F Atomic Rocket: Space War: Weapons

Space War: Weapons

Artwork by Russ Mannings for "Captain Johner and the Aliens", 1963
Based on "First Contact" by Murray Leinster, 1945

When it comes to weapons, it looks like three main types: beam weapons, kinetic weapons, and missiles. Beam weapons are lasers and particle beams. Kinetic weapons are coilguns, railguns, and shrapnel weapons. Missiles are, well, missiles. Ken Burnside compared it to a policeperson armed with a service revolver, a shotgun, and a police dog. The revolver (beam weapon) cannot be dodged or outrun, but can miss. The shotgun (kinetic weapon) is more likely to hit, but with reduced lethality. The dog (missile) can be dodged or outrun (or shot, that would correspond to point defense), but the blasted thing will chase you, and will always hit unless you actively prevent it.

(Holger Bjerre begs to differ. He points out that kinetic weapons are less likely to hit since it can be dodged, beam weapons lose lethality with range just like shotguns, and kinetic weapons do not lose lethality with range just like revolvers. Well, no analogy is perfect...)

Dave Bryant has his own analysis of spacecraft weaponry here. I'm not sure I agree with all of it, so do your own research.

Artwork by Malcolm Smith, Imagination Magazine, October 1953

From Space Dreadnoughts edited by David Drake.

One of the problems with figuring out how ships are going to fight in space (assuming that we have ships in space, which isn't as likely as I wish; and, that we're still fighting when we get there, which is unfortunately more probable) is that there are a lot of maritime models to choose from.

It's also true that some of the maritime models came from very specialized sets of circumstances; and a few of them weren't particularly good ideas even in their own time.

And it's also true that some of the writers applying the models have a better grasp of the essentials than others. For example, I recall two essays which were originally published about fifty years ago in Astounding.

In the first of the essays ("Space War", Astounding Science-Fiction, Aug 1939), Willy Ley, a very knowledgeable man who had been involved with the German rocket program, proved to my satisfaction that warships in space would carry guns, not missiles, because, over a certain small number of rounds, the weight of a gun and its ammunition was less than the weight of the same number of complete missiles. The essay was illustrated with graphs of pressure curves, and was based on the actual performance of nineteenth-century British rocket artillery ("the rockets' red glare" of Francis Scott Key).

As I say, the essay was perfectly convincing ... until I read the paired piece by Malcolm Jameson ("Space War Tactics", Astounding Science-Fiction, Nov 1939).

Jameson's qualifications were relatively meager. Before throat cancer force him to retire, he'd been a United States naval officer -- but he was a mustang, risen from the rank, rather than an officer with the benefit of an Annapolis education. For that matter, Jameson had been a submariner rather than a surface-ship sailor during much of his career. That was a dangerous specialty -- certainly as dangerous a career track as any in the peacetime navy -- but it had limited obvious bearing on war in vacuum.

Jameson's advantage was common sense. He pointed out (very gently) that at interplanetary velocities, a target would move something on the order of three miles between the time a gun was fired and the time the projectile reached the end of the barrel.

The rest of Jameson's essay discussed tactics for missile-launching spaceships -- which were possible, as the laws of physics proved gun-laying spaceships were not. Ley could have done that math just as easily. It simply hadn't occurred to him to ask the necessary questions.

(Ed note: Malcolm Jameson wrote another essay with the intriguing name "Space-War Strategy" in Super Science Novels Magazine, March 1941. )

From Manna by Lee Correy (G. Harry Stine) 1983:

"This is a training flight to trans-lunar space with landing at Dianaport. Request permission to pass within five kilometers of you."

"Training flight? Hah!" Omer exclaimed. "Chung has give us an escort!"

"Yes, but why?" I wanted to know. "What's going on dirtside that we should know about?"

Omer shrugged. "Let Chinese escort us. It will discourage more hassle."

If the Chinese cosmolorcha wanted to escort us, there was nothing we could do about it. It was armed. Cis-lunar space is no place to get whanged; it's a long time to anywhere.

"Permission granted, Heavenly Lighting," I replied. "Be advised you are within our zone of damage if we should have a catastrophic failure." The last was pure bluff, but nobody wanted to be near a space vehicle if it catoed, regardless whether it was due to an internal or external cause.

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.

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 paintjob and blind a few sensors. And in space a kilometer is pretty close range.

Eric Rozier has an on-line calculator for nuclear weapons.

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 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. There won't be any EMP, unless there is a Terra strength magnetic field and a tenuous atmosphere present.

