Fun With: Rail/Coil Guns

[rhoenix]

Here's the basic scenario:

Given a railgun/coilgun hybrid concept, wherein both technologies are used to accellerate matter to high percentages of c, I'd like to hash out the different scenarios wherein one may or may not want to have any special ammo, or ammo somehow modified beyond "it's a rock that's sensitive to electromagnetism, so we stick it in the gun and it shoots."

This is where I'd like your input. I already have my own opinions about each, but I'd like to see this from your eyes. With that in mind, please also explain your answer.

Scenario 1: Space naval ship-mounted. Range is meant to be long.

Scenario 2: Infantry rifle-type. Given that the infantry carrying it will have hearing protection and sealed NBC armor.

Scenario 2: Infantry armor-mounted type. Given that the tank with this hybrid rail/coil-gun will likely used in siege tactics, as well as precision bombing/strikes.

[CaptainChewbacca]

I'm pretty sure any infantryman with a fractional-C weapon would be incinerated when the bullet hit the atmosphere. I may be overestimating rail capabilities, though.

[rhoenix]

Actually, as this was just pointed out to me in the chat (and subsequently seeing your post as I was about to clarify), so I'll amend Scenario 2 to the following:

REVISED Scenario 2 (to account for physics going "lol"): Infantry rifle-type. Given that the infantry carrying it will have hearing protection and sealed NBC armor, and the gun firing approximately 10% c on average.

[kinnison]

Infantry coilgun; if the projectile is doing 10% c then it has to be very small indeed regardless of the wearing of power armour - otherwise the user will be blown backwards.

One solution, used for plasma and fusion weapons in the old Traveller game, is gravitic support; i.e. part of the energy of the shot is used to spread the momentum exchange into the planetary gravity field instead of a conventional recoil. However, this weapon now needs three technologies, none of which we have yet and one of which we don't have any idea how to create; a coilgun that can fire repeatedly without vapourising the weapon itself, a compact high-density power supply (electrostatic-confinement fusion?) and gravity generators.

Even if you can manage all that, you still need some way of keeping the weapon barrel in vacuum, otherwise the projectile will evaporate before it leaves the barrel - and in any case it would evaporate when it hits air, if you use it in atmosphere.

A coilgun might well work though, but the projectile mass needs to go up and the velocity down.

Incidentally, if you can find a way of accelerating small masses to fractional-c velocities in a small, light unit - you have yourself a great reaction drive.

[Ford Prefect]

Scenario 1: Space naval ship-mounted. Range is meant to be long.

You better not get planets in their way. Even relatively small (about the size of your average living room) projectiles could do horrific things to large areas of a planet. See The Killing Star for details.

Scenario 2: Infantry armor-mounted type. Given that the tank with this hybrid rail/coil-gun will likely used in siege tactics, as well as precision bombing/strikes.

Moutains? What are those?

[Lord of the Abyss]

For nullifying recoil on the infantry version, there's a solution that doesn't require more than the tech required to build the supergun itself. Design the thing so it fires backwards, as well as forwards, with the backwards projectile being something that disintegrates rapidly, so you don't blow up something ten miles behind you. Just don't stand behind the thing when someone fires it. And you will want to be wearing that armor when you fire the thing.

Also, such a weapon would be more powerful the lower the atmospheric pressure, since they could survive firing it at greater velocities. This has the interesting effect of making ground forces on, say, the Moon much stronger than those on Earth, and those on Mars somewhere in-between.

Naval version : Make the slug out of metallic hydrogen or with a hydrogen core, so it undergoes fusion when it hits the target at near-c, or antimatter if you can make it. This should help prevent it just passing through the target without transfering much energy to it.

As for why you use the gun instead of a missle; assume really good point defense, that the gun's velocity and relative stealthiness ( no rocket flare ) can overcome.

[Patrick Degan]

Naval version : Make the slug out of metallic hydrogen or with a hydrogen core, so it undergoes fusion when it hits the target at near-c, or antimatter if you can make it. This should help prevent it just passing through the target without transfering much energy to it.

