Conservation of Momentum in Explosive Warheads
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Conservation of Momentum in Explosive Warheads
In the ICS series and elsewhere, SW warheads are said to be directionally-focused explosives. With turbolasers, ion cannons, and solid impactor-based weaponry, the mass-energy content carries momentum with it that is conserved through the incident recoil upon the firing mechanism, and ultimately, upon the ship or emplacement itself. However, a proton torpedo detonating against a hull cannot direct 99% of its yield forward without violating Conservation of Momentum. How can this be reconciled? Is the physical missile vaporized and repelled opposite the vector of directed-yield to conserve momentum? Perhaps the warheads channel a "jet" of weakly-interacting particles like sterile neutrinos that carry away equivalent momentum such as the ones presumably emited by the active radiators in the defensive systems of SW vessels?
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What source say's that it directs 99% in one direction? And if not, what would be the lower limit that it could direct in one direction with out violating CoM?
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Imagine that we shoot a piston-shaped bomb at an enemy ship at 3 kms-1. A few milliseconds before it impacts, we detonate an explosive that accelerates both halves of the bomb in opposite directions at 2 kms-1. The front half will now impact at 5 kms-1 and the back half will impact at 1 kms-1. The original KE at impact was 4.5e6m, the new KE at impact is ((12.5e6+0.5e6)/2)m = 6.5e6m and the (kinetic) energy added by the explosion was ((2e6+2e6)/2)m = 2e6m. Thus this is in principle a lossless way to channel the warhead's explosive energy into the impact point (though not all into KE - waste heat from the explosion will presumably stay inside the piston until impact), under the unrealistic assumption of a perfectly rigid piston that does not allow any of the gasses to escape. Of course the total momentum of the projectile remains constant.
Real warheads will obviously be rather less efficient, but driving projectiles or a high velocity plasma jet forwards into the target is generally a much more effective way to use explosive energy to defeat armour than a simple omnidirectional explosion. Using the stored energy to simply drive the whole projectile to a higher impact velocity in the first place is generally even more efficient, but this may be limited by engine cost/complexity or (particularly for fighter-scale weapons) the difficulty of guiding faster projectiles effectively.
Real warheads will obviously be rather less efficient, but driving projectiles or a high velocity plasma jet forwards into the target is generally a much more effective way to use explosive energy to defeat armour than a simple omnidirectional explosion. Using the stored energy to simply drive the whole projectile to a higher impact velocity in the first place is generally even more efficient, but this may be limited by engine cost/complexity or (particularly for fighter-scale weapons) the difficulty of guiding faster projectiles effectively.
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The warhead must throw a great deal of mass in the opposite direction, very slowly, so you get a jet forward and a puff backward. Momentum goes as mv, whereas kinetic energy (and therefore, damage delivered to the target) goes as mv²/2 = p²/2m.
Let R = M/m, where M is the mass of the low-velocity "puff", while m is the high-velocity jet, where both the jet and the puff are treated as two solid masses. Then, E, the total energy of the explosion, becomes
E = p²_1/2m + p²_2/2M
Of course, M = Rm and p_1 and p_2 are identical in magnitude, so, E = p²/2m(1+1/R), where p = p_1 = p_2. Solving for R, we get R = (2mE/p² - 1)^{-1}. p² = 2rmE, where r = 99/100, so this works out to R = (1/r - 1)^{-1} = 99 here.
The velocity of the puff, V, satisfies mv = MV, which is V = v/R, so the average velocity of the puff is 1/99 that of the jet.
The mass ratio is undoubtedly higher, to absorb losses of energy due to inefficiencies (explosion that is directed forward but not in the jet, explosive debris sideways, ect). Of course, 99% is assumed to refer to the usable yield of the explosion, rather than the total yield. I haven't seen the text, so I don't know.
Let R = M/m, where M is the mass of the low-velocity "puff", while m is the high-velocity jet, where both the jet and the puff are treated as two solid masses. Then, E, the total energy of the explosion, becomes
E = p²_1/2m + p²_2/2M
Of course, M = Rm and p_1 and p_2 are identical in magnitude, so, E = p²/2m(1+1/R), where p = p_1 = p_2. Solving for R, we get R = (2mE/p² - 1)^{-1}. p² = 2rmE, where r = 99/100, so this works out to R = (1/r - 1)^{-1} = 99 here.
The velocity of the puff, V, satisfies mv = MV, which is V = v/R, so the average velocity of the puff is 1/99 that of the jet.
The mass ratio is undoubtedly higher, to absorb losses of energy due to inefficiencies (explosion that is directed forward but not in the jet, explosive debris sideways, ect). Of course, 99% is assumed to refer to the usable yield of the explosion, rather than the total yield. I haven't seen the text, so I don't know.
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Actually its supposed to be a charge of plasma/protons shot at the target (which kinda makes sense, we know SW has nukes so why would they need to invent fancy names for something that already exists?) In that case I suspect the proton torpedo "yield" tends to be treated more like ion engine exhaust, with the momentum of the exhaust acting on (and possibly destroying) the remnants of the missile during the act of channeling it.Howedar wrote:Since I expect the explosion of a proton torpedo is primarily radiation, I don't see a problem here. The momentum of such a huge amount of light is non-negligible but is certainly much less than a big chunk of torpedo parts.
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Why should it, when it can counter the momentum by exploding in the right way?Marko Dash wrote:Couldn't the missile just keep a small reserve of fuel that it burns through to counter the momentum generated by the charge?
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I'd object to that on the grounds of what happens when the thing does not explode in the right way; while I have no experience with the matter, I would imagine that getting fired like a torpedo does would do bad things to any system like the one you describe. Of course, I'm not an expert in this field, so I am probably wrong.Wyrm wrote:Why should it, when it can counter the momentum by exploding in the right way?Marko Dash wrote:Couldn't the missile just keep a small reserve of fuel that it burns through to counter the momentum generated by the charge?
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We already have anti-tank missiles that get fired out of tank guns and guided missiles that accelerate at over 100g. Hardening warheads and seekers to take this sort of thing is a nontrivial but fairly routine engineering challenge.darthbob88 wrote:while I have no experience with the matter, I would imagine that getting fired like a torpedo does would do bad things to any system like the one you describe.