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Has anyone ever came across an issue in the CE version of UFO in which a landed UFO gets assaulted and my guys leave the Power Source intact, however, at the summary screen at the end - there is no elerium, although the power source appears?? I'm confused - I don't understand how that could have happened! I've only seen it once so far - could have been data corruption... not sure :s - Phoenix

Elerium is the very last item to be spawned in a battlescape map (first X-Com equipment, then alien equipment, then Elerium).

If you brought too many items to the battle site, it's possible that there wouldn't be room in memory for the Elerium. Bring along enough stuff and you can also deprive the aliens of their equipment.

Was the Elerium visible during gameplay (little purple stun bomb thingy sitting on the power supply, or white + on the radar)? If so, try picking it up (the power supply won't explode if you just shoot it).

-Bomb Bloke

Cheers for that Bomb Bloke - I haven't seen it happen again, as there has always been elerium on the missions that I expect it. After having a read through this site, about the object table overflowing, I can understand how that can happen. I'll keep an eye out for it in the future. Cheers again! :) - Phoenix

If you used explosives (grenades or cannons) near the power source, it's possible the Elerium was destroyed (damage=20) while the power source remained intact (damage=50) --Ethereal Cereal 21:33, 8 May 2006 (PDT

How much Elerium is “1 Elerium“?

Tequila, I've been away a while and am just noticing your "1 Elerium" section. Very interesting thoughts! Thanks for that bit of armchair science!!

But I can think of a couple of issues... on the one hand, surely at least the Avenger is space-worthy, which could mean it may fly in little or no atmosphere. This is also probably at least potentially true for all the researched craft, since they all use elerium engines and alien alloys, and are originally designed based on researching UFOs (all of which are space-worthy). Also known as, why not make is space-worthy, if you're designing something strong enough to face UFOs. (Even 1990s fighters could fly very high in the atmosphere, with a principle reason for not going to space being there's no air for their jet engines... but Elerium does away with that concern.) On the other hand, you left out of your calculations the price to be paid for fighting off gravity. That's surely energy expensive! (Look how big rockets have to be.) So you might consider toning down the drag factor... and introducing a big gravity factor, if you care to have another go at it.

I also really like the alternate approach to playing XCOM on your User:Tequilachef page. A few small conceptual constraints which make a huge difference in game play (a.k.a. there's always a real risk of losing).

- MikeTheRed 17:34, 10 October 2007 (PDT)

You are right I guess. I formerly had the gravity issue included by that: "Now we take that keeping the craft at max speed only uses 95% of required energy per mission..." I now changed that to 75%, which seems more likely to fit but is still far from exact. In reality, the aspect of overcoming gravity would create a VERY complex mathematical problem. Flying higher lowers the atmospheric density and therefore atmospheric drag, but raises fuel requirements for obtaining flight height. Considering that complex flying maneuvers might be necessary for interception and that the starting height might vary from base to base no absolute solution exists. Constant calculations by computers would be a necessity. Remember: Both atmospheric density and gravity depend on height (or distance from gravity source) and are both differential equations. If anyone reads this and has loads of time, feel free to work out that one. Else, I would prefer those educated guesses ;)

