Idea for orbital unit.

Discussion in 'Backers Lounge (Read-only)' started by midnite111, April 17, 2013.

  1. midnite111

    midnite111 New Member

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    To start of with, this all started from some spare moments of mind wandering and a movie.

    So over the past weekend I went to see the second G.I.Joe movie, Retaliation or something like that (not that great a story IMO but that's not important) but anyway I was really grabbed by the Idea of the villan's WMD Basically it looked like this.
    [​IMG]

    It was called Project Zeus, I remember it by its slang name "Rod's from god" its an orbital satellite that drops tungsten rods from orbit that impact at high velocity, doing lots of damage.

    Now I know that even thinking about designing orbital units is a ways away for Uber as the game is not even in alpha yet but I just wanted to throw this unit idea out here. see if anyone else was thinking about this.
  2. shandlar

    shandlar Member

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    TBH I find this somewhat likely. KEW's dont have to all be astroids. This would be a very powerful weapon for sure, but there are lots of ways to balance it...

    1) You are dropping big chunks of metal, so you should have to 'build' each Sabot before firing. Make the metal cost high and it becomes a game of attrition and strategy instead of just god-mode.

    2) Slow reload/projectile build time.

    3) Delayed landing. Send firing orders, takes 20 seconds to land from orbit.

    Should work out decently I think.
  3. ToastAndEggs

    ToastAndEggs Member

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    This is actually in production.

    Declassified files show a Kinetic Bombardment project is in the works, likely in conjunction with the rail gun project.

    As for in the game, its more fun than orbital nukes so yes, same effect really.
  4. BulletMagnet

    BulletMagnet Post Master General

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    Hell, we currently have bombs that do this. They spray out flechettes and welp into (read:through) buildings with them.

    Great for demolition without the unappealing explosive collateral.
  5. Devak

    Devak Post Master General

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    a Rod From God can be comparable to a tactical nuke (low KT yield).

    they're essentially small nukes without all the hassle of a small nuke (production, storage, radiation, pollution).
  6. BulletMagnet

    BulletMagnet Post Master General

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    I know that, but you seem to have neglected the mass of the rod in your considerations.

    Big rod: big bang. Little rod: little bang.
  7. bobucles

    bobucles Post Master General

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    The power of a rod is based entirely on the power of the planet's gravity well. Low gravity == low boom.
  8. BulletMagnet

    BulletMagnet Post Master General

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    I don't think it's entirely dependent on that (I am of course willing to change my views if you can find me some maths that suggests you are correct).
  9. bobucles

    bobucles Post Master General

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    Gravitational potential energy == Mass * Gravity * Altitude.
    ^ ^ ^ ^ ^ ^ ^
    It kind of says it right there in the name.

    Of course it's not entirely accurate, as gravity diminishes with extreme altitude.
  10. BulletMagnet

    BulletMagnet Post Master General

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    Yeah... it does... especially about the mass bit.

    Or did you accidentally think that was mass of the planet?
  11. shandlar

    shandlar Member

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    Epic nerd-dom to the rescue.

    Assume we are talking about airless planetary bodies because air is a total drag.

    Low orbit (500km) of an airless planet roughly the mass of the earth (10m/s^2 gravity)
    A Sabot around the size of a telephone pole made out of Osmium (nanolathe ftw)

    Lets say 12m long, 0.25m in diameter

    v=(pi)*r^2*h
    v=3.14159*12.5^2*1200
    v=589048.125 cm^3
    589048.125 * 22.61 g (density of osmium per cubic cm) = 13318.38 Kg

    Galilean Equations to the rescue.

    500000m = 1/2(10)t^2
    500000 / 5 = t^2
    t = sqroute(100000)
    t = 316.228 seconds

    velocity = a*t
    v = 10 * 316.228
    v = 3162.28 meters/s at impact

    Energy = (1/2) * m * v^2
    E = 0.5 * 13318.38 * 3162.28 ^ 2
    E = 66,591,900,000 joules

    1 kiloton = 4.184 trillion joules

    0.0665919 / 4.184 = 0.015916 kilotons = 15.916 tons of TNT

    We are talking large tactical missiles NOT small tactical nukes. And it would take over 5 minutes to land.

    But remember, we are going for awesome, not realism.
  12. BulletMagnet

    BulletMagnet Post Master General

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    That was an awesome post!

    I demand more of these in the future.
  13. bmb

    bmb Well-Known Member

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    Why do you feel physics equations need to figure into how much damage a weapon does?

