Wikipedia:Reference desk/Archives/Science/2013 January 18

= January 18 =

V-engine with unequal stroke lengths on different banks
Soviet's V-12 Diesel model V-2 engine has a stroke of 180mm on the left cylinder group and 186mm right cylinder group. Dncsky (talk) 02:00, 18 January 2013 (UTC)
 * 1) Is there a name for this configuration? Googling "unequal stroke length" got me nothing related.
 * 2) What's its purpose?
 * 3) Is there any other engine with this setup?


 * It does seem odd. A quick google search found this, which states that "The V-2-34 features an aluminum alloy body and is meant to be mounted lengthwise in the vehicle hull. Two cylinder banks with 6 cylinders each were placed in an angle of 60 degrees. The pistons are linked to the central crank shaft by wrist connecting rods, which means that only six rods are directly connected to the crank shaft. This special design also results in a slightly lower stroke in both sides of the engine. The right side has a stroke of 186.7 mm and 180mm in the cylinders on the left bank". From the description, the unequal stroke may be a side-effect of the design, rather than an intended feature. The photo of a cutaway V-2 here seems to show a connecting rod (on the left) which is connected to something other then the crankshaft - from the look of it to an extension lug on the side of a conventional connecting rod big-end. This would fit the description, and if it works the way it appears, the reduced stroke for the secondary connecting rod may be a simple consequence of the geometry. AndyTheGrump (talk) 02:28, 18 January 2013 (UTC)
 * Thank you!Dncsky (talk) 02:34, 18 January 2013 (UTC)


 * Looking further, it may be the other way round: using Google translate on one of the sources that the Russian-language article on the engine cites, It appears that the secondary connecting rods have the longer stroke - and in consequence, the cylinders have a higher compression ratio. AndyTheGrump (talk) 02:49, 18 January 2013 (UTC)
 * That would be a physical impossibility, unless the secondary conrod connecting point was very displaced, on a spur off the primary conrod. That would have very severe weight and strength penalty - you wouldn't do it. If you visualise a connecting point on the primary conrod, inline with its logitiudinal axis, that point must, as the crankshaft rotates, describe a circle just above and at smaller diameter than does the big end.  Hence the stroke produced by the secondary conrod must be less than that produced ny the primary conrod.  If I was the designer of this engine, adn for some reason wanted this conrod arrangment (I can't think why I would, although it does allow a large bearing area), I would doctor the intake valve timing so that the effective compression on both sides was equal.  This would enable smooth running while using the greatest commonality of parts.  The impact on power output (rather arbitarily set in a diesel engine) and fuel economy would be negligible.
 * As for other engines using this configuration, radial aircraft engines have been made with one master conrod and several secondary conrods attached in a ring around the master bigend. In a radial engine, there is not the room for separate bigends for each rod to be on the crankshaft.
 * Ratbone 124.178.174.189 (talk) 03:05, 18 January 2013 (UTC)
 * Looking at the cutaway photo, I'd say that the secondary conrod connecting point was displaced, on a spur off the primary conrod, though it is hard to tell. In any case, even with the secondary conrod connecting to the primary on the centreline of the primary, the thrust from the secondary conrod is going to create a bending force, due to the cylinder V configuration. As for the exact geometry, without further information there is no way to be sure. AndyTheGrump (talk) 04:28, 18 January 2013 (UTC)
 * A longer stroke does not imply a higher compression ratio. If you double the stroke and you move the piston down far enough to double the volume at top dead center the compression ratio is unchanged. --Guy Macon (talk) 04:30, 18 January 2013 (UTC)
 * God point, Guy. Anyway, I've found a diagram of the engine here (about 20% down the page): - note the way the secondary conrod 'wrist pin' is offset from the centreline of the primary. As for which cylinder has the longer stroke, I'll leave it for someone else to figure out. AndyTheGrump (talk) 04:41, 18 January 2013 (UTC)
 * A longer stroke with the same compression ratio means that something else has to change, either pistons, and/or cylider head height, remembering that the piston must at top dead centre just about touch the head, or piston squish will not work properly, leading to poor compustion, fuel and combustion products forced past the rings, etc. Cheaper and easier to keep everything the same on both banks, and tweak the intake valve timing as I said.  However, all this appears to be not relevant.  I printed out the cross section view linked by AndyTheGrump, and found by carefull measurement that:-
 * Pistons, and cylinders on both banks are the same; So are the heads, except for being mirror images of each other;
 * The secondary conrods are indeed attached on a spur on the primary rods;
 * The spur is positioned such that, at left bank TDC, the spur bearing is positioned just slightly to the right of crankshaft centre and just slightly below the point at which a line to the spur bearing centre would make 120o to a line from spur centre to crank centre. This means that the stroke for both banks is the same!.
 * Where did the 180 mm / 186 mm dimensions in the Wikipedia article come from? There is no reference for the dimensions cited - it's probably wrong, even though the refrences [16] and [17] (Russian Wikipedia) provided by AndyThe Grump clearly state it's the case. [16] clearly states each bank has a different stroke, but not for all versions of the engine, which does not seem likely.
 * Ratbone 124.178.147.222 (talk) 05:08, 18 January 2013 (UTC)
 * This reference - http://translate.google.com/translate?sl=ru&tl=en&js=n&prev=_t&hl=en&ie=UTF-8&eotf=1&u=http%3A%2F%2Fru.wikipedia.org%2Fwiki%2F%25D0%2592-2, states that the secondary rod side has greater stroke and compression (look at table at bottom), "due to kinematic reasons". Presumably that means to statically balance the engine - which is weird and makes no sense to me.  The spur arrangment would indeed introduce balancing issues, but a difference in stroke will not fix that.  Ratbone 124.178.147.222 (talk)  —Preceding undated comment added 05:39, 18 January 2013 (UTC)

