Wikipedia:Reference desk/Archives/Science/2012 January 13

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January 13[edit]

Plant Identification[edit]

There is a beautiful bush in my yard in Virginia that has glossy red berries right now -- here is a picture of its berries and leaves.

Could someone tell me the name of this lovely plant? At first I thought it was holly, but the leaves don't look like holly leaves at all. 72.218.72.80 (talk) 00:42, 13 January 2012 (UTC)butterfliesarefree[reply]

Not all hollies are evergreen with spikes on the leaves. I think you have a type of winterberry, which is a deciduous holly. Someone else will chime in soon :) SemanticMantis SemanticMantis (talk) 00:57, 13 January 2012 (UTC)[reply]

Thanks, SemanticMantis! I didn't realize there were so many kinds of holly :)72.218.72.80 (talk) 17:21, 13 January 2012 (UTC)butterfliesarefree[reply]

I can't name it right now, which is not very helpful, but the leaves nor the berries are quite right for winterberry. If you Google images for american winterberry you'll see for yourself. Richard Avery (talk) 14:30, 14 January 2012 (UTC)[reply]

meaning of bracketed terms in chemical reaction notation[edit]

I looked up a database of chemical reactions. The following listing is typical: C + O2 → O(1D) + CO where the "2" is the usual subscript 2 indicating a diatomic molecule, and the "1" is a superscript 1. What does the (1D) term mean? The database does not say. Re Keit124.178.172.68 (talk) 04:32, 13 January 2012 (UTC)[reply]

Could you provide a link to the actual database? It may make it easier for us to answer your question if we can read exactly what you are reading. --Jayron32 04:36, 13 January 2012 (UTC)[reply]
The best I can come up with is that it is a Term symbol used to represent the angular momentum state of the monatomic oxygen in the reaction. 1D is a term symbol, where 1 is the spin multiplicity of the atom, and D is the total angular momentum quantum number. It's been years since I studied this stuff, so it may take someone whose used it more recently to explain a bit more. --Jayron32 04:44, 13 January 2012 (UTC)[reply]

Thanks. See NIST database at http://kinetics.nist.gov/kinetics/Search.jsp. This database has a help link on each entry, but it does not work ("webpage cannot be found"). A means of contacting the support staff is provided, but I got no response when I tried it last year. Re Keit124.178.172.68 (talk) 04:53, 13 January 2012 (UTC)[reply]

Yeah, those are term symbols. I looked at other entries in the database, and that's what it looks like they are using in those cases. It is basically a highly condensed way to indicate the electron configuration of the atom. 1D would mean that the oxygen atom in question has a total spin quantum number of S=0 (because the superscript is 1, so 2S+1=1, so S=0, see Term symbol article. Basically this means all individual electron spins are paired) whereas D means L=2; this is the total angular momentum quantum number, so the sum of the individual l values for all of the electrons. You would generally only use the term symbol if the atom was NOT in its ground state, which in this case it is not: the 1D term symbol tells me that the O atom has all spins paired and only two p orbitals occupied; ground state oxygen should have two unpaired electrons and three p orbitals occupied . Or at least, that's what my 15-year-old never-used quantum chemistry memory, jogged by the relevent Wikipedia articles, has brought back. Someone else may correct that. --Jayron32 05:28, 13 January 2012 (UTC)[reply]
Indented line

Thanks Jayron - that's great! Keit58.165.230.28 (talk) 06:24, 13 January 2012 (UTC)[reply]

Cognitive enhancement deadlocks[edit]

Have any experts studied whether radical cognitive enhancement is at risk of deadlocking with other technologies in the future? What I mean is this: If no unenhanced human mind is capable of inventing X, Y or Z, and the only way forward in cognitive enhancement is to apply X, Y or Z, then humans will never invent X, Y or Z on our own. NeonMerlin 04:34, 13 January 2012 (UTC)[reply]

What is an unenhanced human mind? We already use stuff like computers which enhance our ability to do all sorts of creative work... --Jayron32 04:48, 13 January 2012 (UTC)[reply]