- tequilachef

It's worth noting that while the UFO Power Source may be incredibly efficient in Elerium use, there is no guarantee that this is so in regards to the alien weapons. In fact, given the size of a power unit, I'd say it's more likely that the weapons are extremely inefficient in Elerium use. (Especially the heavy plasma clip; 3 times the Elerium for one-third more shots and just over double the killing power of the Plasma Pistol.) We also don't know exactly where that Elerium used in construction of the grenade(or anything else) goes; it's quite possible that only a fraction of it goes into the charge/warhead and some is used in the creation of functional parts. Also, it should be noted that explosions do NOT scale linearly; twice as large a warhead on an atomic or hydrogen bomb does not equal twice the explosive power. In addition, it's been theorized that the explosion from UFO Power Sources is not from the impact; its from trying to start the UFO's engines in order to escape incoming X-COM troops before it's ready(thus why the aliens are killed immediately before the X-COM turn begins, and not when the UFO crashes.)
Of course, since this is fiction, it really doesn't matter, just thought I'd bring a few things to the table since you seem interested in scientific accuracy. Arrow Quivershaft 15:53, 2 November 2007 (PDT)
After running the numbers myself (54000nm in 10hr is, indeed, 10000km/hr - or about 278m/s) I can say that the quoted figures are slightly off. According to some quick research the density of air at that altitude is the same as the density of air at sea level. However, I used the classic formula of the drag equation:
Fd = 1/2ρv2CdA
where ρ is the density of the air, v is the velocity, Cd is the coefficient of friction and A is the surface area. Using this we get Fd = 1/2 1.293 * (2782) * 0.3 * 20 - or about 299,785 Newtons of force. For total power requirement we use the Power Requirement equation P=FdR, where Fd is the result of the preceding Drag equation and R is the range - stated as being 100,000km (100,000,000m). Solving that equation we find that we require 2.99785*1013 Joules of energy. This figure, following the figures, is 75% of the total available power from the Elerium, so the total available power is 3.99712848*1013 Joules. With c equal to 299,792,457m/s plugged into Einsteins famous equation of: E = mc2 we get a result of 4.4*10-4 kilograms. However, this is stated as being 99% of the total, so we have a full load of Elerium fuel for the Avenger of 4.49*10-4kilograms. That is, 4.49 grams of fuel - and as the Avenger is stated as carrying 12 units of Elerium, the result is that each unit is 0.37 grams - that's right, 37 centigrams of Elerium per fuel unit. - Shadow 18:23, 2 November 2007 (PDT)
As for a Terror Ship containing 200 units of Elerium, if my above math is correct (as I believe), the result of that Elerium detonating at 100% conversion, isn't even equivalent to a 2 Megaton nuclear device. (200 units is about 74grams - direct conversion of all that mass would release about 6.65*1015 Joules of energy. 1 Megaton is equivalent to 4.185*1015 Joules. This places a 200 unit, 100% efficient explosive at about 1.6 Megatons in size. IIRC the largest nuclear device ever built was around 200 Megatons. (Note that the edge of non-overpressure damage for a 1 Megaton blast is around 20 Miles) - Shadow 22:12, 2 November 2007 (PDT)
On another note, I don't think that you use that formula for stuff involving Supersonic speeds. In fact, I do believe that you use the equations for Wave Drag... Jasonred 17:59, 25 April 2011 (EDT)
Damn, this is all very cool armchair stuff. I just added a link at Realistic_Equivalents#Elerium-115. If anyone wants to summarize/move all this conjecture there, that's fine, but it sounds like it's still a moving target, as it were. And the E-115 page is a good enough place, anyway. - MikeTheRed 22:42, 2 November 2007 (PDT)
I've found an error in my above math. The velocity is 2778m/s, not 278m/s - this makes the force 29,930,555 Newtons. (call it 2.993*107 for simplicity) This makes it 2.993*1015 for the used power, or 3.99*1015 Joules. That works out to 0.0444kg, but, as stated, this is assumed to be 99% due to the 1% inefficiency - so it's 0.0449kg. Makes it 449 grams of Elerium covering 12 units, or around 37 grams of Elerium per unit. This would give a Terror Ship about 7.5 kilo's of Elerium - at 100% efficiency the amount of generated energy would be 6.74*1017 joules released. That would be equal to about 16 thousand megatons - the "Gigaton" designation comes in at 4.184*1018 Joules. This would be enough to shatter the Earth. However, this assumes that all generators detonate at the same time, there is no "material scattering" effect from the first blast(s) and that the Elerium converts at 100% efficiency. None of these things are likely to be true. - Shadow 22:54, 2 November 2007 (PDT)
Bah, make that around 160 Megatons. This means that there have still been larger nuclear bombs produced on Earth. - Shadow 23:03, 2 November 2007 (PDT)
Very cool. For one thing though, UFO Power Sources do not "chain react", as recently posted at UFO_Crash_Recovery#Power_Source_Explosions_and_Elerium_Recovery. In the extreme case of the Terror Ship, if 1 PS explodes, it wastes all 3 of the others so that they don't explode. However, isolated PSs can all (independently) explode, as in the case of the Battleship. I have to say though I never saw the earth shatter... clearly those UFO walls are pretty heavy armor. ;) - MikeTheRed 23:08, 2 November 2007 (PDT)
