Wikipedia:Reference desk/Archives/Science/2010 October 18

= October 18 =

Train Speed vs Plane Speed
Why are the fastest trains today slower than the average airplane? Shouldn't trains be faster because they don't have to expend energy to stay afloat? What's stopping engineers from making trains that move faster than planes? —Preceding unsigned comment added by Metroman (talk • contribs) 01:13, 18 October 2010 (UTC)


 * They can't use jet engines near inhabited areas, they have to stay on tracks and are restricted by the terrain, and drag is greater on the ground than in the air. Clarityfiend (talk) 01:55, 18 October 2010 (UTC)


 * To clarify the last point, drag is lower in the upper atmosphere because the density of the air is less. On the other hand, the long narrow shape of a train  produces less drag than the wings of an airplane, or at least it would if the same amount of attention was paid to streamlining (which it would be for an ultra high speed train).  On the third hand, with trains you have another form of drag: the friction of the rolling wheels.  (Unless you use a maglev system, but in that case you have trains that can't run onto existing railways, which limits where they can go.)


 * Another factor is that new railways would be needed. The traditional railway lines in most countries are suited to top speeds around 70-100 mph (115-160 km/h).  The present generation of high-speed trains such as the TGV, Shinkansen, ICE, and whatever they call the one in China, which achieve about double those speeds, reach their top speeds only on new railways that were designed for the purpose, with gentler curves and special signaling systems.  (Some of those trains run onto existing railways, but at lower speeds.)  If the top speed was to be doubled again, the same reasons would require a whole new set of railways.  And that's something that costs a lot of money and tends to draw objections from people living near the routes.  With air travel, on the other hand, you only need an airport for each end of the service, not a continuous structure running the whole length of it.


 * --Anonymous, 03:50 UTC, October 18, 2010.


 * let's point out the other obvious factor: a 100+ ton object moving at 600 mph (even under the best circumstances) needs an unbelievable amount of room for turns and deceleration.  A small miscalculation by a pilot in flight will have no serious repercussions, since the nearest physical object might be 10 miles away.  when the nearest physical object may be a few feet away and the train is required to stay precisely on top of a 5' wide path, 600 mph can make even the smallest miscalculation very hairy.  -- Ludwigs 2  06:44, 18 October 2010 (UTC)
 * Just to point out here that the typical TGV runs at 400-800 tons fully loaded, so it might be hairier then you think. Googlemeister (talk) 13:29, 18 October 2010 (UTC)
 * Sorry I don't have the reference on hand, but I once read a paper (from the UK rail authority, I think it was) assessing the economical viability of very high speed trains. The conclusion was mainly that there's an upper limit for the speed of trains above which air traffic is actually more economical on the whole. This upper limit was somewhere around 300 mph, if I remember correctly. As stated above, building new lines is a major factor - and the requirement of gentler curves makes it very difficult to place the new lines (unless you're China, in which case you can just move stuff out of the way as you please). The cost of the trains themselves also goes through the roof. And drag is a drag. Air is much denser at ground level than at 30,000 feet. I don't think you can fly an airliner at 500 mph near the ground.--Rallette (talk) 07:03, 18 October 2010 (UTC)
 * In addition to air density, there's the fact that a train runs next to the ground. No matter how you streamline the train itself, being next to the stationary ground is bound to cause turbulence, which you pay for in drag. An airplane has only air around it for distances many times the plane's own dimensions. That makes it much easier to keep most of the flow around it laminar. –Henning Makholm (talk) 18:05, 18 October 2010 (UTC)
 * Here it is, from the University of Lancaster. It's a review of the European High Speed Network by a Roger Kemp of Alsthom. From the nineties but I suppose the figures are still valid. And the "upper limit" for speed was actually lower than I remembered, at between 300 and 400 kilometres an hour. See page 9, Is there a limit to "high speed"?--Rallette (talk) 08:21, 18 October 2010 (UTC)
 * glancing through that article, it seems that a good part of the speed limitation is that these rail concepts all use traditional wheel/rail friction for propulsion - that's bound to have diminishing returns (slippage) issues as speed increases. by contrast, it's imaginable to have a train that uses (say) a modified jet engine for propulsion and uses wheels only for support and breaking.  That would allow top speeds closer to the speeds of actual airplanes, but would introduce all sorts of ancillary problems (e.g. that small track irregularities at 500 mph might bounce the cars clear off the track). -- Ludwigs 2  09:02, 18 October 2010 (UTC)
 * That article makes it clear that there are problems going above 400 kph whatever the means of propulsion. Firstly, the curves need to be wider (10 km radius at 500 kph) and slopes gentler (1 km run-up and run-down for a bridge over a road). Secondly, the power needed for the train increases as the cube of the speed, so a 500 kph train needs eight-times the power as a 250 kph train. Finally, as the speed increases, you have to reduce the number of trains using the line, because the breaking distance is longer; so there are fewer tickets to be sold for the expensive upgraded service. Another point that's not in the article, but is relevant to jet propelled cars such as Thrust2 (1000 kph), is the problem of keeping the thing on the ground when it's going so fast! Physchim62 (talk) 19:29, 18 October 2010 (UTC)
 * Can't this problem be fixed by using computer-controlled trains? I imagine computer calculations would be extremely precise. Also, since it's a train, it would follow a fixed path. These calculations would only have to be done once. Every other train that traveled the same path could use them with perhaps a slight modification for the number of passengers. Metroman (talk) 01:32, 19 October 2010 (UTC)
 * You're imagining that computer-controlled trains could follow each other along the same track at high speed at a spacing less than their distance to brake to a stop? Maybe in theory, but as soon as you allow that, you introduce new dependencies on the system working perfectly.  In the real world things break down.  For example, if a rail breaks and one train derails, that may be bad, but it's much worse if the following train is so close behind that it cannot stop.  And just because the control system is automated does not mean it will work perfectly.  Consider the Clapham Junction rail crash caused by a wiring error, or the June 2009 Washington Metro train collision caused by an electronic fault -- these are not fully automated systems but are examples of how automated parts of the system can go wrong.  And, of course, the software could also be faulty.  --Anonymous, 06:00 UTC, October 19, 2010.