Obviously, if the unenhanced human mind cannot invent X, Y or Z, then one human will never invent either. On the other hand, as a group of multiple humans might be considered as an enhanced human mind, they are capable of inventing them, thus be able to progress. An unenhanced Human mind is in fact a human mind without any kind of stimuli. It can be described in several ways. In chemical/biological/physical way: No substance(drug, alcohol, medicine etc) has been given or used to stimulate(enhance) the brain; Biolgical Evolution: Without becoming better adaptive to the environment, we would not be the way we are now. Because we kept inventing new ideas and actually created them, we evolved. This would mean that our ancestors had ehanced minds, because they could create X, Y and Z, thus move forward in cognitive enhancement; Ethical/aestethical etc: A single person is just as strong as he is. A sword is only as dangerous as it's wielder. In other words, if the "right" mind is applied to the "right" human, s/he can move forward in cognitive enhancement. Another way to look at it, is a situation where multiple people cooperate. As we notice in daily live, when we work together we can get more work done, get more innovative ideas and make more progress with what we do. In this context, the "enhanced" human mind will be a group of multiple humans, where the "unenhanced" mind is just a single human.

It's not a simple matter to understand, and this is my personal experience after working with several groups studying this matter combined with 18 years of study in the physical/chemical/biomedical sciences, but I hope this helps :-) — Preceding unsigned comment added by 137.120.218.232 (talk) 11:39, 13 January 2012 (UTC)[reply]

Technological advancement is driven by need, and very often those needs are driven by military and industrial competition. They are also driven and made possible by what has come before. Many of the great inventions, discoveries and efforts of the last couple of centuries were being worked on by multiple teams, and the team that got it done first got the patent and the publicity... and the development contract. ←Baseball Bugs What's up, Doc? carrots→ 11:57, 13 January 2012 (UTC)[reply]
'Finally, even basic research can take unexpected turns...' (from Science); just because we haven't realized something is possible doesn't mean it won't be discovered (serendipitously). Dru of Id (talk) 17:13, 14 January 2012 (UTC)[reply]
So far, the human brain has invented many remarkable things... but evolution invented the human brain. With little, if any, cognition involved in that process at all. Wnt (talk) 06:18, 15 January 2012 (UTC)[reply]

Thermonuclear[edit]

Hypothetically, there is a spherical crystal of (neutronium/nihilon/neutron degenerate matter) with a 1 cm diametre. What would be its minimum initial speed, if by friction alone it causes a thermonuclear interaction with the atmosphere as it travels, and how far would it get before subliming? Plasmic Physics (talk) 08:15, 13 January 2012 (UTC)[reply]

I think all the exothermic nuclear fusion reactions that we know of involve either hydrogen or helium nuclei. In your scenario, where does the supply of hydrogen or helium come from ? Are you assuming a hydrogen atmosphere, for example ? Gandalf61 (talk) 13:02, 13 January 2012 (UTC)[reply]
(ec) Thermonuclear reactions are just reactions taking place at a temperature where the thermal kinetic energies of the particles involved are sufficiently high to overcome electrostatic repulsions between positively-charged nuclei. For hydrogen isotopes, this regime is somewhere around 100 million kelvin; for heavier elements (with more-positively charged nuclei) thermonuclear reactions don't become important until you get to even higher temperatures.
Exothermic fusion reactions can take place whenever the fusion products have a greater binding energy (mass deficit) than the reactants. In practice, this means that fusion of elements all the way up to iron is exothermic—provided you can get up to the temperatures required to initiate that fusion in the first place. (Indeed, towards the end of their fusion lifetimes, heavier stars have an iron-rich core surrounded by roughly concentric, overlapping shells of lighter elements fusing; see also helium burning, alpha process, triple-alpha process.)
All that said, the original question doesn't really make a lot of sense. The stability of a little chunk of neutronium like that is rather questionable; without great pressure constantly applied to its surface, neutrons would constantly (and violently, truth be told) evaporate and then decay back into protons and electrons. This process wouldn't require interaction with our atmosphere; it would happen just as easily in space, long before it arrived here.
As well, 'thermonuclear interactions' don't really come into it. Interactions between neutrons and nuclei don't require the million- or billion-kelvin temperatures associated with thermonuclear fusion. Because neutrons are uncharged (and therefore are not repelled by positively-charged atomic nuclei), they're perfectly capable of interacting with other matter at room temperature. Indeed, neutron capture reactions tend to be much more efficient with so-called thermal neutrons: neutrons within a shouting distance of room temperature. Even if a neutronium blob were held stationary in air (and magically didn't violently explode) there would be rapid, vigorous, and spontaneous nuclear reactions occurring as atmospheric oxygen and nitrogen collided with neutrons. TenOfAllTrades(talk) 17:32, 13 January 2012 (UTC)[reply]
Would so small a quantity of neutronium be stable? I'd expect it to expand, become free neutrons and decay as free neutrons do. RJFJR (talk) 17:11, 13 January 2012 (UTC)[reply]

