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Post by luxardens on Apr 4, 2019 17:02:47 GMT
Long story short: I was contemplating what the minimum thickness of steel shell would be to protect a human against a 50 Mt nuke at point blank. I got the thermal, neutron and gamma radiation aspects covered pretty well, but I realized one of the biggest problems to estimating structural requirements and inertial requirements (human inside may not be squished against the walls because of the blast acceleration) was knowing what the force/impulse even looked like.
I figured it's probably easiest to estimate an average speed of ejected material (assuming this is in space), and multiply that by the total ejected mass, then factor in the solid angle that the shield takes up (just multiply by 0.5). However, I'm completely clueless on what the average speed of ejected material could be... So how would you guys estimate the average speed? Or is there an easier way to estimate the momentum change from a point-blank nuke? And is radiation pressure going to be relevant or am I right to forget it?
Bonus discussion: if we go beyond back-of-the-envelope, how is this typically calculated when designing Orion drives and such?
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Post by matterbeam on Apr 4, 2019 22:34:19 GMT
The material ejected from the nuke can be dismissed. What is important is the steel vaporized by the detonation and acting like a rocket on the rest of the steel. We need to know how big the target is to continue.
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Post by luxardens on Apr 6, 2019 14:42:47 GMT
The material ejected from the nuke can be dismissed. What is important is the steel vaporized by the detonation and acting like a rocket on the rest of the steel. We need to know how big the target is to continue. Oh wow, even for a 50 Mt nuke? Would that also hold inside an atmosphere, with air ejected along the same path? Anyway, supposing a target made of low alloyed/carbon steel (0.25% carbon, trace amounts Ni-Cr-Mo-V), of 10 by 10 meters surface area. Thickness can be assumed to be arbitrarily large as far as heat capacity and material available for ablation and such go. |
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Post by airc777 on Apr 6, 2019 22:59:33 GMT
Perhaps this is relevant?
How 'point blank' is 'point blank'? If we're physically touching it then I'm assuming we're also contending with some sort of yottapascal pressure wave propagating through the shield? I'm assuming we've layered our composite armor in order of ablative thermal layer first, then the structural layer(s), then the radiation layer(s), then the inertia dampener?
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Post by AtomHeartDragon on Apr 7, 2019 10:51:24 GMT
The material ejected from the nuke can be dismissed. What is important is the steel vaporized by the detonation and acting like a rocket on the rest of the steel. We need to know how big the target is to continue. Oh wow, even for a 50 Mt nuke? Would that also hold inside an atmosphere, with air ejected along the same path? Anyway, supposing a target made of low alloyed/carbon steel (0.25% carbon, trace amounts Ni-Cr-Mo-V), of 10 by 10 meters surface area. Thickness can be assumed to be arbitrarily large as far as heat capacity and material available for ablation and such go. Inside the atmosphere the effects are dominated by blast wave - all that energy that gets dumped into huge mass of air can confer a lot more momentum - energy is v2, but momentum is only v1.
Outside the atmosphere, you only have the mass of the nuke itself and ablating target.
The nuke is likely going to be negligible, but if you know the temperature it achieves (averaged over the whole nuke) and mass of the whole device you can calculate how fast the material will be going, and how much of it there will be and assume it forms an uniform, expanding spherical shell, so target will be hit depending on solid angle it occupies from nuke's perspective (at point blank it will indeed be 0.5).
The target is trickier. You will need to calculate depth of the vaporization and velocity of the ejecta. I don't think heat conduction will make much of a difference at timescales involved, but there will be many, tricky modes of energy transfer acting there. The momentum transfer also depends on elasticity of collisions. To complicate matter further, your heating won't be uniform at point blank (the ground zero will effectively be at nuke's temperature, dropping off with distance), heat of vaporization matters and you might have fun secondary effects like spallation ejecting whole chunks of non-vaporized matter greatly increasing the impulse (you might also get spallation on the opposite side partly negating the impulse, though of course it's going to be bad news for anyone there, and if there is anything beyond, the negation will only last till the fragments hit the opposite side, spallation is also highly geometry dependant).
Overall the required calculation is way above my skills and pay grade, but you can try estimating from both sides - indestructium plate only showered by elastically bouncing nuke material VS whole plate gets ejected with total absorbed energy as kinetic energy and narrowing down from there.
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Post by luxardens on Apr 8, 2019 10:10:57 GMT
Perhaps this is relevant?
How 'point blank' is 'point blank'? If we're physically touching it then I'm assuming we're also contending with some sort of yottapascal pressure wave propagating through the shield? I'm assuming we've layered our composite armor in order of ablative thermal layer first, then the structural layer(s), then the radiation layer(s), then the inertia dampener?
