Talk:Elerium-115

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

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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

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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

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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)

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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)

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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)

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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)

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)