Railgun recoil: Newton won’t be denied! Pt1

I talked in my last post about the creep of technobabble into TV and movie Sci-Fi, and why I wanted more believability in my written fiction.

Z0053043Back in the early 80s, I used to design ships as part of the game, Traveller. And then I designed how the elements of a multi-ship force would work together for the game expansion called Trillion Credit Squadron (anyone still play that?) Typical ship armaments included meson cannons, pulse lasers and railguns.

Science fiction is fast becoming science fact. It’s a cliché, I know, but it’s happening in my lifetime. One of those exotic theoretical space weapons, railguns, looks likely become the main medium-range ship weapon for the US Navy over the next 20 years. They are already planning how to retrofit existing ships with enough electrical power in readiness.

Armed with a whole load of real technical data on railguns, they were a pick for my weapons in the Human Legion universe. (A special thanks goes out to all those dads who filmed their children’s railgun science projects).

So, what is a railgun?

500px-Railgun-1.svgTo make one, you take two parallel rails made from an electrical conductor (such as copper) and wire them up to a direct current supply. Place another conductor to touch both rails. This armature, as it’s called, must be able to move along the rails. When you supply the electrical power, a magnetic field is induced in the rails which pushes the armature along until it falls off the end of the rails, which breaks the electrical circuit.

In a school science project, the rails might be strips of aluminum foil taped to a board and the armature a steel bar that gently rolls along the board.

Real ones look might use a conducting sabot for the armature. A sabot is a jacket that wraps around a shell or bullet, enabling it to be fired more effectively. Search the ground after one of the fight scenes in my Human Legion books and you will be wading through spent sabot casings.

ASIDE: Coilguns and gauss guns. I’m talking here about railguns, because that’s what I chose as the equivalent of a modern rifle in the Human Legion books. But in another series I write, the Four Horsemen Universe from Seventh Seal Press, I’ve also written about coilguns and MAC cannons (Magnetic Accelerator Cannons).

Magnetic accelerator rifles, Gauss guns, and coilguns are all names for the same thing: a weapon where a sequence of coils around a barrel are switched on and off in a precisely timed sequence, accelerating a projectile. The recoil properties are essentially the same as railguns.

Anyway, back to Traveller and railguns. When I was designing heavy cruisers after school back in the 80s I used to imagine that when a railgun fired you would hear a hum of power build up followed by a whoosh. No bang. I expected it was recoilless too. Maybe that was because there was no explosive charge. With no exploding gases required to push the projectile along a barrel, there would be no recoil pushing back on the gun breech. Right?

Turns out I was wrong on all counts. It seems obvious to me now, but I knew a lot less physics when I was 13, and my misconceptions have stuck with me,

If you watch videos of real railgun test firings [such as this US Navy video below] there is a big bang when it fires. Lots of sound and lots of light. That’s what happens when you suddenly discharge a huge amount of power in an enclosed space, but it is not a chemical explosion as with conventional munitions. You don’t have all those hot, expanding gases pushing back against the breech. So does a railgun have less recoil than a conventional gun?

The answer is that the projectile has exactly the same recoil. In a conventional modern day rifle, this is equivalent to the recoil from the bullet leaving the barrel, and it’s called primary recoil.

But that’s not the whole story, because in a conventional firearm, there’s also a secondary recoil, which overwhelmingly consists of the hot gasses from the propellant escaping from the end of the barrel. In a rifle, the secondary recoil is usually greater than the primary.

You only have to see the muzzle fireball in the video above to see that the railguns also have secondary recoil, even though there’s no chemical propellant. However, the hot gasses in the railgun are, I presume, largely a result of the projectile compressing the air in front of it as it passes along the barrel, like a space capsule on re-entry. It’s a side effect rather than a cause of the shot and so although I don’t have the numbers, I make an assumption that the secondary recoil of a railgun is smaller than the primary recoil, possibly much smaller.

And as commenter, Matt, rightly points out, if the railgun is a spaceborne weapon and the inside of the barrel is a vacuum, then there are no hot, compressed gasses to expel out the tube.(*1) Secondary recoil would be significantly reduced. You might have some secondary recoil from expelled armature sabot sleeves (but only if the armature isn’t integrated with the projectile).

So certainly you have  recoil with a rail gun, but if my assumption that the secondary recoil is much lower than primary recoil (relative to a conventional rifle) then the total recoil will be less than with conventional weapons for a projectile with the same mass and same muzzle velocity.

I emphasized that last point because the recoil of the weapon ultimately depends on choices made by its designer. In the Human Legion Universe, soldiers are usually heavily armored. To penetrate, you might fire heavier projectiles, or fire projectiles similar to modern bullets but at a higher velocity. Or you might prefer to take the railgun’s ‘saving’ in recoil and use it for a higher or more accurate fire rate with some kind of armored piercing shaped charge rounds.

One of the features of the SA-71 personal railgun often seen in the Human Legion books is that it uses a great number of fire mode options and a flexible variety of rounds in their cavernous ammo bulbs. It’s the primary weapon for all theaters of combat, including shipboard and deep space operations, and it needs to be effective against pretty much anything it can come up against.

Having said all that, it still feels intuitively odd that there’s not explosive propellant in a railgun and yet it still has significant recoil kicking back against the breech.

It’s all to do with the law of conservation of momentum. It’s the same law that makes rockets fly into space. Also… It’s a basic law of the universe that you can’t get around by waving a technobabble phrase.

So in going for believability, conservation of momentum is something I can’t ignore.

I’ll cover more about railgun recoil in part2.


*1 The issue of spaceborne railguns is an interesting one. Current railgun testing involves three most common designs for armature. The simplest is where the projectile is made from a conducting material and acts as its own armature. Others use plasma (ionized gas), which allows for a non-conducting projectile, or a hybrid which uses plasma but not exclusively. Using high pressure plasma inside a depressurized gun tube will present additional challenges. Perhaps spaceborne and land-based railguns would develop along different paths for this reason.


  1. […] Railgun recoil: Newton won’t be denied! Pt1 […]

  2. […] The problem with railguns in science fiction is that they are fast becoming science fact. So I thought I’d better get my facts right. I posted part 1 of my recoil and railguns series today. You can read it here. […]

  3. Jean Caroline says:

    Tim, very interesting indeed… thanks. Jean C.

    • Matt says:

      If a rail gun is fired in a vacuum, wouldn’t that stop the secondary recoil?

      • tctaylor says:

        Thanks, Matt. That’s a very good point. I added something to the post on this. Yes, you’re absolutely right that if the gun tube is fully depressurized, there are no gasses to compress and push out the muzzle. But it got me thinking about the armature because that could be a problem in a depressurized tube if the armature is a plasma. In the book series I was planning when I wrote this post, I used a conducting armature sabot that acted as a sleeve to a wide range of largely non-conducting projectiles. I digress, though. Whatever the armature design, secondary recoil will be much less in vacuum and probably negligible if not zero.

  4. Hyperbion says:

    I too am trying to write accurate and believable science fiction. The biggest problems for me to face are the enormous distances in space and the fact that AI are projected to get vastly more intelligent that humans in the future (and thus, remove any need for human protagonists).

    But I got a message from a reader today, saying that he enjoyed my little novel so far, but was annoyed that my starships’ railgun had recoil. “Real railguns,” he said, “do not.”

    I now know that I am right.

    • LOL! Glad to be of service. All you can do is grin and bear it as part of being a writer, and consider that at least you’re engaging your readers. That’s always better than not being read.

  5. robert barber says:

    Actually, any opposing momentum would be related to the mass and speed of the object propelled. An opposing thrust provided by say an ion engine would enable the. rail gun to stay in a predicted orbit. As the mass and speed of the projectile would be predetermined, very reliable and preset counter-thrust could be used for stabilization. In addition, the use of only fissile material projected at mach7 could possibly be enough to detonate it without a trigger.

