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In my sci-fi story there is a man-portable railgun. The gun fires a 4mm-wide tungsten bullet with a muzzle velocity of Mach 23. The inventors of this weapon could have made the projectile faster but were concerned that doing so would cause it to go into orbit and didn't have enough time to fine-tune the velocity. What I'm most worried about is the effects on the nearby atmosphere that such a weapon would cause not only to the individual firing the weapon but also anyone he's fighting along side. Can anyone weigh in on this?

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    $\begingroup$ at that speed something only 2mm across is going to be vaporized almost instantly. $\endgroup$
    – John
    Mar 7, 2023 at 22:03
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    $\begingroup$ @John Hmm, I wonder, though... tungsten melting point is over 3400 C. This answer (and the whole topic there) makes it tempting to think it might work...? Steel has a lower melting point and "estimated 60 km/s" is about mach 174... $\endgroup$
    – JamieB
    Mar 7, 2023 at 22:30
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    $\begingroup$ Speed record for a rail gun is about Mach 8.8, and that thing is a fireball for its entire length. Orbital re-entry speed is in that neighborhood, so I suggest you look at micrometeors. Yes, they would burn up, but I believe they burn up ablatively, ex-gassing in layers, producing something similar to the Leidenfrost effect. $\endgroup$ Mar 7, 2023 at 22:44
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    $\begingroup$ Tungsten rods are great. They slough adiabatically leaving you a similar shaped point to what you started with. Again - length - else it'll vaporise before traveling metres. $\endgroup$ Mar 7, 2023 at 23:15
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    $\begingroup$ Assuming 20 C / 68 F ambient at mach 23, the stagnation temperature is 31,035 C / 44,895 F according to a quick (approximate) calculation. That is about 5.4 times the temperature of the surface of the sun. Not even tungsten is going to survive. The thermal radiation will cook the person firing the gun. $\endgroup$ Mar 7, 2023 at 23:26

2 Answers 2

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You had it right with the 2mm projectile before you edited.

A 20mm wide projectile would have far too much recoil for a man-portable weapon. Let's say it's 30mm long, since it needs to be longer than it is wide. That would give it a mass of 181 grams. By conservation of momentum, if 181 grams goes forward at mach 23 and you weigh 100 kg, you instantly go backwards at 50 km/h. Or in other words, your shoulder is meat pudding and you're going to have to go find where the gun went.

There's a reason why man-portable railguns are often portrayed in fiction as "needlers." If you shoot something much bigger than a needle, you couldn't handle the recoil.

So, 2mm. Say the projectile is roughly a cylinder, 10 cm long and 2 mm diameter. It weighs 6 grams. If 6 grams goes forward at mach 23 and you weigh 100 kg, you go backwards at 1.7 km/h - you'll be knocked a step back with every shot, but it's manageable.

First, would the 2mm projectile vaporize? The short answer is yes - but not before it could hit a target at 100m+ range.

Based on this article, the drag coefficient C_V settles around 0.6 for a properly shaped hypersonic object. Use the formula here to calculate the drag force: 1/2 * (density of air) * (mach 23)^2 * 0.6 * (frontal area of projectile). That gives us 22N of drag force.

Next we need to know how much energy it takes 6g of tungsten to melt. Tungsten melts at 3695 K, so from room temperature we'll say it has to heat up by 3400 K. But once it gets to 3695 K it doesn't instantly melt; there's something called the latent heat of fusion, which means you have to keep dumping more energy into it while it's at 3695 K before it melts. Adding up those effects, it will take 3847 J to melt the tungsten.

So, the drag force is 22N, and we want to know when it has done 3847 J of work on the projectile. Work = Force * Distance, so Distance = Work / Force. 3847 J / 22N = 175 meters. This means that by the time the projectile has gone 175 meters, it could be completely melted, because enough kinetic energy has been dissipated to melt it, if all the lost kinetic energy went into heating the projectile.

In fact, it won't be melted by 175 meters, because much of, perhaps almost all of, the lost kinetic energy will heat the air instead of the projectile. So the projectile could have a range much greater than 175m, especially if it was designed with ablative materials in front of the tungsten, like the Apollo re-entry module.

However, I will proceed on the assumption that it does melt in 175m, simply because it is very difficult to determine how much of the lost kinetic energy actually heats the projectile. It could go ten times that distance. But it will at least go that distance.

It will melt from the front towards the back, like a candle, over those 175 meters. (Presumably it has fins or something to stabilize it through this process.) We can assume that the melted tungsten will immediately stream away from the solid projectile by the force of the air. Once scattered in the air, the melted tungsten will have a much larger frontal area, so its air resistance will be much greater and the liquid will almost immediately come to a stop. As the liquid tungsten stops, it will deposit its energy into the air.

So that means that the high initial energy of the projectile will be dumped into the air at a linear rate, along the entire 175m path of the projectile.

The initial energy of the projectile is 184 kJ. So it will deposit about 1.05 kJ of energy per meter of its flight.

