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::How long would that take to happen? On the night of New Years, I plan to take an hour or two committing multiple "[[enema washes]]" to empty out the bowels as much as I possibly can. If the process, involving only dead cells, would take more than the 4-5 days the trip lasts, and I don't eat any solids, then I don't think it's a problem. So how many days would it take? --[[User:Mount Zynar|Mount Zynar]] ([[User talk:Mount Zynar|talk]]) 08:56, 7 December 2012 (UTC)
::How long would that take to happen? On the night of New Years, I plan to take an hour or two committing multiple "[[enema washes]]" to empty out the bowels as much as I possibly can. If the process, involving only dead cells, would take more than the 4-5 days the trip lasts, and I don't eat any solids, then I don't think it's a problem. So how many days would it take? --[[User:Mount Zynar|Mount Zynar]] ([[User talk:Mount Zynar|talk]]) 08:56, 7 December 2012 (UTC)

== Lung shot ==

Does getting shot in the lung feel like drowning?--[[User:Wrk678|Wrk678]] ([[User talk:Wrk678|talk]]) 09:38, 7 December 2012 (UTC)

Revision as of 09:38, 7 December 2012

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December 3

Chromosomes Question

I apologize if this is a stupid question, but are the chromosomes of every human being different? Also, in regards to the chromosomes in cancer cells, these chromosomes are always duplications or alterations of one or more of the (original) chromosomes of the human being where this cancer now resides, correct? Futurist110 (talk) 03:40, 3 December 2012 (UTC)[reply]

Since the genes of every human being are different, the chromosomes (the material that contains the DNA within it) must also be different. Wikipedia has a (fairly poor) article titled Genetics of cancer which has a little bit of information for your second question. --Jayron32 04:13, 3 December 2012 (UTC)[reply]
The one exception is identical twins, who, theoretically, could have identical DNA. However, even though they may start out identical (or nearly so), there are subtle changes in the DNA which occur throughout our lives, such as telomere shortening (based partly on the environment), causing twins to slowly diverge. In fact, your DNA isn't even completely identical from one cell to the next, due to oxidation damage, etc. StuRat (talk) 06:22, 3 December 2012 (UTC)[reply]
BTW, in case there is any confusion on the part of the OP, Genetic recombination in the form of Chromosomal crossover during Prophase I of meiosis means that the each chromosome is different, people do not inherit chromosomes wholesale from their parents. (Chromosome abnormalities as discussed below in some cases arise from problems occuring during crossing over.) Edit: So yes, those genetic differences reflect differences in chromosomes (genes which are close together in a chromosome will not be inherited independently hence genetic linkage). Probably should also mention independent assortment. See also Mendelian inheritance. (End edit.) Even sex chromosomes generally undergo limited crossing over (mentioned in our article on meiosis and discussed in more detail at Pseudoautosomal region). Nil Einne (talk) 18:44, 3 December 2012 (UTC)[reply]
Most humans have the same number of chromosomes in most of their somatic cells - 23 pairs, so 46 altogether. What differs from one human to another is the contents of those chromosomes - the genes. A small proportion of humans have a non-standard number of chromosomes, usually due to a chromosome abnormality such as Down Syndrome or Turner syndrome. Chromosome abnormalities such as aneuploidy (having an abnormal number of chromosomes) are also often found in cancer cells, although I am not sure whether they are considered a cause of the cancer or a side-effect. Gandalf61 (talk) 10:39, 3 December 2012 (UTC)[reply]
Generally speaking, widespread chromosomal abnormalities – multiple translocations and changes in chromosome number with this sort of ugly karyotype – are a consequence of the overall genomic instability associated with cancer, caused by a combination of loss of cell cycle checkpoints and failure of mechanisms associated with DNA lesion detection and repair. (Of course, this instability increases the heterogeneity of the population of malignant cells, increasing their likelihood of accumulating further harmful mutations and developing drug resistance.)
On the other hand, there are certainly a few specific chromosomal abnormalities that are linked to cancer. The so-called Philadelphia chromosome is probably the canonical example; this translocation of parts of chromosomes 9 and 22 produces an oncogenic fusion protein that is the primary cause of chronic myelogenous leukemia. TenOfAllTrades(talk) 14:31, 4 December 2012 (UTC)[reply]
  • In theory, with a drop of blood from a crime scene, you should be able to clone segments of DNA (by polymerase chain reaction) from regions of V(D)J recombination, which have undergone said recombination (by choosing a short length, which amplifies better anyway). If you know the blood came from one of two identical twins, you could see if the prevalence of B cells matches one twin better than another, which is likely, since each event is random and which events prevail depend on prior exposure to antigens throughout life. I have no idea if this has ever seen a courtroom; you'd probably be paying someone's salary for three months to get the test done and half the time they can't be bothered to do even a simple DNA match for a case. Wnt (talk) 02:01, 5 December 2012 (UTC)[reply]

Are there any blood vessels in the skin? (Skin = Epi, Dermis, Hypo)

thanks. — Preceding unsigned comment added by 109.64.163.33 (talk) 04:06, 3 December 2012 (UTC)[reply]

See the Wikipedia article titled skin. --Jayron32 04:11, 3 December 2012 (UTC)[reply]
Capillaries, certainly, or you wouldn't bleed when you prick your skin. Unless you have spider veins or varicose veins, you probably don't have much larger blood vessels in your skin. StuRat (talk) 06:18, 3 December 2012 (UTC)[reply]

Is this idea accepted as plausible by psychologists, or almost scholars of ancient literature? It seems extraordinary to claim that people were not conscious 3000 years ago, yet I can't find much about how mainstream this idea is. --140.180.249.151 (talk) 06:44, 3 December 2012 (UTC)[reply]

"Is this idea accepted"? No. Is it mainstream? No. If it was there'd be more recent discussion of the question. The scholarly consensus is probably that human cognitive abilities have changed little over the last few millennia - but if it has, it may well have deteriorated, for environmental rather than genetic reasons. AndyTheGrump (talk) 07:05, 3 December 2012 (UTC)[reply]
Thanks for your answer. I didn't know what the most recent discussion of the question was, being completely unfamiliar with psychology. But I do think human cognitive abilities have improved drastically due to increased education, better living standards, and a more advanced world--this is reflected in the Flynn effect. --140.180.249.151 (talk) 18:22, 3 December 2012 (UTC)[reply]
Looks like total BS to me. Very little evolution occurs in 3000 years, certainly not enough to change the core nature of the human mind. StuRat (talk) 07:46, 3 December 2012 (UTC)[reply]
To my knowledge, Jaynes did not hypothesize that genetic evolution played any part in the breakdown of bicameralism. Rather, his thesis is more reminiscent of the evolution of memes, which was curiously proposed in the same year as bicameralism. Someguy1221 (talk) 09:52, 3 December 2012 (UTC)[reply]
This bicameralism theory is from 1976, the meme theory was popularized by Dawkins in a publication of the same year, but it's much older. OsmanRF34 (talk) 17:53, 3 December 2012 (UTC)[reply]
Bicameralism is somewhat akin to the Aquatic Ape Hypothesis in that both are not properly scientific due to being untestable (so far as is known). For that reason alone, beyond whatever anyone actually thinks of these concepts, they're never going to be widely accepted. On the one hand, they are very tempting because they appear to answer so many profound questions at a stroke, while on the other there is frustratingly little rigorous evidence. Matt Deres (talk) 22:42, 4 December 2012 (UTC)[reply]

Binary Black Holes

This is a long one.

Let us assume that we have an observer O, a distant star S, and a Black Hole H between S and O. Thus the light going straight towards O will enter H; however, due to H's gravity, light can go around H in an arc. Within a plane, there are two solutions: one arc goes through X, and one goes through Y.

ASCII art:

. S'
.
.
.              X
.
.S             H             O
.
.              Y
.
.
. S"

So there will be two apparent stars S: at locations S' and S". In the 3D case, O will see a ring with H and the true S at its center (although they won't see H or S). S' and S" will be red-shifted equally in that case, and the amount of redshift will depend on the escape velocity of S and the celestial body O is on. Which can even be negative if, say, S is an asymptotic Red Giant and O is on Mercury (planet) deep within the Sun's gravity well. But whatever, S' and S" will have the same spectrum.

So far, so good. Now let us assume that H is not a single, but a binary Black Hole, rotating clockwise within the plane shown above. Both equal mass, centered at H. I'm interested in the spectra in that case.

My friend argued that because the Black Holes emit gravity waves, S" will have a lower redshift; "its" light passes through Y and hits the gravity wave head-on, which will compress the wave. OTOH, the light from S' will pass through X and move in the same direction as the gravity wave; there will be no collision at all, thus the full redshift will be observed.

I suspect that that's wrong, and that the true way of interaction is completely different; that light does not get compressed like some kind of spring. I have the feeling that S" would exhibit the highest redshift, because the gravity wave takes some momentum from the light when it passes through Y. S' would be the other way around; it would gain a comparable amount of momentum, and thus the wavelength will be the shortest.

Bonus question: Is X or Y closer to H? (I'd guess X, because light passing through X gains momentum, which helps escape the gravity well - thus the light which passes X can make the closer pass and still escape at all).

Is the effect significant? From what it looks like, it takes not only a binary Black Hole, but a quite massive pair, lest the orbit size be negligible to the distances H-X and H-Y. So I guess the pairs which do exist are unlikely to exhibit a significant asymmetry. - ¡Ouch! (hurt me / more pain) 10:13, 3 December 2012 (UTC)[reply]

You may get an effect from a massive spinning black hole using the Kerr metric. An orbiting pair would have s similar effect. However the light rays will have to pass fairly close. If a binary black hole merges into one a significant fraction of mass is converted to gravitational waves, and as they pass your observer O space can be shifted permanently. I expect that this will cause a temporary red or blue shift. Using gravitational waves observers also hope to see variations in pulsar timings. Graeme Bartlett (talk) 10:42, 3 December 2012 (UTC)[reply]
Thanks. Kerr metrics seem to support my point, at least as far as the light paths and the X, Y points are concerned. Particles gain or lose momentum as I predicted, but will that effect red/blueshift light? - ¡Ouch! (hurt me / more pain) 16:06, 3 December 2012 (UTC)[reply]
shifting of the frequency could happen if the observer or source are near the black hole(s) so that they are in the gravitational potential well, and then experience gravitational red shift. But it do not see that one side of the gravitational arc would be shifted differently to the other. The gravitational potential for emitter and observer should be roughly constant for the light traveling via different parts of the arc. Graeme Bartlett (talk) 20:22, 3 December 2012 (UTC)[reply]
Let's assume that S and O are at the same distance from H. - ¡Ouch! (hurt me / more pain) 08:04, 4 December 2012 (UTC)[reply]
Then the gravitational potential due to the black-holes would be the same, no extra gravitational red-shift. Graeme Bartlett (talk) 09:57, 5 December 2012 (UTC)[reply]

Origin of Intelligence and Mental Illness Linked to Ancient Genetic Accident

Or so claims this article. The claim is so sweeping I find difficulty in even taking it seriously. What do you think? --Halcatalyst (talk) 12:32, 3 December 2012 (UTC)[reply]

It looks like they just reprinetd the University press release [1], which can often be pretty far off the mark. I don't have access to anything other than the abstract of the original paper [2], but that is the place to look for the real (although harder to read) version of the story. 209.131.76.183 (talk) 12:40, 3 December 2012 (UTC)[reply]
The role of paleopolyploidy in vertebrate evolution has been known for some time. Most of our genes form a group of up to four different versions that trace back to that event. As a result, double and triple knockout experiments like this are important. This one focused on a single set of genes, Dlg1, Dlg2, Dlg3, Dlg4, and their effects on some touchscreen tests. The university publicity office took that and spun it in a weird way. Note that when the whole genome duplications actually happened, the animals at the time might have seemed bigger or had some other simple differences, but since the genes were all the same these effects would be no more sophisticated than for any other single mutation with pervasive effects. It was not a recipe for instant brilliance. Wnt (talk) 21:16, 5 December 2012 (UTC)[reply]

relationship between conductivity and mean free path

This is a very basic formula found in most textbooks, but I don't have my textbook and the absurdity of this simple relationship not being on Wikipedia is aggravating me. Where can I find it? 128.143.1.238 (talk) 13:44, 3 December 2012 (UTC)[reply]

Are you referring to thermal conductivity or electrical conductivity? douts (talk) 13:52, 3 December 2012 (UTC) Not sure if it's of any use, but a quick Google search found this paper [[3]] entitled "Calculation of the thermal conductiwity and phonon mean-free path" douts (talk) 13:57, 3 December 2012 (UTC)[reply]

Thermodynamics of a microwave

Why is there no heat or work transfer in a microwave heating something or a microwave being used for radiotherapy or sterilising? Thanks. — Preceding unsigned comment added by 138.253.210.27 (talk) 16:32, 3 December 2012 (UTC)[reply]

I'm not sure I understand what you're asking here. Heat transfer does occur in a microwave - the water in the food is heated by the microwaves, cooking the food. Microwaves generally aren't used for sterilising objects - normally either gamma rays or temperatures above 200 degrees centigrade are used. If I've mis-understood the question please let me know. douts (talk) 17:03, 3 December 2012 (UTC)[reply]
Less than 200 °C is enough, as it's the case of an autoclave. OsmanRF34 (talk) 17:30, 3 December 2012 (UTC)[reply]
About 125 °C in a saturated steam environment, a bit more (I've used 160-180ish before) in a dry environment. Fgf10 (talk) 18:37, 3 December 2012 (UTC)[reply]

Watertight compartments

Watertight bulkhead compartments were written of by the Song Dynasty Chinese author Zhu Yu, in his book Pingzhou Table Talks of 1119 AD (written from 1111 to 1117 AD). Watertight compartments were frequently implemented in Asian ships, and had been implemented in the warships of Kubla Khan.[2][3][4] Chinese sea-going junks often had 14 crosswalls, some of which could be flooded to increase stability or for the carriage of liquids.

Anyone can build wood ships with watertight compartments. However, without watertight doors, the compartments are not very useful. You probably have to install many trapdoors on the deck for each compartment so you can climb up to the deck to go to another trapdoor.

I really don't think it's easy to build watertight doors on a ship made by wood. How did Chinese inventors solve this problem? -- Toytoy (talk) 16:36, 3 December 2012 (UTC)[reply]

I don't know about the Chinese but pitch can be produced from plant material and has been used for many watertight application in the past. The introduction of the linked articles talks a little about it. Dauto (talk) 17:07, 3 December 2012 (UTC)[reply]
Don't know if this is the case, but could it not be that the space between the keel (if they had one) and the lowest deck was compartmentalised? So as to compartmentalise at least part to the buoyant volume. In that case the habitable decks wouldn't need watertight doors. Lower compartments could easily be sealed with something like pitch, as suggested above. Fgf10 (talk) 18:36, 3 December 2012 (UTC)[reply]
Also note that the watertight compartments don't need to be 100% waterproof, from the top. Say the boat is swamped by a rogue wave. You only need to keep it afloat long enough to bail out most of the water. If some water leaks into the watertight compartment in that time, it's OK, so long as the boat doesn't sink before you can bail it out. StuRat (talk) 18:42, 3 December 2012 (UTC)[reply]
Yes, with very primitive watertight compartments, you can still buy time to plug the hole and let the sailors bucket brigade the compartment dry. Without compartments, you're dead with the fish.
However, doors made of cheap wood can be very imprecise. They they can absorb moisture. Unless you pay big money for really good woods and treat the parts with tung oil, I don't think it was a good idea to build doors between compartments. -- Toytoy (talk) 04:18, 4 December 2012 (UTC)[reply]

Neutron decay and atomic weapons

So I was just watching this video, http://www.youtube.com/watch?v=o_EBqZPCZdw

And the guy in the video (thunderf00t) said that a neutronium bomb would be very destructive but wouldn't release its energy all at once, but rather over the course of 10-14 minutes because that's the time it takes neutrons to decay (10 minute half-life, 14-15 life-time).

Well in a fusion reaction, about 80% of the energy released is in the form of high energy neutrons, and that made me think of the tsar bomb. According to the article, the bomb was tested without a uranium 3rd stage so almost 97% of the energy released in the test came from fusion alone. But yet the explosion came all at once, instead of being spread out, continuously exploding for 10-14 minutes. I thought about this, and I reasoned that it must have been due to the lead tamper they used to absorb the neutrons. So I thought about the neutron bomb. If the tamper were made out of nickel or chromium, allowing the neutrons to escape, would the explosion be spread out over the course of 10-14 minutes like the video suggests would occur? ScienceApe (talk) 17:47, 3 December 2012 (UTC)[reply]

All neutrons will quickly lose their energy due to scattering off atomic nuclei in the air/water/soil. Then a significant part of the neutrons will be absorbed by some nuclei. The energy that will be released as by radioactive decay of the remaining neutrons is quite small. Ruslik_Zero 19:28, 3 December 2012 (UTC)[reply]
Note that even if neutrons did behave that way — that is, they went until the end of their "life" and then decayed to release their energy — their energy would be incredible diffuse. D-T fusion neutrons travel at a speed of 52,000 km/s; if they survived for 10 minutes they'd be well, well outside the radius of the planet by the time they decayed! It doesn't work that way, obviously. As for the tampers, they do matter with regards to what happens to the neutrons; neutron bombs are just hydrogen bombs where the fusion neutrons are given a direct route outside of the bomb without being use either for fissioning or for just thermal energy. They still all come out in one burst, and still matter for only a fraction of a microsecond. --Mr.98 (talk) 00:25, 4 December 2012 (UTC)[reply]
Also note that even if 80% if the energy were in the radioactivity of neutrons (which is false; much of that energy is kinetic), the initial 20% would equal the first ~4 minutes of decay. Even then, the explosion would hardly be a continuous glow.
Also note that enhanced radiation weapons ("neutron bombs") were not designed to be less physically damaging. The physical effects are somewhat less than those of older nukes of that size, but not by an order of magnitude. They were designed to inflict higher overall damage against armored targets, and since those resist physical force quite well, they maxed their neutron radiation rather than their TNT rating.
Don't know for sure about mean free path of neutrons, though. I doubt many would escape Earth. I would think though that most of them would have been captured by one nucleus or the other, within seconds. - ¡Ouch! (hurt me / more pain) 08:01, 4 December 2012 (UTC)[reply]

Why issue placebos to control groups for objective treatments?

