so I don't have a good idea yet on the limits and possibilities of this mechanism.

There's also the issue that iron heated to 10,000 K, for example, will expand in volume about 150,000 times from its solid phase. So burning a 10 cm wide hole through a 1 cm steel bulkhead would produce a cloud of iron vapor with a volume of about a cubic meter if the final temperature was 10,000 K (note that if the iron was converted to a singly ionized plasma, the temperature would be ten times that much, and you would get ten times the volume). Getting caught in that incandescent cloud simply cannot be healthy.

There's also the ozone and nitrogen oxides and reactive chemicals produced as a consequence of incomplete combustion, which will not be healthy to breathe, but I expect that would be secondary.

Luke Campbell

Luke Campbell said: "My thought was that the shocks could coalesce."

They could if the drilling speed is supersonic. Usually it won't be.

Anthony Jackson

Equations

Now for the dull equations.

"Laser" is an acronym for light amplification by stimulated emission of radiation. A laser beam can cut through steel while a flashlight cannot due to the fact that laser light is coherent. This means all the photons in the beam are "in step" with each other. By analogy, a unit of army troops marching in step can inadvertently cause a bridge to collapse, while the same number of people using the bridge in a random fashion have no effect. Laser light at amazingly low energies can still cause permanent blindness by destroying the retina.

Maximum range will be a few hundred thousand kilometers, otherwise almost every shot will miss due to light-speed lag. You can find more details about light-speed lag here.

Laser beams are not subject to the inverse-square law, but they are subject to diffraction. The radius of the beam will spread as the distance from the laser cannon increases.

RT = 0.61 * D * L / RL

where:

  • RT = beam radius at target (m)
  • D = distance from laser emitter to target (m)
  • L = wavelength of laser beam (m, see table below)
  • RL = radius of laser lens or reflector (m)
BandWavelength (m)
Far Infrared3e-5 to 1e-3 m (30,000 to 1,000,000 nanometers)
Mid Infrared5e-6 to 3e-5 m (5000 to 30,000 nanometers)
Near Infrared7e-7 to 5e-6 m (700 to 5000 nanometers)
Red7.1e-7 m (710 nanometers)
Orange6e-7 m (600 nanometers)
Yellow5.7e-7 m (570 nanometers)
Green5.5e-7 m (550 nanometers)
Blue4.75e-7 m (475 nanometers)
Indigo4.3e-7 m (430 nanometers)
Violet3.8e-7 m (380 nanometers)
Ultraviolet A3.2e-7 to 4e-7 m (320 to 400 nanometers)
Ultraviolet B2.9e-7 to 3.2e-7 m (290 to 320 nanometers)
Ultraviolet C2e-7 to 2.9e-7 m (200 to 290 nanometers)
Extreme Ultraviolet1e-8 to 2e-7 m (10 to 200 nanometers)
X-Ray1e-11 to 1e-8 m (0.01 to 10 nanometers)
Gamma-Ray1e-14 to 1e-11 m (1e-5 to 0.01 nanometer)
Cosmic-Ray1e-17 to 1e-14 m (1e-8 to 1e-5 nanometers)

Note that wavelengths shorter than 200 nanometers are absorbed by Terra's atmosphere (so they are sometimes called "Vacuum frequencies") and anything shorter than 10 nanometers is considered "ionizing radiation" (i.e., what the an average person on the street calls "atomic radiation"). Vacuum frequencies will be worthless for a laser in orbit attempting to shoot at ground targets protected by the atmosphere.

Sometimes wavelengths are expressed in Ångström units, 1.0 Ångström = 0.1 nanometer.

More to the point is the intensity of the beam at the target. First we calculate the beam divergence angle θ

θ = 1.22 L/RL

where:

  • θ = beam divergence angle (radians)
  • L = wavelength of laser beam (m, see table above)
  • RL = radius of laser lens or reflector (m)

Note that this is the theoretical minimum size of the divergence angle, it will be larger with inferior lasers.

Next we decide upon the beam power BP, then calculate the beam intensity at the target (the beam "brightness"):

BPT = BP/(π * (D * tan(θ/2))2)

where:

  • BPT = Beam intensity at target (megawatts per square meter)
  • BP = Beam Power at laser aperture (megawatts)
  • D = range to target (meters)
  • θ = Theta = Beam divergence angle (radians or degrees depending on your Tan() function)
  • π = Pi = 3.14159...

There are a few notes on laser firing rates and power requirements here.

In the US military, the minimum threshold for a weapons-grade is 100 kilowatts.

When figuring the tangent, remember that θ from the beam divergence angle equation is in radians, not degrees (Divide radians by 0.0174532925 to get degrees).

What this means is if you are calculating the Beam Intensity equation with a pocket calculator or the Windows calculator program, the calculator is generally set to degrees and it expects you to punch in the angle in degrees before you hit the TAN key. If you punch in the angle in radians you will get the wrong answer.

If instead you are calculating the Beam Intensity equation with a computer spreadsheet or with a computer program you are writing from scratch, the TAN() function wants the input angle to be in radians.

For comparison purposes, the average beam intensity of sunlight on your skin is about 0.0014 MW/m2.

Please note that the amount of beam power deposited on the target is still BP, the intensity just measures how tightly it is focused. It's like using sunlight through a magnifying glass to burn a hole in a piece of paper (or to incinerate ants if you were one of those evil children). The amount of beam power hitting the paper does not change, it is always BP. But if the magnifying glass is so close that the spot size is large, the paper will just get warm. If you move the glass so the spot focuses down to a tiny dot, the intensity increases and the paper spot starts to burn.

Also note that a laser cannon might have lens/mirror which is larger than strictly required for the desired spot size, due to the fact that otherwise the mirror would melt. The larger the mirror, the more surface area to dilute the beam across, and the less the thermal stress on the mirror.

Example

The good ship Collateral Damage becomes aware of an incoming hostile missile. Collateral Damage has a laser cannon with a ten meter radius mirror operating on a mid-infrared wavelength of 2700 nanometers (0.0000027 meters). The divergence angle is (1.22 * 0.0000027) / 10 = 0.00000033 radians or 0.000019 degrees.

The laser cannon has an aperture power of 20 megawatts, and the missile is at a range of four megameters (4,000,000 meters). The beam brightness at the missile is 20 / (π * (4,000,000 * tan(0.000019/2))2) = 15 MW/m2 or 1.5 kW/cm2.

If the missile has a "hardness" of 10 kilojoules/cm2, the laser will have to dwell on the same spot on the missile for 10/1.5 = 6.6 seconds in order to kill it.

Figured another way, at four megameters the laser will have a spot size of 0.66 meters in radius, which has an area of 1.36 square meters. The missile's skin has a hardness of 10 kilojoules/cm2 so 13,600 kilojoules will be required to burn a hole of 0.66 meters radius. 20 megawatts for 6.9 seconds is 13,600 kilojoules. 6.9 seconds is close enough for government work to 6.6 seconds.

Eric Henry has a spreadsheet that does most of this calculation for you here.

In the game Attack Vector: Tactical, the smallest laser lens is three meters in diameter, the frequency of various models of cannon is from 0.0000024 meters (2400 nanometer) to 0.0000002 meters (200 nanometer) and the efficiency varies from 20% down to 1.5%.

Example

Say you have an ultraviolet (20 nanometer) laser cannon with a 3.2 meter lens. Your hapless target spacecraft is at a range of 12,900 kilometers (12,900,000 meters). The Beam Radius equation says that the beam radius at the target will be about 4 centimeters (0.04 meters), so the beam will be irradiating about 50 cm2 of the target's skin (area of circle with radius of 4 centimeters). If the hapless target spacecraft had a hull of steel armor, the armor has a heat of vaporization of about 60 kiloJoules/cm3. Say the armor is 12.5 cm thick. So for the laser cannon to punch a hole in the armor it will have to remove about 625 cm3 of steel (volume of cylinder with radius of 4 cm and height of 12.5 cm). 625 * 60 = 37,500 kiloJoules. If the laser pulse is one second, this means the beam requires a power level of 37,500 watts or 38 megawatts at the target.

In practice, a series of small pulses might be more efficient, causing a shattering effect and driving chips of armor out of the hole, which of course requires less energy than actually vaporizing the armor.

Efficiency

Note that laser cannon are notoriously inefficient. Free-electron lasers have a theoretical maximum efficiency of 65%, while others are lucky to get a third of that. This means if your beam power is 5,000 megawatts (five gigawatts), and your cannon has an efficiency of 20%, the cannon is producing 25,000 megawatts, of which 5,000 is laser beam and 20,000 is waste heat! Ken Burnside describes weapon lasers as blast furnaces that produce coherent light as a byproduct. Rick Robinson describes them as an observatory telescope with a jet engine at the eyepiece. Laser cannons are going to need seriously huge heat radiators. And don't forget that heat radiators really cannot be armored.

The messy alternative is to use open-cycle cooling, where the lasing gas is vented to dispose of the waste heat. Not only does this endanger anything in the path of the exhaust, it limits the number of laser shots to the amount of gas carried.

