Specific situation question: I have a ship with a 'whipple' shield consisting of 5cm of titanium (or a material similar to it) spaced about 50 meters away from the main hull (which is itself pretty durable, with a much lighter whipple shield to protect from sand-grain size impacts), and that shield gets struck by a tungsten projectile massing about 250kg with a velocity of about 60 km/s. At that speed, it has kinetic energy about equal to 100 tons of TNT. Let's assume the tungsten projectile is about ten times as long as it is wide, which would make it 'about' 10 cm in diameter and about a meter long.

Would the main shield have enough mass and thickness to vaporize the projectile completely, allowing for the secondary shield to handle the debris, or would the impact only destroy the front part of the projectile, allowing the remainder to go on through to the ship?

  • $\begingroup$ To answer exactly, need more engineering knowledge than I have, but a rough answer would require discovering if the round is vaporized by the impact. At 60km/s shield strength is irrelevant; it's all about the masses involved during the impact (round mass vs. shield affected area mass). This means that more mass in the form of weak materials like ice would be better at vaporizing the round. If energy during impact is larger than is required to vaporize the titanium round, the round is now a cloud of plasma, which even at high speeds will probably be stopped by your secondary shield. $\endgroup$ Commented Sep 20, 2019 at 8:08
  • $\begingroup$ At 60km/s, that round is traveling 134216 mph. For modern artillery, 250 kg is roughly 5 times the mass of current 155mm (~95kg) howitzer round, and those have a rough maximum range of 30 km (depending on cannon and charge). The greater the range, the more reaction time to dodge, however, the launch vessel...what's happened to it? Suppose a cannon 6km in length, then you have 0.2 secs to accelerate (average on constant accel is 30km/s velocity), which gives accel as 300 km/s^2, and thus a launch force of 75,000,000 Newtons or 16,860,700 lbs. Suicide... $\endgroup$ Commented Sep 30, 2019 at 13:03

4 Answers 4


To pull up an old but useful formula derived from work on shaped charge jets penetrating tank armour: $$P = L\sqrt{\frac{\rho_j}{\rho_t}}$$

$P$ is the penetration depth, $L$ is the length of the penetrator, $\rho_j$ and $\rho_t$ are the densities of the penetrator and target respectively. Note that this is different from the classic Newtonian penetrator model, because in this case the penetrator is travelling so fast that impact forces will easily overcome any intermolecular bonds and so both the penetrator and armour can be treated as fluids.

Anyway. If you want to stop a metre long projectile made of tungsten, one way to do this would be to have a plate (or multiple plates) of tungsten armour with a total thickness of a little over a metres, then some spacing, then some additional shielding to mop up the high-velocity fragments. If you want less dense armour, such as aluminium, you'll need to increase your armour thickness by $\sqrt{19.25/2.7}$ or 2.6 times. Your 5cm of titanium (twice as dense as aluminium, but far below tungsten) will knock off the front 25mm of the projectile, and all the rest will pass through.


Having read a little more into this, it seems that there has been some thought about the explosive effect of the energy released in this sort of collision. The impact will produce a certain amount of sideways-splattering of the impactor, and a certain amount of damage will propagate up the impactor too. What I've found seems very handwavey, so take this with a small pinch of salt.

We can approximate the volume of the crater carved out by an impact as $V_c = E_p/S_c$ 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. We can imagine your rod to be stationary, with a 10cm wide, 5cm deep cylindrical projectile of titanium striking it. That much titanium weighs 1.77kg, and has a kinetic energy of about 3.2GJ. This gives us a crater volume of about 1.47m3 and assuming this is basically spherical, a crater radius of about 34cm. That's quite a bit more than the 2.5cm the hydrodynamic approximation gave us, which given the huge amount of energy involved isn't really surprising.

What it isn't, however, is enough to blow the whole rod to pieces. The rear two-thirds of the impactor will remain intact and will just keep on trucking, and so absolutely ruin the day of anyone on board the ship.