Reading placebo effect, I saw "The placebo effect is highly variable in its magnitude and reliability and is typically strongest in measures of subjective symptoms (e.g., pain) and typically weak-to-nonexistent in objective measures of health points (e.g., blood pressure, infection clearance)." (Yes, I saw "citation needed", but is the statement really indefensible?). I wonder, if it's an all-but sure bet that sugar pills are *not* really going to be effective against a given cancer, why do testers still do controls with them? It seems about as likely for a control group taking sugar pills to see effective cancer treatment in any of the control group subjects as would be if the control group held water balloons. Is the only purpose in objective (not asking patients how they feel) placebo-using trials to have double-blind studies to keep the doctors honest and from tampering with administration protocols, such as giving more to patients they know are on the real thing so there's a better chance to have success and continued funding? 20.137.2.50 (talk) 18:01, 3 December 2012 (UTC)[reply]

The idea is to make the differences between control and test group as small as possible. The placebo effect is a complex thing, for instance, speaking to a doctor will often also have a beneficial effect. By making the control group as close as possible to the test group, you can rule out as much as possible any effects not due to the drug being tested. An other aspect is that without placebos it would be easy to determine for the trial doctors which group is control, and it's then easy to subconsciously treat both groups differently in for instance bed-side manner. It has been shown (bear with me while I hunt for a ref) that a drug given with a positive talk by the doctor will have a better effect than the same drug given with a disparaging talk. Just to rule out effects like that, we do double-blind placebo-controlled trails. Fgf10 (talk) 18:33, 3 December 2012 (UTC)[reply]
Blood pressure is something I'd expect to respond to placebos, since just relaxing can change it, and somebody thinking they just took some good meds might be more relaxed. Also note that many meds are approved even though they only have a slight benefit, like in 10% of the patients. So, even a tiny placebo effect could have a major effect on such results. StuRat (talk) 18:37, 3 December 2012 (UTC)[reply]
A sugar pill with a smile can rival an alpha blocker with a frown in terms of effectiveness? 20.137.2.50 (talk) 18:53, 3 December 2012 (UTC)[reply]
Maybe - but it's certainly possible that a sugar pill given by a guy in a white lab coat with a stethoscope casually draped around his neck who says "This is an absolute wonder-drug - it's the latest thing and it'll fix you up for sure" - is more effective than an alpha blocker handed to you by an intern who says "I'm this probably isn't going to help, but please take it anyway.". In the case of blood pressure - where stress is a known cause - it's quite obvious that making the patient think that they'll be cured will reduce their stress and therefore help their blood pressure to go down. But there are other placebo effects where the link is not at all obvious. It's been shown that weight lifters can lift more weights if given a course of sugar pills that they are told are really steroids...people who are told they are drinking alcohol when they are not will show signs of intoxication. It's a weird set of effects going on here. SteveBaker (talk) 17:57, 4 December 2012 (UTC)[reply]
As others have stated, placebos can actually have an effect in patients even in cases of cancer. In fact the American Cancer Society has a long discussion of placebos [4] and while it doesn't mention cancer that much it's a good read if your understanding of the placebo effect is poor. [5] and [6] may discuss the placebo effect and cancer somewhat although I haven't actually checked the articles. As Fgf10 said, there are also other advantages to having a placebo, namely in preventing differences in the interaction between the patient and other people. In other words a placebo or something similar makes it possible to do a double blind which is generally considered the gold standard in medical research. However I would note that not all double blind trials are placebo controlled, as our Blind experiment and Clinical trial mention, it's also common to use existing medicines for the condition being treated. Nil Einne (talk) 20:16, 3 December 2012 (UTC)[reply]
I'll look at those sources promptly. But I'm impatiently curious; do those papers even hypothesize what could possibly be going on (in terms of chemistry) in any single case where someone taking a placebo sees a significant improvement in their cancer treatment? 67.163.109.173 (talk) 22:30, 3 December 2012 (UTC)[reply]
BTW the Declaration of Helsinki article which is linked from the clinical trial article goes in to a fair amount of detail about the controversy surrounding the usage of placebos when existing treatment exist. Nil Einne (talk) 20:44, 3 December 2012 (UTC)[reply]
In a case like that, I'd think using an existing med instead of a placebo would make sense. You would then compare the results to find out if the new med is better than the old med. StuRat (talk) 20:51, 3 December 2012 (UTC)[reply]
Yes definitely, when an effective treatment exists for a specific disease, the control group will most likely receive the current treatment, not just a placebo. You aren't necceserily trying to find out if a new treatment is just better then a placebo, you're trying to find out if it's better then the current treatment. Of course "better" might not just be efficacy, it could also be cost, side-effects, how invasive the treatment is, etc... But in most cases where a treatment already exists, it would be considered unetihical to withold it for the purposes of research. Vespine (talk) 21:58, 3 December 2012 (UTC)[reply]
As I mentioned in my post I was replying to with references to our articles, that is often done. But as the Helsinki article attests and Vespine mentioned even if it's generally regarded as unethical to withold an alternative existing treatment (although there is some dispute in cases when the consequences and risks aren't high), that doesn't stop it happening particularly in the developing world. (In case there's still some confusion that's the primary reason I brought the Helsinki article in to the mix after already mentioning that it's common to use an existing medication when one exists instead of a placebo.) Nil Einne (talk) 00:43, 4 December 2012 (UTC)[reply]
which brings up my nagging question; if we reject a treatment that is no better than placebo, but we don't give patients placbos, aren't we therefore undertreating them? Gzuckier (talk) 18:35, 4 December 2012 (UTC)[reply]
Because a measured placebo effect is NOT just 'one' thing. It's not, as most people naively assume, a real physiological change in the patient which improves their condition, or at least it's MOSTLY not that. A placebo is everything except the effect of the substance you are trying to test. That does include things like how the patient reports their condition, but it also includes things like regression towards the mean, statistical bias, reporting bias, experimenter's bias, etc.. Vespine (talk) 02:51, 5 December 2012 (UTC)[reply]
good point. Gzuckier (talk) 06:12, 5 December 2012 (UTC)[reply]
The placebo effect article raises two other important issues with administering placebos as medicine. One being that it would be pretty much impossible without a degree of deception on the doctor's part, which is problematic; the second being that placebo effect is neither consistent or reliable. Not that any medicine is 100% consistent or reliable, but placebo is, pretty much by definition, the very least consistent and reliable treatment available. Vespine (talk) 21:50, 5 December 2012 (UTC)[reply]

Global warming

Is there a way of identifying exactly whether a certain CO2 molecule was created by man or naturally made? Could this identify whether global warming is caused by man or occurring naturally? I've heard that Carbon isotopes can identify CO2 which has been produced by burning fossil fuels but is this an accurate test? Thanks. 138.253.210.27 (talk) 18:01, 3 December 2012 (UTC)[reply]

No. A particular CO2 molecule will contain the carbon-12, carbon-13, or carbon-14 isotope, with the first 2 being stable, and carbon-14 being unstable with a half-life of around 5,730 years. So, if it contains carbon-14, the chances are that it's not a fossil fuel from the age of dinosaurs. However, the molecules containing carbon-12 or carbon-13 aren't necessarily "old", as only 1 in a trillion carbon atoms is carbon-14, even in new CO2. But, if you look at a large sample, then the ratio of carbon-14 will tell you something about the average age of the carbon molecules.
But now for the complication: carbon-14 is being continually generated in the upper atmosphere: [7]. So, the carbon-14 we find up there might be generated in that way, from carbon from burnt fossil fuels, or it might be from wood somebody burned, or it might be from what a human or other animal exhaled. And, of course, the carbon-14 generated up there doesn't stay aloft, it moves throughout the biosphere. StuRat (talk) 18:19, 3 December 2012 (UTC)[reply]
(edit conflict) No, not by individual CO2 molecules. However, that doesn't mean that reliable correlations cannot be drawn to decide if the bulk of CO2 molecules is not from man-made sources. To take an analogy: Imagine you're watching a New York City street and every single person is an exact clone, with the exact clothes. Now, imagine you can only take snapshots of the street, so you don't see people moving, but you can count the number of people at any given time. You don't know where each individual person comes from, but you can correlate the number of people on the street with certain times of day, and you can also correlate other events that occur at those times, so you can say that the increase of people on a specific street at a specific time may be due to a train arriving at a nearby subway stop. You don't need to see the people get off the train to consider that a reasonable conclusion. You don't even need to know which specific people got off the train, the fact that there are more of them at the same time every day, immediately successive to the train arriving, is enough to establish the reasonable conclusion that some of those people got off the train. It's the same thing with CO2 increases. You don't need to know which CO2 molecules came from, say, a volcanic eruptions and which come from burning coal. We know that some increase in CO2 is going to come from burning fossil fuels, because it is patently obvious to anyone with a high school chemistry level knowledge of combustion that burning hydrocarbons makes CO2. We don't need to watch each CO2 molecule individually from the moment it formed and track each one to know that, in the bulk, burning more stuff makes CO2, any more than you need to track every pedestrian to know that the number of people on the street will increase after the train arrives. --Jayron32 18:25, 3 December 2012 (UTC)[reply]
  • There's another aspect to this. As our article on isotopes of carbon points out, plants take up 12C in preference to 13C, and that preference carries over to fossil fuels. The main sources of atmospheric carbon are fossil fuel burning, respiration, and volcanoes. Volcano-derived carbon has a different isotope ratio than biologically generated carbon -- this ratio is known as δ13C. There is also a smaller difference between carbon derived from fossil fuels and carbon derived from recent respiration, because of differences in the atmosphere when the fossil fuels were created. Looie496 (talk) 18:49, 3 December 2012 (UTC)[reply]
  • The effect of fossil fuel burning changing the isotopic ratio of carbon in the atmosphere is known as the Suess effect and was first described in 1955, when its influence on radiocarbon dating was noted. That said, we have many additional lines of evidence that the increase of CO2 is anthropogenic. The simplest is basic chemistry. We (roughly) know how much fossil fuels we burn, and hence how much CO2 we emit. We also can measure the amount of CO2 in the atmosphere. The increase in atmospheric carbon dioxide corresponds to roughly half of what we release (much of the rest is absorbed by the oceans). We can also see a corresponding decrease in atmospheric oxygen. Note that the question of where the CO2 comes from is different from the question of what its effect is. That said, the spectroscopic properties that give rise to the greenhouse effect have also been known for more than 100 years and the effect of adding carbon dioxide was first quantified by Svante Arrhenius around 1900. --Stephan Schulz (talk) 20:41, 3 December 2012 (UTC)[reply]

Travelling to star

Scenario: I pick a star 100 million ly away to travel to. And i can get there because i can live forever. the star is on the very ege of a galaxy 200,000 ly across. When i plug the coordinates of the star into my space craft, HAL, will i travel in a straight line and wind up possible 200,000 ly off my mark? or will i have to adjust my course as i get closer? or something else?165.212.189.187 (talk) 18:40, 3 December 2012 (UTC)[reply]

No, if you travel in a straight line at a star on the other side of the galaxy, when you get there the star will not be there, because the galaxy is in motion: it is moving relative to where it is now in several ways, it is rotating about an axis, and it is moving in a line away from other galaxies. You would need to aim for where the star will be when you plan to arrive, based on where the star is now and where it will move over the course of your trip. This is not trivial, because a) the star is not now where you see it (because light takes time to travel) and b) the star's location needs to be calculated taking into consideration a) your current motion b) the stars motion and c) the complete motion, including acceleration, deceleration, and top travel speed, of your spaceship. This is all hypothetically possible to calculate, so other than the tricky math, there's nothing wrong with calculating a straight line path that should make you arrive at the same point in spacetime as that star some place in the future. --Jayron32 18:52, 3 December 2012 (UTC)[reply]
And this is similar to planning trips within our own solar system, although the other planets are within a few minutes or hours of where they appear to be. If it's a round trip, that makes it even more complex. StuRat (talk) 19:26, 3 December 2012 (UTC)[reply]
Or you could just program your ship's computer to adjust your course every (insert period of time here), so that as you get closer, the error factor will be reduced with each course change. douts (talk) 19:03, 3 December 2012 (UTC)[reply]
You'll have to compensate continually for the near infinite number of disruptions from gravitational influences along the way. Fgf10 (talk) 19:21, 3 December 2012 (UTC)[reply]
Practically yes, but hypothetically, if you could keep track of the entire galaxy in a single computer, and reliably predict the entire gravitational field of the galaxy, as well as changes to that field over time, you could hypothetically have all those corrections made before the trip. That is, any corrections made to your trip would come from incomplete knowledge rather than anything which is physically impossible to know. Any corrections made along the course due to incomplete knowledge could be corrected before the trip if the information had been known ahead of time. --Jayron32 19:24, 3 December 2012 (UTC)[reply]
I don't think that's even theoretically possible. That is, storing the location and vector of every atom in the universe would require a computer made of more atoms than the universe. There's a thought experiment along these lines, but I forget the name. Of course, if your margin for error is large enough, like a solar system, no corrections may be needed. If, on the other hand, your goal is to dock with a spaceship in that solar system, then corrections will be required at some point. StuRat (talk) 19:31, 3 December 2012 (UTC)[reply]
Well, I suppose yes, you are correct. The question is when the corrections need to be made: constantly along the way, or only at the end. I think the analogy of the airliner is appropriate: Airplane pilots basically aim their plane at the airport and don't adjust much during the length of the trip: autopilot is capable of keeping the plane on the correct course, usually a fairly easy-to-calculate great circle course (the Earth equivalent of a straight line), and the pilots only need to be involved in the take off and landing portions. For most of the trip, the pilots don't do much except monitor the plane to make sure nothing goes wrong. For a properly working plane, 99% of the trip doesn't need any adjustments at all. Hypothetically, traveling to a distant star should involve a similar level of involvement: aim and fire, and then adjust the final destination when you get close. You aren't likely to end up ridiculously off course if the proper calculations are made ahead of time. --Jayron32 19:39, 3 December 2012 (UTC)[reply]
Are you sure that autopilot isn't making tiny corrections all along ? StuRat (talk) 19:51, 3 December 2012 (UTC)[reply]
At a more practical level, unless the target star has a circular orbit with precisely known velocity, making predictions will depend on the distribution of dark matter in the target galaxy, which we know only at poor resolution. Looie496 (talk) 19:34, 3 December 2012 (UTC)[reply]
Add to all this above that your target star maybe doesn't exist anymore or disappear during your journey. OsmanRF34 (talk) 22:02, 3 December 2012 (UTC)[reply]

So couldn't the galaxy also be moving not just directly away from us but also some other arbitrary direction? What do you mean "in Aline away from other galaxies" GeeBIGS (talk) 00:44, 4 December 2012 (UTC)[reply]

The expansion of the universe causing movement away from distant galaxies should be the dominant factor at that distance, but there would also be a smaller factor in another arbitrary direction. At shorter distances this factor is dominant. See for example Andromeda–Milky Way collision. PrimeHunter (talk) 01:22, 4 December 2012 (UTC)[reply]
No, gravitationally bound objects like galaxies do not expand due to metric expansion. -- Finlay McWalterTalk 01:32, 4 December 2012 (UTC)[reply]
The thought experiment could be expanded. If the person can live forever, they could travel to a star that is 50 billion light years away. Under that scenario metric expansion might come into play. Would that be correct? Would it still be possible to make almost all calculations before setting out on the journey? Bus stop (talk) 01:58, 4 December 2012 (UTC)[reply]
I don't think any stars (we know about) are that far away. Plus the star would be unlikely to exist when we got there. StuRat (talk) 02:30, 4 December 2012 (UTC)[reply]
[8] Bus stop (talk) 03:04, 4 December 2012 (UTC)[reply]

What I'm trying to get at is that in some cases the star is actually closer than the calculations predict because the arbitrary direction could be directly at us. Right?GeeBIGS (talk) 05:17, 4 December 2012 (UTC)[reply]

At shorted distances the arbitrary direction becomes dominant = the closer you get the more the arbitrary direction matters? So it seems like any path looks like a really long straight line with a sharp turn towards the end of the trip?GeeBIGS (talk) 05:30, 4 December 2012 (UTC)[reply]

How the path looks will depend on the kind of corrections applied.
The computationally least expensive way would be...
1-to fly where you see the star,
2-look for the star and turn towards it,
3-to fly where you see the star then (and so on)...
But that's not only wasteful of time and fuel, it's possible that you lose track of the star and cannot find it because it has changed while you were flying.
A more practical way which does not assume that the star is as unchanging and immortal as the pilot is to keep track of it, and to apply corrections after, say 5% of the distance, then after 5% of the remaining distance, etc.
The former method would look like GeeBIGs assumed, but the latter would be quite close to a least-time intercept course. It would still arc more tightly towards the end but not nearly as much as the first approach. You would be able to account for more and more of the arbitrary components, but quite gradually.
And finally, bright stars don't last for 200 million years. The speed of the ship becomes important, for even a faint star would not last that long if you were going no faster than the Pioneers/Voyagers. But these are probably a moot point, as they would not be fast enough to escape our galaxy... - ¡Ouch! (hurt me / more pain) 08:00, 4 December 2012 (UTC)[reply]

What I'm trying to get at is that in some cases the star is actually closer than the calculations predict because the arbitrary direction could be directly at us. Right?GeeBIGS (talk) 05:17, 4 December 2012 (UTC) — Preceding unsigned comment added by 165.212.189.187 (talk) [reply]

We can detect if a star is moving towards us or away, and measure the speed, using the blueshift or redshift. That's the Doppler effect applied to light. StuRat (talk) 20:41, 4 December 2012 (UTC)[reply]

Then why did others say that the arbitrary direction matters as you get closer? And that it might not be there when you get there? Aren't you only detecting the metric expansion with blue/redshift? It would appear that way but consider a star in some 200,000 ly orbit: we see the light while its at 12 oclock; we travel toward it; when we ge to where it was couldnt it have "come around behind us through 3, 6, 9" ?165.212.189.187 (talk) 21:04, 4 December 2012 (UTC)[reply]

