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October 7

Largest arthropods: land versus sea

See Largest organisms#Arthropods (Arthropoda). It appears to me that the largest arthropods on land are significantly smaller than the largest under-water arthropods. Is this correct? If so, why? Does it have to do with the ability of the exoskeleton to support its weight or access to oxygen or control of water content or vulnerability to predation or what? More generally, what limits the size of arthropods? JRSpriggs (talk) 03:48, 7 October 2011 (UTC)[reply]

For mammals as well, the largest mammal in the sea (the Blue Whale) is orders of magnitude larger than the largest mammal on land (the elephant). The reason for this is buoyancy, and I suspect that for arthropods, the reasoning is similar. --Jayron32 04:04, 7 October 2011 (UTC)[reply]

Do you mean that buoyancy reduces the weight of the animal relative to what it would weigh on land? And wouldn't there be another factor in that the ocean is more vast both in surface area but perhaps more importantly in the depth that it offers? Bus stop (talk) 04:15, 7 October 2011 (UTC)[reply]

Animals on land are limited in size because beyond a certain limit, they would not be able to support their own weight. Animals in the water are bouyed by the water, which allows for larger animals to exist in the water. --Jayron32 04:23, 7 October 2011 (UTC)[reply]

Perhaps nothing illustrates this more clearly then Siphonophorae, some of which can reach 40m-50m in water, but would obviously resemble little more then puddles of goo on dry land. Vespine (talk) 05:45, 7 October 2011 (UTC)[reply]

The size of the largest aquatic and land-living arthropods, does not differ that much. The largest ever aquatic arthropod is the Devonian eurypterid Jaekelopterus rhenaniae (8.2 ft). The largest ever land arthropod is the Upper Carboniferous myriapod Arthropleura (8.5 ft). The largest extant arthropod by mass is the American lobster Homarus americanus (3.5 ft). The largest living land arthropod is the coconut crab Birgus latro (3.3 ft). The Japanese spider crab may be considered larger by some, but only because it has amazingly long spindly legs. However, it is true that land arthropods are smaller than aquatic arthropods on average. And yes, you're probably correct on all counts (though the oxygen factor as a limit is still contentious):

Gravity - exoskeletons cover the exterior of an animal and thus are much heavier than animals with endoskeletons like vertebrates in proportion to size. Proportionally, a smaller arthropod will require a thinner exoskeleton than a larger arthropod to maintain rigidity and motion. Water provides a buoyant support for large aquatic animals. Land animals, however, have to depend on the rigidity of their supports. And in order for exoskeletons to be more rigid it needs to be exponentially thicker and heavier the larger the animal is. There is a point where it becomes mechanically impossible to increase any further (a gravity limit). In addition, exoskeletons need to be molted, they can not grow continuously like endoskeletons. A molting giant arthropod on land would collapse into a squishy mess under its own weight, resulting in deformities when the exoskeleton hardens. Add to this the fact that when molting they are extremely vulnerable (they can't even move as the exoskeletons are the supports for their muscles like our bones are).
Oxygen - arthropods don't breath with lungs but by direct surface absorption into a network of tracheae. Although crustaceans and arachnids do have gills and book lungs, they also have far more primitive circulatory systems than vertebrates. Tracheal respiration are passive systems. The larger the size of an animal or the thicker the exoskeleton, the harder it is to get oxygen into the deeper internal organs. This is proposed to have been why the hyperoxic conditions of the Carboniferous resulted in land arthropods reaching gigantic sizes. Add to this the problems of water balance. Aquatic arthropods do not have to worry about leaks or evaporation.
Predation - Paleozoic land arthropods (and aquatic arthropods a little further back) reached very large sizes due to the lack of competition (compare eurypterids which arose before fish and Carboniferous land arthropods which arose before land vertebrate dominance). As vertebrates started taking over some niches, it became too risky for them to increase in size. Per previous reasons, the larger they are, the weaker they are relatively in comparison to animals with endoskeletons. A flea can jump ~200 times its body length, increase their size, and that proportionally becomes much lower.-- Obsidi♠n Soul 08:20, 7 October 2011 (UTC)[reply]

