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June 29

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Polaroid film powered radio...

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As a boy (perhaps 20 years ago), a relative gave me a small blue radio that was powered by Polaroid film. You popped a cartridge in and it powered right up. Aside from the in-hindsight ridiculous wastefulness (and likely, terrible expense) of the device, I find myself now wondering how exactly it drew power from the film. Has anyone ever seen one of these? Any ideas? 218.25.32.210 (talk) 00:59, 29 June 2010 (UTC)[reply]

As our Polaroid film article describes, such film cartridges often contained a thin battery. -- Scray (talk) 03:09, 29 June 2010 (UTC)[reply]
I had a Polaroid instant camera when I was a kid. It had a flash but no batteries in the camera body. The batteries were in the film cartridge and would power the flash if needed. Dismas|(talk) 10:44, 29 June 2010 (UTC)[reply]
The polaroid film battery by virtue of its slim form can be a component of a Letter bomb. Cuddlyable3 (talk) 11:41, 29 June 2010 (UTC)[reply]
The Polaroid Corporation introduced the Polaroid SX-70 in 1972. The camera was automated and highly motorized, so it was not necessary to pull out the film pack for each picture as in earlier Polaroid cameras. There was a fresh "Polapulse" six volt battery included in each filmpack, so there was not the "60 second disappointment" of weak batteries preventing the consumer from using up a film pack at about a dollar a picture (big bucks in 1972). The [1] was a flat lightweight zinc--chloride thin-film battery in a thin insulated pack, with two exposed metallic contact areas. It had pretty low internal resistance (so the motors would run smoothly). It was developed by Rayovac, and measured 3.5 by 2.75 inches, by 1/8 inch thick. (8.9 cm x 7 cm x 0.3 cm.) It was easily removed from the spent film pack after the 10 or so exposures were used up. It usually had lots of energy left in it. It could be connected via clip leads or wires taped to the metallic contacts to power a 6 volt flashlight or other device for quite a while. It possibly deserves its own article, since it had significant coverage in many books and magazine artilcles. Alkaline batteries today probably fill the niche it once, but it had a very high peak current capability of 26 amperes, decreasing to a still impressive 5 amperes after 30 seconds and 2.5 amperes after 60 seconds. and seemed able to supply high current for a long time without pooping out, from a very low volume and weight battery. It used conventional LeClanche carbon zinc technology, so it had that handicap to overcome compared to alkaline. I would be interested to see what a similar flatpack with alkaline battery or rechargeable lithium ion chemistry could do for applications where weight and size are at a premium. Metal cylinder batteries like AA or AAA are not always a good fit in designs for miniature devices. Edison (talk) 17:26, 29 June 2010 (UTC)[reply]
Take a look at the battery in an iPhone, Edison. --203.22.236.14 (talk) 09:38, 30 June 2010 (UTC)[reply]
There is no info in the article about its size or weight, or its current capability, except to note that it is not easily replaceable and has disappointing performance. Edison (talk) 19:46, 30 June 2010 (UTC)[reply]

Double slit prob

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Someone performs the double slit experiment. The probability that the electron passes through the first slit is 1/4. The probability that the electron passes through the second slit is 1/5. What is the probability that the electron passes through either slit? 1/4 + 1/5 = 9/20?––115.178.29.142 (talk) 01:26, 29 June 2010 (UTC)[reply]

if they are independent, then with 3/4 probability it will miss the firstslit. Also, it has to miss the second slit, which it does with 4/5th probability. If you want to roll two sixes independently, your chances are 1/6 * 1/6, if you want to miss both slits it is 3/4 * 4/5 = 12/20 = 3/5. So you don't go through either 3/5 of the time, then the rest of the time, 2/5 of the time, you must go through either. 92.224.207.131 (talk) 07:39, 29 June 2010 (UTC)[reply]

The issue that the original math misses is that of that 1/4 of the time when it does pass through the first slit, it also passes through the second slit 1/5 of that time. Isn't the whole point of the dual-slit experiment to demonstrate that an electron has wave-like qualities and go through both slits? These are two overlapping/nonexclusive sets of event probabilities (the electron can go through first and/or second), not additive. Say I have 20 scoops of ice cream, of which 1/4 (5) are chocolate and 1/5 (4) have peanuts. That's not 4 of the ones that aren't chocolate--by the strict probabilities, there is one scoop that is chocolate-with-peanuts. So there are 8 not 9 that are [chocolate and/or peanuts]. DMacks (talk) 14:04, 29 June 2010 (UTC)[reply]

The question is insufficiently precise. What circumstance do the 1/4 and 1/5 measure? If they measure the probability of going through slit A given that slit B is covered, and vice versa, then the question will make sense, but it is still not answerable in general. The interpretation of the double slit experiment only makes sense if 100% of the time that the electron appears on the far side of the screen it went through both slits and their wave functions interfered. In general, the probability that the electron is observed on the far side of the screen at all depends on that interference which occurs after the slits (which is just how weird quantum mechanics is). So the short answer is that the probabilities are not additive, and one needs further details about the configuration in order to determine the probability of observing the electron on the far side of the screen. Dragons flight (talk) 18:25, 29 June 2010 (UTC)[reply]

The inteference term should vanish upon integration. By the time the electron is at the screen, we can write the wavefunction foirmally as:
|psi> = |psi1> + |psi2>
where |psi1> and |psi2> are the (unnormalized) states correspond to the electron moving through hole 1 and 2, respectively. Then we have an interference term for the probability as a function of the position on the screen:
P(x) = |<x|psi>|^2 = |<x|psi1>|^2 + |<x|psi2>|^2 +
2 Re[<psi1|x><x|psi2>]
The last term is the inteference term. However when we integrate over x, tis vanishes because |psi1> and |psi2>, being different eigenstates of the observable that measures through which hole the electron went, are orthogonal. Count Iblis (talk) 19:06, 29 June 2010 (UTC)[reply]

Einstein twin paradox confusion

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What is wrong with my logic here?

  1. An astronaut checks her watch and then leaves Earth at some huge percentage of c.
  2. Rather than go out in a straight line, she orbits the Earth, always keeping it in sight.
  3. After her watch advances a few hours, she returns to Earth and lands.
  4. She finds that three months have passed on Earth.
  5. But if the Earth was always in sight, wouldn't she have noticed that the Earth had gone a quarter of the way around the sun?