Seismic morse code
Let's say I was a Soviet spy in Washington, DC, and I wanted to send an SOS message to a seismometer monitored by my contact in Moscow. How large a bomb would I need to use for each dot or dash, and what sort of transmission rate could I get? --Carnildo (talk) 03:08, 18 January 2013 (UTC)
 * You'd need a nuclear weapon for each dot and dash, and your transmission rate would be maybe one character a day (due to the need to travel to a new site and bury the weapon prior to sending each new dot or dash). Of course, with that many nukes, you won't need to send an SOS -- you could just blow up your pursuers (along with anyone else who happens to be within a country mile) and be done with it.  (Of course, doing so will probably lead to a nuclear war between the USA and the USSR, and also setting off nukes on US soil without authorization from the Soviet Prime Minister will most likely get you executed as an "enemy of the state".)  Now can we PLEASE have some SERIOUS questions?24.23.196.85 (talk) 03:28, 18 January 2013 (UTC)


 * Not necessarily. It is a classic question in information theory, sometimes set by lecturers in the subject. In any communications channel, radio, sound in air, seismic, or whatever, there is a tradoff between information rate, transmitter power, and noise level.  Noise level is in this case set by interference at the receiving site by vibrations from vehicles on roads, construction site works, etc.  For a given noise level, how much transmitter power (i.e., size of explosion) you need is proportional to how fast you want to send the message.  Real communications systems generally require more power than is predicted by the theorry - one has to design a encoding system to make the best of the channel.  If you wanted to transmit "SOS" and could accept taking months to send it, you could perhaps employ a building site pile driver, using a code that translates the three letters into a vast number of bangs.  But if you could accept only seconds, you'd probably need a nuclear bomb. See http://en.wikipedia.org/wiki/Information_theory Ratbone 124.178.174.189 (talk) 03:31, 18 January 2013 (UTC)


 * Absolutely not -- the magnitude of the seismic wave drops off as the SQUARE of the distance, so by the time it travels from Washington DC to Moscow, any seismic wave less powerful than that made by a nuclear explosion will have dropped to below detectable levels. This has nothing to do with information theory, this is straight physics. 24.23.196.85 (talk) 03:38, 18 January 2013 (UTC)
 * Sorry, but this is no different to any other communication medium. For example, in radio comms, where information theory has a major application, received power also falls as to the square of the distance.  Info theory tells us why, for example, shortwave broadcast radio requires hundreds of kilowatts of transmitter power to send AM music, but radio hams can communicate around the World using only a few watts using morse code.  The square law makes the transmitter power required much greater than it would be for a linear-with-distance loss system (which never occurs in practice), but the info theory math still applies.  You do need a filter mechanism at the receiver tuned to the coding system in use, so that the noise is discriminated against.  I suggest you read up on info theory before talking about things you are not aware of.  Ratbone 124.178.174.189 (talk) 03:50, 18 January 2013 (UTC)
 * On the contrary, YOU should read up on physics -- a seismic signal simply WILL NOT make it halfway around the world without dropping to undetectable levels. Do you even know WHY a radio signal can be detected at such a distance?  It's NOT because it's so powerful, but because it's focused in just one narrow frequency channel, and because the receiver can efficiently filter out the broadband noise and amplify ONLY the selected frequency -- which is NOT the case with a bomb and a seismometer!  Also, like ALL big cities, Moscow has such a high background seismic noise level on ALL frequencies (from construction work, trucks, buses, trains, trolleys, the Metro, etc.) that even if a weak seismic signal does make the distance, it will be drowned out by the noise and undetectable! 24.23.196.85 (talk) 05:33, 18 January 2013 (UTC)
 * You've sunk your own argument. Yes, indeed radio works by filtering.  But you can filter a mechanical transmission system (eg seismic) too.  There is more than one way to filter.  You can filter on frequency, as in traditional analogue radio, or you can filter on digital coding (as in cellphones and some types of military comms - termed "spread spectrum", although this is only one type of spread spectrum).  Comms systems can, and do, share the same frequency if the digital coding is different, and share the same coding if the frequency is different.  A simple example:  Say the agent makes a thump every 32 hours for a month to send a morse dash, and every 40 hours for a month to make a morse dot.  Moscow could then filter for these thump rates and exclude much noise.  (it would not be smart choosing an integer multiple of 24 hours, or a sub-multiple, for obvious reasons)  There are much better coding schemes but this will do for a simple example.
 * Your bit about Moscow being noisy is of no importance, for two additional reasons: a) it just means the transmitter has to be a bit more powerfull than otherwise, and b) who said the reciving site has to be at Moscow? They can do the same as is done with Embassy radio coms: set up a reciving site at a quiet location, and pass the message on to Moscow via phone or courier etc.
 * I strongly suggest you read up on the subjects of information theory, coding, messaging systems before you goof again. Electronic and comms engineers spend much time studying this very same field in undergraduate courses.
 * Ratbone 124.178.147.222 (talk) 05:57, 18 January 2013 (UTC)