    • Absolutely. I’m sure that keeping a stable firing platform in orbit for kinetic projectiles requires some kind of counter-thrust as you describe. The battlesuits in the Human Legion stories do this for handheld weapons, but we’ve not yet seen how orbital platforms do this. Thanks for commenting 🙂

    • SGT MIKE says:

      For us grunts, it is just about the weapon going BOOM when we point the bang stick in the right direction!! 😛

  6. I’m playing with my new Savage Worlds Rifts RPG books, and was mulling the question myself about railgun recoil. I googled it, expecting to find a bunch of nonsense Rifts fan forums, but thankfully found this page instead. I played a lot of Traveller back in the day, but never “Trillion Credit Squadron” 🙁

    I’ll have to look up “Human Legion,” too.

    • tctaylor says:

      Delighted to help, Patric. After writing that post a few years’ back, my check for any new tech I dream up is whether it obeys conservation of momentum. If I can convince myself that it does, then it’s fit to write. I’ve never played Savage Worlds, but I’ve heard good things. Looks at a cursory glance like it’s core is GURPS with cool dice. Is that fair?

  7. Bob says:

    Thanks for the info! I was curious because I used to play RIFTS, and the Glitter Boy armor had a shoulder mounted railgun and a stabilizer spike that would deploy form the heels into the ground automatically when it was fired to counter the recoil. I wasn’t sure if those were necessary but apparently they were!

    • tctaylor says:

      Stabilizer spike — sounds like a cool idea. I’ve just got back from a writers’ conference last weekend and the topic of railgun recoil came up at the bar. It seems everyone needs railguns 🙂

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  10. Gunner says:

    This is completely incorrect. Although yes a rail gun will have recoil of a force equal to that of what is applied to the projectile, the recoil would be very mild compared to a traditional gun simply because the majority of the recoil generated from a tradional gun is created from gasses escaping the barrel, not the opposing force of the projectile. Most of this takes place after the projectile has left the barrel. Recoil yes, same as regular gun no.

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  13. Master Zed says:

    The thing about rail guns vs conventional guns is not just newton’s second law (equal and opposite reactions) but also HOW that law is applied. While total force is conserved (ie, if the total kinetic energy of the projectile is X kJ, then the launch platform must also experience X kJ of force (so less massive projectile with greater velocity = more massive projectile with proportionally more velocity). However, the manner in which the projectile is accelerated (and therefore it’s reciprocal force) is vastly different, and so too then is the nature of the recoil. Where a conventional bullet is accelerated by a small explosion that delivers all it’s energy in a very short amount of time, the rail gun actually accelerates its projectile along the length of the barrel (or at least the part containing the rail, which I think is fair to assume is most of the barrel length at least). Further, the rail gun’s acceleration is constant along the length of the rail (the velocity is increasing, but the force is generally uniform). This may SEEM a trivial difference, but it isn’t since what we humans experience is more often PRESSURE than TOTAL ENERGY. For example, falling from 6 feet off the ground results in identical kinetic energy being generated at the point of impact. So why then does hitting concrete hurt more than hitting water? Simple, the water decelerates you slower and therefore causes a less “jarring” impact. Rail guns have a similar relationship to conventional chemical propellants. So yes, while delivering a given amount of impact energy at the destination requires a similar amount of energy to be absorbed by the launcher, it doesn’t require the same PRESSURE (or energy over time). And in the same way falling 6 feet into a pool is MUCH better than falling 6 feet onto concrete, I believe a properly designed rail gun would have similar benefits to its recoil.

    • Master Zed says:

      It’s not so much that it’s recoilless, as it is that the recoil is more even and constant and therefore presumably easier to control and manage.

      Also, *less massive projectile with greater velocity = more massive projectile with proportionally LESS velocity — sorry about that

    • tctaylor says:

      I think you’re right. A conventional rifle will get most of it’s kick from the secondary recoil impulse which starts hitting when the hot gas vents out the muzzle. I don’t have definitive timing figures for this other than what I’ve found searching around, which suggests that kicks in something like 1-2 milliseconds after the cartridge ignites. The railgun’s recoil may be more spread out (as you say, like pressing against a wall).

      When I wrote this post (five years ago now!) I kept it simple and have been adding to it over the years. One of the simplifications I’m coming to regret is to talk about momentum. In fact, I think the most important part of the experience of recoil is not so much momentum, but impulse (change of momentum over time). It’s many years since I was integrating all this stuff at school, and my maths is no longer is fully up to it, but I can see that a (relatively) long, steady push will have a lower peak recoil impulse than the sharp kick of the gas venting out the barrel. On the other hand, later on, you make the point that with bullets, it’s the kinetic energy that does that damage rather than the momentum (or impulse), and so if we can increase the velocity of the bullet (or dart in the case of my fictional railguns) then we can reduce the weight of the projectile, its recoil when fired, and fit more into our ammo bulb. It’s science fiction and we’re dealing with civilizations that have worked on these problems for hundreds of thousands of years. So let’s make the muzzle velocity ‘really’ fast.

      However, I think the two ideas are working against each other, particularly with the SA-71 railgun and similar hand-held ‘rifles’. A steady acceleration along the barrel without the explosive gases will mean lower recoil impulse relative to a conventional rifle. Conversely, the faster we’re accelerating the projectile, the shorter the time the projectile takes before exiting the barrel, which increases the recoil impulse.

      I guess there ain’t no such thing as a free lunch 🙂

      1. Weight of projectile. 2. Time taken to accelerate out the barrel. 3. Muzzle velocity. In a fictional setting, I can’t make up the laws of physics (although I cheat them, as you point out) but I can set all three of those variables as I choose. I’ve never put numbers together, but my design had them very light (at least for standard rounds), very short and very fast. Where does that leave the amount of recoil? When I wrote the first bit of Marine Cadet seven years ago, I said it was like blowing bubbles with the recoil dampener active, but when the dampener overloaded then it was a beast to fire. I think that all still sounds plausible, but now I understand it better. Thanks.

  14. Master Zed says:

    Sorry, but all the above was to say that firing a well timed continuous fire rail gun is more akin to pushing against a wall than the “kick” that chemical projectiles create. The force is even and constant, and once braced against would have little to no real effect on the one firing the weapon.

    Further, that reciprocal force is actually being applied to the coils themselves (they’re the drivers) which could be mounted in such a way that they gather and store that reciprocal force and utilize it (to make more electricity for instance – we’ve been turning kinetic energy into electricity for quite some time already…). So there is no law stating that the rail gun needs impart any of that energy to the platform holding/mounting that rail gun – it’s just that our current recoil absorption technology isn’t up to speed with our current ballistic technologies.

  15. Master Zed says:

    Another thought I just had (sorry for the multiple posts) is that there is no reason to assume rail gun projectiles would need anywhere near the kind of kinetic energy that chemical propelled projectiles would. Even modern bullets don’t deliver much of their kinetic energy (unless stopped by a Kevlar vest or the like) or do damage based on their mass and velocity (related to it, but only loosely). They mostly do damage by puncturing vital organs/structures in the body (which again is a function of pressure not total kinetic energy). The bullet’s mass (as I understand it) is largely to help it survive the firing charge and fly more accurately (because greater mass means more gyroscopic stability – so long range rifles use larger calibers). A rail gun then, which can accelerate a mass to much higher velocities with much greater control, and does so by applying an even force along the length of the bullet, may not need to generate as much kinetic force in the projectile to deliver accurate and lethal strikes downrange.

    • tctaylor says:

      Thanks for your insightful comments, Master Zed. I’ve just returned from vacation, so reading your thoughts feels like a little welcome home present to get me back into full science fiction mode. I’ll properly digest and respond during the week. Thank you. Tim.

