Spaceship Damage

[Ulzgoroth]

Originally Posted by Fred Brackin

No blowthrough. It's extremely difficult to come up with scenarios where kinetic weapons don't explode on impact (like meteors) rather than remain solid.

Sure, but unless the target is massive or solid enough to stop the impactor (which, if the thing is doing enough base damage for blowthrough to be an issue, it isn't) that explosion will be passing through the ship very fast and out the other side. The damage pattern is more conical than spherical.

[Fred Brackin]

Originally Posted by Ulzgoroth

Sure, but unless the target is massive or solid enough to stop the impactor (which, if the thing is doing enough base damage for blowthrough to be an issue, it isn't) that explosion will be passing through the ship very fast and out the other side. The damage pattern is more conical than spherical.

The explosion is like every gram of the impactor and the thing it hits exploding as if it were some multiple of its' mass in TNT. 20x at 7 miles per second.

A fragmenting projectile might produce a conical pattern. An aggressively vaporizing projectile not so much. Also, higher energy in the energetic plasma mostly means more thermal radiation and not higher particle speed.

It's pretty easy to get KE hits that look like baby nukes. What you're going to see is points of white light.

[johndallman]

Originally Posted by Fred Brackin

It's pretty easy to get KE hits that look like baby nukes. What you're going to see is points of white light.

Iain Banks coined a good term for this: "weapon blink." Brief intense flashes of light that tell you someone's in trouble.

[Ulzgoroth]

Originally Posted by Fred Brackin

The explosion is like every gram of the impactor and the thing it hits exploding as if it were some multiple of its' mass in TNT. 20x at 7 miles per second.

A fragmenting projectile might produce a conical pattern. An aggressively vaporizing projectile not so much. Also, higher energy in the energetic plasma mostly means more thermal radiation and not higher particle speed.

It's pretty easy to get KE hits that look like baby nukes. What you're going to see is points of white light.

How aggressively is it vaporizing? If it's passing through the target at 10 miles per second and exploding outward at 10 miles per second, that's going to be very much a cone. Superficially it seems that the explosion speed should be lower than the initial impact speed, though maybe not.

If the projectile loses a lot of its relative velocity in the collision, that'll give you a nice roughly sphere-ish 'boom' at the point of impact. If it doesn't, it'll give you an explosion that's moving at interplanetary velocities relative to the target as it expands... This won't make the weapon blink go away, it'll just mean that even more of the energy of it doesn't wind up deposited into the target ship.

[Varyon]

Originally Posted by Ulzgoroth

Sure, but unless the target is massive or solid enough to stop the impactor (which, if the thing is doing enough base damage for blowthrough to be an issue, it isn't) that explosion will be passing through the ship very fast and out the other side. The damage pattern is more conical than spherical.

Assuming speed of sound in the material is around 6,000 meters/second (which seems to be appropriate for most militarily-useful metals), that's a bit faster than 1/3rd the rate at which the 10 mps impactor is moving. When it hits, part of its mass and the target's mass explode at roughly that speed, and as it punches through it keeps shedding more of its mass to this effect, doing the same to the target. The explosion of the impactor would maintain the old velocity, for the conical effect, but I don't think this would be the case for the explosion of the target. If you've got an impactor that's large enough to make it all the way through to the other side (which if I'm not mistaken would require an impactor roughly the size of the vessel itself), the explosion of the impactor has the conical character you've noted, but the rest doesn't. We can probably ignore the impactor's explosion, however - with a (say) 10 cm impactor, you've basically transformed a 10 cm channel through the vessel into REF 7 high explosive, then detonated it. Actually, I think the exploding bits of the impactor, seeing as they're moving faster than the speed of sound, would cause more similar explosions, meaning the channel will widen in a similarly conical pattern. If, as is more likely, your impactor isn't large enough to make it all the way through, that conical channel is going to end short of the other end of the target, leaving it with nowhere to go, meaning the full KE of the impactor is going to get transmitted, one way or another, into the target.

Either way, you've got a massive, nasty explosion inside of the vessel itself. I think by the time you hit a point where blowthrough would be a concern, the vessel is going to be so destroyed it doesn't even matter.

(EDIT: In case it isn't clear, I'm assuming here that the explosions from hypervelocity impacts are at the speed of sound in the exploding material. It makes sense, but I don't know if that's what actually happens.)

[Ulzgoroth]

Originally Posted by Varyon

Assuming speed of sound in the material is around 6,000 meters/second (which seems to be appropriate for most militarily-useful metals), that's a bit faster than 1/3rd the rate at which the 10 mps impactor is moving. When it hits, part of its mass and the target's mass explode at roughly that speed, and as it punches through it keeps shedding more of its mass to this effect, doing the same to the target. The explosion of the impactor would maintain the old velocity, for the conical effect, but I don't think this would be the case for the explosion of the target.

I'd characterize the mass as being in three groups with respect to the collision:
-The impactor, which is passing by real fast and exploding due to the impact.
-The parts of the ship not directly hit, which are almost entirely unaffected by the impact itself, as it goes through too fast for force to transfer.
-The parts of the ship that are directly hit and are also exploding.

Now, my thinking is that the first and the last are engaging in something close to an inelastic collision. The ship mass can't really go slower than the impactor mass, because there's no time for it to be pushed laterally out of the way instead of being driven ahead of the bullet.

So effectively post-impact there's one rapidly expanding mass comprising both impactor and hull, moving through with the impactor's initial momentum but the added mass of the hull section.