But Troy Winchester Campbell brings to my attention a recent news item. In 2004, a company named Alfalight, Inc. demonstrated a 970 nm diode laser with a total power conversion efficiency of 65%. They are working in the DARPA Super High Efficiency Diode Sources program. The goal is 80% electrical-to-optical efficiency in the generation of light from stacks of semiconductor diode laser bars, and a power level of 500W/cm2 per diode bar operating continuously.

W = (1.0 / Ce)

where:

  • We = Waste power percentage
  • Ce = Efficiency of Laser Cannon

Obviously:

CP = BP * We

where:

  • CP = Laser Cannon total power (megawatts)
  • BP = Beam Power at laser aperture (megawatts)
  • We = Waste power percentage

WP = CP - BP

where:

  • WP = Waste Power (megawatts)
  • CP = Laser Cannon total power (megawatts)
  • BP = Beam Power at laser aperture (megawatts)

Getting rid of the waste heat from a laser is a problem if you don't dare extend your heat radiators because you are afraid they will be shot off. A strictly limited solution is storing the waste in a heat sink, like a huge block of ice. "Limited" because the ice can only absorb so much until it melts and starts to boil. If your radiator is retracted and your heat sink is full, firing your laser will do more damage to you than to the target.

Eric Rozier has this analysis of heat sink mass:

One common mistake people make is assuming that lasers are infinite fire weapons. With proper radiators extended, this is true, but with them drawn in, to avoid being shot off, we're limited by the heat capacity of our sinking material, as you well know.

An interesting question to ask is: "Without radiators, how many shots can I get off for some mass of coolant and some sort of laser?"

Given single laser of Bp megawatts at aperture, and an efficiency of eff, duty cycle of dc, and firing time of Tf, we get the waste heat Wh (in MWseconds) as:

Wh = Tf * (Bp/eff * dc) * (1 - eff)

Wh is then the waste heat generated by a single blast from our lasers. To figure out how many times we can fire our lasers we need to perform some calculations based on our coolant, the data of interest is:

  • Mass of coolant dedicated to lasers (Mc) in kg
  • Atomic mass of coolant (Ma) in g/mol
  • Heat capacity of coolant (Hc) in J/(mol * K)
  • Melting point of coolant (Km) in K
  • Boiling point of coolant (Kb) in K

Given this, we can find the number of shots we can fire (S) as follows:

S = ((Mc / Ma) * Hc * (Km - Kb)) / 1000 / Wh

If you do not have the atomic mass of coolant or heat capacity of coolant, you can instead use the specific Heat capacity of coolant. This is useful if the coolant is a compound instead of an element in the periodic table.

  • Specific Heat capacity of coolant (Hck) in J/(kg K)
  • Energy Capacity of coolant in MW seconds (or MegaJoules if you prefer)

Ec = (Mc * HcK * (Km - Kb)) / 1000000

S = Ec / Wh

There is an online calculator for this here.

This assumes the coolant is just melted before firing the laser, and just boiling after firing all available shots. In reality, you want to set Kb at some level below the real boiling point, and Km at some level above the melting point.

As a worked example, a 100MW laser with efficiency of 0.2, 0.5 duty cycle, and 0.1s firing time generates 20 MWseconds of waste heat each time it fires. 1000kg of Lithium, (with about 1140K between melting and boiling) can contain enough heat to fire the laser roughly 204 times.

This, I think, helps show some of the heat limitations of lasers, and constrains them (especially as point defense weapons). You end up having to lug a lot of lithium around if you want to fire them often.

I think this is most interesting when thinking about point defense. Lasers fielded as a CIWS are pretty scary, and if you could fire them infinitely often, they probably keep missiles from hitting you. So in order to constrain you from using lasers for point defense, I simply pull into laser range, threatening your radiators, and forcing you to withdraw them. As such, you can no longer afford to use a laser CIWS, and have to switch to something projectile/missile based, which is liable to be less effective.

Eric Rozier

Attack Vector: Tactical Lasers

Ken Burnside's masterful tabletop wargame Attack Vector: Tactical is fictional, but it was prepared with expert help from real live physicists and other scientists. More to the point, design choices were made to make an interesting game. Which means they would also be design choices that would make an interesting science fiction novel.

In the game, there are various types of lasers of increasingly shorter wavelengths, which due to the diffraction equation have increasingly longer range (by which I mean the spot intensity decreases more slowly). These lasers also have a decreasing level of efficiency of converting power into laser beam, I am unsure if this is due to a physical limit or it is an arbitrary thing used to balance the game.

LaserWavelengthColorEfficiency
Short Range Laser2400 nmNear Infrared20%
Close Range Laser1600 nmNear Infrared16.6%
Medium Range Laser1200 nmNear Infrared12.5%
Extended Range Laser800 nmNear Infrared9%
Long Range Laser600 nmOrange6%
Extreme Range Laser400 nmIndigo3%
Ultraviolet Laser200 nmUltraviolet1.5%

In addition, each laser type comes in seven sizes (with focusing mirrors ranging in size from 3 meters radius to 6 meters radius) and assorted energy requirements. The basic game only has short range and medium range lasers:

LaserMirror
radius
Input
Energy
Effic.Aperture
Energy
Eff
Range
Max
Range
Short Range
Laser 2
3 m3 GW20%0.6 GW80 km300 km
Short Range
Laser 3
3.5 m4.5 GW20%0.9 GW100 km440 km
Short Range
Laser 4
4 m6 GW20%1.2 GW120 km560 km
Short Range
Laser 5
4.5 m7.5 GW20%1.5 GW140 km740 km
Short Range
Laser 6
5 m9 GW20%1.8 GW160 km900 km
Short Range
Laser 7
5.5 m10.5 GW20%2.1 GW160 km1,040 km
Short Range
Laser 8
6 m12 GW20%2.4 GW180 km1,200 km
LaserMirror
radius
Input
Energy
Effic.Aperture
Energy
Eff
Range
Max
Range
Medium Range
Laser 2
3 m2 GW12.5%0.25 GW180 km400 km
Medium Range
Laser 3
3.5 m3 GW12.5%0.375 GW200 km600 km
Medium Range
Laser 4
4 m4 GW12.5%0.5 GW240 km800 km
Medium Range
Laser 5
4.5 m5 GW12.5%0.625 GW280 km1,000 km
Medium Range
Laser 6
5 m6 GW12.5%0.75 GW300 km1,200 km
Medium Range
Laser 7
5.5 m7 GW12.5%0.875 GW340 km1,400 km
Medium Range
Laser 8
6 m8 GW12.5%1 GW360 km1,800 km

The mirror radius is the size of the lens or reflector (RL in the diffraction equation). The input energy is fed as power into the laser, after suffering the horrific effects of typical abysmal laser efficiency the laser beam emerges from the business end containing the aperture energy and leaps out to impale the hapless target. The gigawatts of waste heat are absorbed by the internal heat sink, because extending your heat radiator is just asking for it to get shot off.

The effective range and maximum range are not directly applicable, they are artifacts of the beam damage model used by the Attack Vector: Tactical game. But they do provide some basis of comparison. In the game each "damage point" inflicted upon an enemy ship represents 50 megajoules in an eight centimeter diameter circle inflicted in 1/100th of a second. The effective range is the farthest range that the laser can inflict its full damage. The maximum range is the farthest range that the laser can inflict at least one point of damage. This is all required because Attack Vector is not a computer game, it is an incredible paper and cardboard wargame where all the scientific accuracy and scary mathematics are handled painlessly with cunning player aides.

I would hazard a guess this is the reason for the values chosen for input energy and ranges, to calibrate each laser to 50 megajoules in an eight centimeter spot size.

For our purposes, it might make more sense to use the Brightness equation. Then you can assign hardness values for the target's armor.

Short Range Laser 2
RangeSpot
Dia
Brightness
80 km7.8 cm1.55×109 J/m2
100 km9.8 cm9.9×108 J/m2
140 km13.7 cm5.05×108 J/m2
180 km17.6 cm3.06×108 J/m2
220 km21.5 cm2.05×108 J/m2
300 km29.3 cm1.1×108 J/m2
Short Range Laser 8
RangeSpot
Dia
Brightness
180 km8.8 cm3.06×108 J/m2
200 km9.8 cm2.48×108 J/m2
240 km11.7 cm1.72×108 J/m2
300 km14.6 cm1.10×108 J/m2
380 km18.5 cm6.86×107 J/m2
520 km25.4 cm3.66×107 J/m2
840 km41.0 cm1.40×107 J/m2
1,200 km58.6 cm6.88×106 J/m2
Medium Range Laser 2
RangeSpot
Dia
Brightness
180 km8.8 cm5.09×108 J/m2
240 km11.7 cm2.86×108 J/m2
300 km14.6 cm1.83×108 J/m2
400 km19.5 cm1.03×108 J/m2
Medium Range Laser 8
RangeSpot
Dia
Brightness
360 km8.8 cm1.27×108 J/m2
420 km10.2 cm9.35×107 J/m2
500 km12.2 cm6.60×107 J/m2
620 km15.1 cm4.29×107 J/m2
860 km21.0 cm2.23×107 J/m2
1,220 km29.8 cm1.11×107 J/m2
1,600 km39.0 cm6.45×106 J/m2

Combat Mirror

A more scientifically plausible but much less dramatic laser weapon is the combat mirror. In this scheme, the spacecraft doesn't have a laser, just a large parabolic mirror. The laser is several million miles away, on a freaking huge solar power array orbiting your home planet. You angle the mirror so it will do a bank shot from the distant laser off the mirror and into your target, then radio the laser station to let'er rip. About fifteen minutes later the diffuse laser beam arrives, and your parabolic mirror focuses it down to a megaJoule pinpoint on your target.