 * I strongly agree - even a small source is not 100% undetectable - even if the amount of signal that reaches the detector is amazingly tiny. If the pulses are regular enough - and at a known frequency - then you can detect it given enough time. Suppose we have a hypothetical pile-driver that thumps (or does not thump) at PRECISELY one second intervals and does that (or doesn't do it) continuously to send one bit of information every 32 years.  Let's suppose that the effect of one of those thumps is a signal that appears at the detector that is one part per billion of the 'background' vibrations due to distant volcanoes, cars, footsteps of the experimenter, etc.  Let's have the recieving seismograph be monitored every half second for all of that time and separate out that data into the odd-numbered half seconds and the even numbered half seconds.  (Technically, you'd need to sample a little faster than that...but let's keep things simple for a moment!)
 * If the transmitter sent a "0" bit (the pile driver didn't thump for 32 years) then the difference between the average of the odd and of the even half-second readings measured over 32 years will be very, very close to zero.
 * If the transmitter sent a "1" bit (by thumping once a second for 32 years) then the average vibration at all of the odd-numbered values (when the pile drivers' hammer was being lifted ready for the next thump) over 32 years will be significantly less than the average of the even-numbered half-second values when the pile-driver hit the dirt. The amount of difference in the measurement was one part in a billion - but added up a billion times and divided by a billion - the result should be that the even-numbered reading will be about twice that of the odd-numbered readings...easily noticeable...quite distinctive!
 * The difference between the two is perfectly adequate to receive a one bit signal at a billionth of a Hertz transmission rate. Even if the detector isn't accurate enough to receive that signal, the errors in the detector will average out over enough time - and the signal will show through.  Even if the detector has only one bit of precision - showing a '0' for less-than-average amounts of ground displacement and '1' for greater-than-average, the presence of background "noise" around that cutoff point will result in the pile driver producing a statistically greater-than-average number of 1 bits over 32 years...even if the detector is inherently crappy and noisy - the resulting noise produced in the signal will average out eventually.  There are entire areas of science and mathematics in "communications theory" that relate signal strength, noise ratios, bit rates, detector precision and so forth.


 * Of course this might fail if there is interference from other sources at the same frequency. So if you picked once-per-second (1Hz) as your signalling rate and there is a mechanical clock in the room with the detector - then the clock is a transmitter "broadcasting" on the same frequency and that could easily wipe out the signal that you're trying to detect.  The frequency of our pile-driver clock would also have to be pretty solid - if it was "off" by a bit, or drifted compared to what the detector is expecting, then that would be difficult.  A smart pile-driver-transmitter designer would pick a frequency that's not polluted by noise from man-made sources.  You'd want to avoid 1Hz transmission rates for sure!  Maybe also avoid rates close to that of the footstep-frequency of a typical adult human...avoid rates similar to the density of road traffic near where the receiver is located.


 * The practicality of averaging the results from a seismometer for a billion seconds and getting one bit of information every 32 years is going to be problematic...but it's not impossible.


 * To pick a "real world" example, consider how hard it is to detect a planet orbiting a star that a few hundred lightyears away by the tiny amount of light that's blocked by the planet as it crosses in front of the star for a few days once a year. The amount of variation in the brightness of the star as a tiny planet crosses in front of it's disk is vanishingly small...but analyse the light over a long enough time and you can pull that tiny variation out from all of the background noise in the telescope.  Science does these kinds of amazing tricks routinely.  It's just a matter of technique and time. SteveBaker (talk) 15:48, 18 January 2013 (UTC)


 * All this still won't help the KGB if the signal is so weak as to be below detectable levels (which will be the case for a Washington DC to Moscow seismic signal that uses anything less than maybe a 100-kT nuclear bomb). 24.23.196.85 (talk) 01:04, 19 January 2013 (UTC)
 * Didn't you read anything that Ratbone and SteveBaker wrote? Information theory, as both have outlined, is a well established branch of science, applying to any communication medium, radio, communication by light beam, mechanical systems, seismic communication, etc.  A standard textbook used in many information/communication theory courses for some years is Modern Digital and Analog Communication Systems, B P Lathi, pub: Holt-Saunders.  Chapter 8 covers in depth the principles discused by Ratbone and SteveBaker here.  Coding is covered in other chapters.  If you have good high-school level math, you can study this or similar books and realise you are wrong - given enough time and a good coding system, there is no lower limit on transmitter power. The subject of information theory, which can be regarded as the science of sending a mesage as fast as possible in a finite bandwidth in the presence of noise with the least transmitter power, was established by C E Shannon, R V L Hartley, and H Nyquist papers in the Bell System Technical Journal, in the 1920's. Keit 121.215.159.209 (talk) 01:44, 19 January 2013 (UTC)

I have been poached! μηδείς (talk) 12:30, 18 January 2013 (UTC)


 * You are wrong. Given enough time and a repetitive signal, the concept of "below detectable levels" does not exist. Repeat it enough times on a regular enough basis and you can send a Washington DC to Moscow seismic signal with the power of a firecracker. Of course you will die of old age long before it is received... I haven't run the numbers but I suspect that it would take longer than the current age of the universe. Go back and read SteveBaker's comment again. --Guy Macon (talk) 01:35, 19 January 2013 (UTC)


 * Of course the concept of "below detectable levels" exists -- if the signal is weaker than the seismometer's sensitivity limits, then the instrument won't pick it up no matter how many times it repeats! YOU are the one who's wrong here! 24.23.196.85 (talk) 02:09, 19 January 2013 (UTC)


 * I'm transmitting a "1" with a dim LED in a sealed safe 1km underground at .0187534019265 Hz. The reciever is a silver halide camera on the other side of the Earth. :) Sagittarian Milky Way (talk) 03:28, 19 January 2013 (UTC)