Regarding the metric expansion of space: My understanding is that most galaxies are flying away from each other due to the inertia from the Big Bang, so have a redshift. We expected them to be flying away at a decelerating rate, due to gravitational attraction, but found they are actually flying away at an accelerating rate. The explanation for this is the metric expansion of space. StuRat (talk) 21:10, 4 December 2012 (UTC)[reply]

Mars Curiosity findings

There are reports of interesting chemicals being found on Mars. However, they caution that it might be contamination from Earth. Why didn't they take care of that possibility before they sent it? They had it in a clean room, why didn't they clean the damn thing to preclude the possibility of what they find being from Earth? Bubba73 You talkin' to me? 23:48, 3 December 2012 (UTC)[reply]

I'm sure they tried, but it's quite difficult to keep it sterile. What if an adjustment needs to be made after it's cleaned ? Do you clean the whole thing again ? And do you store it in a vacuum until launch, to prevent contamination by the air ? StuRat (talk) 23:57, 3 December 2012 (UTC)[reply]
This episode of NPR's Science Friday talks about the degree to which a planetary probe is cleaned and sterilised. -- Finlay McWalterTalk 00:18, 4 December 2012 (UTC)[reply]
Thanks for the info and link. But if they are going to send it off not clean, and they have those doubts about their initial results, will they ever be able to have confidence in them? Bubba73 You talkin' to me? 00:53, 4 December 2012 (UTC)[reply]
I recall reading that initial scoops of Martian soil are discarded because of the possibility that they will be contaminated. If I recall correctly, the operation of the mechanism is designed to clean itself with the first few scoops of soil. After that it is not expected to be able to still contain contamination from Earth. But I am just recalling this from memory (faulty). I think I read it in Scientific American. Bus stop (talk) 02:05, 4 December 2012 (UTC)[reply]
Here it is: "But before Curiosity fired up its CheMin (Chemistry and Mineralogy) instrument to analyze the soil, it first had to purify its sample-collection instruments using Martian sand as a cleansing abrasive."[9] Bus stop (talk) 02:21, 4 December 2012 (UTC)[reply]
They go for 50ng/g of sample contamination. If anybody here has a clue how clean that is? To get lower than that you have to built a polymere free instrument only using ceramics and metals. No cables with isolation no electronic boards .... This is impossible. The cleanliness they have is amazing.--Stone (talk) 03:21, 4 December 2012 (UTC)[reply]
To put 50ng/g in perspective, the standard for the purest form of water calls for less than 100ng/g of solid contaminants.Dncsky (talk) 03:58, 4 December 2012 (UTC)[reply]
Bubba: this is how science works. One must always consider alternate explanations for data. It's not that they didn't send the probe off clean, but rather that no matter how clean you think your probe is, if you get striking results that could be explained by contamination, you are going to say that contamination is a possible explanation for the results. Ideally, there will be other tests they can do to rule out the possibility of contamination.--50.196.55.165 (talk) 19:25, 7 December 2012 (UTC)[reply]

What i like best is the fact that curiosity find the same stuff Viking found in the 1970s. Klaus Biemann reported that they fond dichloromethne and chloromethane, but they argued that it was contamination from solvents used for cleaning. But now it looks more like the chloromethanes were produced by the reaction of organics wit the chlorine from the perchlorates.--Stone (talk) 03:25, 4 December 2012 (UTC)[reply]

Perhaps they were using "Earth contamination" as a catch-all for anything that could invalidate the results. Bubba73 You talkin' to me? 18:21, 4 December 2012 (UTC)[reply]


December 4

How long do water molecules last (live?)

If a snowflake falls on a glacier, (3.5 x 10^19 molecules to a snowflake I read)...and we could follow one molecule when the snow melts, the water runs into a river, then to the ocean, and then at some point evaporates becoming clouds, and pushes onshore to fall as snow again...does that one water molecule exist throughout, changing states from solid to liquid to gas, around and around? Forever? Does something break up an H20 molecule at some point? Does a water molecule have a birth, life, death, or will some of them go around and around for years, ages, eons? How does it work? Thanks if you can help me understand and picture it. — Preceding unsigned comment added by 94.208.75.76 (talk) 03:10, 4 December 2012 (UTC)[reply]

Water molecules in liquid and solid change their specific atomic associations over time (exactly which two H are attached to exactly which O). The origin of "neutral water pH is 7" is that some aren't even associated as a 2:1 form at all. See self-ionization of water. DMacks (talk) 03:17, 4 December 2012 (UTC)[reply]
I am wondering though, at any point do H2O molecules cease to be H2O molecules? For instance are there any naturally occurring events or processes causing H2O molecules to cease to exist? Do any H2O molecules drift off as hydrogen atoms and oxygen atoms? Or do any hydrogen atoms and oxygen atoms combine with other atoms or molecules to form new substances? Conversely, are there any naturally occurring processes on Earth which create H2O? Bus stop (talk) 04:40, 4 December 2012 (UTC)[reply]
There are certainly biological processes which do both. Also, an acid plus a base produces a salt plus water, and that reaction happens in nature both inside and outside of organisms. StuRat (talk) 04:46, 4 December 2012 (UTC)[reply]
Also, I'd expect water ice to be more stable than either liquid water or water vapor. On Earth, water ice tends to only last for a few hundred thousand years, in Antarctica, say. However, liquid ice deep inside Pluto or beyond might have formed when the solar system formed, some 4.3 billion years ago, and never melted since (unless tidal forces from Charon (moon) heat it enough to melt the ice). In that case, those molecules should mostly be the same as they were when formed. StuRat (talk) 04:57, 4 December 2012 (UTC)[reply]
  • In quantum mechanics, particles do not have individual identities that persist over time, and molecules, which are arrays of particles, don't either. When two water molecules come close enough together for their wave functions to overlap, it is impossible even in principle to say which is which afterward. So really the question is meaningless, and some of the responses above are too. Looie496 (talk) 05:50, 4 December 2012 (UTC)[reply]
Atoms with individual identities
    • I don't know that you can say that it is impossible to identify individual atoms with individual identities. I mean, this picture does a pretty good job of that. Those bright spots are individual carbon atoms, and the entire structure is a macromolecule known as a carbon nanotube. You can not only resolve the structure as a whole, but individual carbon atoms within it. --Jayron32 07:01, 4 December 2012 (UTC)[reply]
If we consider an oxidanyl group to be the core of a water molecule, how long is the average lifespan above sea level an oxidanyl group with nothing else attached, besides more hydrogen? Plasmic Physics (talk) 06:05, 4 December 2012 (UTC)[reply]
Certainly we can calculate the rate of the reverse reaction of H2O --> 1/2 O2 + H2. It's vanishingly small, but finite at 300K ==> long but finite lifetime. John Riemann Soong (talk) 06:12, 4 December 2012 (UTC)[reply]
Please do. I'd be interested to know if they last thousands, millions, billions, or trillions of years. StuRat (talk) 06:16, 4 December 2012 (UTC)[reply]
Such a calculation is not trivial, as this is not a single step, or elementary ("single hump") reaction. It involves in theory an infinite number of component reactions. However, if only high "engineering" accuracy results are required, you can ignore intermediate forms occuring at less than parts per million concentrations - then the species involved will be limitted to:-
H
H2 (end)
O
O2 (end)
O3
OH
HO2
H20 (start)
H2O2
A total of 136 competing elementary reactions occur between these 9 species. If anyone wants to know what they are, ask, and I'll post.
There are two approaches to the calculation: 1) use the mathematics of dissociation. This requires knowing the enthlapy of formation of each species (no problem, you can look it up in standard tables), writing the dissocation equation, and solving for the roots of the equation. Therein lies a problem - the calculation converges very very slowly on to the roots - days and days of computer time is required (assuming a typical up to date PC), and it is so darm easy to make a mistake in writing the equation. When you get the answer, you'll find the overall reaction rate is so slow, like one molecule every few hours or something, the result can't be valid on relativistic grounds. 2) use reaction rate data in the modified arrhenious equation for each of the 136 reactions, and solve 136 non-linear simultaneous reactions for the overall rate. That has two problems - a) good data is not readily available for all the reactions, so some devious pondering is required, b) solving 136 non-linear reactions is going to take some serious computer time.
However, if you only want to know roughly, you can ignore O3 and cut it down to 40 or so reactions.
Ratbone 120.145.182.171 (talk) 07:00, 4 December 2012 (UTC)[reply]
I like it rough, so please proceed. StuRat (talk) 07:04, 4 December 2012 (UTC)[reply]
Who, me? Or John RS? I accept bribes if they are large enough. But even then, you won't get an answer (even a rough answer) until long after this question has been archived. Ratbone 120.145.182.171 (talk) 07:14, 4 December 2012 (UTC)[reply]
OK, never mind then. StuRat (talk) 20:37, 4 December 2012 (UTC)[reply]
Hmmmm, I think you're looking for the "hydrogen exchange rate" or perhaps "proton exchange rate", "hydrogen-deuterium exchange rate" etc. Looks like [10][11][12][13] might be useful but I haven't accessed; the first abstract says OH- exchange predominates (for water in organic solvents). There are sites like [14] for proteins and surely water is a simpler case. I haven't found your number yet but the truth is definitely out there. Wnt (talk) 01:51, 5 December 2012 (UTC)[reply]

centripetal acceleration and linear acceleration, assuming "constant effort"

Linear plot of the "w1118" genotype
Logarithmic plot of the "fumin" genotype

Following literature concerning velocity and curvature, I decided to explore the power law among my fruit flies by taking their trajectory data and plotting log v against log R for each fly, where R is 1 over the calculated instantaneous curvature at a point. If v = K * R^b, the slope of this plot should be b and the intercept log K. Different genotypes (at least 4 different ones) appear to have per-fly average slopes ranging from 0.465 to 0.499, often with statistically significant differences between the averages: i.e. 95% confidence intervals are on the order of +/- ~0.008 whereas the difference between w1118 and fumin is ~0.03. To the right are two histograms of two genotypes, each a histogram of a collection of 60-75 flies over 4-5 experiments per genotype. One histogram has frequency colored linearly, and the other has frequency colored logarithmically-- I'm doing this to highlight different aspects of my analysis. Despite the differences, the values of the slope imply that b would be almost half, which would be consistent with centripetal acceleration being constant, since if v^2/R = K^2, where K is some constant, then v = K * R^0.5, consistent with the power law. It's the small deviations from this law that are interesting, since I think this is what separates different genotypes from each other.

The Spearman correlation (or even just the plain old Pearson r) is quite decent, on the order ~0.85. However, there is much overlap between the genotypes' distributions, though the means are statistically significantly different probably because of each genotype's sample size being on the order of n = 60-75. I was thinking that part of the spread in the plot could be due to the fact that centripetal acceleration isn't really constant and as linear acceleration will vary too, if we assume that the total "effort" would be nearly constant or fall within a tight range.

I'm actually surprised that my fit is that good, so I'm thinking, how do I get even tighter distributions that can distinguish between genotypes? The thing is, I can't think of what relation would constrain linear acceleration and centripetal acceleration together. I don't think it's really accurate to model linear acceleration + centripetal acceleration = constant, because I think the centripetal acceleration in an arbitrary non-circular path would include portions of linear acceleration. (And centripetal acceleration is a fictitious force and all that, so one might imagine it as a convenient way to talk about linear acceleration in a different reference frame.) Yet I don't think centripetal acceleration = linear acceleration either, even if we model an arbitrary path as the sum of tiny circular paths with ever changing radii (and directions) of curvature. (Ideally, each circular portion is infinitesimal in size, but we sample only at 15 Hz.)

The idea that effort is constant isn't too ridiculous, because the flies' activities are being driven by super-intense blue light (the wavelength they are most sensitive to) and if they rest, they only rest for a brief moment. You can see that to avoid taking the log of 0, and avoid the effects of noise at the same time, I've filtered all the data points where velocity dipped below 0.1 mm/s (that is, all points below log v = -1).

The basic power law relation I'm using predicts that velocity should be infinite for a straight-line path (infinite curvature), and that curvature should be infinite at zero velocity. Some literature offer a correction of R_eff = R / (1 + alpha*R), where infinite R would imply a finite R_eff of 1/alpha, so K*(1/alpha)^b would be the "max velocity". Curiously, I don't appear to have hit the upper limit of this relation, because the straight-line trend without the correction appears to carry to 90 mm/s (log v = ~2, the upper cutoff point, to avoid using data with random tracking errors, which are usually detectable because velocity is some ungodly amount for flies, like 360 mm/s), and using this correction generally makes the overall linear trend worse. In an arbitrary path model, would centripetal acceleration be basically linear acceleration even for seemingly straight line paths? How do I come up with a better "constant effort" model that would also account for the change in intercept (and in the transition zone, slope) at low velocities ? John Riemann Soong (talk) 05:54, 4 December 2012 (UTC)[reply]

One problem with your model is that in an ideal world the fly has to expend no power on a continual basis in order to generate centripetal acceleration whereas to accelerate in a linear fashion requires power. Do you have a model for the L/D ratio of the fly? Do you think the flies perform at their aerobic maximum, or are strength limited? Greglocock (talk) 22:39, 5 December 2012 (UTC)[reply]
I'm measuring walking, not flying. The calculated (supposed) centripetal forces are on the order of ~0.02g. Flight muscles are likely to be stronger than leg muscles (are they actually?) but since a fly has to generate at least 1g of lift during flight, probably on the order of several g for rapid ascent, I think the flies are aerobic-limited? How do I modify the equations for general planar motion if there is friction? Or can I use the equations as is? Are you saying that flies should not be using any energy in centripetal acceleration but my model is not reflecting that? My flies are being confined to a 2D planar arena (they are forced to walk), so I'm trying to wonder what a "real walking" model should include.
If there is friction, tighter turns should require more energy and effort, right? John Riemann Soong (talk) 03:17, 6 December 2012 (UTC)[reply]
Also, there is an observed 1/3 power law in idealized handwriting and eye saccades, and some other movements-- this is in the presence of friction. Both a "minimum jerk" model (a criterion for smoothness) and the viscoelastic model (reducing strain on a nonpoint body) are suggested as origins of the trend. John Riemann Soong (talk) 03:20, 6 December 2012 (UTC)[reply]
In a frictionless world with no losses centripetal acceleration requireds no continual power input, whereas linear acceleration does. That means it is a lousy model for walking. Frankly I doubt that walking in circles uses more power than walking in straight lines. Perhaps it would preferentially tire one set of muscles. Sorry, i can't think of a way to help you further. Greglocock (talk) 22:55, 6 December 2012 (UTC)[reply]

Mauser 98

Does the Mauser 98 give any positive indication when the ammo in the magazine runs out? In other words, when the shooter fires off the last cartridge, how does he (she) know that it's time to reload (other than by counting shots)? 24.23.196.85 (talk) 06:51, 4 December 2012 (UTC)[reply]

While there may be variations - after all the Mauser 98 action was manufactured for a long time and for a great many customers - my M59F1 (built from an ex-german M98k receiver) has a knob on the magazine follower that will hold the bolt back when the magazine is empty. Only way to close the bolt is to fill the magazine or push the follower down with a finger - it is natural to assume that such a device is common to most or all M98 actions. WegianWarrior (talk) 09:16, 4 December 2012 (UTC)[reply]
This article on the The 1896 Swedish Mauser says "After the last round fired and ejected, the follower locks the bolt open for rapid reloading", although this was an earlier design than the M98. However, the Persian vz98/29 Mauser (made in Czechoslovakia for Persia (Iran)) " It’s useful that the magazine platform does not lock the bolt open after the last round has been fired and ejected. As an idiot’s guide it’s probably useful to show some thick squaddy that the gun is empty, but not required for the sort of shooting we do." So it looks a bit variable. Alansplodge (talk) 14:37, 4 December 2012 (UTC)[reply]
Thanks! I was asking specifically about the Karabiner 98k version used by the German Wehrmacht during World War 2. 24.23.196.85 (talk) 22:47, 4 December 2012 (UTC)[reply]
This page has some downloadable instruction manuals for the 98k. This simplified manual for a late-WWII 98k doesn't mention it. Alansplodge (talk) 17:01, 5 December 2012 (UTC)[reply]
Thanks, Alansplodge! I've checked the manual, and it says nothing about the bolt locking back when the magazine is empty. So, I can safely infer that the only way to know for sure is to count your shots -- and if you happened to take the rifle from an enemy who fired it and hadn't reloaded (as my character has done), you're in danger of ending up with an empty mag without knowing it. 24.23.196.85 (talk) 01:41, 6 December 2012 (UTC)[reply]

what's the differences between forged steel and wrought steel?

what's the differences between forged steel and wrought steel?thanks a lot. — Preceding unsigned comment added by 121.15.133.68 (talk) 06:54, 4 December 2012 (UTC)[reply]

Do you mean how they are made, their properties, appearance, or what ? StuRat (talk) 07:00, 4 December 2012 (UTC)[reply]
I've never heard of "wrought steel". There is Wrought iron, which is, I suppose, an ultra low carbon steel. Its properties can be compared to Steel itself, by looking at both of those articles. --Jayron32 07:41, 4 December 2012 (UTC)[reply]
This page describes wrought steel as carbon steel that is shaped by rolling rather than casting. Looie496 (talk) 19:12, 4 December 2012 (UTC)[reply]
The term "wrought" is a little bit confusing, because it refers to both a process of working metal; and also to a chemical recipe for certain types of metal (usually iron) that are soft enough to be useful for that style of metalwork. In this case, wrought steel is referring to steel that has been worked. See also, wikt:wrought. In fact, ASME (an organization that is generally well-regarded for its expertise in the fields of materials science and engineering) has standards related to wrought iron, wrought copper, wrought steel, ... the list goes on and on. I have never personally worked with anything called "wrought copper," but nonetheless, there's at least one ANSI standard for it, (ANSI/ASME B.16), and in fact there are entire trade organizations centered around it. Nimur (talk) 15:27, 5 December 2012 (UTC)[reply]

TEXTABLE LANDLINE TELEPHONES! - Where can I buy one?

If you know the correct article title, you may edit the original dead-end WL to it.