The problem is that metallic hydrogen would have to be kept under supercold and supercompressed conditions. A slug with a hydrogen core would simply pass through the target vessel without initiating —as would the metallic hydrogen slug. Shooting an enemy ship full of holes, however, would certainly render it useless in the course of a battle. A hit on a critical frame juncture could cause structural weakness in that spot and the ship's own motion works thereafter to tear it apart.

[rhoenix]

Well, one idea I had for the naval-mounted version would be to give each shell small thrusters to auto-correct for range, and contain equal amounts of hydrogen and anti-hydrogen separated by a force field. When the shell hits, the thin force field between the two drops, allowing the two to interact with impressive results.

[Starglider]

contain equal amounts of hydrogen and anti-hydrogen separated by a force field. When the shell hits, the thin force field between the two drops, allowing the two to interact with impressive results.

The anti-hydrogen is already slamming into the target at c-fractional velocities. I can't see how having some normal hydrogen in close proximity is going to make any difference to the result. The positrons and anti-protons will annihilate with protons and electrons in the target or the shell casing just fine.

[rhoenix]

The anti-hydrogen is already slamming into the target at c-fractional velocities. I can't see how having some normal hydrogen in close proximity is going to make any difference to the result. The positrons and anti-protons will annihilate with protons and electrons in the target or the shell casing just fine.

I see your point, though my intent was to control the matter-anti-matter reaction a bit better.

[Starglider]

I see your point, though my intent was to control the matter-anti-matter reaction a bit better.

This isn't a particle accelerator. You've got a relatively tiny shell, which has to be robust enough to withstand ridiculous accelerations and cheap enough to mass produce, slamming into a target and flashing into plasma within nanoseconds. There doesn't seem to be much scope or need for fine control to me.

[Admiral Valdemar]

Reminds me of certain high-end, rather dangerous weapons in the Reynolds novels that fire anti-proton streams from an anti-lithium power cell that have the nice kinetic damage along with the "Oh shit that beam is reacting with my ship!" effect you get with AM.

A normal particle beam is nice. For supersized bang, add 30p and go AM.

[rhoenix]

Reminds me of certain high-end, rather dangerous weapons in the Reynolds novels that fire anti-proton streams from an anti-lithium power cell that have the nice kinetic damage along with the "Oh shit that beam is reacting with my ship!" effect you get with AM.

A normal particle beam is nice. For supersized bang, add 30p and go AM.

...Ouch. That does sound like truly good and dirty fun.

The two types of shells I toyed with for the naval-mounted version were only two:

1 - The one I described earlier, which explodes with a big matter / anti-matter boom when it hits something.

2 - One I designated a "Shadowstrike" shell, which 10 kilometers from the target, forcefully causes a matter / anti-matter explosion and with the use of force fields as a last-ditch effort by the shells' fuel, compresses the explosion into a beam about two microns in diameter toward the target. Much fun for precise and surgical strikes from long range.

I wouldn't really call either type of shell "sneaky."

[Starglider]

1 - The one I described earlier, which explodes with a big matter / anti-matter boom when it hits something.

You will not get a conventional explosion when you hit something at a good fraction of c. The plasma that used to be the shell and the armour it hit will exit the back of the target still travelling at a good fraction of c. There will be some blast, but most of the energy will be wasted in overpenetration, and most of the plasma will have a very limited time to couple energy to the rest of the target. The main advantage of sticking some antimatter in the shell is that annihilation produces large quantities of energetic gamma rays which shoot off in every direction, transferring much more energy into the rest of the target than the purely kinetic effects will.

One I designated a "Shadowstrike" shell, which 10 kilometers from the target, forcefully causes a matter / anti-matter explosion and with the use of force fields as a last-ditch effort by the shells' fuel, compresses the explosion into a beam about two microns in diameter toward the target. Much fun for precise and surgical strikes from long range.

That sounds like an even stupider version of this weapon, with all the same problems of not doing much useful to the target and being a wasteful use of technology, but on top of that being plasma based (and hence inheriting all the stupidities of a 10 km range plasma weapon, since you'll need some sort of magic self-propagating forcefield to keep the 'beam' diameter down to two microns 10km downrange of the initiation) and requiring all this magical technology to be packed into a tiny mass-produceable shell that can survive millions of g of acceleration. On top of the usual volatility problems of antimatter weapons. Plus you'll need a trigger that can detonate a precise number of microseconds away from the target (why 10km anyway? what kind of defence is going to stop a c-fractional shell getting any closer?).

