Doomsday Math

Communications Center General Discussion
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Note to viewers linked from IGS, this contains RL information.

Due to this discussion on the IGS, I decided to do some math to figure out the effect the Oblivion doomsday might have if applied to a planet

All of these calculations are supposed to give an order-of-magnitude estimate of the max power of the Oblivion, and should be taken with a grain of salt.

  1. The Oblivion doomsday takes 22,500 GJ of energy to activate
  2. The Oblivion does only kinetic damage, indicating that it uses this energy to accelerate large quantities of mass to high speeds.
  3. If we assume that the Oblivion has an energy efficiency of 100% (A ridiculously high estimate), this means that 22,500 GJ of energy are delivered to the target (The same applies to EM and Thermal weapons. Explosive weapons have different formulas)
  4. 22,500 GJ is also the amount of energy in 5 kilotons of TNT, or a very small nuclear weapon (Little Boy, the bomb dropped on Hiroshima, had a yield of 15 tons for comparison)

But this assumes the only energy to go into the superweapon is the activation energy.

  1. The Oblivion has a powergrid requirement of 100,000 MW. If we assume that all of this energy is stored in a separate capacitor while the weapon recharges (the “cooldown” period) the energy amounts look a bit different.
  2. The Oblivion has a recharge time of 600 seconds
  3. 1 watt = 1 joule per second
  4. 100 GW = 100GJ/sec
  5. 600*100 GJ = 60,000 GJ over the course of the cooldown period
  6. 82,500 GJ is then the total amount of power a ship pumps into the module in one cycle
  7. 82.500 GJ is equivalent to about 20 kilotons of TNT. Or a small nuke. (Fat Man, dropped on Nagasaki, had a yield of about 21 kilotons of TNT)

But the Oblivion also requires fuel! Specifically 75,000 units of Nitrogen isotopes!

  1. Nitrogen isotopes have no explicitly defined mass
  2. However, the mass of Caldari fuel blocks, being solid, can be calculated
  3. Nitrogen ice has a density of approximately 1000 kg/m^3
  4. Caldari fuel blocks have a volume of 5 m^3
  5. If we assume that all of this volume is nitrogen ice, this gives us a mass of 5,000 kg
  6. Each fuel block contains 444 units of nitrogen isotopes
  7. 5000 / 444 = 11.26 kg for the mass of one unit of nitrogen isotopes
  8. Let’s round 11.26 down to 10 for easier math and to partly account for the other components of fuel blocks (it also makes sense for one “unit” to be a round number)
  9. This means that the Oblivion doomsday requires 750,000 kg of nitrogen isotopes to fire
  10. Just for funsies, let’s assume that all 750,000 kg of N is converted directly into pure energy, using E=mc^2, and delivered to the target with zero loss. This gives us a total energy of about 67,407,000,000,000 GJ, 16,000,000,000 kilotons of TNT, 332,000 Tsar Bombas going off at once, or one dinosaur-killing asteroid. Now we can truly call this weapon a doomsday!
  11. A more reasonable-sounding assumption is that the nitrogen fuel is used in a fusion reactor. However, there is no known formula for this, and it is not a common process in stars. Nitrogen is an important part of the Nitrogen-Carbon-Oxygen cycle, but this cycle is used to turn hydrogen into helium, and thus no nitrogen is actually consumed or produced.I could probably calculate the energy if I knew what the products were, but I don’t.
  12. Nitrogen isotopes are explicitly described as stable, and thus fission would not work
  13. Nitrogen does not burn, and trying to store monatomic nitrogen for later recombination is a whole new kind of stupid.
  14. Suddenly direct conversion into pure energy seems much more likely

BUT maybe CCP devs just don’t have a sense of scale. In that case, let’s compare the Oblivion to another weapon in-game: Titanium Sabot M.

  1. The Oblivion doomsday does 2 million points of kinetic damage to whatever it hits.
  2. Titanium Sabot M does 12 points kinetic and 4 points explosive
  3. Titanium Sabot M has a mass of 1 kg.
  4. Since we are focusing on kinetic damage here, I will ignore the shell’s explosive properties
  5. 2 million divided by 12 is approximately 166666.6666666667, which I will round up to 167,000 to make the math easier
  6. In other words, one must fire 167 thousand rounds of Titanium Sabot M to do the same amount of raw kinetic damage as the Oblivion
  7. When the order was given by Tibus Heth during Operation Highlander to fire the Shiigeru’s doomsday at Caldari Prime, the Shiigeru was in “low orbit”, which I will estimate as 200km above the surface (a low estimate)
  8. Assuming acceleration due to gravity at 10m/s^2 , something dropped straight down from a height of 200km will reach a speed of 2 km/s
  9. 167,000 kg of mass moving at 2 km/s has an energy of 334,000,000,000 J, or 334 GJ
  10. BUT the Titanium Sabot M moves faster than 2km/s So let’s crank up the speed to, say, 100km/s. This feels reasonable to me, given that weapons technically hit instantaneously.
  11. For the given mass, this gives us an energy of 835,000 GJ, or 200 kilotons of TNT. Here is a simulation of a nuclear explosion of that magnitude
  12. Of course, this section has the sketchiest math (Because CCP doesn’t give us much to work with here), and 200 kilotons still feels pretty low. So I would consider 1 Chixulub Impactor to be the best estimate for the total damage output of the Oblivion doomsday.
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Keep in mind the actual design of the Oblivion is an enormous cluster of missiles. I’m assuming that the energy required to launch those missiles is actually probably smaller than the energy each of them can deliver in payload.