I should note that as mentioned in Invertebrate trachea, some insects have air sacs at the end of the tracheae, and air can be forced into and out of the system. However, while arthropods ordinarily should be well capable of using muscular movements to suck in air via the tracheae, during molting the situation is different - those movements shouldn't work, but land arthropods take in air by swallowing it.[1] For small arthropods, I assume the air in the stomach provides a source of oxygen (besides, it takes a long, long time to suffocate an arthropod...) - still, it sounds like a problem. But I don't see why very large insects couldn't evolve some method to permit air to leak into the tracheal system or otherwise to distribute itself from the foregut during that time. Wnt (talk) 11:19, 7 October 2011 (UTC)[reply]

Tracheal breathing in bees, grasshoppers, and dragonflies, etc. are actually still heavily mechanical. They only function when the insects are not at rest (e.g. flying, when the metabolic demands are the highest, or when in an alerted state). The rest of the time, it breathes through diffusion. And they did. That's why the largest arthropods today are crustaceans and chelicerates. Because their larger members have gills, branchiostegal lungs, book lungs etc. which do not rely on passive diffusion, and they have blood that can carry oxygen (hemocyanin). But as mentioned, their hearts are still relatively far weaker than vertebrate hearts.-- Obsidi♠n Soul 14:12, 7 October 2011 (UTC)[reply]

e/c: Also no. They can not breathe swallowed air. And I'm curious where you got the 'they take a long time to suffocate' bit. Instances of seemingly long suffocation times are merely the result of them still getting oxygen through cutaneous respiration from what would have drowned a vertebrate lung. See [2] -- Obsidi♠n Soul 14:49, 7 October 2011 (UTC)[reply]

Well, you can "anesthetize" a caterpillar for purposes of macabre manipulation by keeping it under cold water - given the low metabolic rate, further reduced by the cold, they can survive around an hour that way. This is true even though some effort is made to exclude large external air bubbles, though of course small ones remain.

I would be very surprised if no oxygen makes it out of the stomach, when so much is swallowed that it distends the entire insect. Wnt (talk) 17:01, 7 October 2011 (UTC)[reply]

You mention that the lobster and the coconut crab have roughly the same length, but the lobster is more then 4x heavier. I do not see the weights for the prehistoric examples you mention, but the aquatic one looks like it would also have significantly more bulk. Googlemeister (talk) 14:22, 7 October 2011 (UTC)[reply]

Probably, though both Arthropleura and Jaekelopterus were dorsoventrally flattened, unlike coconut crabs and lobsters. Also note that some eurypterids may have been amphibious, some of the first animals to set foot on land, really. They are actually the direct ancestors of modern land scorpions and spiders.-- Obsidi♠n Soul 14:49, 7 October 2011 (UTC)[reply]

Thank you-all for your replies, especially Obsidian Soul. JRSpriggs (talk) 21:31, 7 October 2011 (UTC)[reply]

Collapse of the soft body during molting, or ecdysis provides the critical upper limit in arthropod size. The problem is the regeneration of the entire exoskeleton at once. This does not seem to be mentioned in the article on ecdysis, though. μηδείς (talk) 20:21, 8 October 2011 (UTC)[reply]

Official level of radioactivity in Japan

Where can I find official authoritative radioactivity data of Japan? No problem if it's in Japanese. Wikiweek (talk) 11:26, 7 October 2011 (UTC)[reply]

I would start by reading nuclear organizations in Japan, and decide what specific sort of data you're looking for. Nimur (talk) 16:59, 7 October 2011 (UTC)[reply]

Genetic Population Statistics Question

Where could I find data on what percentage of Filipinos carry a blue allele for OCA2? I once knew a guy with bright blue eyes whose parents were both Filipinos with brown eyes and it struck me as pretty uncommon. His sister had brown eyes like their parents. 20.137.18.53 (talk) 13:38, 7 October 2011 (UTC)[reply]