 * I am totally clueless here, but doesn't transmission require a channel or a medium? The earth will conduct vibrations.  But in Sag.'s example, does the 10,000 degree core of the earth count as a channel for the transmission of light? μηδείς (talk) 04:02, 19 January 2013 (UTC)
 * There is another issue with Sag's example: Silver halide detection requires the received energy to exceed a threshold - below the threshold for each sliver halide crystal nothing happens.  However, this is not important wrt the OP's question, as a linear transducer can be used to detect seismic waves.  Ratbone 121.215.34.98 (talk) 04:09, 19 January 2013 (UTC)


 * The Sun can be seen through thin enough gold leaf. Let's say the sunlight is made at least 10 million times dimmer by passing through the gold leaf (100000 is used for visual filters). The Earth is I don't know a 100 trillion times thicker? If the Sun could be bought, the box would say 35,700,000,000,000,000,000,000,000,000 lumens. Adjust for inverse square-law of the distances. So at least 260000000000000000000 x10000000100000000000000 times dimmer than visual detectability. And if you say the halides won't be activated and there's a bright core in the way, then information theory! (make sure the photos are near the threshold, though) Sagittarian Milky Way (talk) 17:21, 19 January 2013 (UTC)


 * And even a linear transducer will have a certain detection threshold below which it will not detect a signal. If nothing else, the Heisenberg uncertainty principle puts an ABSOLUTE limit on the accuracy of ALL measurements and precludes the detection of an "infinitely small" signal.  24.23.196.85 (talk) 07:04, 19 January 2013 (UTC)
 * But you don't need to detect individual pulses to detect the message. The signal is superimposed on the background noise. Sometimes the sum of noise and signal pulse will breach the detection threshold, and sometimes it wont. But if the signal pulse is there, this sum will breach the threshold more often than if it isn't. You can collect data over a long enough time to get to an arbitrary certainty for distinguishing the "pulse" from the "no pulse" case, no matter how small the difference. Heisenberg does not affect this. --Stephan Schulz (talk) 09:59, 19 January 2013 (UTC)


 * Honestly, all the discussion is coding theory is largely academic, while the original question was phrased in practical terms. A system that takes 100 years to transmit a signal is of no practical value.  So far, no one has really looked at the problem in practical terms.  We would need to know what the transmission efficiency of seismic waves from DC to Moscow actually is.  And we need a frame of reference for how much seismic energy you can generate from an explosive (or any other method).  Till one has those details in hand, you can't figure out if the initial signal to noise ratio is 10:1, 1:1, 1:1000, 1:1000000, 1:1030 or what.  Without such basic details you can't figure out what is needed to build a practical system for seismic communication.  Coding theory can dig you out of the noise, but if you are starting many orders of magnitude below the noise floor then it is unlikely the system can ever be built in a practical way.  Dragons flight (talk) 04:04, 19 January 2013 (UTC)


 * The original question was deliberately posed as an absurd one on the talk page--hence my obscure reference to being poached. Carnildo is having a little joke on us by posting it here.  But I do think the question of a medium which follows from subsequent comments is an interesting one.  For example, "In space, no one can hear you scream."  I take that to mean that ordinary sound waves simply cannot propagate through a vacuum.  (Although supernova shockwaves can through near vacuums, if I understand correctly.)  Doesn't information theory require a suitable medium, and aren't claims for infinite transmissabilty (again, a term I am making up) based on assumption like the perfect gas that don't hold in reality? μηδείς (talk) 04:19, 19 January 2013 (UTC)


 * I reckon I showed that it is not a system you would use in practice in my first post. Using radio is the tried and tested way and is vastly cheaper and more convenient.  The agent could also simply phone and use a coded message.  We used to phone our family in East Germany from time to time, business calls were made, and who is to know that a certain pre-arranged number (not necessarily in any particular country) will be answered by someone trained to react to a certain secret code word by forwarding it on.  However, the OP asked how large a bomb is required for seismic signalling, and what sort of transmission rate could be achieved.  That's a good question to ask, and we've answered it - the transmission rate and bomb size need to be traded off against each other, and the trade off will not be of practical use as you say.  Ratbone 121.215.34.98 (talk) 04:21, 19 January 2013 (UTC)