 * I believe the Japanese experimented with very fast model "trains" that run in vacuum tubes. The reason they are not around is I expect due to the expense of constructing the tubes. 92.29.121.119 (talk) 08:56, 18 October 2010 (UTC)

One point that perhaps hasn't been mentioned so far is the damage caused to trains by the stones of the track bed with the turbulence caused above certain speeds; stone trackbeds are preferable to solid concrete because they allow for thermal expansion without cracking and the resulting damage to rails. Current Spanish highspeed trains run at below their real maximums partly due to this problem. Llocsird (talk) 21:19, 18 October 2010 (UTC)


 * Our OP explained that trains don't have to expend energy to stay afloat. The drag on an airplane caused by the need to stay afloat is lift-induced drag.  However, lift-induced drag is strongest at slow speeds.  (Doubling the speed causes the lift-induced drag to reduce to one-quarter of its former value.)  Consequently, at the speed of an airplane as it cruises towards its destination, the lift-induced drag is remarkably small.  Most of the drag is parasite drag and it is this which limits the cruising speed of the airplane.  So it is reasonable to say the energy required to keep the vehicle afloat is zero for a train, and very small for an airplane.  Dolphin  ( t ) 02:03, 19 October 2010 (UTC)


 * The article Schienenzeppelin may be of relevance. One of the "many problems with the Schienenzeppelin prototype" that the article does not detail is the rather unfortunate effects its slipstream had on the stations (and any loose objects and people in them) through which it passed at full speed. 87.81.230.195 (talk) 14:50, 19 October 2010 (UTC)
 * Could a lot of that have been because the thing had a huge airplane propeller as a means of locomotion? Googlemeister (talk) 16:34, 21 October 2010 (UTC)
 * I believe the time will come when trains will travel as fast - even faster - than jets, and through tunnels with soft vaccuums. The stations themselves will be sort of like air-locks. Without that drag, and on mag lev, you could really be flying. Just wait till the oil starts to run out. I don't see anyone running a 600 passenger airbus on batteries, sunlight or chicken poo. And yes, they will travel across the oceans, across specially built bridges about 100 metres above the water, and through air-evacuated glass tunnels, so that the passengers can see the view! And the trains will be wide, like ships, with everyone having a cabin if they want, and a place to play golf, just like an ocean liner, except an ocean liner takes longer than half and hour to cross the Pacific. (Rem: you read about it here first) Myles325a (talk) 03:33, 23 October 2010 (UTC)

