Talk:Elerium-115

Bug

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:

F_d = ¹/₂ρv²C_dA

where ρ is the density of the air, v is the velocity, C_d is the coefficient of friction and A is the surface area. Using this we get F_d = 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=F_dR, where F_d 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)

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 = 1.2550^kg/_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 ¹/₁₀₀₀^th 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 ¹/₁₀₀₀^th 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 ¹/₁₀^th 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 1.71^m/_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)

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 game. In light of the new information, it can be reverted if you desire. `Arrow Quivershaft 21:56, 31 May 2007 (PDT)

After the events of XCOM, the last Elerium deposit was loaded into an Avenger to return to Cydonia for Elerium. The crew didn't found anything. After TFTD, conventional ships have been sent to Mars and detected a huge Elerium deposit several hundred kilometres from Cydonia, possibly stashed in the first war. Using that reserve, ship research skyrocketed, leading to development of faster-than-light engines. In addition, Interceptor (XCOM 4) says that, during the third war in the Frontier, scientists have eventually managed to create E-115 artificially.

- amitakartok

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

See Also

Elirium