 * Yes, information theory requires a medium by implication, as the system bandwidth is one of teh calculation inputs. For electromagnetic waves (radio, light), the medium can be a vacuum.  Sound can travel in anything that is not a vacuum.  The medium in this case is the earth propagating seismic waves - essentially low frequency sound travelling in an elastic solid the same as it does as a result of geologic disclocations i.e., earthquakes.  Radio waves and light travelling in a vacuum is a lossless system - the strength of the falls of with the square of the distance purely because of the ever increasing volume per unit energy.  In the case of sound (or seismic waves) travelling in a solid, there is a loss above this - because there are frictional losses converting some of the sound energy to heat along the way.  That does not mean a total loss after some distance x though - you just need to increase the transmitter power (bomb size) to compensate.
 * I don't see the question as necessarily absurd or silly - in part because this is a classic assignment question sometimes asked by lecturers in information theory and coding, as I stated in my first post. By specifically asking about bomb size and signalling rate, the OP has displayed some ability to think beyond what many lay people can do (and clearly better than poster 24.23.196.85).  Or he might be just being silly and fluked a good question - I don't know the guy, so I cannot know.  We shouldn't be too ready to judge a question silly - many of us ask questions of an unusual scenario in order to test or increase our understanding of physics.  I've done it myself - it is a very effective technique.
 * Ratbone 121.215.34.98 (talk) 04:39, 19 January 2013 (UTC)
 * The silliness lay in my original talk page suggestion that a spy would send an sos signal by nuclear bomb, not in the pure physics aspect. As a physics question it could simply be reworded as what TNT equivalent would be necessary for a regular signal to be detected on the other side of the Earth.  (I also understand at the antipodes the signal would be focused and hence magnified.) μηδείς (talk) 05:09, 19 January 2013 (UTC)
 * I should point out that Moscow and Washington DC are not geographical antipodes (though they can be considered political antipodes) -- in fact, the distance between them is approximately 5000 miles (8000 km) on a great circle route. As for calculating the size of the bomb needed, the most important piece of info is the seismic wave absorption coefficient of oceanic basalt -- this absorption, along with the inverse-square decrease in intensity I mentioned earlier, will account for most of the signal attenuation over this distance.  (In practice, even this prediction will be overly optimistic, because the seismic waves will be scattered and partly retroreflected back toward Washington DC every time they encounter a geological fault or a boundary between different rocks -- in particular, there will be major attenuation at the Mid-Atlantic Rift, which will mean the signal will arrive in Moscow even weaker than what this theoretical model predicts.) 24.23.196.85 (talk) 07:17, 19 January 2013 (UTC)
 * The same thing happens with any wave propagating through a medium. Electromagnetic waves suffer a degree of attenuation above the square-law-with distance rule in non-vacuum media due to a degree of conversion into heat, and portions get reflected backwards when any change in media offreing different dielectric and/or magnetic properties is encountered.  In all cases, the intensity at some distance x is non zero, and as it is non-zero, it can be used to communicate by means of frequency filtering or pattern filtering.  Indeed, in seismic testing to determine geology for earth resistivity calculations, filtering is used to reduce the size of the bang/thump to that of striking a plate with a sledge hammer a few times - the detection equipment knows when you hit the plate and triggers its signal storage and timing from it.  Ratbone 121.215.17.194 (talk) 09:54, 19 January 2013 (UTC)
 * What kind of seismic testing you are referring to ? I'm not aware of seismic being used to determine resistivity, although I guess it's sometimes used to constrain the earth model for an inversion based on well log data. I don't understand "filtering is used to reduce the size of the bang/thump to that of striking a plate with a sledge hammer a few times". A lot of steps are involved in processing seismic, but I don't recognize that one, although I guess you might be referring to compressing the wavelet via decon. If anything, the opposite of reducing the size of the bang is done to try to minimize the effects of absorption and geometric spreading from the data. The recording equipment for marine streamers and onshore/ocean bottom cables certainly needs to know when the airgun arrays/vibrator sweeps happen or else you will be burning through hundreds of thousands of dollars in acquisition cost with nothing to show for it in no time.  Sean.hoyland  - talk 12:29, 19 January 2013 (UTC)
 * Seismic testing using a sledgehammer and plate as the signal source is common in testing for civil engineering purposes. What I was refering to is the use of the same sort of gear for determing the geo-electric conditions in orfer to design/precalculate earth rod systems for elecrical earthing in high voltage and extra high voltage electricity transmission.  In any AC lectrical transmisssion system, where overheard wires or buried cables transport electrical energy from one location to another, a portion of the return current flows in the Earth.  Under fault conditions this current can be substantial, and the connections to Earth at each end are often substantial engineering projects - they may be multiple metal rods driven 30, 40, 50 metres or more into the ground.  The cost is substantial, so it is important that the Engineer gets his calculations as to just what is required at a given site accurate.  The return current flowing in the Earth flows centered at a depth known as the Carson Depth (after the USA Engineer who first worked out the math, in the 1920's).  The Carson Depth is typically a few hundred meters to a few kilometers down. This means that the Engineer must have an adequately accurate undertanding of the electrical conductivity of the various geological strata in the vicinity, say to within a few hundred metres to a few tens of kilometers of the site.
 * A simple geology is a two-layer geology, e.g., a top layer of dirt having a relatively moderate electrical conductivity and underneath it a water table having high conductivity. Or, a top layer of dirt having moderate conductivity and underneath it at some depth a rock layer having very poor conductivity.  In these cases, one can determine the interface depth by measuring the electrical resistivity on the surface at a progressive set of test distances.  Plotted on a graph, the resistivity vs distance will be an "s"-shaped curve, and by comparison with standard curves, one can determine both the interface depth and the electrical resistivity of each layer.
 * A more common geology, especially in the parts of the World where I practice, is a three-layer geology: A top soil layer having high electrical conductivity, underlying sand having poor conductivity to a depth of a few tens to a few hundred meters, and underneath that a rock layer having very poor electrical conductivity.  This structure can be evaluated by surface electrical measurements, but it is very inaccurate and unsatisfactory - particularly if the rock layer is irregular in depth.  An Engineer can do his tests, decide that x means electrodes to depth y will do, but upon installation there just happend to be some local rock anomally.  Bosses don't appreciate being quoted z dollars and then being told we need 2z dollars after drilling work has supposedly been completed.  However, with seismic testing and electrical testing, the Engineer can get a more accurate picture.
 * Sometimes, the three layer geology is truncated by a geology fault, or a river, or whatever. In these cases electrical testing on its' own is pretty much hopeless, but seismic testing will give an answer.
 * In all these cases, as with geology testing for civil enginnering, the range of the tests is a few hundred metres to a few kilometers, and striking a plate on the ground with a sledge hammer is adequate, and avoids the need for an licenced explosives expert. Often, you are testing in built up areas (towns, cities) where detonations are undesirable or not allowed anyway.  However, ONE strike with a sledge hammer may not be enough to get the signal above the noise from construction sites, trucks on roads, etc.  So, you hit it several times, and the instrumentation integrates the signal from all the hammer blows, and the signal is seen to rise out of the noise.
 * Ratbone 120.145.189.33 (talk) 04:03, 20 January 2013 (UTC)