The extreme spacing of your armour would work against non-solid projectiles (like modern shaped-charge HEAT rounds) because the jet won't remain together over that distance. This isn't necessarily true of a solid tungsten rod though, which will have its tip ablated off but might remain basically intact over that 50m span and then, in all likelihood, tear a huge hole in your ship.

Note that even if the armour did disrupt the projectile, it would still only save you if you had multiple layers of armour of substantial thickness. You've still got most of the 250kg projectile flying towards you at 60km/s, and armour that is intended to "protect from sand-grain size impacts" will absolutely not be up to the task and you'll get totally mangled.

Now, note that if this armour was capable of disrupting the projectile (and I suspect that it is not), then the simplest countermeasure from the attacker's point of view is to fire multiple smaller projectiles, slightly separated along their trajectory. By breaking the single massive round into 10 cylinders, each 10cm wide and tall, it is possible for successive penetrators to travel through the hole left by the penetrator just in front of them. Such a projectile could reasonably punch through 9 layers of armour, defeat clever reactive armour, and deliver a serious punch to the vessel inside.

  • $\begingroup$ It should not be too tough to see something this this incoming. I think successful defense against a proejctile like that described here would have to go to work on the projectile while it was still at a good distance from the ship. AEGIS-style. $\endgroup$
    – Willk
    Commented Sep 20, 2019 at 12:07
  • $\begingroup$ @Willk certainly, it isn't the most efficient use of a kinetic energy weapon, but there are many ways it could be modified to make it substantially harder to intercept (mostly involving shooting multiple projectiles instead of one big one) $\endgroup$ Commented Sep 20, 2019 at 12:27
  • $\begingroup$ I will stick to one big one, but its cross-sectional area is 0.5mm to make it hard to see with radar. $\endgroup$
    – Willk
    Commented Sep 20, 2019 at 19:47

60 km/s is so high, that you can neglect any inter-atom bounds and thermal movement and consider both armor and missile as a set of independent atoms. At first stages of impact missile atoms would pass through atoms of armor. Then scattering of tungsten atoms on tungsten atoms begins. You just can't call it evaporation - it would be an understatement.

Since the materials are the same - scattering would be on an "atom for an atom" basis. So only this 5 cm would be "scattered off" this tungsten rod and 95cm of it would still hit main hull. In that hull this rod would travel at most "95 tungsten-equvalent cm" (it would be twice more for steel) before it all "scatters out".

Thats all means that 5% of rod energy would be released at shield (as 5t TNT explosion) and 95% of energy would be released at and in the hull (95t of TNT explosion).

UPD: the best defence aginst this rods would be counterintuitive: if you make you ship out of thin aluminium with total width in a path of the rod to be about 5mm, only about 0.1% of a rod would deliver energy (100 kg of TNT - but it would be spraded between each surface) and it would just fly through, leaving hole about meters across. Which is much better than almost nuclear-scale explosion inside.

  • 1
    $\begingroup$ While your update raises an interesting point, unfortunately, it is wrong. Ship is not an empty hull, there is stuff inside. And an explosion inside the ship as the rod hits the stuff would be worse than explosion at the hull. $\endgroup$
    – Alice
    Commented Sep 20, 2019 at 9:39
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    $\begingroup$ @Alice, unless you have a hull of a spaceship more then meter of tungsten thick, explosion in both cases would be inside. But if you keep everything made of light and thin materials (wich makes a great sense for spaceship regadless of hypervelocity projectiles) explosion would be much less intensive. $\endgroup$
    – ksbes
    Commented Sep 20, 2019 at 10:02
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    $\begingroup$ For some reason this reminds me of old canvas winged planes. The bullets go in, come out, and unless they hit anything vital the plane keeps flying. $\endgroup$
    – Joe Bloggs
    Commented Sep 20, 2019 at 11:10
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    $\begingroup$ Add in redundancy in your ship systems and you could be riddled with holes, leaking atmosphere, and still be better off than an armoured behemoth that just got hit with the full whack of kinetic energy. $\endgroup$
    – Joe Bloggs
    Commented Sep 20, 2019 at 11:11
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    $\begingroup$ The aluminum solution then runs into the "high velocity sand" problem; fire a shotgun instead of a penetrator round. 1,000 5 mm aluminium pellets at that speed deliver 100 T of TNT (collectively) to the aluminium ship. $\endgroup$
    – Yakk
    Commented Sep 20, 2019 at 17:16

As many other posters pointed out, the Whipple Shield isn't going to do much against a vary large, dense projectile. Its purpose is to absorb the impact of very small objects like dust grains or micrometeors.

