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June 29

Glasses

Why do so many people wear glasses the older people get. In school at early ages, only a few people wear glasses but every year in school this increases and by the time people are in college or start working, so many people wear glasses. Sometimes it seems like over half the population does. Why?
— Preceding unsigned comment added by 90.194.55.177 (talk • contribs) 00:38, 29 June 2014 (UTC)[reply]

First, some forms of nearsightedness (myopia) develop gradually over one to three decades. Also, there have been suggestions that frequent use of near vision can activate a genetic tendency to nearsightedness, and frequent use of near vision is a feature associated with reading in school. Second, after approximately 40 or 45, presbyopia sets in, requiring reading glasses for reading. Robert McClenon (talk) 02:15, 29 June 2014 (UTC)[reply]

One more point: unless eye tests are arranged by the school or the child's parents, myopia might go unrecognized for a while, because the child simply thinks everybody sees that way. I did. How long it takes for someone to realize might depend on how things are done in the particular school, i.e. how often a child has to read written text from a distance. --70.49.171.225 (talk) 05:45, 29 June 2014 (UTC)[reply]

By the way, in a complex human society with division of labor but without glasses, nearsightedness is not actually a disadvantage. Nearsightedness in adolescence would steer a person into a trade or craft involving near vision, such as weaving or woodworking, and a naturally nearsighted person continues to have good near vision after the onset of presbyopia. In a complex human literate society without glasses, nearsightedness may have actually been beneficial, because one could become a scribe, a high-status occupation. So evolution in humans never selected against nearsightedness. Robert McClenon (talk) 02:15, 29 June 2014 (UTC)[reply]

This presupposes that the nearsighted actually see better up close than those with normal vision. Is this true ? StuRat (talk) 15:55, 29 June 2014 (UTC)[reply]

What it presumes is that the nearsighted can see better up close after age 40 than those with normal vision. Robert McClenon (talk) 19:54, 29 June 2014 (UTC)[reply]

Purely OR, but I'm short sighted (as we say over here) and 55; I have been wearing glasses since I was 5. I can easily read tiny text as long as it's only two or three inches from my nose. My previously normal-sighted contemporaries all hold things at arms length to read things now, but I just take my glasses off. Alansplodge (talk) 12:20, 1 July 2014 (UTC)[reply]

Of course, a lot of scribes were nominally celibate. —Tamfang (talk) 03:27, 29 June 2014 (UTC)[reply]

See Human eye#Effects of aging and Presbyopia. Red Act (talk) 02:21, 29 June 2014 (UTC)[reply]

I've read speculation that nearsightedness was more prevalent among long-settled agricultural societies like China and much of Europe than areas where hunting or pastoralism required one to see animals is the distance rather than the crops at your hands and feet. Probably read this in Discover, it's not something I'd have seen in a technical source. My nephew, BTW, was fitted for glasses before he was three, his squinting and refusing to sit as far away as the couch to watch TV was a dead giveaway. μηδείς (talk) 18:06, 29 June 2014 (UTC)[reply]

If limiting career options and one's ability to participate in certain recreational activities isn't a "disadvantage", then I don't know what is. It is a categorical error to confuse what is beneficial to a Society with what is beneficial to an individual. You could (obviously) argue that having no education isn't a disadvantage because some jobs (like toilet cleaner) will always be open to the illiterate and uneducated. Saying that something isn't "necessarily" disadvantageous because there are certain situations where it might be useful is less than honest. I'm near-sighted. I don't especially mind it. My glasses cost me $500-$1000, and contacts also cost money which without my disadvantage I could spend on other things. I have to put them on in the morning, not lose them, not break them, and keep them clean. When I swim, I am certainly at a disadvantage. With water sports, the possibility that I'll lose them is a restriction. When I'm traveling, especially away from "civilization" their loss would be a nuisance at best and life-threatening at worst. By the way, the "disadvantage" post is OPINION. I was told that as children's head grow, some eyes deform leading to an inability of the lens to focus - a simple matter of optics. Most people's lenses also harden with age which also limits the ability of the lens to focus. The human lens is a "bag of gel" which changes focus by stretching and relaxing. http://en.wikipedia.org/wiki/Lens_%28anatomy%29 — Preceding unsigned comment added by 173.189.75.163 (talk) 14:50, 2 July 2014 (UTC)[reply]