Well, maybe a small EMP from x-rays interacting with the hull of the ship. Please, do NOT confuse EMP (electromagnetic Pulse) with EM (electromagnetic Radiation). An EMP can travel through airless space just fine, but it cannot be generated by a nuclear detonation where no atmosphere is present. Note that an EMP can be created in airless space by an e-Bomb, which uses chemical explosives and an armature.

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.

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.

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.

Boom Table

Note in the table below, there is some controversy over the exact values of some of these figures. Note also that the largest SI prefix is "yotta-" which is 1 x 1024. For TNT equivalent, the energy of one gram of TNT was arbitrarily standardized by scientists to exactly 4184 joules (1000 thermochemical calories).

Crewmembers 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 neutron flux is approximately:

Fn = 1.4e12 * (Y/R2)

where:
Fn = Neutron fluence (neutrons/cm2)
Y = weapon yield (kilotons)
R = range from ground zero (kilometers)

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

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.

Eric Henry has a spreadsheet that does nuclear blast calculations, including shaped charges, on his website..

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.

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, found online here. 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!

A scientifically accurate artist conception of a high intensity laser strike. Artwork by Luke Campbell using Bryce.

Laser Cannon

There is a great summary of the various issues of directed-energy weapons at this site. 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. 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:

Luke Campbell:

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.

Anthony Jackson:

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.

Luke Campbell:

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.

Anthony Jackson:

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

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

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.

H = Cm/(78.54 * A2 * (Dk/150,000)4)

 where:
H = maximum percent chance to hit target given light-speed lag(0.0 - 1.0)
Cm = target ship's mean cross section (m2, for a purely convex object this is 1/4 of the surface area)
A = target's acceleration (Gs)
Dk = range to target (km)

Please note that this equation does not work if the target's acceleration is zero (since dividing by zero is mathematically undefined). In that case the target's official status is Sitting Duck and H = 1.0 or 100%. Neither does the equation work if the range is zero, in which the target's official status is At Point Blank Range or Eating The Gun Muzzle, and again H = 1.0 (Thanks to Eric Henry for pointing this out). Just remember that H cannot go over 1.0 and you'll be fine.

How was this equation derived? Well, if H is chance to hit, A is acceleration in Gs, DL is range in light-seconds, and Cm is target's mean cross section:

Scircle = π * Rcircle2
where:
Scircle = surface area of a circle
Rcircle = radius of the circle
π = Pi = 3.14159...
Artwork by Jack Gaughan for "Triplanetary" (1934)
H = Cm / (π * displacement2)
where:
displacement = maximum distance perpendicular to line of fire that the target can move in time between a shot being fired and the shot arriving at target
In other words, take the cross section surface area of the target, divide it by the surface area of the circle the target can move to, and you have your maximum hit chance. e.g., if the target has a surface area of 1, and it can displace anywhere into a circle of surface area 3, then the maximum hit chance is 1/3.

d = 0.5 * a * t2
where:
d = distance (m)
a = acceleration (m/s2)
t = duration of acceleration (s)
which is the classic acceleration equation, assuming a starting velocity of zero. We can assume zero because all we care about is the change in the target's current velocity

Now, to use acceleration equation to calculate displacement:
a = 10 * A (10 is approximately how many m/s2 in one g)
t = 2 * DL (time it takes light from target to travel to laser cannon's targeting sensors plus time it takes for laser beam to travel from laser cannon to target) where DL = range to target (light-seconds)
d = 0.5 * a * t2
displacement = 0.5 * 10*A * 2*DL2
displacement = 5*A * 2*DL2

Inserting displacement equation into hit chance equation and simplifying:
H = Cm / (π * displacement2)
H = Cm / (π * (5*A * (2*DL)2)2)
H = Cm / (78.54 * A2 * (2*DL)4)
Converting range in light-seconds into range in kilometers:
H = Cm/(78.54 * A2 * (Dk/150,000)4)

Note this equation only calculates the percentage chance of missing due to light-speed lag. There are many other factors that can contribute to a miss.

