# Destroying Hypervelocity Rounds With More Hypervelocity Rounds

In a previous question, I asked about using a scaled-up whipple shield to protect against a 250kg tungsten round traveling at about 60 km/s. The round in question is 'about' 10 cm in diameter and 'about' 1 meter long. The conclusion was that, no, the whipple shield would be ineffective because it would only shave off a small part of the tungsten round, which would continue on mostly unscathed to the main hull, fair enough. But what happens if you hit the incoming round with another fairly sizable shot going the other direction?

Say you fire a 10kg steel round at the incoming tungsten round at a comparatively paltry 3 km/s. The combined velocity is about 63 km/s, and by my back-of-the-envelope calculations the impact should release more than ten times the energy needed to vaporize the tungsten round (if the numbers I gave aren't actually quite enough, assume that the mass and velocity of the interceptor round are sufficient to get about this value for kinetic energy). The steel round is gone completely... but is the tungsten round? Will this impact just shave off about 10kg from the tungsten round and leave the rest to continue on to target? Will the tungsten round be vaporized? Will it be blown to fist-sized chunks?

What exactly is left of the incoming round makes a big difference to the target being shot at. A cloud of plasma can be dealt with by a magnetic shield; a cloud of dust can be dealt with by a whipple shield. Fist-sized chunks? You need some pretty hefty armor. Mostly-intact round? Good luck.

• I'm not going to claim this is an entire answer, but here's a simulation of a collision at 15 km/s. Sep 29 '19 at 22:54
• @CortAmmon Preeeeeeety…. And informative! :D Sep 29 '19 at 22:56
• Are we only meant to focus on the actual impact and debris? or the plausibility of intercepting a Hypervelocity round in the first place? We can't even intercept large missiles in the atmosphere reliably... Sep 29 '19 at 23:01
• @Shadowzee Let's assume the problem of hitting something with the cross-sectional area of your fist moving at 60km/s has been solved. We're just talking about what happens after impact. Sep 29 '19 at 23:04
• Instead of hitting the incoming projectile at the dead-center, why not hit it at an angle, this way, even if 100% of the projectile survive the impact, its course will be altered and may miss your ship by miles. Sep 30 '19 at 4:50

In space conditions, tungsten is very brittle. Even monocrystalline tungsten is likely to shatter completely upon impact. This is not so bad for an impactor, unless you're shooting a hardened target (actually it might be desirable, also because tungsten is pyrophoric and a cloud of tungsten shrapnel in a pressurized starship's oxygen atmosphere will supply a decent imitation of a thermobaric bomb).

So if you can hit it, I don't expect pure tungsten impactors to be a problem at all. They will disgregate into dust, and that dust will probably miss the ship by a wide margin.

The impactor might then be made of impactor alloys (tungsten-nickel-iron with 90% of tungsten, or tungsten carbide-doped porous matrix of tungsten - nickel - iron - cobalt with 90% tungsten). The latter has shown a much improved penetrating power against hardened targets, which translates into good resistance to impacts from iron counterimpactors.

The likely outcome is that the front part of the impactor explodes, while the rear section remains mostly unscathed. Even so, in almost all circumstances the projectile will be subject to a significant lateral thrust, which at any realistic distance should ensure it will miss the target.

Much depends about the distance the impactor gets intercepted at.

The impact would surely alter the impactor's trajectory even if it does nothing else; even a small deflection might spell the difference between an impact and the round flying harmlessly by, missing the target. And at a sufficient distance (say, 600 kilometers), a second counterimpactor could maybe still hit the projectile while still .3 seconds from impact.

(I don't know whether a "layered" impactor - an impactor that has been thinned in two points along its length, to divide it into three sections - could be expected to fare better; whether the collision thrust would shatter the link between the sections, leaving the back section almost unchanged in its trajectory while the foremost disintegrated. Better a third of the projectile to make it to the target, than nothing. Probably any attacking ship would have several kinds of impactors to choose from).