  16. Master Zed says:

    Looking forward to your replies. A couple of follow up ideas that I’ve had while designing the rail guns for my setting:

    1) The rails could easily be hooked up to “spring loaded” flywheels (or their high tech equivalents) to turn the linear reciprocal force into rotational energy which could then be used (via computer control or EXTREMELY clever engineering) to actually gyroscopically stabilize the gun. Basically, the longer you fire the gun, the faster the flywheels would spin and the more stable the weapon would become – higher stability the longer you fire. Computer controlled actuators (or carefully engineered systems) would then orient the flywheel(s) to stabilize or even move the barrel via targeting software or the like. This might be more difficult to accomplish in the hand-held version, but would be EXTREMELY useful to ship based weapons. The recoil could be almost entirely absorbed in this manner, ultimately being converted to heat energy or electricity over time after the gun stops firing as friction slowly decelerated the flywheel. It could be used to generate electricity via reverse induction (simply hook a driveshaft into the flywheel and reverse the induction). The electricity could then be used to power the next projectile (there would likely be energy loss to heat from friction – but if carefully designed this could be very minimal) – making such a railgun capable of firing whole clips on something like a small battery that compensates for the energy lost to friction – this battery could easily be built into the clip, making each clip a completely self contained firing solution.

    2) Rifling (bullet spin) is used to stabilize bullets in modern guus and increase their long range accuracy. Railguns have a unique capacity for creating this rifling through a few factors.
    2a) Twisting rails – the rails accelerating the projectile could themselves be twisted along the length of the barrel – imparting rotational velocity as they impart acceleration. This would likely result in some very strange spirals as the twisting may need to become more pronounced near the end of the barrel where the projectile has the most velocity – the best way to do this would require some serious thinking and math to do theoretically, which I just didn’t have the energy to undertake, so I’ll leave it to you if you want to research the details.
    2b) Rotational accelerator in either the loading mechanism (put an already spinning bullet on the rails) or at the end of the barrel (the accelerated bullet will likely have enough charge for a coil arrangement at the end of the barrel to impart significant rotational velocity. Heck, you could even just use normal rifling after the acceleration rails, though this might slow the bullet and create unnecessary waste to friction – whether or not this matters depends on the design.
    2c) A railgun doesn’t necessarily need smooth bullet surfaces and could build stabilization channels directly into the projectile (ie, fire “drill bits”). This, I think, is the least optimal method as it requires an atmosphere to work (although, rifling is used to overcome atmospheric disturbances, so this may not be a problem – won’t spin in a vaccuum, but also won’t need to, so…). It also means less contact surface for the charge to pass between the rails (may mean less efficient acceleration of the bullet, but again, there are compensating factors that could render this point moot, so again it comes down to final design). The one MAIN advatage of this type of rifling is that it may be useful for penetrating armor/targets more efficiently (again though, this would depend heavily on the nature of the target and/or armor – so could be more or less significant depending on situation and setting).

    3) Kinetic Energy and Kelvar vests. As a means of demonstrating the fact that the kinetic force of the bullet isn’t the lethal consideration: a Kevlar vest stopping a bullet means that the target wearing the vest MUST experience the FULL kinetic energy of the bullet (because the vest stops, not deflects, the bullet). Yet this is the defense against bullets – basically, the existence of Kevlar as bullet defense proves that the kinetic energy of the bullet isn’t what makes it dangerous, as experiencing the full kinetic energy is preferable to allowing the bullet to pass through (and experiencing significantly less of that kinetic energy).

    • tctaylor says:

      That’s certainly some intriguing ideas you have there, Master Zed. I spin projectiles by spiraling the rails (or, rather, twisting them into a helix), though I feel there’s probably a much simpler way of applying spin in the absence of rifling. I don’t like the idea of a helix because it doesn’t feel like a robust design to me and the stresses the rails will be subjected to will be severe.

      So, the setting you’re designing for. I’m intrigued. Can you tell me what that is?

      • Master Zed says:

        My setting is still in the very early stages, but I’d be happy to share a link with you (given the usual presumptions of no distribution and non-disclosure for early written works). If you’re interested, shoot me an email at messej_rc_junk@yahoo.com and I’ll get you a link to the WIP doc.

  17. Master Zed says:

    Hmm, as I think more about the rail gun’s recoil the above recoil compensation may not be valid – if the rails are allowed to move (or experience any acceleration in the direction opposite the projectile – even if converted to rotational energy via flywheel absorption) then that actually might just reduce the forward velocity of the projectile. The magnetic field generates a specific amount of force and that force would then be distributed to the rail AND the projectile – meaning any acceleration of the rail takes away acceleration from the projectile, so the rails may need to be “braced” (have a normal force applied) to actually impart forward velocity to the projectile efficiently.

    The question then becomes could you generate this normal force by having flywheels already at speed, effectively bleeding off their rotational velocity to impart linear momentum to the projectile – would this be any more efficient than simply allowing the flywheels to impart their energy mechanically – like a baseball or tennis pitch machine.

    Grrr… Physics…. lol, I just don’t know…

    I am fairly confident that the acceleration is even and (more or less) constant so the pushing against a wall analogy should hold.
    And the points about needing far less mass/total kinetic energy should still be valid.
    But I’m not so sure any more about the rails being separated from the launcher – at the very least conservation of momentum means the system would need to somehow conserve it’s total momentum. In the case of a person on a planet, I guess this is accomplished by stealing/imparting momentum to the planet you’re standing on, but again, this would convert linear momentum to rotational momentum, so why not flywheels? Grr, I guess this is why physicist is an actual profession – lol.

  18. Master Zed says:

    Did some quick unit analysis to see if I could get a feel for this… Figured I’d share it here for those interested.

    Kinetic Energy = 1/2 m*v^2 which is kg m^2/s^2 in SI units. (https://www.chem.wisc.edu/deptfiles/genchem/netorial/modules/thermodynamics/energy/energy2.htm)

    The v^2 is extremely significant here, as it means that velocity is exponentially more important than mass, specifically, this means that if you double velocity, you can QUARTER mass to get the same final kinetic energy. I think this is where the rail gun makes the difference, so let’s continue…

    From (https://science.howstuffworks.com/rail-gun.htm) we can get some rough estimates for velocities of CURRENT chemical and rail technologies.
    Chemical Projectile v = 1200 m/s
    Rail Projectile = 16,000 m/s

    From here we can get a relative velocity of 16,000 / 1200 = 13 1/3 (there is no guarantee this ratio will hold in a futuristic setting, but it is safe to assume this is a fair indicator of the MINIMUM relative velocities possible in a future setting, and we’ll use a lower ratio of 10 going forward to simplify the math anyways).

    Let’s assume our rail projectile has 10 times the velocity of our chemical projectile (well within the 13.33 given above). This means that a rail projectile 1/100th the mass of a chemical projectile can deliver the same kinetic energy to its target (as shown below). (please note this is the mass of the actual projectile, NOT the ROUND that’s loaded into the chamber.)

    KE Chem = 1/2 M.bullet * 1^2 == 1/2* M.bullet/100 * 10^2 = KE Rail

    To simplify later math, let’s go ahead and assume we want equivalent KE (kg m^2/s^2) – this is actually false, since what we really want is equivalent pressure ( kg m/s^2 ) at the point of penetration (so after armor). But since our Rail Projectile is 1/100 the mass, it should be fair to assume it also has a smaller surface of contact, or if we assume the projectile is being stopped/deflected by armor, equivalent KE becomes a fair indicator of stopping power at least and would be roughly equivalent (or better) for the Rail Projectile.

    This means we now have relative mass and velocity for both projectiles assuming we want equivalent Kinetic Energies, as follows (as relative measures, units don’t matter):

    Mass of Chem Pojectile = 1, Velocity of Chem Projectile = 1
    Mass of Rail Projectile = 0.01, Velocity of Rail Projectile = 10

    Using the 0.01 Mass and 10 Velocity of the rail projectile gives some interesting ramifications in terms of conservation of momentum and and reciprocal force:

    Momentum is Mass * Velocity (Mass is kg in SI units, Velocity is m/s in SI units, so kg m/s in SI units).

