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April 10

Unknown drug powder ID

How can I find out what a packet of unknown drug is made of? I'm not sure it's one drug or several, in powder form. It is extremely unlikely to be an illegal drug. I'm more concerned it is a normal prescription drug. Is the Unknown Substance test here [1] the kind of test I should be looking for? Would a doctor or hospital have access to such a service? How would I find a cheap one? How much would it cost? Thanks. 66.108.223.179 (talk) 00:23, 10 April 2012 (UTC)[reply]

Not directly addressing your question: It is of course your own business what you do with this mysterious substance (unless you manage to break any laws by possessing or using it, and are caught), but (and I am not suggesting that this is your intention) it is generally considered extremely unwise to use pharmaceutical drugs (other than "over-the-counter" ones) unless they have been specifically prescribed to treat a condition one currently has, have been kept in the correct conditions where applicable, and are within their "use-by" date (after which they may lose efficacy), which from what you have said doesn't apply. It's also unwise to keep unused and unneeded drugs around (even if their identity is known) in case someone accidentally or through ignorance (a child, an aged adult with dementia) uses them (though again this may not be applicable in your situation), so the prudent course would be to hand it in to the nearest Pharmacist, Doctor's Surgery or Hospital Outpatients Department for safe disposal (tipping it down the drain may have harmful effect on the environment and may well be illegal), which is a routine duty of theirs (in the UK, at any rate). I used to be employed in the pharmaceutical industry (in a minor administrative role) so I'm sensitised to such matters.

It's possible that further circumstances you haven't disclosed may make your retention and investigation of this presumed drug, perhaps purely to satisfy your intellectual curiosity, a reasonable course, but I can't tell that from what you've said. {The poster formerly known as 87.81.230.195} 90.197.66.44 (talk) 02:57, 10 April 2012 (UTC)[reply]

In the UK you are supposed to take such items to a pharmacist who will dispose of them for you. There are public analysts who will, for a fee, undertake the work you are thinking of, but I'm not sure what they will do if they identify that it is an illegal or controlled substance such as diamorphine. --TammyMoet (talk) 09:34, 10 April 2012 (UTC)[reply]

it sounds to me from "I'm more concerned it is a normal prescription drug" that this could be someone "concerned" about this. Whether a teacher or family member, etc, it seems the poster has good reason to say it's "highly unlikely" to be an illegal drug; meaning they know the person or situation in which the bag was found and what the person or people who could be associated with that are likely and unlikely to have access to. It sounds like the worry/concern is prescription drug abuse. 134.255.115.229 (talk) 10:10, 10 April 2012 (UTC)[reply]

I'm not sure what the best option is for you economically for the test, but if you have a significant quantity of the stuff, you could rule out some silly things right away. To do that, try dissolving a trace of it in water, alcohol, acetone, toluene ... whatever chemicals come easily to hand, with one being quite hydrophobic. (Be careful about potential fumes or fires during mixing, if, say, it turns out to be an oxidizer for homemade fireworks!) If the stuff won't dissolve in anything then it might be something like talc/clay/etc. rather than something to worry about. If the stuff has a really, really strong color, even when you add more of your solvent to a drop, then it's more likely (not guaranteed though) to be an art supply. You can also try burning a trace of it in a flame (well ventilated area...) to see if it is flammable; if not, it might be, say, borax to kill roaches. Most modern drugs are carbon-based and would catch on fire. If it gives intense color in a flame (flame test) this might tell you more about it - lithium salts would be red, sodium yellow and so forth. Detergent powder (sodium lauryl sulfate) also has an intense awful smell. Maybe after a few ultra low tech tests like this you'll feel more comfortable to just come out and ask what the stuff is, and find out if it's nothing. Wnt (talk) 16:53, 10 April 2012 (UTC)[reply]

Without an idea of what it might be, such a commercial service is likely your only chance. Checking whether it is a specific substance is much easier. Solubility in different solvents, shape of crystals formed, color etc can give indications, purifying of the active ingredient is done by dissolving, filtering and recrystallising; combined with a melting point measurement would give reasonable certainty. If you're thinking about prescription drugs with abuse potential, it depends on you're experience with and knowledge of drugs and/or chemistry, really. Pharmacies may sell test kits, can't think of a colour test with home chemicals but solubility, crystallisation and melting point usually get the job done. Ssscienccce (talk) 14:11, 12 April 2012 (UTC)[reply]

To clarify the question: The substance is unlikely to be one that would be abused, although obviously I wouldn't rule it out until I know what it is. I got it from a Chinese medicine man. It has significantly alleviated a sinus problem that doctors have been unable to diagnose, let alone solve. I've been looking at commerical services but they focus on abuse substances and I don't know how to find one that'd consider all possibilities. 66.108.223.179 (talk) 14:56, 12 April 2012 (UTC)[reply]

