Wikipedia:Reference desk/Archives/Science/2013 October 15

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October 15 edit

Overview of chemistry - book? edit

Can you recommend me a book that is an overview of chemistry? An overview as in it would have a short description of numerous sub-fields and some of their main results and references. --81.175.225.92 (talk) 00:22, 15 October 2013 (UTC)[reply]

Any high school or introductory college text will probably suffice. The Central Science by Brown, LeMay, and Bursten, or General Chemistry by Kotz and Purcell are two that I have used before. --Jayron32 00:54, 15 October 2013 (UTC)[reply]
Jayron, do you mean Chemistry and Chemical Reactivity by Kotz & Purcell, or General Chemistry by Kotz, Treichel, & Weaver? 121.215.39.252 (talk) 03:25, 15 October 2013 (UTC)[reply]
Both, actually. I've use the two of them, and was trying to recall by memory, and conflated the two. --Jayron32 10:45, 15 October 2013 (UTC)[reply]

ray of light theory of eyes edit

one of our articles says " The first theory, the emission theory, was supported by such thinkers as Euclid and Ptolemy, who believed that sight worked by the eye emitting rays of light."

Let's try to understand how the Greeks could have thought so. Why don't people see in a dark room? (cave etc)? Why does a 'source' of light need to be seen in this case, if eyes cast their own light? Most perplexingly - if eyes cast their own light, why can't we see the 'eyes' of other people glowing in the dark, for example? Could you explain a little the very loose kind of thinking that made this almost kind of make sense in a childlike way? 212.96.61.236 (talk) 00:30, 15 October 2013 (UTC)[reply]