    Therefore the momentum of our Chemical Projectile is 1 * 1 = 1
    And the momentum of out Rail Projectile is 0.01 * 10 = 0.1

    Therefore, in order to deliver equivalent Kinetic Energy, our Rail Projectile has 1/10 as much momentum to conserve and the “recoil” should therefore be 1/10 as much as well.

    We can generalize this as follows:
    Assuming we want equivalent Kinetic Energies in our projectiles ( KE.1 = KE.2 )
    And assuming we’ll accomplish this by lowering mass as velocity increases, then we get:
    KE.1 = KE.2 => 1/2 M1 * V1^2 = 1/2 M2 * V2^2 => M2 = M1 * V1^2 / V2^2

    Assuming M1 and V1 are our unit values (ie we’re interested in proportional effects only) gives M2 = 1 / V2 ^2 (this is exactly what we got earlier, 10 times the velocity = 1/100 the mass, but it matters in out next step)

    Assuming we use conservation of momentum to approximate “recoil” (it’s not the complete story, see pushing on a wall above, but it’s a good start), we can introduce a constant R (for recoil) to determine the relative recoil of two different weapons with equivalent Kinetic Energies but different projectile velocities as follows:

    R = Momentum2 / Momentum1 = ( M2 * V2 ) / ( M1 * V1 )
    Again allowing M1, V1 to be our unit values, and substituting our M2 = 1 / V2 ^2 from before for equivalent Kinetic Energies gives us a final generalized equation of:

    R = ( 1 / V2^2 * V2) / ( 1 * 1 ) = 1/v2, in other words, recoil is inversely proportional to projectile velocity if we’re maintaining Kinetic Energy.

    This may not sound like a HUGE deal, but if we allow future rail technologies to get to something more like 20 or 30 times the velocity, then this means recoil drops to 1/20 or 1/30 for equivalent Kinetic Energy. For a real world reference, consider the following:

    From (https://hypertextbook.com/facts/2000/ShantayArmstrong.shtml) I got “A rifle can shoot a 4.20 g bullet at a speed of 965 m/s.”
    From (https://en.wikipedia.org/wiki/BB_gun) I got that a BB gun has a projectile mass of 0.35g with an upward velocity of 180m/s

    We can do the same kind of math I did above, and the two Momentum calculations are:
    Momentum Rifle = 4.2 g * 965 m/s = 4053 g m/s
    Momentum BB = 0.35 g * 180 m/s = 63 g m/s

    Meaning a rifle recoil is about 64 times that of a BB gun (to give our proportional comparisons a real world feel).

    So, a theoretical futuristic rail gun firing projectiles weighing 4.2 / 64^2 g at a velocity of 965 * 64 m/s would recoil like the above BB gun, but hit its target like the above rifle round. This is an EXTREMELY light projectile (aprox 0.001g) traveling extremely fast (at 61,760 m/s aka Mach 180 aka 0.0002 C), but it demonstrates why and how rail guns could have significantly lower recoil with equivalent performance (and MUCH larger ammo supplies)

    Ok, that’s my math quota for the day… 🙂

    • Master Zed says:

      BTW: The above formulas are simplified formulas that work well enough (very closely approximate) at speeds significantly lower than the speed of light (less than 0.1C). I just mention this because this discussion IS in the context of a scifi setting, so I wanted to make sure someone didn’t try to use these conclusions with 0.5 C projectiles or something. Since the fastest projectile I modeled here (the theoretical velocity * 64 rifle) only hits 0.0002 C, these conclusions should be fine for the purpose they were intended.

      It’s worth noting that even lasers have “recoil” despite their zero mass because they have near relativistic velocities (sometimes refered to as relativistic mass – though even this is a somewhat misleading nomenclature).

  19. Master Zed says:

    Just in case it doesn’t show up for a while (the post says it is awaiting moderation), I did the math to relate kinetic energy to momentum. (Very) Long story short:

    If you want to maintain kinetic energy (KE = M * V^2) you can increase velocity and decrease mass by that increase squared (so 10 times the velocity means 1/100 the mass).

    But momentum is just Mass * Velocity (not squared), which means increasing velocity and keeping Kinetic Energy equal results in proportionally less Momentum to conserve (less recoil) by the same amount.

    For example – if you want to turn something with the kinetic energy of a modern rifle into a rail gun that launches the projectile at 10 times the velocity with 1/100 the mass (very doable from my research), then you end up with 1/10 the momentum to conserve (or 1/100 the recoil).

    • Master Zed says:

      *1/10 the recoil, not 1/100 the recoil as I mistyped at the end of my last post — you get 10 times the velocity and 1/100 the mass, for 1/10 the momentum = 1/10 the recoil

  20. Master Zed says:

    One final note: Newton’s laws apply to a closed system only, so the guns you described as bleeding heat of into a separate dimension would not necessarily apply Newtonian laws anyways…

    This is also why I say a person firing a gun on a planet is conserving momentum by accelerating the planet (quite imperceptibly given the mass ratios, but still if you’re going neutonian, this is technically how you account for it)… Assuming they are braced against the ground (or something that’s braced against the ground, like a wall/etc), then the recoil from a gun goes from gun to arm, arm to feet (via kinetic linking), feet to rotational energy in the planet (because you’re on the surface which is far from the center of mass – so torque). You can say the limb and/or person compresses – but since people don’t shrink from shooting guns we can assume that at some point that energy is transferred to the planet, etc.) But here’s the rub, in reality that recoil isn’t all kinetic energy – the muscles burn chemical energy which contribute forces, the direction of the force arriving on the planet will have a component that is perpendicular to the surface and a component that’s parallel – meaning some of the energy becomes rotational energy and some becomes translational, the ground will compress to some extent probably releasing that energy as heat, etc etc…

    • Master Zed says:

      Sorry, forgot to close the system here, as our bullet is traveling off and unaccounted for… You see, the trick is that at some point the bullet will LAND (hit the Earth, or something connected to the Earth) – returning that borrowed momentum to the Earth and resetting the rotational energy, etc. Actually, if it’s a completely closed system, then you’ll need to account for any air resistance, heat exchange, etc etc. or your system just isn’t closed, and Newton will be denied… Just saying 😉

  21. Master Zed says:

    Another related discussion that helped make this all much clearer. The above is the reason it doesn’t kill you when you fire a gun. If the reactions were truly equal and opposite then a bullet that could kill a target would kill the person pulling the trigger too. So while the gun experiences the same force as the bullet (conservation of momentum), because the gun has much more mass, it ends up with less velocity and therefore transfers less kinetic energy to the attacker. This is also why properly bracing the gun further reduces the damage, holding it snug to your body means that the gun cannot accelerate separately from your own mass and the force must now accelerate your whole mass further reducing the amount of kinetic energy the attacker must absorb – and ideally, being properly set on the ground means that the whole planet’s mass becomes a factor, further reducing the total kinetic energy of the attacker must endure.

    Here’s the reference discussion that does a pretty good job of discussing these factors: https://www.quora.com/What-is-the-difference-between-kinetic-energy-and-momentum

  22. Master Zed says:

    Ok, just saw your response, so I thought I’d get into some of the math formulas if you want to respect the physics.

    Mementum for a BB gun is 63 g m/s, and a BB gun uses expanding air to ire, which should have a similar kind of peek impulse, so I think if you aim for around this number we can assume a pretty easy to fire weapon. So M * V would need to equal something small, like 50 – 100 g m/s. You can go higher if you want to allow for elaborate recoil compensation (like motors or springs that absorb then transfer that energy to prevent it from reaching the occupant of the armor) but this is our first boundary formula. M * V < 100 g m/s (mass of projectile * muzzle velocity – this keeps the momentum to around the BB gun range, you can decide if you want to go higher or lower but this is a pretty negligible recoil imo).