In that case, I would simply ask my doctor to have it analysed. But that's in my situation in belgium. Your IP is somewhere in NY, I don't know what doctors can and cannot do over there... Ssscienccce (talk) 22:17, 12 April 2012 (UTC)[reply]

In this case the substance is probably organic and you're just wondering what species are involved. Start, of course, by getting the name of the formulation and any ingredients listed - Chinese herbal medicines are specific mixtures sold in international trade. It might not be what they say, but they may not know that... see [2] Supposedly for $35 you can do this, but that's not a retail price. Wnt (talk) 23:45, 13 April 2012 (UTC)[reply]

I bet the labor doesn't come cheap. I doubt it's herbal medicine because it's been so effective and started working so quickly. 66.108.223.179 (talk) 01:14, 14 April 2012 (UTC)[reply]

Mass spectrometry and related techniques can identify most chemical substances, Most tests are automated, and for a pure substance, little work would be needed. You could contact one of the commercial services, ask about the possibilities. Interpreting the results or deciding what tests to run requires some knowledge, more than just how to insert a sample and mail the results... Ssscienccce (talk) 04:46, 14 April 2012 (UTC)[reply]

postcognition

what is the minimum time between when something occurs and when we can react to it? (meaning behave in a way that is statistically differentiable from if it did not occur, based only on normal sensory input and our behavior to it). subconscious reactions OK. 188.157.112.202 (talk) 00:30, 10 April 2012 (UTC)[reply]

The human reaction time depends on the stimulus and activity performed. See the article for a variety of different measured values. --Mr.98 (talk) 01:01, 10 April 2012 (UTC)[reply]

Could a satellite really pack the energy to knock out the entire US's power grid?

See from 1:35 on this game trailer, a satellite knocking out the entire grid for the US.

After, in the Homefront reality, Korea reunifies under the Northern regime, annexes several countries on the Pacific Rim, and our infrastructure / society already deteriorates due to massive national debt and hyperinflation, we get our power knocked out by a single satellite.

(Then we get invaded by their forces up to the western side of the Mississippi River.)

Is our power grid really not that robust? How much energy must be packed in an EMP blast to take out our power from sea to shining sea?

Moreover, what safeguards can we put on our electrical grid today so that we do not go dark from an EMP attack in the future? Thanks. --Tergigress (talk) 06:14, 10 April 2012 (UTC)[reply]

I would expect that multiple nuclear detonations in the ionosphere would be needed. The best defense is probably just to bury all our electrical lines, as shielding them all would be prohibitively expensive. (Burying also has other advantages, like protecting them from weather.) Individual buildings might still be vulnerable, but the surge typically needs to build up over a long length of exposed wire to cause damage. Surge protectors would help, too. StuRat (talk) 06:19, 10 April 2012 (UTC)[reply]

(edit conflict)As far as safeguards go, there is the Farraday cage, which is basically an enclosing mesh or box made of conducting metals. It eliminates any danger from EM pulses, though I suspect it would be difficult (and very, very costly) to implement such a protection on the entire United States power grid. But I would say that the chances of a solar flare-induced EMP causing problems with our electrics is far greater than any hostile attack, seeing as how the GBMD exists and stuff. I actually heard once that a single detonation could wipe on US-based electrics if it went off fairly high up above Kansas. As far as the North Koreans go, the worst they could do (assuming they were suicidal) with their current missile technology is, with luck, cause a ~~rolling~~ black-out or two on the west coast, best-case scenario being as far east as Las Vegas (I would guess). Burying power lines would certainly help, and I've often wondered why this isn't standard practice like it is for water infrastructure. Evanh2008 (talk) (contribs) 06:32, 10 April 2012 (UTC)[reply]

Shallow burial offers little protection against EMP (except perhaps if you are burying the cables inside some other conductor like a metal pipe). During Soviet era tests, buried power lines exposed to EMP also experienced severe surges. Dragons flight (talk) 16:42, 10 April 2012 (UTC)[reply]

For the reasons why most high tension electrical infrastructure is not underground, see Underground transmission — Preceding unsigned comment added by 112.215.36.182 (talk) 08:24, 10 April 2012 (UTC)[reply]

I get that there would extra costs associated with it, but it just seems like those costs would be made up for eventually by not having to replace lines every time a major storm blows through. I dunno. Evanh2008 (talk) (contribs) 08:27, 10 April 2012 (UTC)[reply]

Underground cables can be more expensive to maintain than above-ground lines because you have to dig them up to investigate faults. --Colapeninsula (talk) 08:43, 10 April 2012 (UTC)[reply]

In areas with a risk of salt-water flooding and areas with frequent freeze-thaw cycles, it is typically much more expensive to keep your lines underground because these are factors which can damage underground lines, and as mentioned above there is an extra cost (not to mention extra repair time) when they must be maintained and/or replaced. Also in areas with dense soil such as clay the initial installation cost goes up even further. -RunningOnBrains^(talk) 16:35, 10 April 2012 (UTC)[reply]