The advantage is that the spacecraft does not have to lug around the laser, the power supply, the heat radiators, and other massive elements of the laser weapon. The spacecraft can have a higher acceleration or increased payload. The beam can also be of a power level associated with laser equipment that is not considered "portable by spacecraft", if the laser generator is a few miles in diameter your spacecraft could care less.

Disadvantages include the lag time between ordering a shot and its arrival, and the vulnerable nature of the combat mirror (generally little more than a large Mylar balloon).

Mirror Armor

Now I know all you older science fiction fans still remember Johnny Quest and The Mystery Of The Lizard Men where Dr. Quest demonstrates that one can defend oneself against a weapon-grade laser beam with a dressing-room mirror. Sorry, it doesn't work that way in reality. No mirror is 100% efficient, and at these power levels, the fraction that leaks through is more than enough to vaporize the mirror armor. The same goes for "ablative armor." One zap and the impact point is abruptly as bare of armor as a baby's behind.

Inside a laser cannon, a relatively diffuse laser beam is generated. This prevents the beam from vaporizing the cannon's internal optics. At the business end, a parabolic mirror focuses the diffuse beam down to the aforementioned megaJoule pinpoint on the hapless target.

Accuracy

And don't think that lasers will automatically hit their targets either. There are many factors that can cause a miss. Off the top of his head, Dr. John Schilling mentions:

  • Uncertain target location due to finite sensor resolution
  • Uncertain target motion due to sensor glint or shape effects
  • Sensor boresight error due to finite manufacturing tolerances
  • Target motion during sensor integration time
  • Analog-to-digital conversion errors of sensor data
  • Software errors in fire control system
  • Hardware errors in fire control system
  • Digital-to-analog conversion errors of gunlaying servo commands
  • Target motion during weapon aiming time
  • Weapon boresight error due to finite manufacturing tolerances
  • Weapon structural distortion due to inertial effects of rapid slew
  • Weapon structural distortion due to external or internal vibration
  • Weapon structural distortion due to thermal expansion during firing

And we haven't even begun to include target countermeasures...

Turret

Airbourne Laser

What about a laser turret? It can be so inconvenient to have to move the entire ship in order to aim the blasted beam. As it turns out, the US Air Force has a solution created for their Airborne Laser project.

I hear you ask "but why doesn't the beam slice up the inside of the turret?" The key is power density.

For instance, a naughty little boy will find that sunlight does not do much to his skin except warm it up a bit. However, if you whip out a magnifying glass you can focus the sunlight to a white-hot pinpoint that will easily incinerate ants. The magnifying glass increases the power density of the sunlight. So inside the turret, the weapon beam is something like 20 centimeters in diameter which means a power density too low to fry the internal mirrors. At the end, the beam expander mirror evenly shines the laser beam over the primary mirror. That mirror then acts like the magnifying glass in the hands of the anticidal little boy, focusing the diffuse laser beam down to an incinerating pin-point on the hapless target.

Isaac Kuo points out that another factor keeping the laser from chopping up the turret is that the internal mirrors are dielectric mirrors. Those babies can be up to 99.999% reflective. Meanwhile if target has conventional mirror plating it will only be 95% reflective, absorbing 5,000 times as much laser energy. Dielectric mirrors would be difficult if not impossible to manufacture in pieces large enough to cover a missile or spacecraft.

The actual US Air Force Air Borne Laser is a megawatt class chemical oxygen iodide laser (COIL) operating at a frequency of 1.315 microns or 1.315e-6 meters (near infrared). With a 1.5 meter mirror, this gives a divergence angle of 1.07e-6 radians. If my slide rule is correct, this means at a range of one kilometer it will have a spot size of one millimeter radius, and a beam brightness of about 300,000 megawatts per square meter. However, I've seen suggestions that the actual spot size is more like several centimeters, demonstrating the room for improvement.

The US Air Force is understandably reluctant to give any figures on the performance of the Air Borne Laser. The best figures I could find suggest that it could destroy a flimsy unarmored hypergolic fueled missile (with fuel still in the tanks) by expending a three to five second burst up to a range of about 370 kilometers. Three to five seconds is an awfully long time to keep the beam focused on the same spot on a streaking missile. The dwell time will have to be longer if the missile is armored or if it uses solid fuel or other inherently stable fuel.

The giant primary mirror will contain adaptive optics (i.e., it will be a "rubber mirror"). This will allow the mirror to change its focus to accommodate the range to target. In diagram "a" to the right, the flexible mirror is laid over a slab of piezoelectric material that changes shape as power is applied to the electrodes. In diagram "b" individual actuators are used. The image on the right is a 19-actuator deformable mirror built by Rockwell International. The mirror is only 40 cm in diameter. The actuator density is about 150 actuators per square meter, so the 1.5 meter ABL mirror would require about 270. (surface area of a circular 1.5 meter mirror is about 1.8 square meters, times 150 actuators per square meters give 270 total actuators)

Luke Campbell's Turret

Luke Campbell has his own design for a laser turret. Cararra 5 was used to create the 3D mesh and to render the images.

Optics

Rick Robinson has a more serious concern. You know how it is a very bad idea to look through a telescope at the Sun? Well, for the same reason it is bad to unshutter your laser cannon optics and point them at a hostile ship which might zap you with its laser. Your cannon's optics would funnel their beam right down into the delicate interior of your cannon. The optics would also concentrate their beam to 10x or 100x the intensity. This means that if your lasers are unshuttered and your opponents are shuttered, you have the drop on them. The instant you detect their shutters trembling you give them a zap. Their shutters will still be opening when your bolt scrags their laser.

However, Ken Burnside says:

I will point out that the likeliest result of "shooting down the barrel of a laser" is to destroy one of the mirror elements on the focal array. Since those elements are likely to be used with adaptive optics, this won't even hurt the laser that much. It's only if the mirrors are hit at exactly the right angle that they'll direct energy back into the Free Electron Laser itself.

Ken Burnside

Anthony Jackson has another messy solution. One can design a laser cannon without a mirror or lens, if one uses a phased array. Currently we can create phased arrays for microwaves and radars, but have no idea how to do it with visible light. It would take a major technological break-through, but it is not actually forbidden by the laws of physics. Another nifty effect of phased array emitters is that they're flat and can fire at any angle (range will suffer at extreme angles), without requiring a turret assembly.

Dr. Yo came to the horrified realization that the logical acronym for PHased Array laSER was ... aiieee!

Eric Henry prefers that particular name for Free-electron laSER.

Bomb-Pumped Lasers

A special type of laser is the bomb-pumped laser. This is generally found as a missile warhead. A "submunition" is a warhead that is a single-shot bomb-pumped gamma-ray laser. The original concept was developed by Edward Teller under the name "Excalibur." Teller and Excalibur were later discredited, but the basic idea wasn't.

Here's the problem: the lasing medium in a laser has to be "pumped" or flooded with the same frequency that the laser emits. This isn't a problem with infrared or visible light, but sadly there are not many good sources of x-rays and gamma-rays. About the only good source is a detonating nuclear device, which has the distressing side-effect of vaporizing the laser. So the idea is to make a laser that can frantically manufacture one good x-ray zap in the few microseconds before it is destroyed by the bomb blast. This is the reason it is "one-shot."

(Yes, in theory, hafnium-178m2 is also a good source of gamma rays, but it has problems.)

The Excalibur units had about one hundred x-ray laser rods mounted on a nuclear device. When the hordes of evil Soviet nuclear missiles climbed into view, all one hundred lasers would lock on to different targets, then the bomb was triggered. John Schilling said that due to inefficiency each laser would emit a pulse of only 5e6 Joules, but they'd have a range of up to one hundred kilometers.

A one megaton nuclear device releases about four billion megajoules, but only a few percent of this will end up in the x-ray laser beams, due to the inherent inefficiency. Call it a total of about 100 million megajoules of x-ray laser.

Bomb-pumped lasers do not use lenses or mirrors (because there ain't no such thing as an x-ray mirror). Brian Smith-Winsemius gently pointed out to me that I do not know what I am talking about, since he works with x-ray mirrors every day.

I happen to work on a EUV (13.5nm wavelength) prototype photolithography tool. So when I read "Bomb-pumped lasers do not use lenses or mirrors (because there ain't no such thing as an x-ray mirror)." I had to stop and write. The tool I work on uses multi-layer mirrors We have to use mirrors since there are no known lenses that work with 13.5nm or x-ray light. For example, the Chandra X-Ray observatory uses a collector mirror assembly which resembles our collector optic.