Well, cat's eyes and some other animals reflect light, and that could be mistaken for emitting light. Perhaps they just thought human eyes emitted "invisible light" (what we might call infrared or ultraviolet), which changed to visible light under the right conditions. StuRat (talk) 00:49, 15 October 2013 (UTC)[reply]
But we clearly don't see in the dark... moreover, put someone in a darkish place and then put a whole chorus full of people in front of it, their eyes all shining into it: that person can't see any better. so....? I mean it just seems so unworkable... 212.96.61.236 (talk) 01:11, 15 October 2013 (UTC)[reply]
I see in the dark, and furthermore my eyes are sensitive to near IR (down to 800 nm) and near UV (up to 320 nm) -- the IR looks a dark reddish-brown, and the UV looks gray. 24.23.196.85 (talk) 05:04, 15 October 2013 (UTC)[reply]
That seems extremely unlikely, unless you are a bird, and even then you wouldn't be able to see IR. The visible spectrum is 390 to 700nm, no reference I've seen shows the outliers being anywhere close to 320-800. That's much further outside the realms of believable than human biology would permit. I would find some sources of true IR and UV light to really test yourself in a double-blind manner (you can start with the LED at the end of a remote control, with someone who isn't you pressing a button in such a way that you can't tell if they're pressing it). If real, go present your self to the Guinness record-keepers. — Sam 63.138.152.139 (talk) 14:24, 15 October 2013 (UTC)[reply]
People (like my mother) who had cataract surgery before the latest generation of implantable lenses became available are able to see a little way into the ultra-violet. The idea that you could see into the infra-red or in total darkness is ridiculous. If User:24.23.196.85 truly has these super-powers then (s)he should go find a reputable laboratory where these capabilities may be investigated. (...or WP:NOR...either way). SteveBaker (talk) 15:16, 15 October 2013 (UTC)[reply]
Biologists make a lot of generalizations. I will not believe the claim without proof, but see no unevadable reason why it couldn't be true. Wnt (talk) 04:18, 17 October 2013 (UTC)[reply]
For the record, I made the measurements myself using a spectroscope and lights (an incandescent light for the first measurement, and several LEDs for the second one), so I'm reasonably sure that my vision does in fact extend beyond the normal limits of the visual spectrum, but I cannot completely rule out experimental error. I'll be glad to have this independently verified, provided you folks tell me where I can have that done and how much it's likely to cost. As for night vision, I never said I could see in complete darkness -- what I said was that I could see in relative darkness (such as on a moonless night) better than most other people, even to the extent of being able to see (some) colors by the light of a full moon. 24.23.196.85 (talk) 05:41, 17 October 2013 (UTC)[reply]
I'm not sure what to suggest (the claim is interesting enough that we probably ought to be able to talk someone into doing it for free, but I'm not sure who to ask). In any case please register an account so people can get back to you if they come up with an idea. My first wild guess is that you could have some rare variation in a crystallin protein. That said... it still is probably most likely that there was something wrong with the test, for example light scattering in the spectroscope. But could you clarify: what kind of LEDs were these? My understanding is that LEDs typically have a pretty narrow range of frequencies output as part of their fundamental physics, so if these are supposed to be "invisible" it would be hard for that to be wrong? Wnt (talk) 18:36, 18 October 2013 (UTC)[reply]
Erm, actually I may have been wrong - looking a little deeper it turns out that the lens of the human eye actually contains pigments, similar to those of a butterfly (kynurenine), which absorb the UV light [1]. So there could be a general reduction in this pathway in some people, which as described in that source could have various positive and negative effects. Biology being what it is, we should expect some genomes to lay their money on different numbers. Wnt (talk) 18:47, 18 October 2013 (UTC)[reply]
Of course it's not possible to fully reconcile reality with the emission idea, which we know to be wrong. People have speculated on the reasons why this idea was so popular -- and there is even discussion on the question of why many people continue to believe it today. That comes down to thinking of ways that the emission theory has intuitive appeal. StuRat mentioned one reason (shiny eyes of animals). Another possibility is the subjective experience of heat or palpable pressure when someone is watching you; this matches the social understanding that a person who's staring at you is doing something to you, rather than you to them. As for the question about the loss of sight in darkness, that was explained either by some interaction of the sunlight with your eye's light, or by the idea that the eye doesn't create the light, but gathers it in and then sends it out again at whatever you're looking at. Here is a discussion of some of the past arguments for the emission idea: [2]. And a paper [3] and short article [4] discussing the question from the point of view of science education. --Amble (talk) 02:15, 15 October 2013 (UTC)[reply]
Are you looking for a source, or chat partners to tell you you are right? We can't tell you how correct you are, but we can recommend you read visual perception, emission theory, and intromission theory. μηδείς (talk) 02:19, 15 October 2013 (UTC)[reply]
I am offering some answers that have been given to the original question, and was slow in adding the sources from which I drew the information. --Amble (talk) 02:45, 15 October 2013 (UTC)[reply]
I am not the boss here, feel comfortable adding what you think appropriate. μηδείς (talk) 02:48, 15 October 2013 (UTC)[reply]
Superman's eyes emit light. But he's a strange visitor from another planet, where apparently things work differently. ←Baseball Bugs What's up, Doc? carrots→ 02:33, 15 October 2013 (UTC)[reply]
  Resolved
See the transactional interpretation of quantum mechanics. You can say that eyes do emit a ray of light, which is propagated backward in time, and interacts with a source. As the ancients were ill equipped to measure the time of flight of the ray, positive or negative, this would have made little difference to them, and so their idea cannot be considered false. (If the ray of light travelling backward in time finds a bright light, star, etc., it is answered by another retracing the path in the opposite direction...) Wnt (talk) 04:16, 17 October 2013 (UTC)[reply]

Resolved? What? Who made the correct answer? The ancient belief in this ray theory operated like so: Your eyes emit rays that "feel" what is around you, reporting that information back to you as sight. What a "source of light" is actually doing is modifying the properties of air/glass/water so that such rays can transmit through them. Unaltered air does not permit your eye-rays to pass, resulting in a black appearance, a shadow, across your eyes. Such an idea was attractive to people long ago, as the concept of "action at a distance" simply did not sit well with them (they were far more comfortable with explanations of the senses that were reminiscent of how we feel things with our hands). It's easy to construct experiments to either disprove the idea or require ever-more-convoluted explanations. My source for this was the history book I read when I took history of science in college. I wish I could tell you that book's title, but I don't recall. Someguy1221 (talk) 04:51, 17 October 2013 (UTC)[reply]

The OP himself knew the answer. He was looking for our approval, not for us to repeat it back to him. Go back and reread the "question". μηδείς (talk) 03:01, 18 October 2013 (UTC)[reply]

binder related to textile. edit

Hi, I want to know that which chemical can decrease the strength or stickiness or can make it completely useless to work.. but I want to know a chemical name, addition of water do decreases its strength but I after a lot of research I havnt got any chemical which can help me out in decreasing the strength of binder.. Pleas help me out. — Preceding unsigned comment added by 182.180.45.104 (talk) 09:44, 15 October 2013 (UTC)[reply]