I think the main disadvantage is that underground lines cannot dissipate heat as effectively as overhead line which get significant air cooling. Hot wires have much greater resistance. 112.215.36.178 (talk) 04:16, 11 April 2012 (UTC)[reply]

For a long discussion of the effects of an EMP and the fact that our civil infrastructure is largely unprotected from it, see, if you have not already, Electromagnetic pulse. Depending on the size of your weapon and the height at which you detonate it, assuming you can put it where you want, you could do quite a lot of damage to the civilian infrastructure. Whether that is worth spending billions to defend against depends on how likely you think that sort of attack would be and how risky it is. Most non-alarmist commentators think it is a pretty unlikely thing, given that setting off a nuke above the United States in such a way — even firing a nuke on such a trajectory — would be a recipe for massive retaliation, and it would not lessen our ability to retaliate (in part because military technology is shielded from EMP, but also because a lot of our nuclear and conventional forces are not located in the continental United States). --Mr.98 (talk) 19:45, 10 April 2012 (UTC)[reply]

Some general responses to comments: Over a span of years, the installation and maintenance costs for an underground transmission or distribution line far exceed those for overhead. Underground high voltage transmission lines are typically in steel pipes, and buried quite a ways below ground, while distribution lines are typically closer to the surface, often installed by "plowing" them in. The steel pipe and deep burial of transmission lines provide some protection against lightning strikes, if not EMP from nukes. The underground transmission pops up out of the ground every few miles for substations or forced cooling pumping plants, and an EMP could fault them there. A gas insulated substation would provide more lightining/EMP protection than an older style substation with outdoor oil circuit breakers, but an EMP would likely still take out the relay protection and communication equipment required for the line to be in service. Such a line is far more expensive per unit length than overhead. If the underground line insulation fails, it is an expensive and time consuming task to locate the fault and repair it, compared to overhead. Air and porcelein insulators are the insulation of overhead lines. The insulation of an underground transmission or distribution line would fail long before the increased resistance from the heated copper or aluminum became a concern. Underground distribution is often struck by lightning, because it is close to the surface and rarely in steel pipe. A pinhole flaw is all it takes to render an underground line faulted and in need of repair. Edison (talk) 19:08, 12 April 2012 (UTC)[reply]

I disagree that the lines would have to get hot enough for the insulation to fail before the increase in resistance would be a concern. When you're transmitting power across hundreds of kilometers, it won't take a very great change in temperature to make a concerning loss of efficiency. Because of the difficulty cooling the lines underground, there will be stronger positive electrothermal feedback. In copper wiring, a 10°C increase in temperature causes a 3.9% increase in power loss due to resistance. Over the useful life of the transmission line, that will be a significant disadvantage compared to air cooled, overhead lines. 203.27.72.5 (talk) 22:36, 13 April 2012 (UTC)[reply]

But that must be compared with ground temperatures averaging around 60F, and varying little down below the frost line, while above ground temperatures vary dramatically, with many days above 100F in many locations. StuRat (talk) 23:28, 13 April 2012 (UTC)[reply]

Well that depends on where the transmission line is located. My point was that the change in resistance due to the reduced ability to radiate heat is an important consideration (probably more so in an environment like the one where I live; a tropical archipelago of high islands). 203.27.72.5 (talk) 00:21, 14 April 2012 (UTC)[reply]

Sexual vs. asexual reproduction vis-a-vis early evolution

Just a thought that I had regarding the theory of evolution and hypothetical applications to theories regarding the origin of life: Are there any theories among biologists as to at what point (and why) sexual reproduction would have come about in the proverbial "primordial ooze"? In other words, what evolutionary pressures would have caused individuals who developed sexual reproduction to become a dominant faction of life? What are the actual benefits of sexual reproduction as opposed to asexual? In addition, in another world, would there have been any fundamental law of science to prevent asexually-reproducing multi-celled organisms from becoming dominant? Thanks! Evanh2008 (talk) (contribs) 08:25, 10 April 2012 (UTC)[reply]

The article Evolution of sexual reproduction discusses this. --Colapeninsula (talk) 08:44, 10 April 2012 (UTC)[reply]

Indeed it does! I should have looked a little harder before posting this question. Many thanks! :) Evanh2008 (talk) (contribs) 08:47, 10 April 2012 (UTC)[reply]

Sexual reproduction allows mixing of genetic material from different individuals to occur, this results in genetic variation at a far higher rate than in the case of asexual reproduction. Genetic variety in asexually reproducing organisms is purely a result of random mutation in individuals, viable cases of such random mutations are very rare. Compare that to the mixing in sexual reproduction which results in a practically unique genetic makeup for every individual organism (except for identical twins of course). More variation allows faster, more effective adaptation to changing environmental pressures. Roger (talk) 09:03, 10 April 2012 (UTC)[reply]