In summary it's a bad idea.

[Ford Prefect]

Reminds me of certain high-end, rather dangerous weapons in the Reynolds novels that fire anti-proton streams from an anti-lithium power cell that have the nice kinetic damage along with the "Oh shit that beam is reacting with my ship!" effect you get with AM.

A normal particle beam is nice. For supersized bang, add 30p and go AM.

Awww, but 'positron cannon' sounds cooler than 'anti-proton beam'.

[The Dark]

Just to use the rifle as an example, since you explicitly stated 0.1c for velocity...

The speed of light is approximately 300,000 kilometers per second (slightly lower, but the rounded number is easier to work with). This means a muzzle velocity of 30,000 kps. At 3 kilometers per second, a projectile's kinetic energy is roughly equivalent to its mass in TNT's explosive energy. This is traveling 10,000 times faster. A 1 gram bullet would be equivalent to 10 kilograms of TNT. A fairly typical bullet might mass 5 grams, or 50 kilos of TNT equivalent per shot. We don't really have enough experience with super-high velocity projectiles of this size interacting with physical bodies to know for sure, but this would either drill a hole through its target and the targets in line of flight behind it for a couple dozen miles, or it would create a very large, very messy explosion somewhere a few hundred meters behind the general vicinity of whatever it hit.

The entire point of a railgun is its kinetic energy is obscene. Get fancy with it, and you've missed the point. That's what lower-velocity projectiles are for.

[Starglider]

At 3 kilometers per second, a projectile's kinetic energy is roughly equivalent to its mass in TNT's explosive energy. This is traveling 10,000 times faster. A 1 gram bullet would be equivalent to 10 kilograms of TNT.

No, kinetic energy is 0.5 * m * (v^2). A 10,000-fold increase in velocity implies a 100,000,000-fold increase in kinetic energy. A 1 gram bullet at 0.1c would have KE equivalent to 100 tonnes of TNT, but would have trouble coupling that energy to non-enormous targets.

[kinnison]

At 3 kilometers per second, a projectile's kinetic energy is roughly equivalent to its mass in TNT's explosive energy. This is traveling 10,000 times faster. A 1 gram bullet would be equivalent to 10 kilograms of TNT.

No, kinetic energy is 0.5 * m * (v^2). A 10,000-fold increase in velocity implies a 100,000,000-fold increase in kinetic energy. A 1 gram bullet at 0.1c would have KE equivalent to 100 tonnes of TNT, but would have trouble coupling that energy to non-enormous targets.

Good point. There is also the point that total momentum of the system (soldier + bullet) remains zero, and momentum is a vector. Assuming the soldier weighs 1000kg (hey, powersuit OK?) this means he flies backwards at 1E-7 c, which is 30 m/s or about 60 mph. The acceleration? Well, the bullet has accelerated to 0.1c or approximately 30,000 km/sec inside a barrel maybe a metre long. The average velocity down the barrel is 15,000 km/sec, so the acceleration takes 1/15,000,000 seconds.

The firer is therefore accelerated to 30 m/s in about 1/15,000,000 sec. The acceleration of the firer is 450,000,000 m/s/s or approximately 45 million gravities. Splat. Or alternatively, the rifle (10kg maybe) hits the firer's shoulder at 6000mph. Some recoil.

The problem of constructing a gun that will stay in one piece when subjected to four and a half billion G is left as an exercise for the reader.

[rhoenix]

Very interesting - a poster here was kind enough to send me a PM with calcs on it for my own use, but the discussion you two are bringing up is corresponding roughly to where I think this has to go for this gun to work and not kill the shooter with.

I'll revise Scenario 2 even further:

Instead of 0.1c, let's have a much more conservative Mach 18, and the gun design altered slightly to that of a coilgun design, for far less wear & tear.

With this alteration, I could perhaps see a few different types of ammo possible.

If this revision is satisfactory physics-wise, what about the other scenarios posited?

