Wikipedia:Reference desk/Archives/Science/2014 June 27
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June 27
[edit]Expensive fish ...
[edit]![](http://upload.wikimedia.org/wikipedia/commons/thumb/a/a3/Chitala_ornata_%28Zoo_Brno%2C_Czech_Republic%29.jpg/220px-Chitala_ornata_%28Zoo_Brno%2C_Czech_Republic%29.jpg)
Clown featherback is in German "Tausenddollarfisch", in English Thousand Dollar Knife Fish.
- 1. Question: Why is the name "Thousand Dollar Knife Fish" not mentioned in the article?
- 2. Question: What is the ethymology of "Thousand Dollar Knife Fish"? (Who, when and why?). Thx for answers! GEEZERnil nisi bene 12:00, 27 June 2014 (UTC)
- It's a knifefish with a row of circular markings on its side resembling coins. But only 5-10 or so, not a thousand. -- Finlay McWalterᚠTalk 12:13, 27 June 2014 (UTC)
- A translation of its german name is just a Cross-reference. Wikipedia is not a dictionary and thus does not want to collect names and terms but information. --Kharon (talk) 17:23, 27 June 2014 (UTC)
- OK, you're off the hook (...) on both accounts. I found the first mentioning in "Die Aquarien- und Terrarien Zeitschrift, Band 30–31, Alfred Kernen Verlag, 1977".
- Case closed. GEEZERnil nisi bene 10:19, 28 June 2014 (UTC)
- A translation of its german name is just a Cross-reference. Wikipedia is not a dictionary and thus does not want to collect names and terms but information. --Kharon (talk) 17:23, 27 June 2014 (UTC)
MH 370
[edit]What is the explanation of the pings detected during the search operation for MH 370.I cannot find any satisfactory explanation for this anomaly .18:06, 27 June 2014 (UTC)~
- MH370: possible black box 'pings' spur on search for missing airliner (The Guardian 6 April 2014) See the article Flight recorder. 84.209.89.214 (talk) 19:53, 27 June 2014 (UTC)
- (ec) If you read transcripts of the press conferences from the Joint Agency Coordination Centre (the Australian and international intergovernment agency that is coordinating the search), they distinguish between two categories of ping: satellite radio "pings", and SONAR "pings". Particularly, the early April acoustic signals that were detected by Ocean Shield were publicized and described. The explanation provided by Angus Houston pretty clearly outlines the technical challenges and uncertainties involved in detecting and interpreting such signals. What exactly do you find anomalous? Nimur (talk) 19:54, 27 June 2014 (UTC)
Voltage divider
[edit]![](http://upload.wikimedia.org/wikipedia/commons/thumb/3/3e/Resistor_cropped.jpg/250px-Resistor_cropped.jpg)
Do I need high power resistor in a voltage divider circuit to measure DC voltage upto 100V electronically ? — Preceding unsigned comment added by 39.42.126.189 (talk) 20:25, 27 June 2014
- Not as long as you're careful to choose resistance values that are high enough. The power a resistor dissipates due to Joule heating is given by P=V2/R. So calculate the maximum V across each resistor, and then pick values for the R's to be at least big enough that the P's don't exceed the power rating of your available resistors. Red Act (talk) 21:19, 27 June 2014 (UTC)
- The article Potentiometer (measuring instrument) is of interest. However exercise caution because you are dealing with a possibly lethal high voltage. Modern Voltmeter instruments have conducting metal parts insulated for safety. 84.209.89.214 (talk) 01:39, 28 June 2014 (UTC)
- If I choose values in Kilo Ohm, will it be safe ? 39.42.126.189 (talk)
- No, that would quite likely not be safe. If you apply 100V across a pair of resistors that total, say, 10kΩ, the resistors would need to dissipate a total of 1W, which is a problem because most resistors are only rated for a fraction of a Watt. If you don't know the power rating of your resistors, it'd be safest to assume that they're only rated for 1/8W, which is the smallest power rating that they're likely to be. (At least, 1/8W is the smallest power rating of the resistors Radio Shack sells.) Assuming most of the power is dissipated from just the larger of the two resistors, at a 1/8W rating and applying a little bit of a safety margin, you'd want the two resistances in the voltage divider to total 100kΩ at the very least, to make sure you don't smoke the larger of the two resistors in the voltage divider. Red Act (talk) 05:50, 28 June 2014 (UTC)
- I should resistor with a power rating of at least 2W or is 1W resistor good enough ? 39.55.175.44 (talk) 12:29, 28 June 2014 (UTC)
- Again, it depends on the resistance. Rearranging the equation I gave above, a resistor with a power rating of P watts that needs to be able to handle up to V volts across it needs to have a resistance of at least R=V2/P ohms. So for example with V=100V and P=1W, the resistance R has to be at least (100V)2/(1W) = 10kΩ. Or as an alternative but compatible example, using the equation in the form I gave originally, if you need a resistor that you've already chosen to be 10kΩ to handle 100V across it, then the power rating of the resistor needs to have a power rating of at least (100V)2/(10kΩ) = 1W. Red Act (talk) 15:25, 28 June 2014 (UTC)
- What will be disadvantage of using a resistor with high power rating, this voltage divider circuit will be permanently left in. Do resistor with higher power rating will draw more power and waste it as heat ?
