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Thread: GI joe retaliation zeus weapon

  1. #11
    Quote Originally Posted by PIRATEONTHERUN View Post
    I have to disagree with this on the basis that it would be extremely ineffective if this was true and considering the military considered this as a weapons system means that it would have to work pretty well, at least theoretically.
    As to its lethality, AFAIK it was generally envisioned as a bunker buster, probably to take out hardened ICBM launch sites. I don't know about their capability to create artificial earthquakes, although the idea seems reasonable, but according to wikipedia Project Thor allowed for a large version which would impact with 11.5 tons of TNT, significantly weaker than a nuclear bomb.
    Actually, you are right...it would be significantly more destructive than the analogy I used. But the question is energy transfer. Essentially it would be transferring all the energy through a fairly small (give or take .5 meter at most??) diameter. I think it would be more of the kind of thing like you described, like a bunker buster or buried/hardened silos etc. It just wouldn't make much sense for the energy to transfer out to the order of kilometers wide, total destabilizing of the earth's crust. Though the collateral would not be insignificant, it wouldn't level a city.

  2. #12
    Quote Originally Posted by PIRATEONTHERUN View Post
    I have to disagree with this on the basis that it would be extremely ineffective if this was true and considering the military considered this as a weapons system means that it would have to work pretty well, at least theoretically.
    As to its lethality, AFAIK it was generally envisioned as a bunker buster, probably to take out hardened ICBM launch sites. I don't know about their capability to create artificial earthquakes, although the idea seems reasonable, but according to wikipedia Project Thor allowed for a large version which would impact with 11.5 tons of TNT, significantly weaker than a nuclear bomb.
    An fast moving rod of some dense metal is an perpetrator, 0.5 to 1 cm rods of tungsten or depleted uranium is the primary anti tank weapon.
    An telephone pole of tungsten would be able to punch pretty deep and would probably to overkill against an hardened ICBM silo, however you has to hit pretty accurate.

  3. #13
    It seems like it would be far simpler to just "drop" a nuclear missile this way. Relatively modern nuclear weapons can actually be fairly clean, leaving the ground zero safe within hours.
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  4. #14
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    Quote Originally Posted by Kimberly View Post
    It seems like it would be far simpler to just "drop" a nuclear missile this way. Relatively modern nuclear weapons can actually be fairly clean, leaving the ground zero safe within hours.
    I don't think they are THAT clean. Do you know any of the warhead models that are that clean by chance?
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  5. #15
    Quote Originally Posted by Kimberly View Post
    It seems like it would be far simpler to just "drop" a nuclear missile this way. Relatively modern nuclear weapons can actually be fairly clean, leaving the ground zero safe within hours.
    I find that very hard to believe. Do you have any examples?

  6. #16
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    Its the equivalent to a very high velocity rail gun fire a tungsten bar at about Mach 25 for reentry.
    http://www.engadget.com/2012/02/29/r...st-fire-video/
    That was the BAE railgun test firing a 40 pound aluminum payload but zues would probably have a much larger payload and hit 5 times as fast.

  7. #17
    For example the neutron bomb and other fusion powered nuclear weapons produce very little fallout in comparison to their explosive power.

  8. #18
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    I did a little of that thing called "math", and it turned out some interesting numbers.

    First off, I went with the assumption that the projectile in the movie is the same size and shape as a telephonepole, and I went with a length of 15 meter and a radius of 12.7 cm, and made up of basic tungsten/wolfram (wichever sounds the coolest in your mind) at 19.25 g/cm3
    Giving a volume of 760 062 cm3
    for a grand total of 14 631 kg.
    And I asumed the speed of a bullet to be 3 times the speed of sound, so 3 X 330 m/s = 990 m/s, and in this case, times 8
    giving a speed at impact of some 7920 m/s.

    So, using basic equation for energy in moving object (1/2XweightXspeed squared) I get:
    1/2 X 14631 X (7920^2) = 458 874 979 200 kJ of energy at impact.

    So, normalisation to TNT equivalents gives roughly:

    458 874 979 200 kj impact energy / 4 184 000 kj per ton TNT = 109 673,752 tonns of TNT.

    So it´s roughly the equivalent of a 110 KT warhead. As that random smiling evil henchman-like guy says in the movie, more powerfull than most nuclear weapons, if you consider "most" as being in the numerical sense, and then include tactical warheads, wich usualy counts in the lower of 25 KT size.

    But, this brings up a small pile of new questions and problems where reality is conserned. Now, I´m not versed enough in the science of aerodynamics, since I only began reading about it properly half a year ago, but, the projectile is pretty dense, the ends, might just be flat, and then produce a shockbow in front, if it was kept stable on it´s descent, wich would give some protetion against heat, and considering it´s speed, if it came striaght down, would "only" spend some 10 seconds or so from the edge of space and down to the surface. And since only the lower 10% of that contains roughly 90% of the atmosphere, at that speed, atmosphere would only get a very short time to slow it down.

    But this is where it gets realy hairy and ugly. First, the initial "drop", as mentioned by others, "physics don´t work that way". I´d be guessing an initial launch velocity of some 8-9 km/s retrograde for it to be able to "just fall". So, in free fall, when will an object falling towards earth reach such a velocity?

    7920m/s / 9.8m/s = 808 seconds free fall.

    808 seconds of free fall towards earth:

    1/2 X 9.8m/s X (808^2) = 3199033.6 meter, i.e. dropped from an altitude of 3199 kilometer, and thereby defying orbital mechanics, since it´s orbital velocity is ignored. So, what would it´s orbital speed be?
    I´ll try and give a rough estimate then, I realy want to see how much delta V we are missing here.

    Semimajor axis:
    Earth radius 6371 km
    Orbital altitude 3199 km
    6371km + 3199km = 9570km

    Vo = sqr(398 600/9570km) = 6.45km/s retrograde "drop speed" then.

    And all this is ignoring aerodynamic resistance for the last few seconds, assuming the pole is able to stay stable all the way down, and hit it´s target. Oh, and since the projectile is made up of a very dense material, it´s disintegration and deformation will be aimed downward. The energy at impact won´t resemble a nuclear detonation at all.
    And I still very much doubt the impact speed until I see some figures taking aerodynamics into account.
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  9. #19
    Quote Originally Posted by aNewHope View Post
    For example the neutron bomb and other fusion powered nuclear weapons produce very little fallout in comparison to their explosive power.
    "very little fallout in comparison to their explosive power" does not mean "almost no fallout". It just means not as much as a normal nuke


    And tnx for the numbers Thanial. 110 KT is still more than I expected from a metal rod. But as you said ofcourse, that thing would just burry itself deep underground, not explode with all that force on impact

  10. #20
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    Yeah, I must say I was surprised at that high yeld myself. It´s at speeds we puny humans may relatively easily manage to accelerate something like that up to. But, we are talking about deep penetrating stuff, projectiles in the 14-15 ton size, wich in turn have to be accelerated a fair bit as well, and I expect them to have to hit pretty dead on, as a miss at that speed won´t inflict much harm far from the impact itself.

    I´m not entirely sure I got the TNT equivalent right, but I am pretty sure I got the energy at impact right. I went over and recalculated those numbers a few times, weeded out a mistake where the rod was merely 15 cm tall and such. So if anyone spots a mistake, feel free to correct me. But I don´t think I´m too far off realy. After all, 24 times the speed of sound is pretty fast, and in this case, it is literaly as if a lokomotive traveling at that speed hits an area the size of a dinnerplate. There just have to be a lot of energy released in that.
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