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November 6

Thermally unstable compounds

Are there techniques for finding properties of a thermally unstable compound (TUS) beyond temperatures at which significant decomposition takes place? For example, even though mercury(II) hydride readily decomposes at approximately -125 degrees Celsius, is it practical to heat a sample faster than it can decompose and still be able to determine whether or not the decomposition temperature coincides with the melting point? How would I go about detecting a phase chance on the molecular scale under such dynamic conditions? Plasmic Physics (talk) 09:52, 6 November 2016 (UTC)[reply]

For higher temperatures this is certainly possible, and you may be able to overheat by some tens of degrees if you do it fast. However the decomposition products will contaminate your product and disturb the properties. Another issue is that heat is released during the decomposition, there will be positive runaway feedback, and it likely explodes. Perhaps you can subject the solid to high pressure, or high pressure under hydrogen to try to reverse the decomposition. Perhaps a thin layer on some substrate will be more stable. You may be able to use X-rays, Raman spectroscopy, optical dispersion, mechanical properties (eg the sound of the explosion), or even Neutron diffraction or Positron annihilation spectroscopy to find some properties. To see if the liquid is intrinsically unstable you could try dissolving your product in suitable liquids (liquid nitrogen, liquid xenon) and see if it can be recrystalised. Graeme Bartlett (talk) 10:52, 6 November 2016 (UTC)[reply]

What if I prepare microcrystals on a monolayer substrate, and heat the substrate in half-degree multiples, under hydrogen pressure? If I can get a good size distribution of microcrystals, I should be able to project behaviour for the bulk. I don't know if reversing the decomposition process is feasible, on account of the fact that atomic mercury requires excitation for insertion into the dihydrogen bond. Ergo, the pressure required may be too high. Yes... exposure of the compound to a pressure range could yield informative results. Recrystallization, if it works could improve the limited understanding of the solid structure. Plasmic Physics (talk) 10:26, 7 November 2016 (UTC)[reply]

Phase changes don't happen instantly. Whether this makes sense depends on a comparison between the time scale of decomposition and the time scale of the phase change. I believe that in most cases the decomposition will happen faster. Looie496 (talk) 14:37, 6 November 2016 (UTC)[reply]

That exactly. Steel is "unstable" at room temperature (it demixes very very slowly into Fe and Fe3C) but there are tons of tables for its mechanical properties. Tigraan^{Click here to contact me} 17:20, 6 November 2016 (UTC)[reply]

HgH₂ mercury(II) dihydride.
Sleigh (talk) 05:01, 7 November 2016 (UTC)[reply]

The mechanism of decomposition appears to be loss of one hydrogen atom, which then reacts with other hydrogen to yield H²: HgH₂→H + HgH ; HgH → Hg + H (HgH is even less stable) ; H + HgH₂→H₂ + HgH; The overall reaction is exothermic, and once you start forming loose hydrogen atoms it will go out of control fast. This from http://pubs.rsc.org/en/Content/ArticleLanding/2005/CP/b412373e . The solid is made from molecules fairly weakly held together. So it is probably not stabilising the molecules much. This suggests that it is not reorienting or randomising the molecules that makes it unstable. So decomposition could be a long way from the theoretical melting point. The molecule is quite symmetrical, and so it can have a high melting point. cf HgCl2 Graeme Bartlett (talk) 06:06, 7 November 2016 (UTC)[reply]

Perhaps the mercurophilic intermolecular forces have a stabalising tuning effect on the Hg-H bonding? You could be right regarding the high melting point, but the dipoles in HgCl2 is very much stronger than in HgH2, and what about carbon dioxide which is also a very symmetrical molecule? Plasmic Physics (talk) 10:26, 7 November 2016 (UTC)[reply]

The Hg to Hg bonding seems to very very weak, and you don't end up with a HHgHgH molecule at any point. That paper only says dispersion forces are holding the solid together, not any extra Hg-Hg stabilisation. Did you look at the Xuefeng Wang and Lester Andrews paper? Graeme Bartlett (talk) 10:47, 7 November 2016 (UTC)[reply]

Yes, I did look at their paper. Metallophilic forces are dispersion forces. I also looked at their paper published one year earlier, titled: Solid Mercury Dihydride: Mercurophilic Bonding in Molecular HgH2 Polymers. Plasmic Physics (talk) 18:11, 7 November 2016 (UTC)[reply]

Carbon dioxide melts at -56° which is quite elevated compared to other molecules with similar or higher mass, but less neat shapes. eg Oxygen difluoride, H2Se, ClF. HHgH has a somewhat similar mass to Polonium hydride which melts at -35°. But HHgH being symmetrical would have a higher theoretical melting point. A theoretical boiling point could be comparable between the two, but is made more complex because of dimers and trimers. Graeme Bartlett (talk) 21:56, 7 November 2016 (UTC)[reply]

The bomb calorimeter is used to get the enthalpy of decomposition on heating of samples of compounds. Heating is supplied by a wire electric heater inside the bomb, so I suppose you could try, depending on the sample's thermal stability, to superheat the sample and get some of the information you're looking for. loupgarous (talk) 02:33, 9 November 2016 (UTC)[reply]

Borrachero tree

Borrachero tree should redirect to Brugmansia aurea or simply to Brugmansia? This article suggests the second option for example. — Ark25 (talk) 10:38, 6 November 2016 (UTC)[reply]

That article describes a plant with yellow or cream flowers, growing in Colombia. When you eliminate the species with red flowers, and those which don't grow as far north as Colombia, you are reduced to either B. aurea or B. arborea. Wymspen (talk) 11:48, 6 November 2016 (UTC)[reply]

Thanks. I don't understand what's with the red flowers though. — Ark25 (talk) 16:57, 6 November 2016 (UTC)[reply]

There are seven species of Brugmansia - several of which have reddish coloured flowers, rather than the yellow flowers of B. aurea. Have a look at Brugmansia sanguinea - the name gives a good clue to the blood red colour. Wymspen (talk) 09:16, 7 November 2016 (UTC)[reply]

What is the scientifical explanation for that phenomenon?