- Every resistor carrying a current I dissipates a power P as waste heat. You need to know Ohm's Law to appreciate that P (in watts W) can be calculated in any of these ways:
- The power rating of a particular resistor is just the maximum power that it can tolerate; the above equation decides the actual power. Choosing a higher power resistor than a circuit really needs is typically a good decision because a big component that only heats up slightly is more reliable in the long term than a small component that heats up a lot. Obviously that strategy should not be taken to extremes, for example a mobile phone that used 20 Watt rated resistors throughout would need a wheelbarrow to carry it around. But it could still work and the actual power of, say, a resistor of 10 kΩ carrying 1 mA would still be only 0.0000000001 W (= 0.0001 μW) by the above equation. 84.209.89.214 (talk) 19:20, 28 June 2014 (UTC)
- Let say, the maximum current which will be following through that circuit is 8A at 100V. what values of resistors should be used ?
- Shouldn't the current flowing through a Voltage divider circuit should be kept minimum ? Afterall, maximum current should be following through load circuit ? 39.42.120.65 (talk) 19:44, 28 June 2014 (UTC)
- No let's not say that you will waste 800 watts (enough energy to heat a room, illuminate a village or blow-dry someone's hair) in a circuit just to measure a voltage. Please pay attention to the references and examples already given. 84.209.89.214 (talk) 02:02, 29 June 2014 (UTC)
- Thanks Got it. 39.42.120.65 (talk) 05:18, 29 June 2014 (UTC)
- No let's not say that you will waste 800 watts (enough energy to heat a room, illuminate a village or blow-dry someone's hair) in a circuit just to measure a voltage. Please pay attention to the references and examples already given. 84.209.89.214 (talk) 02:02, 29 June 2014 (UTC)
- What will be disadvantage of using a resistor with high power rating, this voltage divider circuit will be permanently left in. Do resistor with higher power rating will draw more power and waste it as heat ?
- Again, it depends on the resistance. Rearranging the equation I gave above, a resistor with a power rating of P watts that needs to be able to handle up to V volts across it needs to have a resistance of at least R=V2/P ohms. So for example with V=100V and P=1W, the resistance R has to be at least (100V)2/(1W) = 10kΩ. Or as an alternative but compatible example, using the equation in the form I gave originally, if you need a resistor that you've already chosen to be 10kΩ to handle 100V across it, then the power rating of the resistor needs to have a power rating of at least (100V)2/(10kΩ) = 1W. Red Act (talk) 15:25, 28 June 2014 (UTC)
- I should resistor with a power rating of at least 2W or is 1W resistor good enough ? 39.55.175.44 (talk) 12:29, 28 June 2014 (UTC)
- No, that would quite likely not be safe. If you apply 100V across a pair of resistors that total, say, 10kΩ, the resistors would need to dissipate a total of 1W, which is a problem because most resistors are only rated for a fraction of a Watt. If you don't know the power rating of your resistors, it'd be safest to assume that they're only rated for 1/8W, which is the smallest power rating that they're likely to be. (At least, 1/8W is the smallest power rating of the resistors Radio Shack sells.) Assuming most of the power is dissipated from just the larger of the two resistors, at a 1/8W rating and applying a little bit of a safety margin, you'd want the two resistances in the voltage divider to total 100kΩ at the very least, to make sure you don't smoke the larger of the two resistors in the voltage divider. Red Act (talk) 05:50, 28 June 2014 (UTC)
- If I choose values in Kilo Ohm, will it be safe ? 39.42.126.189 (talk)
- The article Potentiometer (measuring instrument) is of interest. However exercise caution because you are dealing with a possibly lethal high voltage. Modern Voltmeter instruments have conducting metal parts insulated for safety. 84.209.89.214 (talk) 01:39, 28 June 2014 (UTC)
Eating clay as detox
[edit]It is apparently Calcium bentonite. I don't have detox cures in high regard, nor I believe that we are full of toxins. But could this have some benefit in getting rid of metals or fat in the body? It is even used inFuller's earth that has medical uses. OsmanRF34 (talk) 21:39, 27 June 2014 (UTC)