Can you be more specific? There are many different kinds of binder and many different kinds of textile. Water is a chemical, and it can be used to dilute many binders, as can many solvents.--Shantavira|feed me 11:54, 15 October 2013 (UTC)[reply]
I have a hard time following your Q, but if you're trying to dissolve something, then most things which can be dissolved are water-soluble, oil-soluble, or alcohol-soluble. So, one of those will probably work. Peppermint oil, for example, can dissolve lots of adhesives. If none of those work, then a strong acid or a strong base might work. StuRat (talk) 17:38, 15 October 2013 (UTC)[reply]

i am using binder nameing UD BINDER of BASF for textile printing. i have discusses it with many chemist but non of them helped me out.. i had used a lot of acids to decreases its strength but non of them work out even though i had used strong base also but not succeeded but when I add strong base in the paste which is use in printing after 2 days it's make it like rubber but i want some thing like when i add that powder based chemical in that printing paste it become useless.. its strange but i want to make the printing paste totally unworkable and it happens only when binder losses its strength.. so I want powder based chemical which make binder totally use less..

It might help if you explained what you are trying to achieve. Why do you want to make the binder useless? Wouldn't that be the same as just not using a binder? SpinningSpark 22:25, 15 October 2013 (UTC)[reply]
I think the OP must have got some printing paste on his/her clothes, and is trying to find some chemical that will remove it. What I don't know is why he/she is using strong acids and bases, despite the danger of ruining the clothes in question altogether. 24.23.196.85 (talk) 23:27, 15 October 2013 (UTC)[reply]

there is no danger for fabric because its completely polyester based fabric so strong scid or base aint gona effect the fabric all I want to know which powder base chemical will effect the printing paste.

Thermal radiation edit

I have read the articles on Thermal radiation and Black-body radiation, and I am still struggling to understand the actual mechanism that causes radiation to be emitted from energetic atoms. The explanation in Thermal radiation just says:

These atoms and molecules are composed of charged particles, i.e., protons and electrons, and kinetic interactions among matter particles result in charge-acceleration and dipole-oscillation. This results in the electrodynamic generation of coupled electric and magnetic fields, resulting in the emission of photons, radiating energy away from the body through its surface boundary

I have two conflicting models in my head, from forgotten Physics classes. One or both may be completely incorrect. They are

  • If you take a dipole magnet and vibrate it, you will produce an EM wave. If you were able to vibrate it really really really fast, the frequency of that EM wave would be that of visible light, so it would emit visible light. In a warm body, each atom is like a tiny dipole magnet that vibrates, producing EM waves.
  • In a warm body atoms are colliding with each other, occasionally causing electrons to jump energy levels. When they return, they may emit photons. Thermal radiation is caused by these emitted photons.

Is either explanation close to correct? — Sam 63.138.152.139 (talk) 14:08, 15 October 2013 (UTC)[reply]