I saw in this video on youtube (and also on facebook) a group of Indian people who makes some things which look like dangerous things (such as: getting a lot of beats which normal body can not survive after them and getting an heavy hammer into the chest or one of them was run over by car and motorcycle at the same time - on the torso and the head and laying on standing nails with naked upper bodies), but after that in the past I asked here about phenomenon of putting hand into boiled oil by Indian you linked me the trick behind it and then I saw that every one can actually do that, so then I started to cast doubt in any exceptional phenomenon. What is the explanation in this case? 93.126.88.30 (talk) 15:18, 6 November 2016 (UTC)[reply]

The usual explanation for many of these things is the difference between force and pressure. The human body can withstand a great deal of force, so long as it is distributed over a wide area. The bed of nails provides many not-so-sharp nails, to spread the force over a wide area. The hammer also spreads the force out. Replace that with a pick, then apply the same force, and it would cause major damage. StuRat (talk) 15:35, 6 November 2016 (UTC)[reply]

See also our Bed of Nails article. CodeTalker (talk) 16:31, 6 November 2016 (UTC)[reply]

And see Leidenfrost effect for the "boiling oil" phenomenon. Tevildo (talk) 22:13, 6 November 2016 (UTC)[reply]

You need strong core muscles to be able to do this safely. Your core muscles will absorb the shock waves, and deflect them so that your organs don't get injured. To deal with more static pressures, as StuRat points out, what matters is that the force exerted on your body will be spread out, but to make sure this indeed happens when a car is running over your belly, you need to be able to keep your core muscles tight. If these exercises are not too easy for you, you probably won't be able to do that. Count Iblis (talk) 03:12, 7 November 2016 (UTC)[reply]

Main Battle Tank

Hello , what is the highest speed of a target that tanks with the best fire control system can hit ? 149.200.231.84 (talk) 16:21, 6 November 2016 (UTC) In other words can tanks hit the opponent tanks driving at 45 km/hr speed ? 149.200.231.84 (talk) 16:45, 6 November 2016 (UTC)[reply]

In theory a tank can hit a target going at any speed - even a jet aircraft flying past at low level. However, the faster the target, the greater the element of chance in actually scoring a hit. Wymspen (talk) 16:52, 6 November 2016 (UTC)[reply]

Quantitative details about the capabilities of modern weapon systems are not generally released to the public. It is not likely that you, or anyone, will be able to find a reliable published source that provides very exact information. For example, read about how the term "technical data" is defined and used in this context; and read all about the rules that pertain to it. You can find out, for example, that a commercially-available motor exists which can accurately track some angular velocity ...but you won't reliably be able to determine that weapon-system-X uses motor-model-number-Y ... and so on.

You can try to "guess" at the limitations by estimating engineering parameters from first principles of physics and from generic engineering concepts, but if you're honest with yourself and your math, you'll probably end up with an answer "somewhere in the range between DC to light" - or, to use a little less jargon, somewhere between negative-infinity and positive-infinity.

Some great resources to read include Jane's numerous publications, and the website of the Federation of American Scientists. Both resources publish perspectives on defense systems, and often link to specific programs provided by contractors like Raytheon's AFATDS and to public data from DTIC, like this public program review (2005) of the technology development during the development of the Abrams tank. That report, written by some very well-credentialed authors, includes the startling and fascinating claim that the upgrade to the M1A2 digital computer fire control system quantitatively degraded fire rate and fire accuracy compared to the earlier M1A1.

To quote one other author who holds notoriously strong opinions on the efficacy of digital computers, "I have not yet made up my mind on how I feel about relying on a digital computer during critical phases..."

Nimur (talk) 16:03, 7 November 2016 (UTC)[reply]

Thank you very much 188.247.77.185 (talk) 20:21, 7 November 2016 (UTC) Great information , but I supposed an idea and I tried to test it , that if battle tanks can protect themselves by driving at high speed , so we can say that the differences in fire range and accuracy between different types of tanks can be cured by this technique .92.253.23.28 (talk) 04:36, 8 November 2016 (UTC)[reply]

Well, fire control systems function by taking the following and predicting where you need to aim the gun: Gunner's speed and heading; target's speed and heading; round velocity; distance to target; weather conditions; elevation difference. So the question of accuracy is a question of how well these parameters can be measured or (in the case of round velocity) predicted. And I completely agree with Nimur, that how accurately this can be done is going to be a state secret. Driving faster and faster will decrease the gunner's accuracy, since discrepancies between measured and actual velocity translate to larger distances between the target and the round's trajectory as target velocity increases. But without technical information about an MBT's fire control system and the fidelity of a round's nominal velocity, we can't know how fast you'd have to go to seriously undermine a tank's accuracy. I'd imagine a more effective technique, if possible, would be a jinking maneuver, which we don't seem to have an article on. Basically, this consists of making random changes to your direction of travel, and is (or at least was) taught to combat pilots as a method of avoiding direct-fire anti-aircraft weapons. Imagine I'm fleeing from an M1 Abrams at a range of 3000m, which is a bit over 2 seconds of travel time for its APFSDS round. If I'm traveling at 30m/s, and I turn 90 degrees at the moment the tank fires, then by the time that 2 seconds has passed, I'll be about 85 meters from where I would have been, had I continued in a straight line, and that round will almost certainly miss. There's nothing an unguided weapon can do about this maneuver if my movements are truly unpredictable - given a certain opponent, there is always a range and speed I can stay at where I'll always be able to dodge his projectiles. Of course, if I screw up once my vehicle is destroyed. Or my opponent can lob high explosives and hope the shrapnel gets me. Or if he has a lot of friends, they can spread out their fire to take into account my possible choices of direction. But I suppose if you're just trying to reach cover, or you're faster than the tank and trying to get away, this might be your best chance. Someguy1221 (talk) 05:05, 8 November 2016 (UTC)[reply]

Very great , thank you all 37.202.116.92 (talk) 04:09, 9 November 2016 (UTC)[reply]

November 7

Are there any dinosaur museums in BC?