Brian Smith-Winsemius

According to The Star Wars Controversy: An "International Security" Reader (edited by Steven E. Miller and Stephen Van Evera, 1986), in order to calculate the beam divergence angle of a bomb-pumped laser, use the following:

θ = 2 * (w / l)

where:

  • θ = beam divergence angle (radians)
  • w = width of lasing rod (meters)
  • l = length of lasing rod (meters)

A practical maximum length of a single laser rod is no more than five meters. Making the rod thinner decreases the divergence angle, but this is limited by diffraction, just like in more conventional lasers. Make the rod too narrow and diffraction actually makes the divergence angle larger. The width limit is:

1.22*L/l = 2*w/l

where:

  • L = wavelength of laser beam (meters)
  • w = width of lasing rod (meters)
  • l = length of lasing rod (meters)

For an x-ray laser rod of one nanometer wavelength and rod length of five meters, the optimum rod width is 0.06 millimeters. The beam divergence angle will be 20 microradians.

This relatively huge divergence further degrades the laser performance. Our 100 million megajoules are now diluted into a 20 microradian cone. On a target at ten megameters, it would deposit about 300 kJ/cm2 over a spot 200 meters wide.

Note the consequence of the absence of x-ray mirrors: each laser rod will fire a laser beam out both ends of the rod. The majority of the beam will exit from the end of the rod farther from the nuclear blast, however (i.e., most of the beam will travel in the same direction as the x-rays from the blast). If the rod is perpendicular to the blast, equal beams will emerge from both ends.

A bigger draw-back is the fact that while a laser cannon requires a targeting system, Excalibur requires a targeting system for every single laser rod. Such systems are not cheap.

A more minor problem is "bomb-jiggle." Many types of fission devices use conventional explosives to squeeze the core into a critical mass. While the nuclear blast is far too swift to jog the laser rods off their targets, the conventional explosives are not. They might cause the rods to miss-aim, so when the nuclear blast triggers the x-rays, the beams are off-target. This might be avoided by using a laser-initiated fusion device.

Footfall

There is a variant on the bomb-pumped laser in Larry Niven and Jerry Pournelle's classic novel Footfall, which is arguably the best "alien invasion" novel ever written. They noticed that bomb-pumped lasers is a concept that merges seamlessly with Orion drive spacecraft. In this case the submunitions do not need a bomb. They are thrown below the pusher plate, they take aim at the enemy, then the next propulsion bomb pushes the ship and simultaneously pumps the submunitions. There is a diagram of the ship from Footfall here (design by Aldo Spadoni, president of Aerospace Imagineering).

Impulsively Driven Laser

Andrew Presby found an interesting document entitled "On The Feasibility of an Impulsively Driven Gamma-ray Laser" (1979) at the Federation of American Scientists website. Please note this is for gamma-ray lasers, not x-ray lasers like the discussion above. That is probably why the x-ray laser rods had a maximum length of 5 meters while these graser rods have a length of 0.05 meters.

I wish I'd found the dumb thing years ago when I taking my graduate school lasers class and looking for physics papers on bomb pumped GRASERS. The Nevada experiment described herein sounds suspiciously like the bomb pumped XRASER (xray laser) experiments in the 70s/80s codenamed Excalibur that started the chain of events that got Teller in so much trouble. Thing I cannot figure is that the device described herein seems to produce GAMMA RAYS in the 6-8 MeV range (~0.002 Ångström) which is 10000 times higher photon energy than the stuff I've found in the literature that is available on Excalibur (which was in the ~14 Ångström range).

I've never heard if this worked or not... but there you go.

Andrew Presby

The document suggest using Tantalum-180 dissolved in Lithium-7 for the lasing rods, about one part in four thousand. Alternatives are Cobalt-109 and Molybdenum-99.

The design uses the Mössbauer effect, the recoil-free emission and absorption of gamma ray photons by atoms bound in a solid form. This is important. Laser light is coherent light, where all the photons are in perfect lock-step. The trouble with x-ray and gamma-ray emission is that they are powerful enough to make the excited atom recoil in reaction. This throws off the synchronization, so that the beam is not coherent, and thus not a laser beam. The Mössbauer effect prevents this by locking the lasing atoms in a matrix of anchor atoms, thus dealing with the recoil.

It was estimated that the grasing transition energy densities of tens of kilojoules per cubic centimeter. This means a one megajoule graser could fit in a breadbox, sans bomb of course. A laser beam composed of gamma rays impacting on, say, an incoming Soviet nuclear warhead would produce a flood of neutrons generated by gamma-ray/neutron recations, burning a nice hole. And the high-energy Compton-scattered electrons would create an enormous EMP, frying the warhead's electronics.

The document describes a test for the concept. A cylindrical package five centimeters long by five centimeters in radius would be packed with 20,000 lasing needles 25 µ diameter by 5 centimeters long (I assume that µ means micrometre or micron). The needles would be composed of Lithium-7 with 0.025% Tantalum-180. The needles would be aligned in parallel with 100 µ spacing between their axes, and arranges so that the centers of no three needles would be in a straight line.

The rod assembly package would be insulated from the bomb by insulating and moderating material (from the bomb: 15 cm of space, 7 cm of lead, 20 cm of heavy water, 5 cm to the center of the rod assembly). This will ensure that only the proper radiation strikes the assembly, and to allow the assembly to survive for the few microseconds required to create the graser beam. The lead [1] attenuates the gamma radiation from the bomb, [2] slows the debris motion, [3] and blocks the x-rays that would destroy the package. The heavy water moderates the neutron output.

The beam divergence is determined by the aspect ratio, which for this package is on the order of 0.5 milliradian. This is above the diffraction limit (about 8 milliradian).

In the proposed test, a one kiloton device would be detonated to pump the graser. The five centimeter needles have a calculated gain of 2 x 104. About 9% of the nuclear energy in the grasing transition will actually escape the needles, due to the short pathlength for 6.3 keV gamma rays. The energy available is 7.3 x 1016 MeV cm-3, which means the graser beam will be a piddling little 2.6 kilojoules. Keep in mind that is was intended as a test rig, not a functioning weapon.

Non-Bomb-Pumped Lasers

Laser guru Luke Campbell thinks it not impossible to make an x-ray laser which does NOT require a nuclear device to pump it. In theory a Free Electron laser can produce any wavelength. It is possible approximate an x-ray lens by having the rays make glancing blows off dense materials.

Bottom line is an x-ray laser is technologically very challenging, but if you manage to make one you have an Unstoppable Death Ray of Stupendous Range.

Let's take a 10 MW ERC pumped FEL at just above the lead K-edge. This particular wavelength is used because lead is pretty much the heaviest non-radioactive element you can get, and at just above the highest core level absorption for a material you can get total external reflection at grazing angles - so no absorption or heating of a lead grazing incidence mirror. We will use a 1 meter diameter mirror. The Pb K-edge x-ray transition radiates at 1.4E-11 m. This gives us a divergence angle of 1.4E-11 radians. At 1 light second, we get a spot size of 5 mm, and an intensity of 5E11 W/m2.

Looking at the NIST table of x-ray attenuation coefficients, and noting that 1.4E-11 m is a 88 keV photon, we find an attenuation coefficient of about 0.5 cm2/g for iron (we'll use this for steel), 0.15 cm2/g for graphite (we'll use this for high tech carbon materials) and 0.18 cm2/g for borosilicate glass (a very rough approximation for ceramics). Since graphite has a density of 1.7 g/cm3, we get a 1/e falloff distance (attenuation length) of 4 cm. Iron, with a density of 7.9 g/cm3, has an attenuation length of 0.25 cm. Glass, density 2.2 g/cm3, has an attenuation length of 2.5 cm.

At 1 light second, therefore, the beam is depositing 2E12 W/cm3 in iron at the surface and 7E11 W/cm3 at 0.25 cm depth; 1.2E11 W/cm3 in graphite at the surface and 5E10 W/cm3 at 4 cm depth; and 2E11 W/cm3 in glass at the surface and 7E10 W/cm3 at 2.5 cm depth. Using 6E4 J/cm3 to vaporize iron initially at 300 K, we find that iron flashes to vapor within a microsecond to a depth of 0.9 cm. The glass, assumed to take 4.5E4 J/cm3 to vaporize (roughly appropriate for quartz) will flash to vapor within a microsecond to a depth of 4 cm within a microsecond. Graphite, at 1E5 J/cm3 for vaporization, will flash to vapor to a depth of 0.7 cm within a microsecond (the laser performs better if we let it dwell on graphite for a bit longer, we get a vaporization depth of 10 cm after ten microseconds).

Net conclusion - ravening death beam at one light second.