Sorry, I'm sure this is elementary Einstein physics, but I just don't get it. Is this different than the typical "astronaut-leaves-earth" scenario because my astronaut is not going in a straight line? So is actually in several different inertial frames? Wknight94 talk 01:27, 29 June 2010 (UTC)[reply]

Well, the astronaut isn't in an inertial frame since you have to accelerate to go in a circle, but that isn't an issue since the astronaut in the original version isn't in an inertial frame either since they turn around and come back (that's the solution to the paradox). The time dilation is compensated for by length contraction, so each observer sees everything as being consistent. I'm not exactly sure what would happen in this case (non-inertial frames are complicated), but if you do the maths it will all work out right in the end. I guess the astronaut will see the Earth's orbit as much smaller than the Earth sees it, so it isn't at all odd for it to do a quarter of an orbit in a few hours. The orbit ought to change shape as the astronaut changes direction, though, so I'm not exactly sure how it would work. --Tango (talk) 01:44, 29 June 2010 (UTC)[reply]
If the astronaut went fast enough - in a circle - and waited long enough, billions of years could go by and the sun would burn out, etc., right? Wouldn't the astronaut see all that in the short time she experiences? Wknight94 talk 03:20, 29 June 2010 (UTC)[reply]
The answer is yes, if this thing somehow happened, she would have noticed that the Earth had gone a quarter of the way around the sun. If she got up to a sufficient fraction of the speed of light, she would see the Earth whirling around the sun at a furious pace. I don't see any contradiction anywhere here. Looie496 (talk) 03:31, 29 June 2010 (UTC)[reply]
Just a note though, if you are close enough to actually see earth, then the acceleration from traveling at that speed in a circle is going to kill you. You need to have a radius of 1 ly in order to keep it down around 1g, but of course you can't go around in a circle that big. Googlemeister (talk) 13:51, 29 June 2010 (UTC)[reply]
What if you had a really thick mattress? Heh, thanks. I guess the ridiculous-sounding result stems from my ridiculous scenario. Wknight94 talk 14:20, 29 June 2010 (UTC)[reply]
Actual observations of physical phenomena do indicate that things progress at unseemly accelerated rates. For example, the decay constant for a relativistic muon observed from an observer on Earth is "not correct" as a result of its high velocity - the infamous Muon Experiment - in fact this is a common lab experiment for physics students. See Muon Lifetime or Measurement of Muon Lifetime and Mass Using Cosmic Ray Showers. Decay constants are a little harder to put in "layman's terms" than, say, seasonal changes on Earth - but yes; we actually can see that the rate of passage of time differs because of the relative relativistic speed of the observer and the muon. If you were moving relatively to the Earth with relativistic velocity, you would similarly see an actual change in the rate that things occur - things like how quickly seasons change, how quickly the Earth orbits Sun, and so on. Nimur (talk) 23:19, 29 June 2010 (UTC)[reply]
This can be understood with General relativity. If she orbits the Earth at this rate, she must accelerate towards it very quickly at all times. This causes time to pass faster on the Earth, so she won't see anything unusual about the Earth clocks moving so fast. She spends half her time accelerating away from the sun, but she's further away from it when she's accelerating towards it. As such, it moves forward in time faster when she's accelerating towards it than backwards when she's accelerating away, so she won't see anything unusual about it burning out so quickly. — DanielLC 04:20, 30 June 2010 (UTC)[reply]
For the astronaut to notice the Earth orbiting around the Sun, she would need a frame of reference other than the Earth and Sun, so she would have to notice the positions of other planets and quite possibly the moon to realize how much time has gone by on Earth. ~AH1(TCU) 21:59, 2 July 2010 (UTC)[reply]

Sumo

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Why don't sumo wrestlers have heart problems? --75.25.103.109 (talk) 02:24, 29 June 2010 (UTC)[reply]

First, perhaps we should ask whether sumo wrestlers have heart problems at a rate higher than Japanese men at a similar age? If you have statistics at hand, please cite them. -- Scray (talk) 02:43, 29 June 2010 (UTC)[reply]
There is actually some literature on this. Apparently the story is that because they combine high levels of exercise with high levels of calorie intake, they develop a lot of subcutaneous fat but not much of the more harmful visceral fat; see PMID 7859591. Even so they have higher levels of hypertension than the Japanese public as a whole, and after they retire, their levels of visceral fat and heart disease skyrocket. Looie496 (talk) 03:46, 29 June 2010 (UTC)[reply]

"Chronical outfit of disposition" - jargon or gibberish

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In the article 78391 Michaeljäger an editor recently added the statement "Its chronical outfit of disposition, defined through the Gödel metric, is 0.4206." Is this jargon or gibberish? If the former, what does it mean? -- Tom N (tcncv) talk/contrib 02:36, 29 June 2010 (UTC)[reply]

Well, since the reference makes no mention of anything remotely connected to it and "outfit of disposition" returns zero ghits, I vote for vandalism. Clarityfiend (talk) 02:51, 29 June 2010 (UTC)[reply]
Well, my google search found a couple more Wikipedia articles where similar statements have been added: HD 171028 b and 1801 Titicaca (reverted, but Google hasn't updated their index yet). Since the Goedel metric is to do with clouds of dust in General Relativity (and asteroids and extrasolar planets aren't made of dust - there is significant EM interaction) and "chronical outfit of disposition" doesn't mean anything that I can find, I'm inclined to agree that it is nonsense. It's good nonsense, though! I've removed it from the two articles I can find it still in. If anyone can provide a reference for it making any sense, feel free to revert me. --Tango (talk) 02:57, 29 June 2010 (UTC)[reply]
This IP editor made about a dozen changes to random articles, and they are all clearly vandalism, and have all been reverted as far as I can tell. End of story. Nothing to see here. Move along. Looie496 (talk) 03:35, 29 June 2010 (UTC)[reply]
Maybe they're hoping to get a job as a Star Trek writer? Nil Einne (talk) 09:28, 29 June 2010 (UTC)[reply]

Blended wing bodies

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Are Blended wing bodies stealthy? --The High Fin Sperm Whale 05:50, 29 June 2010 (UTC)[reply]

The article cited mentions the the B-2 Spirit stealth bomber. Cuddlyable3 (talk) 11:29, 29 June 2010 (UTC)[reply]
That still doesn't really answer the question, since the B-2 is only BWB-ish. --The High Fin Sperm Whale 16:52, 29 June 2010 (UTC)[reply]
Along similar lines, it's fair to say that blended wing bodies are stealth-ish. Certainly blended wing concepts can be used in a stealth aircraft (the aforementioned B2), but stealth aircraft can also forego blended wing concepts (the F-117A). Similarly, it would be very easy to make a non-stealthy blended wing aircraft (the exposed engines on the X-48 certainly fit the bill). Throw in that "blended wing" is a term that doesn't appear to be rigorously defined, and I'm not sure that a "yes" or "no" can possibly be provided to the original question. Even "stealth-ish" is fairly useless. By the same arguments, flying wings are also stealth-ish, as are more traditional designs. — Lomn 18:56, 29 June 2010 (UTC)[reply]
Was the YB-49 "stealth-ish"? 67.170.215.166 (talk) 01:10, 30 June 2010 (UTC)[reply]
Yes, it was. --The High Fin Sperm Whale 01:14, 30 June 2010 (UTC)[reply]

Are dogs capable of love?