Would you be more comfortable if I said unobtainium instead of neutronium? Ignoring other factors, at what initial speed does air friction produce along its trajectory, a rapidly expanding fireball, accompanied by a x/gamma ray burst? Plasmic Physics (talk) 00:43, 14 January 2012 (UTC)[reply]

How are we supposed to tell you the physical properties of a fictional substance? They are whatever you make them up to be... --Tango (talk) 01:25, 14 January 2012 (UTC)[reply]

Why are you focusing so much on what the sphere is made of? Plasmic Physics (talk) 04:37, 14 January 2012 (UTC)[reply]

Maybe because your original question said "... it causes a thermonuclear interaction with the atmosphere ..." which implies that the sphere itself is somehow involved in the reaction. However, as TenOfAllTrades said, your original question makes very little sense. If you asked a less baroque question such as "what is the minimum temperature at which nuclear fusion will take place", then it could be answered - the core of the Sun has a temperature of about 15 million K (but is at enormous pressure and density), our nuclear fusion article says that deuterium-tritium fusion requires a minimum temperature of around 120 million K and peaks at 800 million K, and pyroelectric fusion takes place at a temperature of about 1,000 million K. Gandalf61 (talk) 14:06, 14 January 2012 (UTC)[reply]

Good, that is exactly what I getting at. At what speed will friction produce a temperature like that? Plasmic Physics (talk) 20:05, 14 January 2012 (UTC)[reply]

It really depends on what the sphere is made of. Most substances will ablate (bits will come off them, carrying heat away) before they get anywhere near the temperatures required for fusion. --Tango (talk) 23:12, 14 January 2012 (UTC)[reply]

Something incredibly dense and smooth, a perfect little ball of something where such processes are negligeable over a split second. Plasmic Physics (talk) 03:07, 15 January 2012 (UTC)[reply]

I don't think your question is really answerable, but I'll do some back of the envelope calculations to give you some idea of the numbers involves. Let's say it's a 1kg ball of iron and we want to get it up to 1 million Kelvin. According to heat capacity, the heat capacity of iron is 0.450 J/g/K (at 25°C, but we'll assume its the same at all temperatures, which is ridiculous, but nevermind). That means we need 0.45 GJ of energy to heat it up. If we assume all the ball's kinetic energy can be converted into heat and kept in the ball (which is also ridiculous, but nevermind) then, using E=1/2 mv2, we find that we need a velocity of 30km/s. To but that in context, escape velocity (and, therefore, the velocity you would get if the ball just dropped from space, since gravity works the same whether the ball is going up or down) is 11km/s, so you see you need to be going pretty fast. My answer is an under-estimate because I've ignored the heat of vapourisation, which will use up some of the energy without heating up the ball. I've also ignored the fact that a lot of energy will go to heating up the air, not the ball. I've ignored the fact that the atmosphere is nowhere near thick enough to stop a ball of iron travelling 30km/s before it hits the ground, so you won't get anywhere near all the kinetic energy converted into heat. And finally, I've ignored the ablation - even at 30km/s, it will take a second or two to travel through the thick part of the atmosphere, during which time the ball will vapourise and be spread out carrying all the heat away so none of ball actually gets beyond a few thousand degrees. The same is going to be true with any realistic substance you want to use. You might get some interesting results with some kind of degenerate matter, but you won't be able to keep it degenerate once you remove the pressure that made it that way, so that isn't at all realistic. --Tango (talk) 03:47, 15 January 2012 (UTC)[reply]

Let me give you some hypothetical properties: a density of 830 Eg m−3, a mass of 3.5 Pg, a specific heat capacity at 25 °C of <<1 mJ K−1 kg−1, a thermal conductivity of <<1 nW K−1 m−1. The sphere is travelling in a tangenial path. Plasmic Physics (talk) 09:56, 15 January 2012 (UTC)[reply]

I could plug those numbers into the formulae I used above (and so could you), but you are still ignoring too many important factors. In reality, it's just never going to happen. You can't get an object that hot without it being under extreme pressure because it will just vapourise and rapidly expand, cooling down as it goes. You would need to get it to those temperatures quickly enough that fusion can occur before it expands, and that just isn't going to happen. Any object at atmospheric pressure is going to vapourise at only a few thousand degrees, maximum. That means you need to get it all the way from a few thousand to a few million in a fraction of a second. There is just no way friction with the atmosphere is going to do that. --Tango (talk) 18:12, 15 January 2012 (UTC)[reply]