Nope, but it was interesting nonetheless! Point blank as in: the bomb literally touches the shield before it goes off. I was doing the estimation for a non-composite, non-spaced armor, just to make calculations easier. |
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Post by luxardens on Apr 8, 2019 10:37:16 GMT
Outside the atmosphere, you only have the mass of the nuke itself and ablating target.
The nuke is likely going to be negligible, but if you know the temperature it achieves (averaged over the whole nuke) and mass of the whole device you can calculate how fast the material will be going, and how much of it there will be and assume it forms an uniform, expanding spherical shell, so target will be hit depending on solid angle it occupies from nuke's perspective (at point blank it will indeed be 0.5).
The target is trickier. You will need to calculate depth of the vaporization and velocity of the ejecta. I don't think heat conduction will make much of a difference at timescales involved, but there will be many, tricky modes of energy transfer acting there. The momentum transfer also depends on elasticity of collisions. To complicate matter further, your heating won't be uniform at point blank (the ground zero will effectively be at nuke's temperature, dropping off with distance), heat of vaporization matters and you might have fun secondary effects like spallation ejecting whole chunks of non-vaporized matter greatly increasing the impulse (you might also get spallation on the opposite side partly negating the impulse, though of course it's going to be bad news for anyone there, and if there is anything beyond, the negation will only last till the fragments hit the opposite side, spallation is also highly geometry dependant).
Overall the required calculation is way above my skills and pay grade, but you can try estimating from both sides - indestructium plate only showered by elastically bouncing nuke material VS whole plate gets ejected with total absorbed energy as kinetic energy and narrowing down from there.
Ah temperature! I hadn't thought of that. That's useful. So for a very inefficient 50 Mt nuke composed of 50 tons of uranium, we have 2E17 Joules. Assuming heat capacity is constant at 2000 KJ/KgK, average temperature would be only 2,000,000 Kelvin and velocity would be: V = sqrt(2 * Kb * T / m). If m is the mass of a uranium atom, then V = 11.8 km/s and imparted momentum is on the order of 300,000,000 kg*s ; probably a fair bit less, so the armor would have to weigh on the order of >1E7 kg or more to prevent ridiculous accelerations. That's... pretty tough. Though a lighter nuke would probably be a lot better. But yea, spalling is not something I'm going to do in a couple of minutes. At least I got one more order of magnitude estimate down. Thanks! |
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Post by AtomHeartDragon on Apr 8, 2019 20:56:52 GMT
Outside the atmosphere, you only have the mass of the nuke itself and ablating target.
The nuke is likely going to be negligible, but if you know the temperature it achieves (averaged over the whole nuke) and mass of the whole device you can calculate how fast the material will be going, and how much of it there will be and assume it forms an uniform, expanding spherical shell, so target will be hit depending on solid angle it occupies from nuke's perspective (at point blank it will indeed be 0.5).
The target is trickier. You will need to calculate depth of the vaporization and velocity of the ejecta. I don't think heat conduction will make much of a difference at timescales involved, but there will be many, tricky modes of energy transfer acting there. The momentum transfer also depends on elasticity of collisions. To complicate matter further, your heating won't be uniform at point blank (the ground zero will effectively be at nuke's temperature, dropping off with distance), heat of vaporization matters and you might have fun secondary effects like spallation ejecting whole chunks of non-vaporized matter greatly increasing the impulse (you might also get spallation on the opposite side partly negating the impulse, though of course it's going to be bad news for anyone there, and if there is anything beyond, the negation will only last till the fragments hit the opposite side, spallation is also highly geometry dependant).
Overall the required calculation is way above my skills and pay grade, but you can try estimating from both sides - indestructium plate only showered by elastically bouncing nuke material VS whole plate gets ejected with total absorbed energy as kinetic energy and narrowing down from there.
Ah temperature! I hadn't thought of that. That's useful. So for a very inefficient 50 Mt nuke composed of 50 tons of uranium, we have 2E17 Joules. Assuming heat capacity is constant at 2000 KJ/KgK, average temperature would be only 2,000,000 Kelvin and velocity would be: V = sqrt(2 * Kb * T / m). If m is the mass of a uranium atom, then V = 11.8 km/s and imparted momentum is on the order of 300,000,000 kg*s ; probably a fair bit less, so the armor would have to weigh on the order of >1E7 kg or more to prevent ridiculous accelerations. That's... pretty tough. Though a lighter nuke would probably be a lot better. But yea, spalling is not something I'm going to do in a couple of minutes. At least I got one more order of magnitude estimate down. Thanks! And remember, a point-blank nuke against your eyeball is still less bright than a supernova from 1AU ( relevant XKCD).  |
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