That's the point I was making. At only 160MT absolute potential for the 200 units of Elerium onboard a Terror Ship, well... That's assuming that all four cores go simultaneously and the Elerium converts at 100% in the "wild" reaction. The truth is that a single core going actually would disrupt the functioning of the other cores - by causing a scattering of the Elerium. As we can assume that each of the four cores contains 50 units, the math shows that each core is capable of somewhere between 39.5 and 40 megatons at the absolute limit. But that is in a perfect reaction - where the Elerium converts 100% to energy. The fact is that an uncontrolled reaction - like that which causes an explosion - is far from ideal, and would be, at most, 90% efficient, if not closer to 50%. And that also assumes that the Elerium's conversion releases the energy in an even mix of heat, pressure and radiation. The more likely result - seeing as how Elerium is capable of being used as a power source for pistols and other compact weapons - is that it releases a lot of easily converted radiation - likely in the form of high-energy beta particles. This isn't to say that Elerium can't have an explosive form of reaction - just that the way it's used in the reactors is probably as a beta-particle and heat generator, and potentially even X-Ray and Gamma-Ray source (both of which can be used with forms of photovoltaics to generate electricty). If they go for the efficient side, then the reaction used in reactors is more than 50 percent focused towards directly convertable forms of energy. Truthfully, I'm guessing that they focus it at 75 to 80 percent "hard radiation" (beta, gamma-ray and x-ray) output. This means that such a reactor going super-critical and exploding wouldn't do a lot of physical damage from the blast, but it would irradiate quite a bit. - Shadow 14:41, 3 November 2007 (PDT)
I thought along similar lines. Your typical old-fashioned Hiroshima-style fission bomb contains like what, 10kg of U-235? Or whatever is the critical mass. But most of it is vaporized and sent flying in all directions as soon as things really get hot. Only a small amount actually reacts before the rest is blown apart. And this in a device that's meant to explode. IIRC, in Chernobyl none of the fissionable material actually exploded – excess heat produced a lot of steam which forced the lid open, then air rushing in allowed the graphite to catch fire and burn for days. In effect all fissionable material was wasted (in the sense of "not recoverable") yet in terms of explosions, it was nothing special. --Schnobs 19:01, 3 November 2007 (PDT)
I've written a program in Python that solves all the equations - including the Barometric function for atmospheric density. On running it and giving it all the above parameters I've learned that even my "corrected" figures are off. An Avenger is carrying about 25.5 grams of Elerium, which makes each unit about 21/8 grams. This would give a Terror Ship a full load of 425 grams of Elerium, and a "perfect conditions" explosive potential of around 9Mt - the MIRV warheads on most missiles in the US arsenal during the 1960's was larger than that. Taking MikeTheRed's reminder that Elerium reactors do not chain-react and Schnobs' reminder that nuclear explosions never use the whole mass of the available material - a "high-end" estimate of material that reacts would be 50% - this would leave us with a single-reactor explosion of 1.25Mt - about four hundred times the total combined destructive potential of the (in)famous "Fat Man" and "Little Boy" bombs that were dropped on Hiroshima and Nagasaki in 1944. However, the quoted figure for the size of the fireball for a 1Mt nuclear device (the US Minuteman Missile) is .96km. So the fireball from a 1.25Mt device is going to be about 1 mile. This says nothing about the area damaged or destroyed by the pressure-wave that a nuke generates. - Shadow 01:21, 4 November 2007 (PST) (corrected later - Shadow 02:24, 4 November 2007 (PST))
Based on the 11 tile blast radius of any power core and the 2.26 meter per tile conjecture we can see that the blast diameter of a power core is approximately 50 meters. With a 960m blast diameter being what is expected from a 1 Megaton bomb, and a 48m one from a 20 kiloton bomb, we find that the power cores detonation is right around 20 kilotons. That means that about 1 gram of material has been consumed for the explosion. In other words, not even a single unit of Elerium detonates. - Shadow 21:09, 4 November 2007 (PST)
I don't know where or how you came by the figures of 960m or 48m respectively, but have some doubt as to what they actually mean. Pictures from the japanese cities are not conclusive (bomb was touched off high above ground, many wooden buildings might have withstood the actual blast but we won't know as they burned to the ground anyway). However, from chemistry class I remembered an explosion in a fertilizer plant (picture) that was rated in kilotons. Wikipedia speaks of 1-2kt and a 90x125m crater, which would be like 40x50 tiles in UFO scale. This explosion happened at ground level, the buildings were brick or concrete. Looking at the picture, I don't think any explosions in UFO, not even Blaster Bombs, are anywhere near kiloton scale. --Schnobs 10:51, 5 November 2007 (PST)
The values come from the Wikipedia article about Nuclear Weapons Yield. On that page is an equation that can determine yield from blast radius and time after the start of the blast and a diagram of blast radius based on yield. In said diagram it lists the blast diameter of a 1 Megaton bomb (W59 - the Minuteman 1) at .96km - ie: 960 meters - and the blast radius of a 20 kiloton bomb ("Fat Man" - the "gun type" uranium fission device dropped on Nagasaki) at 48 meters. Note that this is the size of the fireball and not the damage radius caused by a nuclear weapons overpressure event.
The city of Hiroshima sits inside a natural "Bowl" or depression in the landscape. The "Little Boy" bomb is estimated to have only been 13 kilotons and the damage effect was multiplied because of the airburst (IIRC, "Little Boy" detonated some 100 feet about the ground) and the damage was magnified by the shockwave from the detonation reflecting off the hills surrounding the city. It was nearly the same for the higher yield "Fat Man" device that was used against Nagasaki. (In fact, if I remember my High School history classes correctly, it was the geography of the two locations that was the reason for them being chosen as targets).