Now lets look at one light minute. The beam is now 30 cm across. This is much deeper than the attenuation length in all cases, so we will just find the radiant intensity and the equilibrium black body temperature of that intensity. We have an area of 7E-2 m2, and an intensity of 1.4E8 W/m2. You need to reach 7000 K before the irradiated surface is radiating as much energy away as heat as it is receiving as coherent x-rays. The boiling point of iron is 3023 K, the boiling point of quartz is 2503 K, and the sublimation temperature of graphite is 3640 K. All of these will be vaporized long before they stop gaining heat. At this range, the iron is subject to 5.6E8 W/cm3 at the surface, the graphite to 3.3E7 W/cm3 at the surface, and the glass to 5.6E7 W/cm3 at the surface. Using the above values for energy of vaporization, we get about 0.1 milliseconds before the iron starts to vaporize, 0.8 milliseconds before the glass starts to vaporize, and 3 milliseconds before the graphite begins to vaporize (because of its long attenuation length, once it begins to sublimate, graphite sublimates rapidly to a deep depth, while you essentially have to remove the iron layer by layer).

Net conclusion - still a ravening death beam at one light minute.

What about at one light hour? The beam is 18 meters across. The equilibrium black body temperature is 900 K. This is well below the melting point of most structural materials. Ten megawatts, however, is a lot of ionizing radiation. Any unhardened vehicle will be radiation killed at these ranges.

Luke Campbell

However, he goes on to note that in order to boost electrons to the velocities required for an X-ray free electron laser, you will need an acceleration ring approximately one freaking kilometer in diameter. So this X-ray laser would only be suitable for exceedingly huge warships, orbital fortresses, and Death Stars.

Since the time he wrote the above, Luke Campbell has reconsidered the use of lead grazing incidence mirror. Now he favors using diffraction.

I have since come to realize that at x-ray energies this high, matter cannot act as a mirror even at grazing angles (the x-rays have such a short wavelength that they interact with the atoms individually, rather than seeing them as a flat sheet - and you can't really get grazing incidence off of an individual atom). This is why I now prefer diffraction for focusing.

Luke Campbell

Particle Beams

Particle beam weapons use a similar principle to the one being utilized in the computer monitor aimed at your face right now (unless you are one of those lucky people who has a flat-panel monitor) those ancient CRT monitors and TV screens they used to use in olden times. Electrons or ions are accelerated by charged grids into a beam. They work much better in the vacuum of space than in an atmosphere, which is why there is no air inside the cathode-ray tube of your ancient monitors. Laboratory scale electron beams can have efficiencies up to 90%, but scaling up the power into a weapon-grade beam will make that efficiency plummet.

Particle beams have a advantage over lasers in that the particles have more impact damage on the target than the massless photons of a laser beam (well photons have no rest mass at least. The light pressure exerted by a laser beam pales into insignificance compared to the impact of a particle beam). There is better penetration as well, with the penetration climbing rapidly as the energy per particle increases. Particle beams deposit their energy up to several centimeters into the target, compared to the surface deposit done by lasers.

They have a disadvantage of possessing a much shorter range. The beam tends to expand the further it travels, reducing the damage density ("electrostatic bloom"). This is because all the particles in the beam have the same charge, and like charges repel, remember? Self-repulsion severely limits the density of the beam, and thus its power.

They also can be deflected by charged fields, unlike lasers. Whether the fields are natural ones around planets or artificial defense fields around spacecraft, the same fields used to accelerate the particles in the weapon can be used to fend them off.

Particle beams can be generated by linear accelerators or circular accelerators (AKA "cyclotrons"). Circular accelerators are more compact, but require massive magnets to bend the beam into a circle. This is a liability on a spacecraft where every gram counts. Linear accelerators do not require such magnets, but they can be inconveniently long.

Another challenge of producing a viable particle beam weapon is that the accelerator requires both high current and high energy. We are talking current on the order of thousand of amperes and energy on the order of gigawatts. About 1e11 to 1e12 watts over a period of 100 nanoseconds. The short time scale probably means quick power from a slowly charged capacitor bank, similar to the arrangement in a typical camera strobe. You want a very thin beam with a very high particle density, the thinner the better and the more particles the better. The faster the particles move the more particles will be in the beam over a given time, i.e., the higher the "beam particle current" and the faster this current flows, the more energy the beam will contain.

The power density is such that the accelerator would probably burn out if operated in continuous mode. It will probably be used in nanosecond pulses.

Protons are 1836 times more massive than electrons, so proton beams expand only 1/1836 times as fast as electron beams and are 1836 times harder to deflect with charged fields. Of course they also require 1836 times as much power to accelerate the protons to the same velocity as the electrons.

It is possible to neutralize the beam by adding electrons to accelerated nuclei, or subtracting electrons from negative ions. While this will eliminate electrostatic bloom, the neutralization process will also defocus the beam (to a lesser extent). As a rough guess, maximum particle beam range will be about the same as a very short-ranged laser cannon.

For a neutral particle beam, the divergence angle is influenced by: traverse motion induced by accelerator, focusing magnets operating differently on particles of different energies, and glancing collision occurring during the neutralization process. The first two can be controlled, the last cannot (due to Heisenberg's Uncertainty Principle). The divergence angle will be from one to four microradians, compared to 0.2 for conventional lasers and 20 for bomb-pumped lasers.

The source of the particles for the beam come from sophisticated gadgets with weird names like "autoresonantors", "inertial homopolar generators", and "Dundnikov surface plasma negative ion sources".

Dr. Geoffrey A. Landis had this to say:

Particle beams disperse for a lot more reasons than laser beams, unfortunately, so it's harder to give a simple formula. It will depend on things like magnetic and electric fields in the region between the source and the target (if the particles have spin, for example, they will couple to the magnetic field gradient even if they are neutral).

However, for a neutral particle beam traversing empty, field-free space, the dispersion is proportional to the temperature of the beam. Using, for the sake of a simple example, a mercury ion beam (dispersion decreases proportional to square root of atomic mass, and mercury is a convenient high-mass atom that ionizes easily), the lateral (spreading rate) velocity of the beam is:

V = 1.4 SQRT(T) m/sec, for T in Kelvins

To calculate the actual angular spread of the beam, you need to know the beam velocity. For a quick calculation, you could say it's no more than the speed of light, 300,000,000 m/sec. So the dispersion in nano-radians is 5 SQRT(T).

So, for a beam with an effective temperature of, say, 1000K, dispersion for mercury is 150 nR, or 0.15 micro-radians. Dispersion at a distance of 100,000 km would be 0.015 km, or 15 meters. A hydrogen beam would disperse SQRT(80)= 9 times more.

[note that if the beam is actually relativistic, you have to apply a relativistic correction, which I'll ignore here.]

Dr. Geoffrey A. Landis

I'm not sure I have this correct, but to put this in useful form:

θ = (5e-9 * Sqrt[BT]) * Sqrt[80/Bn]

where:

  • BT = beam temperature (Kelvin)
  • Bn = atomic number of element composing the beam (Uranium = 92, Mercury = 80, Zirconium = 30, Calcium = 20, Neon = 10, Hydrogen = 1)
  • θ = Beam divergence angle (radians)

RT = Tan(θ) * D

where:

  • D = distance from particle beam emitter to target (m)
  • RT = radius of beam at target (m)

...making sure that Tan() is set to handle radians, not degrees. Or as one big ugly unified equation:

RT = Tan((5e-9 * Sqrt[BT]) * Sqrt[80/Bn]) * D

...again making sure that Tan() is set to handle radians, not degrees. I must stress I derived this equation myself, so there is a chance it is incorrect. Use at your own risk.

Electrostatics, Neutrons, and Space Charge

While particles cannot travel at the speed of light, they can get close enough that it is hard to tell the difference. Unfortunately, particle beams do obey the inverse-square law.

A beam of neutrons does not suffer from electrostatic bloom since they have no charge, nor could they be deflected by charged fields. However, this also means it is difficult to accelerate the neutrons in the first place (and if you discovered a new way to do it, chances are it too could be used as a defense). Without electrostatic bloom neutron beams are only limited by "thermal bloom". Brett Evill says this will give a neutron beam an effective range of 10,000 km, but he doesn't mention the details of this estimate. Nelson Navarro is of the opinion that a science fictional heavy neutron beam could be produced by a science fictionally efficient method of breaking up deuterium nuclei.

Another problem is one shared by ion drives, the "space charge." If you keep shooting off electron beams you will build up a strong positive charge on your ship. At some point the charge will become strong enough to bend the beam. And the moment your ship tries to dock with another it will be similar to scuffing your shoes on the rug and touching the doorknob. Except instead of a tiny spark it will be a huge arc that will blow all your circuit breakers and spot-weld the ships together.

Don't try to neutralize the charge by firing off positively charged proton beams. John Schilling warns that space is filled with an extremely low-density, but conductive, plasma. You try to eject charge from your ship, and the ship itself becomes part of a current loop. Not only is the current flowing through the hull (or trying to) likely to cause problems, but all those electrons or protons being sucked in produce X-rays on hitting the hull.

Bremsstrahlung

Powering up a particle beam to the point where it can cut armor is difficult. But there is another option: death by "Bremsstrahlung".

Consider the x-ray tube in your dentist's office. It is basically an electron beam striking a metal target. Now, what if the electron beam was a particle beam weapon and the metal target was the hull of the enemy spacecraft? A hypothetical observer on the far side of the ship could make a nifty x-ray photo revealing the skeletons of crew members dying in agony of radiation poisoning.