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There’s a number of flaws with your math. Let’s start with your first set of numbers:

In this set, there’s the assumption of perfect energy transfer, which isn’t possible. 22.5k GJ is the amount drained from the capacitor to launch “a storm of missile fire capable of neutralizing almost any target”. So what’s that mean? Good question. For all we know, it takes 22,500 GJ to trigger all of the spark plugs on a volley of 22,500 tiny little missiles that fire off in a stream, each of them carrying a warhead containing 1 gram of anti-matter.

Which means we can’t draw any conclusions from the capacitor hit.

Next, on to the PG math. Here, you assume that the powergrid requirement is discharged all at once into the weapon… despite, you know, the weapon being, again, “a storm of missile fire capable of neutralizing almost any target”. It’s far more likely the PG requirement is for the “targeting and tracking control system more advanced than any other in existence”, missile safe-storage systems, etc etc. So really, we can’t draw any conclusions about the yield of the missiles from this number, either.

Then there’s your numbers about nitrotopes.

We know that all of the mass of fuel blocks isn’t the topes, because it takes other things to make fuel blocks, too. So right off, your numbers are off. Your estimate for the mass of 1 unit of nitrotopes is skewed by this problem, and then more skewed with the idea that there’s no mass differential between a pure nitrogen isotope, and one bonded with something else. Unfortunately, again, we can’t actually guarantee that, so it’s impossible to evaluate.

Again, you seem to be working on the premise of a direct energy weapon here. But worse… it’s nitrogen isotopes. It’s an isotope of nitrogen. This means we can be reasonably sure that the mass of 1 isotope is right around the atomic mass of nitrogen, multiplied by however many atoms are in the volume listed for 1 unit of nitrotopes. Not a perfect match, mind, but it’s right around there.

Finally, there’s the damage comparison.

Well, first, an Oblivion DD on a Leviathan does 1.5M, not 2M.

Second, you go from comparing a weapon to comparing ammunition. Sure, you can look at the damage stats listed for a round of Titanium Sabot M… but that’s a total of 16 points of damage, and a round of Titanium Sabot M fired from a 720mm Howitzer Artillery II on a Hurricane inflicts a volley of 1,475 damage, presumably 75% of which is Kinetic, 25% of which is Explosive. Now, since all of your math is based on the damage and energy projections of a shell fired at the relativistic speeds we all know and love from our insane death-salami… don’t you think all that energy is kinda reflected in the actual damage numbers?

Which means you’re off by a factor of 1,475/16, or 92.185. So instead of 166,666.67, you’re looking at more like 1,808 shells per doomsday.

and then you throw away all of your math completely and say 'after showing all of this and coming up with a 200 kiloton impact as my top end, we’re just going to assume an impact exceeding 27,000,000 megatons (Chicxulub has been estimated at around 1.15E+23j of energy, or 27,485,659.655832 megatons).

So why go through all of the math to show a 5 kiloton, then 20 kiloton, then 16.000,000 megatons (still short of Chicx, and for this you literally needed to do a perfect matter->energy conversion!), then 200 kilotons… if you’re just going to throw it all out anyway and go ‘HANDWAVIUM! IT’S A DINO-KILLER!!!’?

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I think a better math is to take a determine what’s the max amount of ehp that can be wiped out entirely by a single Oblivion blast, figure out how many shots of fusion s is going to wipe out that same amount of ehp, under the assumption that the fusion s rounds are % of Hiroshima blasts, use that to convert into Joules and work from there.

Or instead, use an Avatar’s Doomsday for the math.

Also, there are more ways than one to end the world than to drop a Chicxulub on the planet. You could just nuke the damn place so utterly that everything dies from radiation and nuclear winter.

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Possibly relevant maths from the previous forum? Specifically these are calculating the ‘Judgement’ Doomsday due to the example being Reschard V and effects on it when fired, but yeah, the numbers are far more impressive there (not sure if accurate though). And one can assume that other doomsdays have a similar yield.

https://forums-archive.eveonline.com/default.aspx?g=posts&t=422905

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The 200kt damage wasn’t my top end. That was just an alternate reckoning. And as you pointed out, it was very flawed. I also have no actual experience with projectile weapons, so there’s that too.

Also I did use perfect energy conversion because I was trying to get an upper bound on the total potential energy of the weapon. The damage and other stats were pulled from here, so that’s another thing.

Also I failed to realize that it was a missile weapon. Oops.

So, yeah, mistakes were made. I did calculate the power of the Judgement using similar math and got a theoretical max output of 6.7 trillion GJ (assuming direct matter-to-energy conversion) and more likely max of 4.5 billion GJ (using helium fusion)

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I think the safest math involved in all of this is ‘CCP employs real physics well enough that none of your ships have forward-facing thrusters to stop them, in space’.

So really… don’t even try. A doomsday is a doomsday. vOv

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No ships have thrusters in EVE, only PEGs and thermal exaust ports :slight_smile:
See:

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It’s cute when engineers try math.

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Holy latenight nerdy game math. Love it.

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Introduce engineers to math and you get a magical experience.

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I teach math to engineers at university. In one group there is always 1-3 students that actually have a clue and rest of them just try to memorize everything :smiley:

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