UFOpaedia - User contributions [en]

Talk:Elerium-115

2007-11-08T15:42:19Z

Shadow:

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 - [[User:Phoenix|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).

-[[User:Bomb_Bloke|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! :) - [[User:Phoenix|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) --[[User:Ethereal Cereal|Ethereal Cereal]] 21:33, 8 May 2006 (PDT)

----

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.

- [[User:Bomb Bloke|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. `[[User:Arrow Quivershaft|Arrow Quivershaft]] 21:56, 31 May 2007 (PDT)

----

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).

- [[User:MikeTheRed|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. [[User:Arrow Quivershaft|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. - [[User:Shadow|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) - [[User:Shadow|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. - [[User:MikeTheRed|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. - [[User:Shadow|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. - [[User:Shadow|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. ;) - [[User:MikeTheRed|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. - [[User:Shadow|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. --[[User:Schnobs|Schnobs]] 19:01, 3 November 2007 (PDT)

:I've written a [http://shadowwolf.keil-draco.com/solver.py.txt 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 [[User:MikeTheRed|MikeTheRed's]] reminder that Elerium reactors do not chain-react and [[User:Schnobs|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. - [[User:Shadow|Shadow]] 01:21, 4 November 2007 (PST) (corrected later - [[User:Shadow|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. - [[User:Shadow|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 ([http://www.bufata-chemie.de/reader/ig_farben/pics/1-4-3_01_oppau-big.jpg picture]) that was rated in kilotons. [http://en.wikipedia.org/wiki/Oppau_explosion 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. --[[User:Schnobs|Schnobs]] 10:51, 5 November 2007 (PST)

::::The values come from the [[wikipedia:Nuclear weapon yield|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. - [[User:Shadow|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?! - [[User:Bomb Bloke|Bomb Bloke]] 22:29, 4 November 2007 (PST)

:::It came from [[User_talk:Danial|here]]. [[User:Arrow Quivershaft|Arrow Quivershaft]] 22:38, 4 November 2007 (PST)

::::Correct. [[User:Danial|Danial]] has estimated that each tile is approximately 2.26m2 on his [[User talk:Danial|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. - [[User:Shadow|Shadow]] 15:30, 5 November 2007 (PST)

::::If [[User:Bomb Bloke|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. - [[User:Shadow|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 [[User:Danial|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. - [[User:Shadow|Shadow]] 17:03, 5 November 2007 (PST)

::::Using [[User:Danial|Danials]] estimate of 2.26m per tile the result is a bit more — 34077375066.3 J - about 8.1 tons. - [[User:Shadow|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 [[wikipedia:Geoffrey Ingram Taylor|one scientist]] has calculated it's power at about [[wikipedia:Nuclear weapon yield#Calculating yields and controversy|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 - [[User:Shadow|Shadow]] 17:55, 5 November 2007 (PST)

::::Sadly, [[User:Arrow Quivershaft|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 [[User:Danial|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. - [[User:Shadow|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. - [[User:Shadow|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. --[[User:Schnobs|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. - [[User:Shadow|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. --[[User:Schnobs|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. - [[User:Shadow|Shadow]] 07:42, 8 November 2007 (PST)

Talk:Elerium-115

2007-11-07T03:23:35Z

Shadow:

Talk:Elerium-115

2007-11-06T02:57:21Z

Shadow:

Talk:Elerium-115

2007-11-06T02:29:00Z

Shadow: correct myself again :(

Talk:Elerium-115

2007-11-06T01:55:50Z

Shadow: add a note about the "yield from radius" equation and the test done to make sure the transformation from "radius from yield" was correct

Talk:Elerium-115

2007-11-06T01:18:00Z

Shadow:

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 - [[User:Phoenix|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).

-[[User:Bomb_Bloke|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! :) - [[User:Phoenix|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) --[[User:Ethereal Cereal|Ethereal Cereal]] 21:33, 8 May 2006 (PDT)

----

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.