Considering the original question of does it feel like more people are wearing glasses in this day and age, the answer is yes. The advancements in detecting disorders as well as the societal changes in this day and age play a huge part. Advancements in medical detection [1] have enabled us to help diagnose common diseases such as Presbyopia [2] which is present in many of us and can have very mild affects to our vision later in life. This is most of the cases of people getting glasses when they get older in life as there Presbyopia progresses. The other main reason for the increase in people who wear glasses in today's society is social acceptance and awareness. For instance it was difficult in the 1800's-1900's to convince men to wear glasses simply because it wasn't seen as a masculine thing to do as well as a sign of weakness. People with vision impairments would think that its normal as well because it is the only way they have seen their entire life. When there is an impairment in vision there has been studies [3] that have shown that people can sometimes have heightened senses such as hearing that can be advantageous for them and they sometimes feel that they do not need glasses. DGG8018 (talk) 01:27, 4 July 2014 (UTC)DGG8018[reply]

Note: I removed the <ref> tags in your response (while letting the citation URLs contained within them remain) because they were causing the wiki software to display a big red "Cite error: There are <ref> tags on this page, but the references will not show without a {{reflist}} template (see the help page)" error message. —SeekingAnswers (reply) 02:23, 4 July 2014 (UTC)[reply]

NML Cygni

Move from Portal talk:Astronomy by -- Moxy (talk) 08:17, 29 June 2014 (UTC) ...user notified[reply]

i've got a question concerning the radius of NML Cygni: this article http://arxiv.org/abs/1207.1850 has been quoted as the source of the radius of 1650 R_☉. however, i could not find 1650 R_☉ in the article. all i could find was the following sentence on page 10: NML Cyg’s stellar size of 16.2 mas from Blöcker et al.(2001) was derived using the Stepan-Boltzmann law, adopting Teff=2500 K and a distance of 1.74 kpc. Rescaling this stellar diameter with our distance of 1.61 kpc gives 15.0 mas. well, mas are milli arc seconds, i suppose.

using http://www.wolframalpha.com/input/?i=1.61+kpc*sin%2815+milliarcseconds%29 i get a diameter of 3.613 billion km. this is far from the 2.29 billion km quoted for NML Cygni. can anyone explain, how the 1650 R_☉ were calculated? many thanks --Agentjoerg (talk) 08:01, 29 June 2014 (UTC)[reply]

Note to readers who don't recognize the symbol next to R (or whose browser doesn't render it): it`s the Sun; R_☉ is its radius, about 700,000km on a sunny day.

The Stepan[sic]-Boltzmann law (yahoo only found it in that source) is the Stefan–Boltzmann law.

The 2001 figures return 4.2 billion km, or about 6000R_☉; a factor of 1.8 compared to 2*1650R_☉. - ¡Ouch! (^{hurt me} / _{more pain}) 09:23, 30 June 2014 (UTC)[reply]

Shortcut to balance redox equations

I am trying to find a shortcut method to balance redox equations.

(There is one such method for the simple (non-redox) chemical equations. source: http://www.nyu.edu/classes/tuckerman/adv.chem/lectures/lecture_2/node3.html) (I know the original method to balance an equation by oxidation numbers, but just trying to find if there exists an shortcut method for this. This is not my homework. I can balance these equations, but the original method is too long and boring.)

The problem with this type of equation is - we are not always given H₂O and H⁺ on any of the sides. If we were given all of the resultants and the products, we could solve the equation simply by the algebraic method. (see the link above)

I think that the main thing I have to figure out is - how to determine the side, on which H₂O or H⁺ is, at first sight.

e.g. 1) S + HNO₃ ---> H₂SO₄ + NO
   2) P₄ + NO^3-  ---> PO₄^-3 + NO₂
   3) FeS + H₂O₂ ---> FeO + SO₂

The answers to the above equations is respectively -- 1) S + 2HNO₃ ---> H₂SO₄ + 2NO 2) P₄ + 20NO^3- 8H⁺ ---> 4PO₄^-3 + 20NO₂ + 4H₂O 3) FeS + 3H₂O₂ + 5Fe⁺² + 2H₂O ---> 6FeO + SO₂ + 10 H⁺