• In a situation with point defence, an attacker would be better off firing many smaller projectiles in an effort to saturate the point defence than to try to develop and implement a larger projectile that can tolerate a point-defense hit. Sep 30 '19 at 0:06
• I'm not sure that the physics at play when talking about making better modern-day impactors will apply when talking about rounds going at 60 km/s. Modern-day railguns get you about 3 km/s... and this one has 400 times as much kinetic energy per kilogram. Sep 30 '19 at 0:33
• @FlyingLemmingSoup absent other factors such as atmospheric attrition, the physics should hold at least up to the first percents of the speed of light. I think I've read that beyond 1-2 km/s, any impactor mostly already behaves as a liquid, while at relativistic speeds, to misquote XKCD's Randall Munroe, "you basically have a blob of particles flying in close formation" (i.e., the nature of the thing ceases to matter much). Sep 30 '19 at 6:32
• Exactly as @LSerni says: The end of the rod won't even notice that the tip has been vaporized, it'll pass through the cloud of steel debris before the pressure wave that is traveling up the rod has reached it. And even if the rod did shatter, you still have a blob of fragments flying in close formation with negligible lateral motion. 60km/s is really damn fast, you'd have to hit it while it's still a few tens of kilometers away to give the cloud of fragments to spread enough to get less dangerous. Sep 30 '19 at 17:42
• Btw: $250kg\cdot \frac{60000^2m^2}{2s^2} = 450GJ$ This is about 110 tons of TNT. It won't matter if that is spread across an area of one or ten meters, you'd need to spread it at least into the 100m range to have any hope of absorbing it gracefully. Sep 30 '19 at 17:50

Say you fire a 10kg steel round at the incoming tungsten round at a comparatively paltry 3 km/s

This is no different to having a static steel Whipple shield, and an incoming projectile travelling at 63km/s. The problem is still that hypervelocity impacts do not behave like low velocity ones, and as a result the effects of collisions are counterintuitive. This is why more sedate collisions (say, 15km/s) are modelled as jets of fluid splashing off each other, for example... intermolecular bonds don't mean anything given the impact forces involved.

This means that all the blurb below is just as applicable to whipple shields as it is to interceptor projectiles.

Will the tungsten round be vaporized? Will it be blown to fist-sized chunks?

It isn't entirely clear what will happen to it. Worst case scenario is that a short length of the impactor is ablated off, and the rest just carried on coming. Using the same hydrodynamic penetration approximation as I did last time... if your steel projectile has the same diameter as the incoming projectile it'll be 4cm thick, and ablated about 2.5cm off the front of the impact. That's not so good, from the target's point of view.

I've recently been reading up on "crater strength", a notion that handles the expansion of a crater in a solid object. I'm not entirely certain how this applies to hypervelocity impactors... it doesn't gel nicely with the idea of hydrodynamic jets, that's for sure, but does handle the notion of something exploding, as you might expect releasing an awful lot of energy in a short period of time. It was also suggested by Luke Campbell, who knows a fair bit more about this sort of thing than I do, and has put more thought into it. So with all that said, take this with a small pinch of salt.

The cratering approximation define crater volume $$V_c = E_p/S_c$$ where $$E_p$$ is the kinetic energy of the projectile and $$S_c$$ is the cratering strength of the material involved, handwaved to be three times its yield strength. The yield strength of tungsten is 750MPa, so its cratering strength is defined as 2.25GJ/m3. If we imaging the tungsten impactor as stationary, and the steel interceptor is coming in at 63km/s, it'll have a kinetic energy of nearly 20GJ. That gives a cratering volume of 8.82m3, and hence a cratering depth (as defined as the radius of a sphere with that crater volume) of about 2.1m.

With that approximation, the impactor is indeed blown to pieces. Hooray! (it also suggests your whipple shielding in the previous question is better than originally anticipated, so I'll revisit that at some point).

However.

Given the hydrodynamic penetration depth assumption, the steel interceptor will be more or less "used up" in the front couple of centimetres of the impactor. The energy of the collision must, therefore, be transmitted along the impactor by plain old atoms bumping together. If that happens at the speed of sound in tungsten, 5.2km/s, it'll take 1/5200th of a second for the impactor to fully disintegrate, during which time it will have travelled 11m. If the interceptor hits the impactor any closer to the ship than that, you're still in big trouble (this also suggests your 50m whipple shielding gap in your previous answer is probably a sensible spacing). I'm not sure what speed the debris will expand at, as that requires working out energy budgets and things and I'm feeling too lazy for that. The speed at which the projectile expands should inform you as to how far away you must intercept the projectile, and how much armour you'll need to mop up the bits.

Next, there's a lot of momentum in a quarter-tonne slug travelling at 60km/s. Your little steel projectile may deliver enough energy to break it up, but those bits are going to be quite big and will maintain most of their original speed and direction. This isn't so much "spalling" as a terrifying shotgun blast of doom. Not as much doom as a metre long rod of tungsten, but you might still need to deal with a cloud of fist-sized 60km/s projectiles (it won't matter if they're solid or melted, incidentally).

A cloud of plasma can be dealt with by a magnetic shield; a cloud of dust can be dealt with by a whipple shield. Fist-sized chunks? You need some pretty hefty armor. Mostly-intact round? Good luck.