- [[User:Bomb Bloke|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. `[[User:Arrow Quivershaft|Arrow Quivershaft]] 21:56, 31 May 2007 (PDT)

----

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).

- [[User:MikeTheRed|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. [[User:Arrow Quivershaft|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. - [[User:Shadow|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) - [[User:Shadow|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. - [[User:MikeTheRed|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. - [[User:Shadow|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. - [[User:Shadow|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. ;) - [[User:MikeTheRed|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. - [[User:Shadow|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. --[[User:Schnobs|Schnobs]] 19:01, 3 November 2007 (PDT)

:I've written a [http://shadowwolf.keil-draco.com/solver.py.txt 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 [[User:MikeTheRed|MikeTheRed's]] reminder that Elerium reactors do not chain-react and [[User:Schnobs|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. - [[User:Shadow|Shadow]] 01:21, 4 November 2007 (PST) (corrected later - [[User:Shadow|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. - [[User:Shadow|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 ([http://www.bufata-chemie.de/reader/ig_farben/pics/1-4-3_01_oppau-big.jpg picture]) that was rated in kilotons. [http://en.wikipedia.org/wiki/Oppau_explosion 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. --[[User:Schnobs|Schnobs]] 10:51, 5 November 2007 (PST)

::::The values come from the [[wikipedia:Nuclear weapon yield|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. - [[User:Shadow|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?! - [[User:Bomb Bloke|Bomb Bloke]] 22:29, 4 November 2007 (PST)

:::It came from [[User_talk:Danial|here]]. [[User:Arrow Quivershaft|Arrow Quivershaft]] 22:38, 4 November 2007 (PST)

::::Correct. [[User:Danial|Danial]] has estimated that each tile is approximately 2.26m2 on his [[User talk:Danial|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. - [[User:Shadow|Shadow]] 15:30, 5 November 2007 (PST)

::::If [[User:Bomb Bloke|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. - [[User:Shadow|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 [[User:Danial|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 1/100th 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. - [[User:Shadow|Shadow]] 17:03, 5 November 2007 (PST)

::::Using [[User:Danial|Danials]] estimate of 2.26m per tile the result is a bit more — 34077375066.3 J - about 8.1 tons. - [[User:Shadow|Shadow]] 17:18, 5 November 2007 (PST)

Talk:Elerium-115

2007-11-06T01:03:26Z

Shadow:

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 - [[User:Phoenix|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).

-[[User:Bomb_Bloke|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! :) - [[User:Phoenix|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) --[[User:Ethereal Cereal|Ethereal Cereal]] 21:33, 8 May 2006 (PDT)

----

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.

- [[User:Bomb Bloke|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. `[[User:Arrow Quivershaft|Arrow Quivershaft]] 21:56, 31 May 2007 (PDT)

----

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).

- [[User:MikeTheRed|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. [[User:Arrow Quivershaft|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. - [[User:Shadow|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) - [[User:Shadow|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. - [[User:MikeTheRed|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. - [[User:Shadow|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. - [[User:Shadow|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. ;) - [[User:MikeTheRed|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. - [[User:Shadow|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. --[[User:Schnobs|Schnobs]] 19:01, 3 November 2007 (PDT)

:I've written a [http://shadowwolf.keil-draco.com/solver.py.txt 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 [[User:MikeTheRed|MikeTheRed's]] reminder that Elerium reactors do not chain-react and [[User:Schnobs|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. - [[User:Shadow|Shadow]] 01:21, 4 November 2007 (PST) (corrected later - [[User:Shadow|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. - [[User:Shadow|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 ([http://www.bufata-chemie.de/reader/ig_farben/pics/1-4-3_01_oppau-big.jpg picture]) that was rated in kilotons. [http://en.wikipedia.org/wiki/Oppau_explosion 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. --[[User:Schnobs|Schnobs]] 10:51, 5 November 2007 (PST)