I've been searching high-and-low for landline phones that can send and receive SMS, but even Best Buy has none. Why are they so hard to come by???

If they're not available in the good ol' US of A, then what are some GREAT textable landline phone models that I can order from overseas? (Or if there are some textable landline models sold from some online specialty stores that are physically located in a US warehouse, that's even better due to reduced shipping costs!)

I vow NEVER to get a landline until I find one that sends and receives texting. Why are they so hard to find in America in the first place? Thanks kindly. --70.179.167.78 (talk) 08:29, 4 December 2012 (UTC)[reply]

Once you have the phone, then what? Have you evidence that American phone companies have the ability to send text messages to a land line phone? I've never heard anything about that being possible. Which isn't to say it's not but it's not that popular. Dismas|(talk) 09:08, 4 December 2012 (UTC)[reply]
The American landline phone standard, POTS, was designed decades ago and while they have expanded the service options over time, the system generally does not support text messaging (i.e. SMS). Some of the landline alternatives, such as home VoIP through an internet service or digital phone from a cable company, may offer SMS options but they are still rare in my experience. You should verify that the phone service you want to use supports text messaging at all, before worrying about where you can find compatible phones. Dragons flight (talk) 10:54, 4 December 2012 (UTC)[reply]
If you can use a landline, you must be in a building. And if you are in a building, you can use a PC to access the internet. And if you can do that, great, because many phone companies offer a website that lets you send an SM to a mobile phone on their network. Nobody can send an SMS to you though. Why not use email? Or just phone them? Ratbone 124.182.41.70 (talk) 11:32, 4 December 2012 (UTC)[reply]
The people above are correct - the land line phone system doesn't support SMS. SMS is only possible on mobile phone networks. You can also email texts to cell phones - each carrier has a different format for the address associated with a phone number. You can also send a text message to an email address via SMS on the phone. There are devices that connect to the mobile phone network and let you plug a landline phone into them, but I can't recall what they are called. Perhaps one of those has SMS built in, but you would have to access it from the base, not the landline phone plugged into it. 209.131.76.183 (talk) 13:15, 4 December 2012 (UTC)[reply]
actually, IIRC, texting goes via the same path as the caller id; which is at least optionally available on landlines. so in principle, it should be possible? Gzuckier (talk) 18:40, 4 December 2012 (UTC)[reply]
No, it's very different. In North America the CID burst is an analog frequency-shift-key encoded sequence (a Bell 202 variant) transmitted in-band (and if the phone is off-hook, preceded by a DTMF-like alert tone); arrangements in different phone systems are similar. SMS is an out-of-band digital message type sent as a Mobile Application Part message on a GSM digital connection. There is no agreed upon standard (at least that I've ever heard of anyone actually implementing) for wrapping SMS in an in-band (FSK, QAM, whatever) burst. ISDN, which does support wrapping additional protocols in messages in a D channel, can support SMS, and some ISDN handsets do (but I don't know how standard and interoperable this really is). -- Finlay McWalterTalk 18:55, 4 December 2012 (UTC)[reply]
Fax it: send and receive text. --VanBurenen (talk) 19:21, 4 December 2012 (UTC)[reply]
May I ask, what about a land-line do you actually want ? The more comfortable headset ? Not having to charge it ? Or something else ? Because, if you have a cell phone already, a land line does seem like an unnecessary expense these days (although I have one, since it's bundled with my Internet). StuRat (talk) 20:35, 4 December 2012 (UTC)[reply]
Sound quality is usually better on land lines. --Trovatore (talk) 00:29, 5 December 2012 (UTC)[reply]
I asked pretty much the same thing Wikipedia:Reference desk/Archives/Science/2011 December 30#What'll it take to upgrade landline networks to allow texting to and from landline phones? but never received a response from the OP (which doesn't seem uncommon for them). Me and Astronasut made a similar point to Dismas and most other respondents basically touched on similar ideas as to the above discussion but this doesn't seem to have been take on board either. Nil Einne (talk) 23:00, 4 December 2012 (UTC)[reply]
It does seem like the companies offering landlines have concluded that their extinction is inevitable, and thus decided to invest the bare minimum to keep them operational until that time. So, any R&D, such as into supporting texting, is off the radar. StuRat (talk) 00:22, 5 December 2012 (UTC)[reply]
wandering offtopic here, but the reason i keep the old landline is that it was built to survive nuclear war, so the odd hurricane won't phase it. whereas isdn lines and cell towers are more fragile. of course, eventually, there will be nobody for me to talk to call from my cold and dark house in the disaster area. Gzuckier (talk) 06:05, 5 December 2012 (UTC)[reply]
I'll talk to you! I hope landlines don't go away. I like being able to make calls indoors - I see people standing out in their yard so they can get reception. Also, if I dial 911 they know where I am. Also, it is so easy to use: just pick it up to talk, put it back down to hang up - no extra buttons to press. And my cable goes out so often - if I relied in them for the phone it would be out of order a considerable part of the time. And the landline doesn't drop out! And, perhaps most of all, is the sound quality. Cell phones have terrible sound quality. I think it is bad manners to call someone with a cell phone unless there is no alternative. Bubba73 You talkin' to me? 06:22, 5 December 2012 (UTC)[reply]
I don't think any of those are absolute limits of cell phone technology:
1) A cell phone network can be made more reliable than land lines, since, if one tower goes down, it can glom on to the next. If the telephone wire leading to your house is cut, though, that's it, no hope of using it until a repairman is sent out. I just had my land-line go down, but, other than reception problems, have never had a cell phone go dead on me.
2) I actually have better sound quality on my cell phone than land line.
3) Flip phones can be made to answer when you flip them open and hang up when you close them. Clam-shell phones could do the same.
4) Reception can be excellent, if they have enough cell towers.
5) A cell phone with GPS could direct emergency personnel to your location, even if you are cowering up a tree with a grizzly bear below.
There may be one absolute limit of cell phones, though, and that's on comfort. They really can't provide a full-sized headset and still have it be as portable as they want. A compromise might be to place your cell phone in a docking station when home, which could charge it and also tie in to the house phones, and add a bigger antenna, to boot, and maybe a full-sized QWERTY keyboard for texting. StuRat (talk) 06:38, 5 December 2012 (UTC)[reply]
Landlines won't go away - they'll just change into something better. The trouble with cellphones and any other sort of radio technology is that it won't scale well. The more users use it for data (internet, video on demand, etc), the more the number of cellphone base stations (towers) and/or the radio bandwith required becomes impossible. More of the current problems with landlines (call quality, reliability, bandwidth) is due to the fact that the Customer Access Network (CAN) ie the copper cable network that connects dwellings and offices to the phone company exchange is old. A lot of cabling is 30 to 50 years old and is just about stuffed - in Australia at any rate. Image the car you might have bought 50 years ago - it would have been reliable back then. But now its rusty and worn out. But they have started rolling out Fibre-To-The-Home (FTTH) to replace the CAN. Because it's new it should be reliable - like a new car. And the communication is via VOIP digital instead of voice frequency analog, so upgrades to what your fixed phone can do will be a lot easier. Ratbone 120.145.196.13 (talk) 06:58, 5 December 2012 (UTC)[reply]
I'd argue that land-lines don't scale well, since, if you need to dramatically increase the capacity, at some point you need to lay more wire (or fiber) to every house. That's a lot more work than upgrading the equipment at the cell-towers. StuRat (talk) 16:56, 5 December 2012 (UTC)[reply]
I'd avoid bundling like the plague. If your landline is part of the cable service, you have to use the mobile to call for repairs. One less level of redundancy.
Mobile services are generally not very reliable. I've had 4 outages during November, that is one per week (although that's above average) and there was no reception from any secondary tower. Too loose a spacing, I guess. I've never had my landline fail, only my phone.
Mobiles are more prone to failure, too: made in China, and their batteries can fail.
Even if there is a secondary tower, I doubt that the landline would fail more easily than both towers at the same time. I doubt that "more towers" can be extended on demand. The aether can only take so much. That's no problem with landlines; signals will not carry over from a line to the other.
Mobile is susceptible to flooding. A single jammer used by terrorists could take a large chunk out of a mobile service. It's not that easy to kill landlines.
Flooding 2.0 - If too many users are on, the service degrades. At least twice a year (xmas, new year), mobile would go down on me.
Poor sound quality on a landline seems to be an issue related to the phones themselves, not the lines, or the line is already damaged and needs replacement.
"the companies offering landlines have concluded that their extinction is inevitable": Except T-mobile. They extincted their mobile service for half a day.
Phones with GPS, that is actually useful. OTOH, the towers could easily add similarly useful info which would be spoof-proof (tower ID + direction + approximate distance).
Texting vs. faxes: At least faxes have the advantage that you can "print" documents to faxes. No need to type them using a keypad which is no better then your average remote. If you have the time to text, you could as well call and save some time. Even Apple are expected to support phone calls by 2030. - ¡Ouch! (hurt me / more pain) 09:51, 5 December 2012 (UTC)[reply]
RE: "the companies offering landlines have concluded that their extinction is inevitable", you seem to have misinterpreted my comment to refer to outages. I mean that they think land-lines will no longer exist at some point. In the US, at least, there's the additional issue that all the taxes go on the land-lines, so you get a minimum of like $30 a month, even if you don't use it. My cell phone, on the other hand, is $7 a month.
As for cell phone networks being overwhelmed by traffic, this has happened on land-lines, too, during holidays. It seems to be less of a problem, lately, though, presumably because cell phones have taken much of this burden off the land-line network. So, this is not an inherent problem with cell phones only.
The advantage of texting, just like an e-mail, is that it doesn't require an immediate response. So, if you're in a meeting, and your significant other wants to ask when you will be home, would you rather have your phone ring, or have a text come in, which you can answer after the meeting ?
The advantage of a full-sized QWERTY keyboard, as in a PC, is that I can touch type, dramatically increasing my speed. Those on a cell phone are just too small to permit this. StuRat (talk) 16:38, 5 December 2012 (UTC)[reply]
Where I live I have never experienced a mobile network outage, not even when everything else has gone down. I suspect it's more down to how competitive service providers are in your area. --83.84.137.22 (talk) 16:22, 5 December 2012 (UTC)[reply]
Another problem with cell phones is there is a long delay whereas a landline is essentially instantaneous. I often hear myself being echoed back. And on a landline you can talk and listen at the same time, just as in a normal face-to-face conversation. Cell phones don't allow that. Those two things together make talking on a cell phone like the awkward remote TV interviews where there is a delay before the other person starts responding so the interviewer starts talking and at that time the remarks of the first person come back, etc, etc. Bubba73 You talkin' to me? 16:32, 5 December 2012 (UTC)[reply]
I wonder if the 2-way versus 1-way at a time issue is an inherent limitation of cell phones or just a poor implementation. It might be necessary to turn off the speaker when you're talking, to limit echoes, since the speaker and microphone are too close together. I wish they had tri-fold phones which could still be small yet expand out enough so the microphone is actually by your mouth and the speaker is actually by your ear. The increased distance, and sound damping from your head, should make it possible to enable simultaneous 2-way communication without echoes. StuRat (talk) 16:43, 5 December 2012 (UTC)[reply]
There are 2 reasons why cell phones do not have two-way simultaneous talking ("duplex working" is the technical term). Firstly, two simultaneous directions requires two radio channels, doubling the bandwidth. Bandwidth is precious and MUST be conserved. Secondly, the transmit power power is up to 1 x 1011 times the recieve level, even though it is only around 200 mW. The tranmist energy overloads the sensitive recieve circuits, even if not on the exact same radio channel. It is readily possibly to fix this problem with what is known as discrete directional filtering, but such filters take up a lot of space, and would take the size and weight of a cellphone back to that of a housebrick, or aborb a lot of power, or both. There is a third, minor, reason: The same computer-like chips in the phone are used to process both transmit and recieve signals. To work in both directions simultaneously, you'd need increased processing grunt, increasing the battery drain. Battery capacity is precious too. Feedback from speaker to microphone is not a problem. The success of the better types of loudspeaking conference phones demostrates the success of the electronic enginner's bag of tricks (hybrids and VOGC's) in that regard. Keit 120.145.164.75 (talk) 01:21, 6 December 2012 (UTC)[reply]
Re: "cell phone network can be made more reliable than land lines" back there - I've had a land line for 40 years. I think there have been three times when the service was out. Possibly four times, but I don't think so. So about one outage per 10 years. How many times do I get "no service" on my cell phone? Many. How many times to I call a number and NOTHING happens? Many. How many times are there dropouts on a cell phone? Many. Bubba73 You talkin' to me? 19:31, 5 December 2012 (UTC)[reply]
You're somewhat comparing apples and oranges there if you compare your land-line, used in one location, with a cell phone, used at multiple locations. If you just compare your cell phone, used at home, with your land-line, that's more fair. And, again, just because some cell phone service is crap doesn't mean it always must be, due to limits of the technology. A similar argument could be made on cars, early on: "Those darn horseless carriages break down once a week, but my horse has been reliable for 20 years, so cars will never replace horses". StuRat (talk) 19:38, 5 December 2012 (UTC)[reply]


I don't use my cell phone at home because (1) I want to be able to talk indoors, (2) cell phone is too unreliable, (3) sound quality is bad, (4) it is too hard to use. Bubba73 You talkin' to me? 21:06, 5 December 2012 (UTC)[reply]
And another thing - with a landline, you get in the phone book! Very important. Bubba73 You talkin' to me? 22:22, 5 December 2012 (UTC)[reply]
About landlines not overwhelmed any more: If broadband is possible via landline, they can transmit 1000's of connections via the same line. And if they can serve 10% of the broadband users at the same time, they can handle 5 times as many telephone calls, too. No need to share the burden with mobile networks.
"The advantage of texting, just like an e-mail, is that it doesn't require an immediate response": often cited but answering machines exist for a reason. On top of that, there is e-mail, which is actually standardized, unlike SMS. Characters like € or ñ get mangled, not to mention ¡, even if you can produce them on a phone.
About outage frequency and redundancy being "down to how competitive service providers are in your area" - bloody hell, I can be 20,000ft ASL in the himalaya and "click here to find single women in my area"! If they can pull that feat, they can as well maintain the damn network...J/K
I doubt that a cell phone is not full duplex. You cannot flood a phone by a simple countinuous signal like a beep.
I think the pause between transmitting and receiving is caused by the packeting. Cell phones cannot simply transmit the sound samples uncompressed; they have to compress a packet of 1024 samples or so into a 2kbit data packet. (The numbers may be WAY off but the reasoning is the same.) That's an average of only two bits per sample; if they quantized each sample down to 2 bits, the quality would be extremely poor. The downside of packeting is that they must record 1024 samples, encode them, and transmit. Thus the receiving phone can only start playback when the entire packet has been encoded. At 8k samples per second, we have that kind of lag in both directions, and it sums up to at least .256 seconds each. Add to that lag the actual encoding, transmission, and decoding, and it can easily double.
"You're somewhat comparing apples and oranges there if you compare your land-line, used in one location, with a cell phone, used at multiple locations." I don't see that. Apples and samsungs, maybe... and the cell phone is not meant to be used in one place only. If it were, it would have a more reliable connection (I'm not an engineer but I'd suggest copper), it would do away with most electronics, and it would draw its power from a more reliable source than a battery. See where I'm going?
Ahem. The point is that mobile networks don't care about issues with mobile telephones the way they should. Part of that is expected at $7 a month. If there were taxes, many of those issues would probably get resolved. Why not? If there were taxes on mobile calls, there would at least be fewer verbal pollutants walking-and-talking. Result. - ¡Ouch! (hurt me / more pain) 08:33, 6 December 2012 (UTC)[reply]
Yesterday I bought two years on my Tracfone for $160, so $6.67 a month, unless you include the $50 phone purchase, then it's $8.75. It's a decent cell phone (Samsung S425G) and service, although I rarely use a phone. If I used it often, I'd need a more expensive plan. The only problem I've had with Tracfone is how incredibly painful it is to transfer a phone number, service days, and minutes from one phone to another. This involved hours on the phone with customer service people who barely speak English, waiting for them to mail me a new SIM card (which required a signature, so they couldn't deliver it with me not home), having to re-enter all the phone numbers into my contact list, etc. You also had to figure out just what chain to follow on their phone machine hell. If you picked the wrong path they would say "Sorry, we are experiencing unusually high call demand, please call back later". If you picked the right path, they would answer immediately.
Of course, having just had an outage on my land-line, it was almost as bad. I didn't have a dial tone, and, since I rarely use the phone and it was before the US elections (meaning, if fixed, it would ring off the hook with politicians, who exempted themselves from the no-call list, bothering me) I decided to just wait and see if they'd fix it on their own. They didn't. A month later I decided to go to their website and request a repair, which involved me going outside to connect to the line out there, to test whether the problem is outside the house or inside. Can you imagine if a 90-year old woman had to do that ? So, customer service is iffy either way. StuRat (talk) 17:57, 6 December 2012 (UTC)[reply]
I have Tracfone too, but I don't use it at home. I normally leave it in my car, except when I bring it in to charge. If it rings at home, I don't answer it because I know it is a wrong number, since I don't give out my cellphone number. Bubba73 You talkin' to me? 17:20, 7 December 2012 (UTC)[reply]
I hope you live in Hawaii or someplace else with a year-round moderate climate, since temperature extremes can damage portable electronic devices. Also, even if you don't give your number out, when you call people from it, they get your number by caller ID, and may add it to their list of contacts. StuRat (talk) 00:57, 8 December 2012 (UTC)[reply]
And speaking about the reliability of the landline - the last time mine was out, it was because of a broken wire in my house. And the time before that was because of water in my outside box after heavy rains. Bubba73 You talkin' to me? 17:32, 7 December 2012 (UTC)[reply]
Well, those are two examples of failure modes not possible on a cell phone. StuRat (talk) 00:59, 8 December 2012 (UTC)[reply]


And that is it. About one failure per decade. I can't count the number of times I had no signal, there were dropouts, or I called and nothing happened on the cell phone. And I don't use it that often. Bubba73 You talkin' to me? 02:52, 8 December 2012 (UTC)[reply]
Wow, I just saw an ad for the home cell phone docking station I advocated above. It's the "Straight Talk wireless home phone". I doubt if it does all the things I suggested, but it's a start. StuRat (talk) 00:52, 8 December 2012 (UTC)[reply]

House wall mold sensor?