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I have two of the sweetest, most wonderful dogs in the world. Do they love me, or am I just projecting human emotions on them? A Quest For Knowledge (talk) 10:04, 29 June 2010 (UTC)[reply]

Definitely. Love is strong affection and attachment to someone or something. Dogs and many other animals are capable of love. 82.43.90.93 (talk) 10:35, 29 June 2010 (UTC)[reply]
Your opinion, anon. --Chemicalinterest (talk) 10:40, 29 June 2010 (UTC)[reply]
A dog is by ancestry a pack hunter. It's emotions are better viewed in terms of Pair bond and pack dominance and subjugation than human emotions. However it's ok for us to love them. Cuddlyable3 (talk) 11:20, 29 June 2010 (UTC)[reply]
Love is a very vague term. Since you're on the science desk I suggest you read chemical basis for love, and of course puppy love, and Cupboard Love. And how about Greyfriars Bobby?--Shantavira|feed me 11:15, 29 June 2010 (UTC)[reply]
Or Hachikō... Physchim62 (talk) 11:22, 29 June 2010 (UTC)[reply]
Rolls eyes skywards. Reading puppy love tells nothing about dogs. Cuddlyable3 (talk) 11:25, 29 June 2010 (UTC)[reply]
Get a cat. Then you know that that feeling is simple condescension with a bit of pity... --Stephan Schulz (talk) 11:51, 29 June 2010 (UTC)[reply]
There's actually a long debate between people who argue that dogs really like humans, and those who say, "oh, they've just evolved to simulate what humans consider love, in order to get fed." I tend to think that if it looks like love, it probably is love, with full reference to the fact that we don't have a strong scientific definition of "love" that isn't muddled down by centuries of cultural baggage. My dog seems to genuinely enjoy my company even when the food supply is not an issue. --Mr.98 (talk) 11:50, 29 June 2010 (UTC)[reply]
I have a good pet named Peacy. My Peacy will do anything in its power for me and will never get tired. It can sing for hours straight without a murmur. It never falls asleep when I am playing with it. It is vulnerable to viruses though; an occasional shot of anti-virus should help that. It eats a food known as Electric Power Pet Food. It drives hard for perfection by means of an internal hard drive. (WARNING: Do not take this content seriously). --Chemicalinterest (talk) 12:47, 29 June 2010 (UTC)[reply]
Right, but we understand exactly how a machine works in this respect. We don't really understand an animal quite so well. And at some level, inquiring into the "does this other highly evolved mammal really love/think/whatever" gets into a question of whether we "really" do any of those things either, and what one really means by that. I haven't seen any good evidence to suggest that dogs are incapable of such things. Dogs do seem to go fairly "above and beyond" what would be required for a minimal level of opportunism, but there's no way to quantify that. --Mr.98 (talk) 13:55, 29 June 2010 (UTC)[reply]
Its mostly a matter of opinion. We can't understand dogs' feelings (whether they have them or not). --Chemicalinterest (talk) 14:07, 29 June 2010 (UTC)[reply]
But there seems little reason to assume they don't have feelings, which would seem to be the harder case to argue. We know that we have feelings, whatever that means, and we see fairly similar behaviors in dogs (excitement, eagerness, friendliness, moping, shame), though we of course don't know if these are "real" feelings or if they are either conditioned reactions to our expectations or just projections. The "they just fake it" argument seems rather unproven to me, the kind of cynical response that people give because they don't like the fact that people seem to get enjoyment out of pets. The "projections" argument seems highly possible and indeed there must be a lot of projection going on, but considering that some of these behaviors seem extremely ingrained to both dogs and wolves ("play" behavior, apparent "joy" at meeting a sibling, "sulking" behavior), it seems unlikely to me that they are totally manufactured by humans. It seems to me to be a larger stretch to posit that dogs lack feelings than it does to say that they have them in some form or another (and obviously there has got to be a lot of variation in the "some form or another"). And again, at some point, depending on how we define these things, it gets tricky to even say whether humans act the way we idealize as (is there really "love" between humans that is not just rooted in some kind of evolutionary drive?). --Mr.98 (talk) 15:01, 29 June 2010 (UTC)[reply]
Since evolution selects for a reasult whether or not the subject has real feelings or not, this suggest that the functional AI hypothesis is correct. The moment you have a system that acts as if it has feelings, it has real feelings. Otherwise it would be huge accident that 4 billion years of natural selection happens to have led to people with real feelings instead of zombies that act as if they have real feelings. Count Iblis (talk) 15:12, 29 June 2010 (UTC)[reply]
Yes, dogs are absolutely capable of love, because they are fuzzy, cute, playful and have oxytocin in their brains. TheGoodLocust (talk) 23:04, 29 June 2010 (UTC)[reply]

free radicals

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Do free radicals actually cause cancer, or is this just a myth that I've heard perpetuated enough times to sound plausible? If so, what kinds?

Thanks - flagitious (talk) 10:55, 29 June 2010 (UTC)[reply]

I accidentally double posted, can someone delete one of these for me - I don't know how. flagitious (talk) 10:58, 29 June 2010 (UTC)[reply]
Done. See Radical (chemistry)#Free radicals in biology. The bits about the harmful effects are all unsourced...which is unhelpful. Vimescarrot (talk) 11:07, 29 June 2010 (UTC)[reply]
Thanks. Yeah, I read that and I was really wondering because the section seems to be screaming [citation needed]. To the google I go! flagitious (talk) 11:58, 29 June 2010 (UTC)[reply]
There's also Reactive oxygen species which does contain some sources. Essentially free radicals will react randomly with DNA in a cell which could potentially damage important genes (like p53) which normally kill cells with damaged DNA through apoptosis. If the damaged cell can't be killed then they can potentially grow into a tumour. 86.7.19.159 (talk) 12:08, 29 June 2010 (UTC)[reply]
Free radicals produce reversible damage to body cells. Normally your body can heal the damage. If there is too many, it can form tumors and other malfunctions of the cell. They contribute to cancer risk, but they probably don't cause it. Antioxidants or reducing agents absorb the shock of the free radicals themselves, protecting your body. --Chemicalinterest (talk) 13:06, 29 June 2010 (UTC)[reply]

However, if some random skin or health food product promises to fight free radicals, be highly skeptical. The best way to fight free radicals is to eat tons of Vitamin C and Vitamin E(particularly a la Linus Pauling). There are very few edible and bioavailable radical scavengers. John Riemann Soong (talk) 16:36, 29 June 2010 (UTC)[reply]

