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January 28

Artemisia absinthium

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.

I'm not going to argue this one, because my feeling was that we couldn't have answered it anyway. For most uncommon herbal remedies too little research has been done to tell whether the herb itself is safe or effective, let alone in combination with specific pharmaceuticals. Wnt (talk) 06:27, 28 January 2011 (UTC)[reply]

Chemistry and Calculus

Hello. I am in intermediate chemistry at my high school right now. Unlike most of my peers, however, I have already got a solid grasp of calculus. What is a good book for understanding chemistry with the involved calculus (for example, in entropy or enthalpy)? I would like a book to cover chemistry from the very basics (down to Lewis theory even) up because I am only in intermediate chemistry right now. I have had some physics but also not from a calculus perspective. Thanks. 24.92.70.160 (talk) 02:35, 28 January 2011 (UTC)[reply]

Sounds like what you need is a thermodynamics textbook. The one I used in college for chemical engineering was by Smith, Van Ness, and Abbott. --shoy (reactions) 03:52, 28 January 2011 (UTC)[reply]

Any college-level physical chemistry book should be chock-full-o-calculus. Find the nearest used college book store, or look online for a used book; you can get one on the cheap. --Jayron32 04:16, 28 January 2011 (UTC)[reply]

Hadron collisions

Would the collisions be the same (or show the same information)if, instead of TWO atoms colliding AT each other linearly, there were four or eight atoms colliding in the same symmetrical way? —Preceding unsigned comment added by 98.221.254.154 (talk) 05:30, 28 January 2011 (UTC)[reply]

In theory it would be the same, as long as you had balanced collisions so there was no leftover momentum. But in practice the messier the collision the harder it is to extract good data from the experiment. Plus it would be nearly impossible to get the timing exactly right. Ariel. (talk) 06:58, 28 January 2011 (UTC)[reply]

Does having one linear collision as opposed to a "multi-atomic implosion" also limit the data you can gather? —Preceding unsigned comment added by 165.212.189.187 (talk) 14:41, 28 January 2011 (UTC)[reply]

I'm not sure I understand the difference between this and the original question. Essentially, the LHC is all about high energy and high luminosity (collision rate). They collect insane amounts of very noisy data and plug it into subtle statistical models to search for new phenomena. The type of collision doesn't matter too much, except that (as Ariel said) the more complicated it is, the noisier the data you collect. Proton-proton collisions are already complicated and a "multi-atomic implosion" would be worse. Also, in practice it's impossible to arrange collisions of three or more particles. The LHC doesn't aim particles at each other; it just crosses two particle beams and lets them collide as they may. If you crossed three or more beams, the overwhelming majority of collisions would still involve only two particles. -- BenRG (talk) 04:59, 30 January 2011 (UTC)[reply]

Firing a gun underwater

What factors determine whether a gun (pistol, assault rifle, whatever) can be successfully fired underwater, or will simply explode as soon as the trigger is pulled? --Kurt Shaped Box (talk) 10:04, 28 January 2011 (UTC)[reply]

Mythbusters episode 51 tested four modern firearms. "The entire gun had to be completely submerged in water—all pockets of air must be removed—in order to prevent a possible explosion when fired." No gulls were harmed in the testing of this myth. Clarityfiend (talk) 10:22, 28 January 2011 (UTC)[reply]

I have been told that a firearm will explode (burst the barrel or the breach, with danger to the shooter) if anything but air is in the barrel: a bullet which did not leave the barrel due to poor discharge of powder, snow, mud, someone's finger. Mythbusters "disproved" this for a shotgun they tested, but rifles and pistols are not modern shotguns, and older weapons may not have been made to the same standard. WW2 veterans have told me that keeping barrels clean was highly emphasized, to prevent the barrel bursting. Water is so much denser than air that when a bullet traveling 1000 feet per second strikes it it would seem to be as immoveable as concrete. So it is surprising that the bullet somehow instantly pushed the water out of the way without a pressure buildup which bursts the barrel, if the other plugs mentioned would cause the barrel to burst on some firearms. Rifles and pistils sometimes blow up for no known reason, so water filling the barrel would push a marginal barrel over the edge. Edison (talk) 17:07, 28 January 2011 (UTC)[reply]