- Only before the fats or the heavy metals are actually absorbed into the body -- once in the bloodstream, fat can only be burned off through physical activity, and heavy metals require chelation therapy (which has all sorts of nasty side effects) to remove from the body. 24.5.122.13 (talk) 01:00, 28 June 2014 (UTC)
- Note that some animals eat clay after they get indigestion (or ahead of eating problematic foods, to prevent it). Some parrots and primates, as I recall. StuRat (talk) 01:44, 28 June 2014 (UTC)
- Well, how do you know that they're not eating dirt just because they're very hungry and can't find anything else to eat (as was the case, for example, with some Gulag inmates who were forced to do hard labor on a starvation diet[1]?) 24.5.122.13 (talk) 08:14, 28 June 2014 (UTC)
- Maybe sometimes they do, but the article has all kinds of references to support it being a deliberate, non-starvation situation. Matt Deres (talk) 11:43, 28 June 2014 (UTC)
- "Well, how do you know that they're not eating dirt just because they're very hungry and can't find anything else to eat (as was the case, for example, with some Gulag inmates who were forced to do hard labor on a starvation diet" The Parrot Gulag, by Budginitsyn? Priceless μηδείς (talk) 22:40, 28 June 2014 (UTC)
- The official advice of the English NHS, which is hardly a hotbed of quackery, is "Kaolin is an adsorbent that helps to absorb toxins from the gut". At least in Britain, this is a common treatment which can be bought (almost invariably mixed with morphine) over the counter. 84.13.54.182 (talk) 22:48, 28 June 2014 (UTC)
- I think it's even more familiar as Kaopectate, though apparently the U.S. pulled their license to sell clay. Wnt (talk) 03:34, 29 June 2014 (UTC)
- The key word here, though, is from the gut (as opposed to from the bloodstream). 24.5.122.13 (talk) 04:18, 30 June 2014 (UTC)
- I understand that some long lived Sardinians in the Blue Zone ate dirt decades ago--heard it on CBC's Ideas--and yeah, they were hungry--might have been a way to get minerals.199.119.232.218 (talk) 04:36, 1 July 2014 (UTC)
References
[edit]- ^ Solzhenitsyn, Alexander (1980). The Gulag Archipelago, vol. 2. Paris, VT: YMCA-PRESS. pp. 129–130. ISBN 5-265-01557-4.
Is a true one-way mirror really impossible?
[edit]A Google search finds sources that say that such a mirror would violate the laws of thermodynamics. But I question that argument since any resulting heat transfer can be offset by conduction, or possibly convection around the mirror as well, so the heat transfer always follows the temperature gradient.
Even if a one-way mirror for the entire electromagnetic spectrum cannot exist, could it still be possible to make a one-way mirror for e.g. the visible spectrum only?--Jasper Deng (talk) 22:46, 27 June 2014 (UTC)
- Sure - you just need a really thick wall to block the photons, and somebody to open a doorway whenever they see a photon coming from the correct direction! Nimur (talk) 22:58, 27 June 2014 (UTC)
- In particular, it would violate Clausius's statement of the Second Law - "Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time." Allowing the heat to flow back to the colder body after it's gone in the illegal direction doesn't fix things - the "at the same time" is an equally important part of the statement. Tevildo (talk) 23:13, 27 June 2014 (UTC)
- There is no theoretical obstacle from a thermodynamic perspective, if you consider a circulator or even an optical circulator system. It requires an extra "face" to the mirror, so that a black body can be coupled into the system. Light goes from A to to the black body; thermal radiation only goes from the black body to B, and light goes from B to A. So A can observe B, but B can only see the heating effect of A on what is hopefully a perfect black body. We don't need Maxwell's demon to solve this one. —Quondum 23:35, 27 June 2014 (UTC)
- See also Optical isolator, where the thermodynamic topic is also mentioned. —Quondum 23:43, 27 June 2014 (UTC)
- Isn't a camera on one side of a wall and a monitor on the other side logically a one-way mirror ? StuRat (talk) 01:45, 28 June 2014 (UTC)
- Well, it's not a mirror, but other than that, sure.