Is there any substance that can detect mold growth inside walls? such that when mold start to grow it will also grow on a sensor that will alert the house manager? which can then preemptively take action. This is different from measuring the conditions such as temperature an moisture which doesn't indicate actual mold. Electron9 (talk) 13:25, 4 December 2012 (UTC)[reply]

In my area, the humidity runs about average and mold grows on moist foods just about anywhere and everywhere you happen to leave it out. I'm fairly sure that the mold spores are widely dispersed from soil and foliage here. Thus, if you taped sterile food or growth medium to a wall or inside it you would quickly grow plenty of mold in a day or two, but that wouldn't indicate anything. I haven't lived in any of the more arid areas of the country though, where the spores might be less of a problem. In addition, the available moisture within the wall will likely differ in different locations. So if you placed dried bread high in the wall, it may remain relatively mold free for a while, until it absorbed enough moisture. But even then, its hygroscopicity will likely differ from the wall materials, such that the moisture conditions differ, especially in areas near window sills, bathrooms and basement walls. Low levels of mold are frequently present anyway, which is why storing clothes in such darkened moist areas is generally not a good idea. Modocc (talk) 14:02, 4 December 2012 (UTC)[reply]
That said, the odor of mold/mildew is usually fairly strong and detectable when it does become a problem, thus its likely that an electronic nose will be a very useful maintenance tool. Googling on mold electronic nose appears to bring up some relevant literature. --Modocc (talk) 18:44, 6 December 2012 (UTC)[reply]

I was reading this article but I don't exactly understand what it is. Is it a psychological disorder or a physical disorder of the nervous system? Thanks. 82.132.210.65 (talk) 14:48, 4 December 2012 (UTC)[reply]

Neurasthenia was a 19th century disease concept. It describes a combination of symptoms used by doctors to make the diagnosis. The main symptom was fatigability so severe that the patient felt unable to engage in normal work and normal family and social obligations. The doctors knew that there were no objective structural abnormalities in the body (i.e., no "pathology"). There were many hypotheses of cause. These days patients with those same symptoms are likely to be diagnosed with chronic fatigue syndrome. Both of these conditions are good examples of attempts to provide a social role and medical response for symptoms that are not at all understood at the organ or tissue or cellular level. alteripse (talk) 15:23, 4 December 2012 (UTC)[reply]

December 5

Subjective time

When hiking for four miles in and four miles out in a relatively unfamiliar area, my perception is that a landmark which I remember from "about halfway" on the way in is at little more than 1 mile in on the map; thus invariably when hiking I find that most of the landmarks I've remembered come out in rapid succession "close to" the end of the return trip. I take this to support an idea that subjective time, whether on a short or lifetime scale, depends on novelty (or retroactively from memory?) and runs according to the 1.7 or 1.8th root of clock time (it is relatively easy to derive a square root relation mathematically). Is there any literature on a subjective time formula that is more believable than the IMHO absurd inverse proportion mentioned in the article? Wnt (talk) 01:26, 5 December 2012 (UTC)[reply]

Instead of highway hypnosis...hiking hypnosis.GeeBIGS (talk) 02:43, 5 December 2012 (UTC)[reply]
I suspect that it's not a fixed formula, but depends on what you encounter on the way. In general, the things you encounter on the first half of a trip will be "newer", so you will notice them more (like the 1st weeping willow tree, perhaps), but notice subsequent appearances of similar objects less. So, it will seem like you encountered more during the first part of the trip than the last. However, if the first half was boring and then you hit some fantastic scenery, that might alter your perception. StuRat (talk) 06:25, 5 December 2012 (UTC)[reply]
The memorability of landmarks certainly depends on their novelty which decreases as one penetrates further into the previously unknown area. Since this observation depends so heavily on unpredictable subjective factors and previous experiences, it seems impractical to derive a meaningful mathematical relation to time (or distance) passed. SkylonS (talk) 11:22, 5 December 2012 (UTC)[reply]
I have always noticed that the first trip to a new destination always takes longer than the return and the second trip there even less time than that, tending to a constant rate on about the third and later trips. Is this (at least similar to) what you are talking about? μηδείς (talk) 16:46, 5 December 2012 (UTC)[reply]
Not really. In this instance worry about deer season had pushed me to a simple, straight path, which I'd already checked on the map before going, along the bottom of a valley with regular distance markers. I didn't even need to note landmarks, and my rate of travel was very nearly constant. Even so, I paid more attention to the distance markings on the way back, and was "predictably surprised" to find the thing I'd thought of as being halfway out not far from the first mile marker. Wnt (talk) 20:48, 5 December 2012 (UTC)[reply]

Buttons

Are all silver metal buttons on pants nickel?--Wrk678 (talk) 06:07, 5 December 2012 (UTC)[reply]

I doubt it. Some might be painted silver, with steel or some other metal underneath. Some might be copper or nickel-plated. And, for expensive high-fashion clothes, they may actually use silver (solid or plating). There may not be any which are all nickel. StuRat (talk) 06:21, 5 December 2012 (UTC)[reply]

I'm wondering because I'm allergic to nickel. Is it usually nickel plated. They are just 30 dollar pants. — Preceding unsigned comment added by Wrk678 (talkcontribs) 06:47, 5 December 2012 (UTC)[reply]

I've seen pants with plastic and cloth-covered buttons, but the zipper and any rivets might be a problem. You can get pants without rivets, but the zipper might be a problem. I've seen plastic zippers on coats, but not pants, since they need to be larger. There are button-fly jeans, but I don't know if you can get those in nickel-free buttons. Velcro would be ideal. Have you considered having pants custom made ? StuRat (talk) 07:37, 5 December 2012 (UTC)[reply]
Here's a site which sells nickel-free buttons and rivets, so you can have a tailor replace the ones on your pants: [15]. No mention of the zippers, though. They also have a test kit, so you don't have to trust the salesman who claims their pants are nickel-free: [16]. StuRat (talk) 07:42, 5 December 2012 (UTC)[reply]

I'm just wondering if most silver metal buttons are nickel plated.--Wrk678 (talk) 08:54, 5 December 2012 (UTC)[reply]

Well on [17] I see about one in six of the jeans metal buttons tested contained nickel. You might get away with just coating them with clear nail enamel just in case. Dmcq (talk) 09:45, 5 December 2012 (UTC)[reply]
Could you just paint the buttons with something like clear nailpolish? Vespine (talk) 21:33, 5 December 2012 (UTC)[reply]

Light traveling to a star

As I understand it, while the universe is 13.7 billion years old, the actual current distance from here to any star (let's talk about red dwarfs, which last a long, long time) that we've ever seen is more like 45 billion lightyears, due to expansion of the universe that has occurred since the light we see was emitted. As I also understand it, there could be some stars that we cannot see because their light could never reach us, and that are unreachable by light that we send out.(Correct me if I've got these things confused.)

Suppose there's a visible star that's x lightyears from us in terms of when it emitted the light that we currently see. Based on what we know now about the past and future history of expansion of the universe, are there stars that we currently see that we could never reach with a beam of light? If so, what is the critical value of visible distance x above which the star is unreachable by light that we currently send? Duoduoduo (talk) 15:55, 5 December 2012 (UTC)[reply]

Given that it's impossible for a star to move at the speed of light, if we sent a beam of light to x star, at some point in the future (billions of years perhaps) the light would reach that star (assuming it isn't swallowed up by black hole or distorted in some other way). I've got no idea of the rate the universe is expanding at, but again, assuming that it's less than the speed of light - a beam of light would, in theory, eventually reach the most distant part of the universe. Conversely, if the universe is expanding at the speed of light, then it wouldn't as light cannot travel faster than the speed of light. I may have got that completely wrong, I'm no expert at physics this advanced, so if I have, any astro-boffs feel free to correct me. douts (talk) 16:45, 5 December 2012 (UTC)[reply]
I thought it's established that there are stars that cannot communicate with us, and I thought that that was due to the expansion rate of the universe being greater than the speed of light. If so, my question stands. Or is it known that the expansion rate was only greater than the speed of light during the inflationary period, and not now, and never again? I can't tell from the article Metric expansion of space. Moreover, even if a star is moving away from us within space at of course less than light speed, and even if the universe itself is expanding at less than light speed, couldn't the sum of those two exceed the speed of light, making the star unreachable by our emitted light? Duoduoduo (talk) 17:03, 5 December 2012 (UTC)[reply]
I dont know is the honest answer, I'm not clued up enough on astrophysics to be able to give you a good answer. My response above is based my very limited physics knowledge. I'm sure there are other users who are much more clued up on this than I am. Sorry I couldn't be of more help. douts (talk) 17:23, 5 December 2012 (UTC)[reply]
They think the expansion rate was greater than the speed of light at some time in the past, but is not now. thx1138 (talk) 17:40, 5 December 2012 (UTC)[reply]
No, the expansion was never faster than light (that's a common misconception). More precisely, the rate of the expansion doesn't have units of speed, so it can't be compared to the speed of light. (The units are time−1, often given as kilometers per second per megaparsec.) -- BenRG (talk) 22:16, 6 December 2012 (UTC)[reply]
... but the expansion is speeding up at present, and we can currently observe galaxies with a red-shift greater than 1.5 which we believe are currently "receding" from us at a separation speed greater than the speed of light. (Please note that this doesn't mean that these galaxies are "travelling" faster than light.) Also galaxies at an estimated current distance of about 4,740 megaparsecs (146 million million million million miles) will soon disappear forever from our view because their light will never reach us. Galaxies further away than this (assuming that they exist) can never be observed by us (unless we find "wormholes" or some such trick to move without travelling). I don't have a figure for your "x" distance for light to reach the star from us, but I'm sure there are some experts here who can calculate it. Dbfirs 18:20, 5 December 2012 (UTC)[reply]
Galaxies do not disappear from our view. Ruslik_Zero 18:41, 5 December 2012 (UTC)[reply]
What makes you think that? The article Future of an expanding universe claims that in another two million million years all galaxies will have become undetectable except for those in our local cluster. Dbfirs 23:56, 5 December 2012 (UTC)[reply]
Technically they are still "in our view" but they are redshifted to the point of invisibility. It's the same as a black hole event horizon: you see a finite part of the galaxy's future redshifted over an infinite time. -- BenRG (talk) 22:16, 6 December 2012 (UTC)[reply]
  • See observable universe, which discusses these issues. It says some galaxies exist which we can see at early development, greatly redshifted, but light from them from later times will never reach here. How many can we see in their present form? Well, none, of course (unless you use a speed of light frame of reference...) Wnt (talk) 20:56, 5 December 2012 (UTC)[reply]
Thanks for the link. Like most people, I struggle to get my brain round the complexities the current understanding of the universe, but I think the answer to the OP's "x" is 16 million million light years (94 thousand million million million miles) as measured now. No doubt someone will correct me if I have misunderstood the symmetry. I'm not a cosmologist! Dbfirs 23:56, 5 December 2012 (UTC)[reply]
It's 16 billion (= thousand million, not million million) light years if ΛCDM is correct. ΛCDM is the simplest model that fits the data so far, but it could easily turn out to be wrong. In general the distance is , where a(t) is the scale factor of the universe. Since this distance depends on the scale factor at all future times, the number you get is quite sensitive to the model. If the cosmological constant ever decays to zero, the integral becomes infinite (meaning any distance is conceivably reachable). -- BenRG (talk) 22:16, 6 December 2012 (UTC)[reply]

Variable displacement with open exhaust valves

According to variable displacement, when a cylinder is deactivated all of its valves are closed. The piston then continuously work against the trapped exhaust gases infinitely until that cylinder is reactivated again. Is there any engine out there that keep the exhaust valve open continuously during deactivation? It seems by keeping the exhaust valve open less work will be wasted.Dncsky (talk) 16:25, 5 December 2012 (UTC)[reply]

I think I see a problem with that. You would then be pumping exhaust air (from the operational cylinders) in and out of the cylinder, and I bet that would cause deposits to accumulate in the cylinder in question. StuRat (talk) 17:05, 5 December 2012 (UTC)[reply]
Doh! How did I not see that before. Thanks for your help.Dncsky (talk) 17:07, 5 December 2012 (UTC)[reply]
Resolved
Besides, the work done on the gas trapped in a contained cylinder is conservative (parasitic losses ignored). In this case, the parasitic losses would be... gas escaping by leaking out. So, opening the valve actually is detrimental, as it makes the engine more lossy and less efficient. By keeping the system closed, any energy spent by the engine during compression is recouped in the next phase of the cycle. The problem is then reduced to an engineering technicality, the objective being to smooth out the power variations over a cycle. Nimur (talk) 17:10, 5 December 2012 (UTC)[reply]
The heat of compression would be transferred from the gas to the cylinder walls and to the radiator. If the exhaust valves were open there would be no compression and thus no heat loss due to compression.Dncsky (talk) 17:16, 5 December 2012 (UTC)[reply]
If the cylinder is perfectly closed, upon decompression the exact same amount of heat would go right back from the cylinder walls and into the gas inside the cylinder. Now, as I said above, we have an engineering practicality to consider: which parasitic loss is worse - the energy lost to turbulent convective air movement, or the energy lost to thermal conduction to the outside world? It seems that the engineers who design engines have carefully considered the problem, and decided to use a closed cylinder. That would seem to indicate that wicking away of thermal energy by the engine block is a less lossy process than turbulent airflow, which makes intuitive sense to me. In your own engine, with its own size, shape, and material properties, your mileage may vary, literally and figuratively. Nimur (talk) 18:10, 5 December 2012 (UTC)[reply]
Under full load conditions, the amount of heat lost through the cylinder walls to the coolant, and through the piston into the oil, is of the order 6% of the total combustion heat. Note that the total heat rejected to coolant is much much higher than this, as it includes heat lost in the cylinder head exhaust passages and exhaust valve (around 15% typically) and losses due to piston ring and bearing friction (which ends up as heat in the coolant). See R F Taylor's or Harry Ricardo's classic textbooks on high compression engines for typical figures. The 15% loss in the head will not happen if the exhaust valve does not open. 6% is a low value to start with. However, if the cylinder is not firing, the gas temperature will only be a small fraction of the combustion temperature, lowering the heat conducted to the cylinder wall and piston. Further, this heat flow is aided by gas turbulence within the cylinder. Some turbulence is driven by the shape of the piston top, but a lot is driven by the the gas flow past the intake valve. If the intake valve does not open, turbulence will be less, lowering the heat loss still further. My rough estimate (to complex to set out here) is that when not firing, the cylinder heat loss to coolant will drop to less than 8% of the full throttle value, which was only 6% of combustion heat anyway. Keit 120.145.164.75 (talk) 01:04, 6 December 2012 (UTC)[reply]
Using the numbers you provided the other day:
Brake Power Output:........265 kW (39%)
Total Loss to Coolant:....164 kW (24%)
Loss to exhaust:..............233 kW (34%)
Loss from engine surfaces..19 kW ( 3%)
Loss from closed valve variable displacement: 683kW * 6% * 8% = 3.3kW
3.3kW is admittedly negligible, but it's still larger than if the exhaust valves were open. But like Sturat mentioned opening the exhaust valves is unfeasible. Dncsky (talk) 01:50, 6 December 2012 (UTC)[reply]
I think you are right - loss with valves open should be less. StuRat was concerned about deposits, but I not sure this is a real concern. The Caterpillar 3516 engine (diesel, 2500 kW max output) with electrically controlled fuel injection turns off cylinders at low speeds and loads (unless this feature is disabled at customer option) by not pulsing the injectors in selected cylinders. The intake and exhaust valves continue to operate normally. Soot will not build up in the "off" cylinders as they are still pumping air through. It also picks different cylinders to turn off each time it decides to do it. In a gasoline engine, soot should not be forming. StuRat also realised that there would be new pumping losses in and out the cylinder. It is hard to accept that these would exceed the loss with valves closed, as you can view it as reducing compression, thus reducing the conversion of mechanical effort into heat to be conducted away. Keit 120.145.164.75 (talk) 03:03, 6 December 2012 (UTC)[reply]
But.... if the heat of compression is conducted away, then won't the loss of heat from expansion be refilled from conducting heat in? Gzuckier (talk) 02:00, 6 December 2012 (UTC)[reply]
Some of the heat will be conducted back to the gas, some of it won't. The part that doesn't goes into the coolant and then outside the car. Dncsky (talk) 02:45, 6 December 2012 (UTC)[reply]
Heat conducted into the gas during the low pressure part of the compression-expansion cycle will be less than the heat conducted out of the gas during the high pressure part of the cycle, with the geometric mean gas temperature equal that of the cylinder walls. This will modulate the piston pressure resulting in a net mechanical loss. Keit 120.145.164.75 (talk) 03:08, 6 December 2012 (UTC)[reply]
Here's another thought on this. Sorry, I know of no references to refuse or support it. Many people think that the exhaust gasses are driven out of the cylinder throughout the exhaust stoke (approx 180O crank rotation), but this isn't so. What actually happens is a blowdown phase when the valve opens, followed by a relative calm equalisation phase. The blowdown phase typiclly lasts 5 to 15 degrees of crank rotation, and during it the bulk of the exhaust gasses leave at a veclocity equal to the speed of sound (that's what makes the whack whack whack noise if there's no muffler). A lot of heat is transfered to the exhaust valve due to the presure and velocity. But during the remaining 700 degrees or so of crank rotation in the 4-stroke cycle, the valve can loose its heat via the valve stem into the guide, and for approx 540 degrees it is in contact with the valve seat. Even so, exhaust valves run hot, red hot.
If the exhaust valve is held open on a non-firing cylinder, with the intake valve closed, hot exhaust gases will be pumped past it thoughout the complete cycle. The gasses won't be a full exhaust temperature, as some cooling between cylinders will occur, but as it is continuous on the dead cylinder valve, and not a mere 10 degrees or whatever, it could mean the valve will run hotter than during firing, and could be past its' limit.
This is not a problem with diesel engines (eg Cat 3516) because cylinders can be shut down by ceasing fuel injection - the valves operating normally and pumping air through. But you can't do that on a conventional gasoline engine as the fuel is mixed with the air before the intake valves, so fuel would be wasted. But if you had a separate throttle body and fuel system for each cylinder???
Ratbone 121.215.53.208 (talk) 04:07, 6 December 2012 (UTC)[reply]