From looking on YouTube (of course there are going to be videos of guys firing guns underwater on there!), various Glocks seem to manage just fine, as does the AR-15 and the Ruger LCP - though the bullet's range is something like 4 feet, or less. The Mythbusters ep (also on YouTube) proved that being shot underwater at this range would probably still be lethal (I guess that you'd have to go for a contact shot, or close to it to be certain). Heh, wonder if SpecOps ever do this? --Kurt Shaped Box (talk) 17:35, 28 January 2011 (UTC)[reply]

Naturally there is an article APS_amphibious_rifle --80.176.225.249 (talk) 18:27, 28 January 2011 (UTC)[reply]

It is perfectly possible (if not especially advisable) to fire a well-made gun underwater without an explosion, as is stated above. Firing a gun that is only PARTIALLY filled with water might be more of an issue. On the other hand, even the mythbusters tests didn't use frangible rounds: many (if not most) civillian bullets are meant to expand when they hit flesh. Flesh being very similar in density to water, I'd suspect a hollowpoint or JHP round might start to mushroom in the barrel underwater. 65.29.47.55 (talk) 00:55, 29 January 2011 (UTC)[reply]

When a gun or rifle of any kind is fired the propellant burns completely (or almost completely) before the projectile leaves the barrel. While the projectile is still in the barrel the barrel is being subjected to maximum pressure available from burning the propellant, and yet the barrel doesn't explode. Clearly, barrels are sufficiently strong to withstand maximum pressure available from burning the propellant. Inserting water, mud or any other solid object into the barrel does not somehow increase the energy available from burning the propellant so there is no reason to expect that doing so will cause the barrel to explode. Dolphin (t) 05:33, 30 January 2011 (UTC)[reply]

Full name for invertebrate zoologist "von Linstow" active late C19, early C20?

Resolved

Hi all,
Have I mentioned that my pet peeve is binomial authorities without corresponding biographies? OK. So, "von Linstow" is the binomial authority for Telosentis exiguus. He also appears to be the binomial authority for nematodes hosted in "Salties" Crocodylus johnstoni and Crocodylus porosus.
All that said, who is this "von Linstow" guy? --Shirt58 (talk) 11:57, 28 January 2011 (UTC)[reply]

According to this book, his initials are OFB. Hope it helps. --Albval (talk) 12:03, 28 January 2011 (UTC)[reply]

And after some more google-searching the guy should be this Otto Friedrich Bernhard Von Linstow (page in Finnish) --Albval (talk) 12:09, 28 January 2011 (UTC)[reply]

The above link is to a Finish book shop, for ISBN 9781103982332. Both the Finish bookshop and Amazon indicate that the book is in English, however, the title and front-cover are in a Germanic language. CS Miller (talk) 12:57, 28 January 2011 (UTC)[reply]

The book is about helminths, which would agree with Shirt58's original request. CS Miller (talk) 13:02, 28 January 2011 (UTC)[reply]

(EC)This gives his dates and the dates of publication of his main work Compendium Der Helminthologie: Ein Verzeichniss Der Bekannten Helminthen, Die Frei Oder in Thierisch. Mikenorton (talk) 13:06, 28 January 2011 (UTC)[reply]

His book is on Archive.org, oddly enough. He is listed as a "Stabsarzt," which I believe is an Army medical officer. Seems like an obscure guy who happened to write one of these obscure compendiums that named a few new species? --Mr.98 (talk) 13:43, 28 January 2011 (UTC)[reply]

I think so too; he was probably an army doctor with an interest in helminths. Here are some more publications of his; he wrote more than that one book. Ucucha 14:15, 28 January 2011 (UTC)[reply]