::::The values come from the [[wikipedia:Nuclear weapon yield|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. - [[User:Shadow|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?! - [[User:Bomb Bloke|Bomb Bloke]] 22:29, 4 November 2007 (PST)

:::It came from [[User_talk:Danial|here]]. [[User:Arrow Quivershaft|Arrow Quivershaft]] 22:38, 4 November 2007 (PST)

::::Correct. [[User:Danial|Danial]] has estimated that each tile is approximately 2.26m2 on his [[User talk:Danial|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. - [[User:Shadow|Shadow]] 15:30, 5 November 2007 (PST)

::::If [[User:Bomb Bloke|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. - [[User:Shadow|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 [[User:Danial|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 1/100th 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. - [[User:Shadow|Shadow]] 17:03, 5 November 2007 (PST)

Talk:Elerium-115

2007-11-06T00:21:57Z

Shadow:

Talk:Elerium-115

2007-11-05T23:30:25Z

Shadow: explanations of figures and a rebuttal

Talk:Elerium-115

2007-11-05T05:09:28Z

Shadow: just a bit of conjecture about the size of a reactor core explosion

Realistic Equivalents

2007-11-05T04:52:59Z

Shadow: /* Elerium-115 */ note that a reactor detonating like a bomb is not necessarily nuclear, nor does the blast have to be caused by all the available reaction mass

Realistic equivalents from military and other sources, to add impact to your fanfiction.

== Weapons ==

*[[Rifle]] - [http://remtek.com/arms/hk/mil/g36/g36.htm HK G36] was discussed but it could really be any modern battle rifle - H&K G3, FN FAL, M14...

*[[Pistol]] - SOCOM use .45 ACP, aka [http://remtek.com/arms/hk/civ/mark23/mark23.htm HK MK23]. Very similar to the game pistol in magazine capacity and power (compared to the game rifle, anyway).

*[[Auto Cannon]] - The new 25mm XM307 [http://www.globalsecurity.org/military/systems/ground/ocsw.htm OCSW] could rapidly fire AP and HE grenades. No rotary barrel though it does have automatic fire. Manville 25mm is an obscure make of 18-shot revolver gun.

*[[Heavy Cannon]] - [http://global-military.info/weapons/mgl-mk1.html MGL-MK1], revolver 6-shot 40mm [http://world.guns.ru/grenade/gl14-e.htm grenade launcher], with HE, AP, and In grenades. & (Perhaps upgraded versions could fire high-velocity grenades like the MK19 automatic grenade cannon.) See also Russia's RG-6 (6G30).

*[[Rocket Launcher]] - Predator Antitank Missile has been suggested. The M136 AT-4 is a disposable one-shot. The reloadable 84mm [http://www.globalsecurity.org/military/systems/ground/m3-maws.htm Carl Gustav (MAAWS) launcher] is perhaps a closer equivalent.

*Laser weapons - no modern equivalent. [http://www.cndyorks.gn.apc.org/yspace/articles/battlelaser.htm General Electric]?

== Equipment & Explosives ==

*[[Medi-Kit]] - Stimulants and painkillers aside, [http://www.hemcon.com/ Chitosan] battlefield bandages clot severe bleeding wounds instantly. (Secret ingredients: ground shrimp shells and vinegar. From Slashdot.) Also [http://www.simplerlife.com/quikclot.html QuikClot] - In real life the Air Force gives its pilots amphetamines in order to keep them awake, I'm sure that this would use something similar

*[[Electro-flare]] - Invented because burning flares would set fires. Used as [http://www.uai.ca/sales/images/meprolight/electroflare.htm emergency road markers] or [http://www.uai.ca/sales/images/bartan/lightgrenade.htm Electric Light Grenade]

*[[Grenade]] - Standard M67 fragmentation grenade. Note the nearest thing to our timer is "cook-off" time. Electronic fuses are possible and reliable.

*[[Proximity Grenade]] - no equivalent. Claymore tripwires and ultrasonic sensors are possible. [http://www.fas.org/man/dod-101/sys/land/slam.htm "SLAM Mine"] appeared in Splinter Cell as the Wall Mine.