Ravishankar Joshi

The algebraic method expresses the problem as a set of simultaneous equations. Your boring job of solving them can be delegated to a computer programmed to perform Gaussian elimination which is a well known routine. Balancing any given chemical equation makes a nice exercise in any programming language; simple BASIC will do because calculating speed will be insignificant. 84.209.89.214 (talk) 11:27, 29 June 2014 (UTC)[reply]

If you're allowed to use both H⁺ and H₂O, then because the hydrogen ions are an oxidizing agent, you can use them to balance the equation. What I remember about them is that on the left side, you use them if the solution the reaction is in is acidic (it's been a while since I've done this).--Jasper Deng (talk) 18:34, 29 June 2014 (UTC)[reply]

The steps for balancing these are as follows:

Step 1) Split the reaction into 2 half reactions (an oxidation and a reduction)

Steps 2-5 apply for each half reaction separately

Step 2) Balance all NON-H and NON-O atoms with coefficients for each half reaction
Step 3) Balance the O with extra water molecules. Simply find the side of the reaction that has too few O atoms, and add that number of water molecules (H₂O) to that side.
Step 4) Balance the H with extra hydrogen ions (H⁺). Simply find the side of the reaction that has too few H atoms, and add that number of H⁺ ions to that side.
Step 5) Balance for charge by adding extra electrons to the side whose charge is too positive.
Step 6) Multiply the half-reactions by some whole number ratio to get an equal number of electrons in the two reactions
Step 7) Recombine the two half reactions, and eliminate any common items on either side of the reaction arrow.

Easy peasy, lemon squeezy. Here's how to do it for #2 above, just for example:

Step 1

split to half reactions

Oxidation half reaction: P₄ --> PO₄^-3

Reduction half reaction: NO₃^- --> NO₂

Step 2

balance for non-O and non-H atoms:

Oxidation half reaction: P₄ --> 4PO₄^-3

Reduction half reaction: NO₃^- --> NO₂

Step 3

Balance O using water

Oxidation half reaction: P₄ + 16H₂O--> 4PO₄^-3

Reduction half reaction: NO₃^- --> NO₂ + H₂O

Step 4

Balance H using hydrogen ions

Oxidation half reaction: P₄ + 16H₂O--> 4PO₄^-3 + 32H⁺

Reduction half reaction: NO₃^- + 2H⁺--> NO₂ + H₂O

Step 5

Balance for charge using electrons

Oxidation half reaction: P₄ + 16H₂O--> 4PO₄^-3 + 32H⁺ + 20e^-

Reduction half reaction: NO₃^- + 2H⁺ + e^- --> NO₂ + H₂O

Step 6

Equalize electrons by multiplying

Oxidation half reaction: P₄ + 16H₂O--> 4PO₄^-3 + 32H⁺ + 20e^-

Reduction half reaction: 20NO₃^- + 40H⁺ + 20e^- --> 20NO₂ + 20H₂O

Step 7

Recombine and eliminate common terms

P₄ + ~~16H₂O~~ + 20NO₃^- + 40 8H⁺ + ~~20e^-~~ --> 4PO₄^-3 + ~~32H⁺~~ + ~~20e^-~~ + 20NO₂ + 20 4H₂O

What you are left with is:

P₄ + 20NO₃^- + 8H⁺ --> 4PO₄^-3 + 20NO₂ + 4H₂O

There ya go. --Jayron32 19:04, 29 June 2014 (UTC)[reply]

Should "12" be "16" at steps 3 and 4? Tevildo (talk) 20:22, 29 June 2014 (UTC)[reply]

So fixed. Wikimarkup makes debugging rather difficult on the fly. --Jayron32 23:15, 29 June 2014 (UTC)[reply]

No, I think that I have not properly explained the question I have. I already know to balance the equation by Jayron32's method. I just want to know if it is possible to determine the number of H⁺ or H₂O (and their side) just by looking at the equation. Please tell me if that is possible. Ravijoshi99 (talk) 10:43, 2 July 2014 (UTC)[reply]