Antidepressants

How can pharma. cos. and the dr. and scientists they employ to test these drugs be so sure that they "may cause suicidal ideation", yet not also cause homicidal ideation? Logically, presuming that "this may caues you to care less than you currently do about your own life" it would follow that it might also "cause you to care less than you currently do about a random stranger's life"165.212.189.187 (talk) 18:56, 5 December 2012 (UTC)[reply]

It is based on monitoring patients who take the drugs. Pharmaceutical companies are required to collect information from patients and doctors about potential adverse effects of all the drugs they sell. Suicidal ideation is observed, but homicidal thoughts are not. 148.177.1.210 (talk) 19:02, 5 December 2012 (UTC)[reply]
Suicidal thoughts are a typical symptom of depression. So is a lack of motivation. If an antidepressant helps with motivation when the suicidal thoughts are still there, the risk of suicide goes up. Homicidal behavior is generally not associated with depression, so there are unlikely to be homicide-related effects of antidepressants. 148.177.1.210 (talk) 19:05, 5 December 2012 (UTC)[reply]
The premise is wrong. As the word "may" indicates, there is no certainty that these drugs cause suicidal ideation -- there is only some limited evidence to support that possibility. There is also no certainty that they don't cause homicidal ideation -- there just isn't any significant quantity of evidence to support that possibility. Looie496 (talk) 21:57, 5 December 2012 (UTC)[reply]
A person may find it less objectionable to admit to having urges to kill oneself than to admit to having urges to kill another person. Bus stop (talk) 22:23, 5 December 2012 (UTC)[reply]
A person might be given the possibility of reporting anonymously. OsmanRF34 (talk) 14:21, 6 December 2012 (UTC)[reply]
Using your reasoning suicide rates and homicide rates must be highly correlated in every country then. Dncsky (talk) 02:00, 6 December 2012 (UTC)[reply]
I don't understand your point. Bus stop (talk) 02:21, 6 December 2012 (UTC)[reply]
He's using reductio ad absurdum in order to demonstrate to the OP why his assumption is wrong. 146.87.49.23 (talk) 08:56, 6 December 2012 (UTC)[reply]
That still doesn't make sense. The OP is not saying that antidepressants cause suicide and homicide ideation. He's only asking if they are tested for that. Even is antidepressants caused both, the correlation wouldn't be between suicide and homicide, but between use of antidepressants and the combining number of suicide and homicide. OsmanRF34 (talk) 14:21, 6 December 2012 (UTC)[reply]
Did you read what the OP said? They specifically said 'Logically, presuming that "this may caues you to care less than you currently do about your own life" it would follow that it might also "cause you to care less than you currently do about a random stranger's life'. Nil Einne (talk) 16:24, 6 December 2012 (UTC)[reply]

So the question becomes: is a person who has suicidal thoughts more likely to also have homicidal thoughts than a person who has neither?165.212.189.187 (talk) 13:18, 6 December 2012 (UTC)[reply]

I don't think the reference desk can give a meaningful answer to that question. Obviously drugs (antidepressants) are tested by pharmaceutical companies for potential side effects. The question immediately above does not even mention antidepressants. Is that your intention—to eliminate antidepressants and other pharmaceuticals as a factor in the question that you are posing? I don't think we can just determine whether there is a correlation, and what that correlation might be, between thoughts of suicide and thoughts of homicide. It should be noted that homicide and suicide can occur together. Bus stop (talk) 18:53, 6 December 2012 (UTC)[reply]

Sure, if it helps you to conceptually compartmentalize the components and then retro fit the two later. Once you answer the q immediately above you can use logic: A therefore B therefore C = A therefore C. Right?165.212.189.187 (talk) 19:49, 6 December 2012 (UTC)[reply]

No. I am not trying to "compartmentalize" anything. I think you asked a great question. I'm referring to your first post. I think most of the responses were faulty in some ways. But I may have felt that most of the responses that other editors gave to your initial post were faulty because I perhaps misunderstood the thrust of your first post. After other editors devoted much discussion to teasing out the final iota of meaning from your first post, you re-posted a question. At that point I felt compelled to respond to your newly formulated question. As I indicated in my first post, I wonder about how one would go about collecting this information, and whether there may be a differential between the ease of collecting information on homicidal and suicidal thoughts. Bus stop (talk) 21:02, 6 December 2012 (UTC)[reply]
Sounds plausible. OsmanRF34 (talk) 14:21, 6 December 2012 (UTC)[reply]
Wellbutrin comes with a warning of a potential side effect of "homicidal ideation"[18]. Bus stop (talk) 17:59, 6 December 2012 (UTC)[reply]

why is it that advantageous for RBCs not to have nuclei?

Wouldn't it be more cost-effective if RBCs could repair themselves? It seems to me that the cost of constantly making RBCs would be more than the benefit of the extra space. Or is the risk of oxidative damage and subsequent cancer so great that all bloodmaking is confined to the marrow? 71.207.151.227 (talk) 20:21, 5 December 2012 (UTC)[reply]

The big advantage is that you can pack a lot more hemoglobin into the same space, and the cells are much more flexible and able to squeeze through tight capillaries whike suffering less mechanical damage. That outweighs any benefit you get from self-repair. By the way, mammalian RBCs lack nuclei, whereas those of fish, amphibians, reptiles and birds have nuclei. Dominus Vobisdu (talk) 20:33, 5 December 2012 (UTC)[reply]
There is a correlation between nucleated red blood cells and poor prognosis in intensive care [19] though this doesn't speak to the direction of the causality. Our article red blood cell contains the interesting information, which I hadn't known, that enucleation of red blood cells has evolved three times independently in vertebrates. I would be tempted speculate about the increased need for circulation in warm-blooded mammals, except... birds do all that and more. Fundamentally, saying why evolution went a certain way is a rather speculative enterprise. Wnt (talk) 20:40, 5 December 2012 (UTC)[reply]
Sometimes speculative, sometimes not. Often, it's pretty easy to determine how an innovation is advantageous. You're quibbling with the word "why", which has two distinct meanings, one teleological, and one purely explanatory, closer in meaning to "how". The former lies outside of the scope of evolutionary science. The latter does not. It's rather easy to explain why (= how) a four-chambered heart is advantageous, or why (= how) enucleated RBCs are an improvement over nucleated ones. The problem starts when you start attaching a teleological sense to the word "why". Dominus Vobisdu (talk) 00:40, 6 December 2012 (UTC)[reply]
Enucleated RBCs are smaller, kind of not quite donut shaped rather than fried egg shaped, and it would seem to me can negotiate the tight constriction of the capillaries better. A lousy comparison, but look at the trouble sickle cells have. Even if not directly causing clots, rbcs with a larger cross section would require lower blood velocity, require larger capillaries, or get a lot more wear and tear bashing around the tight turns. "Cold-blooded" critters don't have the same oxygen requirements we do and presumably can deal with reduced blood flow, while birds on the other hand have a more efficient lung design which can presumably provide sufficient oxygen with reduced blood flow from the "oxygen push" side. Plus the raw material of rbcs is largely recycled anyway. (Entirely guessing all this on my part, i should mention) Gzuckier (talk) 02:10, 6 December 2012 (UTC)[reply]
  • Nuclei are metabolically costly and exist to (1) express proteins, and (2) undergo cell division. But red blood cells basically serve as a substance meant to convey oxygen. They don't need to reproduce since they are produced copiously by the marrow, and they don't need to express proteins because they are already packed full of hemoglobin and do their job best just by transporting oxygen as efficiently as as possible until they wear out and are replaced. μηδείς (talk) 03:21, 6 December 2012 (UTC)[reply]
One could argue that with the loss of the nucleus and the related cell maintenance and division functionality, they are no longer real cells at all - just passive "packaging" for the heamoglobin. Roger (talk) 09:56, 6 December 2012 (UTC)[reply]
That's why I called them a substance. μηδείς (talk) 16:46, 6 December 2012 (UTC)[reply]
That is a position some people take, who believe in blood substitutes - unfortunately, this leads, over and over again, to people rewriting the rules so they can experiment in injecting cut-rate substance into trauma patients, who die much more frequently than those receiving real blood. All so some corporations can save money/make money and avoid paying blood donors... which would be "unethical". I don't believe it for a minute. Red blood cells are alive - they are able to control their shape actively. [20][21] They have given up control of many processes via transcription, indeed even by translation after the reticulocyte stage, but for example our neurons don't use those processes for split second decisions. I suspect their ability to exhibit varied behavior is actually greatly underappreciated. Wnt (talk) 19:39, 6 December 2012 (UTC)[reply]
My comment was not any sort of advocacy. μηδείς (talk) 20:56, 6 December 2012 (UTC)[reply]

Phylogeny

This is more of a conceptual question about phylogeny. Looking at phylogenies, how do scientists know that ancestral species is still alive or not? What if the species is an marine invertebrate that was only present shortly and that was in an evolutionary arms race and became quickly extinct but its daughter lineages survived because those lineages could adapt to changing times and evade predation? What if we have no traces of the actual ancestor so we erroneously infer the "actual" ancestor as a extant species? How did adaptations evolve? Exaptation may explain how tetrapods may have colonized the land, but did adaptations initially arise spontaneously and by sheer luck those traits were favored than others and underwent adaptive radiation and diversified into many different forms? Were the initial mutants "hopeful monsters" and somehow the "hopeful monsters" were lucky enough to find mates because the mutations were rather common? Sometimes, evolution by natural selection can seem, well, miraculous. It's like "Ding! You got a new adaptive trait that was beneficial to you and you could successfully breed as much as you want!" 75.185.79.52 (talk) 23:14, 5 December 2012 (UTC)[reply]

To answer your questions in order:
  1. Most phylogenies present an implicit assumption that the ancestor is long gone.
  2. It is actually the usual case that there is no archaeological or other direct evidence of the ancestor.
  3. "We" don't declare an extant species to be the ancestor, ever. It would more usually be stated something like "Species A is the closest living relative of Species B."
  4. Randomly, although some organisms experience mutation faster than others, and some organisms may mutate faster under certain forms of stress.
  5. See natural selection. Genetic changes that have no benefit to a species may become common anyway (see genetic drift and founder effect).
  6. Generally the initial mutants are not suspected to have been "hopeful monsters". Rather, changes occur that are incrementally beneficial to an organism. See evolution of the eye for an example of how something quite complicated can come into being gradually.
  7. It's true that more mutations are harmful than beneficial, and even more are just completely pointless (see silent mutation, for example). But when you have organisms multiplying in great number all over the planet for countless generations, it's OK to be terribly inefficient. If you mutate enough animals, it's simply inevitable that one of them will get some kind of improvement. Someguy1221 (talk) 00:01, 6 December 2012 (UTC)[reply]
Good job I checked before I posted my mini-essay of an answer :) Yours is a much more concise answer! lol. douts (talk) 00:15, 6 December 2012 (UTC)[reply]
A related matter; debating evolution wih a person of normal intellect who nevertheless was an evolutionary skeptic, it became clear that his position was that in cases where the sequence of speciation is recorded, such as the evolution of the horse from eohippus, that in fact that is a species evolving within itself, rather than new species being created and the parent species becoming extinct. Does it all really depend on our definition of speciation? I mean, we argue that Great Danes are the same species as the ancestral canis familiaris that first ate our garbage, but the possibility that they could interbreed is still speculative, no? Gzuckier (talk) 02:20, 6 December 2012 (UTC)[reply]
I'd call it an educated guess with a rather convincing argument. It's the case that most extant breeds to descend from the ancestral Canis lupus familiaris can successfully interbreed, so it stands to reason that most/all of those extant breeds could hypothetically interbreed with that common ancestor. The extant breeds can even mate with familiaris's nearest neighbor, the Grey Wolf, which is rightly considered the same species as the domesticated dog. It would be rather remarkable, nay, paradoxical, if two subspecies could interbreed with one another, but not with an ancestor that post-dates the MRCA. Someguy1221 (talk) 02:34, 6 December 2012 (UTC)[reply]
Along those same lines, I am of the opinion that when everyone agrees that three organisms (one ancestral and two extant descendants) are the same species, they are probably correct. Far more difficult than arguing two organisms could interbreed is arguing that they could not. To this day, it's not clear if Neanderthals were actually a distinct species from humans; everyone agrees that reproductive isolation between humans and neanderthals was at least the norm, but that can arise from geographical isolation and selective breeding just as easily as it can arise from actual speciation. Someguy1221 (talk) 02:44, 6 December 2012 (UTC)[reply]


December 6

heavy cream in a blender

I blend heavy cream for 1 minute on high on a really good 1000W blender with the top open -- I can see the liquid sloshing around. I stop the blender. When I look inside the jar, the cream is essentially solid. When I try blending the cream again, it no longer sloshes around. What's the mechanism? John Riemann Soong (talk) 01:28, 6 December 2012 (UTC)[reply]

Not sure of the chemistry, but we say "the cream has set". StuRat (talk) 01:30, 6 December 2012 (UTC)[reply]
It seems that it could not set while it was kept in constant motion. Bus stop (talk) 01:35, 6 December 2012 (UTC)[reply]
Agglomeration (sticking together) of the butterfat particles? 24.23.196.85 (talk) 01:48, 6 December 2012 (UTC)[reply]
The article on butter (which is essentially what you've made) covers it, as does churning (butter). You've inverted the emulsion from oil-in-water to water-in-oil. Acroterion (talk) 02:43, 6 December 2012 (UTC)[reply]
Well, it's not butter, it's still soft. I wonder why it can't set while it's in motion, and why this setting is not reversible. If I constantly blend water at -5C, won't ice eventually form? Or would I create supercooled ice? John Riemann Soong (talk) 03:50, 6 December 2012 (UTC)[reply]
You may not have completed the inversion, but butter is commonly made in small batches in a food processor. As noted farther down, incorporating air stiffens it up too, and there are degrees of similarity between whipped cream and whipped butter. Butter made in that manner, since it's unsalted, will taste bland. The action of the processor will keep the buttermilk suspended: in finished butter it would be pressed out and the mass rinsed.Acroterion (talk) 15:49, 6 December 2012 (UTC)[reply]
I don't know about -5°C, but keeping it moving should prevent ice from forming down to some temperature. The molecules need to remain adjacent long enough for the ice to form. StuRat (talk) 04:40, 6 December 2012 (UTC)[reply]
I wonder how low before ice would form anyway. If we had a blender that sloshes the water quickly enough and is not itself affected by extremely low temperatures, what's to stop us having a blender mixing liquid water that's 20C below freezing? Or 40C below? Or ... Is it purely a function of how fast the blender can be made to go? -- Jack of Oz [Talk] 07:59, 6 December 2012 (UTC)[reply]
You just made whipped cream: "Cream containing 30% or more butterfat can be mixed with air, and the resulting colloid is roughly double the volume of the original cream as air bubbles are captured into a network of fat droplets. If, however, the whipping is continued, the fat droplets will stick together destroying the colloid and forming butter; the remaining liquid is buttermilk". Don't any of you guys ever cook? --Saddhiyama (talk) 13:39, 6 December 2012 (UTC)[reply]
Yes, but your answer does not address the OP, which asked why the whipped cream only sets when still. StuRat (talk) 17:34, 6 December 2012 (UTC)[reply]
reminds me of simple liquefaction, no?GeeBIGS (talk) 02:46, 7 December 2012 (UTC)[reply]
More like the reverse. StuRat (talk) 05:33, 7 December 2012 (UTC)[reply]
obviously I meant while the blender was still on the only thing keeping it from solidifying was the rigorous agitation.GeeBIGS (talk) 06:17, 7 December 2012 (UTC)[reply]
There are other substances that behave differently when still and when disturbed. See Thixotropy and its See also[22] list. Bus stop (talk) 05:42, 7 December 2012 (UTC)[reply]

how can waterproof camera lenses work both underwater and above water?

If the refractive indices of the media are different, I don't get how one could design a lens to robustly work and focus both above water and under water. John Riemann Soong (talk) 04:20, 6 December 2012 (UTC)[reply]

If an underwater lens housing has a flat glass port then refraction at the glass/water interface is not an issue for narrow-angle lenses. For wide-angle lenses a dome port is designed to correct distortion, and there are wet-coupled accessory lenses that can be added or removed underwater during a dive. See Underwater photography#Equipment. SkylonS (talk) 08:56, 6 December 2012 (UTC)[reply]

Which substance in urine smells like wheat?