However, it is possible to take this principle and apply it as a form of active armour. Rather than a fixed plate, the ship can carry batteries of small rockets and an active radar system which fires the rockets at the incoming projectile. Each rocket upon launch can deploy an umbrella-like Whipple Shield and manouevres in position in front of the projectile, and the entire flock of rocket Whipple Shields will arrange themselves in a line, so the projectile will end up flying through multiple layers of shielding.

The desired outcome is the projectile breaks apart after multiple impacts and the smaller pieces are either absorbed by the terminal armour plating of the ship, or fly past harmlessly.

The entire arrangement would have to be somewhat like the Israeli "Iron Dome" system, capable of tracking incoming rounds and only launching when it calculates that the projectile will actually impact the ship. This conserves ammunition and also adds uncertainty for the aggressor, they will not be able to clearly determine if the system has expended all ammunition or not, and then must carry and fire additional rounds at every target in order to ensure they can overcome it. Since there are multiple layers of defense in a space battle, from lasers to ECM to counter missiles to terminal defense, the enemy spaceship will either run out of rounds, or the expense of building additional spacecraft and missiles will mean they have to give up some other capability (maybe in the larger civilian economy).

The primary purpose, then, isn't to defend the ship, but rather induce enough uncertainty in the adversary's tactical, operational and strategic calculations that they are deterred from attempting aggression in the first place.

  • $\begingroup$ Given that the incoming projectiles are quite dumb, using expensive and expendable active defenses against them is a losing proposition in a war of attrition. $\endgroup$ Commented Sep 20, 2019 at 14:22
  • $\begingroup$ This is actually pretty much my plan already, I was just wondering if I could make any form of passive defense practical. Also, the interceptor round doesn't have to be particularly expensive. It could be fired from a railgun at 2 or 3 km/s for instance, designed to expand out into an umbrella shape after firing. $\endgroup$ Commented Sep 20, 2019 at 15:35
  • $\begingroup$ This is essentially Explosive Reactive Armour transposed into a space environment, but because of the difference in scales (speeds are measured in kilometers per second and impact energies are measured in tonnes or kilotonnes of TNT), you need to expand the idea greatly. $\endgroup$
    – Thucydides
    Commented Sep 20, 2019 at 19:40
  • $\begingroup$ @Thucydides No. Explosive Reactive Armor explodes on impact. The 60km/s tungsten rod needs to be destroyed when it's still far, far away from the ship. To be effective, an interceptor needs to hit the impactor when its still a few tens of kilometers away, or the debris cloud will be deadly. Which means, given the probable speed differences, that the interceptor needs to be fired while the impactor is still hundreds of kilometers away. Good luck trying to a) detect the impactor in time, b) aim and fire the interceptor in time, and c) actually hitting the impactor with the interceptor... $\endgroup$ Commented Oct 1, 2019 at 3:53
  • $\begingroup$ Please re read the comment. ERA works by disrupting the jet while it is being formed. The change in scale means you need rockets or similar devices to place your disruptor ahead of the incoming round. $\endgroup$
    – Thucydides
    Commented Oct 1, 2019 at 15:59

What about the effects of adding an electromagnetic charge to the shield? While it might not do much to negate that kinetic energy, maybe it could deflect the rod or its fragments in harmless directions...

  • $\begingroup$ This is actually similar to what I was planning on doing. I don't think a magnetic shield would protect from the rod itself... but if the rod has been vaporized by the impact with the 'whipple' shield, then what continues on to strike the ship is basically plasma, and in theory can be deflected by a strong enough magnetic shield. But it would require the rod to be completely vaporized first, which doesn't sound likely. $\endgroup$ Commented Sep 22, 2019 at 0:51

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