? "Just by looking" implies you think that intelligence is unnecessary in the solution of a problem. BTW, the word is "reactant" not "resultant". I am not sure why you think (IF you do) that Red-Ox reactions all involve H(+) and H2O ?? H(+) → H2O is NOT a redox reaction, which I hope you understand. Hence they have no net contribution to a redox process (although can be part of the chemical reaction chain). You also seem to think that WE can tell you whether you can, just by looking, know what the roots are for the polynomial equation (x-1)² = 0... or x²-2x = -1...Perhaps a moment's thought will lead you to the realization that it depends on YOUR mental acumen. The best way to improve your acumen is to practice the boring stuff. My suggestion is to NOT spend a lot of time on this, as virtually no one does that type of equation balancing for a living. Don't misunderstand, a chemist SHOULD be able to balance a chemical (redox) equation, and an scientifically educated layperson should be able to recognize when an equation is not balanced, but developing speed at this trivial task is probably not time well spent. I can tell you about several times in my career as a chemist where other chemists got it wrong. Once, a company I worked for sold a test kit to analyze water and the reaction they THOUGHT was occurring with their chromate redox reaction, was WRONG! (They thought the product was CrO4(=) and not Cr2O7(=) ) They'd been wrong for years, until I pointed it out. It took several weeks for them to actually realize that I was right. Changing someone's mind is either hard or impossible, unless you go about it diplomatically. They had to change all of their literature, instructions and what a mess it was. As I said, its important that you can mass balance, ballance atoms, and electrons but the reason why you are being drilled is so that you'll retain enough of the skill so that 10 or 20 years from now you won't just accept what some know-it-all says is "right". Without more information on what exactly the reactants and products - all of them - are, how could we possibly answer you? You could always include electrons, e(-) as a reactant in the reduction half-equation and as a product in the oxidation and balance them separately, but as far as I know there's no fool-proof short-cut (nothing is fool-proof, fools are too clever by far). I'm guessing your equations use a lot of H2 → 2H(+) because the book (or instructor) is too busy (or too lazy) to give you more variety. Try this one ClF3 + F2 → ClF5 (product chlorine pentafluoride). — Preceding unsigned comment added by 173.189.75.163 (talk) 15:24, 2 July 2014 (UTC)[reply]

Organs alive outside the body.

I'm curious to know if organs are able to be supported outside the body on a long term basis? We know that organs require a healthy blood supply and circulation in order to survive. With modern medicine and science could this not be achieved? Perhaps technology could go a step further in the form of a machine that actually simulates the environment of the body.

Building on this, could we then support a fully functioning female reproduction system? Infants could be spawned without the need for a human host. How feasible would all all this be?

And whilst we're at it, I know during brain surgery the brain is actually exposed with the patient conscious. So what would happen in a scenario with a large portion of the skull missing, exposing the brain. Would the said individual be able to function normally for a period of time (beside the distress of having your brain exposed of course) If not, what would be the cause of death?

Liver dialysis, artificial hearts, and even hemodialysis remain crude stopgaps that range from inadequate to laughable as substitutes for the actual organs. Parenteral nutrition is not so good either. Wnt (talk) 12:50, 29 June 2014 (UTC)[reply]

This chap (the Daily Mail, I'm afraid, but it's reliable enough in this sort of situation) has survived for a year without half his skull, and isn't at imminent risk of death. He's by no means alone in his predicament, as a Google image search for "missing skull" will confirm. See also human skull, trepanation and craniectomy. Tevildo (talk) 13:16, 29 June 2014 (UTC)[reply]

Note that scientists are working on ways to grow replacement organs for implantation, and this, of course, requires that those organs survive outside the body for an extended time. There was the famous pic of a human ear being grown on a mouse's back, for example. StuRat (talk) 17:09, 29 June 2014 (UTC)[reply]

Don't know if it counts as an example to your satisfaction, but a hernia in the abdominal wall can allow theintestines to escape the inside of the wall and roll around just under the skin of the belly like sausage links on the run. That's a very dangerous condition, but yo can live that way indefinitely. Advertisements for Truss (medicine) were common when surgery was less available. μηδείς (talk) 17:57, 29 June 2014 (UTC)[reply]

You might want to read this. A lot of scientific detail on organ preservation, but not too incomprehensible to regular people. InedibleHulk (talk) 18:18, 29 June 2014 (UTC)[reply]