In the animal kingdom, do females compete over males?
My friend and I were discussing the human phenomenon of "mean girls" and social bullying. Then I began to wonder if females in the animal kingdom ever compete with one another over males either by physically fighting one another or somehow conspiring to make the other females less desirable to eligible males?--24.188.235.80 (talk) 01:49, 18 October 2010 (UTC)


 * How about the spotted sandpiper?
 * A google search for "female competition" monogamous gives a variety of other interesting leads. –Henning Makholm (talk) 02:49, 18 October 2010 (UTC)
 * Quote-characters are not part of the formal URL character-set--see urlencoding for information about the situation, Help:URL for various solutions on Wikipedia, and for a handy solution specifically for google-search links. DMacks (talk) 04:11, 18 October 2010 (UTC)
 * Oops, I was looking askew at my ascii table and tried to encode " as %32 instead of the correct %22; then assumed that MW was choking at the percent sign. Fixed now. –Henning Makholm (talk) 05:11, 18 October 2010 (UTC)


 * In some canids, there is usually a dominant pair of female and male (alpha female and alpha male). I couldn't find any info on how the alpha female esblishes her position but I did find references to her killing the pups of subordinate females when food is scarce and that the pair share dominance (eating first for example) and control over the pack. And in fact out alpha (ethology) mentions the alpha male doesn't always have exclusive access to the alpha female. Note as well Territory (animal) mentions how in non pair bonding animals, female and males territories are independent and female will defend their territories again other females (and males against males) apparently budgies is one example ehow.com/about_6193661_male-_amp_-female-budgies-fight_.html (blacklisted site) and something which could be considered a form of fighting for males. Display (zoology) also mentions some species where the female actively compete for males I think referring to species that practice Polyandry. Nil Einne (talk) 07:57, 18 October 2010 (UTC)


 * Meerkats have a matriarchal culture where the alpha female controls breeding such that only her pups survive. If a lesser female gets pregnant by any of the males in the group, the pups are usually killed/removed to die by exposure. Occasionally the female will set out to form her own clan. Even less frequently, the new pups are tolerated. This is all nicely depicted by Meerkat Manor. Worth watching if you're curious about this topic. The Masked Booby (talk) 09:20, 18 October 2010 (UTC)

Beatrice being buried alive in Kill Bill Vol. 2
In Kill Bill Vol. 2, the Bride (aka Beatrice) is buried alive in a pineboard coffin. After a panic attack, and a flashback to her kung fu training sessions on top of a misty mountain, she regains her composure and "one inch punch" -es her way out of the coffin, and then (implied but not shown) digs herself up through the loose soil and ultimately escapes. Yes, I know it's just a movie, but that doesn't stop me from wondering about the real scientific parameters of such a situation, namely: (1) how many minutes consciousness would someone have if trapped in your average coffin, airtight of course? (2) is it possible to dig yourself out of a hole filled with loosely packed soil? My understanding is that dirt in any large quantity is far heavier than most people realize, and that if you were submerged there would really be no way to go up, as you wouldn't be able to move the soil around...? The Masked Booby (talk) 05:04, 18 October 2010 (UTC)
 * (2) Not a chance, Kiddo. Mythbusters measured the force of such a punch of a master martial artist, created a robot to duplicate it and, 600 punches later (by which time they calculated the air would have long since run out), the coffin board was still mostly intact. Even assuming a person could punch through, they also showed that it was impossible to work your way up past 6 feet of heavy, heavy dirt (which collapsed into the coffin anyway). Clarityfiend (talk) 06:37, 18 October 2010 (UTC)
 * Your Northern Kung-Fu is weak! --Stephan Schulz (talk) 07:21, 18 October 2010 (UTC)
 * Wouldn't the classic "six feet" be to the bottom of the hole, so there would be 3 or 4 feet above the coffin? Still challenging, if dirt has a density of 120 pounds per cubic foot, or 1922 kg per cubic meter. A coffin  with even 3 feet of dirt above a 12 square foot lid would have 4320 pounds of dirt pressing down.  Punching a hole in a small area of a flimsy lid would just let dirt cascade in.  Edison (talk) 18:59, 18 October 2010 (UTC)
 * Dirt is also a pretty fantastic shock-absorber. It's the preferred construction-material for sandbags, because its loose composition disperses force and inelastically distributes shock - in other words, it's very very hard to "punch a hole" through dirt - no matter how hard you hit.  (That's why sandbags and earthen berms are a better barrier than concrete for military bases!  Even a high-powered rifle can not penetrate 8 inches of dirt).  Intuitively, this makes sense - if you were underneath dirt, and did succeed to "punch a hole" upward, new dirt would immediately begin to cave in and re-fill the hole.  An identical process occurs at smaller ("intra-grain") scales, dispersing the energy and making any effort to dig upward very energy-wasteful.  Here's a fantastic free online textbook, Fundamentals of Particle Technology, with an entire chapter on the mathematics and physics of powder flow. Nimur (talk) 19:17, 18 October 2010 (UTC)
 * As far as (1) is concerned, the volume of a coffin is roughly (6 ft) x (2 ft) x (1 ft) = 12 ft^3, minus the volume of the person trapped inside. I'm willing to bet anyone would pass out within a few minutes as the oxygen level decreases and the carbon dioxide level increases.  You could try sitting in a coat closet for awhile and see how long you last (make sure someone is monitoring you from the outside). Hemoroid Agastordoff (talk) 20:01, 18 October 2010 (UTC)
 * So Buffy would've stayed buried then? 67.243.7.240 (talk) 21:51, 22 October 2010 (UTC)