In conclusion I'd like to say that I doubt that the reactor explosions themselves are nuclear in nature. It is much more likely that they originate from sources very similar to the cause of the explosion at Chernobyl. - Shadow 15:30, 5 November 2007 (PST)
While it doesn't change the ultimate conclusion of "not even a single unit detonating", I'd say any more then 1 meter per tile is being more then a bit generous. 2.26 meters suggests the average unit is over a meter and a half wide, and somewhere over three meters tall. Where did that value come from?! - Bomb Bloke 22:29, 4 November 2007 (PST)
It came from here. Arrow Quivershaft 22:38, 4 November 2007 (PST)
Correct. Danial has estimated that each tile is approximately 2.26m2 on his talk page. I have no interest in this, really, other than of an academic nature, since it can be used to estimate the yield of the device that created the fireball. - Shadow 15:30, 5 November 2007 (PST)
If Bomb Blokes 1.6x1.6x2.4 meter size of a single tile is correct, then a Reactor has a blast diameter of 35.2 meters. Using the "Radius from Yield" equation of R = (Et2/ρ)1/5 we can solve to find the actual yield of the device. I'll run that equation later, but I'm estimating that the yield is less than 10 kilotons. - Shadow 16:21, 5 November 2007 (PST)
Okay, assuming that the explosion is at sea-level (ρ = 1.2550kg/m3) and the explosion covers the entire 11 tile radius at the end of a TU (using Danials 1.71m/s speed estimate we can see that a TU is 18.7s based on the 20 tiles a soldier with 80TU's can cover) we can solve the above equation for bomb yield. (that is, E=R5ρ/t2) and arrive at a yield of about 6060.71 Joules of power - about 1.5 microtons. This seems to indicate that the estimate of the length of a TU is wrong, or that the explosion does not consume an entire TU. If we accept that the TU length is correct and that the explosion does not consume an entire TU - highly unlikely that an explosion would take 18.7 seconds to occur anyway - and run with a figure of 1/1000th of a TU (0.0187s for the explosion to occur) we find that the explosion is much more powerful 6060714855.6 J - or about 1.4 tons. And the figure will continue to rise the shorter the time-span for developing to that diameter is. - Shadow 17:03, 5 November 2007 (PST)
Using Danials estimate of 2.26m per tile the result is a bit more — 34077375066.3 J - about 8.1 tons. - Shadow 17:18, 5 November 2007 (PST)
Note: The Trinity test has been officially stated as having been 20 kilotons. Based on examination of the released videos of that blast one scientist has calculated it's power at about 22 kilotons. (The numbers he used were: R = 140m, t = 0.025, ρ = 1 - using those same numbers we can see that he arrived at a value of 86051840000000 J – about 20.6 kilotons) What the preceding means is that the above stated equation (E=R5ρ/t2 is correct :P - Shadow 17:55, 5 November 2007 (PST)
Sadly, Arrow Quivershaft has pointed out an error in my math regarding the length of a TU. I used the 20 tiles in 80 TU's as a base for the equation, multiplying the '20 tiles' by the distance of 1.6 meters per tile. However, I must have been out of my mind in reporting 18.7 seconds per TU. Running the numbers by hand (on a piece of paper!) I've found that 20*1.6*1.71 makes that 80 TU turn approximately 55 seconds long. Dividing that by the 80 TU's in said turn we find that each TU is about 0.69 seconds long. Re-running the above stated equation with the reactor detonation taking 1 TU and 1/1000th of a TU we get the following results:
  • 1 TU: 3,572,876 J - not even 1 ton
  • one thousandth of a TU: 3,572,875,919,360 - 854 tons
If we use Danials 2.26m per tile figure we find that 1 TU is 0.97 seconds long. Solving the above equation using the 2.26 meters and 0.97 seconds per TU figure we have the following numbers for the size of a reactor cores explosion:
  • 1 TU: 10,302,987 J - again, not even 1 ton
  • one thousandth of a TU: 10,302,987,085,467 J - 2.5 kilotons
As you can see, the size of a tile not only affects the length of a TU (0.69s for a 1.6mx1.6mx2.4m tile, 0.97s for a 2.26mx2.26mx2.4m tile) but also the apparent yield of a reactors explosion. Now, as noted before (I think I mentioned it anyway), 1 gram of matter converting to energy creates an explosion of around 20 kilotons. This means that, if a reactors explosion is nuclear in nature, it's using about 1/10th of a gram of Elerium to create the result.
Finally, I did forget to mention that the figure only applies if the reactor detonations fireball creates the 11 tile damage radius and it is not the result of other features of a nuclear detonation. - Shadow 18:29, 5 November 2007 (PST)
Stepping away from all previous figures and estimating a tile as being 1.71m2, and using the bog standard human walking rate of 1.71m/s we find that, since it takes a rookie an average of 4 TU's to cross one tile that a single TU is 1/4 of a second. Using that together with the above stated 1.71m2 tile size we get the following results:
  • 1TU: 37,675,928 J - about 0.009 tons
  • one thousandth of a TU: 37,675,928,185,077 - about 9 kilotons
Again, this assumes that all the damage seen inside that 11 tile radius was caused by the fireball. - Shadow 18:57, 5 November 2007 (PST)
These "initial fireballs" are a somewhat artificial concept, allowing you to gauge the yield of a nuclear weapon. But equalling it to any distinguishable radius of destruction is nonsense. I suggest you refer to wikipedia:Effects of nuclear explosions#Direct effects instead. --Schnobs
I see your Wikipedia article and raise you the one that spawned the equation I used - [[wikipedia:Nuclear weapons yield#Calculating yields and controversy. I never said "Initial Fireball" and that isn't an "Artificial Concept". Each Nuclear Detonation creates a "Fireball" of varying size, simply because, in pure fission reactions, it causes the air to superheat and in Fusion reactions, because that is what the insanely hot, reacting plasma is - a mass of super-hot plasma. - Shadow 19:23, 6 November 2007 (PST)