*[[High Explosive]] - M183, 20lb C4 satchel charges.
*[[Stun Bomb]] - Advanced version of [http://www.pepperball.com/faqs.asp Pepperballs]

== Vehicles & Armament ==

*[[Tank/Cannon]] - M242 Chain Gun, 25mm, [http://www.sfu.ca/casr/101-m242.htm "Bushmaster"] (modified for single shot) mounted on [http://www.globalsecurity.org/military/systems/ground/talon.htm "Talon"] robot chassis.

Or something larger, like an [http://www.magnatrac.com/ electric mini bulldozer].

*[[Tank/Rocket Launcher]] - [http://www.inetres.com/gp/military/infantry/flame/M202.html M202 "Flash" Quad Launcher], 66mm. (2 per turret @ 4 shots each) Originally developed to disperse incendiaries but fits M-72 LAW rounds, apparently.

*[[Interceptor]] fighter - USAF Lockheed [http://www.geocities.com/nicscics/YF22.htm F-22 Raptor] scheduled for release 2004 will replace the F-15. [http://en.wikipedia.org/wiki/F-35_Lightning_II F-35 Lightning II] hit production 2006.

*[[Skyranger]] transport - [http://de.wikipedia.org/wiki/Dornier_Do_31 DO-31 VSTOL Transporter], [http://www.aeronautics.ru/archive/vvs/an72-01.htm Anotov AN-72] (Cadmus) or Osprey CV-22 have the right size and range.

The german Do-31 concept was intended as a combat zone transporter carrying up to 36 fully equipped troops over a mission radius of about 900 km. It had the ability of taking off and landing vertically with the help of wing mounted engine pods and was able to land on unprepared surfaces. Two flying prototypes were built in Germany during the 1960s.

"Payload considerations have also been a sore point with the CV-22. Planners will have to take into consideration its reduced cabin volume and lift capacity. Not only does it carry fewer troops (18 vs 27), it is also incapable of transporting any of the armored, wheeled vehicles currently used by SOF teams. Procurement of a new vehicle is still pending..."
CV-22 Tiltrotor Osprey compared to Pavelow helicopter in Special Forces deployments, 1 June 1998. Could that new vehicle be [[Heavy Weapons Platforms|HWP]]-sized?

== Unit size, speed, game turn length, etc. ==

See Danial's interesting [[User_talk:Danial|page]] on various game measurements compared to real life. ''(If anyone wants to dive into this deeper, make a new page, copy his stuff over, and have a link at the top of the new page giving kudos back to Danial's page... due to the fact it's '''his''' page, it's not really a place for public revision.)''

== Elerium-115 ==

After solving both the [[wikipedia:Barometric equation|Barometric equation]], the [[wikipedia:Drag formula|Drag formula]] and [[wikipedia:Einstein|Einsteins]] [[wikipedia:Mass-energy equivalence|mass-energy equivalence]] equation we arrive at the mass of a unit of Elerium-115 being approximately 2.13 grams.

Terror Ships have 4 reactors with 50 units of Elerium-115 each. If all the E-115 in the reactor detonated at once, it would have an explosive force of about 2.3 megatons. This value is almost exactly the size of two [[wikipedia:B83 nuclear bomb|US B83 thermonuclear devices]] at their maximum yield. (It is highly unlikely that 100% of the E-115 in the reactor would detonate in this way we would need to assume a figure for the amount of E-115 that goes into an uncontrolled reaction - or estimate the amount based on the displayed blast radius.)

There are some interesting armchair conjectures on the power contained in Elerium, both on the [[Elerium-115#How_much_Elerium_is_.E2.80.9E1_Elerium.E2.80.9C.3F|Elerium-115]] page, and on its [[Talk:Elerium-115|Discussion]] page.