--Inspector (talk) 05:02, 6 December 2012 (UTC)[reply]

The body excretes nitrogen in the form of Urea which, itself odorless, gives off ammonia in the presence of water. Ammonia has a pungent odour which may be what the OP perceives as the smell of Wheat. Otherwise there are specific diseases and foods that give particular urine odour, see the referenced section.SkylonS (talk) 08:45, 6 December 2012 (UTC)[reply]
Hmmmm... my assumption is that most odorants of relevance in wheat will be big hydrophobic molecules that come out in wheat germ oil; if you can smell some of this you can determine if this is the scent you're thinking of. It is possible to "refine" wheat germ oil to remove the odor.[23] This suggests some obvious experiments which I expect have already been done. Maybe PMID 22747466 touches on it, probably not, haven't read the article. Searching "wheat germ oil" on PubMed actually yields a nice set of old free articles going all the way back to 1935, which is nice, but I don't think any of them say what either. I suspect the research into these basics might have been done even well before that, but I backed off and went to Google, and after a few tries wheat, scent, compound got me a paper about rice scent [24] which says "For example, oxidation products l-octen-3-ol, 3-methylbutanal, 2-methylbutanal, hexanal, 2-hexenal, 2-heptenal, 2-nonenal, and decanal were identified as key aroma compounds in 12 barley cultivars based on odor thresholds in water (Cramer et at 2005). In whole meal and white wheat flour, (E)-2-nonenal, (E,Z)- and(E,E)-2,4-decadienal, 4,5-epoxy-(E)-2-decenat, and 3-hydroxy-4,5-dimethyl-2(5H)-furanone were odor-active based on AEDA (Czerny and Schieberle 2002). Most of these compounds are also odor-active in rice." Which, first off, tells us what we should know already, that the smell of wheat will be complex and depend on multiple compounds in a ratio to distinguish it from rice, and second, points us to a reference. Which, as luck will have it, is not listed at the end of the paper, but NCBI quickly gives up PMID 12405784 when prodded with the two names. But I won't even go in there because we already have enough chemicals - the question now is, what might be excreted unaltered? That's a big project and I don't think I'll actually do it, but for example 1-octen-3-ol comes up in PMID 15641996 as being present in bovine urine - and physiologically active in attracting flies to cattle. There's lots of fun to be had here, and some opportunities for fresh research. Wnt (talk) 17:24, 6 December 2012 (UTC)[reply]
when I eat Cheerios my p smells like Cheerios.GeeBIGS (talk) 00:38, 7 December 2012 (UTC)[reply]
Do your nuts smell like honey? μηδείς (talk) 01:54, 7 December 2012 (UTC)[reply]
how did you....? Is that some kind of medical condition?GeeBIGS (talk) 02:04, 7 December 2012 (UTC)[reply]
OK, now that's funny. --Jayron32 03:37, 7 December 2012 (UTC)[reply]

kitchen water vs. bathroom water from the tap

I know people who won't drink water from a bathroom tap. I wonder if it is different from the water that comes out of the kitchen tap. — Preceding unsigned comment added by 98.207.158.141 (talk) 07:16, 6 December 2012 (UTC)[reply]

It's the same until it gets to the faucet, but the kitchen tap is more likely to have an aerator and screen on it, both of which can improve the quality a bit (not as much as a charcoal filter, though). StuRat (talk) 07:43, 6 December 2012 (UTC)[reply]
No it's not the same. One may revisit water but it is always different water. The philosopher Heraclitus 535 – c. 475 BCE had this insight which Simplicius of Cilicia 490 – c. 560 expressed as 'Panta rhei, everything flows'. SkylonS (talk) 08:30, 6 December 2012 (UTC)[reply]
In many houses in the UK, the bathroom tap is fed from a header tank in the attic, so the water has possibly been contaminated there. This is why many people prefer to use a kitchen tap for drinking-water because this is almost always supplied from the incoming mains supply. Dbfirs 09:09, 6 December 2012 (UTC)[reply]
Agreed, although for some years it's been compulsory for new houses to have mains feed to all cold taps. There are stories of people finding dead pigeons in their tank, so you wouldn't want to be drinking that. Alansplodge (talk) 09:28, 6 December 2012 (UTC) Alansplodge (talk) 09:26, 6 December 2012 (UTC)[reply]
Some people have even found Barcelonan pigeons in their water tanks. This elusive species is known for its unusual call, which sounds like "oink, oink". Dominus Vobisdu (talk) 09:40, 6 December 2012 (UTC)[reply]
I do believe you're having a tin bath. Alansplodge (talk) 13:26, 6 December 2012 (UTC)[reply]
They don't speak cockney in Torquay. They speak English. They learn it from a book. Dominus Vobisdu (talk) 21:42, 6 December 2012 (UTC)[reply]
Funny, I only drink from the bathroom tap. Because of the way the pipes flow, the water from my bathroom is more likely to be very cold (which I prefer). The kitchen sink pipes are next to the dish washer and other heat sources and the sink itself is more frequently used for hot water as well. Matt Deres (talk) 18:02, 6 December 2012 (UTC)[reply]
Yes, and bathroom water might actually be better, in certain houses, if the pipe leading to the kitchen tap is rusty, for example. StuRat (talk) 18:04, 6 December 2012 (UTC)[reply]
Given people's habits of peeing in bathroom sinks, but not usually the kitchen sink, I could see germphobics having a problem with this. Doesn't bother me. μηδείς (talk) 20:54, 6 December 2012 (UTC)[reply]
Do you have a source for "people's habits of peeing in bathroom sinks"? HiLo48 (talk) 21:44, 6 December 2012 (UTC)[reply]
Contributing to the plural of 'anecdote', whatever it may actually be: by the time I'm at the bathroom sink and needing to pee, I'm no longer concerned about having to settle for a sink. — Lomn 22:17, 6 December 2012 (UTC)[reply]
What do you do when the wife's already on the toilet and your daughter's in the shower? Pee out the window? Having witnessed otherwise I would never assume a public bathroom sink, men's or women's, had not just been peed in by a drunken busboy or three. μηδείς (talk) 01:53, 7 December 2012 (UTC)[reply]
When we had to turn off the water to the house to make plumbing repairs, I peed in a jar and dumped it out in the garden. I wonder if it killed any slugs. :-) StuRat (talk) 07:24, 7 December 2012 (UTC) [reply]
@ Medeis, why the perpetuation of the myth that urine contains "germs"? Urine in the huge majority of the population is virtually sterile. When urine contains pathological bacteria or viruses the producer of the urine will probably be aware of it by well known signs and symptoms. Caesar's Daddy (talk) 08:25, 7 December 2012 (UTC)[reply]

take off and landing on moon or planets

How can escape the moon or planet landing module from those gravity field?--Akbarmohammadzade (talk) 07:41, 6 December 2012 (UTC)[reply]

The Moon only has 1/6th Earth's gravity, so it takes a lot less fuel to reach escape velocity from it. The lunar landers tended to have just enough held in reserve. Mars has over 1/3 Earth's gravity, so more fuel would be required. This is one complication making manned trips there difficult. Venus has over 9/10ths Earth gravity, and a thick atmosphere to blast through, but no human would want to land there, so it's not an issue, unless we want a robotic probe to grab samples and return with them. StuRat (talk) 07:47, 6 December 2012 (UTC)[reply]
You might also wonder about the possibility of obtaining fuel on the planet or moon. Some do have methane, which would burn nicely, provided we also had oxygen. However, planets rich in methane tend to be oxygen-poor. So, we'd need to bring oxygen (or some other oxidizer) with us. StuRat (talk) 08:02, 6 December 2012 (UTC)[reply]

Then how could Russian module return from hot and dense atmosphere of Venus?--Akbarmohammadzade (talk) 08:12, 6 December 2012 (UTC)[reply]

The lander must reach a certain Escape velocity to break free from the planet or moon's gravitational field without further propulsion. The referenced article gives a formula for escape velocity. SkylonS (talk) 08:14, 6 December 2012 (UTC)[reply]
No Russian modules have returned AFAIK. See Observations and explorations of Venus#Observation by spacecraft. They go in, but they don't come out. Clarityfiend (talk) 09:52, 6 December 2012 (UTC)[reply]

And about landing on moon ,it has not atmosphere then non of helicopter or umbrella system work there.--Akbarmohammadzade (talk) 12:31, 6 December 2012 (UTC)[reply]

Correct. DMacks (talk) 13:04, 6 December 2012 (UTC)[reply]
Starships find landing on tilted planets impossible, except at the North Pole
By "umbrella system" you probably mean a parachute. Since that won't work on the Moon, they use retrorockets, instead. StuRat (talk) 17:31, 6 December 2012 (UTC)[reply]
  • Unexpectedly, and unfortunately, it has been found that almost all planets bearing intelligent aliens are tilted on their side, meaning star ships can only land on their north poles. Luckfully, transporter beams can rotate an object by 90, or 270 degrees, allowing people to materialize on the side of the planet. μηδείς (talk) 16:40, 6 December 2012 (UTC)[reply]
If anyone else finds this problematic they are invited to hat it or remove the small section. I have always worried about poor Captain Kirk trying to land sideways on a planet. μηδείς (talk) 20:50, 6 December 2012 (UTC)[reply]
OK, I get it now. You aren't talking about the planet being tilted, but rather the Enterprise's orientation being tilted relative to the surface below it. StuRat (talk) 05:31, 7 December 2012 (UTC) [reply]
This xkcd comic (wow, two very relevant links to xkcd in one day!) illustrates the situation in a very easy to understand graphic. It's one thing to see the numbers; it's something else entirely to see such a clear visual representation. Matt Deres (talk) 01:22, 7 December 2012 (UTC)[reply]
That is one cool effing graphic! μηδείς (talk) 01:44, 7 December 2012 (UTC)[reply]

plant morphology

what type of structures are found in Bryophyta and what are they used for? — Preceding unsigned comment added by 113.199.156.26 (talk) 10:03, 6 December 2012 (UTC)[reply]

Read Bryophyta. DMacks (talk) 13:05, 6 December 2012 (UTC)[reply]

showing who is online.

Here we can give a line of information that who is online ,why don't you do it ?--Akbarmohammadzade (talk) 12:34, 6 December 2012 (UTC)[reply]

I think you are referrring to the "Users currently online" feature of many online forums. That seems to me it would't "fit" with the way wikipedia reference desks works (ie with edits and so on as opposed to "posts"). Other wikipedians will probably be able to explain why wikipedia has chosen not to add this feature. — Preceding unsigned comment added by 80.254.147.164 (talk) 13:18, 6 December 2012 (UTC)[reply]
Wikipedia cares about privacy, but does not care about tracking users to serve them an ad. OsmanRF34 (talk) 14:59, 6 December 2012 (UTC)[reply]
Wikipedia is not very safe in many ways about privacy, but this feature is burdensome to implement. For the closest thing, look at the "contributions" for a user e.g. Special:Contributions/Wnt and compare the time of the last edit to the actual time. You can get this from the menu at left under "contributions" when you are on someone's user or user talk page. Wnt (talk) 17:33, 6 December 2012 (UTC)[reply]

Pocket self-contained "GPS"

I mean, instead of connecting to a satellite, you could calculate your position with a gyroscope and an accelerometer, and you'll know where you are. I'm sure that would drag less battery power. Was it ever implemented in consume electronics? OsmanRF34 (talk) 14:12, 6 December 2012 (UTC)[reply]

This image is geared towards avionics use, but the GPS and INS lines are nicely illustrative regardless. GPS error is about 9 m, while INS error quickly builds -- 650 m of uncertainty after just one hour since a known position fix.
What you're talking about is an inertial navigation system. They have many uses, but also substantial limitations, and I'm not aware of any being developed for the consumer electronics market. They're certainly small enough, but GPS is also small, cheap, reasonably battery efficient (I've no idea the relative power requirements of the two, but neither is excessive), and more error tolerant. Currently, INS tends to be used for "backup in the event of GPS blackout", but that's not a scenario the average consumer really worries about. Even then, you're talking INS on top of GPS rather than instead of. — Lomn 14:29, 6 December 2012 (UTC)[reply]
There was, in fact, one developed for consumer use in cars, before GPS was deployed. I remember seeing an article about it sometime in the last year. I can't remember the name, and the article, at a quick scan, doesn't seem to mention it. Maybe someone else remembers more specifics? --Trovatore (talk) 17:44, 6 December 2012 (UTC)[reply]


Clarifying: note that a GPS receiver isn't "connecting" to a satellite. There's only a radio receiver listening to satellite signals. That's why I note above that GPS is reasonably power efficient. Contrast with something like a satellite phone, which does have to make those transmissions. Alternately, consider cell phone specs for "talk time" (transmitting) vs "standby time" (not transmitting). So for both GPS and INS, you're looking at power for computation and display, but not transmission.
Yes, that's the name,INS, I couldn't remember it. I suppose ships and aircrafts absolutely need a backup system to the GPS. I was talking of INS instead of GPS, not on the top of it. I don't know how efficient GPS is, maybe it's doing a lot for little battery consumption, but in a pocket device like a smartphone, GPS can't be operated for more than 24 hours. Take also into account that GPS has its accuracy artificially reduced, can be jammed, and doesn't work well in places like caves and tunnels. So, at least in some circumstances INS would be superior to GPS. OsmanRF34 (talk) 14:49, 6 December 2012 (UTC)[reply]
Although when I see the picture to the left, I see that INS won't be of any use in its present form for most civil uses. OsmanRF34 (talk) 14:50, 6 December 2012 (UTC)[reply]
GPS' "selective availablility", the means of artificially increasing position error, has not been used in 12 years. New GPS satellites do not even possess the capability. Additionally, the constellation is currently being upgraded to possess the same jamming-resistant attributes that military receivers currently field. Could it still be jammed on top of that? Sure, but I submit that if that's the case, you've got bigger problems than losing turn-by-turn direction.
As for the "24 hours usage" bit, the key reason for that is continuously processing position updates. This isn't strictly necessary; it's of course straightforward to only turn a GPS on when you need a position fix, and that allows you to stretch your battery life almost indefinitely. INS, on the other hand, requires that continuous processing or it's meaningless. Under power-conservation scenarios, GPS will nearly always beat INS. — Lomn 15:00, 6 December 2012 (UTC)[reply]
GPS can be jammed locally, and apparently this is fairly hard to trace, though there are some (ahem) trying to co-opt people's cell phones to get them involved in the hunt, supposedly voluntarily. [25] Wnt (talk) 17:39, 6 December 2012 (UTC)[reply]

Jump Starting a Car

I had to jump start my wife's car yesterday and it got me thinking. The instructions clearly warn you to connect positive to positive and negative to the engine block or chassis, not negative to negative which could cause an explosion. What I can't understand is why that would be. The negative terminal is just a ground (earth) terminal, right? So wouldn't it be the same electrically to connect to either the negative terminal or the frame? Also, why is shorting across the terminals so bad? Please explain as simply as possible, while I consider myself reasonably well educated electrical circuits and electricity have always seemed like magic to me! Tobyc75 (talk) 19:03, 6 December 2012 (UTC)[reply]

The final connection, which closes the circuit, is prone to creating a large spark. Car batteries can potentially outgas (hydrogen, I think?) which could result in that spark triggering a fire. Thus, the last connection should be made away from the batteries. Now, the positive connections basically have to be made on the batteries, so they're out for being last. The negative connection, though, can be made elsewhere on the frame (and many modern cars have a specific "connect negative here" bolt elsewhere in the engine compartment. As for shorting across terminals, short circuits tend to spark and create a lot of heat (since batteries aren't actually perfect conductors). We've already covered why spark is bad, and heat just makes that situation worse. — Lomn 19:28, 6 December 2012 (UTC)[reply]
And, as I understand, the first connection has to be positive pole of the bad battery to the positive pole of the good battery, right? Would this connection, if made the other way round, also be able to cause a spark? OsmanRF34 (talk) 19:51, 6 December 2012 (UTC)[reply]
The two positives should be connected first, but can be safely connected in either order. The first connection will put the other end of the positive cable at +12 V (or near enough; the dead battery likely still has some voltage), but since the cars are insulated from each other, "+12 V to what?" applies and there's no real effect. You can touch the far end of the positive cable to basically anything that isn't the original car with zero effect. Once you connect the second end of the positive cable, the two ends flopping around are both on the negative cable, disconnected from everything, and so still can't do any damage. Contrast with making your first two connections to the terminals of a single battery -- now you can easily get a big spark (and short the battery) by letting the two floppy ends touch. I've done it by accident, and it's rather impressively scary if you're not expecting it. As best I can tell, the biggest reason to specify "Dead Positive, then Live Positive, then Live Negative" is so that you don't need to walk to the other car between the second positive connection and the first negative connection. For that matter, I think even which negative is battery vs frame (or if you put both on the frame) should be irrelevant electrically; it's just a matter of reducing spark risk. — Lomn 21:39, 6 December 2012 (UTC)[reply]
Alright, thanks for the excellent response. I hadn't even thought of producing hydrogen and that causing the explosion. I definitely saw a big spark when I connected the final clamp to the engine, so that would have been bad. Tobyc75 (talk) 20:44, 6 December 2012 (UTC)[reply]
IMPORTANT SAFETY NOTE: Cars of American and Asian design origin are standardised on battery negative connected to earth (chassis). However, cars made in Britain were often positive to earth. Never use jumper leads on veicles that have differing nominal battery voltages (eg never use a 24V truck to start a 12V car & vice versa; never use a 12V car to start a 6V volkswagen - push it instead). Wickwack 124.182.154.215 (talk) 01:57, 7 December 2012 (UTC)[reply]

Accesing the journal Science from home via institution

Hi, Do you know if it is possible to access the journal Science with an institutional login from my home computer? Most other journal websites (Science Direct/Elsevier, Nature etc) allow me to do this by pressing "Institutional login" (or similar) and logging in using my university credentials, but I can't find it on Science. The article in interested in [26] can only let me login if I have personal (purchased) access. I know I can use remote desktop to connect to the campus computer, but it doesn't seem to be working at the moment. Thanks, 90.203.22.59 (talk) 20:40, 6 December 2012 (UTC)[reply]

Most educational institutions have some sort of proxy service through which you can access google scholar and from there fulltext of any journal your university has a subscription to. Assuming the institution being referred to is a university, I'd start with the library homepage and then look for something labeled 'databases'. Alternatively Google Scholar may have a direct institutional login, I've never explored that option myself. (+)H3N-Protein\Chemist-CO2(-) 21:11, 6 December 2012 (UTC)[reply]
You should check with the e-Resources librarians at your institution. They should know what they are or aren't paying for, whether you can use your institutional login at home, or what. --Mr.98 (talk) 21:36, 6 December 2012 (UTC)[reply]
At some institutions, you can sign into the library webpage and then access those resources through that page. ----Mattmatt1987 (talk)

power with a highly damped, frictionful system

Walking is highly damped as following the removal of force, velocity ceases quickly. The power consumption due to friction is F*v, where F is the frictional force. Hence when the animal is not accelerating, power is constant and dependent on velocity. But sometimes, the animal has a net acceleration, and force applied is more than just friction. What I can measure from the video feed is net acceleration. How do I calculate power consumption in that case?