It's not just force, right? Isn't it also impulse? John Riemann Soong (talk) 21:35, 18 October 2010 (UTC)

linear diode
What is linear diode?? does it work as --Anxious angel (talk) 09:27, 18 October 2010 (UTC)??


 * 1) Common semiconductor diodes have a non-linear exponential (current)/(forward voltage) characteristic. A two-terminal linear diode can be constructed using a sensing switch that has two states: i) For reverse voltages there is no connection between the terminals, hence no current. ii) For forward voltages, a fixed resistance is connected between the terminals so current flows following the linear Ohm's Law relation I = V / R.
 * 2) A diode with a piecewise linear I/V can be a convenient simplification for approximate solution of an electrical circuit.
 * 3) Another meaning is in the name Linear diode array (see article) which is a row of photodiodes. Cuddlyable3 (talk) 09:43, 18 October 2010 (UTC)

Galvanic corrosion
How is the electricity generated? Is it a redox reaction? --Chemicalinterest (talk) 15:41, 18 October 2010 (UTC)


 * Yes, it is a redox reaction. -- k a i n a w &trade; 15:50, 18 October 2010 (UTC)


 * I understand that the more anodic metal is oxidized, but what is reduced? Oxygen in the air? Does this function similarly to a zinc-air battery? --Chemicalinterest (talk) 15:55, 18 October 2010 (UTC)


 * You can think of it as an inherent potential energy gradient. Electrons would rather jump from one metal to the other, because the crystal structure is more "comfortable" for electrons in the cathode than in the anode. So this manifests as a voltage. If current is allowed to flow, the electrons start moving to one metal; but this leaves "openings" in the anode - permitting chemical reactions with the surrounding environment. That can mean oxidation in air, or dissolution of the anode into salt-water, or binding of some other environmental chemical to the anode, or so on, all depending on the specific metal(s) and environments they are in. Nimur (talk) 16:13, 18 October 2010 (UTC)


 * I thought metal anions are not stable. How can electrons move to a metal? --Chemicalinterest (talk) 17:21, 18 October 2010 (UTC)


 * A metal crystal is quite a different environment from a single atom. The outer orbitals hybridize to form a crystal-spanning conduction band with lower energies (and, thus, room for more electrons than the individual atoms would be able to hold on to by themselves). That's why the atoms form a crystal in the first place. See metallic bond. –Henning Makholm (talk) 17:54, 18 October 2010 (UTC)


 * Why do metals lose electrons (e.g. corrode) so easily if they gain electrons so easily also? --Chemicalinterest (talk) 19:33, 18 October 2010 (UTC)
 * Metal anions are not stable if the charge is concentrated. Metals basically have a hyper-form of something called resonance (chemistry). Thus the moment the electron deposits at a metal atom at a cathode, the electronic charge is distributed not only at that atom, but simultaneously all over all the atoms in that metal. This electronic charge is readily absorbed at the other end (such as possible oxides of the cathodic metal). Thus for example, if you're using zinc to protect iron, iron oxides are readily reduced back to iron through zinc. So here oxygen serves as the "terminal electron acceptor" -- it's like the iron serves as a circuit between zinc and oxygen. John Riemann Soong (talk) 19:54, 18 October 2010 (UTC)
 * That's what I was thinking of. Is oxygen needed for galvanic corrosion to happen? --Chemicalinterest (talk) 20:21, 18 October 2010 (UTC)
 * Well sort of there's no rust then there's no net "sink" for electrons to go to. Of course this is ideal. BUT in the total absence of oxygen (and reducible protons!, e.g. no alcohols or water), you might see a steady build up of iron anions, either in a polar aprotic solvent or maybe in the bulk material, I would hazard. After all, there are such known things as capacitors (which is what forms when you try to make a voltaic system without a counterion bridge) and ferricyanide. John Riemann Soong (talk) 21:23, 18 October 2010 (UTC)