    Our next major formulas are kinetic energy (how much energy the impact has) and the projectile's cross section (to determine pressure).
    KE = 1/2 M * V^2 (g m^2/s^2)

    Pressure = Force / Area (g / ms^2)

    Area is the Cross section of the bullet and is in m^2 and would be VERY small (smaller is better since pressure divides by area which means that as area decreases, presure increases and is partially why bullets hurt alot more than baseballs. This is a variable you have alot of control over as follows: First, mass translates to volume based on the density of the material used for the bullet – higher density materials gives less volume for the same mass – the molar mass of the material used in the projectile should give a fair idea of the total volume needed – but alloys, forging methods, material sciences, etc can vary this a fair amount – especially in a scifi setting. Second: the shape of the projectile – the modern bullet shape is superior to the old musket balls because of its shape. The tip has a much smaller cross section for the same mass because its a cylinder rather than a sphere and comes to a roughly parabolic point. So the materials and engineering techniques ultimately determine this value, softer materials (like lead) will flatten on impact and increase the area and decrease pressure – which is why spherical musket balls were ok – they flatten on impact anyways, so the shpe in the barrel and in flight was relatively insignificant. If you want to go super scifi have your railguns fire something like a carbon nanotube (which I do believe is conductive) with a cross section only a few atoms across (for carbon this is around 7e-10 m^2 – http://hyperphysics.phy-astr.gsu.edu/hbase/Particles/atomsiz.html) meaning that it'll increase pressure by 10 orders of magnitude, which is HUGE and would allow it to basically penetrate anything – however, this may not always be the ideal – see more later for possible drawbacks.

    Force = (g m/s^2) and translates from our kinetic energy as one less meters in the numerator, so while you'd need complicated calculus, a diagram of the bullet path through various layers, and some serious materials research to get a completely accurate answer, we can use the distance over which the projectile transfers its energy as a fair average approximation which works well enough for a sort of average/general comparative purpose. So basically, F = KE / Distance traveled in the target (which I estimated to about 0.1m for a humanoid target's chest – but this would be HIGHLY variable depending on individual, where and how struck, where/if the bullet stops in the target, exact path throught that target, armor material thickness and shape, etc. but long story short – the less distance it travels the greater the force in a given moment and therefore pressure/penetrative power at any given point – but see below for some more important considerations)

    So, our Kinetic Energy transforms to a force (and therefore pressure) as it is travels through the target. Basically, when the bullet hits its target it starts decelerating, that deceleration uses up some available kinetic energy to move through the target and displace the material its traveling through (it'll also inevitably lose some to to heat, etc through friction, etc. – so this isn't 100% efficient, but again gives a general feel for relative effectiveness). Here are the primary scenarios. Momentary pressure, nature of the target, etc determines which of these cases occur and the details, but these give the general idea of the range of possibilities and can be used to approximaet the distribution of the energy in each case.

    1) (The optimal case for "squishy" targets) Bullet penetrates a bit more than half way through its target (this is the reason for hollow points which are designed to have a small cross section at the surface – where there are fewer vital tissues – and through the air to penetrate them easier and expend less kinetic energy in these less desirable areas, then mushroom as they travel through the body to increase surface area decrease pressure and deliver more kinetic energy/damage to surrounding more vital tissues. This means that the bullet comes to a stop (all kinetic energy delivered) in a vital location (maximized tissue disruption).

    2) The bullet goes clean through (in all honestly, this is BY FAR the most likely case for a rail gun projectile unless VERY heavy, probably reactive, armors are involved). Since the bullet is still moving on the other side of our target, we know that not all of our kinetic energy was delivered – but there IS a hole where the bullet went through the target, which is more or less bad depending on where this happens. In a battlefield where armor and cover are major considerations, this is kind of the ideal situation as the extra kinetic energy basically translates to penetration power (the ability to go through more armor/cover). However, the damage is now more or less independent of kinetic energy and becomes dependent on cross section and path traveled (it'll make a hole roughly the shape of the cross section of the bullet along the path the bullet travels through the target. If we're minimizing the cross section to maximize pressure and penetration, this now means potentially less damage to the target as the hole is smaller – hence the balancing act that must be done regarding bullet cross section – increase it and you increase the width of the wound track, but decrease penetration – and vice versa). Considering it's VERY likely our railgun will be a VERY high rate of fire weapon (might as well, low recoil and light ammo makes this kind of perfect), it's not too big a deal to allow cross section to decrease and make this our ideal scenario, since a whole bunch of small holes means a good chance you've opened some holes in a fairly vital location – where even small holes are fatal, and the increased penetration power means the bullets can travel further, lose less kinetic energy doing so, and penetrate alot more cover/armor.

    3) Bullet is "caught" (stopped completely) by armor (kevlar vest scenario). This has the effect of turning our kinetic energy back into momentum (recoil), but likely does so over a VERY short distance/time (kick instead of pushing a wall), so the impact will be more jarring to the target, who (hopefully) also won't be as well braced against the direction of the impact further knocking them off balance which is far more disruptive to target than attacker. All kinetic energy is absorbed (since the bullet stops) and is transferred to the target as heat and momentum (recoil) primarily (how this happens depends GREATLY on the material and type of armor). But long story short, our super light high velocity projectiles would lose their v^2 advantage and go back to straight m*v and it'd be like trying to knock someone over with BB's (non penetrating BBs) — so least ideal for our rail gun case. Even a very high rate of fire isn't going to make much difference, in fact, the "plink" of the projectile hitting the armor and startling the target to trigger uncontrolled reflexive actions (flinching from the sound/surprise of the strike) is likely to be more disruptive than the actual impact of the projectile.

    4) The bullet is DEFLECTED by the armor (it hits the armor, but then continues on, possibly in a different direction). This is the worst case, as it's basically the above scenario, but with only a fraction of the kinetic energy getting transferred – so yeah, basically nothing for our low mass rail projectiles. Since its most likely to happen at the edges of the armor, we do get some advantage from leverage (the surface of the body is a distance from the center of mass, so leverage) which will multiply the force and translate it as torque – but you're basically shooting a BB (smaller and faster, but same momentum) at a carnival target, so it's not going to make enough difference to matter in most cases.

    So basically, our two ideals are scenario 1 and 2.

    Against an unarmored target without cover, we'd want to use something like a bolo/dumbell shot which gives up velocity and kinetic energy and decreases range/accuracy a little (since drag will eat up more of it in transit), but greatly increases cross section and distance traveled inside squishy meat targets as the weights snag and rotate as they pass bones and different tissue densities – lengthening the wound track and increasing the chance of striking something vital.

    Against an armored target or one with cover, or if we're trying to deliver a more accurate strike at long range, we'd want a more aerodynamic low cross section high velocity penetrating round which might only make a small hole straight through its target, but which can do so with ore accuracy.

    Here then are our ideals, and if you allow for computer assisted aiming, our rail gun now becomes WAY superior to our chemical weapon. Since the computer targeting system can calculate distance to target, composition of target, orientation of target, take all of these into consideration and load the appropriate projectile and set the current in the rails to give the exact velocity needed to penetrate armor/cover and stop inside the target. Basically, the computer can make it scenario 1 in all situations (or at least maximize probability of scenario 1) since our rails are actually variable force generators (unlike a charge of black powder which is determined by the amount of powder in the shot) and a computer could control the current and do the calculations to turn current into damage to target between bullet fires and then easily maximize that equation – it could further use approximations to account for things like armor and cover penetration based on sensor data. Our low mass ammo comes in doubly handy here since the high rate of fire will allow the computer to "guess and check" it's approximations across a VERY large sample size, subtly adjusting the velocity of the projectiles to maximize damage based on the results from the previous shot(s). Basically, a very simple AI and moderately complex computer (even by today's standards – much less future standards) could make a rail gun CONSIDERABLY more dangerous than a chemical weapon. And should the AI/computer assistant fail, the weapon could always default to "full power" and still do some pretty good damage just by turning people into perforated pin cussions via high rate of fire.