I would like to see some dinosaur fossils. Drumheller in Alberta is kinda far. Swancat (talk) 01:55, 7 November 2016 (UTC)[reply]

Type into Google "British Columbia dinosaur museum", and a number of links come up. Loraof (talk) 02:37, 7 November 2016 (UTC)[reply]

The Royal British Columbia Museum in Victoria, British Columbia, has a collection of paleontology specimens - the listing of dinosaur fossils is pretty short, but bear in mind that those are authentic specimens, not bronze- or plaster- or plastic-resin casts.

If you're looking for big exhibition halls, in the style of the Field Museum's famous T. rex, it looks like you'll have to travel farther afield.

Nimur (talk) 15:46, 7 November 2016 (UTC)[reply]

If you're close to the US border (as many in BC are), you might consider the Burke Museum in Seattle: [1]. StuRat (talk) 18:21, 7 November 2016 (UTC)[reply]

Based on this user's self-description, geolocation, and interests, she is almost certainly the universally banned User:Neptunekh and I plan to file a checkuser ASAP. μηδείς (talk) 06:57, 8 November 2016 (UTC)[reply]

How accurate is the Einstein’s light clock?

A light clock consists of two mirrors / plates in between which a pulse of light is back and forth. A photon doesn’t have acceleration as it always travels at light speed "c" in vacuum from the moment of its creation but a pulse has to halt before it changes its direction (reflected) or be absorbed and re-emitted back. Thus how does a photon (either in particle or wave form) maintain its speed “c” continuously in the light clock during this transit period and what would be its effect on time dilation?2001:56A:7399:1200:433:A9AD:63FC:B7CB (talk) 20:38, 7 November 2016 (UTC)eek[reply]

Here is an explanation of the light clock which brings up some of the more practical aspects of it. Specifically, in this case, this was not ever intended to be a real clock one would build, instead it seems to have been intended as one of Einstein's Gedankenexperiments utilizing an idealized reflective surface rather than any true mirror made of out real material; in physics this is known as the spherical cow approximation, or sometimes a Fermi estimate. In other words, Einstein isn't proposing you build such a clock, merely to consider the implications for the nature of light could you actually build such a clock (which may or may not be actually possible). --Jayron 32 20:50, 7 November 2016 (UTC)[reply]

Think you will find that his description of a light clock was a simple analogy to make his theorem easier to understand. It was not meant to describe a practical clock. --Aspro (talk) 20:55, 7 November 2016 (UTC)[reply]

As an aside. A friend who works at the NPL told me that they now have a 'clock' that is so accurate, that from the time of the big-bang, there would only be one second adrift between the two (which equates to something like 10¹⁸. So asked him which is out by this 'second' since the dawn of creation. The NPL clock or the universe? Took a good minute of a complex explanation before he realized I asked with my tongue in cheek. Yet think about it. Modern atomic clocks are so stable that to take such a clock on vacation it will show a different time. Place one on the International Space Station and the gravitational effect of the earth, moon and sun will effect its time keeping.--Aspro (talk) 21:24, 7 November 2016 (UTC) [reply]

Daniel Kleppner wrote an interesting article on this subject called "Time too good to be true". -- BenRG (talk) 19:03, 8 November 2016 (UTC)[reply]

Thanks for the replies – simplicity is fine but at least we can avoid broken chain in imagination, which creates discontinuity.

...but a pulse has to halt before it changes its direction... - no. Most likely, the reasoning behind this is Newton's second law, which says momentum is differentiable and in particular continuous and subject to the intermediate value theorem. But going from momentum to speed is only true under the assumption that a photon has a mass, which is probably false. Tigraan^{Click here to contact me} 08:48, 8 November 2016 (UTC)[reply]

fracking, Project Mohole

If scientists are worried that fracking could cause earthquakes, then could Project Mohole also have caused earthquakes, perhaps more severe or even volcanoes? — Preceding unsigned comment added by 144.35.45.82 (talk) 22:37, 7 November 2016 (UTC)[reply]

All Mohole extracted were cores (and chips) of the substrate they were drilling through. Fracking is different. To cause a volcano it would have needed to be perpendicular above a bloom. The Mohole Projects did not drill into this type of geology. There was little point in drilling in places like Yellowstone Park as they would have just hit magma. The project was to explore the transition between the crust and mantle. Err. what the WP article doesn’t mention is that the Soviet Union achieved it. So the article needs updating by several decades to counter Wikipedia:Systemic bias--Aspro (talk) 23:26, 7 November 2016 (UTC)[reply]

If you are thinking of the Soviet's Kola Superdeep Borehole, then that didn't reach the mantle either. It was substantially deeper than Mohole, but where it was drilled the crust was thicker so it still remained far from the boundary. As far as I know, no one has yet penetrated the boundary. Dragons flight (talk) 00:13, 8 November 2016 (UTC)[reply]