    As you said, we are in fact not going for realism.
  14. BulletMagnet

    BulletMagnet Post Master General

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    Awesome and realistic aren't mutually exclusive.

    [​IMG]

    Personally, I find the Merkava more attractive but there's bound to be an 'mericahn on here who will take exception.
  15. menchfrest

    menchfrest Active Member

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    You have not accounted for the initial velocity the rod has. You assumed it started at rest and accelerated down, that's not how rods from god would work, you're decelerating from 8km/s to 0 so gravity can bring it up to 3 km/s which is way more expensive than dropping it's speed from 8 to 6 or 7 km/s (exact math on this would depend on relative target location and orbital mechanics) and letting it hit the ground later. It's now just a suborbital weapon, which can get to really high speeds.

    How it could be done with conservation of energy:
    Start with kinetic energy (slightly less than orbital velocity so it's falling) plus potential energy of orbit. Then your second point is potential energy of the earths surface and the remaining energy is your new kinetic energy that it impacts with.

    you should get a number several times (like 2x) for velocity what you had gotten, which is a 4x + increase in energy, which is still like 60 tons of TNT, but that is starting to put it above the LARGEST fuel-air/thermobaric weapons in use, which the Russians are claiming they will use to replace some small nukes in inventory. Using tungsten only reduces your yield by 85% because yield will scale linearly with mass.

    I've already spent too much time on this, but I may break out a calculator and my orbital mechanics textbook over lunch.
  16. rabbit9000

    rabbit9000 Member

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    Merkava is no longer going to be supported by the military, in place of fast attack vehicles.
  17. shandlar

    shandlar Member

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    You are right indeed. This is actually a pretty cool thought experiment. Since we are assuming airlessness here you are pretty much just degrading your orbit by slowing down until your 'orbit' is below the surface of the planetary body. I didn't specify the diameter or density of the body itself so we can't really do anything without them but still, that's fun stuff. It would prolly result in a velocity at least twice that of my numbers and really not have too much to do with 'falling' at all.

    Cause science is cool as hell when you translate it into fictional scenarios (like dropping telephone polls of osmium from orbit vs accelerating an asteroid from across the solar system into a planet)

    Let us do that too! I enjoy being non-productive at work.

    10km diameter spherical nickle/iron asteroid
    1,000,000 Saturn V rockets capable on indefinite continuous thrust
    250m kilometers travel distance

    volume = (4/3) * (pi) * r^3
    v = (4/3) * 3.14159 * 500,000cm^3
    v = 523,598,775,598,298,873 CM^3 * 8g (density of nickle/iron meteorites) = 4,188,790,204,786,391 kg

    1 million Saturn V rockets would provide a continuous 34,000,000,000,000 Newtons of thrust.

    a = f / m
    a = 34,000,000,000,000 / 4,188,790,204,786,391
    a = 0.0081169 meters/s^2

    250000000000 meters = 1/2 * 0.0081169 * t^2
    t = 7,848,558 seconds = 90.84 days of continuous thrust

    v = a * t
    v = 0.0081169 * 7848558
    v = 63,706 meters per second (0.000212c :lol: )

    E = 1/2 * m * v^2
    E = (1/2) * 4,188,790,204,786,391 kg * 63706 m/s^2
    E = 8,500,000,000,000,000,000,000,000 joules

    8,500,000,000,000 / 4.184 = 2,031,548,757,170 kilotons = ~20 times the estimated KT extinction event

    That's pretty big, but prolly not enough to break apart the earth.

    Granted a Saturn V only burns for 150 seconds. So we would need over 52 billion Saturn Vs to burn in sequence to achieve such an effect.
  18. bmb

    bmb Well-Known Member

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    It's also impossible to balance.
  19. xenomorph555

    xenomorph555 New Member

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    I want a laser satellite, just so I can scream "IMMA FIRING MA LAZOR".
  20. Devak

    Devak Post Master General

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    i am pretty certain G (9,81 m/s^2 ish) starts to vary at that distance. You'd need the equation for the gravitational pull between masses.

    ( i believe it was F=G * (M1*M2)/(R^2), M1 and M2 being the masses and R the distance between the mass centerpoints and G is the gravitational constant )

    Additionally i believe you would need to integrate this function for the entire fall as R changes significantly over time.

    It's a nice approximation tho. (haven't done the actual math myself so i can't confirm the accuracy).


    Also, science is awesome.

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