I'm not concerned with the sphere getting hot, just the air that it makes contact with. Plasmic Physics (talk) 22:50, 15 January 2012 (UTC)[reply]

I would expect the air to be even cooler than the sphere, since the sphere is moving through the air so it different bits of air getting heated up, meaning the heat is spread out through quite a lot of air. You certainly aren't going to get the air up to temperatures where oxygen and nitrogen can fuse - that doesn't even happen in stars until they die. --Tango (talk) 12:25, 16 January 2012 (UTC)[reply]

What about an exchange in momentum, where molecules of gas are accelerated to high energies by close encounters with the sphere in transit? Plasmic Physics (talk) 05:24, 17 January 2012 (UTC)[reply]

Newton's third law applies to air as much as to anything else - if the air is slowing the sphere down, the sphere must be speeding the air up. That's not going to cause fusion, though. Travelling fast is different to being hot (heat is a bulk property involving lots of molecules moving around, one molecule moving around is a different matter altogether). --Tango (talk) 12:13, 17 January 2012 (UTC)[reply]

hologram[edit]

I WANT TO MAKE A HOLOGRAPHIC IMAGE OF A MOLECULE IN MY HOME IS IT POSSIBLE ?Rikisupriyo (talk) 10:10, 13 January 2012 (UTC)[reply]

There is no need to shout, we can hear you just fine. Yes, it is possible to make a holographic image in your home, if you have the right equipment. Plasmic Physics (talk) 10:27, 13 January 2012 (UTC)[reply]

HOW IS IT POSSIBLE PLEASE TEL ME .Rikisupriyo (talk) 11:01, 13 January 2012 (UTC)[reply]

You're still shouting, but you could be interested in Holography#Recording a hologram. If you are looking for specific construction instructions, then Wikipedia is not the place to look. A dedicated website for holography hobyists would be more useful. Plasmic Physics (talk) 11:10, 13 January 2012 (UTC)[reply]
How large of a molecule? – b_jonas 16:53, 13 January 2012 (UTC)[reply]
Real molecules are far too small to be imaged by holograms. However, you could make a model of a molecule, and take a hologram of that. There are kits for making holograms in your own home. APL (talk) 22:38, 13 January 2012 (UTC)[reply]

relation between video format and video codec[edit]

what is the relation between video format and video codec? Why some of the files like .3gp etc are not run in DVD players? — Preceding unsigned comment added by Laxmi bsnl (talkcontribs) 11:02, 13 January 2012 (UTC)[reply]

If you have heard about cryptography, then you could understand this: in encryption, plaintext (video) is transformed into ciphertext (video file), using a cipher (video codec), however a key (video format) is required to access the cipher. The key is useless, unless the cipher is known, thereby the video format is useless, unless you have the codec to transform the video file back into a video. If any part of that recipe is missing you can't play the video. Without a right cipher (video codec), the key (video format) has nothing to unlock. Plasmic Physics (talk) 11:31, 13 January 2012 (UTC)[reply]
Normally, you can just find the appropriate codec online and install it as an add-on for your media player on your conputer. However, since you are using a dedicated DVD player, the only thing that I can recommend is to download a program that has the ability to translate that format into something that the DVD player can recogise. Plasmic Physics (talk) 11:41, 13 January 2012 (UTC)[reply]
A codec is used for data compression, not encryption.
There is a lot of confusing terminology (and there is a lot of misinformation on the internet). The two articles you should read are:
  • Digital container format, which specifies the way a file is laid out - which bytes are where, and so on;
  • Data compression, which explains the way imaging scientists reduce the amount of data and still convey most of the same amount of information.
One of the most confusing concepts (and the source of much misinformation on the internet) is that most container formats are codec-agnostic: for example, inside an AVI file, you can store any type of video code-stream: H.264, MPEG-2, or whatever. Often, people use a video player, under the impression that it can open "any file that ends in .avi," which is incorrect for two reasons:
  • The file extension may not accurately represent the true contents of the file; any file can be renamed to call itself an "something.AVI" without actually complying to the format specification.
  • Even if the container is well-formed and valid, the video data may be compressed using a codec that is unrecognized or unsupported.
In the old days, "DVD player" meant exactly one thing: "can play DVD file-formatted discs that contain DVD-Video H.262/MPEG-2 Part 2 code-streams." Nowadays, "dvd player" means "just about any optical disc reader that can interpret around 50 to 90% of the commonly-used file-formats and decode around 10 to 25% of the commonly used video compression codes." Some device makers will print a list of all the file formats and codecs that they can use; but chances are, the product marketing will not allow such technical data to be leaked. Furthermore, there's a good chance that even the engineers don't really know the canonical list of every file format that can be played!
The OP asked, "Why some of the files like .3gp etc are not run in DVD players?" You may be wondering: why don't the designers of a DVD player just add __ format? If I may wax anecdotally... I used to work for a video-camera-chip maker, and the guys upstairs made the CPUs that went inside DVD players (hardware and software video decoders)... DVD is a tough business. The devices sell for cheap, the competition is fierce, and keeping high-quality engineers who know anything about video processing is nontrivial (and not cheap!) Let's just say, the QA process was lackadaisical, the open-source software compliance was tricky, and fewer than half the engineers even knew what software actually went out the door on the shipping products. Nonetheless, our codec guys worked hard to validate and design at least a semblance of product functionality. Of course, once we let the chips into the hands of system-integrators overseas, they'd slap on all kinds of extra software and hardware peripherals that were out of our company's control, throw it in a crummy plastic box with metal veneer logotype and call it a "Major Name-Brand Electronic Appliance DVD Player." Perhaps no surprise, that chip designer went out of business... the company that acquired it lost some 80% of its stock valuation, etc... DVD is a tough business. Nimur (talk) 18:11, 13 January 2012 (UTC)[reply]