Note that the article points out that many of the benefits of sexual reproduction can be had from "sexual non-reproduction", e.g. bacterial conjugation. The origins of this latter phenomenon (horizontal gene transfer, really) are probably ancient indeed - I would think that they should predate the existence of species and individuals entirely, in some era when genetic material still freely reacted and interacted in the open before some means was invented to partition it. Sexual reproduction simply combines a thorough horizontal gene transfer with the process of reproduction, so that the process is only needed and effects are only tested in progeny at the earliest stage of development. Wnt (talk) 16:37, 10 April 2012 (UTC)[reply]

And I believe that the that longer time between generations of larger organisms also plays a role. That is, with the short time between generations in single-celled organisms, evolution proceeds at a rate quick enough to adapt to changes in the environment, even by asexual reproduction. However, with large, multi-cellular organisms, with a much longer period between generations, asexual reproduction would not be quick enough for them to adapt, and they would go extinct. Similarly, relying on a high mutation rate alone produces a large portion of defective organisms. While this is acceptable for single-celled organisms, it is not for large organisms, which may not have enough offspring to survive so many deaths. StuRat (talk) 16:49, 10 April 2012 (UTC)[reply]

What are the colors of human cone cells?

I haven't formally studied color vision, but I've read Ware's Visual Thinking for Design as well as various Wikipedia articles. Often, in diagrams, we treat the L, M, and S cone cells as if they simply detect the colors red, green, and blue. Those colors are used for the curves in the diagrams on the right, and they show up in schematics of the retina, such as this mosaic (source).

This representation can be useful, but it seems misleading. The simplest reason is that L and M are assigned very different colors when they are actually very similar (which is why the opponent process must subtract their signals to derive perceptually interesting information). So the broad version of my question is, what would be a more "honest" assignment of colors to L, M and S?

A simple approach would be to consider monochromatic light with the same wavelength as the peak wavelength in the responsivity spectrum. Let's say the peaks are at 564, 534, and 420 nm (taken from Cone-response-en.svg). Using the calculator at http://rohanhill.com/tools/WaveToRGB/ (which outputs to an unknown RGB color space), we get the following representations:

L, 564 nm: Red 196 Green 255 Blue 0 RGB Hex: #C4FF00
M, 534 nm: Red 87 Green 255 Blue 0 RGB Hex: #57FF00
S, 420 nm: Red 56 Green 0 Blue 255 RGB Hex: #3800FF

That's already pretty surprising, right? But I'm looking for more than just the peaks, represented as fully saturated hues. I'm also looking for an intuitive understanding of the width of the curves and the amount of overlap. What I'd like to do is take each responsivity curve, multiply it by a standard white-light spectrum, and map this filtered light into sRGB. I don't think the result would have a direct physical or psychological interpretation, but I think it would still be instructive.

I have no idea how to actually do this. Do you? :-) If so, I think these colors might be useful to add to a few Wikipedia articles. Thanks! Melchoir (talk) 09:42, 10 April 2012 (UTC)[reply]

One other way to consider is what experience do you get when those cones are stimulated alone. It is pretty clear from the ends of the spectrum that you get a violet experience from the S cone, and a red experience from the L cone. The green one may be very hard to stimulate alone without also doing either the red or blue cones at the same time. One way would be in very dim green light perhaps the M cones are active and the others not. Another way would be to bleach L and S with bright red or violet light and then look at green. I call this experience ultragreen. A green colour more saturated than spectral green. Adding our new ideas to Wikipedia articles would be original research, so we don't do it. Graeme Bartlett (talk) 10:13, 10 April 2012 (UTC)[reply]

RGB is severely undersampled. To make matters worse, display monitor pixels aren't monochromatic outputs, either: each color element emits a wide range of wavelengths centered around a particular color. So we have a seriously undersampled spectrum in a non-orthonormal basis. (You discovered this by plotting and noting an overlap in sensitivities). What is really needed for color accuracy is a full spectrum sweep: such as is performed in an optical physics laboratory or a surveillance satellite. But, for engineering reasons, we have "RGB" cameras and monitors; so unless we work for the NRO, we stuff all our visual information into three numbers - so there's gonna be some overlap.

Anyway, the standard technique for taking a lot of data (spectral response at all frequencies) and optimally representing it as a three-element vector (R,G,B) is called... principle component analysis. If you begin down this path, you'll soon conclude (like everyone else) that the math is pretty horrible; but keep the concept clear: you're trying to find just three number that accurately represent an entire continuous spectrum. Then you will postmultiply against your derived eigenvectors - idealized spectral responses for R,G, and B.

If this process was easy, camera, film, and digital imaging companies wouldn't spend so much effort on color accuracy and white balance. If you implement a program to compute optimal eigenvectors, ("idealized spectral response"), you will no doubt discover that they vary with light source, scene conditions/brightness; leakage from invisible infrared, ultraviolet, and other out-of-band light....