I've never actually been clear on that. Would drilling a small hole into a magma reservoir actually mean a volcano, or just a slow seep? Can a thin tunnel filled with magma melt itself to become wider, or would the cold just freeze the magma before it got far? Wnt (talk) 00:01, 8 November 2016 (UTC)[reply]

It is going to depend on the details. A low melting point, high gas content magma trapped beneath a layer of high melting point rock might well be able to push itself out if punctured by a borehole. However, I would guess that the appropriate physical conditions for such an event would be rare. In a more typical case, I would expect the magma to cool and plug the narrow borehole much more rapidly than it can be pushed out. Dragons flight (talk) 00:20, 8 November 2016 (UTC)[reply]

Really needs a volcanologist/petrologist to answer this. As I understand it: The overbearing rock (in a geological time frame) is like plasticine. It flexes and bends without cracking so won't let the magma escape until the upward pressure forces the magma through. Hypothetically speaking ( not theoretically speaking, because I don't think a testable theory has yet come about) a borehole may allow magma to enter the over-bearing rock. As magma has loads of H2O, hydrofluoric acid and other things, it would lower the melting point of the rock around the bore hole and thus widening it. Thus inducing a volcano. But it would require the magma to have upwards pressure (positive lithostatic pressure). However, it will not just spurt out of the drilling well head like an oil gusher. This may still take a few thousand years for the magma to break surface. I’m more than hopeful for a Earth Scientist to add to my off- the-top- of-my-head ramblings. It will also depend on the origin. Lava in many Pacific locations is thin and runny whereas the lava from the west coast of the US is thick with a different chemical composition.--Aspro (talk) 00:52, 8 November 2016 (UTC)[reply]

Definitely beyond my pay grade! a simple-minded notion of a hole would have it at a bit over atmospheric pressure, i.e. far less than any magma reservoir. But a real well usually (always?) has drilling mud to raise the pressure. But that mud isn't as dense as the rock, is it? So the pressure at depth might be ... less? Of course, mud and magma don't really mix... drilling into 300 C rock is apparently a technical feat only recently attempted according to one of those articles. I suppose though that drilling into magma or into oil have some common features, that a blowout is possible if the pressure isn't handled correctly. Wnt (talk) 02:15, 8 November 2016 (UTC)[reply]

Can't answer this by point -by-point yet just add other comments. As you say, drilling at 300 causes problems for the traditional bits, because steel looses it hardness at around these temperatures. So one has to employ tungsten carbide and industrial diamond teeth. Perhaps a bigger factor is, at these temperatures the lubricating grease and seals on the cutting head break down. Newer synthetic greases and seal materials has bumped that up to about 500 degrees now. Still a long way though, from being able to bore into the very hot rock immediately overbearing the magma – which is why I said this was purely hypothetical according to the current state of technology. The drilling mud should perform until it reaches the critical point of water, where regardless of how much pressure it is under, the water will become vapour. Oil based drilling mud is already used and Barium sulfate increases the density (after the Iraq War , barium sulfate was often injected into the ruptured well-heads in the Kuwait oil fields because it was so heavy, it could overcome the hydrodynamic forces of the crude gushing out). That density of mud and the flow rate brings the rock fragments up to the surface. In the future, new synthetic oil substitutes for water may over come this. A very high flow rate of cooled synthetic mud may also be able to cool the surrounding rock. All very easy to say but difficult to achieve in practice – and very expensive. The mud may be less dense than the rock but magma is viscous, so it is very unlikely to spurt out – unless it happens in Hollywood.--Aspro (talk) 14:47, 9 November 2016 (UTC)[reply]

Fracking involves injecting large quantities of water-based fluids under the Earth. Under the right conditions, those fluids can lubricate existing faults and transfer pressure in ways that allows earthquakes to occur. Deep exploratory drilling like Project Mohole doesn't involve the injection of pressurized fluids the way that fracking does, and so would be unlikely to trigger any faults in the area. Dragons flight (talk) 00:24, 8 November 2016 (UTC)[reply]

Here's an interesting article on what's been going on in Oklahoma recently.[2] ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:42, 8 November 2016 (UTC)[reply]

The actual fracking appears to have at most a minor effect. However, fracking uses a great deal of water, which becomes nasty polluted water that must be disposed of. the industry disposes of it by re-injecting it into the ground, and these re-injection well are highly correlated with subsequent earthquakes. All of this happens in sedimentary rock, generally nowhere near any magma. -Arch dude (talk) 02:08, 9 November 2016 (UTC)[reply]

November 8

Using microwaves or IR to stabilize a compound?

Generally speaking, I tend to think of incoming photons as making a compound more reactive - for example, promoting triplet oxygen to singlet oxygen. But are there cases where, by promoting an electron in an atomic or molecular orbital to a higher level, a compound can actually be made stable against decomposition or other reactions that normally would take place? Wnt (talk) 00:08, 8 November 2016 (UTC)[reply]

There is a technique called Laser cooling ? That can halt chemical reactions. Might even be able to contain anti-matter in a flask without the whole issue blowing up. --Aspro (talk) 01:07, 8 November 2016 (UTC)[reply]

That's a good one, but it's not really what I meant. I wasn't looking to cool an ensemble of atoms - what I was wondering is if, say, you could raise the outermost electrons in potassium atoms to a higher energy level and then have them not react immediately with oxygen because, oh, they're "too far out". yeah, I know, this idea is too far out... All such examples I think of seem like they should be really unlikely to work, but I don't know there aren't some that would be true. Wnt (talk) 02:05, 8 November 2016 (UTC)[reply]