A value of 200m for the Nagasaki Bomb struck me as a trifle odd. Ain't that a bit small? Also, what's this talk about a precise fireball size? Shouldn't it just get larger and larger until it eventually dissipates? So, when exactly will the fireball have the stated size?
I therefore read that article and a followed a few source links (hint: each picture has an article of it's own). Apparently, the given size refers to the moment when the shock wave will overtake the plasma; the air at the shock front will be heavly compressed, hence also hot and bright, but nowhere as bright as the plasma; also, the shock front is incandescent, eclipsing the fireball inside (I'm really struggling for words here, I hope you still get what I'm trying to tell). It was me who called it an initial fireball because it happens after a few milliseconds. Check that Trinity photo -- it's dated 25ms and depicts the shockfront, not the fireball. Maybe I should have named it not "artificial but an "abstract" concept, though. But I still hold the opinion that it's too simple to equal the burst pattern on a UFO floor to the size the fireball happens to have by the time the shock front catches up. --Schnobs 14:14, 7 November 2007 (PST)
Ah, okay. I can't argue with that. And you are right, that picture is from the moment when the shock front ovtertakes the plasma. If memory serves, it's the shock front that does the damage, so I'm probably wrong in the figures, although thinking about it, there is no real way to apply the equation in the manner that I did. At that, it might be impossible to apply it at all.
As to the small size of the fireball from both the Fat Man and Little Boy devices, you have to remember that both Hiroshima and Nagaski are located in natural depressions in the terrain. They are something like bowls, low plains surrounded by hills and mountains. The shock front was powerful enough that, when they contacted the surrounding terrain, they reflected back into the bowl. So the cities were destroyed by more than the initial shockwave and fireball - they got hit by the shockwave at least twice. And the area directly under "Ground Zero" (both bombs were air burst - Fat Man had a radar proximity trigger) got hit even worse, because it was caught by the shockwave and used as a reflector for the shockwave. - Shadow 07:42, 8 November 2007 (PST)
If you absolutely want to gauge the yield of a power source explosion, I suggest you compare it to the damage done by something we know. How many grenades stacked on top of each other would it take to damage the UFO interior (walls & floor) in the way a power source explosion does? Then look up the amount of explosives in a typical hand grenade and there you go. As to fireball size... there's an old saying, comparing arguments like this to the paralympics. Yet I can not resist.
a) the size stated in that wikipedia article is by no means the final size. It's not as if the fireball would reach a certain size and suddenly vanish. It grows ever larger and starts to rise, becoming the cap of the mushroom cloud (by then it's no longer a ball and merely glowing, of course). Defining the fireball size as "the size at the moment when the shockwave catches up with the fireball" is somewhat arbitrary, but you've got to draw the line somewhere. And this definition refers to a time when it's still most certainly a ball, and it'S size directly related to the yield of the bomb (that is to say, other factors like terrain, wind, air pressure, moisture etc are relatively insignificant). That's alright for an article that deals with yield (as in kt or MT equivalents); damage is something else entirely, covered in another article.
b) "Overtaking" sounds as if the shockwave would start somewhere inside the fireball, but of course the shockwave is not a seperate effect. It's the result of all the superheated air trying to expand rapidly. Some sources say the shockwave "detaches from" the fireball, which seems more apt.
c) I guess you've misunderstood the Mach Stem: the reflected shock front moves faster in air that has already been accelerated by the main shock front; the reflected front will therefore catch up and combine with the main shock front, reinforcing it. (diagram). That's reinforcement, not amplification: the shock wave will not become more forceful because of the Mach effect, but it will lose power at a slower rate (which means same damage over a larger area). Please note that the development of the Mach Stem does not depend on any terrain features (in fact, a flat surface may be as good as it gets). In the past few days I've read more than I ever wanted to know about nuclear explosions, but all sources go along the lines that the valley at Nagasaki served to confine the damage, not a single word about reinforcement or amplification. Maybe destruction in the valley was so complete that a possible reinforcement would have made no difference any more. --Schnobs 13:42, 8 November 2007 (PST)
Great armchair discussion, folks. Now if only you could bring some biology into it, I might jump in. Re: the indentation, I think the general idea is to only indent about 4-6 levels deep, then reset back to one. Also - and I hope I don't seem picky here, I love the flow of ideas - it seems like sometimes super and subscripting might be used better. A simple e.g. "m/s" is acceptable for meters per second, and sometimes it seems like e.g. "R = (Et2/ρ)1/5" might better be R = (Et2/ρ)1/5. On my PC, the extra super- and sub-scripting seems to be messing with line overlap.
All that aside, please keep going! It's an interesting reality check, to check back from e.g. the damage of a standard grenade, to back-evaluate the damage caused. One crazy thing about X-COM explosions is how they have "hard wired" edges, as seen in e.g. Explosions#Playing_With_Fire. That's hard to figure into the parlor computations. The standard grenade is the only explosive that does not have "artificial clipping" of its blast radius performed. - MikeTheRed 17:04, 9 November 2007 (PST)