Wikispecies have what looks to be a very comprehensive list of taxon authorities and a tiny stub on him. (I wish I had known about this last time you asked!) SmartSE (talk) 18:51, 28 January 2011 (UTC)[reply]

A biography says that he was born 17. October 1842 in Itzehoe received his medical phD 1864 in Kiel and worked as military doctor in Hameln later in Göttingen he published his book Compendium der Helminthology in 1878 in Hannover and died 3. May 1816 in Göttingen.--Stone (talk) 21:16, 28 January 2011 (UTC)[reply]

Oh we have an article Otto Friedrich Bernhard Von Linstow.--Stone (talk) 22:06, 28 January 2011 (UTC)[reply]

Resolved. Now, if we can turn our attention to the species H. finlandicus, endemic to Albval's Finland... :-) --Shirt58 (talk) 12:19, 29 January 2011 (UTC)[reply]

Fuel efficiency

I know that natural gas and propane both burn cleaner than gasoline but are they as "fuel efficient" as gasoline (i.e. all things being equal, will one get the same "MPG" with NG and Propane as one would with gasoline)? 74.198.17.84 (talk) 14:34, 28 January 2011 (UTC)[reply]

There is no way for all things to be equal. Which is why we have MPGe. Rmhermen (talk) 15:41, 28 January 2011 (UTC)[reply]

The most similar fuel commonly used to power cars is Liquefied Petroleum Gas, which contains mostly butane or propane: efficiency is around 25% worse with LPG compared to gasoline.[1] --Colapeninsula (talk) 15:56, 28 January 2011 (UTC)[reply]

If we're strictly talking propane and natural gas, and comparing "miles per gallon" (really, "energy per volume"), then the comparison is a no-contest win for Gasoline. Have a look at energy density - there are many ways to measure energy "per ___": energy per mass, per volume, per dollar, and so on. This chart of energy density in materials can be sorted by mass- and volume- density. Nimur (talk) 18:14, 28 January 2011 (UTC)[reply]

LPG is as the poster above says around 25% less efficient, but in the UK it's compensated for (at least financially) by being about half the price of petrol (gasoline) so you still come out ahead in terms of cost per mile. Downside of course is the installation cost and the loss of space for a spare wheel or in the boot. Exxolon (talk) 16:42, 29 January 2011 (UTC)[reply]

Can vets determine the cause of death in a hamster?

Topic says it all. ScienceApe (talk) 15:27, 28 January 2011 (UTC)[reply]

In principle, a veterinarian can conduct or order virtually all of the same diagnostic tests during a hamster necrospy that a pathologist would perform as part of a human autopsy. With a certain number of caveats, there is a lot of common ground among most mammals from the level of biochemistry right up to tissue structure.

In practice, there are limitations caused by the smaller size of the animal (it is more difficult to examine and conclusively describe fractures of the smaller bones, less blood and other tissue is available for destructive tests, etc.); the vet may have less access to certain items of specialized equipment, and the resources available for conducting animal necropsies may be limited; and the body of literature and reference materials for diagnosing cause of death may be more limited in animal forensics than it is for humans. As with humans, some causes of death in animals are easier to conclusively diagnose than others. TenOfAllTrades(talk) 15:53, 28 January 2011 (UTC)[reply]

Pragmatically, hamsters are fairly disposable as pets. At the outside, they live maybe 3 years. A hamster that lived into his fifth year would be positively Methusalah-like on a Hamster scale. While that does nothing to lessen the emotional loss for someone that suffers the loss of a pet, it does explain why, even if technically feasible, it probably doesn't happen with any regularity that one would autopsy a dead pet which, if it was older than a year and a half, was already beating the average... --Jayron32 17:51, 28 January 2011 (UTC)[reply]

In vaguely the same ballpark, my dad had his dead rabbit necropsied. He was very close to her (nursed her back from cancer and an op that involved basically lifting all her innards out, cutting out/off the bad bits, then putting them back in - she lived for another three years after) and he wanted to know what had happened. He just asked the vet to do it. Heart attack, as it turned out. --Kurt Shaped Box (talk) 18:01, 28 January 2011 (UTC)[reply]