Please note that Bremsstrahlung only occurs with charged particle beams, it doesn't happen with beams of neutrons.

The particle beam weapons postulated for Star Wars missile defense were to disable missiles by damaging the sensitive electronics via radiation, not by carving the missiles into pieces. An APS directed-energy weapons study written for the Strategic Defense Initiative estimated that in order to disable an ICBM, a particle beam had power requirements between 100 and 1,000 megawatts, depending on range and retargeting rate.

Anthony Jackson says if you crank up your particles to a few GeV per nucleon they will be in the soft end of the spectrum of primary cosmic rays. Each particle will be highly penetrating, and you no longer need to actually focus the beam. Just apply a couple megajoules per square meter and everything dies (unless it's behind a huge amount of shielding or is basically operating at pre-microchip levels of automation. Neither is an option for a surface mounted weapon turret.). We are talking about a surface radiation level of over 500 grays. Such a cosmic ray beam would require armor with a TVT (for radiation purposes) peaking at 200-300 g/cm2.

Also note that if the particles are moving a relativistic velocities higher than, say, 90% c, you will have about the same energy release if the particles are matter or antimatter. In other words, it is pointless for relativistic particle beam weapons to use antimatter, with all the added complexity due to antimatter manufacture and storage.

Ships that expect to be fired upon by particle beam weapons would be well advised to add a layer of paraffin or other particle radiation armor on the outside of their metal hull, to prevent the beam from generating Bremsstrahlung with the hull.

Kinetic Kill Weapons

Kinetic Kill weapons are unguided missiles that have no warheads. Bullets and artillery shells in other words. They can be a simple as a bucket of rocks dumped in the ship's wake. Since they are basically solid lumps of matter they are much cheaper than a missile. They cannot be jammed, but by the same token they do not home in on the target. The damage they do depends upon the relative velocity between the kinetic lump and the target ship.

A sort of hybrid would be a missile which explodes into a cloud of deadly shrapnel that the enemy ship plows through, screaming.

Go to the Rocketpunk Manifesto, and read Kinetics, Part 1 and Kinetics, Part 2 The Killer Bus.

In case it is not obvious, if the weapon projectile has no rocket engine strapped to it (as do missiles), the weapon is not recoiless. Cannons, coil guns, and rail guns all have recoil due to Newton's third law. In fact, the propulsion system know as a mass driver is basically a coil gun optimized as a propulsion system rather than optimzed as a weapon. This means that these weapons can be used as crude propulsion systems in an emergency.

Equations

The damage inflicted can be calculated by the equation below. The same equations will also apply when one ship rams another, of course with added damage from exploding missile magazines, unstable fuel supplies, and out of control power plants. In a ramming, you will have to calculate the equation twice, once to figure damage inflicted on the rammed ship, the second time to calculate damage inflicted on the ramming ship.

To get some idea of the amount of damage represented by a given amount of Joules, refer to the Boom Table.

Eric Rozier has an on-line calculator for kinetic kill weapons.

Please note that it is relative velocity that is important. If your ship is quote "standing still" unquote, and if the enemy is tearing past you at seven kilometers per second, and if you leisurely toss an empty beer can into the path of the enemy, the relative velocity will be 7 km/s and the beer can will do severe damage to the enemy ship (if the beer can masses 0.1 kilogram, it will do 2,450,000 Joules of damage). So even though the beer can has practically zero velocity from your standpoint, from the standpoint of the soon-to-be-noseless ship the can has the velocity of a bat out of you-know-where.

Ke = 0.5 * M * V2

where:

  • Ke = kinetic energy (Joules)
  • M = mass of projectile (kg)
  • V = velocity of projectile relative to target (m/s)

Wp = Ke * (1 / We)

where:

  • Wp = power required by weapon to fire one projectile (Joules)
  • Ke = kinetic energy of one weapon projectile (Joules)
  • We = efficiency of the weapon (0.0 = 0%, 1.0 = 100%)

Rick Robinson's First Law of Space Combat states that:

Rick Robinson's First Law of Space Combat

An object impacting at 3 km/sec delivers kinetic energy equal to its mass in TNT.

Rick Robinson

In other words there are 4,500,000 joules in one kilogram of TNT (3,0002m/s * 0.5 = 4.5e6). This means a stupid bolder traveling at 2,000 km/sec relative has about 400 kilo-Ricks of damage (i.e., each ton of rock will do the damage equivalent of 2e12 / 4.5e6 = 400 kilotons of TNT or about 20 Hiroshima bombs combined).

Ricks = (0.5 * V2) / 4.5e6

where:

  • V = velocity of projectile relative to target (m/s)
  • Ricks = kilograms of TNT worth of kinetic energy per kilogram of projectile

So a projectile moving at 200 km/sec (20,000 m/s) would have about 4,000 Ricks (4 kilo-Ricks) of damage, approximately the same as a standard one-kiloton-yield nuclear weapon. By that I mean it has the same damage per kilogram as a nuke, counting all the nuke's framework, electronics, fissionable material, and whatnot. (for the projectile to do the same damage as a standard nuke, it would need to be the same mass as a standard nuke, about 250 kilograms) A projectile moving at 3,500 km/sec would have about one mega-Rick, which is the same damage per kilogram as the ultra-compact 475-kiloton-yield W-88 nuclear warhead.

As a rule of thumb, anything with more than 100 Ricks (i.e., over 30 km/sec relative) does weapons-grade levels of damage. As an even more shaky rule of thumb, anything with more than 4,000 Ricks (i.e., over 190 km/sec relative) does nuclear warhead levels of damage. This is based on the assumption that a nuclear weapon has about a 4,000 fold increase in energy per kg released versus TNT.

And if you are thinking in terms of bombarding your enemy with asteroids, as a rule of thumb an asteroid's mass will be:

Ma = 1.47e4 * (Ra3)

where:

  • Ma = mass of asteroid (kg)
  • Ra = radius of asteroid (m)
Example

The wet navy battleship Iowa had 16-inch guns. They fired shells which massed about 2000 pounds (907 kg), carried a charge of 145 pounds (54 kg) of high explosive, and traveled at about 820 meters per second. By the kinetic equation above, they contained about 3.0e8 joules of kinetic energy. There are about 4.184e6 joules per kilogram of TNT (which is different from the value used in Rick Robinson's equation, if this annoys you, take it up with him) so the explosive charge contains about 2.3e8 joules of energy.

This means one 16-inch shell does about 3.0e8+2.3e8 = 5.3e8 joules of damage.

Floyd has spent the last 8.6 boring months in the good scoutship Peek-A-Boo, traveling from Mars to Earth in a hohmann orbit. Suddenly he notices a convoy raider from the Asteroid Revolutionary Navy accelerating from low Earth orbit into a Martian hohmann transfer orbit.

Unfortunately for Floyd, scoutships are unarmed. But since the two ships are traveling in opposite directions at a fair speed, anything Floyd can throw at the raider will be good for quite a few Ricks. How massive an object will Floyd have to hurl in order to inflict the same damage as a 16-inch shell?

For the raider to leave LEO and enter Earth Escape orbit takes about 3.17 km/s. To leave Earth Escape and enter Mars Hohmann orbit takes 2.95 km/s. So the raider has about 6.12 km/s relative to Earth.

Since Floyd is on the opposite leg of an Earth-Mars hohmann, he is also doing 6.12 km/s relative to Earth, but with an opposite vector. So relative to the raider, Floyd moving at 6.12 + 6.12 = 12.24 km/s.

Ke = 0.5 * M * V2

therefore

M = Ke / (0.5 * V2)

Ke = 5.3e8 joules and V = 12,240 m/s so M = 7.08 kg (about 15 pounds). A 15 pound object will do as much damage as a 16-inch shell.

At this speed, anything striking the raider will have 16.6 Ricks!

Sneaky the cat watches with bright interest as a space-suited Floyd carries the cat's litterbox into the airlock, and empties it into the path of the raider...

"Kirklin Mines"

In AV:T are kinetic weapons called "Kirklin mines" (invented by Kirk Spencer). They are dirt cheap chemical fueled anti-missile weapons, specifically anti-Torch missile weapons. The ideas is that they cost a fraction of the price of a missile, yet can scrag it. Using the magic of relative velocity, all they have to do is get in the way (this is why they are used against torch missiles, if the relative velocity isn't large enough the mine might not do enough damage to mission-kill the missile).

Launched at the proper time a Kirklin mine can either take out the incoming missile while it is too far away to damage the targeted ship, or force the missile to miss the ship entirely in the process of avoiding the mine (if the mine is launched too soon the missile has enough time to zig-zag around it and still kill the ship). Since they are cheaper, a given spacecraft can carry several mines for every missile their equivalent opponent ship has.

The current thinking is the only way a torch missile can avoid being neutralized by Kirklin mines is by becoming a bus carrying sub-missiles and decoys. Of course for a modest increase in cost the mines can become buses as well...