This can happen with noble gases to form excimers. An inert gas with an electron elevated to an s orbital looks like an alkali metal, which can can react with halides, or hydrogen. See helium compounds, neon compounds, argon compounds, xenon monochloride and argon fluoride laser for examples. Look for excimer. Even the extremely weakly bonded dihelium can gain quite a strong bond if the molecule is excited. The problem is that these excimers can decay fast. You are lucky to have something last a second. Normally you would need UV, or electric discharge to excite the electron. Exciting K atoms will get you larger alkali like atoms that would resemble Rb, Cs, Fr, eka-Fr etc. So they are still reactive and not too different.Graeme Bartlett (talk) 08:06, 8 November 2016 (UTC)[reply]

Rydberg matter is what you get when many of these kind of atoms get together. Triatomic hydrogen is only stable when in an excited state. It resembled a trihydrogen cation with an electron on an orbit around the whole ion. Graeme Bartlett (talk) 08:26, 8 November 2016 (UTC)[reply]

What about exciting a molecule to a state, which is made more stable by requiring a forbidden transition to return to the reactive state? Plasmic Physics (talk) 10:45, 8 November 2016 (UTC)[reply]

Actually I think there might be a nucleus like this. In a metastable excited state it does not decay, but if you make it go to the ground state it decays in hours. See Isotopes of tantalum with ^180mTa. Also Americium-242m from which you can make mini nuclear weapons. Graeme Bartlett (talk) 11:02, 8 November 2016 (UTC)[reply]

Name of condition when one side of the face is tensed

I don't mean that one side of the face has spasms, just that it's extremely tense all the time, involuntarily, the side is raised. The person with this can't even close his mount shout. Obviously it's a medical condition, the face is disfigured. It's like a crooked smile.--Hofhof (talk) 13:22, 8 November 2016 (UTC)[reply]

It doesn't perfectly fit your description, but I suspect you are talking about Bell's palsy. Looie496 (talk) 14:32, 8 November 2016 (UTC)[reply]

The more general term is Facial nerve paralysis - for which Bell's palsy is the most common cause, but not the only one. Wymspen (talk) 18:52, 8 November 2016 (UTC)[reply]

Given the image is entitled "bellspalsy" it seems that the OP has reversed the phenomena. Bell's palsy is the inability to contract the muscles of one side of the face, not the tetanic rictus of the other. μηδείς (talk) 02:37, 9 November 2016 (UTC)[reply]

Was the image added by the OP - or by Looie496 when he suggested the condition as an answer? Wymspen (talk) 09:26, 9 November 2016 (UTC)[reply]

I added the image. I think Medeis understood that, though. Looie496 (talk) 14:23, 9 November 2016 (UTC)[reply]

Artificial islands near China, earthquake stability

How well will the artificial islands that China is making stand up to an earthquake? I'm sure they thought about it carefully, and the rewards are worth the risks, but I would like to know. thanks. 144.35.45.79 (talk) 17:08, 8 November 2016 (UTC)[reply]

You might also ask about stability during cyclones and tsunamis. If they laid the foundations on bedrock, their structures would be less likely to be washed away than if they are resting on sand. StuRat (talk) 17:14, 8 November 2016 (UTC)[reply]

It's sort of a moot point. The function of those islands is to enable China to claim territory in the South China Sea. Structural stability doesn't have much relevance to that function. Looie496 (talk) 17:36, 8 November 2016 (UTC)[reply]

I disagree. Temporary structures may not have the same claim as permanent installations. StuRat (talk) 17:59, 8 November 2016 (UTC)[reply]

The diplomatico-juridic case for recently-built structures is probably very weak in either case. It is more about maintaining a de facto presence. But surely, there is a cost-benefit analysis at the bottom of all this and the budget to maintain the structures is a part of it, so structural stability does matter (since the cost goes up if you need to rebuild everything from scratch every three months vs. twenty years). Tigraan^{Click here to contact me} 18:07, 8 November 2016 (UTC)[reply]

Scaling of muscular strength

If a person could grow or shrink in size, maintaining exactly the same proportions, and all other factors apart from size remaining exactly the same, would their muscular strength vary with the square of the linear scale factor? — Preceding unsigned comment added by 109.146.248.86 (talk) 21:05, 8 November 2016 (UTC)[reply]

There's a problem with the assumption that everything else could remain the same. Larger land animals need to spend a progressively larger portion of their resources just to support their own weight, breath, cool themselves, circulate blood, etc., which leaves fewer resources for everything else. See square-cube law. StuRat (talk) 21:14, 8 November 2016 (UTC)[reply]

Why do you have to answer every question with pointless and useless replies? 109.146.248.86 (talk) 21:17, 8 November 2016 (UTC)[reply]

He cited the relevant article for you to read. I'm not sure what else you wish him to do, since that's all that this desk exists for... --Jayron 32 21:21, 8 November 2016 (UTC)[reply]

I partially withdraw my comment since there is indeed a relevant sentence in that article, though it does not fully and explicitly answer the question. My assumption that the reply would be irrelevant was based on the first two sentences and on past experience. 109.146.248.86 (talk) 21:27, 8 November 2016 (UTC)[reply]

Thanks, you just saved me a bunch of typing. Andy Dingley (talk) 21:46, 8 November 2016 (UTC)[reply]

The allometric scaling article is more relevant. Though the comparison there between elephant and tiger quoll is impressive, to me it doesn't seem like the tiger quoll is really much different than various mouse skeletons I looked at online, like this deer mouse here. Wnt (talk) 23:32, 8 November 2016 (UTC)[reply]