Wow, never would have guessed that me being bored on a train and trying to kill time by this would start such a long discussion. Well... nice :) - tequilachef

Just in passing, maybe the reason continents don't evaporate when UFOs crash is the same reason there's not a mile-wide crater at Chernobyl - a nuclear meltdown is really nasty, but it doesn't turn the reactor into a bomb. And as for its use in explosives? How much of the material is used to create the bang and how much is used to control the bang? Even a gram of matter converted to energy can level a city - not really a weapon you wanna fire wildly in an atmosphere. - Kalaong

To be more specific, my theory is that a UFO Power Source uses Elerium to generate antimatter in a controlled manner(like Bob Lazar's theorizes), and to generate the anti-gravity field that keeps the UFO flying. Plasma weapons are the same - the Elerium provides energy to generate the plasma, then uses an anti-gravity field to keep the plasma from dissipating as it flies across the battlefield. Fusion weapons(the stuff that goes boom) are the odd man out - they generate anti-matter in an out-of-control manner and use the anti-gravity field to control the resulting explosion. In other words, Elerium(In the game! I don't wanna sound like a conspiracy-obsessed geek, just your garden variety geek!) is like nuclear material-plus. You have to encourage it to make all that energy, and when you stop encouraging it, it only has a relatively small tantrum. You have to get really creative to make use Elerium as a Weapon of Mass Annihilation. The aliens(And X-COM!) purposely limit the destructive potential of Elerium-based munitions and technology, despite this lowering the efficiency of the rare element. Otherwise a single Blaster Bomb would kill a city, leaving nothing for the aliens to abduct - or X-COM to protect. It's also why losing the Cydonia mission ends the game - the aliens could have reduced the Earth to beaded glass at any time,

but never consider it until their home base is threatened.

i recall vaugely that 70% of reacted antimatter becomes neutrinos in an incredibly short time frame--(name here) 14:34, 13 March 2008 (PDT)

P.S. I was hoping somebody with a giant brain would come by and critique me like they did the above stuff? My knowledge of nuclear physics is pretty much limited to the contents of the Wikipedia article and Winchell Chung's "Atomic Rockets" website. - Kalaong 17:51, 11 March 2008 (EDT)

I've just mentioned this discussion on the Talk:Main Page. I think it would be great if you guys made a page trying to explain how Elerium would act on the real world, using all those nice scientific formulas and commentaries. - Hobbes 23:35, 10 March 2008 (PDT)

Funnily enough the main page does not seem to have the in-game UFOPaedia entry :

This element has the unusual property of generating anti-matter power when bombarded with certain particles. This creates gravity waves and other forms of energy. It is not naturally found in our solar system, and cannot be reproduced.

This doesn't say anything about the efficiency of the anti-matter conversion (if that's the process it works by). Is that stated elsewhere?

I would challenge the energy calculations done on the Avenger. We have no idea of the drag coefficient of the aircraft though we could take a guess at something similar to an SR-71 Blackbird (which was no doubt built with scraps scavenged from Roswell). And how exactly did we guess the weight of the Avenger? But that's a minor point, much more to the point is that the alien craft fly by non-Newtonian mechanics - bending gravity waves. Calculating the energy that would be required to do the same motion using Newtonian mechanics does not let you deduce the energy required to do it with Elirium-based anti-gravity. Presumably the point of anti gravity, warp drives, wormholes & the like is precisely that they allow you to get more motion for less energy. How much less, we don't know.

On the other hand, serious physicists could calculate the amount of energy required to bend gravity waves. In Einsteinian terms the amount of mass-energy is immense - you would be much better of just building a big jet/rocket thingy and flying the plane the old fashioned way. So it looks like the alien engines are non-Einsteinian as well as non-Newtonian. Once the antigravity field is working, they might well run on the equivalent of a nine volt battery or the body heat of the operator's hands. Or psi energy.

So I don't think we can deduce much from the mechanics of Avenger flight. If anything, it is a maximum limit on Elirium energy (unless alien tech is LESS efficient than human tech).

What about the argument from weapon and engine explosions? As was noted, weapons are not engines and engines are not weapons. Not by design anyway. If Elerium really was as dangerous as antimatter, it would make sense to 'nobble' Elerium weapons and make them weaker. My take on it is that Elerium is a material which generates microscopic amounts of antimatter, and/or gravity waves, when exposed to exotic particles. The antimatter flux could be orders of magnitude smaller than the particle flux. As the Elerium is used, it goes bad, becomes unusable. It's not necessarily all converted to energy, it's just turned into a useless lump that won't generate antimatter or gravity waves any more.

A good suggestion was to look at the observed effects of the weapons and deduce back from that an equivalent amount of pure conversion. As the Elerium weapons and even the spaceship explosions are small potatoes, even compared to crude fission weapons, we are not talking about converting all the Elerium into energy via E=mc2.

I don't believe the Elerium engine is a volatile antimatter rocket tottering on the brink of explosion. Imagine the scene: A pyramid on Mars. Enthusiastic Sectoid Chief Engineer is bouncing up and down with glee. He has just invented a new type of drive for travelling to alien planets and harvesting their bounty. It's super efficient, incredibly powerful, and just has one slight drawback. A little bit of damage, such as might be sustained from the crude weapons of the primitive humans, can cause the engines to become unstable and explode with supernova force at some random time after the damage. Perhaps after they have teetered in to crash land back at Cydonia base. Right next to the Holy Temple of Our Beloved Brain. Oops...

A swift but lifelong trip to the Pit of Psychic Torment for our intrepid Chief Engineer, methinks. And back to the drawing board. The fact is that dangerous engines will be designed with built in failsafes. And their design will in any event be very different from a weapon or a bomb in the basic principles. In our early years playing with jets and rockets, we humans had trouble telling them from bombs. They still have some similarities when they go wrong but they are different. So anyway I don't think we can deduce anything from the explosions of power sources other than to say they are either low powered, or well dampened by failsafes, or maybe both.

