UFOpaedia - User contributions [en]

Psiclone

2025-10-29T05:43:58Z

Kalaong:

[[Image:Equip-Pedia-PsicloneV3-(Apocalypse).png|right]]
The Psiclone implant is manufactured and distributed by criminal gangs. It allows the user to experience any mental state or images just by imagining them. Its widespread popularity and detrimental effect on the health of young citizens led the Senate to ban the use or distribution of the device. The price of Psiclone implants has subsequently soared and this has resulted in open warfare between the criminal gangs and Megapol.
<table><tr><td>[[Image:Equip-InvIcon-Psiclone-(Apocalyse).png|left]]</td><td>
* Size: 1 × 1
* Weight: 6
* Manufacturer: [[Psyke]]
* Base Price: $4,500
* Battlescape Score: 3</td></tr></table>

A Psiclone is an inert item which X-Com cannot use and has no function except to increase funds when [[Tactical_Combat_Missions_(Apocalypse)#Raid|sold]]. A common drop in buildings belonging to the [[Cult of Sirius]]. Cannot be purchased from Psyke unless buying back whatever was sold earlier.

'''Its a respawning cash cow!'''

{{Equipment (Apocalypse) Navbar}}

[[Category: Apocalypse]]

The Mysteries of X-COM

2011-03-28T07:16:00Z

Kalaong: /* What sort of physical process is used to increase human stats over time? */

Discussion page for some less clear aspects of the series

==General issues==

==='''How fast can alien craft travel in space?'''===

Some alien missions (repeated attacks on X-COM bases, for instance) come daily. This seems to imply that alien craft are able to travel the distance from Mars to Earth in a matter of hours.

: Or maybe, as you suggest in your novels, they have a staging area near Earth, such as the dark side of the Moon. [[User:Spike|Spike]] 15:36, 25 February 2009 (CST)

::Really fast. Consider that they don't need to push aside atmosphere, as well as the fact that momentum is conserved in space, so they can achieve very high speeds with gravity slingshots. (Mars DOES have 2 moons, recall.) Also note that they may be operating a bit closer to home(the far side of the moon, perhaps?) it's simply that the command staff are at Cydonia. [[User:Arrow Quivershaft|Arrow Quivershaft]] 00:19, 26 February 2009 (CST)

Good article in last month's Scientific American pointing out we only rely on gravity slingshots because we still use chemical rockets with pathetic delta-V. Once 2nd and 3rd generation plasma engines come on line (1st gen are in flight now) the gravity slingshot will become an irrelevance. No doubt UFO drives are at least as good as our (future) 3rd gen plasma drives, probably way better since they warp space.

(Reminds me of the old Guild Navigator joke - I just warped space from Ix, and boy is my mind tired. Oh well, you had to be there)

[[User:Spike|Spike]] 13:24, 27 February 2009 (CST)

==='''What happens to the crashed UFO craft and its crew?'''===

Downed alien craft disappear after a few days have passed. No explanation is given to this whatsoever, so what really happens to them?

Possible answers:
* The UFO and its occupants are recovered by other humans.
* The Aliens manage to repair the craft and fly back to space. (unlikely, in the event that the power plant blew up and they have no Elerium)
* The Aliens self-destruct the craft and kill themselves in the process, ensuring their remains will not be recovered by humans.
* The UFO Powerplant eventually suffers a meltdown and explodes, eliminating any vestiges of alien presence.
* The Aliens blow up the UFO and disappear into the countryside.
* Most likely answer, IMHO: The local government/ funding nations give X-COM a limited time window to launch any operation, similiar to what you see in covert ops movies: "Complete the mission within 36 hours, or we initiate Carpet Bombing of the area". This is very likely considering that each nation actually has jurisdiction, and X-com is operating each military op with permission and cooperation by local authorities. [[User:Jasonred|Jasonred]] 13:42, 25 February 2009 (CST)
*Similiarly, I assume that nations which have signed a pact with the aliens launch a rescue operation and assist their alien friends. [[User:Jasonred|Jasonred]] 13:42, 25 February 2009 (CST)
: In the UFO TV show, it's stated that alien craft and bodies degrade quickly in Earth's atmosphere, disappearing completely in hours or a few days. [[User:Spike|Spike]] 15:36, 25 February 2009 (CST)
:: If that was the case, then one of the gases present in the atmosphere would be very toxic to the aliens. They would be restricted on their activities outside their craft, not to mention they would have to terraform the planet to be able to live here. [[User:Hobbes|Hobbes]] 20:32, 26 February 2009 (CST)

:: As Zombie said, doesn't Alien Containment support this idea? [[User:Spike|Spike]] 13:24, 27 February 2009 (CST)

::Those are excellent suggestions. One of the X-COM books detailed that aliens throw up a force field around crashed UFOs to give them time to repair the craft. This would also explain the limited size of the Battlescape (the area of the force field...the field was thrown up before the crash, thus why the craft wasn't always centered in it) as well as why the Battlescape is devoid of human life(the aliens took care of that up front.) Similarly, large scale bombing works as well, as does the local government going in to clean it up themselves. [[User:Arrow Quivershaft|Arrow Quivershaft]] 00:19, 26 February 2009 (CST)
:::If there's a force field around the craft what is it supposed to repel? The atmosphere? Because humans have no problem entering the field and operating inside it. [[User:Hobbes|Hobbes]] 20:32, 26 February 2009 (CST)
::::Of course they do! Have you ever noticed how dumb your units can be? That's because the aliens can't live with too much nitrogen, so they turn some of it in their force field to oxygen, and excessive amounts of oxygen make you act weird. After a while, their power source runs out and the nitrogen returns, dissolving them into E-115. (What did you think it was made of?) AT least, that's what seemes logical to me, ad is a combination of many postulates here. 21:36, 18 March 2010 (EDT)
:::::That would require that the aliens' biochemistry and the alien alloys used in the power sources reacted with nitrogen, which is a mostly inert gas on normal temperature and pressure and a . It's kinda of weird that the aliens didn't bothered with fixing that vulnerability with their craft and bodies during millions of years (what happens if the force field malfunctions while capturing cattle? ooops!) but ok. The increase in the amount of oxygen would probably also turn any kind of fire into large explosions throughtout the force field. [[User:Hobbes|Hobbes]] 21:24, 19 March 2010 (EDT)
::The force field was intended to keep humans out so the aliens were undisturbed. The first major hurdle X-COM had was figuring out a way to bypass those fields so they COULD get troops and aircraft inside. [[User:Arrow Quivershaft|Arrow Quivershaft]] 20:54, 26 February 2009 (CST)

===How do the aliens carry their equipment?===
Like human soldiers, aliens can carry weapons and equipment in locations like legs, belt, shoulders and backpack, regardless of the fact that some of their races even lack those anatomical features.

:Potentially a sticky gel-like area on the limb. Though really, since we were never intended to access alien inventories and the AI does all inventory management internally, this may simply be something that was never considered. Speaking from a slightly different standard, most aliens do not carry excessive amounts of gear; often their equipment would be able to fit in both hands. Also recall that Floaters and Ethereals have capes and robes(which may have inside pockets, or the Floaters could store them in the anti-grav/life support unit or inside surgically created body cavities during the installation, while Ethereals could support their excess gear with telekinesis), Mutons have armor(which may have external straps or adhesive areas), and Snakemen have an armor plate(which could have straps, adhesive, inside pockets, or even a backpack.) [[User:Arrow Quivershaft|Arrow Quivershaft]] 00:19, 26 February 2009 (CST)

:We already know the answer to this one - they cheat! ;) [[User:Spike|Spike]] 13:24, 27 February 2009 (CST)

===Are X-COM transport craft piloted?===

:Almost certainly, since a remotely-piloted craft could be returned to base when the mission was aborted or failed. It's entirely possible that all X-COM soldiers are qualified pilots of the appropriate craft, since it would make no sense for X-COM to waste space on the plane for a noncombatant, or to have a single-point of failure on the mission like that. (The aliens could screw over the entire op by killing the pilot). It also explains why the craft is lost when the mission fails or is aborted with no one inside(lacking a pilot, the aliens are able to easily destroy it.) [[User:Arrow Quivershaft|Arrow Quivershaft]] 00:19, 26 February 2009 (CST)

:: On the other hand, even a tank/hovertank is capable of getting the craft back to base. Perhaps there is an autopilot function. X-com craft are also infamous for choosing strange and bizarre intercept paths, based on latitude lines... almost as if they followed some a few simple lines of code from 1993 programming (bit of 4th wall breakage there...) -[[User:Jasonred|Jasonred]]

:Of course, given the tanks seem to be remotely piloted from an X-COM base, its possible that the tank being in the craft allows the Tank pilot to reroute into the control systems for the dropship and take it over. Limiting this to having the tank inside is a rather good idea. [[User:Arrow Quivershaft|Arrow Quivershaft]] 12:07, 27 February 2009 (CST)

:I recall that the various cutscenes in the PSX version had a pilot, most notably in the "Mission Failure" scene, where it shows the pilot being killed. --[[User:Mabmoro|Mabmoro]] 16:06, 13 March 2009 (EDT)

===What sort of physical process is used to increase human stats over time?===

The increases in some stats are easy explained by experience gained on missions (firing abiility, reactions, etc.). However, in the cases of physical stats (TUs, stamina, strength) the increase must be augmented by an artificial process, since it isn't easily explainable that humans can significantly increase body mass/speed/endurance just by physical activity/exercise.

:Lifting weights will increase strength. Running and cardio exercise will increase endurance, and performing the same task multiple times will allow you to perform it faster. I see no reason natural increase doesn't work. [[User:Arrow Quivershaft|Arrow Quivershaft]] 18:33, 26 February 2009 (CST)
::I assume that it is possible to a human to use those methods to double its physical condition, but that being the case why are X-COM recruits so... undeveloped? Maybe this is a more intriguing aspect. [[User:Hobbes|Hobbes]] 20:32, 26 February 2009 (CST)

What I don't get is that when someone gets blasted that they gain a lot of extra health. For crying out loud, the aliens are throwing around plasma and ridicoulously HUGE explosions. Shouldn't they be suffering from third-degree burns? I would expect them to at least have a major sore spot where they got hit. [[User:Tsunamiatunzen1|Tsunamiatunzen1]] September 24

:There's been arguments over whether the soldiers in X-COM are the Green Berets or equivalents of their various militaries, just average soldiers that volunteered for the job, or if the Council of Funding Nations is corrupt and is using this as an excuse to foist off their most useless soldiers onto the X-COM project. If the latter, it would easily explain their rather poor early stats. [[User:Arrow Quivershaft|Arrow Quivershaft]] 21:06, 26 February 2009 (CST)

::Maybe the stat increases relate to the troops getting more comfortable performing all operations - lifting, running, combat actions - when the aliens no longer scare the cr*p out of them so much. [[User:Spike|Spike]] 13:24, 27 February 2009 (CST)

:: I know what you mean... if this were Jagged Alliance, it'd be like trying to hire Mike and getting Gumpy instead... ouch! ... IMHO, it looks as if the COFN is being funny about this. Clearly, the troops assigned to X-com have had extensive weapons training... each and every one of them can use just about any standard weapon, including Rocket Launchers, incindieries, auto Cannons, etc. However, NONE of them have any combat experience, coming to you as fresh rookies. And their stats look as if they were selected based on a pot luck basis rather than screening among the elite troops... [[User:Jasonred]]

:Health isn't gained from getting shot. But as for the "ludicrously low stats" issue, maybe they're being selected on some other basis, or there's a real shortage of volunteers. [[User:Magic9mushroom|Magic9mushroom]] 04:56, 25 September 2009 (EDT)

:: I think that the X-COM's soldiers have big responsibilities, can't have children. They are sworn in and only then go to the field - if someone can't be trusted to keep a secret, he's getting sacked, given a shitty job (ever wondered who maintained the General Stores?) or worse. - n, 16:47, 16 August 2010 (GMT+2)

In Hotpoint's [http://www.tthfanfic.org/Story-16092/Hotpoint+XSGCOM+Goa+uld+Defence.htm| XSGCOM: Goa'uld Defense], his explanation is that after a few missions of watching their fellow newbies die, the survivors give in and volunteer for an experimental regimen of performance-enhancing drugs. IRL, steroids and such don't result in instant strength enhancement, they just encourage accelerated development of muscles. [[User:Kalaong|Kalaong]] 03:16, 28 March 2011 (EDT)

==Enemy Unknown/UFO Defence issues==
==='''Why isn't Earth overrun by Snakeman/Chryssalids?'''===

According to the UFOPaedia, Snakemen's "Reproduction is asexual, with each snakeman carrying up to fifty eggs inside its body at any one time" adding the ominious conclusion: "Left to its own devices this species would be a severe threat to life on earth."
Moreover, this species is usually accompanied by the Chryssalids, which have a capacity to reproduce themselves very quickly using humans. So, any survivors of crash sites or terror attacks could start reproducing themselves hidden, resulting in large areas being overrun by those aliens later on.

Possible answers:
* Both races have a self-destruct mechanism incorporated into their psysiology to prevent this.
* The entire area is purged by large scale bombing. [[User:Jasonred|Jasonred]] 13:42, 25 February 2009 (CST)

::One thing suggested in fan data was that Snakemen have air tablets in their stomach...which may be different from earth's atmosphere, which would limit their lifespan in earth's atmosphere, also making egg-laying pointless, since the offspring wouldn't be able to breathe or survive. It has also been suggested that Chryssalids have a very rapid metabolism. Though Chryssalids are likely just as, if not more useful, as a threat or a bargaining tool. When attempting to get a nation to capitulate to their demands, the aliens could threaten to employ Chryssalids en masse, or offer to remove a mass infestation in exchange for the government's cooperation. Or even further, it's possible that Chryssalids are under Ethereal control and maintaining the control link at that distance is taxing, thus eliminating mass use of the creatures. [[User:Arrow Quivershaft|Arrow Quivershaft]] 00:19, 26 February 2009 (CST)

Two words: [[Alien Containment]]. That answers everything except for the UFOPaedia articles for the aliens themselves which contradict it. --[[User:Zombie|Zombie]] 00:42, 26 February 2009 (CST)
: For captured aliens, yes. But what survivors of uninvestigated crash sites? [[User:Hobbes|Hobbes]]

::Zombie is commenting that based on the need for an Alien Containment unit, the aliens cannot survive in earth's atmosphere for extended periods, needing special atmospheric blends and/or nutrient pools which earth is unable to provide naturally, thus limiting their operations outside of the craft. (If the aliens won the war, it's likely this would be one of the first things that they would 'correct'.) [[User:Arrow Quivershaft|Arrow Quivershaft]] 20:54, 26 February 2009 (CST)

===What is the propose of the 'disco balls' found inside some UFOs?===

:Given that they explode, they could be storage reservoirs for coolant for the computers or other systems. They could also be circuit breakers or electrical junction boxes, or even a component of the UFO's particle beam they use to fry X-COM Interception craft. [[User:Arrow Quivershaft|Arrow Quivershaft]] 00:19, 26 February 2009 (CST)

Here again, two words: [[Alien Entertainment]]. Even though the spheres are not set to Alien Entertainment in the MCD files, they are almost certainly related to the process somehow. --[[User:Zombie|Zombie]] 00:42, 26 February 2009 (CST)

:Maybe they are for having discos? [[User:Spike|Spike]] 13:24, 27 February 2009 (CST)

:Presumably [[Alien Entertainment]] is psionic in some way, maybe they're Psi-Emitters or something? -[[User:magic9mushroom|magic9mushroom]]

===Who buys those alien bodies/equipment from X-COM?===

:Equipment likely goes to the funding nations or the international black market. No rebel group is going to ask too many questions about being offered guns that can [[Heavy Plasma|slice through the hull of an MBT]] or [[Alien Grenade|grenades that can level a building]] or [[Blaster Launcher|man-portable guided missiles]]; it'd just be cash-and-carry. Similarly, scientists would likely be interested in looking at much of this stuff for their own research. This would also explain the lack of market forces; the funding nations could have a set price for each item, or if X-COM is selling them under the table to rebels and rogue scientists, they can set the price and refuse to budge. The money on corpses could also be an "Alien Bounty" paid by the Funding Nations, as a reward for each alien that X-COM can prove they killed. Or it could be bought by other groups...rumor has it that some fast food restaurants have processes that can make ANY meat, no matter the source, look and taste the same, and a Muton would make a LOT of McBurgers. [[User:Arrow Quivershaft|Arrow Quivershaft]] 00:19, 26 February 2009 (CST)
From the USO (Kasey Chang): XARQUID SUSHI.

===How did the aliens got to Mars?===

There are no indications that UFOs are capable of faster than light speed. So how did they get to Mars in the first place?