From physiological considerations, it seems to me that muscular strength should vary as the cube of body size, not the square. (The contractile force generated by a cubic millimeter of muscle tissue is roughly constant.) Since mass also varies as the cube of body size, the result, applying Newton's formula F=ma, is that animals of all sizes generate similar accelerations. In other words, a mouse, a human, and an elephant can all jump roughly the same distance. It isn't easy to figure out what the literature says about this issue. Looie496 (talk) 14:46, 9 November 2016 (UTC)[reply]

This is a thought experiment that neglects all sorts of toher factors, but it's still interesting. This means that if a man can jump two meters, then he can still jump two meters after he has been shrunk, therefore, the smaller man jumps proportionally further, but not absolutely further. -Arch dude (talk) 18:10, 10 November 2016 (UTC)[reply]

Radiation and cancer

In a nutshell, if ionizing radiation is a known cause of cancer, how is that it's also used in radiation therapy? Specifically, if ionizing radiation can eliminate cancerous cells, how does it outweigh the risk of getting even more serious cancer? At first glance Radiation_therapy#Mechanism_of_action does not clarify this.Thanks.--93.174.25.12 (talk) 21:45, 8 November 2016 (UTC)[reply]

It's a matter of balancing risks. If the therapy saves life, then the small risk of causing a further cancer is considered worth the risk. Great care is taken to concentrate the radiation only on the cancerous tissue as far as is possible. The risk of the cancer spreading is usually a more serious risk than that of causing further cancer. Dbfirs 21:53, 8 November 2016 (UTC)[reply]

This is a question that many oncologists have sought to answer. This is a meta-analysis of clinical trials involving over 40,000 women that looks at both the effectiveness of radiation therapy for breast cancer, and the likelihood that the radiation itself kills the patient. It found that even though radiation therapy measurably increases the risk of a number of deadly diseases, the overall survival of patients who received radiation is better than of those who did not. And further, apparently these side effects have been decreasing over time, presumably as newer techniques and technology minimize risk. Someguy1221 (talk) 22:06, 8 November 2016 (UTC)[reply]

“The risk is considered worth the risk.” I don't think I've seen that construction before. —Tamfang (talk) 17:51, 9 November 2016 (UTC)[reply]

this source describes a few types of radiation therapy; they have different uses. That said, I've never heard much about the precise frequency/wavelength of the photons used. It doesn't seem obvious to me that the frequency best suited for killing a cancer cell is the frequency best suited for mutating DNA or for tumor initiation. For example, high energies can induce a double strand break while much more sedate chemical reactions like thymine dimers from even UV light can lead to the modification of a single base pair. Wnt (talk) 23:43, 8 November 2016 (UTC)[reply]

Gamma and X-rays are chosen not because they're magic tumor-killers, but because they penetrate very deeply into the body. Someguy1221 (talk) 23:57, 8 November 2016 (UTC)[reply]

To treat cancer, you must kill cancer cells, which are similar in most ways to the other cells in the body. Therefore, essentially all cancer treatments are harmful to a patient's non-cancerous cells. Better treatments kill cancer cells more selectively, but as yet there are no perfect treatments. This is why chemotherapy has such nasty side effects. For some cancers, radiation is relatively more selective than other treatments. -Arch dude (talk) 01:53, 9 November 2016 (UTC)[reply]

Discriminating between a normal cell and a cancer cell is hard. But discriminating between killing a cell and causing a carcinogenic mutation doesn't seem obviously impossible, though I'll concede it could be. Wnt (talk) 16:50, 9 November 2016 (UTC)[reply]

Let me rephrase the OP's question so they can understand the answer "If we know that water is the primary cause of drowning, how can we also use it to cure thirst." If the OP can understand the answer to that question, they can understand the answer to their own. --Jayron32 23:23, 9 November 2016 (UTC)[reply]
- Roger Miller said: "Not enough water puts thirst in a man / And just enough water puts joy in his land / But too much water will overflow / And tear the man's land up - so there you go." ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:27, 10 November 2016 (UTC)[reply]
  - Or as Paracelsus said: "The dose makes the poison" --Jayron 32 12:07, 10 November 2016 (UTC)[reply]

Orbicularis oculi

Does the palpebral portion of the orbicularis oculi muscle solely run from the medial superiorly to the lateral, or are there fibres inferior to the eye? Le Sanglier des Ardennes (talk) 22:51, 8 November 2016 (UTC)[reply]

November 9

Wiktionary: "timescape"

At the English Wiktionary, we're struggling to define the physics (astrophysical? cosmological?) sense of the word timescape. One editor put down "A function of time that is dependent on the position of the observer", but this doesn't seem to fit some of the quotations we have found: see "Citations:timescape". David Wiltshire uses the term in his concept of "timescape cosmology". Please help if you can. Thanks. — SMUconlaw (talk) 17:44, 9 November 2016 (UTC)[reply]

You might perhaps look at Gregory Benford's 1980 novel Timescape, which is the earliest use of the word I know of, and predates your current citations by nearly 30 years. Benford was/is of course a practicing scientist, so probably used (or coined) the word with some firm idea of what he meant by it. {The poster formerly known as 87.81.230.95} 90.211.130.104 (talk) 12:25, 10 November 2016 (UTC)[reply]

Thanks, I am aware of this book. However, we are trying to establish the use of the word in a physics rather than a science fiction context (hence the post at this reference desk). — SMUconlaw (talk) 17:12, 10 November 2016 (UTC)[reply]

Well, you could actually ask Benford directly. He is or was a working physicist with a particular interest in time, so he should know both what he meant by it and what meaning it has in current science, and as an active writer he's easy to contact via his website. {The poster formerly known as 87.81.230.195} 90.211.130.104 (talk) 18:18, 10 November 2016 (UTC)[reply]

Honestly, I don't think it's a word in the usual sense. Wiktionary did a bang-up job of getting a lot of quotes about it, and in almost every one, it sounded like it was being coined afresh. The word is an analogy of landscape to time, clearly. Beyond that, each author is saying something different. It lacks the clear definition of something like spacetime. Wnt (talk) 18:58, 10 November 2016 (UTC)[reply]

The book by Ishi Nobu is self-published and appears to be science fiction or mysticism (or nonsense), not physics.