    • Master Zed says:

      So let’s run some numbers… (I’ll keep them simple to make it easy to adjust them to fit your setting and vision)

      Let desired momentum (Recoil) be 100 g m/s (recoil of firing a BB gun).
      Let muzzle velocity be 100,000 m/s (approximately mach 300 (291.5) or 0.00033 C if I did my math right)
      Then our projectile’s mass is 100 g m/s divided by 100,000 m/s = 1/1000g = 1 milligram (that’s TINY – 1,000,000 projectiles in a 1 Kg drum…)
      Let’s use carbon to make our projectiles since it’s abundant and has some other nice features – specifically we’ll basically fire graphine jacketed diamonds… (two allotropes of carbon, one which conducts electricity very well for our railgun, and one which is extremely strong, for surviving acceleration and impact)

      Allowing for a 0.1 meter rail (about 4 inches) this means a bullet is out of the barrel in 0.1 m / 100,000 m/s = 1 microsecond (one millionth of a second). Allowing for a peak possible fire rate of (pinky finger to corner of mouth) one million projectiles per second (that’s 1 Kg of ammo per second – so probably not the fire rate that is typically used, but wanted to give you an idea of potential rate of fire… btw, since this is continuous fire, this would be the equivalent of holding up a 10,000 Kg mass for 1 second in terms of strength/recoil stress fractures and the like — so NOT firing that by hand… that 10,000 Kg mass for 1 second number comes from a really rough approximation for turning that momentum back into weight given a 10 m/s^2 acceleration from gravity – which is off a bit but close enough to 9.8 m/s^2 for super easy math estimation) this sustained fire is how our pushing on a wall analogy works – let me know if you’d like me to work up an exact load formula that you can just plug numbers into to figure out this “recoil weight” based on mass, velocity, and fire rate.

      Since I’ve decided the majority of our projectile is diamond, I’m going to use diamond’s density of 3.51 g/cm^3 to determine the volume of the projectile. Which gives our 0.001g projectile a volume of 0.00028 cm^3 = 2.8e-10 m^3. Assuming a sphere (which isn’t the optimal configuration, but at this scale might not make much difference), this gives our projectile a cross section of 5.2e-7 m^2 (this is kind of a worst case scenario for lazy manufacturing – a dart shape would be far more efficient but much harder to manufacture, and since we’re talking about potentially firing millions of these per second, and we might want to make it possible for soldiers to produce ammo in a field situation, this isn’t a bad compromise.) Actually, to simplify our math later, let’s squish it just a iittle to make it 2.5e-7 m^2 and say it’s more rice grain shaped…

      Now we can figure out the kinetic energy and pressure delivered to our target per projectile (and per second if we wish).
      KE = 1/2 m*v^2 = 0.5 * 0.1 * 0.001 * 100,000^2 = 500,000 g m^2/s^2 = 500 joules of energy per projectile
      500 Mj per second at max fire rate (which we don’t actually use) = 500 MW of energy

      Pressure = Force/Area = KE / penetration distance / cross section = (scenario 1, assuming 0.1m of travel) 500 kg m^2/s^2 / 0.1 m travel / 2.5e-7 m^2 = 2e10 kg / ms^2 = 20 million pascals (for reference… 70 kPa can sever limbs… https://en.wikipedia.org/wiki/Overpressure) — let’s go ahead and assume this means our projectiles will easily travel through basically anything (scenario 1) – i mean that’s still 2 MPa after even 1 meter of travel…

      So now the REAL question becomes, can we just plain cut a running man in half by firing a short burst in front of them?! And if so, how long a burst?
      Given a thickness of less than 0.5 meters let’s say, and our projectile’s diameter of roughly 0.001 m, we’d need 500 projectiles to completely sever the upper torso from the lower torso – we’d literally use a stream of diamonds to chop a man (or basically ANYTHING at those pressures 0.5 meters across) in half… At our 1million rounds per second this is a 5 millisecond burst and consuming 0.5g of ammo.

      (need a break, will do more later)

  23. Master Zed says:

    Quick plug for some wonderful references that I’ve basically been stupidly recreating here – le sigh… 🙂

    Had some inspiration… Instead of estimating all this stuff, let’s use the desired numbers to calculate the realities of the weapon, as follows:

    My Goal: create a projectile that will penetrate armor (and thus be almost guaranteed to penetrate a body completely), and figure out the weapon’s profile from there.

    This brought me to this page (https://physics.stackexchange.com/questions/16817/penetration-of-armor-plate), from which I pulled the formula for penetrating steel of a given thickness (we can now express any future armors or potential cover as “equivalent to T meters of steel plating” to eliminate the need to figure out high tech material sciences)

    The top answer gave us some important information:
    “toughness of steel” = 2000 MPa = 2e9 kg m^-1 s^-2

    Ok, now we can determine our necessary KE based on Cross Section (square meters) and Equivalent Thickness (meters) of Armor.

    KE needed to penetrate armor equivalent to T (thickness) meters of steel with A (area) square meters of cross section= 2000MPa * T meters * A meters^2 = 2e9 kg m^-1 s^-2 * T m * A m^2 = 2e9 T A kg m^2 s^-2 <– Hey, that's the same units as our 1/2 M V^2 formula (convenient…)

    Kinetic Energy for the bullet's mass and velocity = 1/2 M V^2 = 1/2 M V^2 kg m^2 s^-2

    Setting these two kinetic energies equal tells how much Mass (in kg) and Velocity (in meters per second) we'd need to penetrate the equivalent of T meters of Steel with a bullet cross section of A square meters. But this is the KE needed at the point of impact, so we'd also need some KE to get the projectile there, but this is likely to be as trivial as the KE needed to pierce the fleshy bits behind that futuristic armor at these scales, so we'll ignore it for now.

    2e9 T A kg m^2 s^-2 = 1/2 M V^2 = 1/2 M V^2 kg m^2 s^-2 (our units cancel right away, and we can move all our variables to one side, to get)
    4e9 = M V^2 / (T A), or expressed slightly differently to put target on one side and attacker on the other
    4e9 T = M V^2 / A

    Which tells us that the ratio of Mass times Velocity squared divided by Cross Section area must be equal to 4e9 times the (equivalent) Thickness of the armor. Huzzah!

    Next, we can relate the cross section to the mass using the diamond spheres from earlier.

    Mass = density (Den) * volume (Vol)
    volume (Vol) of a sphere = 4/3 pi r^3
    great cross section (A) of a sphere = pi r^2

    Vol = M / Den (for our diamond spheres, Den = 3.51 g/cm^3 = 3510 kg/m^3, so…)
    4/3 pi r^3 = M / 3510
    r = 0.040812 M^1/3
    Side Note: Diameter = 0.08164 M^1/3

    A = pi r^2
    A = pi * (0.040812 M^1/3)^2
    A = 0.0052346 M^2/3

    Plugging this area back into our target/attacker formula gives:
    4e9 T = M V^2 / A
    4e9 T = M V^2 / (0.0052346 M^2/3)
    2.1e7 T = M^1/3 V^2

    Now we can either chose T and V based on setting, determine mass, then calculate the corresponding momentum, fire rate, etc.

    Or we can start with the desired Recoil Load from above, determine velocity and rate of fire and mass, then plug M and V in above to get T – which determines how much armor these guns can penetrate… Hmm, I might see if I can figure out a way to relate these all in a single formula that's basically tell you the recoil load for a "toughness" of target armor. This would actually likely be the most useful in terms of computer controlled/assisted targetting, since the computer would vary the rail power as necessary to hit the target, and therrfore projectile velocity, and therefore recoil load… The safeties would most likely be based on projected recoil load, which would then designate targets as green or red based on whether or not they'd require more recoil load than is safe.

    But I think that's all I can do for now…

    • Master Zed says:

      (previous reply awaiting moderation, so be patient if this response doesn’t seem to make sense, since I’ll be referring to information presented in it)

      I’m going to try to calculate the armor penetration of a railgun that’s designed to be fired with minimal impact on the user – and then calculate the armor penetration capacity of such a weapon. Here are my assumptions

      1) I’m going to set the fire rate, velocity and mass of the gun to be roughly the equivalent to holding 1 kg above your head (this is similar to lifting – and holding for the duration of the weapon’s burst – a one liter bottle above your head). Using approximately 10 m/s^2 for acceleration of gravity, then gives us 10 kg m/s^2 as our target recoil (this is TINY – just lift a liter of soda above your head to see how little force this is to resist).

