Talk:Elerium-115

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

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

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

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

-Bomb Bloke

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

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

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

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

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

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

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

My apologies. I do not have access to X-Com: Apocalypse, and was basing my data merely on what was said in the game. In light of the new information, it can be reverted if you desire. `Arrow Quivershaft 21:56, 31 May 2007 (PDT)

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

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

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

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

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

- tequilachef

It's worth noting that while the UFO Power Source may be incredibly efficient in Elerium use, there is no guarantee that this is so in regards to the alien weapons. In fact, given the size of a power unit, I'd say it's more likely that the weapons are extremely inefficient in Elerium use. (Especially the heavy plasma clip; 3 times the Elerium for one-third more shots and just over double the killing power of the Plasma Pistol.) We also don't know exactly where that Elerium used in construction of the grenade(or anything else) goes; it's quite possible that only a fraction of it goes into the charge/warhead and some is used in the creation of functional parts. Also, it should be noted that explosions do NOT scale linearly; twice as large a warhead on an atomic or hydrogen bomb does not equal twice the explosive power. In addition, it's been theorized that the explosion from UFO Power Sources is not from the impact; its from trying to start the UFO's engines in order to escape incoming X-COM troops before it's ready(thus why the aliens are killed immediately before the X-COM turn begins, and not when the UFO crashes.)

Of course, since this is fiction, it really doesn't matter, just thought I'd bring a few things to the table since you seem interested in scientific accuracy. Arrow Quivershaft 15:53, 2 November 2007 (PDT)

The math on this page does not check. 5400nm in 10 hours is almost exactly 1000km/hr (v = (r*f)/t - r = 5400, f = 1.852, t = 10 - 10000/10 - ie: 1000). That figure, coverted to meters per second, is almost exactly 277m/s.

According to the articles on Atmospheric Pressure and the Barometric Formula, below about 10K feet, you're at 1 standard atmosphere - at 0C that's 1.293, not 0.85. So for the Force, we have (277^2*1.293*20*0.3)/2 - or about 2.98*10^5N. For the equation to result in Joules we have to convert the range from kilometers to meters - 1km == 1000m, so we wind up with 10^8m. That's 2.98*10^13 J. Finally the conjecture is that the power cell is only 75% consumed. That (2.98*10^13/0.75) puts us with at about 3.97*10^13 J of total available power. So when we take that to Einstein (E=mc^2 - so m = E/c^2) we have (3.97*10^13)/c^2. Since C is 299792458m/s^2, we get: 4.4*10^-4 kg. Call it 4.4 grams. (1g is 10^-4kg, no?) If that's 12 units, then each unit is about 37 centigrams (0.37 grams). (Please note that this is probably going to be my only post here - I had a friend point me to this page and the math seemed wrong, so I checked it myself. And that 37centigram per-unit figure is assuming a 100% conversion rate. it becomes 4.5g per ship - or about 38centigrams per unit - at the stated 99% conversion rate. All math figures done using the 'bc' - a very powerful command-line calculator found on Unix/Linux systems). - Shadow 17:48, 2 November 2007 (PDT)