That's probably just a result of mixed ancestry? Some "full Filipinos" tend to not be fully Austronesian. One of my grandfathers had blue-gray eyes (more gray than blue), and one of his sons (my uncle) had gray-greenish brown eyes. But our ancestry is very mixed. And yes, this is very uncommon. -- Obsidi♠n Soul 15:29, 7 October 2011 (UTC)[reply]

Here is an article that defines a single SNP that is exquisitely linked to eye color. The SNP in question is rs12913832 and in the HapMap samples the Asian populations have essentially 100% A/A (brown allele). Compare to J. Craig Venter who is G/G and has blue eyes. --- Medical geneticist (talk) 20:41, 7 October 2011 (UTC)[reply]

rabbits and radiation

Why are rabbits more resistant to radiation poisoning then humans and most other mammals? Googlemeister (talk) 18:56, 7 October 2011 (UTC)[reply]

What makes you think they are? AndyTheGrump (talk) 19:00, 7 October 2011 (UTC)[reply]

They are better at self-repair. Why? I don't know. See radioresistance. -- kainaw ™ 19:04, 7 October 2011 (UTC)[reply]

Note from the article it's not clear rabbits are really better then other mammals. They may be better then humans and dogs but don't see much better then rats or mice. Even compared to humans they only have a LD₅₀ about double although it's not clear to me how reliable the human figure is nor whether the human (or for that matter rats or mice) value is over 30 days. Nil Einne (talk) 20:05, 7 October 2011 (UTC)[reply]

Alright, I adjusted the question to state that rabbits are only better then most other mammals, but am looking for a reason as why rodents seem to handle it better then canines, primates etc... Googlemeister (talk) 20:29, 7 October 2011 (UTC)[reply]

Why etc? The article doesn't give any figures for cats, or ruminants or ungulates etc. (And it only gives one figure for primates i.e. humans and one figure for canines i.e. dogs so it seems a bit quick to decide all primates and canines are comparable.) Do you have some sources suggesting rodents and rabbits are unusually high among mammals? Nil Einne (talk) 21:01, 7 October 2011 (UTC)[reply]

I'm not comfortable drawing the conclusion that rabbits are significantly more radioresistant than other mammals, despite the contents of the table at radioresistance. It's very easy to inadvertently do an apples-to-oranges comparison when trying to estimate LD50 values for ionizing radiation exposure. For one thing, we don't know that all of the test animals used to generate the table were exposed to the same type of radiation. (Indeed, we're certain that the numbers for humans aren't based on systematic testing; they're mostly derived from dose estimates assembled for people bombed at Hiroshima and Nagasaki.)

The pattern of radiation deposition within the body of an animal will depend strongly on the shape of the animal and the energy of the gamma ray photons used. Bones can 'shadow' the tissue behind them if radiation is incident from one side only. Low-energy photons will be strongly absorbed in the first few centimeters of tissue, shielding the organs below; high-energy photons generate showers of secondary electrons that actually deposit more dose a centimeter or two below the surface than is absorbed at the skin. Even though the whole-body LD50 values as presented imply a uniform dose of radiation to the entire volume of the body, in practice such a dose was almost certainly not what was delivered—simplifying radically, the extremities of a larger animal would be 'hotter', while the core would be 'cooler', and the reported figure just an average dose. The mouse would receive a more uniform dose than the dog. TenOfAllTrades(talk) 21:28, 7 October 2011 (UTC)[reply]

As I understand it, there's a tradeoff between cancer resistance and radiation resistance. Humans, with a 70+-year lifespan, are highly resistant to cancer, but at the cost of being unusually vulnerable to radiation poisoning: even minor DNA damage tends to trigger cell suicide. Rabbits and other short-lived mammals don't have the strong anti-cancer mechanisms humans do, so it takes a larger dose of radiation to cause the large-scale cell die-offs of acute radiation poisoning (but at the same time, they're more likely to get cancer from radiation). --Carnildo (talk) 01:03, 8 October 2011 (UTC)[reply]

2D creatures in a 3D space - could not resist taking it up again

Hi I posted this Q in 2006 - and sorru if it offends anybody also reposted in 2010 - well now I got the inspiration to add to the question instead of reposting : Pasting here