light pulse intensity with time
Assume a light pulse of about 1 second interval was shot in a closed spherical room which has its wall reflecting like normal mirrors. How and for how long would the light intensity (or light in general) behave inside this room after the light source was turned off (assume 0 transition time from ON to OFF)? I don't know how to describe the room size and light intensity but want to imagine what happens generally during such moments that human eye can't realize and also don't know if there are some web videos showing such events (high speed cameras).--Email4mobile (talk) 16:10, 18 October 2010 (UTC)


 * It depends how imperfect the mirror surfaces are. Real mirrors do not reflect all light: they reflect most of it, transmit a tiny percentage, and absorb some energy.  This is quantitatively measured by the Reflection coefficient of the mirror.  A good quality silver-on-glass mirror might have Γ = 0.999999 or better; the remainder of the energy is absorbed through lossy processes (typically, photons are absorbed and may be re-emitted as phonons - in other words, "heating" of the mirror surface).  Every time the light bounces, you have to multiply the amplitude by the reflection coefficient, so you have an intensity that decays as Γn after n bounces.  The time-constant for the light amplitude (or intensity) decay is determined by the way the geometry defines the number of bounces.  Two mirrors, a fixed distance apart, will reflect the light pulse n times per second based on their distance and the speed of light.  Because practically-sized rooms are small and the speed of light is fast, millions of bounces occur within just a few moments, and the light intensity decays very fast - too fast to be captured by any type of camera that I know of.  Nimur (talk) 16:19, 18 October 2010 (UTC)


 * Actually, 0.999999 (which I'll take your word for for this purpose) does not sound too forbidding. With a room-sized cavity the ringout time would be on the order of milliseconds. You could capture the fadeout profile on photographic film by spooling it quickly past a pinhole in the wall. On the other hand, the engineering challenges in making sure the chamber is completely covered in mirror (except for a pinhole for observation and another one for injecting light) would be considerable. Any visible gaps between mirrors would cause more loss than the mirrors themselves. –Henning Makholm (talk) 17:35, 18 October 2010 (UTC)


 * Whenever you (or a camera) see light, you absorb it. Light bouncing off a mirror is invisible, except for the person whose eye that light hits.  So, given a perfect mirror, there would be an (imperceptible, even to high-speed photography) delay, and then a flash, and that would be the end of the light beam.  (This assumes that all the opaque parts of the camera (including the inside of the lens) are covered in a perfect mirror.)  Paul (Stansifer) 16:45, 18 October 2010 (UTC)


