Re: Spaceship Damage
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Re: Spaceship Damage
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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. |
Re: Spaceship Damage
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Re: Spaceship Damage
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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. |
Re: Spaceship Damage
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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.) |
Re: Spaceship Damage
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-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. Quote:
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Re: Spaceship Damage
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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. |
Re: Spaceship Damage
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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. |
Re: Spaceship Damage
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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. |
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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. Quote:
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