- It does deserve the term one-way mirror more than the topic of our one-way mirror article does, though. The proper name for that is two-way mirror. Unfortunately the wrong side won that argument. --Trovatore (talk) 01:47, 28 June 2014 (UTC)
- You can also put a duplicate monitor image on the side with the camera, to make it into a "mirror". And, since one camera wouldn't quite mimic a mirror, you could have many tiny cameras each displaying just a few pixels. It should be possible to make it pretty much indistinguishable from a real mirror. If you can make the camera the size of a pixel, you could have each camera surrounded by 6 pixels, where those 6 would display the color detected by the camera (possibly blending the color with colors detected by nearby cameras). StuRat (talk) 01:53, 28 June 2014 (UTC)
- Any thermodynamic argument against such a mirror is going to rely on it being a mirror — that is, a passive device that doesn't need to be powered. Of course, if you supply energy and negentropy to the system, you can (approximately) reconstruct any wavefront you want. --Trovatore (talk) 02:03, 28 June 2014 (UTC)
- I think a passive "mirror" that radiates all the energy that it receives from the other side as pure black body radiation, while transmitting images faithfully in the other direction, qualifies under the intent of the concept "one way mirror" even though it reflects nothing, and satisfies the thermodynamic constraints. Of course, if one wanted it to look like a mirror rather than like a room-temperature black surface, one could just put a partial mirror on the one side. —Quondum 02:37, 28 June 2014 (UTC)
- Any thermodynamic argument against such a mirror is going to rely on it being a mirror — that is, a passive device that doesn't need to be powered. Of course, if you supply energy and negentropy to the system, you can (approximately) reconstruct any wavefront you want. --Trovatore (talk) 02:03, 28 June 2014 (UTC)
- You can also put a duplicate monitor image on the side with the camera, to make it into a "mirror". And, since one camera wouldn't quite mimic a mirror, you could have many tiny cameras each displaying just a few pixels. It should be possible to make it pretty much indistinguishable from a real mirror. If you can make the camera the size of a pixel, you could have each camera surrounded by 6 pixels, where those 6 would display the color detected by the camera (possibly blending the color with colors detected by nearby cameras). StuRat (talk) 01:53, 28 June 2014 (UTC)
- Doesn't even a conventional mirror absorb and/or scatter some percentage of the light that hits it? ←Baseball Bugs What's up, Doc? carrots→ 02:41, 28 June 2014 (UTC)
- A one-way mirror is similar to an irreversible reaction - light flows from an area of high concentration to an area of low concentration. In the case of the chemical reaction, people may omit the back arrow once it becomes a sufficiently complicated exercise in entropy on one side (for example, when you use pyrophosphatase to enhance an ATP-dependent reaction in biology). So an interesting question would be if you can spice up a one-way mirror in some similar way, i.e. by splitting the photon somehow into two or even three weaker photons in a coating on the protected side, and then putting some kind of band pass filter in the mirror so that the weaker photons can't make it back through unless they somehow manage to meet up and recombine. That way even if your perp suddenly whips out a super-mag flashlight to shine through the glass, your protected witness should remain unknown to him. Wnt (talk) 03:55, 28 June 2014 (UTC)
- I already linked to optical isolator above – it already exists, only it is not similar to a one-way reaction. Light goes one way, not the other. No matter that one shines a bright light from the one side. Simply sandwich a 45-degree Faraday rotator between two polarizing filters oriented at 45 degrees. —Quondum 05:24, 28 June 2014 (UTC)
- An isolator is quite different to a (hypothetical) one-way mirror. Both transmit electromagnetic radiation in one direction only (say from side 1 to side 2). However, an isolator absorbs radiation travelling the other way (impinging on side 2), whereas a mirror would reflect it (or it wouldn't really deserve to be called a mirror).
- The mirror would violate conservation of energy (unless it had a power supply), the first law of thermodynamics. Suppose the mirror was illuminated from both sides with radiation of equal wavelength and equal amplitude, then on side 2 of the mirror, the reflected wave would interfere with that transmitted from side 1. If the relative phase of the incoming waves was adjusted so that the interference was constructive, then the combined wave leaving the mirror would have twice the amplitude and carry four times the power of each of the two incident waves. Overall the mirror would be sending out twice as much power as it was receiving. --catslash (talk) 00:07, 1 July 2014 (UTC)
- Well, true, if you demand 100% reflection one way, 100% transmission for the other, and a passive linear system, there'll be no solutions. I find it far more interesting that one can take two polarizing filters with a
sugar solutionsuitable medium between them, and arrange it so that light going one way will be transmitted 50%, and the opposite direction absorbed 100% (admittedly only at a selected frequency with this simple setup). Just allow a little leeway in the interpretation of the question. —Quondum 04:22, 1 July 2014 (UTC)
- Well, true, if you demand 100% reflection one way, 100% transmission for the other, and a passive linear system, there'll be no solutions. I find it far more interesting that one can take two polarizing filters with a