 * We have a real-world example of a chemical explosion in Wyoming that registered on seismographs as far away as Europe. --Guy Macon (talk) 09:18, 19 January 2013 (UTC)
 * That had "only" about 40 tonnes of (unspecified) explosives - or about one large truck full of the stuff, so it gives us an upper bound. And it wasn't focused to produce a seismic shock, but set off by accident, so much of the blast probably dissipated into the air. One should be able to do better. --Stephan Schulz (talk) 09:52, 19 January 2013 (UTC)
 * All right then, I guess a few tons of high explosive will do the trick. Still, that would be completely impractical as a signalling system. 24.23.196.85 (talk) 05:43, 20 January 2013 (UTC)

Reason for glass being transparent face on but opaque edge on
My understanding of why transparent materials (at least SiO2) are transparent (which is informed by what I remember from a semiconductor physics class I took in college) is that the energy difference between the constituent atoms' base state and next available excited state is larger than the energy that photons (of visible light frequencies encountered in everyday life here on Earth) have, so they pass through the material. But I was just thinking about this as I looked at a piece of glass that was transparent when viewed face-on but green when viewed edge-on. And it's not because of thickness: a piece a quarter-inch wide and a quarter-inch thick is still highly transparent through its face and nearly opaque green through its side. Who can explain in as SIMPLE terms as possible what key features of the crystal structure (though I thought glass was often amorphous) cause high transparency at one angle and high opacity at another? 20.137.2.50 (talk) 14:44, 18 January 2013 (UTC)
 * In a word: thickness. Consider that a cube of glass won't have any preferred direction of "transparency": it will be equally transparent in all directions.  Differences in transparency only become apparent in big sheets of glass, which are basically large retangular prisms.  The shortest side is dramatically shorter than the two longer sides, so the amount of material that the light has to pass through is significantly more edge-on.  If the index of refraction of the glass is such that a light beam traversing the glass in a particular direction gets deflected before it reaches the other side, you can't see clearly in that direction.  Picture light going through a piece of glass 1/4 inch thick and 3 feet wide, edge on.  Any light that strikes that edge even at a slight angle will be deflected by refraction enough to never reach the opposite side.  This doesn't happen when you view the pane of glass from the front.  -- Jayron  32  14:51, 18 January 2013 (UTC)
 * I don't have a cube of glass handy to check, but I'm pretty sure it will be equally transparent through all faces, if they are finished the same way. Note that many types of window glass have a different type of finish on the edges, and this can contribute to the thickness effects Jayron describes above. SemanticMantis (talk) 14:59, 18 January 2013 (UTC)
 * (ec) The key bit there is 'if they were finished the same way'. If you're looking at pieces cut from flat sheets of glass (like panes of glass used for windows) I can think of (at least) three reasons why the appearance of the glass might be different from the cut sides/edges as it is from the originally-manufactured flat face.
 * Most plate glass is made using the float glass process. This process produces faces that are extremely smooth and uniform&mdash;and the opposing faces will be very nearly perfectly parallel.  The other surfaces of the glass cannot be subjected to this process (obviously).
 * Cutting the edge of the sheet of glass can (will!) introduce microscopic and macroscopic defects, fractures, and other light-scattering features to the cut edge. Even if polished, these surfaces will not be identical to the float-glass faces.  Opposing cut faces are also unlikely to be quite as parallel as the opposing faces of the float-glass pane.
 * Surface coatings may have been applied to the exposed surfaces of the plate glass. Some may be deliberate and permanent, for reducing visible reflections or heat leakage, others may be incidental, such as adhesive residue left behind from the protective paper often applied to glass for shipping.
 * Any or all three of these factors can and will affect the way that light scatters and reflects (internally and externally) from the surfaces of the piece of glass, and thereby affect the appearance of the light transmitted through the glass. If you were to take a larger lump of glass, and cut and polish it into a cube, I suspect that you would find its properties are pretty isotropic: the same in all directions. TenOfAllTrades(talk) 15:21, 18 January 2013 (UTC)


 * Then I would like to know the details of how finish affects transparency. This picture captures my observational rationale. About in the middle horizontally and at the bottom quarter vertically, for instance, you'll see a fairly trapezoidal piece that closes to a triangular point at its right. So though its width is at or less that of its thickness, it is still green edge on but is transparent face on. 20.137.2.50 (talk) 15:09, 18 January 2013 (UTC)
 * In that image, the camera is at an angle of (let's guess) 60 degrees to the "horizontal" surface of the glass. Light from the ground beneath is refracted going into the bottom of the glass, then again as it emerges from the top and heads towards the camera.  The light is attenuated by maybe a quarter inch of glass.  But the "vertical" sides of the glass are more like 30 degrees from the camera.  The light that emerges from that angle at the sides has been subject to "total internal reflection" and bounced around perhaps dozens of times within the sheet before emerging towards the camera - the total distance travelled through the glass could be a dozen times more than the relatively straight path taken on the horizontal sides.  Hence more attenuation - and it looks green.  If you look at some of the smaller chunks (especially the ones just to the left of the stone column), the horizontals and verticals look about the same color. SteveBaker (talk) 15:19, 18 January 2013 (UTC)


 * Jayron gave you the right answer. You may like to read up on optical fibre.  Optical fibre is used by phone companies to send information via very thin glass strands dozens of kilometers long.  The glass used is essentially quite normal silica glass, equally transparent in all directions.  However, it is thousands of times more transparent than standard silica glass due to extreme purity, so that the signal can go end to end.  Ratbone 120.145.29.226 (talk) 15:31, 18 January 2013 (UTC)
 * More reading: See Total internal reflection, which is exactly what occurs when you view a pane of glass edge on. Light simply can't escape in that direction.  -- Jayron  32  15:56, 18 January 2013 (UTC)
 * That was nice. Sunny Singh (DAV) (talk) 13:18, 19 January 2013 (UTC)