Original 2006 Q link: http://en.wikipedia.org/wiki/Wikipedia:Reference_desk_archive/Mathematics/March_2006#About_2_D_creatures_traveling_on_a_2_D_plane_seen_from_a_3_D_point_of_view

My addition below:

Just googled and found this again - I'm the OP - well now I'm interested in two things. One to see if anybody else but me will notice that I've been here 5 years later, and two : Is there someone out there who can explain in laymans terms why this

“ In this case the maximum factor between the distance of the taxicab traveler and the distance of the Euclidean traveler would be \sqrt{3}, for example going from (0,0,0) to (5,5,5) the taxicab traveler has to move 15 units, as opposed to the Euclidean traveler's 5\sqrt{3} units. But as before, that's a maximum: in going from (0,0,0) to (3,4,12) the taxicab traveler goes 19 units to the Euclidean traveler's 13, but \frac{19}{13}<\sqrt{3}. It's possible to choose a pair of points where the taxicab traveler's distance is exactly \frac{\pi}{2} times as long as the Euclidean traveler's, but that's only because such a pair of points can be found for any number x such that 1 \le x \le \sqrt{3}, which \frac{\pi}{2} happens to satisfy. ”

is right?

As I see it - Going from A to B in 3D is just (always) a straight line. But this C cannot see that so it has to travel from the perspective of the 3D creature in X + Y ie. at the point where it needs to change direction is where the 2D creature actually has traveled the least distance in the X distance (or Y) and then changes direction 90 degrees to be 'the fastest' without changing directions every x/Y mm (or whatever).

So the 3D viewer would, if seeing the 2D creature move along the xy points, percieve the path at some fractional degree from the perfect - Squareroot 2 path - which means the 3D viewer would be able to see the 2D creature's energy expenditure adhere to my postulation.... 2D travelers always expend square root 2 times more energy when the 3D viewer multiply this with hmmm is it cosinus or sinus ? (Perfect angle view that makes a straight line away from the XY plane) That's the hint for ½ PI value. To elaborate : A photon is emitted from somewhere..... this photon is measured at some point and only there you can make the mark that you measured a photon (in whereever space) how it got there (Cab or Helicopter) you don't know - you can just measure the time difference between you emtting the photon - and collecting information about when it hit 'point B' 85.81.121.107 (talk) 19:59, 7 October 2011 (UTC) — Preceding unsigned comment added by 85.81.121.107 (talk)

It looks to me like you are trying to reinvent the basic concepts of ~~Minkowski space~~ Taxicab geometry. -- kainaw ™ 20:08, 7 October 2011 (UTC)[reply]

OP here, it's not the Manhattan distance I'm "looking" for, my original foundation was when I in my teens read all kind of pop physics books - think stuff like schroedingers cat etc. (http://www.benbest.com/science/quantum.html) BTW Fixed Taxi to Helicopter :-)

I am interested in finding out if there is some merit to the idea that 'my' 3D viewer - if it wanted to travel the same distance as the 2D creature - could do so while expending at least squareroot 2 times less energy as the 2D creature - because the 3D creature would ALWAYS travel in the straight AB line not some xyxyxyyyyyxxyyxyy path as the 2D creature had to...perfect conditions would in my mind suggest this squareroot 2 times less - because the fastest 2D creature would only travbel one X and then one Y. Drawing it up on a XY crossed paper this could look like going from 0,0 to 6,4....but the 3D viewer would just 'measure' the AB distance - The view angle of the 3D creature will if it's not perfectly squareroot 2 times the distance be a candidate to a formula where you can figure out the angle where one observer sees the AB distance for the 2d creature as XY = 5,8 but the actual 3D viever just sess this as the perfect 90 degree travel path for the 2D creature.. so the actual distance/time used traveling would be 1,1 times (co)sinus the angle

I wish I could explain it better....and now I introduced a third 'viewer' or maybe planePOV... in the 3d space observing a 2D creature traveling on a fantasy xy crossed paper....angled so it just looks like a straight line to a human....

So there it is .... redshift is all over me now.... ok going trying once more....