Realistic Equivalents

2007-11-05T04:36:02Z

Shadow: /* Elerium-115 */ Move a bit of the conjecture here and provide a real-world equivalent

Realistic equivalents from military and other sources, to add impact to your fanfiction.

== Weapons ==

*[[Rifle]] - [http://remtek.com/arms/hk/mil/g36/g36.htm HK G36] was discussed but it could really be any modern battle rifle - H&K G3, FN FAL, M14...

*[[Pistol]] - SOCOM use .45 ACP, aka [http://remtek.com/arms/hk/civ/mark23/mark23.htm HK MK23]. Very similar to the game pistol in magazine capacity and power (compared to the game rifle, anyway).

*[[Auto Cannon]] - The new 25mm XM307 [http://www.globalsecurity.org/military/systems/ground/ocsw.htm OCSW] could rapidly fire AP and HE grenades. No rotary barrel though it does have automatic fire. Manville 25mm is an obscure make of 18-shot revolver gun.

*[[Heavy Cannon]] - [http://global-military.info/weapons/mgl-mk1.html MGL-MK1], revolver 6-shot 40mm [http://world.guns.ru/grenade/gl14-e.htm grenade launcher], with HE, AP, and In grenades. & (Perhaps upgraded versions could fire high-velocity grenades like the MK19 automatic grenade cannon.) See also Russia's RG-6 (6G30).

*[[Rocket Launcher]] - Predator Antitank Missile has been suggested. The M136 AT-4 is a disposable one-shot. The reloadable 84mm [http://www.globalsecurity.org/military/systems/ground/m3-maws.htm Carl Gustav (MAAWS) launcher] is perhaps a closer equivalent.

*Laser weapons - no modern equivalent. [http://www.cndyorks.gn.apc.org/yspace/articles/battlelaser.htm General Electric]?

== Equipment & Explosives ==

*[[Medi-Kit]] - Stimulants and painkillers aside, [http://www.hemcon.com/ Chitosan] battlefield bandages clot severe bleeding wounds instantly. (Secret ingredients: ground shrimp shells and vinegar. From Slashdot.) Also [http://www.simplerlife.com/quikclot.html QuikClot] - In real life the Air Force gives its pilots amphetamines in order to keep them awake, I'm sure that this would use something similar

*[[Electro-flare]] - Invented because burning flares would set fires. Used as [http://www.uai.ca/sales/images/meprolight/electroflare.htm emergency road markers] or [http://www.uai.ca/sales/images/bartan/lightgrenade.htm Electric Light Grenade]

*[[Grenade]] - Standard M67 fragmentation grenade. Note the nearest thing to our timer is "cook-off" time. Electronic fuses are possible and reliable.

*[[Proximity Grenade]] - no equivalent. Claymore tripwires and ultrasonic sensors are possible. [http://www.fas.org/man/dod-101/sys/land/slam.htm "SLAM Mine"] appeared in Splinter Cell as the Wall Mine.

*[[High Explosive]] - M183, 20lb C4 satchel charges.
*[[Stun Bomb]] - Advanced version of [http://www.pepperball.com/faqs.asp Pepperballs]

== Vehicles & Armament ==

*[[Tank/Cannon]] - M242 Chain Gun, 25mm, [http://www.sfu.ca/casr/101-m242.htm "Bushmaster"] (modified for single shot) mounted on [http://www.globalsecurity.org/military/systems/ground/talon.htm "Talon"] robot chassis.

Or something larger, like an [http://www.magnatrac.com/ electric mini bulldozer].

*[[Tank/Rocket Launcher]] - [http://www.inetres.com/gp/military/infantry/flame/M202.html M202 "Flash" Quad Launcher], 66mm. (2 per turret @ 4 shots each) Originally developed to disperse incendiaries but fits M-72 LAW rounds, apparently.

*[[Interceptor]] fighter - USAF Lockheed [http://www.geocities.com/nicscics/YF22.htm F-22 Raptor] scheduled for release 2004 will replace the F-15. [http://en.wikipedia.org/wiki/F-35_Lightning_II F-35 Lightning II] hit production 2006.