The other two citations use the word only to refer to a specific model called "timescape cosmology". It was published in PRD, which is a prestigious journal, but there's no Wikipedia article on it, and I'm almost sure it has very little buy-in among cosmologists since it (deliberately) violates the cosmological principle which almost all cosmologists believe is correct.

"Timescape" isn't in the standard vocabulary of physics. Any paper that used it would have to explain what it meant by it. Of "timescape cosmology", Wiltshire says "The word 'timescape' captures the idea that it is the relative calibration of ideal clocks in a nonuniform dynamically evolving geometry which is a distinguishing feature of the present cosmology." Based on that, and not really understanding the model, I'd say it means something like "a physical model where the nature of time depends in some way on position." But considering all uses in physics (if there are others), I doubt you could give a definition more specific than "something like a landscape but involving time".

In short, I agree with Wnt. I don't know Wiktionary's rules, but there probably shouldn't be a physics-specific entry for "timescape". -- BenRG (talk) 19:12, 10 November 2016 (UTC)[reply]

Thanks, this is useful. I'll see if anyone else would like to comment here. In the meantime, I'll delete the Nobu quotation. — SMUconlaw (talk) 19:41, 10 November 2016 (UTC)[reply]

90.211.130.104, on a whim I left a message for Benford through Facebook. (The contact form on his website doesn't work.) Let's see if he responds. He doesn't seem to be very active on Facebook – I only see posts from 2014. However, that could be because I am not a Facebook friend of his. — SMUconlaw (talk) 20:03, 10 November 2016 (UTC)[reply]

Can cancer be treated by infecting cancer cells with microbes?

Thinking about the question posted above about why radiation therapy works for cancer led me to an answer that is purely my OR so it's probably not worth posting there. Basically, I think that an offshoot growth that originates from a very complex self regulating system, is typically going to be much more vulnerable to external stresses as it's less adapted for survival under general conditions. This suggests that tumors should be more vulnerable to infections, which leads to the question if cancer can be treated by infecting people with certain microbes. Count Iblis (talk) 21:35, 9 November 2016 (UTC)[reply]

There has been research into using retroviruses to deliver gene therapies that carry "suicide genes," which would induce cell death in cancers during replication. I'm not sure if your adaptation idea bears out, however, or whether cancer cells are "more" vulnerable to infections. --OuroborosCobra (talk) 22:32, 9 November 2016 (UTC)[reply]

See Coley's toxins and Oncolytic virus. There have been anecdotal reports of spontaneous remission after bacterial/viral infection going as far back as before the germ theory of disease. It's an active area of research to this day. It's worth noting that some chemotherapeutics are actually derived from bacterial extracts, such as the anthracyclines. Someguy1221 (talk) 22:41, 9 November 2016 (UTC)[reply]

There are experiments (with mixed success) in which brain tumors are treated by infecting them with the polio virus, which activates the immune system to kill the tumor. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:26, 10 November 2016 (UTC)[reply]

Do you have anywhere the OP can read more about these experiments? --Jayron 32 00:27, 10 November 2016 (UTC)[reply]

This[3] could be a good starting place. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:32, 10 November 2016 (UTC)[reply]

The general term for these types of therapies is immunotherapy. --47.138.165.200 (talk) 10:17, 10 November 2016 (UTC)[reply]

The ultimate cure for cancer may be to program viruses to seek out and destroy the specific genetic mutation in one cancer cell, then let it loose to find others in the body. If the cancer mutates, the virus would then need to be reprogrammed. Once we can program viruses faster than cancer cells can mutate, we should have a cure. StuRat (talk) 18:17, 10 November 2016 (UTC)[reply]

There are a lot of possibilities. There is no subdiscipline of biology that in which intricate cleverness of design and strategy are more important than in immunology! Usually non-replicating viral vectors are preferred, in part so the researcher doesn't get featured in I Am Legend. Just do a PubMed search for viral vector tumor and you'll see some ideas come up. The most recent is "Dipeptide-functionalized polyamidoamine dendrimer-mediated apoptin gene delivery facilitates apoptosis of human primary glioma cells." (see PAMAM) That's a mouthful... but I've seen more complicated ideas. But another recent paper on that list cites some literature suggesting why entirely non-viral options seem more promising to some: [4] Wnt (talk) 19:12, 10 November 2016 (UTC)[reply]

Thanks everyone, this very interesting stuff! Count Iblis (talk) 21:27, 10 November 2016 (UTC)[reply]

November 10

ID of the species

Can anyone help in identifying the species in the picture? Thanks in advance :). Nikhil (talk) 04:23, 10 November 2016 (UTC)[reply]

Assam Roofed Turtle? There seem to be a lot of them in the Kaziranga National Park where this photo was taken. Richard Avery (talk) 07:55, 10 November 2016 (UTC)[reply]

Look like it's turtles all the way down. :-) StuRat (talk) 18:19, 10 November 2016 (UTC) [reply]

Feynman Lectures. Lecture 33. Ch. 33–6 The intensity of reflected light [5]

It is possible to go on with arguments of this nature and deduce that b is real. To prove this, one must consider a case where light is coming from both sides of the glass surface at the same time, a situation not easy to arrange experimentally, but fun to analyze theoretically. If we analyze this general case, we can prove that b must be real, and therefore, in fact, that b=±sin(i−r)/sin(i+r). It is even possible to determine the sign by considering the case of a very, very thin layer in which there is reflection from the front and from the back surfaces, and calculating how much light is reflected. We know how much light should be reflected by a thin layer, because we know how much current is generated, and we have even worked out the fields produced by such currents.