Going from A to B is just start and stop - if you were drawing this path on a piece of paper - it's fairly simple.... just point a and then point B draw a line between these to points.

Ok, so now we introduce a (normal) coordinate system showing XY direction - so the paper now has some kind of information on it that can be used to measure stuff - and we draw a triangle with a 90 degree C angled based on where we put our A and B points - so now there is a triangle with at least one 90 degree angle, independent on the orientation of the 'imaginary' papers xy coordinate lines.

Going from A to B would be straight forward if you could just do it...and here it is - the issue...the 2D creature can only move in XY direction of its own 'coordinate system' being imagined as another virtual coordinate system laying at some angle to what a 3D observer perceives (the '3d' xy paper) so the 2d creatures fastest path is one stop and go in its own coordinate sytem. and in this coordinate system where the 2D creature travels exactly half the distance between point A and B before it does a direction shift. And this half (x2 ) equates to square root 2 of the distance that a 3d creature would use to travel that Ab distance seen from the 3d creatures POV.

So 3d creature can use less time/energy if it should travel the same distance as an observed 2d creature... - it doesn't have to stop and change direction when half the distance is reached...because it just arrives :-)

http://upload.wikimedia.org/wikipedia/commons/thumb/6/69/Coord_system_CA_0.svg/240px-Coord_system_CA_0.svg.png

Picture link inserted above - The 2D creature only perceives the x and y directions so it cannot travel the direct route throug Z (as observed by the 3D viewer) - btw that picture does not satisfy my idea of the 90 degree rule - = half the distance in X direction and the other half in the Y direction - and now that I think about it - the 2d creature would have no good idea about the Z axis.... which we could imagine to be time :-D

Damn forgot that it also follows that you can then for the 2d creature travel 2 paths.... which have the exact same lenght -> through the 90 degree triangle on the x plane or 90 degree on the y plane... That was my hint for the 3d things going through the 4D stuff and double slith experiment...

85.81.121.107 (talk) 20:58, 7 October 2011 (UTC)[reply]

You can think of it in terms of vectors: the "Euclidean" 3-D creature can travel in a direct straight-line distance between all the vectors, while the 2-D creature must travel along all three vectors to get to the same location. ~AH1 ^(discuss!) 22:33, 7 October 2011 (UTC)[reply]

OP here - Found an interesting link http://blazelabs.com/f-p-hds.asp helping to visualize seeing the 'world' from a lower/higher dimensional 'space' and the implications. 85.81.121.107 (talk) 06:54, 8 October 2011 (UTC)[reply]

I'm sorry, I don't understand. What is your question? The Euclidean distance in 3D is

{\sqrt {x^{2}+y^{2}+z^{2}}}

and the Taxicab distance is

x+y+z

. What more do you want to know? --Tango (talk) 12:49, 8 October 2011 (UTC)[reply]

are stars visable from space.

From the surface of earth a vast quantity of stars are visable with the human eye. 1 if you could move outside the earths exosphere would the same quantity of stars be visable with the human eye. 2 if you could move outside the solar system would the same quantity of stars be visable with the human eye. Cubedmass 21:08, 7 October 2011 (UTC) — Preceding unsigned comment added by Cubedmass (talk • contribs)

1) You would see more, because the earth's atmosphere ruins our view of some stars. That's the reason why the Hubble Space Telescope and devices like it are outside of the atmosphere.

2) You would likely see more stars, as the nearby sun causes some light pollution which interferes with seeing some dim stars. --Jayron32 21:14, 7 October 2011 (UTC)[reply]

I recall that around 6000 were visible from Earth. How many could you see where no one can hear you scream? Clarityfiend (talk) 22:15, 7 October 2011 (UTC)[reply]

During the daytime on Earth, diffraction of light in the sky causes stars other than the Sun to become invisible. Even when the Sun is visible in space, the stars can still be seen, and the same is true for the Moon as well. ~AH1 ^(discuss!) 22:20, 7 October 2011 (UTC)[reply]

It's not diffraction, but Rayleigh scattering. --Carnildo (talk) 01:06, 8 October 2011 (UTC)[reply]

Star_catalogue#Full-sky_catalogues suggests that it really depends on what kind of technology you're using, but with good enough terrestrial telescopes you can observe around a billion stars (!!) (e.g. the Guide Star Catalog). --Mr.98 (talk) 01:35, 8 October 2011 (UTC)[reply]

1) Definitely would see more in space.