 * Thanks Nimur and Paul. If I consider the 0.999999 reflection coefficient and 1 meter room radius then I would I expect enough time to capture events with a high speed camera since the new reflected energy intensity would reduce by that factor I guess (Assuming no other absorbing material around including the camera). For instance after 1000 reflections a time of about 6.7 us will have passed which can be recorded by high speed cameras. I also expect intensity to have reduced to 99.9% of original one. I expect it will take around 6.7 ms to reduce intensity to 10% unless my calculation were wrong. The serious problem then would be the camera sensor.--Email4mobile (talk) 17:11, 18 October 2010 (UTC)
 * This problem is not as abstract or hypothetical as you might imagine. In laser physics, it is important to know how perfect your mirrored surfaces are (if you are using an optical resonator).  A simple method of measuring the reflection coefficient of metallic mirrors at a wavelength of 16.6 μ (1974) provides a theoretical basis and an experimental methodology for setting up the measurement in question (decay time for a light-pulse).  In their work, the "unknown" is the mirror quality, and the decay-time is used to measure the reflectivity.  (Actually it appears that they set up a beam splitter and measure differential intensities, which is probably more accurate and less sensitive).  A quick survey of laser research publications will turn up many similar experiments.  Nimur (talk) 19:10, 18 October 2010 (UTC)
 * I think an important consideration that's being overlooked in the above is the OP's specification that the room is spherical. It's a rather different situation than with a laser, in which the two mirrors are essentially perfectly flat, and the interference between the photons is purely constructive.  With a spherical mirror, there are going to be complicated patterns of both constructive and destructive interference, resulting in a broadening of the light's bandwidth to the point where most of the light's energy will very quickly be outside of the visible range.  The visible light isn't going to last a thousand reflections, no matter how close to perfectly reflective the mirror is.  Red Act (talk) 05:44, 19 October 2010 (UTC)
 * I don't know how interference can change the frequency (or bandwidth) of the signal - adding any combination of sine waves with the same frequency can only affect the phase and amplitude. Changing frequency is nonlinear optics.  The parameters of the mirror/glass might actually do some strange stuff, but most mirror surfaces will not have significant nonlinear optical properties.  Nimur (talk) 14:40, 19 October 2010 (UTC)
 * Well, what's going on isn't really a combination of sine waves with exactly the same frequency. It's not like the case of a stationary optical system being continuously illuminated with a coherent beam, in which case the waves all having the exact same frequency would be an extremely good model of the light.  Indeed, if you take the OP's specification "assume 0 transition time from ON to OFF" too literally, you wind up with complete uncertainty in the light's frequency, since from the Heisenberg uncertainty principle, an uncertainty of exactly 0 as to the time when a photon ends would require infinite uncertainty as to the photon's energy (and hence frequency).  The best you can start off with is with photons consisting of a wave packet in which the amplitude of the frequency observable forms a narrow band around some desired frequency.
 * If you then reflect that initial wave packet off of a perfectly flat ideal mirror, the uncertainty of the frequency doesn't increase at all, since the reflected wave packet is exactly identical to the incident wave packet, except for a symmetry transformation. And even if you bounce the wave packet off of a spherical but otherwise ideal mirror once, the uncertainty of the frequency will still only increase negligibly, as long as the mirror is negligibly different from flat within some small area within which the reflection almost certainly occurs.  But with a non-flat but otherwise ideal mirror, the uncertainty of all observables does in general change a bit with each reflection.  And after some number of reflections, the uncertainty of the location at which a reflection occurs is going to become large enough such that the mirror is no longer negligibly different from flat within a minimal area within which the reflection almost certainly occurs.  Once you get to near that point, it will rapidly start becoming the case that a wave packet after a reflection bears little resemblance to the wave packet before the reflection, in which case the uncertainty of all observables will all increase rapidly.  Then after enough reflections, the uncertainty of the observables will each approach some limit, and I think you'll basically wind up with black body radiation.  Red Act (talk) 05:49, 21 October 2010 (UTC)
 * I remember reading (and seeing a picture) of a ray of light that had been shone from a torch of some kind and was HALFWAY across a dark room. It was the first time anything like this had been done. The room had some kind of gas in it so that the light ray could be seen from the side. Awesome! Does anyone remember or have any other details on this? Myles325a (talk) 03:40, 23 October 2010 (UTC)

Mass spectra
Today, I saw a (fairly contrived) mass spectrum for benzonitrile, with only a small peak at 77 normally associated with the breakdown of benzene; instead it had a peak at 76. Was the spectrum wrong? Is C6H4 formed because of the high-energy conditions?

Also, the usual method of (relative abundance of the [molecular weight+1])/(1.1*molecular weight)*100 to find the number of carbons in the carbon chain (assuming you had no easier way of finding this) doesn't work on amines. Is this because of the abundance of the 15N isotope or some other reason? Grandiose (me, talk, contribs) 17:06, 18 October 2010 (UTC)
 * Even the actual spectrum of benzonitrile has only a small abundance at m/z 77 vs much larger at 76. Consider what pieces have to be lost to get the 76 signal, and whether those lost pieces could together form a stable structure (in general, high-abundance fragmentations result when one or both fragments are fairly stable). All sorts of fragmentations can occur: simple one-bond-cleavage, further bonds breaking from that initial cleavage product, McLafferty rearrangements and other concerted changes, etc. Regarding your heuristic for number of carbons, look back and consider why it does work. It's probably based on the isotopic abundances of C (seems to agree with 13C being 1.1% of 12C). Each isotopic combination is unaffected by the others--the peaks are really just the arithmetic of each formula weight times the abundance of a given isotopic mix. 15N is only 0.4% abundance among nitrogen isotopes, so it also affects the abundance at M+1 but "differently" from a carbon atom...maybe enough to throw off your "/1.1" formula calculations. I bet you could actually derive a similar formula for amines though...the result of that calculation you have would give something like "number of C plus one-third number of N" (since an N only contributes about a third as much to the size of the M+1 than C does). DMacks (talk) 17:47, 18 October 2010 (UTC)