The second statement is basically the quantum mechanical version of the first, but then specialized to electrons. The first picture is a classical picture of vibrating charges, but then if you describe this more precisely using quantum mechanics, each such vibrating charge is an (approximate) harmonic potential and there are then energy levels here too. So, it also emits radiation due to the system falling back to a lower energy level.
The relevant processes are spontaneous emission when making a transition to a lower energy level, excitation when absorbing a photon, and stimulated emission. Count Iblis (talk) 14:19, 15 October 2013 (UTC)[reply]
Hmmm, you say the second statement is a QM version of the first, but there seems to be a very critical difference: in my "vibrating magnet" explanation I can make my magnet produce *any* frequency by vibrating it faster or slower. In my QM explanation, my iron magnet could only produce those few precise frequencies defined by the energy differences in the electron shells of iron atoms (the "quantum" of quantum mechanics), right? — Sam 63.138.152.139 (talk) 14:35, 15 October 2013 (UTC)[reply]
If you look more precisely at e.g. the rotation of molecules mentioned by Gandalf below, then there are energy levels there too, but they are so densely packed that it looks like a continuum. Also, you have to take into account the interactions between the different molecules, an N particle system will have energy levels that for large N will be extremely densely packed. So, physically you putting a magnet in your hand and letting it vibrate at seemingly an arbitrary chosen frequency, or an atom emitting a photon are not distinct physical processes. The former involves many more particles and has a far larger number of degrees of freedom, so the emitted photons can have many more possible frequencies. But it is ultimateley the same quantum theory that explains everything (classical mechanics is only an approximation to quantum mechanics; unlike classical mechanics, quantum mechanics is always valid). Count Iblis (talk) 15:23, 15 October 2013 (UTC)[reply]
Both explanations are (more or less) correct, depending on the wavelength of the radiation. Infrared radiation and microwave radiation are caused by vibrations and rotations of molecules. Shorter wavelengths, such as visible light and ultraviolet radiation, are caused by transitions of electrons between energy levels within atoms/molecules. There is a table at Electromagnetic spectrum#Rationale. Gandalf61 (talk) 14:36, 15 October 2013 (UTC)[reply]
Wow. Thank you for that table -- my jaw just dropped. I had simply no idea that the continuum of the EM spectrum was caused by separate distinct processes, rather than a continuum of one process (like vibrating atom faster and faster). Thanks! — Sam 63.138.152.139 (talk) 14:41, 15 October 2013 (UTC)[reply]

Follow-up question, after seeing the table that Gandalf61 linked to. (please excuse me: my brain is trying to crush together two mental models that have always happily lived in separate bins and now need to be entangled together):

For the "vibrating dipole" model, my physics teacher always likened it to a whip: if your bar magnet is sitting on a table, there is a magnetic field line coming straight out of the North pole. Move the magnet and that field line shifts, but the displacement moves away from the magnet like a wave down a whip, traveling at the speed of light. Vibrate it back and forth and you get an EM sine wave. This was my model for thermal radiation, and a key point of this is that the wave has an amplitude in *physical space*, and that amplitude is the displacement of the magnet (or atomic dipole). That is, in magical-theoretical-world, if you put out metal filings and viewed the magnetic field lines like kids do in school, and your filings were absolutely weightless and frictionless etc etc., wiggling the magnet side to side would produce a sine wave of filings that you could even photograph (ignoring the speed of light).

In the QM model, an electron drops to an lower energy level and the energy is lost to a photon that is emitted with a specific frequency based on its energy. But this "frequency" has always seemed to me to be almost metaphorical -- the photon is a packet of energy with an associated wavelength, but it's not like a sine wave wiggling through space with an actual physical amplitude.

Now I find that both models are kind-of correct for different frequencies. Are both photons of exactly the same "kind"? Is the vibrating magnet actually producing a sine wave with an amplitude in physical space? How about the photon emitted by the electron? The two explanations seem so different, yet it seems they both produce the exact same thing. — Sam 63.138.152.139 (talk) 15:06, 15 October 2013 (UTC)[reply]