I don't think it will be dust. And remember, even if it was dust, a quarter tonne of dust travelling at 60km/s requires a pretty substantial whipple shield, and that whipple shield will have a pretty substantial hole in it afterwards.

As for dealing with plasma via a magnetic shield... again, the remnants of the impactor will have considerable momentum, and you have a short period of time in which to exert a considerable force on it in order to deflect it. I'm not going to try and work this out here (magnet maths is hard :-() but it sounds a little dubious.

Finally though, let's revisit those fist sized chunks. At the end of my last answer, I said this:

the simplest countermeasure from the attacker's point of view is to fire multiple smaller projectiles, slightly separated along their trajectory.

This still holds true. Colinear kinetic penetrators will be excellent at defeating defense in depth, though sufficient interceptor railgun fire might be able to swat all the projectiles before they reach the target. The other alternative, many long, slender projectiles (a sort of parallel impactor instead of serial) would require one railgun round each, which may very very quickly overwhelm any plausible defense. A combination of the two approaches sacrifices the sheer one-hit-kill ability of the monolithic tungsten slug for a huge cloud of kinetic death which will be impractical to defend against.

• Actually, considering that the impactor has a kinetic energy of roughly 110 tons of TNT, I think it may actually be safer to not shatter the impactor: If the impactor is whole, it'll just go through the ship at one point, vaporising all in its path, but leaving the rest of the ship intact. If it's spread out, the effect is more like a small nuke detonating next to the ship. Sep 30 '19 at 18:03
• @cmaster ksbes suggested much the same thing in their answer to the last question. You might be OK so long as the trajectory doesn't come too near anything too important, but there's not much scope for "useless space" on a warship. A small nuke going off outside is preferable to one on the inside! Sep 30 '19 at 18:15
• @StarfishPrime: The answers to the last question actually set me off designing a world of space battleships that we’re almost entirely useless space: Kilometer long webs of rapidly replaceable rails with mounted, mobile, multiply redundant pods full of important equipment (or maybe just decoys). The idea being that even if a small nuke did go off ‘inside’ the ship it wasn’t really that big a deal. Oct 1 '19 at 18:36
• @JoeBloggs big, soft ships are no tougher than small, hard ones. Larger numbers of smaller penetrators can be used (eg. by adding a bursting charge to a big kinetic projectile) and casaba howitzers and bomb-pumped lasers can be used from further away because they need to punch through less tough stuff. You can't scale up indefinitely, because you'll have problems turning your ship without the chassis warping or breaking, limiting your manoeverability, too. Oct 1 '19 at 18:45
• @StarfishPrime: Yep! Fun bit about that ship design is that it’s a soft, replaceable, easy to traverse (even when wreckage) framework on which a series of hard nodes crawl. Weapons effective against soft targets aren’t useful against the worrying bits (enemy weapons etc), and the hard nodes can move around enough to dodge without endlessly wasting reaction mass. One of the constraints that leads to such a design is highly limited delta V requiring ground pumped mass acceleration. Basically the battleship gets launched in a compact bundle then unpacked. It’s space warfare by way of Ikea. Oct 1 '19 at 20:43

Try not to hit the incoming projectile dead-on. If you hit it slightly off-axis, you will nudge the trajectory slightly. And then the incoming round will miss.

This is similar to some asteroid defense proposals.

• That would require the intercept to take place several dozen kilometers short of the target. Definitely viable, but the scenario I'm thinking of here is more of last-ditch defensive measure using dumbfire railgun projectiles, which aren't likely to score that kind of hit. Sep 30 '19 at 6:32
• @FlyingLemmingSoup if the counter-shot is spherical, the probability of a dead-on center impact is negligible. At ±20° (as the near-normal incidence) will define a cross-section of about 12% from the total sphere cross-section (sin(20°)²) = 12% chances of a near-normal incidence. To favor a deflecting behavior, use a hail of conic counter-impactors a cone - a null dead-on impact cross-section area. Sep 30 '19 at 12:00
• You will almost certainly not knock it off its axis. Hypervelocity impacts don't work like that. If you can hit the projectile in the side then you will seriously reduce its effectiveness, but that's a tricky proposition, to say the least. Sep 30 '19 at 12:09
• if you can hit it from the side, it's not aimed at you.
– ths
Sep 30 '19 at 18:11
• @FlyingLemmingSoup given the potential variety of projectile types and sizes and fire rates, it isn't obvious that there's any point trying to use any kind of method of shooting down incoming dumb projectiles. Oct 1 '19 at 7:36