Anyway, there's no need for original research. This stuff is well documented in texts and research papers. Color vision is very well understood, both from a physics and from a biological perspective. A great deal is known about color psychology and perception, too. I'll dig up a good introductory text and get back to you. Nimur (talk) 16:03, 10 April 2012 (UTC)[reply]

Direct stimulation of optic nerves

An interesting aspect of the question above is what is the experience of a subject who has only the green cones stimulated? Under normal circumstances this is impossible to achieve because of the overlap with the response of the red cones. I was wondering if any research has been done on this - perhaps by directly stimulating the optic nerve or some chemical stimulus. Would the result be some kind of super green or impossible colour? Spinning Spark 18:26, 10 April 2012 (UTC)[reply]

triplicate carbon copies' carbon footprint

to settle an argument I'm having. What is "greener" (better for the environment): using one triplicate carbon copy document or 100 pages with print from a modern printer/copier? — Preceding unsigned comment added by 165.212.189.187 (talk) 15:08, 10 April 2012 (UTC)[reply]

This question cannot be answered with surety as the content is unspecified. For example, if there is sparse text on each of 100 pages, the modern printer/copier would use minimal toner/ink and suffer minimal wear, and the carbon footprint of making the paper in each case would roughly be the same. Note that this is not about the actaul carbon in either the carbon paper or the modern printer ink or toner - this is not an environmental impact. What you should consider is the cost of carbon emissions cost of electrical power used to make the paper and ink, plus the chemicals used to bleach the paper. But if large areas of solid black are on each of 100 pages, the ink/toner consumption would be huge. The carbon footprint of manufacturing the printer - should this be included? Wickwack120.145.191.82 (talk) 15:19, 10 April 2012 (UTC)[reply]

Ok, regular text on all pages. yes, the footprint to MAKE the paper plus the toner plus the printer vs. that of the carbon copy doc. plus the machine used to make the carbon doc. — Preceding unsigned comment added by 165.212.189.187 (talk) 15:36, 10 April 2012 (UTC)[reply]

It seems to me as with many questions of this sort, it's almost impossible to answer in any meaningful way since the answer will likely depend on the assumptions you make about the particular scenario. For example to use an extreme scenario the carbon footprint of paper could easily vary significantly if we compare a case where a company gets the paper directly delivered from the paper factory next door (who's power primarily comes from low carbon footprint sources like solar, wind, hydroelectric, nuclear, geothermal and wood pulp comes from nearby sustainably managed plantations); to another company who's office manager buys a few packets of paper every fortnight from the shop 150 km away, going there and back in the SUV usually buying nothing else (not particularly effective but perhaps the manager likes the drive), and the paper itself comes from a factory (where it's produced using power primarily coming from coal) many hundreds of kilometers away by truck to the shop. Nil Einne (talk) 16:57, 10 April 2012 (UTC)[reply]

You said "one triplicate carbon copy document". Does that mean a single page of text duplicated 3 times ? If so, I'd expect that to be less of a problem. The carbon paper, after all, will likely end up buried in a landfill, not in the atmosphere anytime soon. StuRat (talk) 16:56, 10 April 2012 (UTC)[reply]

Alright that settles it.165.212.189.187 (talk) 18:32, 10 April 2012 (UTC)[reply]

Don't forget the means of producing typed documents back in the olden days (when I learned to type) was on manual typewriters: no electricity needed, just arms like tree trunks! Very green. --TammyMoet (talk) 19:00, 10 April 2012 (UTC)[reply]

Um, I think you would have to include the CO2 excreted by the typist which would probably far outweigh the carbon footprint of a modern printer due to its energy consumption. Spinning Spark 19:29, 10 April 2012 (UTC)[reply]

And if we're concerned about global warming, let's not forget the methane excreted by the typist. :-) StuRat (talk) 21:34, 10 April 2012 (UTC) [reply]

Maybe we should go back to spirit duplicators, rotary duplicators or hectographs. Dunno how "green" they are, but I like purple – and the smell! Mmmmm. {The poster formerly known as 87.81.230.195} 90.197.66.16 (talk) 22:41, 10 April 2012 (UTC)[reply]

Are there more/fewer/equal quantum fluctuations near massive bodies?