I want to ask again [6] [7]. What does this part of the Lecture mean? How can we use thin layer? In thin layer refracted light beam will infinitely reflect. Second reflected beam will have amplitude $b^{*}\cdot a$ , where $|b^{*}|^{2}={\tfrac {\sin ^{2}\,(r-i)}{\sin ^{2}\,(r+i)}}$ . — Preceding unsigned comment added by Username160611000000 (talk • contribs) 18:11, 5 November 2016 (UTC)[reply]

Feynman Lectures. Lecture 33. Ch. 33–7 [8] . Fig. 33–7

I think image (b) of Fig. 33–7 is not correct, because atoms must oscillate vertically, so secondary waves must be perpendicular to image, producing something like cylindrical surface. There is some component of the field in other directions according to formula (29.1), but it is much smaller. — Preceding unsigned comment added by Username160611000000 (talk • contribs) 18:21, 5 November 2016 (UTC)[reply]

We can split vector E into components situated only in plane perpendicular to direction of propagation (like is shown in Ch. 33–1). We cannot represent vector E as superposition of one vector perpendicular and the second vector parallel to optical axis of crystal. Is it correct? Username160611000000 (talk) 10:21, 6 November 2016 (UTC)[reply]

Antimicrobial resistance

I read part of Antimicrobial resistance. I'm wondering - since bacteria are evolving to be resistant to antibiotics, is it possible that there can be an entirely new type of treatment that attacks the DNA of the bacteria? Bubba73 ^{You talkin' to me?} 01:05, 11 November 2016 (UTC)[reply]

There are indeed antibiotics that attack bacterial DNA, such as furazolidone and nitrofurantoin. There is a particular difficulty in making such a drug specific to bacterial DNA. Nitrofurantoin is able to achieve this specificity by having low toxicity by itself, however bacterial enzymes metabolize the drug into a far more toxic metabolite after it enters the bacterial cell. Now obviously the bacteria can't evolve to not have DNA, but they can evolve resistance to these drugs through other means: The enzyme that activates the drug could be lost or altered; if the drug enters the cell through a transporter, that transporter could be lost or altered; an efflux pump could evolve to force the drug out of the cell; an enzyme could evolve to destroy the toxic metabolites of the drug; a protein could evolve to bind and sequester the drug or its metabolites in a fashion such that it can do no further harm. Someguy1221 (talk) 02:17, 11 November 2016 (UTC)[reply]

As mentioned, there already are. The problem is that you have DNA too. We generally want to avoid harming the patient with antibiotic therapy. The utility of most antibiotics comes from the fact that they selectively target bacteria (and sometimes other microbes as well). Usually this is because they target things that differ between prokaryote and eukaryote cells. DNA, however, is pretty much the same among all known life, so anything that affects bacterial DNA will likely affect your DNA as well. We've discovered some antibiotics that attack DNA, but we use them not for their antibiotic effect, but for chemotherapy. In this instance, their nonspecific effect is useful, although, as with many chemotherapeutic drugs, their usefulness is limited by their adverse effects. --47.138.165.200 (talk) 05:28, 11 November 2016 (UTC)[reply]

What's the name of the star?

This evening at 8:00, in the US Central time zone, I noticed (from right to left) (if I have them right) Venus, Mars, and Jupiter in a straight line pointing to a very bright star, which from my point of view was in the northeast. Which one was it? PS: the local weatherman pointed out the conjunction a few nights ago. The question is asked in good faith. --Halcatalyst (talk) 03:57, 11 November 2016 (UTC)[reply]

According to software, starting closet to the Sun are Mercury, Saturn, Venus, and Mars, but Mercury it probably too close to the Sun to see. But I don't see them pointing at a bright star. Bright stars Altair and Vega are to the north of Mars. Bubba73 ^{You talkin' to me?} 04:17, 11 November 2016 (UTC)[reply]

Not sure you are correctly recognizing planets. The most obvious bright star in the northeast at 8 PM right now is Capella. If you can recognize the Pleiades, Capella will be pretty much directly north of them. Looie496 (talk) 05:12, 11 November 2016 (UTC)[reply]

What is the most polluted?

Hi all,

I'm from France and I've lived in several cities there and other countries. I live now in Davao, Philippines. It's also a city but it's surrounded by the sea and the province side (with a huge quantity of trees). The problem is that the city is filled with old diesel vehicles and I really cannot breath normally every time I walk in a street. Local people seem to have get used to it.

My question is: What is the healthiest place? A city (like Europe) which seemed to be not that polluted (but with almost no forests around and many plants, factories and industrial centers within a few kilometers) or a city like Davao located very near the see and surrounded by green (but filled with polluting vehicles)? Thank you very much for your help! Ericdec85 (talk) 04:46, 11 November 2016 (UTC)[reply]

I've been in this place a few times, the air here is very clean especially if you move a bit away from the town. Count Iblis (talk) 04:51, 11 November 2016 (UTC)[reply]

Not sure if you are referring to the most polluted place in Europe, but worldwide, I would say it would be in Bhutan. The place is a carbon sink, which emits 2.2 million tons of Carbon Dioxide but absorbs 4 million tons a year.[9] JoshMuirWikipedia (talk) 04:54, 11 November 2016 (UTC)[reply]