      2) I’m going to use a 0.1 gram (1e-4 kg) carbon (graphine jacketed diamond) spherical projectile – this is about 1% the weight of a chemical bullet from what I can gather. This gives…
      1 million projectiles per kg of carry weight.
      a projectile volume of 0.1 g / 3.51 g / cm^3 = 0.0285 cm^3 ( 2.85e-8 m^3 )
      a projectile diameter of 0.38 cm (0.0038 m) – less than half a modern bullet
      a projectile cross section area of 0.11 cm^2 ( 1.1e-5 m^2)

      From 1 & 2 we can now calculate maximum allowed acceleration: given 1e-4 kg per projectile, and our target recoil of 10 kg m/s^2, we get an acceleration of 10/1e-4 = 100,000 m/s^2.

      3) I’m going to assume a 0.2m (8 in) long rail. This is a bit longer than a long pistol barrel, but rail length makes a big difference on a railgun’s performance, and given the more narrow barrel, the linear nature of the acceleration which means we can comfortably move more of the rail behind the grip, and the removal of length needed for a modern pistol’s “chamber”/bullet length – after all our 0.1 g bullets are only .4 cm long and don’t need a casing or powder charge this is still around the size of a modern pistol. And if we move the power supply away from the hand (to a hip mounted power pack with a small cable for instance) or have advanced battery tech that makes the power supply tiny this will make it even less unwieldy.

      We can now calculate the final velocity of the bullet leaving the rail:
      given a constant acceleration and no (relative) initial velocity means the time to travel a given distance is t = (2d/a)^1/2 (from d=1/2at^2), for our rail length of 0.2m and our acceleration of 100,000m/s^2 we get t = 0.002 seconds. Plugging that into our velocity formula v = at gives us a final velocity of 200 m/s (this is actually about half as fast as a pistol round).

      increasing barrel length won’t be a big help, since v = (2da)^1/2 tells us that the velocity is increasing based on the sq. root of rail length and acceleration. ie if we quadruple barrel length or acceleration, we only double projectile velocity. And as we can see, there are no other variables acting on the formula, which means MUCH high accelerations and longer barrels are necessary… This might be part of the reason they’re being considered for ship mounted weapons… but let’s press on…

      As a side note, the t = 0.002 seconds means we can comfortably maintain a rate of fire of 500 projectiles per second. For reference, given our 0.38cm diameter, this means we can cut out a line 190 cm long/second though basically anything we can penetrate, which we’ll calculate now… (given an approximate hip width of about 38cm for an average human, allows us to cut a human in half with a burst lasting 0.2 seconds and consuming 38/0.38 = 100 rounds, or 10g of carbon)

      assuming this projectile was being fired at point blank range (so peak velocity), the projectile would hit with:
      KE = 1/2 m v^2 = 0.5 * 1e-4 kg * (200 m/s)^2 = 2 kg m/s^2 = 2 Joules of KE
      Going to the above formulas, we could penetrate
      T = M^1/3 V^2 / 2.1e7
      T = 1e-4^1/3 * 1e5^2 / 2.1e7 = 22.1 m of steel?!

      (Did I mess something up in all these calculations?! – if not, I think we have a killer weapon here o.0)

      • Master Zed says:

        *10,000 rounds per 1kg of ammo
        (each burst does use 10g of carbon though, so a 1kg clip would be enough to chop 100 man sized targets sporting the equivalent of 22m of steel – 11m on each side – in half)

        * I totally messed up that last penetration calculation, I used acceleration instead of velocity (whoopsie)
        The correct calculation is:
        T = 1e-4^1/3 * 200^2 / 2.1e7 = 8.8e-5 m of steel (0.088mm, or 88 micrometers)

        (This makes alot more sense)

        Also note that the more useful version of the thickness of steel formula when designing a gun in this direction is

        T = 4.8e-8 M^1/3 V^2

        Where T is thickness of steel penetrated (in meters), M is the mass of the DIAMOND projectile (in kg) and V is the velocity of the round at point of impact (in meters per second)
        As we can see, mass has very little impact (being to the 1/3 power) and velocity has a huge impact (being squared).
        Though the original shows us that we can see that a 1 kg mass requires a velocity of just (2.1e7)^1/2 = 4,582 m/s to penetrate 1 meter of steel
        A 0.1 kg mass requires roughly 1.5 times this velocity (6726 m/s), and a 1g (0.001 kg) mass requires only roughly 3 times this velocity (14,491 m/s).

        Additional thoughts:
        1) Railgun projectiles have very little mass/inertia/momentum (momentum is just 2 g m/s), therefore they may be easy to deflect – their extremely high pressure (from high velocity + small size) may offset this though as they rapidly dig into any surface making sloped deflection very difficult. For sloped armor to deflect our diamond spheres they’d need to be of equivalent “toughness” (ie diamond/carbon fiber themselves) and/or use clever engineering that allows the projectile to penetrate, then guides it along “channels” inside the armor (very difficult to do against near perpendicular strikes). Kevlar style woven armor would be pretty useless as the diamond projectile does not deform like a normal modern bullet (this is the same reason Kevlar doesn’t stop “armor piercing” rounds). These are purely case 2 projectiles (ideally they pierce clean through their target, delivering very little of kinetic energy to their target).

        The ideal defense would actually be some kind of reactive armor (armor that exploded when hit by the projectile) – since the projectile has very little mass/inertia/momentum, so an explosion with an equal amount of momentum in front of the projectile would literally stop the diamond round in it’s tracks. For reference here, the equivalent of just one gram of TNT can produce 4,184 joules (https://en.wikipedia.org/wiki/TNT_equivalent) of energy, so we’re talking microgram explosives here – A LOT of them, but still – even stopping the full 100 round man-splitting burst (200 joules) only requires less than 0.05g of TNT equivalent (alot less of something like modern C4 and even less of whatever futuristic explosive gets thought of). BTW: These reactive explosions would have the same (or potentially less if engineered correctly) recoil as if the armor stopped the projectile which is roughly the same as the recoil experienced by the shooter (ie, trivial). The problem here though is that this is an ablative defense (it’s used up each time it protects against a projectile), so the extremely high fire rates and ammo capacities of rail guns would make it fairly easy to win such a battle of attrition – so even this defense is more a waiting game.

        As one last point, assuming the graphite jacket is engineered to be just thick enough to “burn off” within a very short time after leaving the barrel, the projectile becomes non-conductive and non0ferrous, making sci-fi style magnetic or EM shielding mostly useless as well. Diamonds are also great heat conductors, which means any kind thermal shielding wouldn’t be too useful either.

        2) These 0.2 second bursts mean Computer Assistance would be extremely useful, but not strictly necessary. A simple gyroscopic, or even mechanical, setup could cause the barrel to rotate the ever so small amount needed between shots to create a 100 round “line” of projectiles each time the trigger is pulled.

        With really good computer assistance (and control gyroscopes, or even just barrel actuation – the recoil is certainly in a range where adjusting the angle of the barrel is perfectly fine), you could fire much shorter, much more precise bursts. For instance, a human heart is in the neighborhood of 8cm across (roughly 20 rounds, 2g of ammo, 0.04 second burst – 10 rounds, 1g of ammo, 0.02 second burst if you’re willing to just cut open one chamber, which is probably enough… Large intestines are about 8cm across as well as well, so a 10 round burst could literally open a hole in the target’s bowels.