Two black hole questions[edit]

Firstly, as I understand it, an particle precisely at the event horizon of a black hole would stay in place as long as it is travelling at c away from it. Is there a region further back where a particle travelling perpendicular to the black hole needs to travel at c in order to maintain a stable circular orbit (assuming no further perturbations of the system)?
Secondly, I remember reading on a wiki article (though I can't remember which one) that there was a "theoretical maximum size" for black holes. That doesn't jibe with my understanding. Seems like you could just always add mass to one. Goodbye Galaxy (talk) 15:14, 13 January 2012 (UTC)[reply]

I don't know much about that kinda thing, my guess is that the "theoretical maximum sixe" means as in volume not mass. Heck froze over (talk) 15:21, 13 January 2012 (UTC)[reply]
Fir the first question, I'm not quite sure but, see photon sphere, and see the photon sphere section of David Madore's Black Hole page. – b_jonas 16:51, 13 January 2012 (UTC)[reply]
Ah, thanks! That's exactly what I was thinking of for the first question. It's a nice feeling to reason that something should exist, and then it turns out it does! Goodbye Galaxy (talk) 17:05, 13 January 2012 (UTC)[reply]
For the second question, are you sure you aren't remembering reading about a theoretical maximum size that a star can be without turning into a black hole? If that might be the case, see Tolman–Oppenheimer–Volkoff limit and Chandrasekhar limit.
As another thing you might be thinking of, there is something that limits the initial size and growth of a supermassive black hole in that if the angular momentum of the matter that the supermassive black hole is initially formed from is too great, the matter tends to wind up in an accretion disk instead of in the black hole. See Supermassive black hole#Formation. But even bigger supermassive black holes can form from the merger of two smaller supermassive black holes during a galaxy merger.
The biggest known supermassive black hole, in NGC 4889, has a mass of about 21 billion solar masses, about 5000 times more massive than the supermassive black hole in Sagittarius A* at the center of the Milky Way. So there doesn't seem to be much of a limit as to how big black holes can get, unless you start getting into cosmological considerations like how much mass there is in the visible universe, and the ultimate fate of the universe and stuff like that. Red Act (talk) 19:29, 13 January 2012 (UTC)[reply]
How about a black hole massive enough for the event horizon to be so large that dark energy becomes a significant factor? Could you get a black hole being ripped apart? Or does the fact that the actual mass in concentrated at the singularity mean spacial extent doesn't come into it? --Tango (talk) 01:31, 14 January 2012 (UTC)[reply]
Yeah, anything large enough that the cosmological constant becomes important gets into the cosmological considerations that I was trying to sidestep. I don't know if dark energy of either the cosmological constant or scalar field variety could rip an enormous black hole apart. Red Act (talk) 03:05, 14 January 2012 (UTC)[reply]

"Genetic analysis" of fossils[edit]

Our article Coelacanth claims, "According to genetic analysis, the divergence of coelacanths, lungfish, and tetrapods is thought to have occurred 390 million years ago." There is a reference that I don't have access to.