:The UFOs are mission craft, used for the legwork. The fighters; we never see the carriers. Given the aliens have been proven to be interstellar, they either Clone-A-Crew as needed when coming the long way to keep the UFOs crewed, or its far more likely that the aliens did have or still do have larger "Carrier" ships, which are capable of FTL travel, that were/are further out in the Solar System that store and dispatch UFOs to mission locations. [[User:Arrow Quivershaft|Arrow Quivershaft]] 18:33, 26 February 2009 (CST)

:: One word: TFTD. The entire city of T'leth was put into cryogenic suspended animation? Or look at X-com Interceptor. X-com and the aliens show the ability to enter hyperspace or whatever it is.
::: T'Leth is another mystery of its own. More to that later on :) [[User:Hobbes|Hobbes]] 20:32, 26 February 2009 (CST)
::... there are no indications that the UFOs are INcapable of FTL... I don't think you would want to perform FTL travel within Earth's planetary atmosphere!
::: There are no indications that they are capable as well. And X-COM scientists don't seem to detect any FTL capabilities in UFOs during their research. And after the war the Elerium stocks dwindled, and it would make sense to perform some sort of interstellar missions to detect and harvest Elerium, however none are mentioned. [[User:Hobbes|Hobbes]] 20:32, 26 February 2009 (CST)
:::: Unless you count the events of X-com Interceptor? [[User:Jasonred]] [[User:Jasonred|Jasonred]] 21:15, 26 February 2009 (CST)
::::: I am refering to the events between Enemy Unknown and TFTD. There is clearly a big distinction between the alien craft on EU and those of Interceptor. [[User:Hobbes|Hobbes]] 09:38, 27 February 2009 (CST)
::::: Enemy Unknown is set in 1999, TFTD in 2040, Interceptor in 2067... looks entirely plausible that they DID begin research into space exploration immediately after the events of Enemy Unknown. These things take time you know. Remember that the universe is a huge place, and Earth had rather limited Elerium Reserves by the end of EU. It takes... what, 30 Elerium just to fly an Avenger halfway across Earth? They could hardly afford to fly around randomly in space HOPING to come across elerium, they had to figure out detections methods, then scan the galaxy sector by sector, possibly partially using non-Elerium based propulsion at times... I can't remember if it's canon or fanfic, but I remember reading that all Elerium on Earth was reserved for space exploration.
As for FTL, Earth does get it for sure sometime between 1999 and 2067. And I'm pretty certain the technology is Elerium based. It's not a huge logic jump to assume that the aliens have access to FTL Elerium based tech.
Though I'm a bit puzzled why all T'leth technology is based on Zrbrite, when the aliens uniformly use Elerium, all the way from Earth to Cydonia to the far reaches of space. -[[User:Jasonred|Jasonred]]

:It takes 12 Elerium to fuel an Avenger, though how much 1 Elerium is is an ongoing debate. As for Elerium, it was reserved for propulsion research when the funding nations divvied up X-COM's resources, and then they blew it all without learning anything more than the original X-COM scientists. And yes, Earth clearly gets FTL after TFTD but before Interceptor. The reason Elerium is not used in TFTD is because Elerium becomes inert and useless upon contact with seawater. Similarly, seawater aggressively corrodes Alien Alloys and eventually completely dissolves them. Zrbite functions similar to Elerium, being gold mixed with alien bio-material. Unfortunately, Zrbite only works when supported by a massive energy grid created by T'leth and becomes inert upon its destruction. [[User:Arrow Quivershaft|Arrow Quivershaft]] 12:07, 27 February 2009 (CST)

OK... take the UNIT of elerium out of the equation... let's say that Earth had 5000 units of elerium, so 6 units get's an Avenger halfway around the world, and 12 units is sufficient to reach Mars. Hardly enough fuel reserves for intergalactic travel then.
2065 On October the 27th, the probe 'Tombstone 1' returns reports to Earth. It's data show that the globular star cluster where it rests, one hundred light-years from Earth, contains many life-supporting planets. Many of the planet's within the probe's scanning range also apparently possess great mineral wealth, including trace veins of elerium-115.
Hmm... So, mankind discovers FTL technology on their own in those 65 years? ah... come to think of it, if they've got non-Elerium based space travel and FTL, and more powerful weapons too, what's the big deal about Elerium in the Frontier? Does mankind even need it anymore?
As for T'leth, it is meant to be over 65 million years old, and CRASHLANDED on Earth due to a solar flare. Was the Ultimate Alien a prophet, thus chose to base T'leth on aqua plastics and Zrbite when T'lth was first constructed? Or did T'leth crash land, followed by frenzied activity where the entire city was replaced part by part, the alien alloys swapped for Aqua Plastics?
...
Actually, come to think of it, it's obviously a massive plothole due to limited timeframe, no point in discussing too deeply. Sigh... [[User:Jasonred|Jasonred]] 13:50, 27 February 2009 (CST)

:It's implied that despite the loss of ability to use the alien technology from the First and Second alien wars, simply being able to see and examine their designs catapulted earth's technology forward at least a few decades...which really is entirely reasonable. Much of the technology can be replicated on earth, and the principles and designs can be reapplied.

:As for the deal with Elerium, its needed to power stronger weapons and is also wonderful for power generation; its efficiency in power generation is what allows Mega Primus to even exist.

:And was there anywhere that specifically said that T'Leth was made of Aqua Plastics? I don't recall. Yes, the rest of their subs are made of aqua plastics, but I'm wondering if something the size of a medium city might perhaps be made of something a bit more durable. PS: Thanks for signing your post! :D [[User:Arrow Quivershaft|Arrow Quivershaft]] 14:17, 27 February 2009 (CST)

It's not a plothole at all. There is no Elerium on Earth, whereas the aliens can manufacture Zrbite on Earth, since there's gold here. Therefore it's obvious why they used Zrbite. Also, there's the fact that it was an Aquatoid colony mission, intended to produce an "aquatic paradise", so using Aqua Plastics instead of Alien Alloys is perfectly justified. [[User:magic9mushroom]]

:I think that the use of Aqua Plastics instead of Alien Alloys is perfectly justified by the fact that T'leth crashed million of years ago while the aliens that came to take a shot or two at earth were two million years ahead technologically. So Aqua plastics was some plastic that was used by aliens before they've got the Alloys. You might argue - it's Aqua Plastics, so Aqua =/= Space. Well, Aliens weren't calling them aqua, and there wasn't anything saying that Aqua Plastics is not suitable for Space Travel. --[[User:Domenique|Domenique]] 11:10, 19 May 2010 (EDT)

... I think Alien Alloys alone would catapult earth's technology forward a decade, and there's no reason humanity can't use those anymore, just not in water. According to timeline, some space pirates manage to make the decommisioned Avengers run on non-Elerium fuel... a large technological step.
By the time of Interceptor, Elerium weapons aren't that powerful. Good point about power generation though.
I would assume that T'leth SHOULD have been constructed out of Alien Alloys, since it was originially an interplanetary vessel? Generally, everything in EU was made of Alien Alloys, everything in TFTD was made out of Aqua plastics. Both of which seemed plenty durable.

::Have you noticed that the UFOs fly however their mission parameters tell them to? The UFOs pretty much IGNORE interceptions by X-com craft... if their mission tells them to make 3 passes, speed up, slow down, speed up... they will follow that pattern exactly, whether X-com craft are firing on them or not.

::Question: How long does it take the Avenger to reach Mars from Earth? [[User:Jasonred]] [[User:Jasonred|Jasonred]] 19:12, 26 February 2009 (CST)
:::No longer than a week, in my opinion. Probably less than 2 days. Since canonically, the design of the Avenger had the Cydonia mission in mind, it would be capable of very high interplanetary speeds. (You could choose to burn 40% of the Elerium in one blast to get to high speed. Or you could burn even more and refuel while it's landed...or it could be a mission with no guaranteed escape for the crew. The lives of the many over those of the few and all that, especially since the war hinges on the mission.) In addition, you can fit a full complement of soldiers on board with no real excess room for supplies, and the longer it takes to get to Mars, the greater the chance the aliens will spot it coming for them and mount a serious defense. [[User:Arrow Quivershaft|Arrow Quivershaft]] 19:20, 26 February 2009 (CST)
:::If you'd burn that much fuel to accelerate the craft then you'd have to use as much again to decelerate it and attain a planetary orbit, otherwise you'll simply overshoot the planet and head towards outer space. This is also another aspect to take into account when thinking about the speed of UFOs. [[User:Hobbes|Hobbes]] 20:32, 26 February 2009 (CST)

::: UFO drives are non-Newtonian so those sort of rocket equations don't necessarily apply. I think filling up a car with gas is a closer analogy. [[User:Spike|Spike]] 13:24, 27 February 2009 (CST)

::: Well, they could put a mini mind shield on the Avenger, shoot down a UFO roughly their size, and then fly to Cydonia when the UFO they shot down was supposed to return based on the instructions found by the hyper-wave decoder. Basically taking the place of the UFO. Shouldn't be too hard since the UFO's are pretty common by the time you research Cydonia or Bust. [[User:Tsunamiatunzen1|Tsunamiatunzen1]] 14 February 2010 (MST)

::That would leave 20% of the fuel to take off the Avenger and land it. Not really that unreasonable. While they're landed, they could potentially refuel the Avenger, or the mission might have been planned as a 1-way trip from the get-go. [[User:Arrow Quivershaft|Arrow Quivershaft]] 21:06, 26 February 2009 (CST)

===What happened to Mars and the alien civilization there?===

According to the Brain, Mars was blooming with life had a alien civilization millions of years ago. However, Mars nowadays is a barren world and the alien civilization seems reduced to the area on Cydonia.

:That may well have been before Mars lost the majority of its atmosphere due to its weak magnetic field. As the atmosphere dissipated, the aliens left or died off. It's also possible that the aliens, shown in the game over to have little respect for planets other than as sites for slaves and resources, they strip-mined the planet dry(and the rust from the machines created the red coloring), and then seeded Earth so that the slave workforce would grow for future extraction of Earth's resources. [[User:Arrow Quivershaft|Arrow Quivershaft]] 18:33, 26 February 2009 (CST)

===Where are the human-alien hybrids referred to on the UFOPaedia?===

:For the ones on earth, probably in hiding or in laboratories for research. For the ones the aliens have, potentially improving the Sectoid gene pool or being used as food or menial tasks. Cloning is alot easier than making genetic hybrids and there's nothing that says their first-generation experiments would be suitable for combat. [[User:Arrow Quivershaft|Arrow Quivershaft]] 18:33, 26 February 2009 (CST)

:: They are babies at the time of X-com, and few in number. Their aren't even that many of them by the time of X-com Apocalypse. [[User:Jasonred]] [[User:Jasonred|Jasonred]] 19:12, 26 February 2009 (CST)

:Not to mention the ones in X-COM: Apocalypse are less-than-fit for battle before extensive training. [[User:Arrow Quivershaft|Arrow Quivershaft]] 21:06, 26 February 2009 (CST)

===Why did the aliens only activate T'Leth after they were defeated?===

On TFTD T'Leth is shown as an entity/city of major power that is capable of conducting a war on its own. But the aliens leave it dormant although they could have used it to speed the process of taking control of Earth.

:Perhaps the Enemy Unknown aliens are legitimately scared of the TFTD aliens and are unsure how long they could trust them. Evil is not monolithic; the TFTD aliens may be more interested in themselves than the alien empire, so they were kept as an ace-in-the-hole. This is the same reason (canonically) that SKYNET did not originally send the T-1000 to assassinate Sarah Connor; SKYNET was scared of what the T-1000 could do and had only a bare minimum of control over it, so it only used it as an option when it had nothing left to lose. [[User:Arrow Quivershaft|Arrow Quivershaft]] 21:06, 26 February 2009 (CST)

T'Leth is in fact so powerful that all it has to do is surface, in order for X-com to be considered to have lost the war.
In fact, looking at the timelines, it takes 40 years for T'leth to wake up from it's slumber... that's one good reason not to use it. By the time it activated, the war would already be over. [[User:Jasonred]]

Because the whole purpose of the Enemy Unknown aliens was to rescue the TFTD aliens from T'leth, and there was no way of seeing whether the T'leth-based invasion in TFTD would even work - from their perspective it's possible that getting T'leth to bootstrap itself could have caused a catastrophe (they don't know whether or how badly it's damaged). Presumably the aliens planned to mount a proper rescue operation after locking down Earth and readying it for the aquatic paradise that was the entire point of the T'leth expedition in the first place. [[User:magic9mushroom]]

T'Leth was a coleny ship sent by the Sectoid/Aquatoid's millions of years earlier. This is why the Aquatoids use electronics to augment there control over other creatures, while there progeny are genetically modified to gain the same control. In TFTD it's implied that the T'Leth had been partialy active for a long time. Thawing out aliens in small groups but never going in full production. UFO aliens may have not intended to start the full awakening cycle until they had a chance to prepare the planet.
OR, given the Ultimate Alien was aquatoid in origin by his looks and there was no other races from the first game involved. And the fact that Sectoid/Aqutoid's are not the top of the food chain with the Ethereal and Brain being more powerful it's possible the brain had decided that the Ultimate Alien was a threat to it's power. It was not until it's death that they tried send the signal. --[[User:BladeFireLight|BladeFireLight]] 17:19, 22 March 2010 (EDT)

===Why did the aliens use limited force during the First Alien War?===

Imagine Independence Day or War of the Worlds: UFO above the major Earth cities destroying the national leadership and any resistance. Or simply announce to Earth that they are now a part of their empire and resistence is futile. Instead, they go 1 mission each day, allowing humans to capture their craft, research their technology, discover their intentions and mount a successful defense. Don't the aliens watch sci-fi movies to see how it should be done?

:Perhaps they don't have the standing forces to do so, and are in the process of building up the forces needed to do so. Perhaps they don't want to wipe out the entire power structure too fast; they want to leave some pieces in place for when they rebuild. Perhaps they're too condescending to think that humanity ever really has a chance; they've probably conquered thousands of other planets without anyone ever successfully resisting them. Perhaps they consider the X-COM project to be a rearguard action that, while a valiant effort and a credible threat, is ultimately doomed to failure because they simply cannot win in the end, which is why they undermine it. Indeed, the reason you need to launch the Cydonia mission in order to win is because X-COM simply cannot stop the aliens in a ground war; the aliens have an effectively infinite supply line and standing forces(though nothing says they're all waiting to swamp the earth), and the only way to win is to kill the command staff(which the aliens believe X-COM will not be able to do, lacking both knowledge of where the Brain is and any practical means to get there.) [[User:Arrow Quivershaft|Arrow Quivershaft]] 21:06, 26 February 2009 (CST)
::The most plausible explanation seems to be the one employed in the Worldwar series, by Harry Turtledove, but that does not seem to mesh with what the Brain says about having been on Mars for a while, since then they could watch the Earthlings perpetually. Of course, what it says is probably a bunch of lies. Thinking about what it says for too long also raises the question of why the aliens attacked when they did, of course... [[User:Vizzydix1|Vizzydix1]] 21:52, 18 March 2010 (EDT)

The aliens are completely lacking in weapons of mass destruction. When you get right down to it, their aircraft are very fast, manuevarable and durable, but they have rotten firepower. Even the battleship is unable to bring down an Interceptor in 1 shot.
The terror missions and X-com Base Defences prove that the Aliens are unable to simply launch orbital bombardments... in fact, they appear to have no Air to Land weapons whatsoever...
When you get right down to it, the aliens are pretty stupid. Also, their scientists seem inferior to Earth's. Seems to me that they only had the advantage of Elerium deposits and thus elerium based research.
X-COM was unable to win in an all-out war with the aliens, but remember that X-com is a small little covert group with several dozen soldiers and a handful of aircraft. Can you imagine the result if the aliens had caused a joint war effort by the UN? You would have Lockheed factories converted to Avenger production, several platoons of soldiers outfitted with Flying Suits, Lasers, Heavy Plasma, thousands upon thousands of Laser Tanks...
I would say that keeping the fight to covert action on both sides was actually beneficial to the aliens, really. [[User:Jasonred]]
* Nope, the aliens could just invoke John's Law and blow up the planet with a kamikaze battleship at .9c. Even failing that, the alien battlefleet could come in numbers sufficent to blot out the sun.--[[User:(name here)|(name here)]] 14:39, 8 November 2009 (EST)
** Lest we forget, X-Com is essentially the Spartans to the aliens' Persian Empire. So they'd just fight in the shade (which would be a blessing in desert missions). --[[User:Guido Talbot|Guido Talbot]] 13:58, 16 July 2010 (EDT)

Aliens first began with smal scouting missions, so maybe all we expirience in X-Com is initial attacks by aliens, maybe the whole base was begining scouting and waiting for the invasion fleet? Aliens problably could be in sense dumber than humans, humans are adaptable and thinking, our technology advances fast, and we are fast and smart enough to stop the invasion before it begins. Aliens problably were not used to it so they thought "oh well, another invasion...". --[[User:Domenique|Domenique]] 11:10, 19 May 2010 (EDT)

==TFTD issues==
===What was the relationship between the aliens from the 1st and 2nd wars?===

Quoting from the UFOPaedia regarding Alien Origins: 'Deep in the oceans there lie ancient
sites used by the Aliens to contact their stellar cousins.' This also has some implications regarding the issue of why T'Leth was only activated when the Sectoids were defeated.

*Gill Men are coopted Terran creatures, Aquatoids are a differently-modified Sectoid breed, Lobstermen are machine soldiers that are manufactured, Tasoths are clone soldiers that are grown.

Aquatoids are the ancestors of the Sectoids. The colony ship was sent out slowing than light hundreds of millions of years ago. Sectoids are a more genetically advanced race. The rest of UFO aliens were picked up after that point. The Brain and Ethereals probably conquered the Sectoids and don't regard them highly. While the Aquatoids that were thawed over the years created, conquered (Gill Man) or manufactured the rest of the allies over time. I picture the Tasoth as probably something they brought with them and have been working on. Since most of the Aquatoids come from suspended animation they have not tinkered with genes much. Instead modifying and using electronics (MC Chip) to control. --[[User:BladeFireLight|BladeFireLight]] 17:28, 22 March 2010 (EDT)

===What kind of materials were 'synomium' and 'adamantium'?===

The first one is mentioned on the name of the alien communication devices and the second one appears at the end when T'Leth is destroyed: 'he twisting hugeness of T'leth begins to rupture.
Flames and smoke spew from its gleaming spires and adamantium halls.'

:Synomium is probably a special material used in the comm. devices, like Stargate's naquadah (universal stuff), naquadria (unstable power source), trinium (hull material) and neutronium (superdense metal). Adamantium is a legendary material in ancient literature that is said to be indestructible, similarly to mithril.--[[User:Amitakartok|amitakartok]] 10:55, 3 November 2009 (EST)

===How deep were X-COM bases located on the seas?===

*Floating bases would be easier to build, repair and supply. However they would have to be tethered to the ocean bed or possess some sort of propulsion to prevent them from drifing with the ocean currents. But it would also allow for easy redeployment of the base.
*Submersible bases could allow for better sonar detection. Same problems regarding ocean currents would apply. In case of hull breaches entire modules would be quickly flooded and any crew present would be crushed by water pressure or drown. Base could be built and then submerged (requires depth control)
*Seabed bases would be the hardest to build and supply. Several other factors could limit their deployment, such as unstable areas (underwater volcanoes, prone to seaquakes, rock avalanches, etc.) and depths.
*Given that the Alien Retaliation missions in TFTD are called "Floating Base Attack", I'd say floating. Also remember that your starting sonar can't see Very Deep, which rules out seabed bases. [[User:magic9mushroom]]
** Floating doesn't always happen on the surface. The surface has to deal with large waves ad bobbing up and down, submerged only has the currents it could be stabilized easier. --[[User:BladeFireLight|BladeFireLight]] 11:57, 14 January 2010 (EST)
* I was recently researching how far you have to be down to not be effected by surface conditions. You have to be submerged 1/2 of a waves lenght (measured crest to crest) Best I can find is that the average wave is 150 yards accross. To not be effected by the waves you have to be 75 Yards below. This is way below the depth needed to not be effected by the bends when surfaceing. I figure they are probably right at the limit of what can be safe for quick surfacing and well anchored to avoid getting [http://www.youtube.com/watch?v=LwKXfc_a4Ag tossed in a storm]. --[[User:BladeFireLight|BladeFireLight]] 17:37, 22 March 2010 (EDT)
**Something else to consider - these 'floating bases' could be large modular submarines/submersibles. If memory serves, submarines basically maintain sea-level pressure regardless of the depth, so that could explain why they'd be able to be 75+ feet below sea level and not suffer the bends when surfacing.
::(Or, y'know, we could just say "a wizard did it".) --[[User:Guido Talbot|Guido Talbot]] 14:13, 16 July 2010 (EDT)
[[Category: Fiction]]

Realistic Equivalents

2009-03-30T03:05:58Z

Kalaong: /* Equipment & Explosives */ YEAH BABY!