You need to note a couple of things. Firstly, merely vibrating a magnet DOES NOT produce electromagnetic radiation. If you mechanically rotate a bar magnet, you will get a rotating magnetic field, that is all, no matter how fast or slow you rotate or move it. To get EM radiation, you must have both a varying magnetic field AND a similarly varying electric field in the proper phase relationship. This is easily demonstrated as it is easy to produce very intense varying magnetic fields by passing large varying currents through a wire coil. However, by other means you can produce radio waves that embody significant energy yet the magnetic filed component may be quite small compared to that produced by a current in a coil.
Secondly, yes a warm body DOES radiate EM radiation (with a continuous spectrum albiet peaked at a given frequency). However, this does not require, and mostly does not involve, collisions between atoms or molecules (or collisions between any sort of particle. Collisions cannot occur in a solid or in a pure crystal. But all non-trasparent substances radiate. For instance, carbon, a solid, is a near perfect black body radiator at all temperatures in which it can exist as a solid, including up to the sublimation point, ~3900K, at which it will glow yellowish-white.
Collisions are relevant when considering gasses. Collisions transfer energy from one molecule to another, by the impact changing the translational and rotational velocity of the molecules concerned. Atoms and molecules must distribute their energy between translation, rotation, and electron orbital configuration. Only the spontaeous changes in orbital configuration contribute to black body radiation. The division of energy between translation & rotation, and electron orbitals, is governed by the emission laws.
[Special:Contributions/121.215.39.252|121.215.39.252]] (talk) 15:12, 15 October 2013 (UTC)[reply]
If you vibrate a magnet you will have a time dependent magnetic field and hence an electric field. Count Iblis (talk) 15:26, 15 October 2013 (UTC)[reply]
So, please Count Iblis, explain then why a radio transmitter needs an antenna, and does not radiate significant energy from its tank coil. After all, considerable energy goes into the tank coil, every half cycle of the carrier frequency, far more (typically 10 to several hundred times) than what leaves the antenna into space. That energy does not leave the coil by going off as EM radiation, it gets passed back and forth to and from that tank capacitor. And as I alluded to ealier, the magnetic component of the EM leaving the antenna my be considerably weaker than the magnetic field near the tank coil. Do not be confused by the fact that any time-varying magnetic field will induce a voltage in nearby conductors, and the electric field thereby created may result in some EM radiation as a secondary effect.
Lastly, consider this: A sinusoidaly varying current in an ideal coil absorbs no energy (as does a sinusoidal current in a capacitor) as the current is 90 degrees out of phase with the voltage. However, EM radiation contains/carries energy, lost to space, which is why an antenna presents an electrical resistance at its terminals (practical antennas may display reactance as well, but resiatnce is always present). Since an ideal coil, which of course does produce a sinusoidal magnetic field, absorbs no energy, there can be no EM radiation. Vibrating a magnet, and any other rythmic mechnical thing you can do to a magnet is not essentially diffrent to driving a periodic current through a coil. In vibrating a mass, you exchange kinetic energy between the mass and the driving device, twice each cycle. In vibrating a magnet in free space, some of the energy gets stored twice each cycle in the magnetic field but it always returns to the driving device, also twice each cycle.121.215.39.252 (talk) 15:45, 15 October 2013 (UTC)[reply]
Let's stick to one well defined example, let's consider the vibrating magnet modeled as an exact dipole magnet and work out this example from first principles in full detail. Here you can't a priori assume that a freely vibrating magnet will execute an exact harmonic motion and will therefore not radiate any energy, as you would then assume what you want to prove. An outline of this is is as follows. What you need to do is solve the Maxwell equations (taking e.g. the case of a frced harmonic motion of the magnet) which leads to an expression for the electromagnetic fields which are given in terms of the retarded potential, so the magnet at position r' and time t contributes to the field at position r at time t + |r-r'|/c, this time lag is going to lead to an 1/r contribution to the asymptotic behavior of the fields. If you ignore this time lag, then there is no 1/r behavior. Then the energy flux is proportional to the square of the fields which behaves as 1/r^2, therefore energy will leak away to infinity (the energy flux through a sphere of radius r is the proportional to r^2*1/r^2 = 1, so this stays finite in the r to infinity limit).
Another way to approach this, which is however not so practical for calculations, is to consider the problem of the self-force in electromagentism. If you consider the freely oscillating magnet, then it will oscillate according to a damped harmonic oscilator. But where does the damping force come from? This is, of course due to the emitted radiation, but the source of that is the magnet itself. How to properly deal with this was only solved recently. Count Iblis (talk) 17:56, 15 October 2013 (UTC)[reply]