As massive gravitational bodies distort space, is it helpful to think of space as being "denser" or "less dense" in areas? Do massive bodies affect quantum fluctuations of the surrounding space? -Goodbye Galaxy (talk) 15:33, 10 April 2012 (UTC)[reply]

No, space doesn't get "more dense" or "less dense". If you're on the inside of some kind of small chamber in freefall, any quantum mechanical experiment you perform within that chamber will behave the same whether you are in deep space or near a strongly gravitating body. By "small" I mean small enough that tidal forces are negligible. I'm also using the strictest sense of the word "freefall", i.e., the chamber isn't being subjected to any air resistance or other external forces. Red Act (talk) 18:54, 10 April 2012 (UTC)[reply]

And if the chamber is subjected to external forces, then the effects of gravity will be completely equivalent to the effects of acceleration (ie. you can't tell if the chamber is sitting on the Earth's surface or if it is out in deep space accelerating at 9.8m/s/s). See equivalence principle. --Tango (talk) 21:44, 10 April 2012 (UTC)[reply]

To an observer far away from a massive body, there would be more quantum fluctuation per unit time on the surface of the body due to time dilation, but as I understand it quantum fluctuations aren't observable since they are by definition virtual particles that exist in less spacetime than a single planck unit. So to answer your question; an unobservable phenomenon would occur at a greater frequency to certain observers...whatever that means. I don't think it's useful to think of space as being more or less dense, but it is useful to think of spacetime as being more or less curved. 112.215.36.179 (talk) 09:02, 11 April 2012 (UTC)[reply]

Identify the fishes

Type 1 and Type 2. --SupernovaExplosion ^Talk 15:54, 10 April 2012 (UTC)[reply]

Hello, could anyone identify the species, or at least the genus? The photos were taken near river mouth of Digha and the fishes are likely marine. --SupernovaExplosion ^Talk 04:26, 11 April 2012 (UTC)[reply]

I think #1 could be a pike conger eel, and #2 looks like Indian sole (a flounder), but I'm no expert. -- Scray (talk) 04:18, 13 April 2012 (UTC)[reply]

Jungle versus machete

Just having watched a movie where a machete was used to slash through a thick jungle, I find myself wondering how far one could go without having to stop to resharpen the machete. StuRat (talk) 17:04, 10 April 2012 (UTC)[reply]

One kilometre apparently. Spinning Spark 17:29, 10 April 2012 (UTC)[reply]

A trip through the jungle at one kilometer a day would get tedious. And I really liked how the guy in the video was sharpening one side of the blade by stroking toward it, a great way to slice your hand. Edison (talk) 15:41, 11 April 2012 (UTC)[reply]

Yes, if the machete was to slip off the tree, it might do rather more damage than a few sliced fingers as he would be falling towards it. I was also amused by the fact that "Machete safety" is only at video number 7, after the student has been practicing the earlier six videos presumably without safety training. He seems to be cheerfully ignoring his own safety rules both before and after the safety video in any case. Spinning Spark 17:12, 12 April 2012 (UTC)[reply]

That's what I was thinking. I imagine you could carry a few spare machetes and sharpen them all at night. If there's a larger group, they could take turns hacking and dulling their machetes and using each other's paths, thus moving more quickly. I imagine you'd get tired quickly doing the hacking, especially in a hot, humid jungle. StuRat (talk) 17:15, 11 April 2012 (UTC)[reply]

Taking turns does not make faster progress (unless a lone hacker were to become exhausted) because only one person can be leading at any one time. Spinning Spark 17:12, 12 April 2012 (UTC)[reply]

Also, it should avoid having to take breaks to sharpen the machete, which is what this post is all about. Presumably each hacker will carry his own, and when it dulls, another hacker, with another machete, can take over. I'm not sure if the followers would be able to resharpen their machetes as they follow. If not, all could resharpen at night, for multi-day trips. StuRat (talk) 17:40, 12 April 2012 (UTC)[reply]

Is there a video on walking through the jungle carrying a big knife and sharpening it rather than watching for tripping hazards, snakes, etc? Edison (talk) 18:54, 12 April 2012 (UTC)[reply]

I'm thinking there are lots of times when the followers must wait for the leader to hack a bit more, so they could sharpen then. StuRat (talk) 05:44, 13 April 2012 (UTC)[reply]

Depends how much agent orange you packed. 110.151.252.240 (talk) 18:20, 11 April 2012 (UTC)[reply]

does cured cyanoacrylate penetrate skin if dissolved in acetone?

I've been working with acetone with my bare hands in the lab, using a kimwipe to clean glass bonded with krazy glue. I didn't realise that krazy glue contained a cyano group, which I realise is non-volatile but I worry about how it will be metabolised. 216.197.66.61 (talk) 17:15, 10 April 2012 (UTC)[reply]

Not all cyano groups are created equally. Organically-bound cyano groups (called nitriles) are fundementally chemically distinct from their inorganic cyanide cousins. This is similar to the wide chasm between how hydroxy groups behave. In a compound like sodium hydroxide they are highly basic. In a compound like ethanol they are essentially neutral, and in compounds like phenol or boric acid they are acidic. The molecule as a whole needs to be considered to understand its property; not just a coincidental organization of atoms. In the case of nitrile, organic cyanides like this are very unlikely to produce free cyanide ions; they more commonly and readily undergo nitrile hydrolysis to form either amides or carboxylic acids. Just as ethanol produces no free hydroxide ions in your body, nitriles like cyanoacrylate do not produce free cyanide ions. Cyanoacrylates are not fully inert in the body, there are some toxicity issues which is discussed in the Wikipedia article, but these are wholly unrelated to cyanide toxicity. Acetone is the recommended prodcedure for removing cyanoacrylate, and if you are concerned about what may happen if you get it on your hands, use impermiable and unreactive gloves of some sort. --Jayron32 17:42, 10 April 2012 (UTC)[reply]