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]?

*Plasma weapons - here are a few [http://en.wikipedia.org/wiki/Plasma_rifle promising theories], especially [http://en.wikipedia.org/wiki/Plasma_channel "plasma channels".]

== [[Craft Armaments]] ==

*[[Cannon]] - The [http://en.wikipedia.org/wiki/GAU-12_Equalizer GAU-12 Equalizer] is the current state-of-the art aircraft-mounted cannon.

*[[Stingray]] - The [http://en.wikipedia.org/wiki/AIM-7_Sparrow AIM-7C/D Sparrow] has the closest equivalent range at 32km.

*[[Avalanche]] - The [http://en.wikipedia.org/wiki/AIM-120_AMRAAM AIM-120D AMRAAM] has the closest equivalent range at 64km.

== 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]. No need for nanites! Stimulants - In real life the Air Force gives its pilots amphetamines in order to keep them awake, I'm sure that the medi-kit 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]] - Possibly an advanced version of [http://www.pepperball.com/faqs.asp Pepperballs]. Other than the fact that the stun bomb causes an explosion which knocks units UNCONSCIOUS, and effects units that do not possess eyes or respitory systems... even inorganic units for that matter. Anyhow, this is an item of alien technology, and should not have an earth equivalent.
*[[Stun Rod]] - Cattle prod or Taser. Not much call for this item in real life since Tasers, Rubber Bullets and Tranquiliser Darts have better range.
*[[Power Suit]] - has arrived! The [http://en.wikipedia.org/wiki/HAL_5 Hybrid Assistive Limb] is available for rent in Japan for $1500 a month. All that is missing are the [[Alien Alloys]] to reinforce it and a [[UFO Power Source]] to power it for extended periods of time.

== 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 [http://en.wikipedia.org/wiki/V-22_Osprey|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.)''

There is also a [[Talk:Craft|discussion]] taking place on the speeds of aircraft and UFOs.

== 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.)

''What if the reactor's are 99% efficient (as stated in the UFOpaedia) because to achieve nearly complete reaction of Elerium requires a certain catalyst? That would also explain why the Elerium in the reactors didn't detonate with such force, the alien Engineers having removed the catalyst to prevent the unstable reaction." -[[User:Amitakartok|amitakartok]]

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.

Alien Artefacts (TFTD)

2009-01-24T09:18:14Z

Kalaong:

Here is a collection of all alien artifact weapons from the [[Second Alien War]]. Alien Artifacts are all forms of equipment, weapons and ammunition used by the aliens. X-Com will only score points for recovered Artifacts that have not yet been researched. The following items are recoverable from USOs, Alien Colonies, Terror Missions and X-Com Base Defense Missions:

*[[Sonic Pistol]]
**[[Pistol Power Clip]]
*[[Sonic-Blasta Rifle]]
**[[Blasta Power Clip]]
*[[Sonic Cannon]]
**[[Cannon Power Clip]]
*[[Thermal Shok Launcher]]
**[[Thermal Shok Bomb]]
*[[Disruptor Pulse Launcher]]
**[[Disruptor Ammo]]
*[[M.C. Reader]]
*[[Sonic Pulser]]
*[[Zrbite]]
*[[Unused Items (TFTD)|Unused Items]]

Alien Artefacts (TFTD)

2009-01-24T09:17:59Z

Kalaong: Okay, how do you fix a page name typo?

Here is a collection of all alien artifact weapons from the [[Second Alien War]]. Alien Artefacts are all forms of equipment, weapons and ammunition used by the aliens. X-Com will only score points for recovered Artefacts that have not yet been researched. The following items are recoverable from USOs, Alien Colonies, Terror Missions and X-Com Base Defense Missions:

*[[Sonic Pistol]]
**[[Pistol Power Clip]]
*[[Sonic-Blasta Rifle]]
**[[Blasta Power Clip]]
*[[Sonic Cannon]]
**[[Cannon Power Clip]]
*[[Thermal Shok Launcher]]
**[[Thermal Shok Bomb]]
*[[Disruptor Pulse Launcher]]
**[[Disruptor Ammo]]
*[[M.C. Reader]]
*[[Sonic Pulser]]
*[[Zrbite]]
*[[Unused Items (TFTD)|Unused Items]]

Realistic Equivalents

2008-09-12T01:58:30Z

Kalaong: /* Weapons */ I'm not that much of a hardware nut, I'm just getting the thread started. Please assist!

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]?

*Plasma weapons - here are a few [http://en.wikipedia.org/wiki/Plasma_rifle promising theories], especially [http://en.wikipedia.org/wiki/Plasma_channel "plasma channels".]

== [[Craft Armaments]] ==

*[[Cannon]] - The [http://en.wikipedia.org/wiki/GAU-12_Equalizer GAU-12 Equalizer] is the current state-of-the art aircraft-mounted cannon.

*[[Stingray]] - The [http://en.wikipedia.org/wiki/AIM-7_Sparrow AIM-7C/D Sparrow] has the closest equivalent range at 32km.

*[[Avalanche]] - The [http://en.wikipedia.org/wiki/AIM-120_AMRAAM AIM-120D AMRAAM] has the closest equivalent range at 64km.

== 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]. No need for nanites! Stimulants - In real life the Air Force gives its pilots amphetamines in order to keep them awake, I'm sure that the medi-kit 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]

*[[Power Suit]] - no equivalent. Probably an advanced version of the BLEEX (Berkeley Lower Extremity EXoskeleton) project with alien materials and energy source.

== 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 [http://en.wikipedia.org/wiki/V-22_Osprey|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.)''

There is also a [[Talk:Craft|discussion]] taking place on the speeds of aircraft and UFOs.

== 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.)

''What if the reactor's are 99% efficient (as stated in the UFOpaedia) because to achieve nearly complete reaction of Elerium requires a certain catalyst? That would also explain why the Elerium in the reactors didn't detonate with such force, the alien Engineers having removed the catalyst to prevent the unstable reaction." -[[User:Amitakartok|amitakartok]]

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.

Cydonia

2008-04-24T20:34:24Z

Kalaong: /* Description */

A two part mission. Note that this is a DO OR DIE mission. If you fail or hit abort at any time, the entire campaign ends, you lose.
To have this mission available, you must research "Cydonia or Bust", and access it by launching an avenger. It'll either give the option to go to intercept or to Cydonia. The "Cydonia" button is at the top of the screen if you have an Avenger selected in the Geoscape.

=Mars: Cydonia Landing=
==Briefing==
Your [[Avenger]] craft has landed in the region of Cydonia on the surface of [[Mars]]. Our information indicates that one of the pyramid constructions contains a green access lift to an underground complex. Once you have assembled all your soldiers in the lift area, continue to the next stage…

==Description==
First part is on the surface of Mars. Lots of Pyramid structures with singular windows, some with aliens inside, some empty. Only one has the entrance to the aliens HQ, the entrance being an area with glowing green floor. Now, this floor has 36 squares, or a 6x6 area. Anyhow, you can step onto that with as many people as possible, hit abort, and it'll bring them and whatever they're carrying to part 2. OR, you can just kill everything on the stage and everyone will go to part 2.

The surface is filled with many many [[Sectoid]]s and [[Cyberdisc]]s. Much Commanders and leaders methinks, as there's Psi all around. I find this mission hilarious when I shoot a Cyberdisc, it blows up, and the chain reaction kills over 5 other Cyberdiscs. Plus many many Sectoids. It then turns into Pyramid hunting.

'' [ [[User:NKF|NKF]]: I think the easiest way to look at it is that this mission is basically a Sectoid Battleship/Base crew in a one-of-a-kind map with a choice of two different winning conditions] ''
''[ [[User:(name here)|(name here)]]'' actually, they replaced the navigators, engineers, and medics with leaders and commanders. important diffrence.

==Alien Deployment (Surface)==
<table {{StdCenterTable}}>
<tr {{StdDescTable_Heading}}><th align="left">Rank</th>
<th>Beg./Exp.</th>
<th>Vet./Gen.</th>
<th>Super.</th></tr>
<tr><td align="left">Sectoid Soldiers</td><td>4</td><td>5</td><td>6</td></tr>
<tr><td align="left">Sectoid Navigators</td><td>0</td><td>0</td><td>0</td></tr>
<tr><td align="left">Sectoid Medics</td><td>0</td><td>0</td><td>0</td></tr>
<tr><td align="left">Sectoid Engineers</td><td>0</td><td>0</td><td>0</td></tr>
<tr><td align="left">Sectoid Leaders</td><td>3</td><td>4</td><td>5</td></tr>
<tr><td align="left">Sectoid Commanders</td><td>3</td><td>4</td><td>5</td></tr>
<tr><td align="left">Cyberdiscs</td><td>2</td><td>3</td><td>4</td></tr>
<tr><th>Totals</th><td>12</td><td>16</td><td>20</td></tr>
</table> 

==Tips & Tricks==
[[Heavy Weapons Platforms|HWP]]'s can follow you onto the lift and the next level if you use the [[Blaster Launcher|tools at your disposal]] to widen the doorway for them.

=Mars: The Final Assault=
==Briefing==
The pyramid lift takes your battle weary soldiers deep below the planet's surface. They arrive in the heart of a large complex of tunnels and chambers. The alien brain is hidden somewhere in the labyrinth. It must be destroyed if the earth is to be saved from alien enslavement.

==Description==
[[Image:Brain.gif|right|The Alien Brain]]Second part is almost like a [[Alien Base Assault]]. It's filled with all manner of creatures: [[Ethereal]]s, [[Sectopod]]s, [[Silacoid]]s, [[Celatid]]s, and [[Chryssalid]]s. Anyhow, you need to destroy at least one portion of a four-part [[Brain|Alien Brain]] (40HPs - pictured on the right) to win the mission (and the game). There's a whole bunch of heavily armed Ethereals defending the Brain Room. Of course, either a [[Blaster Launcher|Blaster Bomb]], or plain mind controlling one of them to shoot the Brain ends the mission easily... to be true to storyline, kill all opposition, then shoot the Brain pointblank with a Plasma-based weapon. Afterwards scramble for the lifts before the whole complex blows up (as shown in the PS movies).

'' [ [[User:NKF|NKF]] : This is UFO's only mixed crew mission. It's mainly an Ethereal crew with a few extras, mainly [[Terror Units]], from the other races. This mission only really needs a Blaster Launcher, and with luck, if you locate the command theatre right away, will be over in less than a few turns.] ''

'' [ [[User:Bomb Bloke|Bomb Bloke]] : If you want to bypass the "luck" element, you can take advantage of the fact that you can see unexplored terrain by moving the cursor over it. The brain room is quite recognisable, and once you've found it, it's simple enough to send Blaster Bombs there on your first turn.] ''

'' [ [[User:Hobbes|Hobbes]] : If you choose to find the Command Center the harder way ("Blaster Launchers? Bah! Too impersonal for my taste") be aware that unlike the Alien Bases there can be dead ends.]

==Alien Deployment (underground base)==
<table {{StdCenterTable}}>
<tr {{StdDescTable_Heading}}><th align="left">Rank</th>
<th>Beg./Exp.</th>
<th>Vet./Gen.</th>
<th>Super.</th></tr>
<tr><td align="left">Ethereal Soldiers</td><td>2</td><td>3</td><td>4</td></tr>
<tr><td align="left">Ethereal Leaders</td><td>1</td><td>2</td><td>3</td></tr>
<tr><td align="left">Ethereal Commanders</td><td>3</td><td>4</td><td>5</td></tr>
<tr><td align="left">Chryssalids</td><td>3</td><td>4</td><td>5</td></tr>
<tr><td align="left">Silacoids</td><td>3</td><td>4</td><td>5</td></tr>
<tr><td align="left">Celatids</td><td>3</td><td>4</td><td>5</td></tr>
<tr><td align="left">Sectopods</td><td>2</td><td>3</td><td>4</td></tr>
<tr><th>Totals</th><td>17</td><td>24</td><td>31</td></tr>
</table> 

==Tips & Tricks==
Go armed for bear. Best agents, best weapons, best armour. Bring many troops. The 80 Item Limit is perhaps your worst enemy.

The surface part of the mission is nothing special; you should however have some means of illuminating the battlefield. The [[Auto Cannon]] with incendiary rounds (14 light sources per clip) can come in handy. Only eqipment carried by your soldiers at the end of that stage will be taken into the base proper, so you maybe want to take it slowly and pick up some ammunition before killing the last alien or activating the lift.

Underground is like any other base assault, only more so. More corridors to watch, more opponents hidden. Most enemies will take several hits to die, so when using conventional weapons you will need a great many seasoned troops. Even with the support of two or three Mind Controllers and Blaster Launchers each (Hovertanks are great as they carry 8 rounds yet count as only one item) you will need at least eight or ten men for the actual killing and may still only advance with utmost care. Also, sort out all known PSI weaklings and then some: With so many Ethereal Commanders around, you may see soldiers mind controlled who previously appeared to be absolutely safe even from panic attacks.

... and for the masochist (cough cough try this) Total loadout.... 1 [[Psi-Amp]] & 1 Soldier with no armour.
This works in every difficulity and the only way you can lose is if you get shot from an alien in its reaction, but with a soldier having 200+ time units you have plenty of time to find the first alien, control it (25 TU not a percentage!!) and find the way to the elevator or the brain.

=See Also=

[[Category:Missions]]

Psionic Strength

2008-04-02T04:15:55Z

Kalaong: The Psionics page says that Psi Strength is half the equation in both defense and offense.

Psi Strength is a measure of the soldier's innate ability to defend against psionic attacks - and to execute them.

==Starting Values==
New recruits will always begin with a value between 0 and 100 (inclusive). Your soldier's Psi Strength will not be revealed until his or her first month of [[Psi_Skill|Psi Lab]] training is complete.

Prior to gaining Psi technology, if you have soldiers who are consistently panicked or mind controlled by aliens, it's a very good bet that they have low Psi Strength (below 60). (Aliens 'peek' and know who's low.) Don't strip such soldiers of their weapons, and use them as scouts or cannon fodder, just fire them and get new soldiers. After one month of Psi training you know who is going to stay and who is going to be sacked.

As a consolation prize, soldiers who have panicked many times are liable to see their [[Bravery]] increase.

==Improvement==
Psi Strength cannot be increased. It remains constant throughout the game.

==See Also==
*[[Psionics]]
*[[Psi Skill]]
*[[Bravery]]

[[Category:Soldiers]]

Cydonia

2008-04-02T02:40:36Z

Kalaong: /* Description */

A two part mission. Note that this is a DO OR DIE mission. If you fail or hit abort at any time, the entire campaign ends, you lose.
To have this mission available, you must research "Cydonia or Bust", and access it by launching an avenger. It'll either give the option to go to intercept or to Cydonia. The "Cydonia" button is at the top of the screen if you have an Avenger selected in the Geoscape.

=Mars: Cydonia Landing=
==Briefing==
Your [[Avenger]] craft has landed in the region of Cydonia on the surface of [[Mars]]. Our information indicates that one of the pyramid constructions contains a green access lift to an underground complex. Once you have assembled all your soldiers in the lift area, continue to the next stage…

==Description==
First part is on the surface of Mars. Lots of Pyramid structures with singular windows, some with aliens inside, some empty. Only one has the entrance to the aliens HQ, the entrance being an area with glowing green floor. Now, this floor has 36 squares, or a 6x6 area. Anyhow, you can step onto that with as many people as possible, hit abort, and it'll bring them and whatever they're carrying to part 2. OR, you can just kill everything on the stage and everyone will go to part 2.

The surface is filled with many many [[Sectoid]]s and [[Cyberdisc]]s. Much Commanders and leaders methinks, as there's Psi all around. I find this mission hilarious when I shoot a Cyberdisc, it blows up, and the chain reaction kills over 5 other Cyberdiscs. Plus many many Sectoids. It then turns into Pyramid hunting.

'' [ [[User:NKF|NKF]]: I think the easiest way to look at it is that this mission is basically a Sectoid Battleship/Base crew in a one-of-a-kind map with a choice of two different winning conditions] ''
''[ [[User:(name here)|(name here)]]'' actually, they replaced the navigators, engineers, and medics with leaders and commanders. important diffrence.

==Alien Deployment (Surface)==
<table {{StdCenterTable}}>
<tr {{StdDescTable_Heading}}><th align="left">Rank</th>
<th>Beg./Exp.</th>
<th>Vet./Gen.</th>
<th>Super.</th></tr>
<tr><td align="left">Sectoid Soldiers</td><td>4</td><td>5</td><td>6</td></tr>
<tr><td align="left">Sectoid Navigators</td><td>0</td><td>0</td><td>0</td></tr>
<tr><td align="left">Sectoid Medics</td><td>0</td><td>0</td><td>0</td></tr>
<tr><td align="left">Sectoid Engineers</td><td>0</td><td>0</td><td>0</td></tr>
<tr><td align="left">Sectoid Leaders</td><td>3</td><td>4</td><td>5</td></tr>
<tr><td align="left">Sectoid Commanders</td><td>3</td><td>4</td><td>5</td></tr>
<tr><td align="left">Cyberdiscs</td><td>2</td><td>3</td><td>4</td></tr>
<tr><th>Totals</th><td>12</td><td>16</td><td>20</td></tr>
</table> 

==Tips & Tricks==
[[Heavy Weapons Platforms|HWP]]'s can follow you onto the lift and the next level if you use the [[Blaster Launcher|tools at your disposal]] to widen the doorway for them.

=Mars: The Final Assault=
==Briefing==
The pyramid lift takes your battle weary soldiers deep below the planet's surface. They arrive in the heart of a large complex of tunnels and chambers. The alien brain is hidden somewhere in the labyrinth. It must be destroyed if the earth is to be saved from alien enslavement.