How would I make a good estimate of the coefficient of friction (kinetic) from experimental data of behavior? 128.143.100.192 (talk) 20:58, 6 December 2012 (UTC)[reply]

I doubt that friction has much if anything to do with it. Could it be friction between the feet and the gound surface? No, as when in contact the feet do not move laterally. Could it be friction within the joints of the legs? Maybe, and this could be measured in a suitable jig with the subject given a spinal block aneasthetic. But I expect such friction will account for only a very small part of the energy consumed in walking. Friction-wise, walking up stairs must be much the same as walking on a horizontal surface, but it sure makes you breathe harder. The answer probably lies in the chemo-thermodynamics of muscle contraction. Work done on the air due to the air's viscosity must be very small.
Why do you think that walking is highly damped (presumably you meant damped by friction)? If you are walking, and decide to stop, you stop instantly without conscious effort. But this does not mean there is no unconscious muscle action. The whole time you are not in bed, the automatic regions of the brain constantly evaluates position sensors in the limbs, the ear canals, and the visual scene and continually sends motor signals to the muscles to keep you stable and upright. This same automatic system means that when you stop walking, while inertia would tend to keep you going forward, the right muscles automatically apply force to overcome this inertia.
Wickwack 124.182.154.215 (talk) 01:47, 7 December 2012 (UTC)[reply]
From our power (physics) article:

And we also know that:

So, it should be possible to determine power usage from mass, velocity, and acceleration. StuRat (talk) 05:23, 7 December 2012 (UTC)[reply]

Purpose of flat tunnel roof

What's the purpose behind the flat internal "roof"[27][28] in tunnels? Since Sasago Tunnel did not collapse after these roof panel fell out, I can only assume they are not load bearing. Dncsky (talk) 22:27, 6 December 2012 (UTC)[reply]

My initial guess, and I know nothing about this, is that the flat roof simply conforms to the flat tops of vehicles. Bus stop (talk) 22:32, 6 December 2012 (UTC)[reply]
But doesn't the roof restrict vehicle height? Dncsky (talk) 22:36, 6 December 2012 (UTC)[reply]
Yes, I see your point. I was just about to revert my initial post. Bus stop (talk) 22:38, 6 December 2012 (UTC)[reply]
Also, is this a Japanese only thing? I've never seen a tunnel with an internal roof like that, though admittedly I haven't been to a lot of places and haven't seen a lot of tunnels. Dncsky (talk) 22:38, 6 December 2012 (UTC)[reply]
This article (which may only be accessible if you visit it from Google) says the space above the panels is for ventilation. A major concern in long road tunnels is where a fire in one place blocks traffic, and the smoke travels back down the stationary jam behind the blockage, choking more people. Some tunnels have ducts, some have impellers that are suspended in nacelles, and some use a concrete ceiling plane. A concrete barrier means that fires and explosions (e.g. from tanker lorries) are less likely to penetrate into this ventilation space. They can also run utilities (lighting, phone, CCTV) through there, and again they're better protected from fire (naturally they could also run these through fire resistant conduits). The Sasago incident resembles the Big Dig ceiling collapse. -- Finlay McWalterTalk 23:00, 6 December 2012 (UTC)[reply]
Thanks. I've seen ducted fans in long tunnels, but never the roofs. Guess I'll keep an eye out next time I travel. Dncsky (talk) 23:10, 6 December 2012 (UTC)[reply]
Resolved
This page of info about the ventilation system in the Gotthard Road Tunnel shows a very similar ceiling construction to the Sasago tunnel, and explains some how this giant tunnel's ventilation is done. -- Finlay McWalterTalk 01:25, 7 December 2012 (UTC)[reply]
From the diagrams, it also looks like the side panels either contain lights or reflect lights (probably painted white, versus grey concrete). StuRat (talk) 05:15, 7 December 2012 (UTC)[reply]

Africans and headlice

At my day program, we were having a discussion on hygiene, and the job coach, who is African-American, mentioned that Black people do not get headlice. The word for "louse" is present in every language in the world, appearing on the Swadesh list, the Dolgopolsky list and the Leipzig-Jakarta list, so presumably it is present in Sub-Saharan African languages too. Is it really true that African-Americans don't get headlice? Given the linguistic universality of the term, I find it hard to believe. If it is true, then how did all those Bantu languages come upon their words for louse? Enzingiyi (talk) 22:33, 6 December 2012 (UTC)[reply]

Well, there are other lice but head lice, so a language could have a word for them, without headlice existing. HiLo48 (talk) 22:38, 6 December 2012 (UTC)[reply]
I also seriously doubt that there's a lexeme for them in every language in existence, though I could see it being generally ubiquitous. Snow (talk) 23:11, 6 December 2012 (UTC)[reply]
In a study in 1985, only 0.3% of African-American children were infested with head lice, in contrast with 10.4% of the non-African-American children in the study.[29] However, according to the same web page, there's an African variety of head lice, which is better adapted to being able to hold on to kinky hair than the European variety of head lice is. Red Act (talk) 23:05, 6 December 2012 (UTC)[reply]
Here are two more studies sharing the conclusion that children of African descent are less likely get headlice.[30][31]Dncsky (talk) 23:17, 6 December 2012 (UTC)[reply]

Interesting to know why your colleague thinks do-rags originated with black men in prison. μηδείς (talk) 01:41, 7 December 2012 (UTC)[reply]


December 7

Why aren't all houses made of brick?

Topic says it all. ScienceApe (talk) 00:26, 7 December 2012 (UTC)[reply]

let me guess...you just re read the three little pigs to your kid. Couldn't helpGeeBIGS (talk) 00:36, 7 December 2012 (UTC)[reply]
There are probably a great many reasons. Just one example — brick houses don't do well in earthquakes, so if you're building where earthquakes happen, brick is a bad choice. --Trovatore (talk) 00:37, 7 December 2012 (UTC)[reply]
Economics often has a lot to do with it. For example, the climate in the major Australian cities is not very different. However, in Perth double brick (brick outside, and rendered brick inside) construction completely dominates, but in other cities brick veneer (brick on the outside, frame&clad on the inside) dominates. The reason is an accident of history. In Perth a couple of clever chaps took advantage of local high quality clay deposits and devised their own factory machinery made from war surplus construction equipment - the result being that Perth was supplied with very cheap bricks. In other words in Perth double brick construction was cheaper but in other cities brick veneer was cheaper.
Double brick construction works very well in mediteranean climates such southern Australia, much of southern Europe. But in colder climates (ambient temperatures routinely below zero C) timber framed timber clad construction offers better thermal insulation, but is not maintenance free.
There are cultural aspects. For example, in many parts of Australia during the 1930's depression and at certain other times, people with low incomes got houses of timber frame asbestos/cement clad construction. Such houses were cheap to build and actually perform quite well in durability and in thermal properties, but ever since have had a cheap and nasty stigma. So, after the post- World War 2 ecomonic boom, anybody who didn't earn much usually stretched themselves and committed to an achievable dream - owning a brick house (even if that meant smaller rooms and paying off a loan for the next 30 years).
I imagine similar economic, environmental, and cultural considerations apply elsewhere in the world. Wickwack 124.182.154.215 (talk) 01:27, 7 December 2012 (UTC)[reply]
Relevant past discussion here. Evanh2008 (talk|contribs) 03:24, 7 December 2012 (UTC)[reply]

Venus in a small telescope

The other day I was looking at Venus with my 6.25" reflector. I was hoping to see it as a crescent. All I could see was a very bright spot, with all three eyepieces. Is Venus just too bright for that? Venustar84 (talk) 00:39, 7 December 2012 (UTC)[reply]

Estimate the current angular size: its diameter divided by its distance yields about 11 arc seconds. In principle, your telescope should be able to resolve about a half an arc-second in visible light. So, your minimum resolvable feature-size is good enough to image the structure of Venus' illumination pattern (its crescent); and because the planet appears bright enough that you should be able to magnify to fairly large size.
In practice, you're only about 10x above the resolution limit. So, any imperfections at all - atmospheric aberration, telescope alignment or mirror collimation issues, and so forth - will blur the image. How good is your scope's primary mirror, and how good are the optic(s) you're putting behind it? You can push it to its limits by measuring the circle of confusion of a bright star, who is well-approximated by a perfect point-source.
If you're collecting too much light - which happens when you're looking at planets - you can create an iris by cutting a hole of whatever size you want, out of cardboard or tinfoil, and shutter over some of your aperture. This dims the image and has the added bonus of improving your focus. Camera people call that process "stopping down the aperture." Telescope stores charge a lot of money for stops, but you can build one out of tinfoil or stiff card paper, without worrying too much about precision. Just try to be mostly circular, mostly centered, and mostly symmetrical. (Other experts recommend off-center stops, especially if your scope is Newtonian, for obvious reasons).
I have a 205 mm Newtonian scope, and lots of very good glass, a not-too-shabby set of cameras, and (formerly) a whole lot of human- and CPU-time to spend post-processing; but I have never successfully imaged any structure on Venus. I have had better luck with Jupiter clouds. Nimur (talk) 01:27, 7 December 2012 (UTC)[reply]
The crescent shape of Venus is visible using even moderately-good binoculars. Unless there's something seriously wrong with your telescope, you shouldn't have any difficulty resolving its shape. (Though if you do find it uncomfortably bright, you could partially obstruct the open end of your telescope aperture, or acquire a neutral density filter to take some of the edge off.)
That said, when exactly was 'the other day'? In recent weeks, Venus has been getting closer and closer to 'full' (see aspects of Venus for dates). Remember that Venus takes a lot longer than, say, the Moon to work through its phases. Based on this calculator widget, Venus was more than 80% illuminated on 1 November, and 88% illuminated on 1 December. Any time in November, Venus would only look a little bit out of round. Venus is also very nearly as far away from Earth as it can get (it is at its furthest when it is 'full', at superior conjunction on the opposite side of the Sun from Earth), and therefore very nearly at its smallest apparent size. At its furthest, its angular diameter is just under 10" (10 arc seconds); at its nearest, it swells to 66" across. TenOfAllTrades(talk) 01:35, 7 December 2012 (UTC)[reply]
As an amateur astronomer, I can clearly see Venus as a crescent (or some other phase) with my 60 mm (2.4 inch) telescope. I haven't tried recently, but if you're having trouble with your 6.25" reflector, and you're sure Venus wasn't near full that day, something is wrong. --140.180.249.232 (talk) 05:46, 7 December 2012 (UTC)[reply]

Speed of light, etc.

All our articles on relativity seem to be a little sparse in this regard, so I hope no one minds me asking some dumb questions. What I am trying to work out is how exactly two objects can have independent velocities that sum to c or greater, without actually being measured by each other as accelerating to the speed of light. I believe the easiest way to put this into simple terms is the following:

Two objects (A and B) accelerate in the same direction at the same speed (for our purposes, let's peg it at 51% the speed of light) relative to an observer in an inertial reference frame. This observer sees two objects, both accelerating toward him at 51% the speed of light. Nothing weird so far, but then, B changes direction. While A continues to accelerate toward our observer, B reverses direction, now accelerating away from the observer at 51% the speed of light.

The observer (as far as I can tell) will not notice anything out of the ordinary. Both objects continue to move at the same speed relative to him/her; one has merely changed direction. Object A, however, has a problem -- according to our observer, B is now accelerating away from A at slightly over the speed of light. Because of special relativity, however, A measures the velocity of B at slightly under the speed of light (I think this is correct). Therefore, if asked to describe B's location in time and space, Object A and the observer will give two completely different answers. Issues like length contraction aside, the object they are observing possesses the same physical qualities, but the two will not agree on its location. How, if it all, is this resolved? Evanh2008 (talk|contribs) 04:08, 7 December 2012 (UTC)[reply]

These sort of apparent paradoxes show up all the time in special relativity problems. See the Ladder paradox for another one. The answer to your question is that both Object A and our observer are perfectly capable of calculating the location and velocity of B and both will arrive at the same answer so long as both work from the same set of data. If they don't get the same answers, then they have left something out of their calculations. It's the same basic problem with observing any far away object. Where we see a star now in the sky isn't where the star is now. An observer on the other side of the galaxy will see the same star in a different location, but given the right information about the star's position and velocity we can both calculate the star's actual position and come up with the same answer. --Jayron32 04:17, 7 December 2012 (UTC)[reply]
You need to think a lot more carefully about how your "observers" are measuring these things. With a radar gun? Then how do you convert a redshift into meters per second, and why? Or by timing how long it takes to go a given distance? Then how are you determining the elapsed time and the distance? The reason no one understood special relativity before Einstein's 1905 paper (even though all the math had been worked out before) is that they imagined that measuring speeds was something you could just do, and the details were irrelevant. Einstein carefully considered what it actually means to measure a speed, and showed that the specialness of the speed of light is quite natural. Unfortunately, most people still don't get it. They talk about special relativity in terms of "observers" who somehow just know the length and speed of everything (though apparently they don't really, since they all disagree with each other). Special relativity seems mysterious when described in this way. It isn't really.
You might want to read this translation of Einstein's paper. It's still one of the best introductions to special relativity. Note that what he calls "observers" in the paper are just scientists who look at things. The strange beast that's now called an "observer" is a later invention. -- BenRG (talk) 04:42, 7 December 2012 (UTC)[reply]

In other words, How much does Aging affected by the Nervous system (and the endocrine system that it governs), In compare to Genetic control? — Preceding unsigned comment added by 109.66.11.190 (talk) 06:42, 7 December 2012 (UTC)[reply]

Genetics determines the form of the nervous system to a very detailed level. Aging only causes some minor deterioration, except in the case of diseases such as Alzheimer's disease. Looie496 (talk) 07:16, 7 December 2012 (UTC)[reply]
Hi. please not that I'M not asking how aging affects the NS.

About not eating for 4 days straight, and subsisting only on fluids...

I only ask for nutritional advice, nothing medical, as no one is allowed to here.

Hello,

I am due to go to a ski resort in Monarch, Colorado on around January 2nd, 2013 and be on the whole trip for 4-5 days.

I'm afraid that at the ski resort, there may be no washlet / bidet-like devices and the resort staff would likely reject my donation of a BioBidet attachment device to install on the toilet of the accommodation I'll stay in. (I'll still bring one in the remote off-chance that they consider my digestive needs and accept it after all.)

Being faced with undue hardship without access to a bidet, I would much rather subsist on fluids during the whole four days. Weighing 206 pounds currently, I would have 25 pounds to lose until reaching the normal-weight threshold anyway.

However, if the resultant hunger makes me ravenous, I may take yogurt in hopes that it's only urinated out, as yogurt is still more liquid than solid.

Therefore, the questions:

1. After explaining my digestive concerns for needing a bidet, would they let me install the bidet-attachment on their toilet in the room I'm in? (It's a donation that I'd like to have remain there after I return home with my ministry group.)

1a. Why might they not allow me to, when bidets clearly have better results than the "old way?" Wouldn't it be like rejecting the donation of a shower stall 100 years ago?

2. If I subsist strictly on fluids for the 4-5 days I'm away from home, how many pounds would I stand to lose? Also, what unpleasant nutritional side-effects would be bound to happen?

3. If the only things I eat are no tougher than yogurt, will I still be compelled to take the #2 before the 4 days, or will only the #1 happen?

4. What (else) can I eat that will not force me to take the #2, no matter how much of it I have? (Someone up to no good from AOLAnswers answered, "Hemlock. Try it." Needless to say, I have thus far, found no non-poisonous forms of hemlock, so it's seemingly not like mushrooms.)

I hope the answers prepare me well for this trip. --Mount Zynar (talk) 07:26, 7 December 2012 (UTC)[reply]

You're approaching this from a totally wrong angle. You should be asking what you can use in lieu of a bidet, and the answer is quite simple. Pop in the shower and wash yourself off down there. Like you, I like my nether regions to be clean, and when at home, I always take a partial shower afterwards. Just takes a minute or two, and it's a lot easier, and healthier, than starving yourself. Dominus Vobisdu (talk) 07:46, 7 December 2012 (UTC)[reply]
Do you want to step on what others leave behind (especially that which comes from "up there?") And do you ever want to stick your foot in the toilet bowl, even after it's flushed clean? Think about it. That's why I would not use that particular shower-option in place of bidets. --Mount Zynar (talk) 08:56, 7 December 2012 (UTC)[reply]
You could clean it, with bleach, before you leave. StuRat (talk) 09:16, 7 December 2012 (UTC)[reply]
Along those lines, request a handicapped access room, as they often have a movable shower hose, which would be a lot easier to use as a bidet. You could also ask if they have any rooms with bidets. If they have international tourist there, it's possible, although those rooms might cost more. StuRat (talk) 07:56, 7 December 2012 (UTC)[reply]
And, if forced to use a toilet, a bit of moisturizer squeezed onto the toilet paper can both assist in cleaning and prevent irritation. You could also just attach the device without asking them. You might want to remove it before the maid arrives each day, as there's a slight chance she might complain to the hotel management. But, most likely, they would just ignore it. I'd remove it when you go, though, or they might very well complain. I myself have made little fixes to hotel rooms I've stayed in, from putting bright CFLs in place of dim incandescent bulbs, to fixing a rattling A/C unit and a toilet that had a lever which stuck down. I never asked permission. StuRat (talk) 07:54, 7 December 2012 (UTC)[reply]
That all liquid diet might also result in diarrhea, so I'd avoid that. But, if you insist, all least do a dry run first (or is it a wet run ?), so you discover any problems it causes while you still have access to a bidet. StuRat (talk) 07:55, 7 December 2012 (UTC)[reply]
Stu, how are we sure that an all-liquid diet won't just cause frequent urination? --Mount Zynar (talk) 08:56, 7 December 2012 (UTC)[reply]
With pure water, you might be right. But other liquids are likely to contain some indigestable material. For example, pulp in orange juice. StuRat (talk) 09:14, 7 December 2012 (UTC)[reply]
You also misunderstand what faeces consist of: it is entirely possible to pass a stool without eating any solid food because of the dead cells the body has to eliminate, which it does in faecal matter. --TammyMoet (talk) 08:23, 7 December 2012 (UTC)[reply]
How long would that take to happen? On the night of New Years, I plan to take an hour or two committing multiple "enema washes" to empty out the bowels as much as I possibly can. If the process, involving only dead cells, would take more than the 4-5 days the trip lasts, and I don't eat any solids, then I don't think it's a problem. So how many days would it take? --Mount Zynar (talk) 08:56, 7 December 2012 (UTC)[reply]

Lung shot

Does getting shot in the lung feel like drowning?--Wrk678 (talk) 09:38, 7 December 2012 (UTC)[reply]