Count Iblis, you cannot just ignore a logical argument and go off somewhere else in gibber-land. You need to show why my discussion above is wrong - and you haven't done that, because I have merely recited facts well known to graduates in electrical and electronic engineering world-wide. There is no difference between a magnetic field established by a current carrying coil (or a straight conductor for that matter) and a magnetic field established by a simple dipole magnet. In free space, an ideal coil or conductor absorbs no energy, and no EM radiation occurs. (In practice, of course, while we can have superconductors, we cannot have completely free space. There is always other (imperfect) conductors somewhere with closed loops. Current will be induced in the closed loops, setting up their own varying magnetic fields, which will do the same to the originating coil. By Lenz's Law (for which the proof is the impossible existence of perpetual motion machines) the induced voltage in the originating coil will always be in a direction/phase that will oppose the originating current, thus synthesising an electrical resistance.) I discussed sinusoidal exitation above to simplify it for the OP, however my argument does apply to any varying exitation, as any engineer will know (think in s-plane). 120.145.145.144 (talk) 00:41, 16 October 2013 (UTC)[reply]
Those text books will ignore the effects leading to radiation being emitted, but they will only tell that much later when they actually treat the subject of electromagnetic radiation, because they have not yet introduced the complete Maxwell equations before that.
Thing is that even an uncharged conducting metal sphere rotating in a perfect vacuum will emit electromagnetic radiation and slow down as a result of that. But this is due to quantum electrodynamical effects. Count Iblis (talk) 01:21, 16 October 2013 (UTC)[reply]
I'll take that as your subtle admission, Count Iblis, that any such EM radiation, if it in fact occurs, coming from a vibrating magnet or a wire/coil carrying a perioic current, is negligible. It has to be, or said textbooks, including what electrical/electronic undergrads have to study on Maxwells' equations (we got Maxwell in 3rd year of a 4-year course), would not have ignored it. Nor could practicing engineers get away with ignoring any such effects, which they universally do. Nor could they ignore it in the design/engineering of mechanical filter resonators, many of which achieve extremely high Q-factors (20,000 and better, which means the knietic energy is >20,000 times what is lost each cycle), well beyond what is practical in LC resonating circuits. They are carefully sized pieces of vibrating metal. And it is something that can be ignored with respect to the OP's questions too. His teacher was wrong; vibrating dipoles are not the source of black body radiation - electron orbital drops are. 120.145.145.144 (talk) 05:40, 16 October 2013 (UTC)[reply]
Infrared and microwave radiation is emitted and absorbed by changes in rotational and vibrational modes of polar molecules - see our articles infrared spectroscopy, vibronic spectroscopy, rotational spectroscopy and rotational-vibrational spectroscopy. Gandalf61 (talk) 08:13, 16 October 2013 (UTC)[reply]
True, but that is not black body radiation, and each applies to specific phases - eg the last 2 you mentioned apply only to the gas phase. The radiation and absorbance in these cases does not conform to the black body emission laws, and black body radiation applies to solids and liquids, and in theory, to gasses. In fact, the known atomic structure is not even required to derive the ideal black body emission curve. 120.145.145.144 (talk) 12:28, 16 October 2013 (UTC)[reply]
The OP asked about the "actual mechanism that causes radiation to be emitted from energetic atoms". The black body model is a theoretical abstraction based on thermodynamic principles. It does not posit a particular emission mechanism, and so it is something of a red herring in answering the OP's question. The OP's teacher was not wrong, although they may have given an oversimplified explanation. Gandalf61 (talk) 13:03, 16 October 2013 (UTC)[reply]
The OP was specifically asking about black body radiation - his/her first sentence is I have read the articles on Thermal radiation and Black-body radiation. And later in his/her question, he/she uses the terms "warm body" and "thermal radiation". So all this nonsense about dipoles and electric fields created by vibrating magnets is a side track. If we are talking about factors affecting the thermodynamic efficiency of gasoline engines, do we concern ourselves about oil drawn past the pistons, just because its calorific value has a tiny tiny theoretical impact? Yes, black body theory itself does not posit a particular emission mechanism - I said that myself. But that only means bodies must radiate with a tell-tale continuous spectrum (the other forms of radiation you mentioned have quite different discrete spectra) - we still need to understand what the actual mechanism is. The OP's teacher was wrong, and wrong in the same sense that you and I would be wrong by saying the energy of a gasoline engine comes from the lube oil burnt. Only more so. 124.178.48.59 (talk) 14:34, 16 October 2013 (UTC)[reply]