==Description==
[[Image:Brain.PNG|right|Brain]]Second part is almost like a [[Alien Base Assault]]. It's filled with all manner of creatures: [[Ethereal]]s, [[Sectopod]]s, [[Silacoid]]s, [[Celatid]]s, and [[Chryssalid]]s. Anyhow, you need to destroy at least one portion of a four-part [[Brain|Alien Brain]] (40HPs - pictured on the right) to win the mission (and the game). There's a whole bunch of heavily armed Ethereals defending the Brain Room. Of course, either a [[Blaster Launcher|Blaster Bomb]], or plain mind controlling one of them to shoot the Brain ends the mission easily... to be true to storyline, kill all opposition, then shoot the Brain pointblank with a Plasma-based weapon. Afterwards scramble for the lifts before the whole complex blows up (as shown in the PS movies).

'' [ [[User:NKF|NKF]] : This is UFO's only mixed crew mission. It's mainly an Ethereal crew with a few extras, mainly [[Terror Units]], from the other races. This mission only really needs a Blaster Launcher, and with luck, if you locate the command theatre right away, will be over in less than a few turns.] ''

'' [ [[User:Bomb Bloke|Bomb Bloke]] : If you want to bypass the "luck" element, you can take advantage of the fact that you can see unexplored terrain by moving the cursor over it. The brain room is quite recognisable, and once you've found it, it's simple enough to send Blaster Bombs there on your first turn.] ''

'' [ [[User:Hobbes|Hobbes]] : If you choose to find the Command Center the harder way ("Blaster Launchers? Bah! Too impersonal for my taste") be aware that unlike the Alien Bases there can be dead ends.]

==Alien Deployment (underground base)==
<table {{StdCenterTable}}>
<tr {{StdDescTable_Heading}}><th align="left">Rank</th>
<th>Beg./Exp.</th>
<th>Vet./Gen.</th>
<th>Super.</th></tr>
<tr><td align="left">Ethereal Soldiers</td><td>2</td><td>3</td><td>4</td></tr>
<tr><td align="left">Ethereal Leaders</td><td>1</td><td>2</td><td>3</td></tr>
<tr><td align="left">Ethereal Commanders</td><td>3</td><td>4</td><td>5</td></tr>
<tr><td align="left">Chryssalids</td><td>3</td><td>4</td><td>5</td></tr>
<tr><td align="left">Silacoids</td><td>3</td><td>4</td><td>5</td></tr>
<tr><td align="left">Celatids</td><td>3</td><td>4</td><td>5</td></tr>
<tr><td align="left">Sectopods</td><td>2</td><td>3</td><td>4</td></tr>
<tr><th>Totals</th><td>17</td><td>24</td><td>31</td></tr>
</table> 

==Tips & Tricks==
Go armed for bear. Best agents, best weapons, best armour. Bring many troops. The 80 Item Limit is perhaps your worst enemy.

The surface part of the mission is nothing special; you should however have some means of illuminating the battlefield. The [[Auto Cannon]] with incendiary rounds (14 light sources per clip) can come in handy. Only eqipment carried by your soldiers at the end of that stage will be taken into the base proper, so you maybe want to take it slowly and pick up some ammunition before killing the last alien or activating the lift.

Underground is like any other base assault, only more so. More corridors to watch, more opponents hidden. Most enemies will take several hits to die, so when using conventional weapons you will need a great many seasoned troops. Even with the support of two or three Mind Controllers and Blaster Launchers each (Hovertanks are great as they carry 8 rounds yet count as only one item) you will need at least eight or ten men for the actual killing and may still only advance with utmost care. Also, sort out all known PSI weaklings and then some: With so many Ethereal Commanders around, you may see soldiers mind controlled who previously appeared to be absolutely safe even from panic attacks.

... and for the masochist (cough cough try this) Total loadout.... 1 [[Psi-Amp]] & 1 Soldier with no armour.
This works in every difficulity and the only way you can lose is if you get shot from an alien in its reaction, but with a soldier having 200+ time units you have plenty of time to find the first alien, control it (25 TU not a percentage!!) and find the way to the elevator or the brain.

=See Also=

[[Category:Missions]]

Realistic Equivalents

2008-03-31T21:05:24Z

Kalaong: /* Weapons */

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]?

*Plasma weapons - here are a few [http://en.wikipedia.org/wiki/Plasma_rifle promising theories], especially [http://en.wikipedia.org/wiki/Plasma_channel "plasma channels".]

== 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.)''

There is also a [[Talk:Craft|discussion]] taking place on the speeds of aircraft and UFOs.

== 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.

Craft

2008-03-30T00:45:24Z

Kalaong: /* Quick Comparision Table */

Several different types of '''craft''' are available to X-COM Commanders, each has its own strengths and weaknesses. There are two basic roles any X-COM craft can fulfill, the ability to [[UFO Interception|intercept]] enemy [[UFOs]], which requires mounting points for [[Craft Armaments|craft weapons]]. The second potential role is for a troop transport which requires carrying space for [[soldiers]], [[equipment]] (especially [[weapons]]) and potentially [[Heavy Weapons Platforms]]. A couple of the craft with can be [[research]]ed can fill both roles.

=== Interception Craft ===
Interception Craft are used by X-COM to chase after and shoot down UFOs as they travel through Earth's atmosphere performing some [[Alien Missions|mission]]. They need to be fast enough to catch up to the UFO, have sufficient weaponry to take it down, and enough damage capacity to survive long enough to use that weaponry.

*[[Interceptor]] - The only initial interception craft you can use, it is the best humans have at the time but it is comparatively slow and fragile.
*[[Firestorm]] - The first new hybrid of Human-Alien technology applied to craft design, it is a big step up from the interceptor in all but range.

=== Troop Transporters ===
Troop Transporters are used to deliver soldiers to a remote mission location. Ideally they need as high a capacity as possible so that large squads can be sent out where a mission demands it, and sufficient range to get to all potential mission sites. High speed will also help for certain mission profiles, but is of lesser importance for most purposes.

*[[Skyranger]] - A solid piece of equipment that gets the job done for most of the conflict, although it is slow when missions are required at all sorts of places around the world.

===Multi-Role Interceptor/Transporters===
Later research will allow X-COM to field some multi-role craft that can fill both the interceptor and troop transport roles. These craft might supercede the other more specialist craft for general usage, although some commanders may prefer to use them more rigidly in one role or the other, or create entirely new roles that make use of all aspects of these craft.

*[[Lightning]] - This design suffers a little from being too small to do both jobs well, but it can be useful in certain roles, mopping up smaller UFOs with it single weapon and then immediately being able to recover the resulting crash site if a small squad is enough.
*[[Avenger]] - The ultimate hybrid craft created, has almost no weakness - although its fuel is limited, its speed means it still retains an impressive range and gets there much faster than the Skyranger leading to faster mission turnarounds.

==Quick Comparision Table==
<center>
<table {{StdCenterTable}} width="100%">
<tr {{StdDescTable_Heading}}><th>Craft</th>
<th>Max Speed (kts)</th>
<th>Range (nm)</th>
<th>Acceleration</th>
<th>Fuel Capacity</th>
<th>Weapon Pods</th>
<th>Damage Capacity</th>
<th>Cargo Space</th>
<th>[[HWP]] Capacity</th>
<th>Cost/Rent ($)</th></tr>
<tr><td align="left">Interceptor</td><td>2,100</td><td>8,050</td><td>3</td><td>1,000</td><td>2</td><td>100</td><td>None</td><td>None</td><td>600,000*</td></tr>
<tr><td align="left">Firestorm</td><td>4,200</td><td>7,000</td><td>9</td><td>20</td><td>2</td><td>500</td><td>None</td><td>None</td><td>400,000**</td></tr>
<tr><td align="left">Skyranger</td><td>760</td><td>13,500</td><td>2</td><td>1,500</td><td>None</td><td>150</td><td>14</td><td>3</td><td>500,000*</td></tr>
<tr><td align="left">Lightning</td><td>3,100</td><td>7,750</td><td>8</td><td>30</td><td>1</td><td>800</td><td>12</td><td>None</td><td>600,000**</td></tr>
<tr><td align="left">Avenger</td><td>5,400</td><td>27,000</td><td>10</td><td>60</td><td>2</td><td>1,200</td><td>26</td><td>4</td><td>900,000**</td></tr>
</table>
</center>
<nowiki>*</nowiki> This cost is paid upon ordering, and also at midnight on the first of every month once delivered as a craft rental fee 
<nowiki>**</nowiki> These craft also require a number of [[Alien Alloys]], [[UFO Navigation]] and [[UFO Power Source]] to build as well as this price, and take a substantial time to [[manufacture]].

Talk:Craft

2008-03-29T21:07:58Z

Kalaong:

OK! Well past time to theorize on the capabilities of X-Com's air force. The Interceptor, the fastest thing humans can build in X-Com's 1998, has a maximum speed of "2100".

EDIT: The very intelligent Dumas has cleared up my confusion over the "nautical mile", which is the unit of distance a "knot" measures. A vessel traveling at 1 knot along a meridian, covers one sixtieth of a degree geographic latitude in one hour - that is, one nautical mile per hour. A nautical mile is 1,852 meters, 1.150779 English miles, or 6,076.1155 feet.

In dry air with a temperature of 21 °C (70 °F) the speed of sound is 1230 kilometers per hour, 770 miles per hour or 1130 feet per second. Thus, Mach 1 is roughly 669 knots.

Thus, in knots, the Mach numbers of your craft in X-Com: UFO Defense would be;

<center>
<table {{StdCenterTable}} width ="60%">
<tr {{StdDescTable_Heading}}>
<th>Craft</th>
<th>Speed(knots)</th>
<th>Speed(Mach)</th></tr>
<tr><td>Skyranger</td><td>760</td><td>1.1</td></tr>
<tr><td>Interceptor</td><td>2100</td><td>3.1</td></tr>
<tr><td>Lightning</td><td>3100</td><td>4.4</td></tr>
<tr><td>Firestorm</td><td>4200</td><td>6.3</td></tr>
<tr><td>Avenger</td><td>5400</td><td>8.1</td></tr>
</table>
</center>

<center>
<table {{StdCenterTable}} width ="60%">
<tr {{StdDescTable_Heading}}>
<th>UFO</th>
<th>Speed(knots)</th>
<th>Speed(Mach)</th></tr>
<tr><td>Small Scout</td><td>2200</td><td>3.1</td></tr>
<tr><td>Medium Scout</td><td>2400</td><td>3.6</td></tr>
<tr><td>Large Scout</td><td>2700</td><td>3.9</td></tr>
<tr><td>Supply Ship</td><td>3200</td><td>4.6</td></tr>
<tr><td>Abductor</td><td>4000</td><td>6</td></tr>
<tr><td>Harvester</td><td>4300</td><td>6.4</td></tr>
<tr><td>Terror Ship</td><td>4800</td><td>7.2</td></tr>
<tr><td>Battleship</td><td>5000</td><td>7.4</td></tr>
</table>
</center>

Half my original totals. Hideously inflated indeed. But still the kind of thing I want to see in the next "alien invasion" blockbuster. Though those "Dual Pulse Detonation Engines" still make me wonder - those are supposed to be ''hyper''sonic. --Kalaong 13:58 29 March 2008

:Pulse Detonation Engines can have another propose rather than hypersonic speed: efficiency in burning the fuel. Their proposed applications usually refer to their possible use in hypersonic secret aircraft but it is possible that the PDEs are simply used to increase efficiency in fuel comsumption, thus increasing range and loitering time [[User:Hobbes|Hobbes]] 12:25, 29 March 2008 (PDT)

::Which ''would'' explain the insane operational ranges of both the Interceptor and Skyranger. Great point. --Kalaong 16:09 29 March 2008
----

Tempting as it is to say you know nothing about anything, I'm just going to say that you have a hideously inflated idea of how fast these craft should be. The [[http://en.wikipedia.org/wiki/SR-71_Blackbird| Blackbird]] was pretty much designed purely for speed at very high altitude (80000 feet in round numbers). At that altitude,the speed of sound is much much lower than at sea level (Density and temperature effects) and its absolute speed was around 1900 knots.

Since the knot or nautical mile is rather longer than a statute or land mile, a speed expressed in knots is 'lower' than that speed expressed in mph. The speed of sound is closer to 669 knots than 2100. So even taking the Interceptor's listed speed as being in knots, it is ''faster'' than the Blackbird and the Skyranger is supersonic.

In any case, if we want to get away from the game's arbitrary units, the best method is to time various craft as they fly between two given points separated by a well-known distance. Say, crossing the USA from east to west at given latitude.--[[User:Dumas|Dumas]] 20:34, 28 March 2008 (PDT)

:I get all of my numbers from Wikipedia. But the description of nautical miles seems to have thrown me - not just a different unit of measure, but also a different unit ''being'' measured. And what do you mean "hideously" inflated? There's maybe one X-Com base per continent, yet interceptors chase UFOs clear across the planet. Hypersonic speeds seem to be the best way to define that kind of maneuvering - at the very ''least''. Anyway, keep talking. Constructive criticism is why I started this discussion - I'm just a geek punching keys, and I wanted to visualize the kind of craziness these dogfights would be. --Kalaong 0:16, 29 March 2008 (EST)

The nautical mile is defined as 1852 meters, or one minute of latitude along any meridian (neglecting the funky shape of the Earth). A statute mile is somewhere around 1600 meters. For reference, moving at 2100 knots gets you about 35 ''degrees'' of latitude an hour flying north/south, which is rather more than a third of the distance from either pole to the equator.

You seem to be conflating speed and range. Sure, interceptors can go halfway around the Earth, but how quickly are they doing it? Unless we time a few of the buggers, this is just going in circles. The Skyranger takes ''hours'' to get anywhere and even the Avenger takes a while to get from, say, somewhere in China to Indonesia.

Edit: Just timed a couple short flights. A Firestorm takes one hour and fifteen minutes of game time to fly from Equator to Pole (I have a base in Africa that's close to the Equator). An Avenger makes the trip in one hour. 90 degrees of latitude (Equator to Pole) is 5400 knots, so I think that taking the craft speeds as being listed in knots is correct.--[[User:Dumas|Dumas]] 23:02, 28 March 2008 (PDT)

: Better if you make a separate page discussing the real-life issues regarding X-COM's craft than to be adding/changing measure units to the craft's pages. Since the original game does not mention anything regarding which measure applies, extrapolating into reality should be kept separately, otherwise you're adding things to the original game. And my 2c for this discussion: Dumas's in game observations regarding the equator to pole travel time pretty much resolve the issue of which measure should be used. The fun thing is to try to figure out ''why'' X-COM would use subsonic interceptors. Several points could be made: dogfighting takes place at subsonic speeds, supersonic burns a lot of fuel (decreasing range), etc. [[User:Hobbes|Hobbes]] 07:18, 29 March 2008 (PDT)

:Dumas, now that you've explained the concept of a "nautical mile", I just realized that using knots to define craft speed allows them to be mapped right onto the Geoscape, so it makes even more sense. Equator to pole in an hour - passenger jets take ''fifteen'' to get from LA to Hong Kong! I still think some mach numbers would be neat though - everyone has a good mental image of superjets trailing sonic booms. Hobbes, I've linked this page to the [[Realistic Equivalents]] page. And as for dogfights taking place at supersonic speeds? I direct you to the book "Speed is Life, More is Better" By John M. Scanlan. --Kalaong 13:19 29 March 2008

::I mentioned dogfights, not air to air combat in general. Quoting from the Wikipedia's [http://en.wikipedia.org/wiki/Dogfight article]: ''(...)Superiority in a dog fight can depend on a pilot's experience and skill, and the agility of his fighter when flown at minimum air speeds approaching loss of control (causing a danger of stalling)(...)Dogfights are generally contests fought at low airspeeds, while maintaining enough energy for violent acrobatic maneuvering, as pilots attempt to remain within air speeds with a maximum turn rate and minimum turn radius: the so-called "corner speed" that often lies between 300 and 400 knots, depending on conditions(...)'' Interceptors and UFOs don't actually engage in actual 'dogfights' since they simply fire at a distance in the game but my mention to it was merely a hypothesis to justify why the Interceptors are slow (i.e. they were designed with an emphasis on maneuvering rather than speed because it was previously thought that most air to air combat regarding UFOs would consist of dogfights, but operations revealed otherwise - a classic case of pre-war expectations by planners not corresponding to reality. [[User:Hobbes|Hobbes]] 12:15, 29 March 2008 (PDT)

:::Oh, MAN this is cool. Glad I started this up. OK, back to it. The Kiryu-Kai's failed attempts to capture UFOs should have given everyone involved a hint of what they were dealing with. You seem about as sharp as a [[vibroblade]], so I'll get right to what I was hoping to find out in this discussion. All anyone gets in the game when they send an interceptor after a UFO is some cool theme music and a few windows.

:::What do ''you'' think the air-to-air engagements in X-Com: UFO Defense would be like IRL? Total otaku here - I was visualizing crazy hypersonic dogfights that can span continents; the human pilots sweating and grunting under multiple Gs, the UFOs making right-angle evasions with their gravity drives. In your opinion, what would an UFO interception be like? --Kalaong 16:09 29 March 2008

::::One factor crucial to figuring that one out is to determine first the maneuverability of the UFOs. The anti-gravity proporties of their Elerium engines allow them not to be impeded by factors such as the atmospheric pressure or the need to counteract the gravity's effect like jets do. Taking a bit of physics into account, what causes their movement and how it is controlled. From the game there isn't much to imply that UFOs are capable of changing the course 180º all of a sudden with no care for inertia and momentum, simply that they are capable of high-speed accelerations when attacked by a human craft. Instead, they just fly to their target areas and when attacked never move to engage/approach the attacker. This could also lead into a discussion of if they consider human craft nothing more than a nuisance and inconsequential for their grand strategy, but that's another discussion.
::::If they were capable of that unrestricted movement it would be easy for them to avoid any shots fired by human craft, since they would be most likely be able of outmaneuvering anything human fired at them. In that case, no interceptions would be possible, which is not the case. Assuming that the shoot down of an UFO affects their operations (enough for a Battleship to be sent after it happens a few times) to be hardly considered a trivial matter and the UFO crews likely orders only to respond an attack if it the human craft is in range (from seen in the game) then the initial phase of the engagement is pretty much what happens:
::::Human craft moves to reach weapons range and engage, UFO only fires back when it's in range. Then battle is pretty much reduced to a BVR fight, although if it were at hypersonic speeds the human missiles might have a hard time catching a retreating UFO. If both craft aren't shot down, them there would be a dogfight, but the speeds would be greatly reduced.
::::If you wanna have an idea, I wrote one such fight in my X-COM faction, you can read it [http://www.fanfiction.net/s/245417/9/XCOM_The_Unknown_Menace here]. It's the entry for April 7th - [[User:Hobbes|Hobbes]] 13:43, 29 March 2008 (PDT)

:::::Lots of conspiracy lore says UFOs are supersonic at least, and regularly make right-angle turns at speed. I imagined that Interceptors would develop something similar to Age of Sail tactics("Near, near, near, far, ''bracket!''")- hitting something that can corner like that would be impossible unless you force them into a shot. And you've gotta do it without getting hit yourself.