Hypervelocity Weapons

A special type of kinetic weapon is the hypervelocity weapon. These come in two types: rail guns and coil guns.

However, once the speed of the projectile surpasses about 14% the speed of light (42,000 kilometers per second), it is no longer a strict hypervelocity weapon, it has become a relativistic weapon.

Railguns

A railgun is two highly charged rails. When a conducting projectile is introduced into the breech, it strikes an arc between the rails, and is accelerated down the barrel by Lorentz force. The projectile can be composed of anything, as long as the base will conduct electricity. Sometimes a non-conducting projectile is accelerated using a conducting base plate called a sabot or armature. The maximum velocity of the projectile is about six kilometers per second, which is pretty freaking fast. This would give the projectile about 3.8 Ricks worth of damage, e.g., a ten kilogram projectile would have as much striking power as thirty-eight kilograms of TNT.

And when we say "strike an arc", we don't mean "make a tiny spark like scuffing your shoes on the carpet and touching the doorknob." It is more like "incredibly powerful continuous electrical explosion." Those rails are carrying pleny of juice, and quite a bit of it is wasted.

Advantages are simple construction, disadvantage is the severe rail erosion each projectile causes, requiring frequent replacement of rails (some prototypes required replacement after each use). The rails need massive braces, since they are under tremendous force trying to repel the rails from each other.

Remember, since the projectiles are not rocket-propelled, railguns are not recoiless.

US Navy

In 2007, the US Navy demonstrated a railgun prototype. It used about 8 megajoules, but the full scale weapon is designed to use 64 megajoules. By way of comparison, current conventional naval 5-inch guns have the equivalent of 9 megajoules of muzzle energy. The full scale weapon will have a range of 200 to 250 nautical miles, as compared to less than 15 nautical miles for a 5-inch gun. The PR handout said the full scale weapon will have "the punch of a Tomahawk cruise missile", or be the equivalent of "hitting a target with a Ford Taurus at 380 mph." It will also travel the 200-250 nautical miles to the target in about six minutes, as opposed to 8 for a Tomahawk cruise missile. At the peak of its ballistic trajectory, the projectile will reach an altitude of 500,000 feet, or about 95 miles, actually exiting the Earth's atmosphere.

We shall see if these rosy predictions pan out.

I tried to derive some values for the above weapons system and produced the following analysis. It turned out to be totally wrong, I reproduce it here so you can see my mistakes:

225 nautical miles in six minutes is an average velocity of 463 meters per second. The best estimate I could find in a five minute Google search for the mass of a Ford Taurus is 3111 pounds or about 1400 kg. 3111 pounds at 380 mph is 1400 kg at 170 m/s. Ke = 0.5 * M * V2 so the Ford Taurus will hit with about 2e7 joules or 20 megajoules. About the equivalent of 4.5 kilograms of TNT (170 m/s is about 0.003 Ricks of damage). I guess the other 44 megajoules are lost due to wind resistance.

Working the other way, we can take the 463 m/s average velocity and the 64 megajoule power consumption. Ke = 0.5 * M * V2 therefore M = Ke / (0.5 * V2). This means the projectile mass is around 600 kg.

As I said, the above analysis is incorrect. Lucky for me, a gentleman named Thomas Rigby appeared and set matters straight:

I noticed some deficiencies in your analysis of the Navy's proposed 64 MJ railgun system, particularly in your derived velocity. The M1 Abrams main gun fires a FSAPDA round somewhere between 1200 and 1800 m/s (can't remember exactly), so why would the Navy put so much unto a system that only fires at a third the velocity?

I also remember reading a Popular Science article on the new features of the DD(X) project, one of which is the railgun. According to the article the railgun would fire a 40 pound projectile (about 18.2 kg) with a Mach 8 muzzle velocity and Mach 7 velocity at the target. A quick calculation (setting speed of sound a 343 m/s):

KE = ½ (18.2 kg) (2401 m/s)2 = 52.46 MJ

KE = ½ (18.2 kg) (2744 m/s)2 = 68.52 MJ

Which compares much more favorably as a weapon system. Derived values can easily be obtain close to these numbers

We'll take the average range, 225 nmi, for the calculations. Of course we can't just convert 225 straight to meters, since a nautical mile is a bit over 15% longer than a standard mile (about 6076 feet). After converting to miles we can go to meters (or go straight from nmi to meters, if your calculator has a bunch of built-in conversion factors):

1nmi = 1.151mi

225nmi (1.151nmi / mi) = 258.975mi

1mi = 1.609km = 1609m

x = (258.975mi) (1609m / mi) = 416690.775m

Real Value: 416700 m

Dividing by the time (6 min / 360 sec):

Vx = 416700m / 360s = 1157.5 m/s

Which s a far more appropriate velocity for a kinetic kill weapon. However, this is only part of the velocity. The railgun fires in a parabolic arc, getting almost 95 miles up. Assuming the Earth is flat, and the projectile is launched and lands at the same height, this part of the velocity component is easy to calculate. In theory the projectile reaches its maximum height half way through the journey, or at 3 min - 180 s. We can put this into the gravity-displacement equation to determine the speed. A height of 95 miles (500,000 feet) is about 152400 m.

h = -4.9t2 + vtv = (h / t) + 4.9t

Vy = (152400m / 180s) + (4.9 m/s2)(180s) = 1728.67 m/s

Now we can combine the two velocity components to determine the actual velocity, by Pythagorean Theorem.

VT = √(1157.52 + 1728.672) = 2080.41 m/s

Which is much closer to the Mach 7 value that the Navy claims the projectile hits at. Using this value to calculate the kinetic energy:

KE = ½ (18.2 kg) (2080 m/s)2 ≈ 39 MJ

Thomas Rigby

Coil Guns

Coil guns, magnetic linear accelerator, or mass drivers are a series of donut shaped electromagnetic coils (Philip Eklund calls it a "centipede gun", in the Traveler role playing game they are called "gauss guns") A projectile composed of some ferromagnetic material is introduced into the first coil. The coil is energized so it repels the projectile and the next coil is energized so it attracts the projectile. When the projectile reaches the second coil, the second switches to repulsion and the third starts attracting, and so on. Advantages are a much lower power consumption than an equivalent rail gun. Disadvantages are the massive power switches required. Each individual coil needs bracing, as they are under tremendous force trying to expand the coil (actually for "expand" read "explode").

Note that one can use the kinetic energy equation above to see how much power the railgun or coilgun will require for each shot. Since these weapons are nowhere near 100% efficient, you will quickly discover that these weapons are power hogs.

When these weapons are armed they will be carrying plenty of electricity. If they are damaged by enemy weapons fire, there will probably be plenty of high-voltage fireworks, at least inside of the ship. I am unsure if there will be much arcing outside of the ship unless the ship is venting gas by accident (atmosphere through a hull breach) or design (open-cycle cooling gas).

Also note that as the guns get more powerful, the more recoil they will have. Indeed, they will approach being auxiliary propulsion systems. If such a gun was optimized as a propulsion system it is called a "mass driver".

To calculate parameters of your coilguns, Eric Henry has an Excel Spreadsheet. Or you can use Luke Campbell's method:

Here's a quick method to estimate what kind of performance you can get out of a coilgun. Some folks here might find it interesting.

First, decide on the efficiency of your coilgun. Coilguns are linear brushless electric motors, and brushless electric motors have demonstrated efficiencies of 90% to 95%. Superconductive electric motors might have efficiencies of 98% to 99%. Denote this as a decimal, and call it e; that is e = 0.9 to e = 0.95.

Next, decide on the length and radius of your projectile. Decide on what your projectile is made of and find its mass

mass = density * length * radius2 * &pi (and remember to use consistent units).

Also find the projectile cross-sectional area

area = radius2 * π

Decide how fast you want your projectile to be going and find its final kinetic energy

kinetic energy = 0.5 * mass * velocity2 (again remember to use consistent units).

Given the efficiency of your coilgun, you can find out how much your projectile heats up. You might figure that half of the wasted energy goes into the projectile, and thus your projectile will gain a heat energy of

heat energy = 0.5 * (1/e - 1) * (kinetic energy)

Look up the specific heat of the material your projectile is made of, commonly called C. Then your projectile reaches a temperature of

projectile temperature = (heat energy) / (C * mass) (again make sure your units are consistent).

If you are using a synchronous coilgun with a permanent magnet in the projectile, this temperature needs to be less than the Curie point or the projectile will become non-magnetic. If your coilgun projectile is made of superconductors and you are using Meissner effect repulsion, this temperature will need to be less than the critical temperature of the superconductor or your superconductor will become non-superconducting. If you are using an asynchronous coilgun which uses inductive forces on conductive loops, this temperature will need to be less than the melting temperature of your projectile. If the temperature is too high, you will either need to use a material that can handle higher temperatures, make the coilgun more efficient, or accept a lower velocity for the projectile.