In conclusion my bid for 'what is 1 Elerium' is that it weighs exactly what the game says it ways, and after being exposed to 'certain particles' for a while to produce nanoscopic amounts of antimatter and copious amounts of other exotic things, the Elerium is 'burned out' and a useless inert husk of roughly the same weight. Spike 14:38, 24 March 2008 (PDT)

Fairly good arguments. As for your question, the description for the UFO Power Source does indeed claim the efficiency of Elerium; it says the UFO Power Source converts elerium directly into antimatter with 99% efficiency. Arrow Quivershaft 15:21, 24 March 2008 (PDT)

Thanks, and thanks for the help AQ! Looking at that UFOPaedia entry for UFO Power Source (maybe we should have both entries in the main article?), the wording is compatible with the idea that not all the matter is converted into energy. A small amount could be converted at 99% (the part that is struck by the tiny amount of antimatter generated by the Elerium), and the rest of the Elerium is inert. Or indeed the 'reaction matter' might be some other matter (hydrogen for example), and not the Elerium at all - the Elerium may just be the antimatter source, and not the source of the 'reaction matter'. I'm using reaction matter in a different sense than when we talk about rockets and 'reaction mass'. Think of the anti matter as the flame and the matter as the fuel (wood, coal, plutonium, lithium, etc). Or to use another analogy, Elerium is like the platinum catalyst in the mythical Cold Fusion powerplant. Spike 15:38, 24 March 2008 (PDT)

That's entirely possible, IMO. It should be noted that any reactor with a 100% efficiency rating of energy production would violate Newton's Second Law of Thermodynamics, and as such, would demolish a good deal of our current scientific understanding. Not that it's not impossible; the programmers just probably thought it would be a good idea to avoid anything incredibly earth-shattering, science-wise. Arrow Quivershaft 16:24, 24 March 2008 (PDT)
What about this: Power Sources don't detonate because to achieve this high efficiency, a catalyst is required? Alien Engineers would remove it from the reactor to prevent it from going supercritical before impact. Plus, we already know that X-COM hybrid tech is much more efficient than alien tech. For example: A Battleship has four PS. An Avenger has two, yet it is actually FASTER. If you think this is because of the size difference: Firestorms and Medium Scouts are about the same size. Both have one PS, yet the Firestorm is nearly twice as fast (short of 600km/h). In fact, the only known exception to this is the Blaster/Fusion Launcher pair, where the vanilla tech is better.--amitakartok 14:41, 16 November 2008 (CST)

What's wrong with using object weight to calculate mass of elerium? heh. Anyhow, you only need one factor to throw your equation off: Alien Alloys and special construction have highly reduced friction coefficients. Looking at the aerodynamics of an Avenger, I would go so far as to say that the alien alloys and anti-gravity fields most likely create a near frictionless barrier surrounding all Alien Space craft... obviously that must be why Earth's top scientists didn't bother to give a sleek shape to a craft designed to maintain >Mach8 for over 10 hours! Even more interesting, try applying these calculations to a Leviathan Submarine or other TFTD craft. A Baracudda appears (by my rough estimates) to travel over 1/4 the Earth's circumference in under 4 hours. In other words, faster than 2500kmph. Underwater. Using conventional Earth Fuel... jasonred

Bob Lazar claims in an interview that a UFO Power Source required 223 grams of Elerium-115 to function. Not sure how this ties in, but the number is pretty close to the one calculated on the article page. Any ideas? --Zombie 23:10, 24 November 2009 (EST)

About Elerium mining...

Re the "mining near Cydonia" issue. As per the UFOpedia:

"It is not naturally found in our solar system and cannot be reproduced."

However, in X-Com 3, Transtellar mines the stuff from Mars (and brings back regular shipments). Therefore, there must be a reserve near Cydonia.

While there is no official explanation, a meteorite seems to be the most likely cause of this.

- Bomb Bloke 21:41, 31 May 2007 (PDT)

My apologies. I do not have access to X-Com: Apocalypse, and was basing my data merely on what was said in the . In light of the new information, it can be reverted if you desire. `Arrow Quivershaft 21:56, 31 May 2007 (PDT)
It could also be the wreckage of alien craft and structures destroyed during the cydonia attack. they might have had a squadron of defensive ships which went down somewhat intact, maybe shattering inactive ships on the ground. add in elerium in the weapons and armor of the ground troops which was left behind, and you'd have quite a lot of it underground.--(name here) 09:57, 28 April 2008 (PDT)

No, there IS a reserve near Cydonia. The final X-COM cleanup mission in November 2009 didn't found it, it was found in October 2061. Given that it is still mined in X-COM 3, if a X-COM 1 Battleship was powered by 200 units of Elerium, the reserve might be thousands of quadrillions (one quadrillion is 1 000 000 000 000 000 or 10^15). And I believe X-COM 3 is after X-COM 4 in the timeline; X-COM 4 states that there is a synthetic variant which is weapons-grade (at least for space-weapons). As for no official explanation: "In October, 2061, after several weeks of transmissions, the Mars probes show an indication of what could be a huge stockpile of Elerium-115 several thousand kilometers from Cydonia." (Excerpt from the 'Classified - Eyes Only' booklet provided with X-COM: Interceptor - amitakartok 0:56, 31 August 2008

Reserve, not mine. There's a big difference Vizzydix1 22:01, 18 March 2010 (EDT)
Apocalypse and Interceptor mention a Mars Elerium miners' union. Perhaps the stockpile was just the stuff that was already mined, processed and stored? Hobbes 21:39, 19 March 2010 (EDT)

Chuck Norris?