[Hotfoot]

Good point. There is also the point that total momentum of the system (soldier + bullet) remains zero, and momentum is a vector. Assuming the soldier weighs 1000kg (hey, powersuit OK?) this means he flies backwards at 1E-7 c, which is 30 m/s or about 60 mph. The acceleration? Well, the bullet has accelerated to 0.1c or approximately 30,000 km/sec inside a barrel maybe a metre long. The average velocity down the barrel is 15,000 km/sec, so the acceleration takes 1/15,000,000 seconds.

The firer is therefore accelerated to 30 m/s in about 1/15,000,000 sec. The acceleration of the firer is 450,000,000 m/s/s or approximately 45 million gravities. Splat. Or alternatively, the rifle (10kg maybe) hits the firer's shoulder at 6000mph. Some recoil.

The problem of constructing a gun that will stay in one piece when subjected to four and a half billion G is left as an exercise for the reader.

This is actually incorrect. The amount of time the subject undergoes the force of an acceleration, the LESS acceleration he undergoes. If I apply an acceleration of 40m/s^2 to you for 0.1 seconds, you would undergo less acceleration (and have a lower final velocity) than if I applied said acceleration for a full one second.

What you're thinking of is the equation by which you find the amount of acceleration needed to get something from an initial velocity of zero to a final velocity of whatever the case may be and the subsequent equation to see how long said acceleration would take. Here is the correct method for determining the force of acceleration (and final velocity) of the force of recoil on a human being.

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v(f)^2 = v(i)^2 + 2a(x-x(0))

This tells us the acceleration needed. v(f) is the muzzle velocity of the projectile, v(i) is 0, and x-x(0) is the distance covered during the course of the acceleration. We solve for acceleration (a), and then we can move on to the following equation

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t = (v(f)-v(i))/a

This tells us how long the total time of acceleration took, and thus how long it takes for the round to leave the gun, and how long the acceleration from both the bullet and the recoil is. We then move to a very simple equation.

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F = ma

This tells us how much force is needed to accelerate the mass of the round to the speed required. Remember, this doesn't care about time, because the rate of acceleration from the first equation already factors in the distance, and thus the time.

From here, we change the variables. Now we have the force generated by the firing, so we can now apply that force to your bog-standard human. Instead of using the mass of the bullet, we use the mass of the human (with and/or without powered armor) and solve for acceleration.

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a = F/m

And now we work backwards. With the acceleration and the time, we now solve for the final velocity of the subject.

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v(f) = v(i) + at

Now, from here, you can go into the height of the soldier and rotational models to see if he'd land on his ass or go flying back, but you should be able to get a decent idea of if this would or would not be lethal.

[rhoenix]

Just for the record, your mathy wisdom is both awesome and appreciated, Hotfoot.

I see how relativistic speeds for infantry is...wankish and requires too many physics dodges to work. From space-borne navy, on the other hand, it's just a much more scientific way of throwing a rock through someone's window.

[kinnison]

Hotfoot:

There are simple and complex ways to calculate most things, and force calculations are not necessary in this case at all. It's a bit like the way that some physics questions can be solved by calculus or by simple conservation-of-energy arguments. In this case, it's conservation of momentum - the total system's momentum starts at zero and stays that way.

OK. Assume a metre-long barrel and a projectile that at the end of the barrel is travelling at 0.1c or 30,000 km/s (3E7 m/s). Assuming a constant acceleration, the average velocity is half the final velocity. This can be derived from the equation s=(at^2)/2.

The average velocity of the projectile while in the barrel is thus 1.5E7 m/s, and therefore the time in the barrel is the reciprocal of that in seconds. This is also obviously the time available for accelerating the rest of the system - this could be the gun itself or the gun plus firer. And the total momentum is the same, but with a negative sign. Let's use the gun as the other part of the system, and assume that the weapon is weakly coupled to the firer. A reasonable figure for the weight of the gun is 10kg - let's use that. As momentum equals velocity times mass, and the weapon weighs 1E4 times as much as the projectile, its velocity at the end of the acceleration time will be 1E4 times less or 3 km/s. If you assume that all the momentum of the gun will be absorbed by the firer then the firer's eventual velocity will be 30 m/s, as he masses 100 times more than the weapon.

Regarding your post again; it is true that a constant acceleration applied for a shorter time will result in a lower final velocity (not acceleration - the acceleration is the rate of change). However, in this case we know two variables, final velocity and distance available in which to produce it; we are solving for acceleration and total time, which are not independent.