12cm
is a man with 12cm wide feet (at the widest point, the ball of foot) a 4e, 5e or 6e wide width shoe?--Shoes15151617 (talk) 17:12, 18 January 2013 (UTC)


 * That's impossible to know without knowing the length as well. Per the Wikipedia article on shoe size Shoe widths are relative to the length of the shoe, so a size 10/4E width foot will be a different width than a size 14/4E shoe (assuming U.S. measurements, as the "4E" width designation is usually a U.S. designation).  If you have a question, you should visit a shoe store and get measured; most shoe stores have a Brannock Device to measure your foot, and that can give you an idea of your correct size.  -- Jayron  32  17:26, 18 January 2013 (UTC)


 * I hate that device, since it doesn't measure height. As a result, they always recommend normal width for me, when I really need wide shoes to accommodate the additional height of my feet. StuRat (talk) 17:39, 18 January 2013 (UTC)


 * As with any tool, it just gets you started. You, for example, know that your feet run wide from what the device tells you.  So, you're going to tend to find shoes sized somewhat wider than it recommends.  The idea is that, if you know absolutely nothing about your shoe size, it gives you a ballpark to start from.  You should always try on the shoes in question, and several sizes for each style, because styles and manufacturers will all vary somewhat, so even though you're a 10.5/4E in Nike sneakers, you may find you're an 11/D in Timberlands.  The device is useful, but only if you're not unwise in how you use it.  Try the shoes on regardless... -- Jayron  32  18:30, 18 January 2013 (UTC)


 * The device doesn't seem to help. Since I need to try them on to see if they fit anyway, and know my approximate size, why bother with the device at all ? StuRat (talk) 19:07, 19 January 2013 (UTC)

I am a 11 1/2 shoe length with 12cm wide feet. Is that a 4e, 5e or 6e?--Shoes15151617 (talk) 19:52, 18 January 2013 (UTC)

Neurotransmitters - if someone is being hard with dancing in clubs
This is not a medical advice, just a theoretical case to evaluate.

someone go to lot's of dance clubs, and dance very genteelly, he\she could be with someone who "knocks out" in dancing, he\she could sit and the partner will dance near him - but when he\she dances, it's always have to be clumsy, absent, slow, gentle, he/she being hard with this because it potentially hearts him/her - hearts the experience. the one do fell the need to do dancing but something just blocks it from it. alchaol seems to worsen it (by upgrading neural inhibition?)

i think it have to do with levels on neurotransmitters, what do you guys think of such matter? maybe the individual lacks dopamine?, maybe it needs caffeine/xanthins? — Preceding unsigned comment added by 79.176.113.107 (talk) 17:44, 18 January 2013 (UTC)


 * The question is hard to understand because of poor English, so I have to guess at what you mean. If you are asking whether slow and clumsy dancing can be caused by altered neurotransmitter levels, the answer is yes.  But there are other possible causes too. Looie496 (talk) 18:02, 18 January 2013 (UTC)
 * poor english? excuse me?, what neurotransmitters could play a role?, sure that a neuro-informative level could be another option (the individual doesn't have neural information about how to dance), but, i am interested to hear opinions about the neurotransmitters that has to do with dancing. — Preceding unsigned comment added by 79.176.113.107 (talk) 19:04, 18 January 2013 (UTC)
 * Dopamine is the most obvious possibility. A person with Parkinson's disease (caused by loss of dopamine cells) will have great difficulty dancing.  But also a person who is depressed (low levels of norepinephrine and serotonin, among other things) will often be uninterested in dancing. Looie496 (talk) 21:44, 18 January 2013 (UTC)
 * What do you mean by "knocks out" in dancing? What do you mean by "potentially hearts him/her - hearts the experience"? Do you mean "potentially loves him/her"? Do you mean "potentially gives him/her a heart attack"? It's not that your English is poor, it's that there's too much slang in it for those of us with more experience (OK and who are older) to be able to give you a proper answer. --TammyMoet (talk) 09:53, 19 January 2013 (UTC)
 * What Looie was trying to say is that you're English is shit. 78.150.17.65 (talk) 04:53, 21 January 2013 (UTC)
 * Do you mean "your"? - Goodbye Galaxy (talk) 17:58, 21 January 2013 (UTC)
 * I think Developmental dyspraxia is probably as good as you'll get from Wikipedia about such problems. We can't give medical advice or diagnosis of individual cases. As to neurotransmitters there's a large number of them and people aren't at all certain what they all do. Dmcq (talk) 20:00, 18 January 2013 (UTC)


 * Being able to dance requires a sense of rythm. Our sense of rythm arises from certain neural structures - timing neurons.  It is a brain wiring configuration thing, not just an adequate supply of neurotransmitters.  It is also highly variable between individuals.  I was always unable to dance.  At one time I got involved in electronic music, and from there got interested in playing guitar.  That was terribly difficult at first, as a sense of timing/rythm is required for that as well.  But, by persistence, after a while things "clicked" and I was able to play.  I was then able to dance as well. So dancing skill can be acquired. Floda 121.221.210.234 (talk) 00:18, 19 January 2013 (UTC)

Alex Hum
Alex Hum is making revolutionary discoveries in the field of intelligence clothing such as sunglasses that one can watch television. I have found very little on this pioneer, and many around college campus' are calling him the next Steve Jobs. Below is a link from ScienceDirect.com.