Decide the maximum magnetic field your coilgun can handle. If you are using a synchronous coilgun with permanent magnets (probably in the projectile, with the field coils along the barrel) you are limited by a saturation field of around 0.2 to 2 tesla beyond which your efficiency falls off rapidly. If you are using superconductors, your field is limited by the critical field of the superconductor. For conventional BCS-type superconductors this limits you to fields of several tens of tesla or less, for high Tc superconductors you might be able to get to 100 to 200 tesla. If using an asynchronous coilgun that uses induction to launch normally conductive projectiles there is no obvious physical upper limit to the magnetic field strength, although high field strengths will require massive bracing to keep the barrel from exploding.

Now assume that the barrel is filled with field, and that the projectile sweeps the field out of the barrel, turning the field energy into kinetic energy (this is not actually how coilguns work, but it gives the physical upper limit based on energy conservation). The energy density is about 400 kJ/m3/T2 times the square of the magnetic field strength (398,098 J/m3/T2 to six significant figures). Call this value K

K = 400 kJ/m3/T2

You now know the volume needed in the barrel based on how much energy the projectile ends up with

volume = kinetic energy / (K * (magnetic field)2)

Since you know the cross-sectional area of the projectile and thus of the barrel, you know how long the barrel needs to be

length = volume / area

If the barrel is unacceptably long, you will either need to figure out how to get a stronger field in the barrel, make the projectile shorter (if you do the math, you can see that the barrel length will be a multiple of the projectile length for a given field, material, efficiency, and final velocity) or make due with a lower velocity of the projectile.

As an example, suppose we have a synchronous coilgun, and that the coilgun can generate 1 tesla fields (a good number that will not saturate the ferromagnet). Our presumed ferromagnet is probably mostly iron, with about 8000 kg/m3. To reach 100 km/s, you will need 40 TJ per cubic meter of projectile. Since this is 100 million times the energy density of the field, you will need the projectile to sweep out 100 million times its volume in order to accelerate up to the desired speed. This means you need an accelerating track 100 million times the length of your projectile. If the projectile is the size of a dime, with 1mm thickness, you will need a 100 km long track. If 2.5% of the energy goes into the projectile as heat as a result of inefficiencies, you get 100 GJ of heat per cubic meter of projectile, or 12 MJ/kg. This is three times the specific energy liberated by detonating high explosives, so you can expect your projectile to explode like a bomb inside your coilgun barrel. Consequently, this appears to be an unworkable design.

Luke Campbell

Effects

Ken Burnside notes how difficult it is to calculate the damage caused by a solid shell:

In terms of how ships survive taking damage, there is also the matter of rate of deposition to the target and area of deposition.

Basically, you're poking holes in a compartmentalized object. Unlike an aircraft, or a submarine, the outside environment isn't that hazardous. It doesn't take much damage to make a jet fighter unflyable at air combat speeds. Getting hit with a torpedo in a sub can cause the hull to collapse.

Hitting a spaceship won't cause it to pop like a balloon. There's likely a swath of compartments that are uninhabitable at this point...but the ship can still fight.

For example, an M1A2's main gun is about a 5" naval gun -- firing an armor piercing round, at a target that wouldn't quite actually be a full sized Naval compartment. Very rarely does it leave an exit wound in the back of an enemy tank, which is the indicator of what it would do to the NEXT compartment of a ship. It WILL destroy everything in that compartment, unless it's blunted by hitting an engine in the way (like the Merkava design of the IDF).

For point of reference, an M1A2's round has a velocity of about 1600-1700 m/s. Mass between 3.5 and 4 kg, diameter about 2.5 cm.

Quite simply, there isn't a lot known about the interaction dynamics of objects impacting at 1.5+ kips. One field says that they'll turn into a plasma spray (more or less what happens when a tank round hits a tank...), which limits their damage to the compartment hit. Another says they'll get a plasma sheathe and go through multiple compartments shedding a bit of energy (but far less than the total carried by the round) in each, and exit the back of the ship.

Either of these makes for a more interesting fight than "gee, one hit, one kill, no stealth."

Ken Burnside

Isaac Kuo is of the opinion that hypervelocity weapons will have limited penetration. He notes that a projectile has both kinetic energy and momentum. Momentum is what keeps the projectile moving in its direction of motion.

Now, if you look at the equations for kinetic energy and momentum, you will note that as the velocity rises the kinetic energy goes up much faster than momentum (1/2 velocity squared vs just plain velocity).

Ke = 0.5 * M * V2

p = M * V

So Mr. Kuo figures that the greater your ratio of kinetic energy to momentum, the more spherical the resulting explosion and the less penetration into the interior you will get. This means hypervelocity weapons can be stopped (for a while) by a Whipple shield (until it is shot full of holes). Whipple shields are set at some distance from the hull, if the spacing is larger than the radius of the explosion, the shield takes damage but the hull does not.

I'm still looking for more details on this, especially the mathematical relationship between the ratio and the explosion sphericality.

Missiles

Missiles are small drone spacecraft that chase enemy ships and attack them with their warheads. It can have its own propulsion unit, or be launched by a coilgun and just use small guidance jets. It can carry a single warhead, or be a "bus" carrying multiple warheads. Or multiple mini-missiles. Go to The Tough Guide to the Known Galaxy and read the entry "MISSILE".

One of the big advantages of missiles over directed energy weapons is that missiles do not generate huge amounts of waste heat on the firing ship. A missile can be pushed off with springs or cold gas. Once clear of the ship, the missile's propulsion system ignites. But then all the waste heat is the missile's problem, not the ships.

By the same token, the disadvantage is that missiles are expendables, unlike laser bolts (as Anthony Jackson puts it: "If you're willing to have expendables, you can also have expendable coolant."). When the missile magazine runs dry, the launcher will just make clicking noises. But a laser cannon can fire as long as it has electricity.

The second advantage of missiles over directed energy weapons is that (depending upon the warhead) most missiles are not subject to the inverse square law. Laser bolts grow weaker with distance but a nuclear warhead has the same strength no matter how far the missile travels. However, laser bolts cannot be neutralized by point defense.

The warhead is generally a nuclear weapon but others are possible. One possibility is a single-shot coilgun firing a kinetic weapon. Another type of warhead is an explosive charge coated with shrapnel, designed to deliver a cloud of kinetic kill masses into the path of the target spacecraft. A third type is the "submunition".

Of course the simplest is no warhead at all, making the structure of the missile an impromptu kinetic kill weapon. According to the first law of space combat, above about a three km/s relative velocity difference a chemical explosive warhead is superfluous. Rick Robinson says that at these speeds the only reason for conventional explosives is for the bursting charge on a shrapnel cloud.

Rick Robinson suggested that the term "torpedo" be used for a missile that has acceleration capacities comparable to a spacecraft, while the term "missile" or "torch missile" be used for those that have somewhat more acceleration than spacecraft. In GURPS: Transhuman Space they use the term "Autonomous Kill Vehicle" (AKV) instead of torpedo.

To be an effective weapon, missiles have to have acceleration abilities at least as good as the target ship. Rick Robinson says "Basically you have to make your ship drive, or something comparable to your ship drive, small enough and cheap enough for a one-shot weapon." Some drive technologies cannot be squeezed down since they have a minimum size.

Rick also notes that missiles have stupendous range. If your spacecraft can cross the solar system, so can your missiles.

Ken Burnside did the math and found that it is worse than Rick realized.

Missiles Will Always Hit

There is a temptation to make a game where torch missiles can be run out of propellant. The problem with this is that when you do the geometry of the shot, you assume two things:

  1. The target is pointed exactly away from the inbound missile bearing.
  2. The target is using its maximum thrust.

This is the worst case for the person launching the missile; you subtract the target's acceleration from the missile's acceleration, and build a reference frame where all the velocity is on the missile — this may result in the missile overcoming a velocity away from the target.

At that point, you calculate delta-v. Unless the target has some way to leave the battle, you do a simple calculation of delta-v over time overcoming the initial shot velocity; if the missile can overtake the target in a stern chase, you'll know before the missile gets launched.

Once I built this up for Attack Vector: Tactical, I did the math for the torch missiles Rick loved so dearly...and it gets very bad; because missiles can afford drop tanks more readily than spaceships.

In the real world, missiles also have sensors for autonomous homers, and those sensors have batteries — the batteries tend to be good for roughly twice the "expected" fuelled flight parameter for redundancy. I suspect powering onboard sensors for a torch missile may also be the real limit — sure you can make your fusion torch missile also self-power off of the fusion rocket, but that increases the cost.

Of course, you're in a society that throws away a hundred-kW fusion motor away as an expendable munition, so that cost may not be a factor at all.

There's a reason why Attack Vector: Tactical missiles ended up being a more advanced solid fuel rocket: Cost and ease of maintenance. You need to think about how your spacers — who if Air Force enlisted personnel are any indication — have high school or two-year degree equivalents are going to keep those missiles in launch readiness for multi-month cruises. Rocket propellants tend to have a shelf-life...

Ken Burnside

Drones

3D artist Scott Halls has made an amazing website illustrating technical information about Peter F. Hamilton's Night's Dawn trilogy. Above are the "Combat Wasps", which are a sort of armed drone. Left to right are the Kinetic Harpoon, Electronic Warfare, Fusion Torpedo, and Particle Beam Cannon Wasps. You can read all the details here.

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