Okay, back to my second question regarding brain exposure, I mean literally an exposed brain, no tissue or flesh covering it. Is it possible to function normally in this condition? Obviously, aside from the somewhat obvious mental distress. If not, what would be the cause of death?~~

Infection - specifically, encephalitis - would be the big risk in that situation. However, unless Jeffrey Dahmer was responsible for your treatment, some sort of scar tissue would form over the exposed brain (probably from the dura mater, if that was damaged - see this fascinating article about an experiment with rats) to provide an adequate barrier to bacteria. Tevildo (talk) 21:03, 29 June 2014 (UTC)[reply]

Maybe goes without saying, but you'd want to keep large objects off of it, too. And the sun, rain and snow (at least until your scar grows in). Not sure of the physiology, but it seems like a recipe for strokes and seizures, especially in autumn. Unless your hypothetical guy can wear a helmet. Then it's mainly the dirt and swelling. InedibleHulk (talk) 21:22, 29 June 2014 (UTC)[reply]

This equally-fascinating article would appear to indicate that keeping the sun off isn't a major consideration. Incidentally (and perhaps one for the language desk), I note that vivisectionists were using "sacrificed" as a - euphemism? - in 1949. When did this usage first come in? Tevildo (talk) 22:34, 29 June 2014 (UTC)[reply]

Hey, if a priest can get the genetics ball rolling, it's only fair that scientists can appease the Lord of Light every 28 days, if that's their hobby. But yeah...a bit weird. Maybe we can still trust them with our exposed brains, though. Good find! InedibleHulk (talk) 00:49, 30 June 2014 (UTC)[reply]

What would happen if you were sweating in the sun, and it leaked in? Saltwater's no good for metal electronics. Same deal? InedibleHulk (talk) 00:58, 30 June 2014 (UTC)[reply]

Aside from irrelevant aquarium stuff, a quick Google finds this charming brain-eating amoeba. Seems freshwater isn't great, either. InedibleHulk (talk) 01:01, 30 June 2014 (UTC)[reply]

(see primary amoebic meningoencephalitis for that --catslash (talk) 00:17, 1 July 2014 (UTC))[reply]

There's an interesting report http://www.bbc.co.uk/news/health-28061265 on how to chill livers to keep them viable for longer, up to 3 days instead of the normal 24hours. It also hints that the organ is supplied with oxygenated fluid. However, I do not have a Nature subscription to read the full article. CS Miller (talk) 12:25, 1 July 2014 (UTC)[reply]

How to remove rust from stainless steel

I have a pair of spoons that, well, "spooned" in the dishwasher. One apparently has a flaw in the surface that allowed iron to escape, and the area between the spoons stayed wet and formed a rust ring on both spoons. I tossed out the one with the flaw, but can the other be saved ? So far I tried using steel wool on it, which removed some, but not all, of the rust stain. Obviously I want to avoid damaging the surface. StuRat (talk) 17:01, 29 June 2014 (UTC)[reply]

Try this. --Jayron32 17:06, 29 June 2014 (UTC)[reply]

(ec) You could try citric acid. - Lindert (talk) 17:09, 29 June 2014 (UTC)[reply]

Toothpaste is a very fine abrasive. You could polish out the corrosion pits on top with it if these spoons are worth the work. --Kharon (talk) 20:57, 29 June 2014 (UTC)[reply]

No, the spoon with the corrosion pit is in the trash. It's the other one I'd like to save, so my set of flatware isn't so short of spoons that I have to buy another set (you can never find an exact match to replace the missing ones). StuRat (talk) 14:58, 30 June 2014 (UTC)[reply]

I dread to think how long they were left like that for a rust ring to form, weeks at a guess. Anyway the sad thing you have learned is that cheap stainless steel isn't stainless. Greglocock (talk) 22:23, 29 June 2014 (UTC)[reply]

As with cheap copper-based utensils with a coating to make them look like silverware. Used by low-income Irish, when the veneer wore off and the underlying metal became corroded, the folks would discuss "The greening of the ware". ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:26, 29 June 2014 (UTC)[reply]

^{[clarification needed]} Bugs, you're going to have to explain; I don't get it. Nyttend (talk) 02:58, 30 June 2014 (UTC)[reply]