Alex Hum, voted as one of the outstanding people of the 20th century by the Cambridge International Biographical Centre, Cambridge, England, has achieved much of his world-wide attention because of his aim and vision of infusing wireless technologies into everyday high-tech devices and, now into, haute-couture. Dr. Hum is managing the international i-Wear consortium (sponsored by Adidas, Energizer, Seikon–Epson, France Telecom, Levi's, Samsonite, Bekintex, Recticel, Philips, Siemens, WL Gore, Courreges, Vasco Data Securities, and more). i-Wear is a multi-disciplinary research and industrial consortium to invent the deep future of intelligent clothing. In addition, he is the Chief Scientist in Starlab Research Laboratories nv/sa in Belgium. He was a scholar and obtained a direct Ph.D. in RF/Microwave Engineering. He was a Member of Technical Staff at the Centre for Wireless Communications, with experience in project management and industrial collaboration with top technology companies. His specialities are RF/microwave circuit and system design, wireless technology, 3G cellular systems, antenna systems, and the design of RFIC and MMIC integrated circuits. A member of IEEE and IEE – Dr. Hum publishes regularly in internationally-refereed journals and frequently addresses international conferences. He holds several patents. Dr. Hum is listed in Who's Who in Science and Technology as well as Who's Who in the World. — Preceding unsigned comment added by 199.120.31.20 (talk) 19:11, 18 January 2013 (UTC)

mrsa
I heard if someone gets a mrsa infection they carry mrsa for life is that true? Is that also true for non mrsa staph?--Shoes15151617 (talk) 19:54, 18 January 2013 (UTC)


 * Do you mean MRSA?--Shantavira|feed me 20:15, 18 January 2013 (UTC)

yes--Shoes15151617 (talk) 20:53, 18 January 2013 (UTC)


 * In general, "going dormant" only to emerge later and cause more problems, is more a behavior I associate with viruses than bacteria. StuRat (talk) 02:45, 19 January 2013 (UTC)


 * Bacteria can colonize people; for example, MRSA can colonize the nose without causing disease for long periods of time, then under certain circumstances it can cause disease in that same person. Example:  and    -- Scray (talk) 03:08, 19 January 2013 (UTC)


 * Staph aureus is very often found on the skin, and as Scray said, also in the nose. MRSA is just a strain of Staph aureus which is resistant to methicillin. So if someone has had lots of S. aureus infections which have been treated with antibiotics (note that minor issues such as boils and sinusitis can be caused by S. aureus infection) it is likely that there will be some bacterial cells on the skin or in the nose which are methicillin resistant, and they can happily stay there without causing any issues. douts (talk) 13:54, 19 January 2013 (UTC)


 * Hmmmm... I'm having an unexpected amount of trouble trying to get a straight answer about the risk of reinfection (as opposed to infection rate in those colonized). Some sources like  are interesting.  Perhaps the problem is that conceptually a person who is "colonized" is infected, making the question philosophically invalid; yet I feel like there should be a distinction between a few bacteria hiding out intracellularly and a visible, active infection that would make such a number possible to obtain. Wnt (talk) 17:38, 20 January 2013 (UTC)

heatwave?
can anyone tell me why this http://www.bbc.co.uk/news/world-asia-21072347 says it was a "heatwave" in sydney today of 46C but this http://www.weather.com/weather/today/Sydney+ASXX0112:1:AS says it only reached 24C ? --Shoes15151617 (talk) 20:59, 18 January 2013 (UTC)


 * The 46C reading is reiterated by the Sydney Morning Herald here, by the Nine Network here, and by Weather Underground here. -- Finlay McWalterჷTalk 21:14, 18 January 2013 (UTC)


 * The cooler temperature is for the 19th - the BBC story (from the 18th) mentions the expected sharp drop in temperature "However, meteorologists have forecast a dramatic change in weather overnight in Sydney, with thunder storms expected to bring a rapid drop in temperatures". Mikenorton (talk) 21:31, 18 January 2013 (UTC)


 * Yes, it's a forecast for the 19th. It says "Today" but that's local time where it's already the 19th. The "Yesterday" link for the 18th says "115°F High". That's 46°C. PrimeHunter (talk) 21:36, 18 January 2013 (UTC)


 * Incidentally, the fire in Victoria that killed one man at Seaton yesterday (the same one that's cut off Licola) was only 20km from where I live, and day was like night here (Maffra) yesterday from all the smoke. The imminent danger has passed but we remain on high alert, because the fire's predicted to spread up into the high country where there's even more abundant dry fuel, gather a great deal more energy, and possibly turn back this way.  Australia has been enduring its worst heatwave and bushfire season on record, with almost no part of the country untouched, and temperature records falling like flies.  See 2012–13 Australian bushfire season. --   Jack of Oz   [Talk]  21:54, 18 January 2013 (UTC)


 * And it's not just the heat. An absence of rain for over a month hasn't helped. My vegie garden is very sad. But it's worth noting that fatalities from the fires so far have been very low. Just luck? Not sure. HiLo48 (talk) 21:58, 18 January 2013 (UTC)
 * We learnt a lot from the Black Saturday bushfires in 2009, which killed 173 people. --   Jack of Oz   [Talk]  22:30, 18 January 2013 (UTC)
 * True, but we still have opposing opinions on what's the safest approach when fire is on its way, and whether fuel reduction burns are a good idea, and how and when to conduct them, with the proponents of each view 100% certain that their view is correct. Still, the ongoing debate keeps the issue of fire safety in peoples' minds, and that can only be a good thing. HiLo48 (talk) 04:13, 19 January 2013 (UTC)