Yes, they can, if you want to pay a lot of money. Small mammals have been dissected and studied micropically for a long time. Robert Koch in the 1870's would give mice anthrax, then dissect them and examine the liver, lungs, etc microscopically to determine the cause of death. Today medical researchers use mice, hamsters, guinea pigs, etc to study all sorts of infections, cancer, and other ailments, and routinely dissect them postmortem and study sections microscopically and in comparable ways to what a human autopsy involves. It is just not done that often for dead pets, due to the cost. Edison (talk) 21:08, 28 January 2011 (UTC)[reply]

Though if you're a hamster breeder, or have a large collection of fancy hamsters, it may be worthwhile to order (and pay for) a necropsy if your beasties start dropping dead inexplicably at a young age. --Kurt Shaped Box (talk) 00:40, 29 January 2011 (UTC)[reply]

If the gas pressure decreases what happens with temperature?

What happens with the temperature when the pressure of a gas decreases? I'd like to know this for my job, but it doesn't make sense for me. For example if we have steam at 1.3 MPa and 192 Celsius (just above saturation), according to WolframAlpha, its enthalpy is 2788 kJ/kg (http://www.wolframalpha.com/input/?i=enthalpy+water+at+1.3+MPa+and+192+Celsius). If the pressure would decrease, and I assume the temperature is constant, to 1.0 MPa its enthalpy would be 2809 kJ/kg (http://www.wolframalpha.com/input/?i=enthalpy+water+at+1.0+MPa+and+192+Celsius). So the energy increases which seems strange to me. I thought the energy would decrease when the gas loses pressure. Will the gas temperature actually drop for the energy to be constant or decrease? In my example it is steam in a power plant that passes through a valve and there the pressure drops. Assuming no loss of heat to the surroundings. Wikifantast (talk) 15:36, 28 January 2011 (UTC)[reply]

See Ideal gas law. Looie496 (talk) 18:44, 28 January 2011 (UTC)[reply]

An assumption is wrong; unless the valve does work on the steam, the energy in the steam can't increase. So, either the steam is no longer at 192 celsius, or work was performed on it (in the form of heat transfer, or some other method). If exactly zero energy was transferred, and there was absolutely no change in volume, the temperature will change (per the pressure-temperature law; with that temperature, you should find an exactly equal enthalpy on both sides of the valve. In reality, energy should be lost: work was performed by the steam as it passes through the valve: some percentage of steam cavitated, liquified, condensed, changed volume, and lost heat to the walls of the valve and pipe. So energy was lost through that process. We should expect a temperature change to account for both the adiabatic- (ideal gas law) and non-conservative work (energy loss to surroundings). Why do you believe the temperature would be constant on both sides of the valve? Have you instrumented the steam-temperature on both sides? Nimur (talk) 19:27, 28 January 2011 (UTC)[reply]

Thank you. I think then that both the volume and temperature will change too. At 184 Celsius Wolfram Alpha says the energy is the same[2]. I have no data of the temperature on both sides. But intuitively it seemed to me that the temperature could stay constant although the pressure dropped through the valve; pV/T could still be constant if the volume increased (density decreased). Wolfram Alpha also calculates the density.

In reality of course there will always be heat loss to the ambient (non-conservative work) but I guess you can assume it is zero. Wikifantast (talk) 21:37, 28 January 2011 (UTC)[reply]

please explain binary stars in simpler language than the article

I am trying to understand how binary star systems work with planets. 1. how close are the two binary stars? would it be like the sun and jupiter, or the sun and pluto, or the sun and alpha centori? 2. would the planets orbit both stars or just one? 3. in stars wars on luke's planet there are two suns in the sky of the same size not too far from each other. does this mean the stars are very close or one is a lot bigger than the other? would luke's planet be orbiting one or both of them? 4. if you are on a planet orbiting just one of the stars, and the other one is far away, does it look like another star in the sky? is there a way you could tell it is your binary sun? for example, i know you can tell other planets from stars because they sometimes move backwards in the sky. would the sun you are not orbiting do the same? Many thanks. —Preceding unsigned comment added by 74.14.13.241 (talk) 15:43, 28 January 2011 (UTC)[reply]