Note that as said in the article, it can actually be used in surgery with good results. Cyanide can be found at low levels in some foods; it really isn't a problem until you get too much. Wnt (talk) 20:37, 10 April 2012 (UTC)[reply]

Isn't hydroxide a worse leaving group (and hence a stronger nucleophile) than cyanide? Can't hydroxide ions perform SN2 attacks on the tetrahedral center? 216.197.66.61 (talk) 02:37, 11 April 2012 (UTC)[reply]

OH- ions do perform SN2-type attacks (not strictly SN2, but similar idea), but you are not going to break the C-C bond. Instead, what you do is progressively substitute C-O bonds for C-N bonds until you convert the nitrile into a carboxylate/carboxamide. The nucleophilic OH- attacks the carbon which is triply bonded to the nitrogen because that is the most electron-deficient carbon. That's exactly how nitrile hydrolysis works (see link above). If you want more, look up "base-mediated (or catalyzed) nitrile hydrolysis mechanism" in google for all the details. --Jayron32 02:49, 11 April 2012 (UTC)[reply]

There's another problem in 216's premise: nucleophile-strength and leaving-group-quality do not correspond to each other as opposite trends. Halides are good nucleophiles and good leaving-groups (and going down that column on the periodic table they even become better at both modes in parallel). The two modes of reaction involve the reverse mechanistic arrow but the cause of the change and the stability/reactivity differences are not due to the same underlying atomic/molecular properties. DMacks (talk) 15:43, 11 April 2012 (UTC)[reply]

You probably shouldn't be working for very long with your hands in acetone. It's not good for your skin. For limited exposure, get a box of cheap nitrile disposable gloves at your local hardware store. If you are going to be working with it a lot and the parts are big, a pair of non-disposable gloves are better. Either way, use nitrile as it is more resistant to acetone than other glove materials. Don't rely on disposable gloves to protect your hands from long exposure or full immersion.--Srleffler (talk) 03:54, 14 April 2012 (UTC)[reply]

Sun dogs

What about sundogs, the halo around the hot Arizona sun. There could not be any ice crystals, could there? Just saw a beautiful one today, all morning, still going on, Tucson, April 10, 2012. — Preceding unsigned comment added by 24.255.30.15 (talk) 17:23, 10 April 2012 (UTC)[reply]

See the Wikipedia article titled Sun dog. --Jayron32 17:33, 10 April 2012 (UTC)[reply]

Note in particular that the ice crystals responsible for sun dogs occur high in the atmosphere, where the temperature is cold and generally independent of ground-level temperatures. — Lomn 17:38, 10 April 2012 (UTC)[reply]

In fact, because the surface air is dry, you have less diffusion, and therefore a better view of the high atmosphere (whatever conditions may exist at altitude, including sundog-causing ice). For this reason, a lot of interesting aeronomy and physics is practical in the high desert. Tucson is great for amateur astronomers, as is the entire region. The higher you go, the better the view. For example, Lowell Observatory is in Flagstaff; Kirtland, (elevation 5,300'), is home to an Air Force aeronomy and optics center. You can see some amazing photos on Wikipedia that are possible due to the clear air in the troposphere, enabling a clear view at optical bands all the way to the mesosphere. Nimur (talk) 17:48, 10 April 2012 (UTC)[reply]

Carnot efficiency

Is there mathematical proof for why the efficiency of a heat engine cannot exceed the Carnot efficiency? — Preceding unsigned comment added by Clover345 (talk • contribs) 20:55, 10 April 2012 (UTC)[reply]

Have you looked at Carnot efficiency and Carnot engine? The maximum efficiency is derived directly from the second law of thermodynamics. Spinning Spark 21:08, 10 April 2012 (UTC)[reply]

Why the south is hotter?

