I read a book once where the primary weapon was a "Zip" gun that utilized a spool of wire as ammunition and a battery for power. There weren't any heat issues and the spool of wire lasted quite a while. The wire was snipped to a designated length and propelled out of the gun via electromagnetism, similar unto a rail gun.

In the book, a direct hit hurt (possibly fatally) but didn't immediately kill unless it was a perfectly placed head shot.

I've considered using a weapon like this in one my stories but it just doesn't seem effective as defined in this book. I've had nightmares where I fired into an enemy with no apparent effect. No thank you.

If I had a setup similar to the one described, instead using a 3" length of 18 gauge steel wire weighing approximately 1.6 ounces, with an automatic weapon that could fire 3 shots a second, propelling the wire at around 2500fps leaving the barrel

  • What would my range be?
  • How devestating would it be?
  • What kind of armor could stop it?

Lastly, would a different material work better than steel?

  • $\begingroup$ Your biggest issues in railguns is almost always the kickback of the device, which is magnified as railgun is rarely a oneshot weapon and relies on a high rate of fire to do it's damage. Calculate the energy the projectile has leaving the device and apply it as a kickback to the weapon holder...1.6 ounces at 2500fps one direction can be thought of as 160 ounces at 25fps the other direction. $\endgroup$
    – Twelfth
    Aug 18, 2016 at 19:17
  • $\begingroup$ Take a look at the kenetic energy of the projectile, and graph that next to conventional bullets. $\endgroup$
    – JDługosz
    Aug 18, 2016 at 19:22
  • $\begingroup$ Your weapon delivers about 14kJ energy per missile. in Poland, 17J (not kJ) is a border - above that and it is a weapon, not toy. So... $\endgroup$
    – Mołot
    Aug 18, 2016 at 19:22
  • $\begingroup$ Hmm, my quick calc indicates that a piece of 18 gauge copper wire that weighed 1.6 ounces would be more than 20 feet long. So maybe a numerical adjustment is in order. $\endgroup$
    – Seeds
    Aug 18, 2016 at 21:24

1 Answer 1


Raw Power

Short version: Too much.

Your 1.6oz piece of wire with a muzzle velocity of 2500fps carries more than 13,000 Joules of energy with it as it leaves the weapon. For comparison, the ever-popular AK-47 firing the 7.62×39mm round at ~2400fps carries only 2100 Joules out of the barrel. The massive difference in energy, despite roughly equivalent velocities, is owed to the huge increase in mass of your projectile -- the bullet fired by the AK-47 weighs in at a mere 0.3 oz.

And this is a big problem for the practicality of this weapon. Even ignoring power requirements and heat dissipation, in order to fire that projectile downrange, Newton's Laws dictate that the same force must likewise be pushed back the other way, i.e. into the shooter's shoulder (in shooting terms, we're talking about the recoil). Give-or-take a margin of error depending on the exact length of the barrel (and thus the amount of acceleration imparted to speed up the projectile), we're talking about recoil approximately in the range of the powerful .50 BMG -- a weapon that is virtually never fired except from the prone position, while mounted on a bipod, and utilizing a "muzzle break" device to reduce the amount of force shoved backwards into the shooter. Since the physics of the railgun make a "muzzle break" device impossible (there are no expanding gasses that can be redirected backwards), even though raw energy of the bullet is a bit higher in the .50 BMG rifle it's very probable that your shooter will feel more recoil with your railgun than with a .50 BMG rifle. (NB: Not saying that .50 BMG rifles can't be shoulder-fired, just that they tend not to be because of the weight and recoil.) If your railgun ends up being significantly lighter than a .50 BMG rifle (an oft-cited benefit of these weapons in science fiction, though realistically probably not going to be the case), then your shooter feels even more recoil.

This massive recoil is also why .50 BMG rifles tend to come solely in bolt-action varieties (notwithstanding mounted weapons): You simply cannot fire effectively in automatic or even semi-automatic operation, because the recoil prevents you being able to follow up so quickly. 3 of these rounds per second? Not happening. Not effectively, anyway -- that first one might be on target, but none of the rest will be anywhere close.

This will not be an effective handheld weapon without significantly augmenting the strength of the shooters. As a sniper/anti-materiel rifle or a vehicle-mounted weapon it could work, but within the parameters of your question such a weapon just is not going to be usable.

However, continuing on anyway...


Assuming a rifled barrel to impart a stabilizing spin on your projectile, it's reasonable to assume an effective range similar to that of the .50 BMG, so probably around 2500 yards (about 2400 meters), give or take. The increased mass of the wire will actually help here, because it means more momentum. The big question is how the weapon's wire-snipping apparatus works, specifically on what shape it winds up putting on the tip of the wire -- if you snip a wire with standard wire cutters, you end up with a curved, flat nose, which in a projectile will act as a destabilizing aerodynamic element that may dramatically reduce the effective range.

There's also the fact that a coil of wire, even when pulled into the weapon's firing chamber, snipped, and fired through the straight barrel, will still have a tendency to want to coil back up on itself, resulting in a curved projectile rather than a straight one and significantly increasing tumble, which will decrease range, accuracy, and penetration.


Your projectile has approximately the same energy as a .50 BMG bullet, however it is confined to a diameter that is a full order of magnitude smaller -- approximately 0.04 inches, versus 0.5 inches. This means that it will be more effective against armor than the .50 BMG, and likely means that there's simply no way to protect personnel against this weapon. It might even be effective against light armor (something the .50 BMG cannot match), but heavy tanks will still shrug this off.


Steel's a good choice. It will resist deformation on impact, meaning it's more likely to penetrate armor than a softer metal (such as copper) would. I doubt you'll find a better material that can be carried as a coil of wire, unspooled and trimmed on demand to be fired from your weapon.

  • 3
    $\begingroup$ " Newton's Laws dictate that 13 KJ must likewise be pushed back the other way, " NONONONONO That's false. It must be the same momentum pushed back. Not energy! Momentum is mass * velocity. Energy is mass * velocity squared / 2. $\endgroup$
    – Mołot
    Aug 18, 2016 at 19:57
  • $\begingroup$ Excellent analysis. Based on the comments I started to pull together a lot of these numbers; though making sense of them has been difficult. If the projectile was 2", it would be doable, and at 1" it's about the power of a .308. Assuming the technology is available to do this in the first place, I'm thinking heat and power wouldn't be much of an issue. $\endgroup$ Aug 18, 2016 at 19:57
  • $\begingroup$ Momentum vs energy means that average male soldier (85kg) would need to be pushed back with less than half a meter per second velocity, and that gives energy of less than 10 J. For gun itself, well. It'll be more as the mass difference is less. but energy vs momentum is big, big error. $\endgroup$
    – Mołot
    Aug 18, 2016 at 19:59
  • $\begingroup$ How effective would this be though, really? Would it penetrate and just keep going, doing little damage? And just becasue the weapon produces 2500fps muzzle velocity, doesn't mean that would be the velocity at impact. $\endgroup$ Aug 18, 2016 at 20:04
  • 1
    $\begingroup$ @Mołot You're right that I erred, but it's not momentum per se -- it's actually mass × acceleration that I should have discussed. Will fix this, but ultimately I think the results will stand vis-a-vis this weapon's recoil being comparable to that of a .50 BMG rifle. $\endgroup$
    – Kromey
    Aug 18, 2016 at 22:01

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