        In fact, firing a 10 round (0.02s) back to back burst like this with just a basic mechanically controlled “line” would be roughly the equivalent of getting stabbed clean through by a diamond sword or spear with a stabbing profile roughly 4 cm by 0.4cm… So, if you cannot afford a diamond sword, then perhaps consider this railgun…

        3) This is an absurdly low recoil, think of holding a fresh 1 liter bottle of soda for just 0.2 seconds… even catching a falling soda bottle would present more force to your arm as the bottle would have some velocity. You won’t even need to brace against this weapon – and that’s with ZERO recoil compensation built into the weapon or its firing platform…

        The momentum of the whole 100 round burst is 10g * 200m/s = 2 kg m/s. Assuming a 60kg person firing a 100 round burst in space (so not a soldier in great shape, no weight for armor, spacesuit, or even the weapon or its ammo, etc), this burst will cause them to “fly backwards” at 2 kg m/s / 60 kg = 1/30 m/s (0.067 mph).

        So we can probably increase this significantly.

  24. Master Zed says:

    Ok, wet full nerd and made a gul durn spreadsheet to address these problems. You can view and comment on the spreadsheet here:

    Before I get into the conclusions from my spreadsheet, here are a few more useful links:


    I modeled 3 different rail gun ideas. All three use a 1 NANOGRAM graphine jacketed spherical diamond projectile (I figured using the same ammo for all of them would be conveient in the world, but this doesn’t have to be the case). You literally can’t easily handle an individual “bullet” without tweezers or the like – they are about 82 micrometers across – for reference, a human hair is about 25 micrometers across, so these are about 3 hairs wide… I’m also assuming mechanical or computer assisted “aiming” to create lines of projectiles that essentially do damage by cutting a gash in a target. These tiny projectiles do not do damage like a conventional gun, instead they fully pierce their target, but the high rate of fire of the railgun allows the attacker to chain them together to carve out a gash similar to getting stabbed by a knife (2cm long slice), sword (4 cm long slice), or fully chopped in half (0.5m long slice).

    1 kg of ammo gives each of the guns the following number of “attacks”:
    Knife: 4billion+
    Sword: 2billion+
    Cut in half: 160Million+

    One thing that I did not do, but which could allow much higher velocities is use some kind of “shock absorbers” to spread the recoil over longer periods of time. Since the cut a man in half burst is only 6/1000 seconds long, a basic spring type recoil compensator could distribute this force over even 0.1 seconds and significantly reduce the forces applied to the arm – or allow increases in the rail’s acceleration forces.

    The first is a “pistol” style railgun, with a short (0.3m) barrel that can be held in the hand and fired basically like a normal pistol. I limited the recoil force to the equivalent of holding yourself off the ground in a push-up position. This isn’t nothing, but you can easily do it for several minutes if needed, and it’s not going to cause any kind of “knock-back” or even aiming difficulties – you can hold the gun in place with arm muscle alone and you can move the gun while firing without any resistance.

    Can fire up to 1 million rounds per second
    Projectile Velocity: Mach 1,166
    Penetrates 0.76 meters of Steel (ie, can easily penetrate unarmored or lightly armored foes)

    The second is an arm mounted “rifle” with a 0.5m long barrel that sticks out past the elbow and past the fingertips. It would sit on the arm kind of like a stowed tonfa. It can be strapped to the arm, or held by a handle in the hand. Ideally this would be mounted in the arm of a piece of armor. It uses heat build up in the gun to vent hot air in opposition to the force of firing, offsetting some of the recoil and allowing the rails to apply more force without the user having to experience it. Basically, it is partially recoil compensated through what amounts to a rear facing rocket that keeps it in place and easy to manage. As a result though, the back of the barrel should always be kept from facing anything you care about, like your ribs….

    Can fire up to 1 million rounds per second
    Projectile Velocity: approx Mach 3,000
    Penetrates 4.78 meters of Steel (ie, can easily penetrate cover and well armored foes, can cut a car in half – easily)

    The third is an over the shoulder version that stows on the back similar to a medieval sword in a fantasy movie. The barrel pivots on a special shoulder mount that is worn and designed specifically for the weapon. It is technically possible to wield two such weapons, but why would you need to?! The barrel is pulled into position when needed, then allowed to settle against the back when not in use. It vents gasses out the back of the barrel (which is several inches from your head) to FULLY compensate for any recoil (like a modern recoiless rifle). It is aimed with computer controlled actuators at the shoulder mount if using computer assisted aiming, or via a handle at the front of the barrel if manual fire mode is preferred.

    Can fire up to 2 million rounds per second
    Projectile Velocity: 1% C (Mach 9,329)
    Penetrates 48.9 meters of Steel (ie, can easily penetrate pretty much anything, could be used to sever a building from its foundations, or cut a large suspension bridge in half… lengthwise!)

    See the linked spreadsheet for more data. (Finally found some gun designs that actually do the job 🙂 )

  25. Master Zed says:

    *The pistol sports a 0.2m long barrel

  26. Master Zed says:

    Updated the spreadsheet to do some rebalancing, and added a tab for Ship Based Railguns

    Some quick Stats on the Ship Based Railguns:
    All use diamond spheres as ammo, but each type of gun uses a different sized diamond (ammo is not standard across all guns like with the personal railguns)
    The smallest gun uses a 1 Femtogram sphere (aprox 50million carbon atoms). The medium gun uses a bullet 10 times that mass, and the spinal rail uses one 100 times that (0,1 picograms – or about 1/10 the mass of a single strand of hummingbird DNA – ie, VERY small). At these sizes, there may be an argument to be made for simply using a particle weapon…

    If you look at the spreadsheet, you can see that all three weapons fire their projectiles at 0.99C (99% the speed of light). I have adjusted for relativistic effects (ie, used Einstein’s equations to ensure the projectiles exit with enough kinetic energy to actually reach 0.99C even when accounting for their relativistic mass — I don’t believe I did the math wrong, but well – that’s some advanced math, so no promises 🙂 )

    You’ll also notice that all 3 weapons have a recoil force of 550N. This is about the force that a small woman standing on the surface of the Earth – it is TINY – so there is very little concern that the weapon will experience any sort of problem with continual fire. This is also the recoil “thrust” that will have to be compensated during the weapon’s fire, so 550N of extra thrust per railgun will balance the recoil.

    The longest burst duration is the Wild Burst for the Small Turret (which really isn’t intended for that weapon anyways – it’d be the kind of attack a single 5m drone would have to use against an aircraft carrier sized ship) @ 0.23 seconds long – which is (as the name of burst suggests) a “Wild Burst”. The shortest is the Large Turret 1m gash at 0.01seconds. So these bursts lengths are usable at 1ls ranges (targets can’t detect and “dodge” them, though they may be able to reduce damage from Wild Bursts slightly by moving the impact point slightly – wld bursts are meant to fire a line so long that they will cut an aircraft carrier in half, this also means that the attacker can instead cut a zigzagging swath across all of smaller ship’s trajectories, possibly making avoidance impossible… (This is like the shotgun approach I discussed in another post, where even 5% of a 100m line is enough to chop a 5m diameter probe in half, so a “grid” pattern of projectiles could potentially cover all potential locations of the distant craft and secure a definite kill…

    Ammo is not really a concern for these weapons, the largest guns can fire 400Million aircraft carrier destroying “wild bursts” on just 1kg of ammo.

    all 3 guns can pierce thousands of meter of steel, however, the low mass may mean they are easily deflected. Given their velocity and impact pressure, it is unlikely this could be used as an actual defense, but it may mean the ship is severely “swiss-cheesed” instead of actually cut in half. Such an occurrence would probably be worse for the ship in question though, as bulkheads and airlocks would probably save much of the ship and its passengers after being cut clean in half, but this is very unlikely in the swiss-cheese scenario as the incredible amount of kinetic energy and small cross section means the could bounce around for quite some time piercing vital components and people at random. Also note that their extreme penetration and low momentum means they will do very little to “move” the enemy target. These weapons COULD split an incoming target in half, but they’re not likely to stop those two halves from continuing on to smashing into you – unless the target was pressurized of course, in which case the ship’s depressurization may save you from a collision. (basically, I’m saying lasers would make better point defense weapons… 😉 )

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