As you can see the south is hotter than the north along the year, and I ask why? Exx8 (talk) 21:34, 10 April 2012 (UTC)[reply]

I'm replacing this post which was accidentally deleted [3]. There is more land in the north. Large bodies of water take a long time to heat up and cool down, so they tend to reduce extremes of temperature. That means the southern hemisphere doesn't experience as much seasonal change as the northern hemisphere. I'm not sure the southern hemisphere is actually hotter on average. It is hard to tell from that image, but I think the northern summer is hotter than the southern summer and the northern winter is colder than the southern winter, ie. the north is just more extreme rather than colder. --Tango (talk) 21:40, 10 April 2012 (UTC)[reply]

More open ocean. Plasmic Physics (talk) 21:41, 10 April 2012 (UTC)[reply]

Can you please expand on your answer a bit, Plasmic Physics? Does it have to do with the higher albedo of snow-covered land?Anonymous.translator (talk) 22:31, 10 April 2012 (UTC)[reply]

I'd suspect it is less due to albedo, and more to do with the higher heat capacity of water, compared to land. Climate science is very complicated and there are many interacting factors, so it's difficult to definitively attribute an observed effect to one single cause. Nimur (talk) 23:18, 10 April 2012 (UTC)[reply]

I think you're all overlooking the obvious. The lower latitudes (closer to the Equator) get more-direct sunlight and thus more insolation. --Trovatore (talk) 23:21, 10 April 2012 (UTC)[reply]

Why does Exx8 say the south is hotter than the north? Is this based solely on the diagram, or on something else? The diagram steps through the year, one month at a time. It clearly shows that the northern hemisphere is hotter than the southern during the northern summer, and that the southern hemisphere is hotter than the northern during the southern summer, but it is debatable whether this diagram shows one hemisphere is hotter than the other on an annual cycle. Dolphin (t) 23:25, 10 April 2012 (UTC)[reply]

I was assuming he meant that the southern part of the Northern Hemisphere is warmer than the northern part. --Trovatore (talk) 23:27, 10 April 2012 (UTC)[reply]

This discussion is about why the southern hemisphere as a whole is hotter than the northern hemisphere as whole, when averaged over the entire year. Plasmic Physics and Nimur have the right answer, which is also the explanation given in our article on the Southern Hemisphere: "Climates in the Southern Hemisphere overall tend to be slightly milder than those in the Northern Hemisphere at similar latitudes except in the Antarctic which is colder than the Arctic. This is because the Southern Hemisphere has significantly more ocean and much less land. Water heats up and cools down more slowly than land."Anonymous.translator (talk) 00:10, 11 April 2012 (UTC)[reply]

Milder doesn't mean warmer. It just means less variation. 112.215.36.178 (talk) 04:25, 11 April 2012 (UTC)[reply]

(Multiple ECs) I'm not really seeing a South v North hemisphere discrepancy in that graphic. The main difference apparent to me is that the South's temperature zones are a little more stable through the year, which I'd assume is because the land/sea structure is somewhat simpler in the Southern hemisphere, but the areas appear to balance out either side of the equator fairly well. In terms of notionally inhabited areas, there are perhaps more hotter ones in the South than the North, but that's just an artefact of the distribution of landmasses (since we don't think of the oceans as "inhabited"), with more land in the North Temperate zone and more ocean in the South Temperate zone. If anything, I'd guess from the graphic (and this can doubtless be checked with articles elsewhere) that the South averages a little colder because of the polar-positioned Antarctic continent leading to colder Winter extremes – see the 4th map in the Geographical zone article. A contributory factor will be that the Earth's perihelion occurs during the Northern Winter and Southern Summer, which must mildly ameliorate the North's extremes and exacerbate the South's. {The poster formerly known as 87.81.230.195} 90.197.66.16 (talk) 23:29, 10 April 2012 (UTC)[reply]

Look at how cold the North gets in January and how far down it gets cold compared to the Southern hemisphere in July. There is a massive difference. --122.111.0.88 (talk) 13:21, 11 April 2012 (UTC)[reply]

It's only massive in the sense that the land is what is primarily what is changing color, and there is less land in the bottom reaches of the Southern hemisphere than there is in the North. This is what people mean about the water being the important factor. The water temperature is about the same for both. --Mr.98 (talk) 13:38, 11 April 2012 (UTC)[reply]

HCl NaOH titration

Does a buffer form in the HCl - NaOH titration? — Preceding unsigned comment added by 150.203.114.37 (talk) 22:55, 10 April 2012 (UTC)[reply]

See Buffer. You need a conjugate pair of a weak acid or a weak base (that is, a weak acid and its conjugate base, or a weak base and its conjugate acid). In order to answer your homework question, you first need to identify what the weak acid or weak base is in your mixture. If you don't have one, you don't have a buffer. --Jayron32 02:43, 11 April 2012 (UTC)[reply]

So the answer is no because HCl is a strong acid, NaOH is a strong base, and the products NaCl and H2O are not acids or bases at all. — Preceding unsigned comment added by 150.203.114.37 (talk) 02:50, 11 April 2012 (UTC)[reply]

Damn skippy. --Jayron32 02:58, 11 April 2012 (UTC)[reply]

Oh yeah? Well to hell with Peter Pan too! DMacks (talk) 15:34, 11 April 2012 (UTC)[reply]

For those who were as puzzled as I was about Jayron's "Damn skippy" reply, I've done the googling: Here's the Urban Dictionary entry. And thanks for the laugh, DMacks. --NorwegianBlue^talk 05:13, 12 April 2012 (UTC)[reply]