Either the gun goes from zero to 3km/s in about 60 ns (4 billion g or so), or the firer is subjected to approximately 40 million g of acceleration. Unless his suit has inertial compensation, I respectfully suggest that he won't be good for much afterward.

Force calculations are not required; however, the average force can be simply derived. 4 billion G is 4E10 m/s/s. f=ma, therefore the force on the gun (and on the bullet) is 4E11 N, or in older units forty million tons.

Your method of sorting this all out will work, but IMHO is unnecessarily complicated. Excuse me if there are some minor errors; I'm doing this in my head and it's 1:15 AM here.

Incidentally, we can do a power calculation as well. The kinetic energy E is equal to mv^2, or 0.001 x (3E7)^2 for a 1 gram bullet, in SI units. This equals 9E9J, and it is all put into the round in 1/15,000,000 sec - which means that the power is 9E11 x 1.5E7 or 1.35E19 W. Or to put it another way, 13 million terawatts. I hope the weapon has good capacitors!

[Hotfoot]

The problem is that force is necessary here, because the force applied to the round is equal to the force applied to the person firing the gun. Newton's third law, remember? Moreover, you have not provided the math for figuring out how long the acceleration would take. I have provided the equation.

Here is the math involved

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v(f)^2 = v(i)^2 + 2a(x-x(0))
30,000ms^2= 0^2 + 2a(1-0)
900,000,000 = 2a
450,000,000 = a
acceleration = 450,000,000 m/s^2

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F = MA
F = 0.001*450000000
F = 450,000

THAT is the amount of force applied to the shooter and the projectile. Now, instead of a one gram projectile, let's look at a 100kg human.

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F/M = A
A = 450000/100
A = 4,500m/s^2

That is the force of acceleration applied to the human. Now we take into account the amount of TIME that acceleration is applied over.

How much time is that? Well, we follow the equation I've given above:

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t = (v(f)-v(i))/a
t = 30,000/450,000,000
t = 0.000067 seconds

Now we figure out the velocity this will accelerate a 100kg human to.

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v(f) = v(i) + at
v(f) = 0 + 4500*0.000067
v(f) = 0.30015 m/s

This is why you do ALL the steps in an equation. You DO NOT KNOW final velocity of the human firing the projectile, you only know the final velocity of the projectile, from there you can figure out the force applied to the person, the acceleration that the individual undergoes, and how long that acceleration is applied for.

Now, the recoil for this IS a bitch, but if you're working with a metric ton of powered armor, then it becomes much less of an issue than if you're dealing with your average grunt.

[Kuroneko]

A projectile with a significant fraction of c in the atmosphere simply won't work. At those speeds, any appreciable atmosphere might as well be an impenetrable barrier--in the amount of mass intercepted per unit time (initially), a 0.007c projectile at sea level is comparable to a slightly supersonic projectile hitting an iron wall, all else being equal. Obviously, the specifics of those two interactions will be different, but at that scale, it's still clear that we won't have a usable weapon.

I'm not sure where your infantry is going, but anywhere but hard vacuum is out, and that's ignoring all the other issues.

[Kuroneko]

If we have a large d = 1.0m barrel, a m = 1g projectile at v = 0.10c has a kinetic energy of about E = 4.5e11J, meaning the average force would be F = E/d = 4.5e11N, which is about 46 million gees. On the other hand, the momentum of the bullet is p = 3.0e4Ns, so a 100kg armored human would have a final velocity p/100g, a bit over 300m/s--pushing around Mach 1 in Earth atmosphere--which means that Mr. Kinnison was off by a factor of 10. The time for the acceleration, although not directly relevant here, is t = vm/F = 66ns.

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v(f)^2 = v(i)^2 + 2a(x-x(0))
30,000ms^2= 0^2 + 2a(1-0)
900,000,000 = 2a
450,000,000 = a
acceleration = 450,000,000 m/s^2

Er, yes, but v_f = 3.0e7 m/s, since the scenario had 0.1c; this gives acceleration a = 4.5e14 m/s². You have v_f = 0.0001c instead. I think you misread 30,000 km/s to be 30,000 m/s. Although frankly, 0.0001c is more than enough for anyone's needs.