Let's simplify the problem first. 1) Any two objects orbit the Barycenter of the two objects. The barycenter is the center of mass of the two object system. For the earth-sun system, the barycenter is NOT the center of the sun. It is inside the sun, but off-center by a small amount. So it is technically incorrect to say that any one object orbits another; the two objects oribit around a point on a line between their individual centers of mass, which is located relative to their relative masses. The effect is really obvious when the objects are close enough in mass so that the barycenter lies outside of either object; this happens in the Pluto - Charon system. This works exactly the same regardless of the composition of the two objects; so two chunks of rock (like Pluto-Charon or Earth-Moon), a planet-star system (like Earth-Sun) or a binary star system. The way the two objects move is the same. 2) The major problem with adding a planet to the system is that you create an n-body problem, which is not stable; that is there is no way to predict the relative motion of all three bodies in the long term, even given their initial locations and velocities. --Jayron32 16:14, 28 January 2011 (UTC)[reply]

(e/c) Binary systems vary a lot in terms of the size of stars and the distance between them. This page discusses a stable system similar to that around Tatooine where stars are on average 23 AU apart, which is slightly more than the distance from Earth to Uranus.

The Wikipedia page on binary stars says it is possible for a planet to either orbit a single sun or to orbit both, depending on how close the stars and planet are. The example I mention above assumes the planet orbits only one star, at a distance of less than 3 AU (a bit more than the distance between the sun and Mars). Orbitting a single star would probably be a bit more hospitable for human life.

The distance from the planet to the farther star would be far less than between the sun and the next nearest star (our sun is 250,000 AU from the next-nearest star), so assuming the farther-away star was a similar size to the sun, it would look much brighter than any star but much smaller and dimmer than our sun. --Colapeninsula (talk) 16:23, 28 January 2011 (UTC)[reply]

Thank you and thank you. that helps to visulize it. Jayron do you mean that a planet like tattoone is not possible because it would keep moving a different distance from the suns and get too hot or cold for life? Because colapeninsula says it can be stable? That is interesting you would always be able to tell you are in a binary system because there are two things brighter than the stars. --74.14.13.241 —Preceding unsigned comment added by 74.14.13.241 (talk) 17:39, 28 January 2011 (UTC)[reply]

You can carefully construct 3-body problems like Tatooine so that they can be stable over fairly long periods; so yes, it is technically possible to do so. It would be rare, but of course not impossible, that such an arrangement could arise naturally. --Jayron32 17:47, 28 January 2011 (UTC)[reply]

Sorry, I just read colapeninsula's link that says the planet like tattoine would have billions of years before it moved enough to make it unhabatable. thanks again for the help. (talk) —Preceding undated comment added 17:43, 28 January 2011 (UTC).[reply]

Can we keep Tatooine from getting too close to a star by putting it in a Lagrangian point? I guess if you want to see both suns in the sky at the same time it would have to be points L4 or L5, unless it's possible to orbit around the L2 point (the article implies that it's possible). —Preceding unsigned comment added by 205.193.96.10 (talk) 21:03, 28 January 2011 (UTC)[reply]

The "orbits" around L1-L3 are not completely stable. They require small amounts of station keeping, which is easy for a spacecraft like the Solar and Heliospheric Observatory, but impossible for a planet. The other problem is the relative masses of the objects. L4 and L5 are only stable if the larger of the two objects (in this case, the two stars) is 25 times more massive than the smaller. That can happen if you have a binary system of a blue supergiant and a red dwarf, but you aren't going to get two sun-like stars with that kind of difference in mass. The other problem is that the object at the L4 or L5 point has to have a very small mass - small enough not to have any noticeable effect on the other two bodies (the stars). I'm not sure quite sure how small the body must be, but I suspect a planet would be much too large, especially since one of the stars is, because of the 25 ratio thing, required to be very small. --Tango (talk) 20:56, 29 January 2011 (UTC)[reply]