:::::As for the scene; ''really'' cool, but I'm reminded of something I read somewhere, what was it... "Instants of terror separated by long periods of boredom". You spend a lot of time chasing the target, but the engagements themselves last seconds. --Kalaong 18:08 29 March 2008

Talk:Craft

2008-03-29T20:09:43Z

Kalaong:

Talk:Craft

2008-03-29T20:08:44Z

Kalaong:

OK! Well past time to theorize on the capabilities of X-Com's air force. The Interceptor, the fastest thing humans can build in X-Com's 1998, has a maximum speed of "2100".

EDIT: The very intelligent Dumas has cleared up my confusion over the "nautical mile", which is the unit of distance a "knot" measures. A vessel traveling at 1 knot along a meridian, covers one sixtieth of a degree geographic latitude in one hour - that is, one nautical mile per hour. A nautical mile is 1,852 meters, 1.150779 English miles, or 6,076.1155 feet.

In dry air with a temperature of 21 °C (70 °F) the speed of sound is 1230 kilometers per hour, 770 miles per hour or 1130 feet per second. Thus, Mach 1 is roughly 669 knots.

Thus, in knots, the Mach numbers of your craft in X-Com: UFO Defense would be;

<center>
<table {{StdCenterTable}} width ="60%">
<tr {{StdDescTable_Heading}}>
<th>Craft</th>
<th>Speed(knots)</th>
<th>Speed(Mach)</th></tr>
<tr><td>Skyranger</td><td>760</td><td>1.1</td></tr>
<tr><td>Interceptor</td><td>2100</td><td>3.1</td></tr>
<tr><td>Lightning</td><td>3100</td><td>4.4</td></tr>
<tr><td>Firestorm</td><td>4200</td><td>6.3</td></tr>
<tr><td>Avenger</td><td>5400</td><td>8.1</td></tr>
</table>
</center>

<center>
<table {{StdCenterTable}} width ="60%">
<tr {{StdDescTable_Heading}}>
<th>UFO</th>
<th>Speed(knots)</th>
<th>Speed(Mach)</th></tr>
<tr><td>Small Scout</td><td>2200</td><td>3.1</td></tr>
<tr><td>Medium Scout</td><td>2400</td><td>3.6</td></tr>
<tr><td>Large Scout</td><td>2700</td><td>3.9</td></tr>
<tr><td>Supply Ship</td><td>3200</td><td>4.6</td></tr>
<tr><td>Abductor</td><td>4000</td><td>6</td></tr>
<tr><td>Harvester</td><td>4300</td><td>6.4</td></tr>
<tr><td>Terror Ship</td><td>4800</td><td>7.2</td></tr>
<tr><td>Battleship</td><td>5000</td><td>7.4</td></tr>
</table>
</center>

Half my original totals. Hideously inflated indeed. But still the kind of thing I want to see in the next "alien invasion" blockbuster. Though those "Dual Pulse Detonation Engines" still make me wonder - those are supposed to be ''hyper''sonic. --Kalaong 13:58 29 March 2008

:Pulse Detonation Engines can have another propose rather than hypersonic speed: efficiency in burning the fuel. Their proposed applications usually refer to their possible use in hypersonic secret aircraft but it is possible that the PDEs are simply used to increase efficiency in fuel comsumption, thus increasing range and loitering time [[User:Hobbes|Hobbes]] 12:25, 29 March 2008 (PDT)

::Which ''would'' explain the insane operational ranges of both the Interceptor and Skyranger. Great point. --Kalaong 16:09 29 March 2008
----

Tempting as it is to say you know nothing about anything, I'm just going to say that you have a hideously inflated idea of how fast these craft should be. The [[http://en.wikipedia.org/wiki/SR-71_Blackbird| Blackbird]] was pretty much designed purely for speed at very high altitude (80000 feet in round numbers). At that altitude,the speed of sound is much much lower than at sea level (Density and temperature effects) and its absolute speed was around 1900 knots.

Since the knot or nautical mile is rather longer than a statute or land mile, a speed expressed in knots is 'lower' than that speed expressed in mph. The speed of sound is closer to 669 knots than 2100. So even taking the Interceptor's listed speed as being in knots, it is ''faster'' than the Blackbird and the Skyranger is supersonic.

In any case, if we want to get away from the game's arbitrary units, the best method is to time various craft as they fly between two given points separated by a well-known distance. Say, crossing the USA from east to west at given latitude.--[[User:Dumas|Dumas]] 20:34, 28 March 2008 (PDT)

:I get all of my numbers from Wikipedia. But the description of nautical miles seems to have thrown me - not just a different unit of measure, but also a different unit ''being'' measured. And what do you mean "hideously" inflated? There's maybe one X-Com base per continent, yet interceptors chase UFOs clear across the planet. Hypersonic speeds seem to be the best way to define that kind of maneuvering - at the very ''least''. Anyway, keep talking. Constructive criticism is why I started this discussion - I'm just a geek punching keys, and I wanted to visualize the kind of craziness these dogfights would be. --Kalaong 0:16, 29 March 2008 (EST)

The nautical mile is defined as 1852 meters, or one minute of latitude along any meridian (neglecting the funky shape of the Earth). A statute mile is somewhere around 1600 meters. For reference, moving at 2100 knots gets you about 35 ''degrees'' of latitude an hour flying north/south, which is rather more than a third of the distance from either pole to the equator.

You seem to be conflating speed and range. Sure, interceptors can go halfway around the Earth, but how quickly are they doing it? Unless we time a few of the buggers, this is just going in circles. The Skyranger takes ''hours'' to get anywhere and even the Avenger takes a while to get from, say, somewhere in China to Indonesia.

Edit: Just timed a couple short flights. A Firestorm takes one hour and fifteen minutes of game time to fly from Equator to Pole (I have a base in Africa that's close to the Equator). An Avenger makes the trip in one hour. 90 degrees of latitude (Equator to Pole) is 5400 knots, so I think that taking the craft speeds as being listed in knots is correct.--[[User:Dumas|Dumas]] 23:02, 28 March 2008 (PDT)

: Better if you make a separate page discussing the real-life issues regarding X-COM's craft than to be adding/changing measure units to the craft's pages. Since the original game does not mention anything regarding which measure applies, extrapolating into reality should be kept separately, otherwise you're adding things to the original game. And my 2c for this discussion: Dumas's in game observations regarding the equator to pole travel time pretty much resolve the issue of which measure should be used. The fun thing is to try to figure out ''why'' X-COM would use subsonic interceptors. Several points could be made: dogfighting takes place at subsonic speeds, supersonic burns a lot of fuel (decreasing range), etc. [[User:Hobbes|Hobbes]] 07:18, 29 March 2008 (PDT)

:Dumas, now that you've explained the concept of a "nautical mile", I just realized that using knots to define craft speed allows them to be mapped right onto the Geoscape, so it makes even more sense. Equator to pole in an hour - passenger jets take ''fifteen'' to get from LA to Hong Kong! I still think some mach numbers would be neat though - everyone has a good mental image of superjets trailing sonic booms. Hobbes, I've linked this page to the [[Realistic Equivalents]] page. And as for dogfights taking place at supersonic speeds? I direct you to the book "Speed is Life, More is Better" By John M. Scanlan. --Kalaong 13:19 29 March 2008

::I mentioned dogfights, not air to air combat in general. Quoting from the Wikipedia's [http://en.wikipedia.org/wiki/Dogfight article]: ''(...)Superiority in a dog fight can depend on a pilot's experience and skill, and the agility of his fighter when flown at minimum air speeds approaching loss of control (causing a danger of stalling)(...)Dogfights are generally contests fought at low airspeeds, while maintaining enough energy for violent acrobatic maneuvering, as pilots attempt to remain within air speeds with a maximum turn rate and minimum turn radius: the so-called "corner speed" that often lies between 300 and 400 knots, depending on conditions(...)'' Interceptors and UFOs don't actually engage in actual 'dogfights' since they simply fire at a distance in the game but my mention to it was merely a hypothesis to justify why the Interceptors are slow (i.e. they were designed with an emphasis on maneuvering rather than speed because it was previously thought that most air to air combat regarding UFOs would consist of dogfights, but operations revealed otherwise - a classic case of pre-war expectations by planners not corresponding to reality. [[User:Hobbes|Hobbes]] 12:15, 29 March 2008 (PDT)

:::Oh, MAN this is cool. Glad I started this up. OK, back to it. The Kiryu-Kai's failed attempts to capture UFOs should have given everyone involved a hint of what they were dealing with. You seem about as sharp as a [[vibroblade]], so I'll get right to what I was hoping to find out in this discussion. All anyone gets in the game when they send an interceptor after a UFO is a cool theme and a few windows.

:::What do ''you'' think the air-to-air engagements in X-Com: UFO Defense would be like IRL? Total otaku here - I was visualizing crazy hypersonic dogfights that can span continents; the human pilots sweating and grunting under multiple Gs, the UFOs making right-angle evasions with their gravity drives. In your opinion, what would an UFO interception be like? --Kalaong 16:09 29 March 2008

Skyranger

2008-03-29T18:00:51Z

Kalaong: /* General Information */

[[Image:skyranger.png|right|Skyranger Troop Transporter]]
The '''Skyranger''' is a troop transporter with Vertical Take-Off and Landing (VTOL) capabilities. It is one of your most critical assets, as it the only way to get your [[soldiers]] to the scene of any [[Alien Missions]] (apart from when aliens [[Base Defense|attack your base]]). The Skyranger is also the only way for you to retrieve all the [[Alien Artefacts]] needed to [[research]] the production of newer types of [[craft]].

==General Information==
The Skyranger is a supersonic transport capable of transporting up to fourteen X-COM [[soldiers]] to any point on the globe. It can also carry [[Heavy Weapons Platforms]] in the place of four soldiers each.

Each Skyranger has a $500,000 price tag: this is both an order price and a monthly fee, making it expensive to keep several Skyrangers at your base(s). It is usually best not to order a new Skyranger near the end of a month, as you will pay a full months' rent even if it has only been at your base for a day. However, if the aircraft hasn't arrived yet, no rent will be due. Because it takes 72 hours for a Skyranger to be delivered, you can order them up to 71 hours before the end of a given month in order to avoid that month's rent.

Unlike transferring employees, you still pay the rental for Skyrangers that are transferring between bases at the end of the month, although you probably do get to [[ExploitsA#Free_Wages|save the wages]] of any soldiers on board at the time.

==Fuel Usage and Refuelling==
While airborne, the Skyranger consumes fuel at the rate of 7 units (0.466%) every ten minutes, at top speed. This gives the Skyranger a maximum (radial) range of 13,553 nautical miles, while being airborne for 17h, 50m. Its maximum (distance) range would be 27,106 nautical miles, while being airborne for 35h, 40m (nearly one and half days).

While patrolling in an area, fuel consumption is cut to 3 units every ten minutes, allowing for a maximum stationary patrol time of 83hrs, 20m. This gives it the most fuel capacity of any ship in the game, and as it can stay in the air for a maximum of around 3.5 days the Skyranger remains the best craft for performing sweeps for suspected [[Alien Missions#Alien Bases|alien bases]] throughout the game. Compare to the [[Avenger]], for example, which can remain in the air for a maximum of 10 hours.

After a flight, the Skyranger refills its tanks at 3.33% every half hour, on the half hour. (Since you don't see decimals, [[refueling]] follows a pattern of +3%, +3%, +4% fuel.) Thus it can take up to 15 hours (30 half hours) to refuel, if on empty. It will then perform a [[rearming]] step, even though it has no weapons. It will drop out of rearming status and be ready to fly when the clock is next straight up on an hour. Thus, rearming will either take a half hour (if fueling ended at XX:30) or no time at all(if fueling ended at XX:00).

==Vital Statistics==
<table>
<tr><td>Maximum Speed:</td><td>760</td></tr>
<tr><td>Acceleration:</td><td>2</td></tr>
<tr><td>Fuel Capacity:</td><td>1,500</td></tr>
<tr><td>Weapon Pods:</td><td>None</td></tr>
<tr><td>Damage Capacity:</td><td>150</td></tr>
<tr><td>Cargo Space:</td><td>14</td></tr>
<tr><td>Initial/Monthly Fee:</td><td>$500,000</td></tr>
<tr><td>[[:Category:Heavy Weapons Platforms|HWP]] Capacity:</td><td>3</td></tr>
</table>

== See Also ==
*[[Lightning]] - New Fighter-Transporter
*[[Avenger]] - Ultimate Craft
*[[Data Canister: Skyranger]]

[[Category:Crafts]]

Talk:Craft

2008-03-29T17:57:59Z

Kalaong:

Talk:Craft

2008-03-29T17:24:44Z

Kalaong:

OK! Well past time to theorize on the capabilities of X-Com's air force. The Interceptor, the fastest thing humans can build in X-Com's 1998, has a maximum speed of "2100".

As Arbitrary Game Balance Units that is obviously incalculable so I will never use AGBUs again. Instead, I will translate variations of English and Metric units into oh-so-fun Mach numbers - the speed of sound. In dry air with a temperature of 21 °C (70 °F) the speed of sound is 1130 feet per second, 1230 kilometers per hour, 770 miles per hour or 344 meters per second. (I know the Skyranger page says that its speed is in "knots", but Mach 1 is 2277 knots. That makes both craft ''subsonic''. I'm gonna go change that ''right now''.) As for the other numbers, in order of increasing speed...

2100 feet per second doesn't break Mach 2. Come ON! The SR-71 Blackbird(the current record holder for manned supersonic flight) was built in 1964, and it broke Mach 3! (Hmm... the F-100 Super Sabre broke Mach 1 only a decade earlier in 1953... The Roswell crash was in 1945... what if ALL jet aircraft are based on alien technology... Hey! Get that straight jacket off me!)

2100 kilometers per hour STILL doesn't break Mach 2.

2100 miles per hour doesn't break Mach 3. A single-seater pseudo-Blackbird after thirty years of development. Joy. Maybe it's realistic; that would make Lockheed about as good at R&D as another government contractor - NASA. Thirty years to build a better toilet.

2100 meters per second breaks Mach 6. Finally! That's around the flight regime you'd expect from "dual pulse detonation engines".

Thus, in meters per second, the Mach numbers of your craft in X-Com: UFO Defense would be;

Skyranger: 2.2

Interceptor: 6.1

Lightning: 9

Firestorm: 12.2

Avenger: 15.7

The UFOs clock in at;

Small Scout: 6.4

Medium Scout: 7

Large Scout: 7.8

Supply Ship: 9.3

Abductor: 11.6

Harvester: 12.5

Terror Ship: 14

Battleship: 14.5

I have GOT to get me some of these! Somebody call Roland Emmerich! I WANNA SEE THESE BIRDS FIGHT!

----

Tempting as it is to say you know nothing about anything, I'm just going to say that you have a hideously inflated idea of how fast these craft should be. The [[http://en.wikipedia.org/wiki/SR-71_Blackbird| Blackbird]] was pretty much designed purely for speed at very high altitude (80000 feet in round numbers). At that altitude,the speed of sound is much much lower than at sea level (Density and temperature effects) and its absolute speed was around 1900 knots.

Since the knot or nautical mile is rather longer than a statute or land mile, a speed expressed in knots is 'lower' than that speed expressed in mph. The speed of sound is closer to 669 knots than 2100. So even taking the Interceptor's listed speed as being in knots, it is ''faster'' than the Blackbird and the Skyranger is supersonic.

In any case, if we want to get away from the game's arbitrary units, the best method is to time various craft as they fly between two given points separated by a well-known distance. Say, crossing the USA from east to west at given latitude.--[[User:Dumas|Dumas]] 20:34, 28 March 2008 (PDT)

:I get all of my numbers from Wikipedia. But the description of nautical miles seems to have thrown me - not just a different unit of measure, but also a different unit ''being'' measured. And what do you mean "hideously" inflated? There's maybe one X-Com base per continent, yet interceptors chase UFOs clear across the planet. Hypersonic speeds seem to be the best way to define that kind of maneuvering - at the very ''least''. Anyway, keep talking. Constructive criticism is why I started this discussion - I'm just a geek punching keys, and I wanted to visualize the kind of craziness these dogfights would be. --Kalaong 0:16, 29 March 2008 (EST)

The nautical mile is defined as 1852 meters, or one minute of latitude along any meridian (neglecting the funky shape of the Earth). A statute mile is somewhere around 1600 meters. For reference, moving at 2100 knots gets you about 35 ''degrees'' of latitude an hour flying north/south, which is rather more than a third of the distance from either pole to the equator.

You seem to be conflating speed and range. Sure, interceptors can go halfway around the Earth, but how quickly are they doing it? Unless we time a few of the buggers, this is just going in circles. The Skyranger takes ''hours'' to get anywhere and even the Avenger takes a while to get from, say, somewhere in China to Indonesia.

Edit: Just timed a couple short flights. A Firestorm takes one hour and fifteen minutes of game time to fly from Equator to Pole (I have a base in Africa that's close to the Equator). An Avenger makes the trip in one hour. 90 degrees of latitude (Equator to Pole) is 5400 knots, so I think that taking the craft speeds as being listed in knots is correct.--[[User:Dumas|Dumas]] 23:02, 28 March 2008 (PDT)

: Better if you make a separate page discussing the real-life issues regarding X-COM's craft than to be adding/changing measure units to the craft's pages. Since the original game does not mention anything regarding which measure applies, extrapolating into reality should be kept separately, otherwise you're adding things to the original game. And my 2c for this discussion: Dumas's in game observations regarding the equator to pole travel time pretty much resolve the issue of which measure should be used. The fun thing is to try to figure out ''why'' X-COM would use subsonic interceptors. Several points could be made: dogfighting takes place at subsonic speeds, supersonic burns a lot of fuel (decreasing range), etc. [[User:Hobbes|Hobbes]] 07:18, 29 March 2008 (PDT)

:Dumas, now that you've explained the concept of a "nautical mile", I just realized that using knots to define craft speed allows them to be mapped right onto the Geoscape, so it makes even more sense. Equator to pole in an hour - passenger jets take ''fifteen'' to get from LA to Hong Kong! I still think some mach numbers would be neat though - everyone has a good mental image of superjets trailing sonic booms. Hobbes, I've linked this page to the [[Realistic Equivalents]] page. And as for dogfights taking place at supersonic speeds? I direct you to the book "Speed is Life, More is Better" By John M. Scanlan. --Kalaong 13:19 29 March 2008

Realistic Equivalents

2008-03-29T17:19:16Z

Kalaong: /* Unit size, speed, game turn length, etc. */

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.)''