Squirrel locomotion
I see a lot of squirrels around my neighborhood. They appear to hop around, keeping their fore legs and hind legs together. It's as if you could tie their fore- and hind legs together and they would not have any trouble moving around. Are there other animals that run in a similar manner? Hemoroid Agastordoff (talk) 18:13, 18 October 2010 (UTC)
 * Many -- it's called an asymmetrical gait (strangely enough). Looie496 (talk) 18:31, 18 October 2010 (UTC)
 * As an Aussie, I immediately thought of kangaroos. HiLo48 (talk) 03:29, 19 October 2010 (UTC)
 * You mean when they are not hopping? I wonder if their tail counts as an extra limb for gait classification when kangaroos and wallabies do their strange slow speed "walk".  Googlemeister (talk) 13:23, 19 October 2010 (UTC)

What we see is a prediction
I just heard on tv that what we see when catching a ball etc is actually a prediction by our brains rather than the actuality. This is because our brains are too slow to provide the information in real time. Where can I read more about this? Thanks 92.15.28.219 (talk) 20:41, 18 October 2010 (UTC)
 * The way I understand it is that when the ball is moving, your eyes and brain predict its motion. So when you catch the ball you catch where it will be, not where it is. It's not a matter of "seeing" though. Also a lot of the information the eyes provide is actually interpolated based on expectations, rather than what's actually happening. I don't have a particular link for you though, sorry. Maybe one of the many vision articles? Ariel. (talk) 22:26, 18 October 2010 (UTC)
 * See Flash lag illusion. --Dr Dima (talk) 00:57, 19 October 2010 (UTC)
 * Hmmm, I looked at that article, and the example it cites, but calling it a "flash lag" seems a weird way to describe it. The illusion is that a blue circle which flashes yellow doesn't appear to be entirely yellow when viewed obliquely, but only partially yellow at the lagging edge.  But the yellow doesn't leak out the back edge of the circle.
 * To put it directly, areas that go blue->yellow->blue look blue; areas that go blue->yellow->black->white look yellow. Areas that go blue->yellow->blue(very short time)->black->white may look yellow also.  To me it doesn't seem like perception is unusually fast or slow, just that the periphery sometimes simplifies the apparent series of events in a set of brief flashes.  There should be further tests that can be drawn up to verify or disprove this. Wnt (talk) 19:13, 19 October 2010 (UTC)
 * The particular example you link to is not a standard flash-lag illusion but a (very nice) extension of it. The standard flash-lag illusion is as follows. Imagine you have a steadily moving visual stimulus A and a flash B in your field of view. Visual input takes a few tens of milliseconds to be processed by your retina, thalamus, and primary visual cortex. During that time, A shifts by a certain distance, so ostensibly you should see it where it was and not where it is. To compensate for that, our visual system extrapolates where A is now from the available information (that is, from the trajectory of A up to few tens of ms ago). Imagine that B flashes exactly as A is passing it by. For few tens of ms you perceive A but not B - your visual system did not process the flash B yet. After few tens of ms you perceive B where it WAS; but you perceive A where it is NOW; that is, your brain is tricked into thinking that the flash lags behind the moving stimulus by a few tens of ms. That is the standard flash lag illusion. The Flash lag illusion article lists alternative explanations, too; I just gave you the simplest one. There is an enormous literature on all sorts of optical illusions related to the flash lag illusion; there is a short list of references in the Flash lag illusion article, which may be a good place to start. Hope this helps. --Dr Dima (talk) 22:34, 19 October 2010 (UTC)

When a balloon pops, does it go supersonic?
it sounds to me like the crack of a whip... do either of these two things really go supersonic? 92.229.12.100 (talk) 21:33, 18 October 2010 (UTC)