Be skeptical. Ironic that you would then recommend to "eat tons of Vitamin C and Vitamin E". Where is the evidence that taking excessive amounts of vitamins would be any more helpful than "some random skin or health food product"? In fact, the evidence for C and E supplementation is pretty ambiguous. For example, a large randomized, double-blind, placebo-controlled factorial trial of vitamin E and vitamin C showed no benefit for cardiovascular disease and no benefit for preventing cancer. --- Medical geneticist (talk) 17:48, 29 June 2010 (UTC)[reply]
Because it's actually backed up by science? Those studies do not supplement Vitamin C at even the same magnitude of a level that other mammals without our enzyme deficiency would make. Besides, the benefit is lifelong. Naturally if you take it at age 50, where the genetic mutation that will trigger cancer down the road has already occurred, then it is often too late.
People who study Vitamin C megadosage to disprove it don't seem to get the idea: the idea is to make up for our enzyme deficiency. Among other things, the dose could be administered inefficiently -- perhaps direct delivery to the cells is most efficient, seeing as Vitamin C is quite a sensitive reducing agent. It's not just about taking vitamins at the store. John Riemann Soong (talk) 03:24, 30 June 2010 (UTC)[reply]
Could you cite a few references, please? Also, what do you mean by "direct delivery to the cells"? --- Medical geneticist (talk) 19:26, 30 June 2010 (UTC)[reply]

Manganese

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Is the punny name a coincidence? --76.77.139.243 (talk) 12:59, 29 June 2010 (UTC)[reply]

??? Don't under stand what you mean. --Chemicalinterest (talk) 13:03, 29 June 2010 (UTC)[reply]
Is it a WP:Troll? --Chemicalinterest (talk) 13:03, 29 June 2010 (UTC)[reply]
If you want data, look at Manganese#History. --Chemicalinterest (talk) 13:10, 29 June 2010 (UTC)[reply]

No,manganese has nothing to do with the language of mangas... Physchim62 (talk) 13:14, 29 June 2010 (UTC)[reply]

If it's unrelated to manga, where does it come from? --76.77.139.243 (talk) 14:19, 29 June 2010 (UTC)[reply]
You have already been pointed to Manganese#History. Why you think that an ages old metal should bear the name of a relatively new art form is puzzling. --Tagishsimon (talk) 14:31, 29 June 2010 (UTC)[reply]
Not even the first time someone has asked that here. 76.77, you ask a *lot* of questions, at least some of which are easily answered by reading the very obvious articles and/or searching the ref-desk archives or wikipedia/google--please help conserve the limited time of ref-desk participants. DMacks (talk) 15:41, 29 June 2010 (UTC)[reply]
Or ask me on my talk page. My talk page is the haven for all chemistry questions. --Chemicalinterest (talk) 17:04, 29 June 2010 (UTC)[reply]

Making Fatty Acid Methyl Esters

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Hi all. Recently I've been doing fatty acid extractions and I've been wondering how my method works. In short the method involves putting a biological sample + methanol + acetyl chloride + hexane into a test tube and cooking it for a while. As far as I can tell I'm breaking existing ester bonds (triglycerides) and making new ones (methyl esters). Why is the methyl ester the prefered form? Why do the existing ester bonds break? I thought you needed a basic enviroment for that. —Preceding unsigned comment added by 137.224.252.10 (talk) 14:12, 29 June 2010 (UTC)[reply]

"As far as I can tell" how do you know (or at least suspect) that's what's happening?--are you following some published procedure and seeing predicted positive results, or just seeing some physical change that you are trying to rationalize? (I'm not trying to be rude here, just trying to to figure out what your level of experience/background is here.) Our ester article has a section about reactions, including notes relevant to your "Why do the existing ester bonds break? I thought you needed a basic enviroment for that." concern. DMacks (talk) 15:34, 29 June 2010 (UTC)[reply]
The method is published and relatively old (oldest paper I can find that fully describes it is from 1984). The trick is to extract fatty acids (actually triglycerides) and subsequently methylate them. The methylated fatty acids are then analysed using GC. It's a method that's been used in my lab for ages however when I ask the researcher who first started using it how the method actually worked (the chemistry behind it) he just smiled and said " it just works ". My background is not in chemistry as I'm more on the biology side of things. I did have some introductionary organic chemistry courses. Enough to understand a reaction mechanism, just not enough to figure this out. —Preceding unsigned comment added by 137.224.252.10 (talk) 16:00, 29 June 2010 (UTC)[reply]
OK, you're going to need a bit of thermodynamics to understand why the reaction goes in the direction it does... Imagine a triglyceride, something like a letter E with three long dangly bits hanging off. Now those "long dangly bits" (the fatty acid chains) want to move around in solution, and they do, but they are tied together at one end. If you methanolyze the glyceride linkage to form a methyl ester, the chains can move around even more, because they're free at both ends. In technical terms, they have more conformational degrees of freedom. And more degrees of freedom means a higher entropy (positive ΔS). ΔH is as near to zero as makes no difference, because the bonds being broken and the bonds being formed have about the same strength, so the positive ΔS leads to a negative ΔG and the reaction goes towards the methyl esters.
As for the mechanism, it is an acid-catalyzed nucleophilic acyl substitution: the acid comes from the acetyl chloride you add, which reacts with a bit of the methanol to release HCl. Physchim62 (talk) 18:11, 29 June 2010 (UTC)[reply]

Can I ask why the acetyl chloride is so necessary? It seems pretty hazardous for someone who doesn't understand the chemistry to handle. Why not just add up straight up HCl? I don't think nucleophilic or Lewis acid catalysis is involved.

Also, it's enough to think about kinetics without even thinking about thermodynamics. A heavy polyol that already has partially been esterified will find it difficult to nucleophilically attack your methyl ester, while methanol being light and numerous will find it easy to attack triglycerides. It can be a simple case of numbers: if you have enough methanol, you will end up with mostly methyl esters. I bet a lot of tri/di/mono glyceride is left behind though.

Your lab's method is pretty outdated and hazardous and I think needs to be changed. The trend in vogue today is to add some peptide coupling reagents. You just stir in powder and that's it. Far more selective, less harsh and has a much higher yield. You don't have to deal with HCl gas. Ugh.