How is "genetic analysis" performed on fossils? Fossils are stone. Comet Tuttle (talk) 20:02, 13 January 2012 (UTC)[reply]

There are live examples of coelacanths and lungfish and tetrapods. DNA changes in roughly constant ways, so one can broadly calculate the age of the Most recent common ancestor based purely on known rates of genetic change and difference between the genetic code of modern specimens. Most_recent_common_ancestor#Time_to_MRCA_estimates is a very brief explanation of how it is done, and it is very human-centric; however conceptually it works for any two organisms. --Jayron32 20:12, 13 January 2012 (UTC)[reply]
Doh, the obvious answer. My excuse is that that sentence in the article is directly below the section header "Fossil record". I have added a clarifying phrase to the article. Thank you! Comet Tuttle (talk) 20:16, 13 January 2012 (UTC)[reply]

One-point implosion[edit]

Would it be possible to create a nuclear weapon with a "one-point implosion" system, where the density of the material surrounding the inner explosives is continuously altered so as to make the shock wave from the single detonator impact the inner explosives all at the same time? Whoop whoop pull up Bitching Betty | Averted crashes 23:15, 13 January 2012 (UTC)[reply]

That information is classified. Expect a visit from some gentlemen in conservative suits and dark glasses Real Soon Now. {The poster formerly known as 87.81.230.195} 90.197.66.91 (talk) 07:40, 14 January 2012 (UTC)[reply]
If you're talking about an explosive lens, then yes (see the article). -- BenRG (talk) 08:09, 14 January 2012 (UTC)[reply]
It makes no mention of a lens which would be useful for "one-point" implosion, which would require the shock wave to converge to a single point within the lens, instead of outside of it. Whoop whoop pull up Bitching Betty | Averted crashes 15:36, 14 January 2012 (UTC)[reply]
A 'one point' egg shaped explosion shell can (and has been) be employed to compress a core to criticality.--Aspro (talk) 17:44, 14 January 2012 (UTC)[reply]
Refs please. Whoop whoop pull up Bitching Betty | Averted crashes 19:09, 14 January 2012 (UTC)[reply]
Medical and legal questions are not answered here on the Reference desk because the professions of clinical and legal has managed to censor WP on the premise that it is 'dangerous' to let OP's discover such information (that which they alone can proffer in-return for oodles of money ...) Therefore, in the same spirit, I don't feel I can give you references to anything that is so more potentiality dangerous than any knowledge that they covet -so jealously. You will have to be satisfied with Wikipedia's little mention of the egg shell:Teller–Ulam_design#W88_revelations. However, if you are willing to offer to deposit oodles of money into a certain Swiss bank account, then I think that you will have the means to ensure your neighbours will no longer sleep easy in their beds.--Aspro (talk) 19:47, 14 January 2012 (UTC)[reply]
Eyes rolling at your "I have secrets" nonsense. The W-88 primary is reported universally as being two-point. I don't see any reason to think it's one-point. --Mr.98 (talk) 19:54, 14 January 2012 (UTC)[reply]
Two-point implosion — definitely. One-point? I've never heard of it. It's inherently unsafe — one-point detonation means that all you need is one mistake for a full nuclear blast. (Little Boy is a one-point detonated device, but not implosion.) In other words, it inherently fails one-point safety. Is it possible? I don't see why not, in principle. You can do very clever things with the hydrodynamics of explosives. But it strikes me as something that would be very finicky. But I'm not a weapons designer, to say the least. I don't think it would involve changing the density of the materials in real time, though. (I don't know how you would do that — sounds implausible. The density changes would have to propagate at a fantastic speed.) I would think of a one-point compression to be something like two-point linear implosion, but instead of it being symmetrical, it would be one-sided compression. Something like a cross between a gun and implosion device — e.g. compressing a tube of plutonium lengthwise. --Mr.98 (talk) 19:54, 14 January 2012 (UTC)[reply]
The W-54 Davy Crockett was a 'one point' device. A fast 6000 foot/sec explosive lens ensures that the compression sphere, detonates symmetrically in order to compresses the core to criticality. Oh, I'm so glade that you et. al. can only guess at its principles, I can sleep easy to-night.--Aspro (talk) 21:17, 14 January 2012 (UTC)[reply]
That's not one-point. It's two-point. There are distinctly two detonators in the description given there. It's the same as the two-point description on our design page previously linked to. You seem to be ignorant about the distinction. Two-point implosion means you create a compression based on only two detonators (as opposed to the Fat Man, which was had 32 detonators). One-point would be creating implosion with one detonator. Most nukes are in fact purposefully tested to be one-point safe — if one detonator randomly goes off, it will not cause a nuclear yield. If you had read the links I had given earlier, all this would be quite clear. I am happy to believe that your assertion to the contrary is based merely on ignorance, but your defense of this with allusions to secret knowledge I find offensive because it is clearly just false. --Mr.98 (talk) 03:26, 15 January 2012 (UTC)[reply]