 * I understand the crack of a whip can cause the tip of the whip to reach a critical Mach number. See Whipcracking. For an expanding sphere of air to be at sonic speed requires a certain pressure (that can be calculated.)  However, it is a much higher pressure than can be achieved in a party balloon, so I suspect a popping balloon is not associated with sonic speed.  Dolphin  ( t ) 21:39, 18 October 2010 (UTC)


 * The expanding air surely doesn't go supersonic as Dolphin pointed out above, but the retracting rubber probabily does and that's the sound we hear. 67.78.137.62 (talk) 21:46, 18 October 2010 (UTC)


 * On second thoughts, sound does travel at the speed of sound! A bursting balloon causes a sharp-edged series of sound waves which obviously travels at sonic speed, so perhaps the best answer is yes.  Other sounds, like clapping hands, don't cause a sharp-edged series of sound waves, so they don't cause us to register the same cracking sound.  Dolphin  ( t ) 21:57, 18 October 2010 (UTC)


 * Every supersonic shock wave has an associated sonic wave-front. But not every loud "burst" or "crack" is due to a supersonic shock-front.  Sometimes, that is just the waveform of the sound source.  In the case of a popping balloon, I can't think of any reason to assume there would be a supersonic shock.  The burst is just due to the sudden loss of confinement of the air, which is at marginally higher pressure than the surrounding air; and the retraction of the rubber.  Nimur (talk) 22:28, 18 October 2010 (UTC)


 * See this video . The stated frame rate are 2700 Hz, the sound travels 340/2700 m = 12.6 cm per frame. If the framerate is correct the maximum speed of the rubber are between 50% and 120% of the speed of sound but i do not know if this is the source of the sound. Maybe someone can analyze the video closer. The rip expands faster than the rubber moves, this can maybe be interpretated as the speed of sound in the rubber membrane, I do not know how that relates to the speed of sound in the air. The bulk of the airflow are much slower.
 * --Gr8xoz (talk) 00:01, 19 October 2010 (UTC)

Nepal timezone
Is there an interesting reason to explain why the country of Nepal decided to use UTC+5:45 when selecting its time zone (which is a very odd offset compared to timezones in the rest of the world)? I thought that it might be mentioned in the Nepal Time article, but this is a mere stub at present. --Kurt Shaped Box (talk) 21:45, 18 October 2010 (UTC)


 * UTC+05:45 (which probably does not need its own separate article) explains that the local mean time, which is purely geographically derived (longitude), is 5:41:16 ahead of UTC. Nimur (talk) 22:24, 18 October 2010 (UTC)


 * According to the tz database, until 1986 Nepal used Indian time (UTC+5:30). I've seen someone say somewhere that they changed for political reasons during a period when Nepal and India were unfriendly, but I don't have a cite.  --Anonymous, 06:11 UTC, October 19, 2010.


 * I thought I've read something similar but couldn't find it until now, see UTC+05:40. It was actually Kathmandu mean time they used until they turned to UTC+05:45. See also. It's perhaps worth remembering there are a whole lot of "odd" (odd being relative of course, from the POV of a Kathamandu meridian UTC is odd) historic timezones, I guess in the age before easy international communication it didn't really matter that much. (You may also notice a lot of the places which had the odd timezones are either Commonwealth or with strong links, I suspect there's something there.) Nil Einne (talk) 09:48, 19 October 2010 (UTC)


 * Well, that contradicts the tz database, which is a pretty reliable source, but perhaps less so for historical data. The database says they changed from +05:41:16 to +05:30 in 1920 and then, as I said, to +05:45 in 1986. I've seen +05:40 in some sources myself, but I don't remember when; I just assumed it was a mistake for +05:45.  Maybe the 1920 change was an attempted change that didn't take hold with the public, or something.  --Anonymous, 11:06 UTC, October 19, 2010.

Can acid reflux irritate the tonsil?
Thanks. 67.243.7.240 (talk) 23:10, 18 October 2010 (UTC)
 * Yes, acid reflux is when stomach acid rises up out of the stomach, sometimes all of the way to the back of the throat. Because of this, the acid may cause damage and/or inflammation/irritation of the tonsils fairly easily. Tyrol5   [Talk]  23:53, 18 October 2010 (UTC)
 * If you have a problem that you think may be caused by acid reflux you should see a qualified medical practitioner and not take the advice of an anonymous stranger. Richard Avery (talk) 08:03, 19 October 2010 (UTC)


 * Yes. See, , & . Axl  ¤  [Talk]  09:19, 23 October 2010 (UTC)