Btw, base is used when you want saponification. When you use base, carboxylates are not electrophiles so they don't react further so there is usually no equilibrium reaction. However ions don't undergo GC very well... being well, ions that will probably stay in the liquid phase. John Riemann Soong (talk) 18:56, 29 June 2010 (UTC)[reply]

Ooof! If you can't handle acetyl chloride, you shouldn't be in a chemical lab. full stop end of sentence. The method is a standard method for determining fatty acids in triglycerides by gas chromatography, in use for at least forty years now.
And the reason you add acetyl chloride is exactly to avoid using (expensive, fiddly) hydrogen chloride gas. You can't add hydrochloric acid, because you don't want to add any water to your system of fats and hexane, that doesn't really go very well, now, does it. On the other hand, if you add a bit of acetyl chloride (cheap, available in most decent chemical labs, and a substance that any chemistry student worth their salt will already have handled), it will both scavange any stray water in your methanol and it will produce enough HCl in solution (and a controllable quantity, much easier than working with a cylinder) to provide the acid catalyst to protonate the ester carbonyl group and promote the transesterification. Physchim62 (talk) 19:08, 29 June 2010 (UTC)[reply]
Umm I'm sure acetyl chloride is an even more dangerous reagent than HCl. Most biological labs I know do not have it. And the reaction is just so UGLY though! There's so much opportunity to end up with a mix of products. Why not a mild reagent, like DMAP, and a Lewis acid catalyst you can recycle? John Riemann Soong (talk) 19:33, 29 June 2010 (UTC)[reply]
Btw, I'm curious why you don't think acyl chlorides are intimidating reagents to work with. Plus inventory-keeping and storage are such a pain in the ass. The fat-solubility of acetyl chloride is also another issue, making it more hazardous than even concentrated HCl or sulfuric acid. John Riemann Soong (talk) 19:38, 29 June 2010 (UTC)[reply]
Please tell me you joking? or just trying to wind me up? Physchim62 (talk) 19:46, 29 June 2010 (UTC)[reply]
Um no? I don't like reagents that react whenever you add some random nucleophile to it...and God forbid if your fatty acid has a hydroxyl group somewhere like in a lot of natural plant products. John Riemann Soong (talk) 20:19, 29 June 2010 (UTC)[reply]
Come on, acetyl chloride isn't THAT reactive. Sure, it will give you a burn if you get it on your skin, so don't get it on your skin and, if you do, wash it off straight away. It forms very irritating fumes, so use it in a fume hood. End of story. DMAP, on the other hand, is a central nervous system toxin by skin absorption, with a LD50 of just 13 mg/kg by skin absorption [2]: now, for me, that's a reagent to be treated with respect! And you didn't say what Lewis acid you'd be using (nor how you'd stop the Lewis acid being gobbled up by the DMAP): you have plenty of toxicity and corrosivity problems with common Lewis acids.
Inventory-keeping and storage for acetyl chloride are no worse than for any other chemical – just keep the bottle tightly closed, and lock the lab door when there's no one around. Oh, and don't order 50 litres of the stuff when you've just come back from holiday in central Asia, or the DEA might start getting suspicious! In fact, don't order 50 litres of the stuff, full stop: 500 mL bottles are usually more than adequate even if you're using it fairly routinely.
Acetylation of chain hydroxyls shouldn't be a problem, given the large excess of methanol present (methanol is much more nucleophilic than a long chain secondary alcohol). The acetyl chloride is completely consumed, and your only side product is a small amount of methyl acetate, which will fly through your GC column. And therein is the great beauty of methanolysis: it's simple, it's cheap, it's relatively safe and doesn't produce products which need special treatment for disposal AND you can take your reaction mixture and inject it straight onto your GC column to analyse it, no need for work-up. That's why methanolysis is still a routine technique after forty-odd years. Physchim62 (talk) 21:23, 29 June 2010 (UTC)[reply]
comment - if any 'spare' OH groups are exist, then acetylating them will almost certainly smooth the subsequent chromatography.83.100.252.42 (talk) 11:52, 30 June 2010 (UTC)[reply]
The reaction is Transesterification, the driving force is almost certainly an excess of methanol. (though Physchim's comments about entropy of fats are also valid)83.100.252.42 (talk) 11:52, 30 June 2010 (UTC)[reply]

General anesthetic

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What's the fastest-acting general anesthetic? --76.77.139.243 (talk) 14:44, 29 June 2010 (UTC)[reply]

Try Googling fastest general anesthetic. --Chemicalinterest (talk) 14:52, 29 June 2010 (UTC)[reply]
I tried already, and I couldn't find anything. --76.77.139.243 (talk) 15:01, 29 June 2010 (UTC)[reply]
Blunt impact to the head? —ShadowRanger (talk|stalk) 14:58, 29 June 2010 (UTC)[reply]
I'm no expert on anesthesia, from reading General anaesthesia I gather that there are different stages of anesthesia so the answer to your question could be complicated by this. After reading General_anaesthesia#Induction_of_anaesthesia and the links from there I'd have a stab at a volatile anaesthetic such as sevoflurane being pretty fast acting. This paper says it takes between 37 and 70 seconds for "loss of eyelash reflex" to occur depending on the age of a patient. This paper (not free) found it was faster acting than thiopental (an intravenous anesthetic), taking on average 42 seconds to cause loss of eyelash reflex compared to 45 seconds. 86.7.19.159 (talk) 16:40, 29 June 2010 (UTC)[reply]

Why does it list a reaction between tin(II) chloride and hydrogen sulfide to form tin(II) sulfide and hydrogen chloride? I thought that the reaction went the other way. --Chemicalinterest (talk) 15:05, 29 June 2010 (UTC)[reply]

Depends on the concentrations, as with many chemical reactions. Tin(II) sulfide is very insoluble in soluble in water, which means that H2S will precipitate it even from mildly acidic solutions of tin(II): this sort of reaction used to be very important in qualitative inorganic analysis (tin is in group II, cations that form sulfides in acid solution). It takes concentrated hydrochloric acid to push the equibrium back in favour of SnCl2 and H2S. Physchim62 (talk) 16:34, 29 June 2010 (UTC)[reply]
So only concentrated HCl will reverse the reaction? --Chemicalinterest (talk) 16:53, 29 June 2010 (UTC)[reply]
Actually, looking at the figures, bench dilute hydrochloric acid would probably do it. But don't forget that, when you are precipitating tin sulfide, you are only making a stoichiometric amount of HCl, that's rather different from adding hydrochloric acid onto tin sulfide precipitate (when acid will be in large excess). Physchim62 (talk) 19:25, 29 June 2010 (UTC)[reply]

Why doesn't it have an article? --Chemicalinterest (talk) 16:53, 29 June 2010 (UTC)[reply]