2) I don't think light pollution is a significant problem, as long as you are in a shadow, like on the dark side of the Moon when it's not positioned to receive Earthshine. Therefore, I wouldn't expect leaving the solar system to make much difference versus that. (The one exception would be stars near our Sun from your POV. Obviously the farther away you are, the fewer stars our Sun will block, but then again, just moving around within our solar system also fixes that.) StuRat (talk) 22:23, 7 October 2011 (UTC)[reply]

Billions and billions. μηδείς (talk) 22:31, 7 October 2011 (UTC)[reply]

Light pollution often reduces limiting magnitude by 1 or 2 apparent magnitudes in suburban areas, and up to 5 in city centres. Light pollution-free zones usually get to 6th magnitude visibility for those who have average eyesight. ~AH1 ^(discuss!) 23:11, 7 October 2011 (UTC)[reply]

Right. In my answer to part 2 I was talking about light pollution from the Sun, which isn't a problem everywhere in the solar system. Light pollution is more of a problem on Earth, though, due to the large number of light sources and the atmosphere which reflects and refracts that light back into out eyes. StuRat (talk) 00:05, 8 October 2011 (UTC)[reply]

Light pollution is a term with a specific meaning; it doesn't apply to natural light sources. You're going to confuse readers if you misuse it.

Even beyond the Earth's atmosphere, however, there are effects of sunlight scattering from dust particles in the solar system. (These effects are visible even to naked-eye observers on Earth under very dark-sky conditions as zodiacal light and gegenschein.) TenOfAllTrades(talk) 02:47, 8 October 2011 (UTC)[reply]

Bortle class 0? The best we can get on Earth is Class 1:

Class 1: Excellent dark-sky site. The zodiacal light, gegenschein, and zodiacal band (S&T: October 2000, page 116) are all visible — the zodiacal light to a striking degree, and the zodiacal band spanning the entire sky. Even with direct vision, the galaxy M33 is an obvious naked-eye object. The Scorpius and Sagittarius region of the Milky Way casts obvious diffuse shadows on the ground. To the unaided eye the limiting magnitude is 7.6 to 8.0 (with effort); the presence of Jupiter or Venus in the sky seems to degrade dark adaptation. Airglow (a very faint, naturally occurring glow most evident within about 15° of the horizon) is readily apparent. With a 32-centimeter (12½-inch) scope, stars to magnitude 17.5 can be detected with effort, while a 50-cm (20-inch) instrument used with moderate magnification will reach 19th magnitude. If you are observing on a grass-covered field bordered by trees, your telescope, companions, and vehicle are almost totally invisible. This is an observer's Nirvana!

Count Iblis (talk) 04:25, 8 October 2011 (UTC)[reply]

1) You would not see significantly more. On a clear night the atmosphere blocks about 1% of visible light. The Hubble Space Telescope is not in space to get that 1% back; it is there for two reasons: to avoid atmospheric diffraction (does not greatly reduce the number of stars you see, merely makes them twinkle), and to see in non-visible-light frequencies - as mentioned in the HST article. The lost 1% of light can be much more cheaply compensated for by increasing photograph exposure time by 1%.

2) Very very little more. The Sun affects your viewing only if you are looking directly at it. Since there is no atmosphere, there is very little to scatter light from the Sun when looking in any other direction. However, there are places in the galaxy where you would see more stars, such as the galactic center if you are willing to travel that far. 88.112.59.31 (talk) 08:50, 8 October 2011 (UTC)[reply]

1) I can see why the atmospheric diffraction wouldn't be an issue with bright stars, but it does seem likely to make a dim star on the edge of perception no longer consistently visible. StuRat (talk) 17:31, 8 October 2011 (UTC)[reply]