You might find HD 188753 interesting. Astronaut (talk) 16:23, 29 January 2011 (UTC)[reply]

ear piercing

When you pierce your ears is anything dying or is it just being pushe to a new location?Accdude92 (talk) 16:35, 28 January 2011 (UTC)[reply]

Surely at least some cells are ruptured by the act of piercing, though that article doesn't discuss piercing trauma very much at all. Comet Tuttle (talk) 17:46, 28 January 2011 (UTC)[reply]

Agree. In addition to the many types of skin and subcutaneous tissue cells that would be directly crushed during piercing, trauma to blood vessels will result in some bleeding, and the cells leaking from the laceration will die quickly. -- Scray (talk) 19:25, 28 January 2011 (UTC)[reply]

Hmmm. Why is it that ear piercings done with a piercing gun don't tend to bleed (or if they do, in such a tiny volume as to be unnoticeable)? This is my experience, as well as that of others I've discussed it with. --Kurt Shaped Box (talk) 19:29, 28 January 2011 (UTC)[reply]

I would guess that they are designed to apply pressure on both sides of the puncture, in a standardized way. The challenge has been to do so with an instrument that can be cleaned properly between ~~victims~~customers. -- Scray (talk) 23:50, 28 January 2011 (UTC)[reply]

Personally, I'd recommend that anyone who fancies a piercing goes to an *experienced*, *competent* (shop around, ask questions - they're pros, they don't mind) piercer who uses a needle and an autoclave. I know it's a highly politicized subject (some piercers basically consider those guns to be the devil's spunk and seem genuinely furious that they exist) - but I'm simply stating this from my own personal experience with ear and eyebrow piercings (been considering a bridge piercing on and off for a couple of years) - and the healing thereof. Plus, you're not restricted to a stud for your first piece of metal. --Kurt Shaped Box (talk) 00:58, 29 January 2011 (UTC)[reply]

Also, is the a time when closing of the piercing is impossible?Accdude92 (talk) 17:26, 31 January 2011 (UTC)[reply]

Physics, conservation of linear momentum-collisions

2 bodies make elastic head-on collision on smooth horizontal table kept in car. Do u expect change in result, if the car is accelerated on a horizontal road because of the non-inertial character of the frame? Does the equation " Velocity of separation = velocity of approach" remains valid in accelerating car? Does the equation " Final momentum = initial momentum " remain valid in the accelerating car?

.....with proper reasons.

—Preceding unsigned comment added by 122.169.145.208 (talk) 17:20, 28 January 2011 (UTC)[reply]

Please do your own homework.

Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. DMacks (talk) 17:29, 28 January 2011 (UTC)[reply]

No this isn't a home work question! I saw this question while I was going through the book "Concepts of physics 1" by H.C.Verma. I know that in inertial frame, momentum and kinetic energy are very much conserved in this case but I couldn't decide myself whether momentum is conserved or not. I think its not conserved as pseudo force acts. But just I need clarification whether my thinking is correct or not. I am not confident about my answer, so I put this question here, but not for the sake of completing the homework! Hope you answer this ....

Yes, you are right, the pseudo-force acts, but the collision takes a very short amount of time to happen and since the change in momentum is given by dP = F dt, unless the pseudo-force is very strong (very high acceleration), the change in momentum is likely small enough that it can be considered negligible in which case all the usual rules of thumb for solving collision problems still apply. 71.101.41.73 (talk) 16:30, 30 January 2011 (UTC)[reply]

Aside the obvious answer

How can I find out if a sheet of glass is toughened (without actually breaking it)? --BozMo talk 17:31, 28 January 2011 (UTC)[reply]

A polarizing filter will help you see if there are strains in the glass. Here's a reference, The Identification of Toughened Glass using Polarized Light (1973). You can also use a web-search for comparative images: here's a sample Google query. Nimur (talk) 18:00, 28 January 2011 (UTC)[reply]