What makes you think it exists? Physchim62 (talk) 17:56, 29 June 2010 (UTC)[reply]
Any other metal oxide exists. --Chemicalinterest (talk) 20:24, 29 June 2010 (UTC)[reply]
Hg(0) + Hg(II) is more stable than 2 Hg(I), as can be seen with the reactivity of mercury(I) chloride. As I imagine, it quickly decomposes even shielded from catalytic agents. Any Hg(I)-O-Hg(I) is probably covalent in nature; the oxygen pulls electrons from one mercury(I) atom and returns another pair of electrons to the other. John Riemann Soong (talk) 20:43, 29 June 2010 (UTC)[reply]
Second question: What makes certain substances more stable while others disproportionate easier? --Chemicalinterest (talk) 20:54, 29 June 2010 (UTC)[reply]
Are you asking why Hg(I) chloride doesn't disproportionate spontaneously? I'm guessing it's because the oxygen is actually rather "unhappy" with the whole Hg-O-Hg arrangement and can get much better electron density by proceeding with disproportionation. Whereas chlorides are relatively 'content'. John Riemann Soong (talk) 21:04, 29 June 2010 (UTC)[reply]
This is complete speculation, but with Hg2O, the oxygen does the reducing and oxidising, which it seems apt to do (oxygen can be both electron-donating and electron-withdrawing at times). With Hg(I) Cl, the mercury atoms have to reduce/oxidise each other directly ... which is difficult without the help of say some UV radiation. John Riemann Soong (talk) 21:06, 29 June 2010 (UTC)[reply]
As an aside, Chemicalinterest, you might want to have a look at Frost diagrams, which graphically represent the stability of an element's oxidation states and makes a link between stability and electrode potentials. Our article is pretty scanty, but Google throws up some good examples. If you're after a textbook, check an undergraduate-level textbook, such as Shriver and Atkins' Inorganic Chemistry, for a more in-depth look at them. Brammers (talk/c) 22:44, 29 June 2010 (UTC)[reply]
But why do some substances disproportionate, eg solid NaClO, while others, eg O2, don't disproportionate into O2+ and O2-? Thank you. --Chemicalinterest (talk) 00:01, 30 June 2010 (UTC)[reply]
The favourability of disproportionation is a result of the stabilities of the disproportionation products relative to the disproportionating species (try saying that after a few beers!). O2+ has a stronger bond (check out the O2 MO diagram but with one antibonding electron removed) but bears a positive charge, and O2-, while having a less unfavourable negative charge also happens to have a weaker bond due to the presence of an additional antibonding electron. Hand-wavey, I know, but that's the best I can do. As for your NaClO question, I'm going to use an example with electrode potentinals from Shriver and Atkins:
5 HClO → 2 Cl2 + ClO3- + 2 H2O + H+ is the overall disproportionation
4 HClO + 4 H+ + 4 e- → 2 Cl2 + 4 H2O Eo = +1.63 V
ClO3- + 5 H+ + 4 e- → HClO + 2 H2O Eo = +1.43 V
so Etotalo = 1.63 - 1.43 = +0.20 V. ΔG = -nFEo so ΔG is negative, and so HClO's disproportionation is thermodynamically favourable (note that this doesn't deal with kinetics, which is tougher to predict). On a Frost diagram, you'd see that HClO's point would lie above a line drawn between its two neighbours: this indicates that it disproportionates. Similarly, if a species lies in a 'trough' on a Frost diagram, the adjacent species will (assuming kinetics are OK) comproportionate to give that species. Brammers (talk/c) 08:57, 30 June 2010 (UTC)[reply]
Wrong reaction there buddy. See potassium chlorate; it is formed by the reaction of sodium chlorate with potassium chloride. The sodium chlorate is caused by the disproportionation of the hypochlorite into chloride and chlorate in the high electrolysis temperature. Thank you. --Chemicalinterest (talk) 10:42, 30 June 2010 (UTC)[reply]
And? That's exactly the reaction that Brammers described. You get sodium chlorate at high electrolysis temperatures because the higher temperature increases the rate of disproportionation of the hypochlorite which is initially formed. The thermodynamic products are chloride and perchlorate, but the production of perchlorate by disproportionation is negligeably slow so, in practice, the system stops at chlorate. Potassium chlorate is less soluble than sodium chlorate, which is why it can be prepared by the reaction you describe. Physchim62 (talk) 10:49, 30 June 2010 (UTC)[reply]
It shows the reduction of hypochlorite to chlorine, not chloride. I never can find a standard potential for reduction to chloride. --Chemicalinterest (talk) 14:54, 30 June 2010 (UTC)[reply]
Generally electrode tables only show a reduced set of potentials from which you can work out most others = for ClO3-/Cl- you probably need to combine ClO-/Cl2 (A) and Cl2/Cl- electrode potentials (B) - the value in this case will be A+B/2 - it's related to the Gibbs energy, if you don't know how to combine electrode potentials that would be a separate question...83.100.252.42 (talk) 21:23, 30 June 2010 (UTC)[reply]
Hang on a minute. Hg2O possibly does exist , see http://antoine.frostburg.edu/chem/senese/101/inorganic/faq/missing-mercurous-compounds.shtml
One important thing to note is that "Hg+" is in fact Hg22+, there's a covalent bond. There's probably not an article because it's not stable and not useful.. Search for it in google books, there's some info on it there... It appears it decomposes in light, and even when ground in a mortar.83.100.252.42 (talk) 21:23, 30 June 2010 (UTC)[reply]

what are these strange holes in the ground in afghanistan?

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Here are coordinates,31.63779,65.050017. The town is "Pir Zadeh". Look it up on google maps. There are lines of deep looking holes crossing the terrain, it looks like perforated paper. These lines of holes continue up that river all the way into the mountains. I don't think they are for storing water because they occur in the mountains where there are no fields or orchards. What are these holes? —Preceding unsigned comment added by 75.164.144.81 (talk) 18:45, 29 June 2010 (UTC)[reply]

They're Quanats, an ancient and effective method of channeling water from hills to areas where it is needed for irrigation etc. Mikenorton (talk) 19:22, 29 June 2010 (UTC)[reply]
31°38′16″N 65°03′00″E / 31.63779°N 65.050017°E / 31.63779; 65.050017, fwiw. --Tagishsimon (talk) 21:04, 29 June 2010 (UTC)[reply]

Thank you, thats cool. —Preceding unsigned comment added by 75.164.144.81 (talk) 23:34, 29 June 2010 (UTC)[reply]

Actually it's spelt qanat (as the redirect shows), one of the few words in Scrabble you can spell using a Q but no U. Zunaid 08:27, 30 June 2010 (UTC)[reply]

Soreness from a Shot

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What is the origin of residual pain in your arm after a vaccine injection? Is the soreness from damage to the tissue by the needle or is it something to do with whatever is being injected into your arm? Not a request for medical advice, just idle curiosity. —Preceding unsigned comment added by 72.85.199.192 (talk) 19:40, 29 June 2010 (UTC)[reply]

It likely depends on what vaccine and what soreness you're referring to. The Bacillus Calmette-Guérin vaccine for example is noted for the swelling and scar that results (and I can say from personal experience swelling can be fairly sensitive for a fair while). As noted in our artice, this arises from the body's reaction to the vaccine. Nil Einne (talk) 21:14, 29 June 2010 (UTC)[reply]

Where did this stupid idea that investment is a trade-off between risk and returns come from?