For comparison, an M80 firecracker has around 12 kJ of explosive energy in it. So the projectile will release 1/12 of an M80 firecracker for each meter it travels. That will be loud, but not too dangerous; it won't hurt the operator of the railgun.

Because it is continually melting, the damage the projectile does when it hits depends on the range. If you hit an enemy at close range, it will release most of the 184 kJ, which will do damage comparable to a hand grenade. By the very end of the 175m range, it will be more like a rifle bullet.

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  • $\begingroup$ How fast would heat dissipation be, depending on the current speed (so the current volume of air passing), surface area and heat? That would be valuable info as well. $\endgroup$
    – Demigan
    Mar 8, 2023 at 5:26
  • $\begingroup$ @Demigan that radiant heat dissipation is quite easy to calculate hyperphysics.phy-astr.gsu.edu/hbase/quantum/radfrac.html (the hardest part is probably finding a suitable value for the emissivity of tungsten at such a high temperature). By that I estimate it will be radiating at 3kW once it's up to the melting point. But Mach 23 is nearly 8km/s so it only radiates a fraction of a J for every metre travelled, compared to a kJ lost to ablation. $\endgroup$
    – Chris H
    Mar 8, 2023 at 10:10
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    $\begingroup$ The energy loss won't be spread linearly over 175m. The firer will get a bit of respite because before the material melts and can be ablated, it has to heat up, and most of the work goes into raising the temperature (rather than into latent heat). It would take a full (and rather difficult) thermal model to work out how much the heat spreads back from the tip. Tungsten has a decent thermal conductivity, close to that of aluminium, but the heating is very fast. $\endgroup$
    – Chris H
    Mar 8, 2023 at 10:14
  • $\begingroup$ "If you hit an enemy at close range, it will release most of the 184 kJ" -- erm, I assume that it would overpenetrate. Unless your enemy is a tank. In any case, the projectile will send ripples to run though the enemy's body during traversal. $\endgroup$
    – Klaws
    Mar 8, 2023 at 12:34
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    $\begingroup$ @Klaws No, it won't overpenetrate. We are talking about a hypervelocity impact here where atomic bonds become irrelevant to the impact physics. Basically, both the projectile and the body it hits will act like a bag of atomic balls. In this setting, the penetration depth is always only as deep as it needs to go to encounter enough atoms to match its own mass. These two bullet masses together instantly form a dense, superheated plasma that subsequently expands thermically and rips the target apart. The effect is more that of a meteorite impact than a bullet hole. $\endgroup$ Mar 8, 2023 at 16:19
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2mm projectile?

It's turned to plasma near instantaneously and unless the round is very long, the lack of mass means that it's unlikely to have enough energy from being turned into plasma to do any meaningful damage to anything nearby.

And yes, I do include downrange in that statement.

Speed is good, but mass is also important. I'd suggest increasing the round size to something more like 20mm - and using the friction and speed as a limiting factor: In a standard atmosphere it can travel X distance before it is totally consumed by vaporization.

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    $\begingroup$ You can't just say it turns into plasma instantly without showing any numbers for drag heat. Humans have already had a few vehicles doing mach 20 in (tenuous) atmosphere; there's the HV-2 and there's the Apollo missions on both launch and re-entry. $\endgroup$
    – causative
    Mar 7, 2023 at 23:38
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    $\begingroup$ Meaningless without numbers. How many joules per second is the 2mm projectile getting hit with from drag? If it's 10cm long, how long does it take for that much tungsten to melt when subjected to that heat influx? How far will the projectile go in that time? $\endgroup$
    – causative
    Mar 8, 2023 at 0:00
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    $\begingroup$ You didn't account for the latent heat of tungsten and you didn't actually calculate the drag heat. Well, I just actually did the numbers. The drag coefficient C_D would be about 0.6 for a hypersonic projectile. The drag force on the projectile is 22N, and it takes 3.847 kJ to melt it, which means it could travel about 174 meters before melting. That could be an effective engagement range. This neglects radiative heat loss from the projectile, which would extend the range. $\endgroup$
    – causative
    Mar 8, 2023 at 0:43
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    $\begingroup$ @causative - You are right, I didn't - because 174 Metres at Mach 23 is 'Near Instantaneous' - Thank you for proving me correct. 174 Metres is not an effective Engagement range for a projectile doing Mach 23. It's pointless. The whole reason for a Rail Gun is accuracy over long range (less bullet drop, wind deflection etc.) or hitting fast moving targets. I'd be more worried about being shot with a regular rifle at that distance than this weapon. $\endgroup$ Mar 8, 2023 at 5:02
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    $\begingroup$ @TheDemonLord a smaller projectile would be better? Look at Rods from God. You don’t make wide projectiles, you make them thin and long so it presents the least frontal surface area and generates the smallest amount of heat. Fatter is worse, always go longer $\endgroup$
    – Demigan
    Mar 8, 2023 at 5:07

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