The color copper turns when it oxidizes + "The Wearing of the Green". (Blame Johnny Carson for this one.) ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:20, 30 June 2014 (UTC)[reply]

I thought maybe you were getting at the oxidising, but I'd never heard of "The Wearing of the Green". I thought maybe it was "wear" in the sense of "wow, that's worn; it needs replacement". Nyttend (talk) 03:32, 30 June 2014 (UTC)[reply]

I was told that, at least with acetic acid (vinegar), you have to heat the reactants to start the acid-base reaction. I'm not sure of it, but it may also be possible to use electrolysis to reduce the iron back to the free element (the main problem I can think of is that water might be too easily reduced instead); my book mentions the untarnishing of antique silverware this way.--Jasper Deng (talk) 03:13, 30 June 2014 (UTC)[reply]

I use Bar Keepers Friend to clean rust off my stainless steel sink and burnt spots off of cookware. It's a mild abrasive and an acid. I've never tried it on utensils though. Mr.Z-man 19:00, 1 July 2014 (UTC)[reply]

How to identify quality stainless steel ?

This brings up the obvious Q, how do I find stainless steel flatware that won't have flaws in the surface ? Or do I just buy a really expensive brand and hope they are good ?

1) Is there no way to tell before I make the purchase ?

2) Is there a test I could do immediately when I get them home, so I could return them if defective ? (I'm thinking submerge them in bleach so any corrosion will happen far sooner.) StuRat (talk) 14:58, 30 June 2014 (UTC)[reply]

No, there is no easy way to tell. I'd use either salt water or whatever it you usually use to rot your utensils. Even with engineering grade stainless steels, such as 304, their life in seawater is supplier dependent. 316 is generally pretty good wherever it comes from. Greglocock (talk) 22:16, 30 June 2014 (UTC)[reply]

What does 304 and 316 mean ? Also, are some stainless steel items solid, as opposed to just the surface layer ? StuRat (talk) 16:42, 1 July 2014 (UTC)[reply]

The numbers relate to the ratio of metals in the alloy; see: [4] -Or this interesting PDF —71.20.250.51 (talk) 17:18, 1 July 2014 (UTC)[reply]

Per Stainless steel#Types of stainless steel, stainless steel flatware is usually made from austentitic steel, which is not magnetic. Mild steel (your basic cheap steel), on the other hand, is magnetic. If it doesn't stick to a magnet, it's probably stainless. --Carnildo (talk) 23:38, 2 July 2014 (UTC)[reply]

RGB colour model

In the diagram here, is "Area of triangle / Area of grey shape" a sensible estimate of the proportion of the colour space of human vision that can be represented by the RGB colour model? 86.179.117.18 (talk) 20:23, 29 June 2014 (UTC)[reply]

No, that diagram isn't perceptually uniform. This diagram, which uses CIELUV instead of CIE xy, is better. Surprisingly, I couldn't find an appropriate image on Commons. Note that there are many color spaces called RGB. The diagrams above show sRGB, which is the de facto standard. Adobe RGB and CIE RGB cover more colors than sRGB. -- BenRG (talk) 22:20, 29 June 2014 (UTC)[reply]

Thanks, by the way, in the theory of RGB colour models, is it assumed that the intensity of a colour can vary continuously from zero to any desired brightness? (I understand, of course, that in any practical implementation the intensity of light is limited to what the device can physically pump out). 86.179.117.18 (talk) 22:59, 29 June 2014 (UTC)[reply]

The RGB color model is typically applied to a set of 3 primary color lights whose intensities are each separately controllable in 256 steps from zero to a maximum whose absolute intensity need not be specified. However the relative peak intensities represented by [RGB] = [255 255 255] are in proportions that give a reference White point. 84.209.89.214 (talk) 00:52, 30 June 2014 (UTC)[reply]

sRGB has a nonlinear intensity curve which is only defined for values from 0.0 to 1.0 (or 0 to 255). If you use a linear (energy) scale then the values can be arbitrarily large or even negative. If they are allowed to be negative then any set of three primaries is equivalent to any other, since they are just different bases for the same space of colors. -- BenRG (talk) 02:26, 30 June 2014 (UTC)[reply]