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where did this stupid idea that investment is a trade-off between risk and returns come from? I mean, I see how it applies to someone with a very poor understanding of the market, industry, and the factors, such as reputations and technology involved, but this adage (risk/return tradeoff) seems to be stated as though it applies to everyone, even people who understand, correctly, exactly what they are doing and why it will get a high return, in that case, without any particular risk. Thank you. 92.230.66.154 (talk) 20:58, 29 June 2010 (UTC)[reply]

Lots of people thought that they had a good understanding of the market, but got their fingers burnt! Perhaps Warren Buffet is one of the very few who got it right by getting out! Dbfirs 21:15, 29 June 2010 (UTC)[reply]
CAPM 92.29.119.46 (talk) 21:16, 29 June 2010 (UTC)[reply]
warren buffet is a good example. obviously he doesn't trade off between risk and return (for example he doesn't ever invest in technology at all - he doesn't know for sure what will happen). so, now that we know that it is false to say, in general, that there is a trade-off between risk and return, can we get to my question: where does this stupid idea come from? 92.230.66.154 (talk) 21:21, 29 June 2010 (UTC)[reply]
Sure, he trades off between risk and return, but the key thing he does is make better estimates of return than other people do (which is why he doesn't invest in technology - he isn't a technology expert so doesn't feel his estimates would be better than other people's). He compares his estimates of risk and return to work out a fair price and then buys if the market price is lower than what he has calculated as a fair price. This is called value investing. --Tango (talk) 02:25, 30 June 2010 (UTC)[reply]
This isn't exactly my field and I would question if it's a science question anyway but my impression is even those with great experience and knowledge of the market generally agree that there is connection between risk and return, and particularly you should considered whether your expected return is worth the risk you are taking (and whether you can accept that risk). Note that when it comes to the more informed discussion, I've never heard it claimed that all risky investments have a high return or that all low return investments have a low risk, in fact it's very commonly acknowledged that there are plenty of cases when the risk doesn't correlate well with the return (always in the wrong direction of course meaning the return isn't worth the risk or you can get a better return for about the same risk) and that unfortunately the more clueless investors are often fooled by such thing. People who understand the market well can generally better understand the risk, and importantly may be able to reduce it (meaning low risk investments may not be of great interest to them). Perhaps most importantly (even more so after recent events), I don't think many of those people who do have such knowledge would agree that it's possible to get a high return (all being relative of course for them a high return may be higher then for an 'average' person) without any risk (actually to be blunt it sounds like the sort of thing those inexperienced with the market seem to think usually to their peril). Nil Einne (talk) 21:23, 29 June 2010 (UTC)[reply]
It is just your opinion. Mine is that it is true. --Chemicalinterest (talk) 22:43, 29 June 2010 (UTC)[reply]
Perhaps that stupid idea originated with the fact that some small startup companies have a product that may or not catch on or a small research firm thinks they maybe onto a possible cure for cancer and if they are correct their stock will quickly multiply in value, but if they are wrong their investors may lose everything - such investments are inherently more risky than those in established companies and investments, even in large firms, are generally, more risky than investing in government bonds, which, in turn are less "risky" than investing in material goods like gold. Of course, a lack of growth may be considered a risk as well and I'm also speaking in general terms. TheGoodLocust (talk) 01:53, 30 June 2010 (UTC)[reply]
It isn't a stupid idea. It's true. Consider the following choice: A) I give you £1000, B) I toss a coin and if it's heads I give you £2000, if it's tails I give you nothing. Both those options have the same expected return, but B is much riskier than A. If you present that choice to a lot of people, most will choose A (it doesn't always work for small amounts of money because people have different priorities when it comes to amounts they consider trivial). In order to get the same number of people choose A as B, I'd have to increase the expected return of B. --Tango (talk) 02:25, 30 June 2010 (UTC)[reply]
Oooh careful...if I remember game-theory studies, for the identical average payoff over many trials, many will pick the option of a larger prize (with the risk of none, vs sure-thing smaller prize) but many will pick sure-thing smaller cost (vs chance of none or larger cost). Humans aren't rational! See risk aversion. As to the original question, if there's a greater chance (less likelihood of winning) of failure (loss of investment), there needs to be a greater prize (return on investment) if you do happen to win in order to reach the break-even point to bother investing at all. To "win" in the long run (i.e.m over many investments in a portfolio) you need to find opportunities whose average payout is higher than the cost (as I understand how someone else proposed Buffett analysis) or else you need to have enough information to (on average, correctly) think that the likelihood of success is much higher than others do (so I can buy their futures that they think are doomed). DMacks (talk) 08:49, 30 June 2010 (UTC)[reply]
It sounds to me like a rhetorical question. Wikipedia will be able to supply lots of good information about the merits and limitations of this idea, but I seriously doubt anyone will be able to identify where it came from. (Also, I don't think the questioner really wants to know where it came from.) This page is not the place for rhetorical questions. Dolphin (t) 02:57, 30 June 2010 (UTC)[reply]
Investments are scarce. People will invest in the ones that are the most appealing. People like investments with low risk and high returns. As such, all the ones with low risk high return, medium risk and very high return, and very low risk and medium return will tend to already be taken. People are unlikely to even consider the really bad investments such as high risk low return. This leaves you with the trade-off of risk and return. — DanielLC 03:02, 30 June 2010 (UTC)[reply]
May be because many people know that there is correlation between risk and returns, but not many are aware that Correlation does not imply causation. - manya (talk) 07:03, 30 June 2010 (UTC)[reply]
This "stupid idea" comes from the capital asset pricing model, developed by Nobel prize winner William Sharpe and others. Like all models, CAPM is built on certain assumptions. One of the assumptions of CAPM is that all investors will act rationally. So one method of "beating the market" is to exploit irrationalities of other investors. Another assumption of CAPM is that all information is available at the same time to all investors. So another way to "beat the market" is to use information that is not available to other investors - but there are often laws against that. Gandalf61 (talk) 09:21, 30 June 2010 (UTC)[reply]
CAPM was the second reply given above. 92.29.114.87 (talk) 11:48, 30 June 2010 (UTC)[reply]
As for Buffett, he buys at a significant discount to the value he calculates the investment has -- he has stated that this "margin of safety" is his form of "risk/return." As for the concept in general, there's a jargon term: "upside down ratio." What is your "upside"? What is your "downside"? The ratio is the upside down ratio. If you think the investment could conceivably go to zero during your time horizon, then your ratio is infinite, so you better be able to conceive of a very high return in the same period. But if you think the worst-case scenario is e.g. a loss of half, but the best-case is tripling, then the ratio is 6 (3 over 1/2). The OP's question is predicated on the idea that a great investor would know that the risk is minimal and the potential return is great, but investors have diversified portfolios and it is never possible to know what sort of "black swan" event might happen to any (or all?) of the positions. Risk/return is best exemplified in the 90-day note, which is considered the proxy for a "risk-free" return: right now, it returns something like 0.15%. You have a 99.99999% chance of getting the return OF investment, in exchange for a tiny return ON investment. But if the OP is so certain that great investors needn't consider risk/return, he/she should try to find an example of ONE investor who has not made many losing investments even after decades of experience and intensive study of the positions prior to investing. 63.17.87.23 (talk) 04:28, 3 July 2010 (UTC)[reply]

Name these plants?