There is also a [[Talk:Craft|discussion]] taking place on the speeds of aircraft and UFOs.

== 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

2008-03-29T17:19:03Z

Kalaong: /* Unit size, speed, game turn length, etc. */

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.)''

There is also a [[Talk:Craft|discussion] taking place on the speeds of aircraft and UFOs.

== 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.

Talk:Craft

2008-03-29T17:18:44Z

Kalaong:

OK! Well past time to theorize on the capabilities of X-Com's air force. The Interceptor, the fastest thing humans can build in X-Com's 1998, has a maximum speed of "2100".

As Arbitrary Game Balance Units that is obviously incalculable so I will never use AGBUs again. Instead, I will translate variations of English and Metric units into oh-so-fun Mach numbers - the speed of sound. In dry air with a temperature of 21 °C (70 °F) the speed of sound is 1130 feet per second, 1230 kilometers per hour, 770 miles per hour or 344 meters per second. (I know the Skyranger page says that its speed is in "knots", but Mach 1 is 2277 knots. That makes both craft ''subsonic''. I'm gonna go change that ''right now''.) As for the other numbers, in order of increasing speed...

2100 feet per second doesn't break Mach 2. Come ON! The SR-71 Blackbird(the current record holder for manned supersonic flight) was built in 1964, and it broke Mach 3! (Hmm... the F-100 Super Sabre broke Mach 1 only a decade earlier in 1953... The Roswell crash was in 1945... what if ALL jet aircraft are based on alien technology... Hey! Get that straight jacket off me!)

2100 kilometers per hour STILL doesn't break Mach 2.

2100 miles per hour doesn't break Mach 3. A single-seater pseudo-Blackbird after thirty years of development. Joy. Maybe it's realistic; that would make Lockheed about as good at R&D as another government contractor - NASA. Thirty years to build a better toilet.

2100 meters per second breaks Mach 6. Finally! That's around the flight regime you'd expect from "dual pulse detonation engines".

Thus, in meters per second, the Mach numbers of your craft in X-Com: UFO Defense would be;

Skyranger: 2.2

Interceptor: 6.1

Lightning: 9

Firestorm: 12.2

Avenger: 15.7

The UFOs clock in at;

Small Scout: 6.4

Medium Scout: 7

Large Scout: 7.8

Supply Ship: 9.3

Abductor: 11.6

Harvester: 12.5

Terror Ship: 14

Battleship: 14.5

I have GOT to get me some of these! Somebody call Roland Emmerich! I WANNA SEE THESE BIRDS FIGHT!

----

Tempting as it is to say you know nothing about anything, I'm just going to say that you have a hideously inflated idea of how fast these craft should be. The [[http://en.wikipedia.org/wiki/SR-71_Blackbird| Blackbird]] was pretty much designed purely for speed at very high altitude (80000 feet in round numbers). At that altitude,the speed of sound is much much lower than at sea level (Density and temperature effects) and its absolute speed was around 1900 knots.

Since the knot or nautical mile is rather longer than a statute or land mile, a speed expressed in knots is 'lower' than that speed expressed in mph. The speed of sound is closer to 669 knots than 2100. So even taking the Interceptor's listed speed as being in knots, it is ''faster'' than the Blackbird and the Skyranger is supersonic.

In any case, if we want to get away from the game's arbitrary units, the best method is to time various craft as they fly between two given points separated by a well-known distance. Say, crossing the USA from east to west at given latitude.--[[User:Dumas|Dumas]] 20:34, 28 March 2008 (PDT)

:I get all of my numbers from Wikipedia. But the description of nautical miles seems to have thrown me - not just a different unit of measure, but also a different unit ''being'' measured. And what do you mean "hideously" inflated? There's maybe one X-Com base per continent, yet interceptors chase UFOs clear across the planet. Hypersonic speeds seem to be the best way to define that kind of maneuvering - at the very ''least''. Anyway, keep talking. Constructive criticism is why I started this discussion - I'm just a geek punching keys, and I wanted to visualize the kind of craziness these dogfights would be. --Kalaong 0:16, 29 March 2008 (EST)

The nautical mile is defined as 1852 meters, or one minute of latitude along any meridian (neglecting the funky shape of the Earth). A statute mile is somewhere around 1600 meters. For reference, moving at 2100 knots gets you about 35 ''degrees'' of latitude an hour flying north/south, which is rather more than a third of the distance from either pole to the equator.

You seem to be conflating speed and range. Sure, interceptors can go halfway around the Earth, but how quickly are they doing it? Unless we time a few of the buggers, this is just going in circles. The Skyranger takes ''hours'' to get anywhere and even the Avenger takes a while to get from, say, somewhere in China to Indonesia.

Edit: Just timed a couple short flights. A Firestorm takes one hour and fifteen minutes of game time to fly from Equator to Pole (I have a base in Africa that's close to the Equator). An Avenger makes the trip in one hour. 90 degrees of latitude (Equator to Pole) is 5400 knots, so I think that taking the craft speeds as being listed in knots is correct.--[[User:Dumas|Dumas]] 23:02, 28 March 2008 (PDT)

: Better if you make a separate page discussing the real-life issues regarding X-COM's craft than to be adding/changing measure units to the craft's pages. Since the original game does not mention anything regarding which measure applies, extrapolating into reality should be kept separately, otherwise you're adding things to the original game. And my 2c for this discussion: Dumas's in game observations regarding the equator to pole travel time pretty much resolve the issue of which measure should be used. The fun thing is to try to figure out ''why'' X-COM would use subsonic interceptors. Several points could be made: dogfighting takes place at subsonic speeds, supersonic burns a lot of fuel (decreasing range), etc. [[User:Hobbes|Hobbes]] 07:18, 29 March 2008 (PDT)

:Now that Dumas has explained the concept of a "nautical mile", I just realized that using knots to define craft speed allows them to be mapped right onto the Geoscape, so it makes even more sense. Equator to pole in an hour - passenger jets take ''fifteen'' to get from LA to Hong Kong! I still think some mach numbers would be neat though - everyone has a good mental image of superjets trailing sonic booms. Hobbes, I've linked this page to the [[Realistic Equivalents]] page. --Kalaong 13:19 29 March 2008

Talk:Craft

2008-03-29T04:22:35Z

Kalaong:

Talk:Craft

2008-03-29T01:19:15Z

Kalaong:

Skyranger

2008-03-28T22:56:23Z

Kalaong: /* General Information */ If the numbers are in knots, the Interceptor can't even break the sound barrier. Check X-Com Craft Discussion.

[[Image:skyranger.png|right|Skyranger Troop Transporter]]
The '''Skyranger''' is a troop transporter with Vertical Take-Off and Landing (VTOL) capabilities. It is one of your most critical assets, as it the only way to get your [[soldiers]] to the scene of any [[Alien Missions]] (apart from when aliens [[Base Defense|attack your base]]). The Skyranger is also the only way for you to retrieve all the [[Alien Artefacts]] needed to [[research]] the production of newer types of [[craft]].

==General Information==
The Skyranger is capable of transporting anywhere up to fourteen X-COM [[soldiers]] to any point on the globe. It can also carry [[Heavy Weapons Platforms]] in the place of four soldiers each.

Each Skyranger has a $500,000 price tag: this is both an order price and a monthly fee, making it expensive to keep several Skyrangers at your base(s). It is usually best not to order a new Skyranger near the end of a month, as you will pay a full months' rent even if it has only been at your base for a day. However, if the aircraft hasn't arrived yet, no rent will be due. Because it takes 72 hours for a Skyranger to be delivered, you can order them up to 71 hours before the end of a given month in order to avoid that month's rent.

Unlike transferring employees, you still pay the rental for Skyrangers that are transferring between bases at the end of the month, although you probably do get to [[ExploitsA#Free_Wages|save the wages]] of any soldiers on board at the time.

==Fuel Usage and Refuelling==
While airborne, the Skyranger consumes fuel at the rate of 7 units (0.466%) every ten minutes, at top speed. This gives the Skyranger a maximum (radial) range of 13,553 nautical miles, while being airborne for 17h, 50m. Its maximum (distance) range would be 27,106 nautical miles, while being airborne for 35h, 40m (nearly one and half days).

While patrolling in an area, fuel consumption is cut to 3 units every ten minutes, allowing for a maximum stationary patrol time of 83hrs, 20m. This gives it the most fuel capacity of any ship in the game, and as it can stay in the air for a maximum of around 3.5 days the Skyranger remains the best craft for performing sweeps for suspected [[Alien Missions#Alien Bases|alien bases]] throughout the game. Compare to the [[Avenger]], for example, which can remain in the air for a maximum of 10 hours.

After a flight, the Skyranger refills its tanks at 3.33% every half hour, on the half hour. (Since you don't see decimals, [[refueling]] follows a pattern of +3%, +3%, +4% fuel.) Thus it can take up to 15 hours (30 half hours) to refuel, if on empty. It will then perform a [[rearming]] step, even though it has no weapons. It will drop out of rearming status and be ready to fly when the clock is next straight up on an hour. Thus, rearming will either take a half hour (if fueling ended at XX:30) or no time at all(if fueling ended at XX:00).

==Vital Statistics==
<table>
<tr><td>Maximum Speed:</td><td>760</td></tr>
<tr><td>Acceleration:</td><td>2</td></tr>
<tr><td>Fuel Capacity:</td><td>1,500</td></tr>
<tr><td>Weapon Pods:</td><td>None</td></tr>
<tr><td>Damage Capacity:</td><td>150</td></tr>
<tr><td>Cargo Space:</td><td>14</td></tr>
<tr><td>Initial/Monthly Fee:</td><td>$500,000</td></tr>
<tr><td>[[:Category:Heavy Weapons Platforms|HWP]] Capacity:</td><td>3</td></tr>
</table>

== See Also ==
*[[Lightning]] - New Fighter-Transporter
*[[Avenger]] - Ultimate Craft
*[[Data Canister: Skyranger]]

[[Category:Crafts]]

Talk:Craft

2008-03-28T22:46:05Z

Kalaong: New page: OK! Well past time to theorize on the capabilities of X-Com's air force. The Interceptor, the fastest thing humans can build in X-Com's 1998, has a maximum speed of "2100". As Arbitrary G...

Talk:Elerium-115

2008-03-14T05:29:54Z

Kalaong: /* Chuck Norris? */

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

==How much Elerium is “1 Elerium“?==

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

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

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

- [[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)

::::::::::If you absolutely want to gauge the yield of a power source explosion, I suggest you compare it to the damage done by something we know. How many grenades stacked on top of each other would it take to damage the UFO interior (walls & floor) in the way a power source explosion does? Then look up the amount of explosives in a typical hand grenade and there you go. As to fireball size... there's an old saying, comparing arguments like this to the paralympics. Yet I can not resist.

::::::::::a) the size stated in that wikipedia article is by no means the final size. It's not as if the fireball would reach a certain size and suddenly vanish. It grows ever larger and starts to rise, becoming the cap of the mushroom cloud (by then it's no longer a ball and merely glowing, of course). Defining the fireball size as "the size at the moment when the shockwave catches up with the fireball" is somewhat arbitrary, but you've got to draw the line somewhere. And this definition refers to a time when it's still most certainly a ball, and it'S size directly related to the yield of the bomb (that is to say, other factors like terrain, wind, air pressure, moisture etc are relatively insignificant). That's alright for an article that deals with yield (as in kt or MT equivalents); damage is something else entirely, covered in another article.
::::::::::b) "Overtaking" sounds as if the shockwave would start somewhere inside the fireball, but of course the shockwave is not a seperate effect. It's the result of all the superheated air trying to expand rapidly. Some sources say the shockwave "detaches from" the fireball, which seems more apt.
::::::::::c) I guess you've misunderstood the Mach Stem: the reflected shock front moves faster in air that has already been accelerated by the main shock front; the reflected front will therefore catch up and combine with the main shock front, reinforcing it. ([http://de.wikipedia.org/wiki/Bild:Mach_effect_sequence.png diagram]). That's reinforcement, not amplification: the shock wave will not become more forceful because of the Mach effect, but it will lose power at a slower rate (which means same damage over a larger area). Please note that the development of the Mach Stem does not depend on any terrain features (in fact, a flat surface may be as good as it gets). In the past few days I've read more than I ever wanted to know about nuclear explosions, but all sources go along the lines that the valley at Nagasaki served to confine the damage, not a single word about reinforcement or amplification. Maybe destruction in the valley was so complete that a possible reinforcement would have made no difference any more. --[[User:Schnobs|Schnobs]] 13:42, 8 November 2007 (PST)
:Great armchair discussion, folks. Now if only you could bring some biology into it, I might jump in. Re: the indentation, I think the [[Talk:Main_Page#Discussion.2Ftalk_page_proposed_format|general idea]] is to only indent about 4-6 levels deep, then reset back to one. Also - and I hope I don't seem picky here, I love the flow of ideas - it seems like sometimes super and subscripting might be used better. A simple e.g. "m/s" is acceptable for meters per second, and sometimes it seems like e.g. "R = (Et2/ρ)1/5" might better be R = (Et2/ρ)1/5. On my PC, the extra super- and sub-scripting seems to be messing with line overlap.
:All that aside, please keep going! It's an interesting reality check, to check back from e.g. the damage of a standard grenade, to back-evaluate the damage caused. One crazy thing about X-COM explosions is how they have "hard wired" edges, as seen in e.g. [[Explosions#Playing_With_Fire]]. That's hard to figure into the parlor computations. The standard grenade is the only explosive that does not have "artificial clipping" of its blast radius performed. - [[User:MikeTheRed|MikeTheRed]] 17:04, 9 November 2007 (PST)

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

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

:To be more specific, my theory is that a [[UFO Power Source]] uses Elerium to generate antimatter in a controlled manner(like Bob Lazar's theorizes), and to generate the anti-gravity field that keeps the UFO flying. Plasma weapons are the same - the Elerium provides energy to generate the plasma, then uses an anti-gravity field to keep the plasma from dissipating as it flies across the battlefield. Fusion weapons(the stuff that goes boom) are the odd man out - they generate anti-matter in an ''out-of-control'' manner and use the anti-gravity field to ''control'' the resulting explosion. In other words, Elerium(In the game! I don't wanna sound like a conspiracy-obsessed geek, just your garden variety geek!) is like nuclear material-plus. You have to encourage it to make all that energy, and when you stop encouraging it, it only has a relatively small tantrum. You have to get really creative to make use Elerium as a Weapon of Mass ''Annihilation''. The aliens(And X-COM!) purposely limit the destructive potential of Elerium-based munitions and technology, despite this lowering the efficiency of the rare element. Otherwise a single [[Blaster Bomb]] would kill a city, leaving nothing for the aliens to abduct - or X-COM to protect. It's also why losing the [[Cydonia]] mission ends the game - the aliens could have reduced the Earth to beaded glass at any time,
but never consider it until their home base is threatened.

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

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

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

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

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

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

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

- [[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)

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

- amitakartok

==Chuck Norris?==
There is a grave error on this page: Chuck Norris is GOD! He can't be killed ;) [[User:Hobbes|Hobbes]] 14:01, 11 March 2008 (PDT)
: What about if Steven Seagal, Bruce Lee and Mike Tyson ganged up on him and jumped him from behind after he'd had a few beers at the end of a hard day defending America from evil drug lords? [[User:Spike|Spike]] 16:41, 11 March 2008 (PDT)
::This is entertainingly silly, but now has nothing to do with the game. Start theorizing on how said demigods would fare against Mutons or ''somebody's'' going to have to delete this thread;)! Kalaong 16:28 13 March 2008 (EDT)
::: Mutons would be able to crush Chuck Norris beneath their endless legions of sucky terror units
:::: If one of X-COM's soldiers was Chuck Norris, he would have a Melee attack with a strength of about 60, requiring 10% TUs per attack. His stats (including firing accuracy) would be maxed out. However I like to believe that Bruce Lee's Melee damage would be nearer to 200, allowing him to punch through UFO doors. (Are we back on topic now?) [[User:Spike|Spike]] 15:23, 13 March 2008 (PDT)

[[Image:X-COMChuckNorris.png]]

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

==See Also==

[[Elirium]]

Talk:Elerium-115

2008-03-13T20:29:55Z

Kalaong:

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

==How much Elerium is “1 Elerium“?==

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

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

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

- [[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)

::::::::::If you absolutely want to gauge the yield of a power source explosion, I suggest you compare it to the damage done by something we know. How many grenades stacked on top of each other would it take to damage the UFO interior (walls & floor) in the way a power source explosion does? Then look up the amount of explosives in a typical hand grenade and there you go. As to fireball size... there's an old saying, comparing arguments like this to the paralympics. Yet I can not resist.