There is a grave error on this page: Chuck Norris is GOD! He can't be killed ;) Hobbes 14:01, 11 March 2008 (PDT)

What about if Steven Seagal, Bruce Lee and Mike Tyson ganged up on him and jumped him from behind after he'd had a few beers at the end of a hard day defending America from evil drug lords? Spike 16:41, 11 March 2008 (PDT)
This is entertainingly silly, but now has nothing to do with the game. Start theorizing on how said demigods would fare against Mutons or somebody's going to have to delete this thread;)! Kalaong 16:28 13 March 2008 (EDT)
Mutons would be able to crush Chuck Norris beneath their endless legions of sucky terror units
If one of X-COM's soldiers was Chuck Norris, he would have a Melee attack with a strength of about 60, requiring 10% TUs per attack. His stats (including firing accuracy) would be maxed out. However I like to believe that Bruce Lee's Melee damage would be nearer to 200, allowing him to punch through UFO doors. (Are we back on topic now?) Spike 15:23, 13 March 2008 (PDT)
"Bruce Lee's Melee damage would be nearer to 200"??? Don't be silly, he would be well over 9000 so Battleships would hide in shame XD -amitakartok 23:32 28 August 2008


The reason why Mutons are green is because they used to be red (like in the game introduction) but they turned to that colour the first time they saw Chuck Norris. Hobbes 16:51, 13 March 2008 (PDT)
Exxxcellent. Smithers, hire that man as my personal bodyguard and send whatshisname the Safety Inspector from Sector 7-G to Cydonia in his place>;PPPP Kalaong 1:30 14 March 2008 (EDT)

See Also



Seems I can't edit the talk page right now, so I'll "comment" here. Um, I'm no nuclear physicist, but I'm pretty sure that there's a difference between "blowing something up" and "making something blow up".
That is to say, if you could get a horse to combust, you'd probably get quite a bit of energy out of it. - Bomb Bloke 09:26, 21 March 2009 (EDT)
Actually, you can get horses to combust... just set them on fire. duh. There are actually cases of horse corpses exploding, through intestinal gas... cute? Jasonred 10:32, 21 March 2009 (EDT)

Separate pages for EU2012 and the Bureau?

It might be better to use the already existing Elerium (EU2012) page for the stuff present on Enemy Unknown. Hobbes 07:48, 3 September 2013 (EDT)

There was. It's a blank/stub now. (same thing with the linked pages I could find for Weapon Fragments, too). I don't remember if they were big enough to warrant their own pages: apparently not, given that those pages were wiped. Plus, would that mean copy-pasting the "Elerium in Reality" part two more times? --Xuncu 15:02, 3 September 2013 (EDT)
There's enough content to fill those pages (namely the descriptions on the Research (EU2012) page, the problem is the work involved. I've started today collecting images for all alien artifacts and designing those pages but that's going to take a while. Hobbes 16:01, 3 September 2013 (EDT)
I'm pretty sure there was already a page for Weapon Fragments set up, but didn't you delete it to make room for the move of the other fragment page that was made.--Ditto51 17:07, 3 September 2013 (EDT)
And I can't even find the other page(s). All links I checked go to that stub.--Xuncu 23:01, 3 September 2013 (EDT)
On the delete Log (Either the 5th or 6th down) is the Weapon Fragments Page that had the information on before it was deleted for some unknown reason--Ditto51 03:23, 4 September 2013 (EDT)
Hm, I'll leave that to you then, Hobbes, while I'm currently focusing on getting screenshots from The Bureau -- fyi: I don't know how to get stuff directly out of the program files. Like in the Area 51 level, there was a Periodic Table that also had a full atomic analysis on Elerium that'd be great for any Elerium page-- well, what I was thinking, since Elerium is the franchse McGuffin, it could go on one page for the whole series, while/since the different iterations of it have not been as different as, say, the Sectoids have been (different stats and abilities in both EUs, and as NPCs in Apocalypse). For example, the Master Sword has had different designs, strenghts, and backstories, but only one page in eanch of the two big Legend of Zelda wikias for the whole francise. --Xuncu 22:26, 3 September 2013 (EDT)
With EU there's an editor developed for the Unreal 3 engine that allows to extract the textures to png or bmp formats (but editing them for use here is very work intensive) but with the Bureau the only option is Print Screen, I'm afraid.
There was already a discussion on this wiki about using the same page for certain elements that appear throughout the series but at the time we decided to use separate pages but linking them together. On the Civilization wiki they actually use both, one page for the 'Settler' unit, then separate pages for the Settler unit on each of the 5 games. That might be a way to tackle this issue later on.Hobbes 13:09, 5 September 2013 (EDT)
Hm, interesting method. At any rate, I've been collecting screenshots, at least enough to start developing a list of what I should and haven't yet, and think I have a way to publicly post at least a .rar; I guess I'll post to that in the Community pages when I'm done, so that anyone could go through them and help develop a more organized system than the half-rp, half meta ranting that I've been writing. Though seeing that the reviews have fallen between 3/10ish and 6/10ish, I'm not too confident about seeing an influx of writers like EU got. --Xuncu 15:47, 5 September 2013 (EDT)


Apocalypse isn't mentioned on this page. This is rather odd, considering that Elerium features in Apocalypse. Magic9mushroom (talk) 00:12, 3 April 2015 (EDT)