2066

Stationed on Mars to quell a rebellion

Become side door gunner for atmospheric dropship.

No miniguns or gatling cannons, just some metal brick with a pipe on one end.

Get sent in to extract some wounded.

Reach the evac zone and come under attack.

Hoard of rebels charging in with their new plasma guns and compact rocket launchers.

Let loose a stream of bullets.

The sounds of the rebel’s screams are nearly drowned out by the heavy “Kachunk chunk chunk chunk” of the machinegun.

The wounded are loaded up and returned to base.

Inspect MG afterwards.

Thing was made in 1942.

Tunisia, Italy, and Germany are scratched onto the gun.

Scratch “Mars” on with a knife.

Gunpowder is self oxidizing when ignited. As long as they used a non-evaporating lubricant on the actual mechanical components like graphite, and included a barrel liquid cooling loop, the M2 would operate totally fine in space.

See also Brigador: apart from the various lasers, exotic ballistics, and nightmarish chemical weapons it includes, there’s also the prosaic “Mãe Dois.” The tech entry leaves no doubt about what it is:

My understanding is that this weapon not only predates the colonies, but space travel entirely. If that is the case then it’s a truly venerable design, and one I’m told will continue to serve for the foreseeable future.

The year is 2734. Humanity has conquered death, cured every disease (even the common cold), and colonized the stars. Yet on battlefields from Earth to Europa, the M2 Browning still thunders, the BUFF rumbles through alien skies, the M1 Abrams fires its 120mm cannon, and every Marine is issued a KA-BAR.

Once fired, the bullets would just retain their muzzle velocity.

I don’t know what kind of muzzle velocity increase would happen – no air means that they’d be moving somewhat-faster. I’d think that that’d depend on a number of things, probably fluid dynamics and stuff. But maybe you can back-of-the-napkin it by figuring that any acceleration is roughly bounded by the energy required to accelerate the mass of air involved to muzzle velocity. I don’t know exactly how much air that is. Certainly the air inside the barrel, but also some of the air outside the muzzle.

https://en.wikipedia.org/wiki/Density_of_air

Air has a density of approximately 1.225 kg/m 3 (0.0765 lb/cu ft)

https://oow-govmil.com/firearms/50-m2hb-qcb-2/

Barrel Length – 45 inches (114.3 cm);

https://en.wikipedia.org/wiki/.50_BMG

Bullet diameter: 12.98 mm (0.511 in)

So that’s about (.511/2)^2*3.1415*45 =9.22850 in^3, or 0.00015122811 m^3, so .00018525 kg, so 0.1g of mass of air in the barrel.

https://barrett.net/products/accessories/ammunition/50bmg/

Bullet Weight: 661 gr

~15.4g per grain, so ~43 grams. So I figure that the mass of the air in the barrel probably isn’t a huge factor, and I don’t know how to compute the effective amount of air that needs to be accelerated outside the barrel and how much…that’s probably a fluid dynamics question .

Let’s just say that it’s three times that amount of air. Even if so, that’s a pretty miniscule factor compared to the mass of the bullet, like under half a gram. So I figure that the muzzle velocity probably isn’t all that much higher in space.

Are we moving fast enough to do much in terms of orbit change?

https://en.wikipedia.org/wiki/Low_Earth_orbit

The mean orbital velocity needed to maintain a stable low Earth orbit is about 7.8 km/s (4.8 mi/s), which translates to 28,000 km/h (17,000 mph). However, this depends on the exact altitude of the orbit. Calculated for a circular orbit of 200 km (120 mi) the orbital velocity is 7.79 km/s (4.84 mi/s), but for a higher 1,500 km (930 mi) orbit the velocity is reduced to 7.12 km/s (4.42 mi/s).[10] The launch vehicle’s delta-v needed to achieve low Earth orbit starts around 9.4 km/s (5.8 mi/s).

https://www.gd-ots.com/wp-content/uploads/2017/11/M2HB-50-Caliber-Heavy-Machine-Gun.pdf

Muzzle velocity: (M33) 3,050 feet per second (930 meters per second)

Hmm.

https://space.stackexchange.com/questions/15349/how-can-the-delta-v-to-a-specific-altitude-in-earth-orbit-be-calculated

Depending on T/W and other factors, delta V to LEO can vary. But once you’ve reached LEO, gravity loss and atmospheric drag are no longer factors. So I’ll give you some delta Vs going from a 300 km altitude circular orbit to higher altitude orbits.

300 km to 500 km - .11 km/s

300 km to 1000 km - .38 km/s

300 km to 2000 km - .83 km/s

300 km to 4000 km - 1.51 km/s

300 km to 8000 km - 2.37 km/s

300 km to 16000 km - 3.22 km/s

300 km to 32000 km - 3.83 km/s

300 km to 64000 km - 4.1 km/s

300 km to 128000 km - 4.13 km/s

300 km to 256000 km - 4.02 km/2

300 km to 512000 km - 3.87 km/s

300 km to 1024000 km 3.72 km/s

So, the International Space Station is at 340 km. So figure that our astronaut is acting as gunner on a vehicle in LEO at 300 km, can use the above table.

The upper end of LEO is 2000 km.

So you’ve got about 1 km/s in terms of delta V to work with there. So if our astronaut is acting as roof gunner on a Humvee in LEO, that’s actually quite a lot of ability to reach. Given sufficiently-accurate aim, and maybe willingness for a bullet to do a sufficient number of orbits before it collides with a target, he’s got the range to hit anything in low earth orbit.

He can’t hit GPS/Galileo/GLONASS satellites (19k km to 23k km). And he can’t hit escape velocity, so that Model S that Elon Musk launched into an eccentric path wandering the solar system is probably safe. But he can hit a lot of stuff.

Okay, the image is of the Moon. How about acting as a gunner there?

https://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html

Escape velocity: 2.38 km/s

No. From the surface of the Moon, his bullets are going to come back to the Moon. So he can fight lunar battles, but he can’t be engaging targets on other celestial bodies or in their orbit, like the Earth.

reminds me of the best patch I’ve seen in a long time - Shuttle Door Gunner

So… I don’t see that anyone’s brought this up yet so here goes:

If we are talking space space, you know, in orbit, transiting between planets…

You have to account for the recoil of each shot potentially sending your craft into a rotational spin, which can alter its trajectory.

Depending on the mass of your firing platform, you may have to expend a considerable amount of monoprop fuel, or energy in batteries to run multi axis centrifugal stabilizers…so that you maintain your desired trajectory.

Also, even if you are not in space, but on a moon or planet with significantly lower mass than earth… you’re gonna have to do extra work to keep your tripod from dislodging from the ground, even knocking you backward with a sustained burst.

It might not even be practical to do a sustained burst or sweep, it might end up making sense to mod a slower cyclic rate into the thing, jerry rig a Kriss Vector counter weight style contraption to lower recoil.

Basically, the weight of everything involved goes down as G goes down… but the recoil energy remains the same… if you are in 0.5 G, basically, the recoil is twice as intense, as its pushing against half as much weight.

Each single shot would be more precise and travel farther in a vacuum or thin atmosphere and lower G… but sustained fire would be less accurate.

…Or maybe I’m overstating this effect? I’m too tired and lazy to attempt the actual physics calculations for recoil force.

The space shuttles were originally designed with 6 M2s mounted in the wings. It was mostly a tradition thing, but they cut the armament to save weight or something stupid. Fast forward to the Columbia disaster, and where does the fatal foam piece strike? Right in the firing line of old ma deuce.