Not in time–in space (the variations in density would be built into the warhead during the process of machining the egg-shaped explosive layer). Whoop whoop pull up Bitching Betty | Averted crashes 20:20, 14 January 2012 (UTC)[reply]
I'm really not picturing what you're describing by the "continuously altered" density, and your description there isn't helping much... --Mr.98 (talk) 21:05, 14 January 2012 (UTC)[reply]
If you don't follow this, then why are you including yourself in the conversation?--Aspro (talk) 02:07, 15 January 2012 (UTC)[reply]
Because I actually understand the general point he's trying to make, unlike you, but I'm having a hard time visualizing the specifics of the implementation he's trying to articulate. --Mr.98 (talk) 03:34, 15 January 2012 (UTC)[reply]
It's basically an egg-shaped high explosive lens which is shaped in such a way as to cause the shock wave to become a circle converging towards a certain point within the lens; the material within this circle could be replaced with the basic innards of a fusion-boosted, hollow-pit nuclear weapon to produce a nuclear bomb which works via one-point implosion. Whoop whoop pull up Bitching Betty | Averted crashes 21:26, 14 January 2012 (UTC)[reply]
Two-point linear implosion
OK, that I understand, and I would chalk it up to the "you can do very clever things with explosion hydrodynamics in theory," but it strikes me that it would be quite, quite difficult. You're talking about setting off an explosion on one end of a spheroid, using some very fast explosives to get it around to the other sides of it, and then somehow having all of that compress the central portion more or less symmetrically. That strikes me as a lot of work for an exploding blast wave to do, a lot of timing to get just right. Two-point makes more sense, given the geometry of it — it allows you to have each explosion only do half of the geometry.
The linear implosion illustration in the article illustrates this problem well — the explosions are radiating outward, and that is what leads to the eventual implosion. How are you going to do the work of both sides, the left and the right, with just an explosion on the left? It's going to be tough making the explosion on the left create an explosion on the right. You can do very clever things with shaping the geometry of the explosion — the wave shaper is a great example of that, taking the single detonator and turning it into two wave fronts — but it strikes me there are probably limits to that. Can you imagine a scheme in which a single wave shaper makes four wave fronts that converge on a central point? I almost can but it seems like a Rube Goldberg sort of bomb. My idea above would be to replace the right half of that linear implosion design with some sort of very dense wedge, and hope that it would substitute for a true implosion on that side, if plutonium was compressed into it. But it probably wouldn't get very good compression, so it probably wouldn't work. Certainly not as good compression as an honest-to-god explosion.
Even two-point seems not very efficient at its compression (and seems only to be used in situations where space premium is more important than yield); one-point would be dreadfully difficult for no apparent gain over two-point. --Mr.98 (talk) 03:34, 15 January 2012 (UTC)[reply]
I don't know anything about this, but I assume some arrangement of detcord could transfer the explosion from a single detonator to many points around a bomb? But I suppose an electronic system would be more reliable, as electricity tends to be faster? Wnt (talk) 06:15, 15 January 2012 (UTC)[reply]
I'm not sure I'd consider that one-point detonation, but if you wanted to consider it that, perhaps it would work. The tricky thing would having the explosion transferred evenly enough to produce implosion. This is why electricity is used as the wiring, as it can be made to transfer the signal within a very tight level of tolerance. In the Fat Man style bombs, you needed a level of simultaneity in the sub-millisecond range. That's hard to do without using electricity. --Mr.98 (talk) 15:35, 15 January 2012 (UTC)[reply]
What if you covered the entire bomb with extremely thin foil, placed one pole of a capacitor at one end, the other pole at the antipode of the first pole, and sent a huge jolt of electricity through it to vaporise all the foil at once and simultaneously initiate the detonation of the entire surface of the explosive? Whoop whoop pull up Bitching Betty | Averted crashes 21:37, 15 January 2012 (UTC)[reply]
It would probably depend on the ability of the exploding foil to achieve the necessary simultaneity and uniformity. I kind of doubt it would work in practice. What's the goal of this kind of investigation, dare I ask? I mean, why try to think up increasingly inefficient, increasingly Rube Goldbergish ideas? You could put a nuclear weapon inside of an elephant if one wanted to, but what would be the advantage? --Mr.98 (talk) 00:59, 16 January 2012 (UTC)[reply]
Disguising it. Whoop whoop pull up Bitching Betty | Averted crashes 02:26, 16 January 2012 (UTC)[reply]
Well, it would make it a mammoth of a problem to detect, I'll grant you that. --Mr.98 (talk) 20:28, 16 January 2012 (UTC)[reply]
Slapper detonator? --Sean 20:05, 19 January 2012 (UTC)[reply]