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I've seen the this plant, which was about four feet high, in many places in the southern UK, both gardens and wild, and have always wondered what it was. It reminds me of a mallow, but its a lot bigger than a mallow. http://img134.imagevenue.com/img.php?image=45121_DSCF0002_122_247lo.JPG http://img186.imagevenue.com/img.php?image=45114_DSCF0001_122_462lo.JPG

This one is interesting too, about a foot high: http://img232.imagevenue.com/img.php?image=45106_DSCF0003_122_185lo.JPG http://img134.imagevenue.com Thanks 92.29.119.46 (talk) 21:11, 29 June 2010 (UTC)[reply]

General Alert. On the first linked picture, an apparent advert for wind turbines pops up which, when closed, opens a soft porn link. 87.81.230.195 (talk) 21:27, 29 June 2010 (UTC)[reply]

Really? All I see is plants. 92.29.119.46 (talk) 21:30, 29 June 2010 (UTC)[reply]

No really. I got the lovely Jasmine, who might want to go see a dermatologist to get some of those blemishes checked out. --Tagishsimon (talk) 22:00, 29 June 2010 (UTC)[reply]

The image counter indicates people are seeing the plant photos, unless its fake. 92.24.183.236 (talk) 22:57, 29 June 2010 (UTC)[reply]

They doubtless are were seeing the plant photos as well (and as I did), The LiveJasmine porn stream opens opened in a new window which is was initially minimized on to the lower task bar (this is a common ploy of this particular site, which I've encountered before in, ahem, other circumstances ;-) ). Note that all this refers only to the first linked picture; I didn't bother clicking on the second one. Your particular setup, 92, might be automatically blocking this uninvited link.
[Update, as of this posting, the UFO (Unidentified Flirting Object) seems to have desisted.] 87.81.230.195 (talk) 00:34, 30 June 2010 (UTC)[reply]
The foot-high plant is a sedum, I think sedum Autumn joy. Robinh (talk) 07:17, 30 June 2010 (UTC)[reply]
The plant in the first series is a hollyhock, no sign of any 'Jasmine'! Richard Avery (talk) 07:43, 30 June 2010 (UTC)[reply]
I got the same thing as 87, livejasmine popup after I closed the Flash advert for wind power in the main window Nil Einne (talk) 13:55, 30 June 2010 (UTC)[reply]

Stopping time in a fall on concrete

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I want to demonstrate the importance of stopping time to my physics students. I can calculate how much time it takes to slow a high diver to zero when he lands in a pool of water, by calculating how fast he is when he reaches water level and using the pool's depth as an upper limit to the distance the diver covers in stopping. I want to contrast this with what would happen to the diver if he landed on concrete instead of water. I figure the stopping time on concrete is very very small, but I'd like to have at least an order of magnitude. Inkan1969 (talk) 21:36, 29 June 2010 (UTC)[reply]

I doubt anyone will venture a guess as to stopping time, but it is easy to make a guess as to stopping distance - assume the diver lands flat on his back and his center of mass stops in a distance of a centimetre or half an inch. I suggest a more satisfactory illustration can be achieved by considering a mass falling onto springs, and calculating the deceleration during the stopping phase. With a small spring constant k the stopping distance and time are greater, and the deceleration is smaller, than with a large spring constant, illustrating that it is better for the diver to find the water than the concrete! Dolphin (t) 22:54, 29 June 2010 (UTC)[reply]
It is so easy to reading "stopping time" as a referring to a time machine of some sort. But you mean how long it takes the object to decelerate. I don't think it's possible to calculate it on cement, a person is flexible, so the skin might decelerate much faster than other body parts. And if you set the number too low you can easily achieve insanely high numbers. The elasticity of the object plays a very large role in how much force it experiences. I agree with Dolphins suggestion - assume a rigid person, and let a spring act as the elasticity. Ariel. (talk) 23:10, 29 June 2010 (UTC)[reply]
If you have the final velocity before impact v and the stopping distance d, you can estimate the stopping time t by assuming linear deceleration: t = 2d/v. Physchim62 (talk) 23:25, 29 June 2010 (UTC)[reply]
Dolphin51 suggests stopping distances of 1 cm to 0.5 in. Does that sound reasonable to others? If so, I can figure out the stopping time from there, like how Physchim62 notes. I also found out on the Diving discussion board that if you bellyflop from 10 m the water will stop you in only 1 ft distance. That's another good illustration of the importance of stopping time.Inkan1969 (talk) 19:17, 30 June 2010 (UTC)[reply]
You might find research on automotive collisions helpful here. There was a question some time back that I remember finding a reference for about the speeds of collision and its effects on the human physiology. Ah here it is - Impact acceleration stress, mentioned during this unfortunately grisly discussion of motorcycle accidents from last year. This book has some very useful quantitative graphs of experimentally measured impact decelerations (literally, from slamming into concrete), and their effects on humans. "We can say that mechanically the lumbar vertebrae can stand a static stress of 800 kp or a dynamic stress during 0.006 sec of 1,300 kp." (Page 26, discussion of maximum durations of impact stresses before mechanical failure of ... human). Beyond this level of impact, humans no longer behave as intact rigid bodies and will deform as a fluid; you will have to model individual stopping-distances and stopping times for individual parts. Nimur (talk) 23:47, 29 June 2010 (UTC)[reply]
Thanks for the suggestion, Nimur. I'll look up the reference. Inkan1969 (talk) 19:17, 30 June 2010 (UTC)[reply]

What are the worms that create this red moss formation

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The red moss formations are actually worms. They are in about 10cm of water and are about 1mm by 2-4cms. The red "fuz" will dissapear if touched. If some of the red formation is turned over, the mud will boil for a while and then turn red if undisturbed. What kind of worms are they? Is this a larval stage? File:C:\temp\red moss worms.jpg —Preceding unsigned comment added by 205.175.250.27 (talk) 23:19, 29 June 2010 (UTC)[reply]

Sorry, you've linked to a file on your local hard disk, and we don't have access to that. Could you try uploading to a service like Flickr and then point us to that photo? Comet Tuttle (talk) 23:27, 29 June 2010 (UTC)[reply]
Or you can upload the image to Wikipedia's servers, if it complies with our requirements and you are willing to license it for free re-use. Nimur (talk) 23:58, 29 June 2010 (UTC)[reply]
It sounds like you are talking about tubifex worms? eg http://www.google.co.uk/images?hl=en&q=tubifex&um=1&ie=UTF-8&source=og&sa=N&tab=wi 83.100.252.42 (talk) 11:38, 30 June 2010 (UTC)[reply]
Tubifex worms typically move in a "wavy" motion, but often recoil if disturbed or eaten by fish. I've seen a video of tubifex worms in a sewer moving as if they were a single muscle undergoing contractions (it was simply disgusting). ~AH1(TCU) 21:42, 2 July 2010 (UTC)[reply]