Art by Frank R. Paul for Air Wonder Stories (1929)

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)
Band Wavelength (m)
Far Infrared 3e-5 to 1e-3 m (30,000 to 1,000,000 nanometers)
Mid Infrared 5e-6 to 3e-5 m (5000 to 30,000 nanometers)
Near Infrared 7e-7 to 5e-6 m (700 to 5000 nanometers)
Red 7.1e-7 m (710 nanometers)
Orange 6e-7 m (600 nanometers)
Yellow 5.7e-7 m (570 nanometers)
Green 5.5e-7 m (550 nanometers)
Blue 4.75e-7 m (475 nanometers)
Indigo 4.3e-7 m (430 nanometers)
Violet 3.8e-7 m (380 nanometers)
Ultraviolet A 3.2e-7 to 4e-7 m (320 to 400 nanometers)
Ultraviolet B 2.9e-7 to 3.2e-7 m (290 to 320 nanometers)
Ultraviolet C 2e-7 to 2.9e-7 m (200 to 290 nanometers)
Extreme Ultraviolet 1e-8 to 2e-7 m (10 to 200 nanometers)
X-Ray 1e-11 to 1e-8 m (0.01 to 10 nanometers)
Gamma-Ray 1e-14 to 1e-11 m (1e-5 to 0.01 nanometer)
Cosmic-Ray 1e-17 to 1e-14 m (1e-8 to 1e-5 nanometers)

Note that frequencies below 200 nanometers are absorbed by Terra's atmosphere (so they are sometimes called "Vacuum frequencies") and anything below 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.


Artwork by Jack Gaughan for "Galactic Patrol" (1937)

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.

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.

From the anime "GunBuster" (1988). In the first picture, look at it sideways. The two rails down the center are handrails, which end in a hatch. In the last picture, two cadets are working off demerits by the dirty job of window-washing the laser cannon lenses.

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.

Artwork by Ed Valigursky

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

Martin-Marietta design for laser anti-missile station. Note open-cycle cooling ports along sides venting red-hot gas.
Artwork by Kelly Freas for "Crown of Infinity" (1968)

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.

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.

This is the turret on the nose of the aircraft. The brown circle on the front of the olive drab sphere is the laser emitter. Two spoon-shaped holders grip the ball on either side. The situation is similar to you holding a golf ball between your thumb and middle finger.

Note how they tend to have the beam aimed sideways. This is prevent bugs and debris from damaging the emitter.

The olive sphere can pitch up and down, while the black spoon holder can roll around the plane's long axis. The pitch is limited so the turret cannot inadvertently shoot itself. The turret can cover a solid angle of almost 120 degrees of sky, targeting anything in that volume.

The key to preventing the laser beam from slicing the turret into pieces is to feed the beam into each rotating segment along the axis of rotation. So the olive ball has to be fed the laser beam along the pitch axis, and the black spoons fed along the roll axis.

images courtesy Lockheed Martin Missiles and Space

Here's a cutaway view. The lime green sphere is the olive ball, flanked by the two spoons. The laser generator is deep in the bowels of the plane. The beam is shot along the roll axis, entering from the left of the picture. It hits three mirrors, and enters the olive ball along the pitch axis. The dog-leg passage is technically called a "Coude path", from the French word for "elbow".

The beam bounces off the mirror on the far side of the pitch axis, hits a small angled mirror in the center, and then hits a small convex mirror (the Beam Expander) supported in the center of the emitter hole by three struts.

The beam expander spreads the beam over the giant 1.5 meter gold-coated primary mirror. The primary mirror sends the beam out the emitter opening (the Conformal Window), on its way to drill a hole through an unlucky enemy spacecraft. Note how the system of mirrors will work regardless of how the turret pitches or rolls.

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

Isaac Kuo has some interesting observations on the placement of 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.

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.

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.

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."

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). To calculate their beam divergence angle, 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) / w = 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.

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.

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. And while there ain't no such thing as an x-ray mirror, 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.

Luke Campbell:

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.

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 kilometer in diameter. So this X-ray laser would only be suitable for exceedingly huge warships, orbital fortresses, and Death Stars.

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 aim the mirror so it will do a bank shot from the distant laser to your target and tell the laser crew 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.

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...
Artwork by Timmins for Recoil by George O. Smith (1943)
Note the vertical and horizontal deflection coils

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). 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 monitor. 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. 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.]

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.

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.

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.

Triplanetary by E.E."Doc" Smith. The Nevian star ship has sliced Costigan's ship and is tractor beaming the segments aboard.

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.

Dorsal railgun, USS Sulaco, Aliens (1986) Image from Starship Modeler

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.

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.

Project Thor. Orbital bombardment kinetic energy weapon. Artwork by John MacNeill

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

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)
Two frames from the webcomic Nip and Tuck, written and drawn by rhjunior.

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...

Floyd and Sneaky the insouciant cat

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.

RAIL GUNS

A rail gun 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.

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.

Railgun projectile and conducting sabot casing

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.
Popular Science Magazine

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:

Thomas Rigby:

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 heig