Originally Posted by Varyon

If you've got an impactor that's large enough to make it all the way through to the other side (which if I'm not mistaken would require an impactor roughly the size of the vessel itself)

This is a mistake - it requires the impactor to have mass comparable to the cross-section of the vessel that it is punching through, not the entire vessel.

Originally Posted by Varyon

(EDIT: In case it isn't clear, I'm assuming here that the explosions from hypervelocity impacts are at the speed of sound in the exploding material. It makes sense, but I don't know if that's what actually happens.)

That might make sense? I'll buy it pending somebody who knows the right physics filling us in, at least. The speed you went with would suggest that conventional or even EM gun rounds would pretty much just go boom, but grav guns or some missiles might not...

[Fred Brackin]

Originally Posted by Ulzgoroth

How aggressively is it vaporizing? .

Um, "No two atoms of the impactor will remain in contact with each other."?

As to inelastic collisions the first collision the first collision between an atom of the impactor and it's opposite number will conserve the vector of the two original atoms but only as a pair and only once you take into consideration energy emitted as photons which can be considerable.

The problem is that atoms will be separated from each other on a scale of the rough order of 10 nanometers or so but a velocity of 10,000 meters per second can also be expressed as 10,000 nanometers per nanosecond. There will be a total number of collisions that will be a really rather large number in a very short period of time.

There might be an overall bias after the first "round" of collisions but there will be thousands of "rounds" of those collisions and any bias gets averaged out. If you were filming the collision at a frame rate that let you see each layer of the impactor and hull annihilating each other the bias in direction you might see would be lateral as the path of least resistance is away from in between the two colliding surfaces. I believe this is why meteor craters are usually wider than they are deep.

So I think you would see what is almost the opposite of blowthrough.

As a final note the I don't believe that the mass of the impacted hull section has to be equal to the impactor. I believe it only has to be enough that the resulting explosion from impact is enough to completely vaporize the impactor.

[Ulzgoroth]

Originally Posted by Fred Brackin

Um, "No two atoms of the impactor will remain in contact with each other."?

As to inelastic collisions the first collision the first collision between an atom of the impactor and it's opposite number will conserve the vector of the two original atoms but only as a pair and only once you take into consideration energy emitted as photons which can be considerable.

The problem is that atoms will be separated from each other on a scale of the rough order of 10 nanometers or so but a velocity of 10,000 meters per second can also be expressed as 10,000 nanometers per nanosecond. There will be a total number of collisions that will be a really rather large number in a very short period of time.

There might be an overall bias after the first "round" of collisions but there will be thousands of "rounds" of those collisions and any bias gets averaged out. If you were filming the collision at a frame rate that let you see each layer of the impactor and hull annihilating each other the bias in direction you might see would be lateral as the path of least resistance is away from in between the two colliding surfaces. I believe this is why meteor craters are usually wider than they are deep.

So I think you would see what is almost the opposite of blowthrough.

As a final note the I don't believe that the mass of the impacted hull section has to be equal to the impactor. I believe it only has to be enough that the resulting explosion from impact is enough to completely vaporize the impactor.

Conservation of momentum means that the stuff emerging from all those collisions is going to have the same momentum as the stuff going in. It's either going to go through the ship, or go until it hits enough ship to slow it down to speeds where material strength of the remaining ship can stop it, regardless of how much exploding it does.

The question about vaporization isn't how thoroughly it'll happen, it's how fast the stuff expands laterally from the point of collision, in terms of actual speed as compared to the speed with which the collision products' center of mass proceeds along the impactor's path.

[Fred Brackin]

Originally Posted by Ulzgoroth

Conservation of momentum means that the stuff emerging from all those collisions is going to have the same momentum as the stuff going in. .

Coming out of each collision you will see those two particles conserve the _average_ momentum and vector of the two particles before the collision. Individual momentum can and will vary quite a bit and there are those losses due to radiated heat. Actual direction of travel(vector) of any individual particle will be quite random.

The next set of collisions will only conserve the momentum and vector of the particles involved in it and not whatever momentum or vector the particles had two collisions ago. Atoms do not have memories. Molecular bonds of a solid object will keep all the particles travelling in the same direction but we're talking very post solid here.

You also don't seem to have grasped the lesson of the meteor craters. They aren't deep and narrow with the atoms involved remembering the momentum of the original meteor. They are wide and shallow with non-random motion following a path of least resistance.

[Ulzgoroth]

Originally Posted by Fred Brackin

Coming out of each collision you will see those two particles conserve the _average_ momentum and vector of the two particles before the collision. Individual momentum can and will vary quite a bit and there are those losses due to radiated heat. Actual direction of travel(vector) of any individual particle will be quite random.

The next set of collisions will only conserve the momentum and vector of the particles involved in it and not whatever momentum or vector the particles had two collisions ago. Atoms do not have memories. Molecular bonds of a solid object will keep all the particles travelling in the same direction but we're talking very post solid here.

Momentum is a vector, and total momentum is conserved at all points throughout the entire event. Atoms don't have memory, but that doesn't let you ignore basic conservation laws.

The momentum the impactor brings into the picture all has to either be transferred to the ship as a whole, or carried away on material that blows through. It can't be redirected, that would fail to conserve the momentum vector.

Originally Posted by Fred Brackin

You also don't seem to have grasped the lesson of the meteor craters. They aren't deep and narrow with the atoms involved remembering the momentum of the original meteor. They are wide and shallow with non-random motion following a path of least resistance.

Meteor craters are what you get when the impactor encounters an effectively infinite thickness of material. That changes things considerably.