Nuclear tank ammunition[edit]

If they can be made small enough to be deployed from recoilless rifles, why have nuclear weapons about the size of the Davy Crockett never been used as ammunition for a main battle tank? Whoop whoop pull up Bitching Betty | Averted crashes 23:21, 13 January 2012 (UTC)[reply]

There's not really a technical barrier, so much as a philosophical one. The main battle tank's main gun is principally a direct fire weapon—you're going to fire it straight at stuff that you can see, generally less than three kilometers away (and sometimes much closer). Your tank is going to be in the line of fire; if the tank can see and shoot at the enemy, the enemy can fire right back. Not only that, but one's own forces are likely to be close by, and may be less than pleased by exposure to your close-range nuclear fire.
In other words, if the other guy fields nuclear tanks, you don't respond with nuclear tanks of your own—you instead lob nuclear shells from your self-propelled artillery, ten or fifteen kilometers before the guy with the tanks can even get close enough to fire. (Nuclear artillery shells were widely deployed by Cold War armies. NATO forces still use M109 and M110 howitzers – self-propelled, tracked guns – along with an assortment of towed artillery pieces that can launch nuclear shells, including the W33 and the W48.) TenOfAllTrades(talk) 00:20, 14 January 2012 (UTC)[reply]
Desirability of such a weapon aside, the necessary further reduction in size might not be so easy. The Davy Crockett round was 28 cm at its widest point, compared to the 90 to 130 mm caliber most common on main battle tanks. Of course, nothing but common sense is keeping you from mounting an M-29 on an MBT. -- ToE 01:04, 14 January 2012 (UTC)[reply]
On the other hand, the W48 artillery shell is a 155 mm caliber, so one could probably shave another 25% off that (though probably at some cost of efficiency).
Our article on nuclear artillery mentions testing of a single 5-inch (127 mm) nuclear fusion(!) artillery shell in 1958; apparently the fission primary worked but the fusion secondary didn't, giving a yield of only 190 tons. There are reports that the Navy considered developing 5-inch 1-kiloton shells for naval guns, as well. While neither of these devices were mass-produced or deployed, there didn't seem to be any insurmountable technical barrier to their construction. TenOfAllTrades(talk) 01:32, 14 January 2012 (UTC)[reply]
Thanks Ten. I was thrown of by W54: "The W54 was the smallest nuclear warhead deployed by the United States.". W48's final paragraph compares the two warhead's dimensions and weight. The W48 achieved its much smaller diameter (155mm vs. 270mm) at the expense of a greater length (845mm vs. 400mm) and weight (55kg vs. 23kg), due to its much less efficient linear implosion design. -- ToE 04:01, 14 January 2012 (UTC)[reply]

Tactical nuclear weapons that have fission devices have limited utility due to the undesirable nuclear fallout they produce. It would expose friendly troops to dangerous radiation. In the context of total nuclear war, they would probably be used eventually though. The US Military is working on 4th generation nuclear devices that do not use a fission warhead to trigger a fusion secondary stage. Out of all the possibly triggers, it was decided that if small amounts of antimatter could be produced at an improved level of efficiency (and by improved I mean at least .01% efficient which is a lot more efficient than our current efficiency of producing antimatter), then small fusion warheads with an antimatter trigger are feasible. This would produce a very powerful device in a very small package while producing very little radioactive fallout. ScienceApe (talk) 10:55, 19 January 2012 (UTC)[reply]