Petroleum Refinary

There is a fire seen above petroleum refinaries in maxmum photographs of a refinary.can any one help me with what exactly is that fire called .is there any use of that fire?why do we waste energy like that?couldnt that fire be put off? thanks in advance 117.230.71.121 (talk) —Preceding undated comment added 22:18, 7 October 2011 (UTC).[reply]

That's intentional, to burn off natural gas which accompanies petroleum. I agree that it's wasteful. Apparently they think the cost of capturing it and selling it is more than they would get. It's burnt because most of the combustion products will be water and carbon dioxide, which are less dangerous in the atmosphere than natural gas. See gas flare. You'd think that if they are going to have a continuous flame like that, that they could at least recover some energy from it to run the refinery. StuRat (talk) 22:26, 7 October 2011 (UTC)[reply]

Actually, they mostly do [disclosure: some years ago I worked at this refinery, albeit at a desk job].

The gas burning at the top of refinery flare stacks (as opposed to those sometimes seen at well heads, particularly offshore oil rigs) is not merely the discarded natural gas component of crude oil: such gas is usually also collected and processed at refineries, as can be seen in this diagram, but this and the other distillation and treatment processes going on collectively produce a surplus of mixed inflammable gases, which where I worked were (if I recall correctly) called "refinery gas."

The bulk of this refinery gas is indeed used to generate on-site power, often by burning it to heat water to produce steam which is then piped around the refinery so that its contained heat can be used (and some steam is used directly in some of the chemical processes). However, the distillation and other processes going on in all those distillation towers and other bits of plant are quite sensitive to slight fluctuations in temperature, pressure, and the precise chemical makeup of the crude, so the volume of refinery gas being produced over the whole refinery is subject to unpreditable fluctuations and occasional major "surges." When those occur the sudden large excess of gas cannot be captured (think of an accidentally shaken bottle of beer), and has to be burnt off through the flare stacks.

The quiet, flickering flames normally seen on flare stacks are merely the pilot lights of the stacks. If a surge occurs, the resultant flames are powerful, large and loud (they may be heard for miles) - it's much like the difference between the flickering flame and the roaring flame on a Bunsen burner, except that the 'roaring flame' on a stack is much taller, not smaller, than the 'flickering flame' because it's caused by (much) greater gas volume, not merely by admitting more air to the same volume.

Such surges (at least at the refinery I worked at) are relatively rare, so that the resultant burn-offs can cause some alarm to those in the vicinity unaccustomed to them: they are, however, quite harmless and are indeed a designed safety feature of the system. {The poster formerly known as 87.81.230.195} 90.197.66.142 (talk) 23:50, 7 October 2011 (UTC)[reply]

I would like to see some sources about the quantities involved. I spoke with someone today who said that the plant where he worked had to flare off more than 2.5 million cubic feet of refinery gas per day above what they were using to power the plant because even when there was a market for it, which was rare, logistical problems were involved with transmitting it by pipeline. 70.91.171.54 (talk) 01:14, 8 October 2011 (UTC)[reply]

I have to wonder if they had to pay the cost of all that pollution, say via carbon credits, if the economics would then favor finding a way to capture it. StuRat (talk) 17:21, 8 October 2011 (UTC)[reply]

Simply remove the fossil fuel subsidies. Dualus (talk) 03:30, 9 October 2011 (UTC)[reply]

Well, as I understand it, under carbon credits, the refineries would start off with free carbon credits for all their current activities - if they then make one of these natural gas burning stacks, they are transforming methane, a more potent greenhouse gas, into carbon dioxide, which is worth a few extra credits, which they can sell to homeowners who want to purchase the privilege of running a fireplace, a precious keepsake to pass down to future generations... Wnt (talk) 13:38, 10 October 2011 (UTC)[reply]

That doesn't seem right. That would mean anyone could get extra credit for leaving the gas tap open with a fire lit above it all the time. Eternal flames at cemetaries should collect lots of credit, then. StuRat (talk) 00:14, 11 October 2011 (UTC) [reply]

Only if their previous releases of methane were "grandfathered" into the system with an issuance of free carbon credits - which is not an option for just anyone. Wnt (talk) 19:56, 11 October 2011 (UTC)[reply]