::::::::::a) the size stated in that wikipedia article is by no means the final size. It's not as if the fireball would reach a certain size and suddenly vanish. It grows ever larger and starts to rise, becoming the cap of the mushroom cloud (by then it's no longer a ball and merely glowing, of course). Defining the fireball size as "the size at the moment when the shockwave catches up with the fireball" is somewhat arbitrary, but you've got to draw the line somewhere. And this definition refers to a time when it's still most certainly a ball, and it'S size directly related to the yield of the bomb (that is to say, other factors like terrain, wind, air pressure, moisture etc are relatively insignificant). That's alright for an article that deals with yield (as in kt or MT equivalents); damage is something else entirely, covered in another article.
::::::::::b) "Overtaking" sounds as if the shockwave would start somewhere inside the fireball, but of course the shockwave is not a seperate effect. It's the result of all the superheated air trying to expand rapidly. Some sources say the shockwave "detaches from" the fireball, which seems more apt.
::::::::::c) I guess you've misunderstood the Mach Stem: the reflected shock front moves faster in air that has already been accelerated by the main shock front; the reflected front will therefore catch up and combine with the main shock front, reinforcing it. ([http://de.wikipedia.org/wiki/Bild:Mach_effect_sequence.png diagram]). That's reinforcement, not amplification: the shock wave will not become more forceful because of the Mach effect, but it will lose power at a slower rate (which means same damage over a larger area). Please note that the development of the Mach Stem does not depend on any terrain features (in fact, a flat surface may be as good as it gets). In the past few days I've read more than I ever wanted to know about nuclear explosions, but all sources go along the lines that the valley at Nagasaki served to confine the damage, not a single word about reinforcement or amplification. Maybe destruction in the valley was so complete that a possible reinforcement would have made no difference any more. --[[User:Schnobs|Schnobs]] 13:42, 8 November 2007 (PST)
:Great armchair discussion, folks. Now if only you could bring some biology into it, I might jump in. Re: the indentation, I think the [[Talk:Main_Page#Discussion.2Ftalk_page_proposed_format|general idea]] is to only indent about 4-6 levels deep, then reset back to one. Also - and I hope I don't seem picky here, I love the flow of ideas - it seems like sometimes super and subscripting might be used better. A simple e.g. "m/s" is acceptable for meters per second, and sometimes it seems like e.g. "R = (Et2/ρ)1/5" might better be R = (Et2/ρ)1/5. On my PC, the extra super- and sub-scripting seems to be messing with line overlap.
:All that aside, please keep going! It's an interesting reality check, to check back from e.g. the damage of a standard grenade, to back-evaluate the damage caused. One crazy thing about X-COM explosions is how they have "hard wired" edges, as seen in e.g. [[Explosions#Playing_With_Fire]]. That's hard to figure into the parlor computations. The standard grenade is the only explosive that does not have "artificial clipping" of its blast radius performed. - [[User:MikeTheRed|MikeTheRed]] 17:04, 9 November 2007 (PST)

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

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

:To be more specific, my theory is that a [[UFO Power Source]] uses Elerium to generate antimatter in a controlled manner(like Bob Lazar's theorizes), and to generate the anti-gravity field that keeps the UFO flying. Plasma weapons are the same - the Elerium provides energy to generate the plasma, then uses an anti-gravity field to keep the plasma from dissipating as it flies across the battlefield. Fusion weapons(the stuff that goes boom) are the odd man out - they generate anti-matter in an ''out-of-control'' manner and use the anti-gravity field to ''control'' the resulting explosion. In other words, Elerium(In the game! I don't wanna sound like a conspiracy-obsessed geek, just your garden variety geek!) is like nuclear material-plus. You have to encourage it to make all that energy, and when you stop encouraging it, it only has a relatively small tantrum. You have to get really creative to make use Elerium as a Weapon of Mass ''Annihilation''. The aliens(And X-COM!) purposely limit the destructive potential of Elerium-based munitions and technology, despite this lowering the efficiency of the rare element. Otherwise a single [[Blaster Bomb]] would kill a city, leaving nothing for the aliens to abduct - or X-COM to protect. It's also why losing the [[Cydonia]] mission ends the game - the aliens could have reduced the Earth to beaded glass at any time, but never consider it until their home base is threatened.

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

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

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

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

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

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

- [[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)

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

- amitakartok

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

: What about if Steven Seagal, Bruce Lee and Mike Tyson ganged up on him and jumped him from behind after he'd had a few beers at the end of a hard day defending America from evil drug lords? [[User:Spike|Spike]] 16:41, 11 March 2008 (PDT)

::This is entertainingly silly, but now has nothing to do with the game. Start theorizing on how said demigods would fare against Mutons or ''somebody's'' going to have to delete this thread;)! Kalaong 16:28 13 March 2008 (EDT)

==See Also==

[[Elirium]]

Talk:Elerium-115

2008-03-13T20:29:17Z

Kalaong: /* Chuck Norris? */

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

==Item Spawning==
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

==How much Elerium is “1 Elerium“?==

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

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

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

- [[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)

::::::::::If you absolutely want to gauge the yield of a power source explosion, I suggest you compare it to the damage done by something we know. How many grenades stacked on top of each other would it take to damage the UFO interior (walls & floor) in the way a power source explosion does? Then look up the amount of explosives in a typical hand grenade and there you go. As to fireball size... there's an old saying, comparing arguments like this to the paralympics. Yet I can not resist.

::::::::::a) the size stated in that wikipedia article is by no means the final size. It's not as if the fireball would reach a certain size and suddenly vanish. It grows ever larger and starts to rise, becoming the cap of the mushroom cloud (by then it's no longer a ball and merely glowing, of course). Defining the fireball size as "the size at the moment when the shockwave catches up with the fireball" is somewhat arbitrary, but you've got to draw the line somewhere. And this definition refers to a time when it's still most certainly a ball, and it'S size directly related to the yield of the bomb (that is to say, other factors like terrain, wind, air pressure, moisture etc are relatively insignificant). That's alright for an article that deals with yield (as in kt or MT equivalents); damage is something else entirely, covered in another article.
::::::::::b) "Overtaking" sounds as if the shockwave would start somewhere inside the fireball, but of course the shockwave is not a seperate effect. It's the result of all the superheated air trying to expand rapidly. Some sources say the shockwave "detaches from" the fireball, which seems more apt.
::::::::::c) I guess you've misunderstood the Mach Stem: the reflected shock front moves faster in air that has already been accelerated by the main shock front; the reflected front will therefore catch up and combine with the main shock front, reinforcing it. ([http://de.wikipedia.org/wiki/Bild:Mach_effect_sequence.png diagram]). That's reinforcement, not amplification: the shock wave will not become more forceful because of the Mach effect, but it will lose power at a slower rate (which means same damage over a larger area). Please note that the development of the Mach Stem does not depend on any terrain features (in fact, a flat surface may be as good as it gets). In the past few days I've read more than I ever wanted to know about nuclear explosions, but all sources go along the lines that the valley at Nagasaki served to confine the damage, not a single word about reinforcement or amplification. Maybe destruction in the valley was so complete that a possible reinforcement would have made no difference any more. --[[User:Schnobs|Schnobs]] 13:42, 8 November 2007 (PST)
:Great armchair discussion, folks. Now if only you could bring some biology into it, I might jump in. Re: the indentation, I think the [[Talk:Main_Page#Discussion.2Ftalk_page_proposed_format|general idea]] is to only indent about 4-6 levels deep, then reset back to one. Also - and I hope I don't seem picky here, I love the flow of ideas - it seems like sometimes super and subscripting might be used better. A simple e.g. "m/s" is acceptable for meters per second, and sometimes it seems like e.g. "R = (Et2/ρ)1/5" might better be R = (Et2/ρ)1/5. On my PC, the extra super- and sub-scripting seems to be messing with line overlap.
:All that aside, please keep going! It's an interesting reality check, to check back from e.g. the damage of a standard grenade, to back-evaluate the damage caused. One crazy thing about X-COM explosions is how they have "hard wired" edges, as seen in e.g. [[Explosions#Playing_With_Fire]]. That's hard to figure into the parlor computations. The standard grenade is the only explosive that does not have "artificial clipping" of its blast radius performed. - [[User:MikeTheRed|MikeTheRed]] 17:04, 9 November 2007 (PST)

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

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

:To be more specific, my theory is that a [[UFO Power Source]] uses Elerium to generate antimatter in a controlled manner(like Bob Lazar's theorizes), and to generate the anti-gravity field that keeps the UFO flying. Plasma weapons are the same - the Elerium provides energy to generate the plasma, then uses an anti-gravity field to keep the plasma from dissipating as it flies across the battlefield. Fusion weapons(the stuff that goes boom) are the odd man out - they generate anti-matter in an ''out-of-control'' manner and use the anti-gravity field to ''control'' the resulting explosion. In other words, Elerium(In the game! I don't wanna sound like a conspiracy-obsessed geek, just your garden variety geek!) is like nuclear material-plus. You have to encourage it to make all that energy, and when you stop encouraging it, it only has a relatively small tantrum. You have to get really creative to make use Elerium as a Weapon of Mass ''Annihilation''. The aliens(And X-COM!) purposely limit the destructive potential of Elerium-based munitions and technology, despite this lowering the efficiency of the rare element. Otherwise a single [[Blaster Bomb]] would kill a city, leaving nothing for the aliens to abduct - or X-COM to protect. It's also why losing the [[Cydonia]] mission ends the game - the aliens could have reduced the Earth to beaded glass at any time, but never consider it until their home base is threatened.

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

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

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

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

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

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

- [[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)

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

- amitakartok

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

: What about if Steven Seagal, Bruce Lee and Mike Tyson ganged up on him and jumped him from behind after he'd had a few beers at the end of a hard day defending America from evil drug lords? [[User:Spike|Spike]] 16:41, 11 March 2008 (PDT)

::This is entertainingly silly, but now has nothing to do with the game. Start theorizing on how said demigods would fare against Mutons or ''somebody's'' going to have to delete this thread;)! Kalaong 16:28 13 March 2008 (EDT)

==See Also==

[[Elirium]]

Talk:Elerium-115

2008-03-12T18:54:58Z

Kalaong: /* About Elerium mining... */

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

==Item Spawning==
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

==How much Elerium is “1 Elerium“?==

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

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

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

- [[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)

::::::::::If you absolutely want to gauge the yield of a power source explosion, I suggest you compare it to the damage done by something we know. How many grenades stacked on top of each other would it take to damage the UFO interior (walls & floor) in the way a power source explosion does? Then look up the amount of explosives in a typical hand grenade and there you go. As to fireball size... there's an old saying, comparing arguments like this to the paralympics. Yet I can not resist.

::::::::::a) the size stated in that wikipedia article is by no means the final size. It's not as if the fireball would reach a certain size and suddenly vanish. It grows ever larger and starts to rise, becoming the cap of the mushroom cloud (by then it's no longer a ball and merely glowing, of course). Defining the fireball size as "the size at the moment when the shockwave catches up with the fireball" is somewhat arbitrary, but you've got to draw the line somewhere. And this definition refers to a time when it's still most certainly a ball, and it'S size directly related to the yield of the bomb (that is to say, other factors like terrain, wind, air pressure, moisture etc are relatively insignificant). That's alright for an article that deals with yield (as in kt or MT equivalents); damage is something else entirely, covered in another article.
::::::::::b) "Overtaking" sounds as if the shockwave would start somewhere inside the fireball, but of course the shockwave is not a seperate effect. It's the result of all the superheated air trying to expand rapidly. Some sources say the shockwave "detaches from" the fireball, which seems more apt.
::::::::::c) I guess you've misunderstood the Mach Stem: the reflected shock front moves faster in air that has already been accelerated by the main shock front; the reflected front will therefore catch up and combine with the main shock front, reinforcing it. ([http://de.wikipedia.org/wiki/Bild:Mach_effect_sequence.png diagram]). That's reinforcement, not amplification: the shock wave will not become more forceful because of the Mach effect, but it will lose power at a slower rate (which means same damage over a larger area). Please note that the development of the Mach Stem does not depend on any terrain features (in fact, a flat surface may be as good as it gets). In the past few days I've read more than I ever wanted to know about nuclear explosions, but all sources go along the lines that the valley at Nagasaki served to confine the damage, not a single word about reinforcement or amplification. Maybe destruction in the valley was so complete that a possible reinforcement would have made no difference any more. --[[User:Schnobs|Schnobs]] 13:42, 8 November 2007 (PST)
:Great armchair discussion, folks. Now if only you could bring some biology into it, I might jump in. Re: the indentation, I think the [[Talk:Main_Page#Discussion.2Ftalk_page_proposed_format|general idea]] is to only indent about 4-6 levels deep, then reset back to one. Also - and I hope I don't seem picky here, I love the flow of ideas - it seems like sometimes super and subscripting might be used better. A simple e.g. "m/s" is acceptable for meters per second, and sometimes it seems like e.g. "R = (Et2/ρ)1/5" might better be R = (Et2/ρ)1/5. On my PC, the extra super- and sub-scripting seems to be messing with line overlap.
:All that aside, please keep going! It's an interesting reality check, to check back from e.g. the damage of a standard grenade, to back-evaluate the damage caused. One crazy thing about X-COM explosions is how they have "hard wired" edges, as seen in e.g. [[Explosions#Playing_With_Fire]]. That's hard to figure into the parlor computations. The standard grenade is the only explosive that does not have "artificial clipping" of its blast radius performed. - [[User:MikeTheRed|MikeTheRed]] 17:04, 9 November 2007 (PST)

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

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

:To be more specific, my theory is that a [[UFO Power Source]] uses Elerium to generate antimatter in a controlled manner(like Bob Lazar's theorizes), and to generate the anti-gravity field that keeps the UFO flying. Plasma weapons are the same - the Elerium provides energy to generate the plasma, then uses an anti-gravity field to keep the plasma from dissipating as it flies across the battlefield. Fusion weapons(the stuff that goes boom) are the odd man out - they generate anti-matter in an ''out-of-control'' manner and use the anti-gravity field to ''control'' the resulting explosion. In other words, Elerium(In the game! I don't wanna sound like a conspiracy-obsessed geek, just your garden variety geek!) is like nuclear material-plus. You have to encourage it to make all that energy, and when you stop encouraging it, it only has a relatively small tantrum. You have to get really creative to make use Elerium as a Weapon of Mass ''Annihilation''. The aliens(And X-COM!) purposely limit the destructive potential of Elerium-based munitions and technology, despite this lowering the efficiency of the rare element. Otherwise a single [[Blaster Bomb]] would kill a city, leaving nothing for the aliens to abduct - or X-COM to protect. It's also why losing the [[Cydonia]] mission ends the game - the aliens could have reduced the Earth to beaded glass at any time, but never consider it until their home base is threatened.

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

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

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

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

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

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

- [[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)

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

- amitakartok

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

: What about if Steven Seagal, Bruce Lee and Mike Tyson ganged up on him and jumped him from behind after he'd had a few beers at the end of a hard day defending America from evil drug lords? [[User:Spike|Spike]] 16:41, 11 March 2008 (PDT)

==See Also==

[[Elirium]]

Talk:Elerium-115

2008-03-12T18:54:22Z

Kalaong: /* Item Spawning */

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

==Item Spawning==
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

==How much Elerium is “1 Elerium“?==

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

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

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

- [[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)

::::::::::If you absolutely want to gauge the yield of a power source explosion, I suggest you compare it to the damage done by something we know. How many grenades stacked on top of each other would it take to damage the UFO interior (walls & floor) in the way a power source explosion does? Then look up the amount of explosives in a typical hand grenade and there you go. As to fireball size... there's an old saying, comparing arguments like this to the paralympics. Yet I can not resist.

::::::::::a) the size stated in that wikipedia article is by no means the final size. It's not as if the fireball would reach a certain size and suddenly vanish. It grows ever larger and starts to rise, becoming the cap of the mushroom cloud (by then it's no longer a ball and merely glowing, of course). Defining the fireball size as "the size at the moment when the shockwave catches up with the fireball" is somewhat arbitrary, but you've got to draw the line somewhere. And this definition refers to a time when it's still most certainly a ball, and it'S size directly related to the yield of the bomb (that is to say, other factors like terrain, wind, air pressure, moisture etc are relatively insignificant). That's alright for an article that deals with yield (as in kt or MT equivalents); damage is something else entirely, covered in another article.
::::::::::b) "Overtaking" sounds as if the shockwave would start somewhere inside the fireball, but of course the shockwave is not a seperate effect. It's the result of all the superheated air trying to expand rapidly. Some sources say the shockwave "detaches from" the fireball, which seems more apt.
::::::::::c) I guess you've misunderstood the Mach Stem: the reflected shock front moves faster in air that has already been accelerated by the main shock front; the reflected front will therefore catch up and combine with the main shock front, reinforcing it. ([http://de.wikipedia.org/wiki/Bild:Mach_effect_sequence.png diagram]). That's reinforcement, not amplification: the shock wave will not become more forceful because of the Mach effect, but it will lose power at a slower rate (which means same damage over a larger area). Please note that the development of the Mach Stem does not depend on any terrain features (in fact, a flat surface may be as good as it gets). In the past few days I've read more than I ever wanted to know about nuclear explosions, but all sources go along the lines that the valley at Nagasaki served to confine the damage, not a single word about reinforcement or amplification. Maybe destruction in the valley was so complete that a possible reinforcement would have made no difference any more. --[[User:Schnobs|Schnobs]] 13:42, 8 November 2007 (PST)
:Great armchair discussion, folks. Now if only you could bring some biology into it, I might jump in. Re: the indentation, I think the [[Talk:Main_Page#Discussion.2Ftalk_page_proposed_format|general idea]] is to only indent about 4-6 levels deep, then reset back to one. Also - and I hope I don't seem picky here, I love the flow of ideas - it seems like sometimes super and subscripting might be used better. A simple e.g. "m/s" is acceptable for meters per second, and sometimes it seems like e.g. "R = (Et2/ρ)1/5" might better be R = (Et2/ρ)1/5. On my PC, the extra super- and sub-scripting seems to be messing with line overlap.
:All that aside, please keep going! It's an interesting reality check, to check back from e.g. the damage of a standard grenade, to back-evaluate the damage caused. One crazy thing about X-COM explosions is how they have "hard wired" edges, as seen in e.g. [[Explosions#Playing_With_Fire]]. That's hard to figure into the parlor computations. The standard grenade is the only explosive that does not have "artificial clipping" of its blast radius performed. - [[User:MikeTheRed|MikeTheRed]] 17:04, 9 November 2007 (PST)

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

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

:To be more specific, my theory is that a [[UFO Power Source]] uses Elerium to generate antimatter in a controlled manner(like Bob Lazar's theorizes), and to generate the anti-gravity field that keeps the UFO flying. Plasma weapons are the same - the Elerium provides energy to generate the plasma, then uses an anti-gravity field to keep the plasma from dissipating as it flies across the battlefield. Fusion weapons(the stuff that goes boom) are the odd man out - they generate anti-matter in an ''out-of-control'' manner and use the anti-gravity field to ''control'' the resulting explosion. In other words, Elerium(In the game! I don't wanna sound like a conspiracy-obsessed geek, just your garden variety geek!) is like nuclear material-plus. You have to encourage it to make all that energy, and when you stop encouraging it, it only has a relatively small tantrum. You have to get really creative to make use Elerium as a Weapon of Mass ''Annihilation''. The aliens(And X-COM!) purposely limit the destructive potential of Elerium-based munitions and technology, despite this lowering the efficiency of the rare element. Otherwise a single [[Blaster Bomb]] would kill a city, leaving nothing for the aliens to abduct - or X-COM to protect. It's also why losing the [[Cydonia]] mission ends the game - the aliens could have reduced the Earth to beaded glass at any time, but never consider it until their home base is threatened.