(EC) As per Toughened glass#Properties, the strain pattern resulting from tempering can be observed with polarized light or by using a pair of polarizing sun glasses. Red Act (talk) 18:05, 28 January 2011 (UTC)[reply]

In case anyone doesn't have access to the article I linked, here's a summary of the "methods" section: place the glass-under-test between two crossed polarizer filters, and illuminate from behind with white light. Rotate the polarizer filters. If you see an iridescent color-behavior that has a periodicity with each 90-degree rotation of the polarizer-filter, you have toughened glass; if you simply see a variation in light intensity, you have annealed glass. A few caveats exist, related to whether your glass has any fracture surfaces, that are described in detail in the paper I just linked and elsewhere on the web. Nimur (talk) 18:08, 28 January 2011 (UTC)[reply]

I'd consider the obvious answer to be to contact the manufacturer. I doubt anyone would sell toughened glass without advertising it as such. What are the circumstances you're trying to figure it out under? — DanielLC 03:03, 30 January 2011 (UTC)[reply]

Obviously he's asking out of doing research for a breaking and entering robbery job. In this case, calling the building's owner and ask whether he had toughened windows installed is not exactly an option... 62.166.201.27 (talk) 13:31, 31 January 2011 (UTC)[reply]

A hydrogen bomb in an undersea volcano= add new land?

If a fusion bomb was placed in an undersea volcano, would it induce an eruption so massive that this volcano will spew out enough lava to permanently break the sea level and become an island?

Surtsey happened naturally, and Japan and Taiwan is in an epic land crunch for real estate so this would be an instantaneous way to reclaim more real-estate from the sea, wouldn't it?

By the way, I think the results might be better if the bomb was made to explode upside-down. Instead of a mushroom cloud, how about a mushroom laccolith under the volcano's base? Therefore, how can a bomb be made to explode a mushroom cloud into the earth instead of up in the sky? --129.130.32.220 (talk) 20:07, 28 January 2011 (UTC)[reply]

I'll add that to the list of crazy ideas I've seen on those pages over the years. BTW, there is no such thing as an upside down explosion. The mushroom goes up because hot air rises. 71.101.41.73 (talk) 20:37, 28 January 2011 (UTC)[reply]

Yes, the idea of an upside-down explosion is just confused. The explosion of a nuclear blast is just a sphere of fire and blast. It rises because it is hot.

You could shape the blast, probably, to fire more energy in one direction than another, though. These sorts of schemes were considered in developing Project Orion. It strikes me as probably doable, though whether that would hep or not, I don't know.

As for the geology of it, I have no clue. There were worries in the 1980s that sufficiently large underground nuclear tests could induce seismic and possibly volcanic activity (see, e.g. Amchitka). But this didn't happen. But it wasn't meant to happen, either — it's not clear to me that you couldn't do it on purpose. It should be noted, though, that Japan is probably the last place on Earth that would ever embrace Plowshares-like projects. Their disinclination to mess with nukes goes pretty deep!

There is nothing on this topic in The constructive uses of nuclear explosives (1968), which is sort of a catch-all book of proposals for "fun" and "peaceful" things you could do with nukes. --Mr.98 (talk) 20:55, 28 January 2011 (UTC)[reply]

How could one prevent the bomb from melting or being disrupted by the heat while placing it "in a volcano?" Place it in a shell of Unobtanium? Edison (talk) 21:01, 28 January 2011 (UTC)[reply]

Heh, heh, Edison, Tungsten has a higher melting point than the temperature of lava, IIRC. We just have to get it thrown in there deep enough (by a tungsten drill, maybe?) and have it detonate by a timer or a switch. --70.179.181.251 (talk) 06:03, 29 January 2011 (UTC)[reply]

Presumably you wouldn't put it in the lava itself, but near it. E.g. to the right of the label "3" in this diagram, or something like that. But I am not a geologist. --Mr.98 (talk) 21:05, 28 January 2011 (UTC)[reply]