*[[Skyranger]] transport - [http://de.wikipedia.org/wiki/Dornier_Do_31 DO-31 VSTOL Transporter], [http://www.aeronautics.ru/archive/vvs/an72-01.htm Anotov AN-72] (Cadmus) or Osprey CV-22 have the right size and range.

The german Do-31 concept was intended as a combat zone transporter carrying up to 36 fully equipped troops over a mission radius of about 900 km. It had the ability of taking off and landing vertically with the help of wing mounted engine pods and was able to land on unprepared surfaces. Two flying prototypes were built in Germany during the 1960s.

"Payload considerations have also been a sore point with the CV-22. Planners will have to take into consideration its reduced cabin volume and lift capacity. Not only does it carry fewer troops (18 vs 27), it is also incapable of transporting any of the armored, wheeled vehicles currently used by SOF teams. Procurement of a new vehicle is still pending..."
CV-22 Tiltrotor Osprey compared to Pavelow helicopter in Special Forces deployments, 1 June 1998. Could that new vehicle be [[Heavy Weapons Platforms|HWP]]-sized?

== Unit size, speed, game turn length, etc. ==

See Danial's interesting [[User_talk:Danial|page]] on various game measurements compared to real life. ''(If anyone wants to dive into this deeper, make a new page, copy his stuff over, and have a link at the top of the new page giving kudos back to Danial's page... due to the fact it's '''his''' page, it's not really a place for public revision.)''

== Elerium-115 ==

After solving both the [[wikipedia:Barometric equation|Barometric equation]], the [[wikipedia:Drag formula|Drag formula]] and [[wikipedia:Einstein|Einsteins]] [[wikipedia:Mass-energy equivalence|mass-energy equivalence]] equation we arrive at the mass of a unit of Elerium-115 being approximately 2.13 grams.

Terror Ships have 4 reactors with 50 units of Elerium-115 each. If all the E-115 in the reactor detonated at once, it would have an explosive force of about 2.3 megatons. This value is almost exactly the size of two [[wikipedia:B83 nuclear bomb|US B83 thermonuclear devices]] at their maximum yield.

There are some interesting armchair conjectures on the power contained in Elerium, both on the [[Elerium-115#How_much_Elerium_is_.E2.80.9E1_Elerium.E2.80.9C.3F|Elerium-115]] page, and on its [[Talk:Elerium-115|Discussion]] page.

Talk:Elerium-115

2007-11-04T10:24:07Z

Shadow: have to correct myself - it's a lot later than I like and my mind kept telling me that "Fat Man" and "Little Boy" both were about 600kilotons rather than the average of 17kilotons

Talk:Elerium-115

2007-11-04T09:23:42Z

Shadow: fix the ' where there should be a ' that I seem to have these days

Talk:Elerium-115

2007-11-04T09:21:41Z

Shadow: hopefully final corrections made

Talk:Elerium-115

2007-11-03T21:41:12Z

Shadow: a bit of "armchair science" about the possible reasons Elerium reactor detonations don't cause the damage the math seems to indicate they are capable of

Talk:Elerium-115

2007-11-03T06:03:58Z

Shadow: another math correction

Talk:Elerium-115

2007-11-03T05:54:49Z

Shadow: correct myself. 10^5km/h is 2778m/s, not 278m/s

Talk:Elerium-115

2007-11-03T05:12:02Z

Shadow: add a note about the maximum size of a terror ships detonation

Talk:Elerium-115

2007-11-03T01:23:18Z

Shadow: fix my math, do some nice formatting to the information and make it available

Talk:Elerium-115

2007-11-03T00:52:20Z

Shadow: Undo revision 13882 by Shadow (Talk) (Based some equations on 5400nm rather than the 54000nm that was quoted.)

Talk:Elerium-115

2007-11-03T00:48:09Z

Shadow: note problems with the displayed math, provide (hopefully) correct math