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

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

==About Elerium mining...==

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

- amitakartok

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

: What about if Steven Seagal, Bruce Lee and Mike Tyson ganged up on him and jumped him from behind after he'd had a few beers at the end of a hard day defending America from evil drug lords? [[User:Spike|Spike]] 16:41, 11 March 2008 (PDT)

==See Also==

[[Elirium]]

Talk:Elerium-115

2008-03-11T21:56:13Z

Kalaong: /* How much Elerium is “1 Elerium“? */

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

==Item Spawning==
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)

==Elerium Sources]]
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)

==How much Elerium is “1 Elerium“?==

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

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

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

- [[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)

::::::::::If you absolutely want to gauge the yield of a power source explosion, I suggest you compare it to the damage done by something we know. How many grenades stacked on top of each other would it take to damage the UFO interior (walls & floor) in the way a power source explosion does? Then look up the amount of explosives in a typical hand grenade and there you go. As to fireball size... there's an old saying, comparing arguments like this to the paralympics. Yet I can not resist.

::::::::::a) the size stated in that wikipedia article is by no means the final size. It's not as if the fireball would reach a certain size and suddenly vanish. It grows ever larger and starts to rise, becoming the cap of the mushroom cloud (by then it's no longer a ball and merely glowing, of course). Defining the fireball size as "the size at the moment when the shockwave catches up with the fireball" is somewhat arbitrary, but you've got to draw the line somewhere. And this definition refers to a time when it's still most certainly a ball, and it'S size directly related to the yield of the bomb (that is to say, other factors like terrain, wind, air pressure, moisture etc are relatively insignificant). That's alright for an article that deals with yield (as in kt or MT equivalents); damage is something else entirely, covered in another article.
::::::::::b) "Overtaking" sounds as if the shockwave would start somewhere inside the fireball, but of course the shockwave is not a seperate effect. It's the result of all the superheated air trying to expand rapidly. Some sources say the shockwave "detaches from" the fireball, which seems more apt.
::::::::::c) I guess you've misunderstood the Mach Stem: the reflected shock front moves faster in air that has already been accelerated by the main shock front; the reflected front will therefore catch up and combine with the main shock front, reinforcing it. ([http://de.wikipedia.org/wiki/Bild:Mach_effect_sequence.png diagram]). That's reinforcement, not amplification: the shock wave will not become more forceful because of the Mach effect, but it will lose power at a slower rate (which means same damage over a larger area). Please note that the development of the Mach Stem does not depend on any terrain features (in fact, a flat surface may be as good as it gets). In the past few days I've read more than I ever wanted to know about nuclear explosions, but all sources go along the lines that the valley at Nagasaki served to confine the damage, not a single word about reinforcement or amplification. Maybe destruction in the valley was so complete that a possible reinforcement would have made no difference any more. --[[User:Schnobs|Schnobs]] 13:42, 8 November 2007 (PST)
:Great armchair discussion, folks. Now if only you could bring some biology into it, I might jump in. Re: the indentation, I think the [[Talk:Main_Page#Discussion.2Ftalk_page_proposed_format|general idea]] is to only indent about 4-6 levels deep, then reset back to one. Also - and I hope I don't seem picky here, I love the flow of ideas - it seems like sometimes super and subscripting might be used better. A simple e.g. "m/s" is acceptable for meters per second, and sometimes it seems like e.g. "R = (Et2/ρ)1/5" might better be R = (Et2/ρ)1/5. On my PC, the extra super- and sub-scripting seems to be messing with line overlap.
:All that aside, please keep going! It's an interesting reality check, to check back from e.g. the damage of a standard grenade, to back-evaluate the damage caused. One crazy thing about X-COM explosions is how they have "hard wired" edges, as seen in e.g. [[Explosions#Playing_With_Fire]]. That's hard to figure into the parlor computations. The standard grenade is the only explosive that does not have "artificial clipping" of its blast radius performed. - [[User:MikeTheRed|MikeTheRed]] 17:04, 9 November 2007 (PST)

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

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

:To be more specific, my theory is that a [[UFO Power Source]] uses Elerium to generate antimatter in a controlled manner(like Bob Lazar's theorizes), and to generate the anti-gravity field that keeps the UFO flying. Plasma weapons are the same - the Elerium provides energy to generate the plasma, then uses an anti-gravity field to keep the plasma from dissipating as it flies across the battlefield. Fusion weapons(the stuff that goes boom) are the odd man out - they generate anti-matter in an ''out-of-control'' manner and use the anti-gravity field to ''control'' the resulting explosion. In other words, Elerium(In the game! I don't wanna sound like a conspiracy-obsessed geek, just your garden variety geek!) is like nuclear material-plus. You have to encourage it to make all that energy, and when you stop encouraging it, it only has a relatively small tantrum. You have to get really creative to make use Elerium as a Weapon of Mass ''Annihilation''. The aliens(And X-COM!) purposely limit the destructive potential of Elerium-based munitions and technology, despite this lowering the efficiency of the rare element. Otherwise a single [[Blaster Bomb]] would kill a city, leaving nothing for the aliens to abduct - or X-COM to protect. It's also why losing the [[Cydonia]] mission ends the game - the aliens could have reduced the Earth to beaded glass at any time, but never consider it until their home base is threatened.

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

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

==About Elerium mining...==

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

- amitakartok

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

Talk:Elerium-115

2008-03-11T21:25:06Z

Kalaong: As per the warning, I am breaking this page into smaller sections. Please change the section titles if you find them unsuitable.

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

==Item Spawning==
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)

==Elerium Sources]]
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)

==How much Elerium is “1 Elerium“?==

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

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

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

- [[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)

::::::::::If you absolutely want to gauge the yield of a power source explosion, I suggest you compare it to the damage done by something we know. How many grenades stacked on top of each other would it take to damage the UFO interior (walls & floor) in the way a power source explosion does? Then look up the amount of explosives in a typical hand grenade and there you go. As to fireball size... there's an old saying, comparing arguments like this to the paralympics. Yet I can not resist.

::::::::::a) the size stated in that wikipedia article is by no means the final size. It's not as if the fireball would reach a certain size and suddenly vanish. It grows ever larger and starts to rise, becoming the cap of the mushroom cloud (by then it's no longer a ball and merely glowing, of course). Defining the fireball size as "the size at the moment when the shockwave catches up with the fireball" is somewhat arbitrary, but you've got to draw the line somewhere. And this definition refers to a time when it's still most certainly a ball, and it'S size directly related to the yield of the bomb (that is to say, other factors like terrain, wind, air pressure, moisture etc are relatively insignificant). That's alright for an article that deals with yield (as in kt or MT equivalents); damage is something else entirely, covered in another article.
::::::::::b) "Overtaking" sounds as if the shockwave would start somewhere inside the fireball, but of course the shockwave is not a seperate effect. It's the result of all the superheated air trying to expand rapidly. Some sources say the shockwave "detaches from" the fireball, which seems more apt.
::::::::::c) I guess you've misunderstood the Mach Stem: the reflected shock front moves faster in air that has already been accelerated by the main shock front; the reflected front will therefore catch up and combine with the main shock front, reinforcing it. ([http://de.wikipedia.org/wiki/Bild:Mach_effect_sequence.png diagram]). That's reinforcement, not amplification: the shock wave will not become more forceful because of the Mach effect, but it will lose power at a slower rate (which means same damage over a larger area). Please note that the development of the Mach Stem does not depend on any terrain features (in fact, a flat surface may be as good as it gets). In the past few days I've read more than I ever wanted to know about nuclear explosions, but all sources go along the lines that the valley at Nagasaki served to confine the damage, not a single word about reinforcement or amplification. Maybe destruction in the valley was so complete that a possible reinforcement would have made no difference any more. --[[User:Schnobs|Schnobs]] 13:42, 8 November 2007 (PST)
:Great armchair discussion, folks. Now if only you could bring some biology into it, I might jump in. Re: the indentation, I think the [[Talk:Main_Page#Discussion.2Ftalk_page_proposed_format|general idea]] is to only indent about 4-6 levels deep, then reset back to one. Also - and I hope I don't seem picky here, I love the flow of ideas - it seems like sometimes super and subscripting might be used better. A simple e.g. "m/s" is acceptable for meters per second, and sometimes it seems like e.g. "R = (Et2/ρ)1/5" might better be R = (Et2/ρ)1/5. On my PC, the extra super- and sub-scripting seems to be messing with line overlap.
:All that aside, please keep going! It's an interesting reality check, to check back from e.g. the damage of a standard grenade, to back-evaluate the damage caused. One crazy thing about X-COM explosions is how they have "hard wired" edges, as seen in e.g. [[Explosions#Playing_With_Fire]]. That's hard to figure into the parlor computations. The standard grenade is the only explosive that does not have "artificial clipping" of its blast radius performed. - [[User:MikeTheRed|MikeTheRed]] 17:04, 9 November 2007 (PST)

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

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

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

==About Elerium mining...==

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

- amitakartok

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

Elerium-115

2008-03-11T21:22:32Z

Kalaong: /* How much Elerium is „1 Elerium“? */

Elerium, otherwise known as element 115, is the energy source for all alien techonologies. Mined from reserves near [[Cydonia]], it has a variety of uses. However, it cannot be found on Earth...

== Stats ==
<table><tr><td>[[Image:BIGOBS56.GIF|left|64 px]]</td><td>
*Size: 1 high x 1 wide
*Weight: 3
*Sell Price: $5000</td></tr></table>

== Obtaining Elerium ==
Elerium cannot be synthesized on Earth; it can only be obtained from [[UFO Assault]]s or [[Alien Base Assault]]s. Elerium is housed within [[UFO Power Source]]s, and is very fragile. If upon starting a UFO assault you find a [[UFO Power Source]] to have been destroyed (as a result of a crash landing), its Elerium will surely have been destroyed as well. Be careful about using explosives near intact UFO Power sources, as the Elerium within is easily toasted!

At the end of a successful mission, any Elerium contained within intact UFO Power Sources will be automatically collected and returned to your base. If for some reason you need to end a mission prematurely, you can still collect Elerium by hand, by shooting open a UFO Power Source (generally with a plasma weapon). For more on securing Elerium in this way, see [[UFO Power Source]]s.

Perhaps the most efficient way to gain a stock of Elerium is to find an Alien Base and attack the [[Supply Ship]]s as they land nearby. Leave the base intact to attract more supply ships, as the Elerium is worth more than the score penalty (150 units per intact Supply Ship).

UFO Power Sources can also be found in ''some'' Alien Bases, if they have a particular base "module". You can raid such a base without having to win the mission, and have a soldier carry the Elerium back to the entry lift and then abort (see [[Alien Base Assault#Smash And Grab]] for more details). Attacking the base's Supply Ships is likely to be a safer and more reliable source of Elerium, however; see [[UFO Recovery Values#Alien Base Assaults]] for more details.

On the map, Elerium appears as a purple ball (the same graphic as a [[Stun Bomb]] on the ground). This object is worth 50 points of Elerium. During [[Base Defense]], some of your own Elerium from stores may appear on the battlescape. This is a Good Thing if you win the mission, as each unit is represented as a pod and can be recovered for ''fifty'' units. This is probably a bug and appears to be absent from the Collector's Edition.

== Uses ==
'''Don't sell Elerium.''' Although you do have the option to do so, this is to be highly discouraged. It is used to fuel the [[Firestorm]], the [[Lightning]], and the [[Avenger]]. It is also used, in varying amounts, to [[manufacture]] most [[Alien Artefacts]], [[Flying Suit]]s, and various other technologies. Given that you can only obtain Elerium in the battlefield, selling it is not a good idea. If you choose to, it will fetch you $5,000 per unit.

== Elerium In Reality ==
[http://en.wikipedia.org/wiki/Ununpentium Ununpentium], the temporary name for synthesized element 115, was only created briefly in atomic laboratories in 2004 and is not the subject of much research. However UFO 'researcher' Robert Lazar made several [http://www.beyondweird.com/element115.html claims] about E-115 in the 1980's. For example, it would be the only element that had a gravitational field extending past it's own atomic 'shell'. This deviance from the normal rule would be so significant that perhaps the technologies seen in the [[X-Com Series]] could become a reality - anti-gravity devices used for both flight and particle acceleration.

== How much Elerium is “1 Elerium“? ==
Since some information about the physical laws that rule Elerium-related technology is given in the game this question can be answered to some extend. To fit elerium into the laws of physics, some basic facts have to be “chosen”. The only way to equal an amount of elerium to an amount of power is to calculate the power needed by elerium-propelled crafts. I chose the “Avenger” for this calculation. Since it is the most advanced craft built by X-Com we can assume it has maximum efficiency, which eases the maths a bit.

I also assumed that the weight of the elerium item that can be picked up in combat (3 weight units) can not be used to deduct the actual mass of elerium in this item.

First we have to find the “Avenger”s maximum speed in SI units.

r * f / t = v

With r as range (54000 nautic miles), t as time that the craft needs for that range (10 hours) and f as factor translating nautic miles into kilometres (1.852) the maximum speed v is almost exactly 10,000 km/h.

The required power for maintaining this speed can be deducted solely from atmospheric drag.

F = v^2 * RHO * A * C_d /2

P = F*r

With v as velocity (10,000 km/h), RHO as density of atmosphere (see below), A as area where the drag applies to the craft (for the “Avenger” this would be roughly estimated 20 m^2), C_d as drag coefficient (since the “Avenger” should be constructed quite well I estimated 0.3) and r as range in kilometres (100000), F is the force applied by atmospheric resistance and P the total power over total range that is used to overcome this resistance.

I took 5000 ft. as average flight height to estimate the atmospheric density. This means that RHO roughly equals 0.85.

The velocity has to be [m/s] and the range has to be [m] to fit for the following formulas.

Therefore follows: F = 1.968 * 10^7 kN and P = F * 10^9 = 1.968 * 10 ^16 kNm

Now we take that keeping the craft at max speed only uses 75% of required energy per mission, since combat situations as well as turning, starting and landing (overcoming inertia) use extra power. Now we use Einsteins most popular formula to calculate the required mass from our power requirements:

1,968 * 10^16 / 0.75 = m´ * c^2

Therefore follows: m´ = 291.555 g

Taking into account that “only” 99% of mass is converted into energy:

m = 291.555 / 0.99 = 294.5 grams , which is the mass carrying the “Avenger” for 100.000 kilometres!!!

Since an “Avenger” is fuelled with 12 units of elerium, one unit wheights roughly 25 grams, which means one UFO power source carries about 1.25 kilogram of elerium. This also means that the explosions of UFO Power Sources after crashing are BY FAR too small. An explosion in which less than 0.1% (around 1 gram) of the contained Elerium is transformed into energy had more explosive force than the Hiroshima atomic bomb. Shooting down a terror ship with 4 power sources destroyed would therefore almost certainly end the battle for Earth... by completely vaporizing it :/ And keep in mind: An alien grenade contains 25 grams of elerium, enough to destroy the USA (or kill Chuck Norris).

[[Category:Equipment (UFO Defense)]]

Talk:Elerium-115

2008-03-11T02:01:50Z

Kalaong:

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)

::::::::::If you absolutely want to gauge the yield of a power source explosion, I suggest you compare it to the damage done by something we know. How many grenades stacked on top of each other would it take to damage the UFO interior (walls & floor) in the way a power source explosion does? Then look up the amount of explosives in a typical hand grenade and there you go. As to fireball size... there's an old saying, comparing arguments like this to the paralympics. Yet I can not resist.

::::::::::a) the size stated in that wikipedia article is by no means the final size. It's not as if the fireball would reach a certain size and suddenly vanish. It grows ever larger and starts to rise, becoming the cap of the mushroom cloud (by then it's no longer a ball and merely glowing, of course). Defining the fireball size as "the size at the moment when the shockwave catches up with the fireball" is somewhat arbitrary, but you've got to draw the line somewhere. And this definition refers to a time when it's still most certainly a ball, and it'S size directly related to the yield of the bomb (that is to say, other factors like terrain, wind, air pressure, moisture etc are relatively insignificant). That's alright for an article that deals with yield (as in kt or MT equivalents); damage is something else entirely, covered in another article.
::::::::::b) "Overtaking" sounds as if the shockwave would start somewhere inside the fireball, but of course the shockwave is not a seperate effect. It's the result of all the superheated air trying to expand rapidly. Some sources say the shockwave "detaches from" the fireball, which seems more apt.
::::::::::c) I guess you've misunderstood the Mach Stem: the reflected shock front moves faster in air that has already been accelerated by the main shock front; the reflected front will therefore catch up and combine with the main shock front, reinforcing it. ([http://de.wikipedia.org/wiki/Bild:Mach_effect_sequence.png diagram]). That's reinforcement, not amplification: the shock wave will not become more forceful because of the Mach effect, but it will lose power at a slower rate (which means same damage over a larger area). Please note that the development of the Mach Stem does not depend on any terrain features (in fact, a flat surface may be as good as it gets). In the past few days I've read more than I ever wanted to know about nuclear explosions, but all sources go along the lines that the valley at Nagasaki served to confine the damage, not a single word about reinforcement or amplification. Maybe destruction in the valley was so complete that a possible reinforcement would have made no difference any more. --[[User:Schnobs|Schnobs]] 13:42, 8 November 2007 (PST)
:Great armchair discussion, folks. Now if only you could bring some biology into it, I might jump in. Re: the indentation, I think the [[Talk:Main_Page#Discussion.2Ftalk_page_proposed_format|general idea]] is to only indent about 4-6 levels deep, then reset back to one. Also - and I hope I don't seem picky here, I love the flow of ideas - it seems like sometimes super and subscripting might be used better. A simple e.g. "m/s" is acceptable for meters per second, and sometimes it seems like e.g. "R = (Et2/ρ)1/5" might better be R = (Et2/ρ)1/5. On my PC, the extra super- and sub-scripting seems to be messing with line overlap.
:All that aside, please keep going! It's an interesting reality check, to check back from e.g. the damage of a standard grenade, to back-evaluate the damage caused. One crazy thing about X-COM explosions is how they have "hard wired" edges, as seen in e.g. [[Explosions#Playing_With_Fire]]. That's hard to figure into the parlor computations. The standard grenade is the only explosive that does not have "artificial clipping" of its blast radius performed. - [[User:MikeTheRed|MikeTheRed]] 17:04, 9 November 2007 (PST)

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

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

---------------------

About Elerium mining...

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

- amitakartok