For perspective, consider reading about human-induced seismicity. Most induced seismic response is the result of many years of continuous changes in the overburden or fluid-pressure, due to extraction of solid or liquid material during mining, water-well extraction, or fossil-fuel extraction. Nimur (talk) 21:04, 28 January 2011 (UTC)[reply]

Even if it could work (I am pretty sure it couldn't, but lets play a game and pretend that it could), what would be the point? I am pretty sure that the lava would become impregnated with lots of highly radioactive fallout from the bomb you just set off; great, so you have new land but its so radioactive to be unlivable. What good is that? That of course ignore the point that even the most powerful thermonuclear devices pale in comparison to the power of geologic events. A single volcanic eruption or earthquake packs many times the wallop of an H-bomb. I'm pretty sure if you put the entire world's nuclear arsenal in one place and set it off all at once, it wouldn't even be as big as the 1980 eruption of Mount St. Helens, which while impressive was not uncommonly so, at least on a geologic scale. The main blast which caused the collapse of the north face of the mountain moved about 1/20th of a cubic mile of earth, an unimaginably large chunk of solid rock to be pulverized to dust. I just can't see a bomb doing that kind of work. --Jayron32 21:14, 28 January 2011 (UTC)[reply]

To be fair, nobody is saying that the bomb would have the force of a volcano, just that it could be used to trigger one. I'm not sure that's not impossible. The trick about energy release is that it matters how much is released in how small a volume. An earthquake releases a huge amount of energy but over long distances. The sun releases an unfathomable amount of energy but it does so over such a diffuse distance and time scale that we perceive it as a warm bath, not a searing flame. I think there are probably geological situations where a properly placed, sufficiently large thermonuclear detonation could act as a catalyst for a much larger volcanic eruption. It's not necessarily the case that fallout would be mixed into the magma itself — there might be clever ways to make the eruption contain the bomb blast (e.g. bury it even while it erupts upwards). But I don't know much about volcanos, to be sure. --Mr.98 (talk) 21:27, 28 January 2011 (UTC)[reply]

Actually Mount St. Helens released "only" about 24 Megatons of energy which is somewhat less than the largest hydrogen bomb ever detonated. The Krakatoa eruption, one of the largest eruptions in recorded history, was about 200 Megatons of energy. So, volcanoes are somewhat larger than but on the same order as hydrogen bombs. Of course the energy is released in very different ways, which does make a significant difference with respect to the impacts it has. Dragons flight (talk) 21:36, 28 January 2011 (UTC)[reply]

It's worth noting that Krakatoa's explosion left less land above the sea than before. And Mt St Helens got smaller when it exploded. Guaranteeing more land would be a challenge. HiLo48 (talk) 21:58, 28 January 2011 (UTC)[reply]

True to a point, but keep in mind that the goal would be to take the internal parts of the mountain and spread them out in order to create a structure with more surface area and less height. Still, Mt. St. Helens is a good example in many ways. Even if it had been surrounded by water, and somehow did end up with water above sea level, mixed ash is not really an ideal building material. And it strikes me that for all of the effort, the amount of land is still not going to be that useful by human standards. (And again, I'm wary about comparing nukes to volcanos in terms of explosive force, anyway, because the time scales are usually quite different, and the effects that correspond are quite different.) --Mr.98 (talk) 20:12, 29 January 2011 (UTC)[reply]

I'm not certain that the demand for new land in Japan is any higher than it is in New York. Though they have enormously high population densities in the biggest cities, due to most of the land being unusably steep (Japan#Geography), they also have a falling population size, and 6,852 little islands that already exist. The most significant reason for the high population of cities is that a large number of people want to live close together. Is Izu Ōshima so crowded that they need a second one? Land reclamation does say that 20% of Tokyo Bay has been reclaimed, but land that's contiguous with Tokyo must be worth a lot more than land which is offshore and radioactive. 81.131.22.166 (talk) 23:25, 28 January 2011 (UTC)[reply]