Surgical Caps and Shoe Covers edit

Hello. How do you visually distinguish between surgical caps and shoe covers? They look very similar. Thanks in advance. --Mayfare (talk) 16:13, 15 October 2013 (UTC)[reply]

Er, well, I'm not sure how to answer this. If you were physically presented with them, the difference would be obvious. If you're looking at a photo, it's hard to come up with solid criteria, because shoe covers are often crumpled and folded in a way that makes their shape hard to recognize. Basically hair covers are round, about a foot in diameter, and relatively thin; shoe covers are foot-shaped with the opening on one end, and pretty robustly constructed. Looie496 (talk) 16:44, 15 October 2013 (UTC)[reply]
It may depend on which part of the world you are living in (you don't give your county of origin) but I would say that visually 'shoe covers' are just large enough to encapsulate the foot and 'caps' are larger enough to cover the head, hair (and Tin foil hats for those quacks that feel they need ware need them). --Aspro (talk) 16:52, 15 October 2013 (UTC)[reply]
Head covers and shoe covers are generic medical supplies, used in vast quantities (along with gloves, masks, and gowns), and I think they're probably the same shape all over the world. In the veterinary facilities where I've worked, the most obvious difference was that the shoe covers were bright blue and the head covers were white. Looie496 (talk) 17:04, 15 October 2013 (UTC)[reply]
If you do a Google Image search for both items you will see there are big differences between the two. The shape to begin with, round for the head and narrow for shoes.Shoe CoverSurgical CapHope this helps! Mike (talk) 16:57, 15 October 2013 (UTC)[reply]
As a practical issue, there's no need to distinguish between them, as they come in labelled boxes. - Nunh-huh 02:39, 16 October 2013 (UTC)[reply]

Domestic waste solder edit

I have a handful of bits of solder, mostly crap sucked up with my desoldering tool and bits that ran off during tinning my soldering iron. I live in Edinburgh, Scotland. Should I make an effort to dispose of this in a special way or is that just for commercial enterprises producing large quantities of waste? --2.97.26.56 (talk) 20:12, 15 October 2013 (UTC)[reply]

Around here, (in California, in the United States), you would get in touch with your local county's Household Hazardous Waste program and determine the best way to dispose of those types of materials. If you're actually in Edinburgh, here's the website for your city government waste service. Nimur (talk) 20:39, 15 October 2013 (UTC)[reply]
It's WEEE, something that ideally you wouldn't put in the landfill waste stream. They do accept WEEE at Edinburgh's community recycling centres (it goes in the "small electrical"). But obviously a wee freezer bag full of WEEE (ahem) isn't worth driving out to e.g. Sighthill. Personally I keep a ziploc freezer bags of the little nasty stuff that they don't collect at the kerb (batteries, CF bulbs, WEEE, paint, etc.) and take it to the recycling place only when I'm taking something large. -- Finlay McWalterTalk 20:40, 15 October 2013 (UTC)[reply]
Okay, thanks. I'll hold onto it until my next visit to the recycling centre. 2.97.26.56 (talk) 21:22, 15 October 2013 (UTC)[reply]
By the way, you should no longer be using lead containing solder unless it is for maintenance of equipment that predates the ROHS directive. SpinningSpark 22:17, 15 October 2013 (UTC)[reply]
While it may come in scope of local regulations, I would not be concerned about a mere handfull. You need to keep things in perspective and understand the partly political motivation for the European Lead-Free Directive. Lead is ubiquitous in the environment. Those that frame laws and regulations seem not to understand how and why. Lead was used in all manner of things, including paint. That contributes to lead dust everywhere. Another source is the used of lead sheathing in power and telephone cables for about 80 years, until satisfactory plastic sheaths were developed in the 1970's. I was involved in the installation and testing of lead sheathed cables in the 1960's and 1970's. The sheath was about 3 mm thick and the cable pressurised with air, so as to enable detection of sheath damage and keep out moisture. Those cables still in use or just abandonned and left in the ground (which is most of them) have become porous, constantly leaking air. In many cases the lead has, over the intervening 40 to 80 years, become paper-thin. Where has the lead gone? Leached into the soil generally of course - where it can be further distributed whenever someone disturbs the soil for building construction or whatever. Authorities became concerned about the lead levels in the blood of childen 30 or so years ago. They thought that lead in gasoline was the problem, so various countries around the World banned lead in gasoline. That improved things a bit in the USA because of their high population densities, considerable use of private cars, and low use of diesel engines in busses and light trucks. But Australia and Europe, which have always used diesel engines in any sort of truck, didn't see much change in blood levels. So Europe decided to ban the use of lead altogether - at least that will mean lead levels don't get any worse, and help countries like Australia where some of the environmental contamination comes from dust released in the mining, processing, and transport of lead.
In any case, the lead in solder is pretty much trapped with the tin and rendered harmless. There has never been much concern about electronics technicians and electronics factory staff being affected by lead from solder - though it has always been standard to caution workers to wash hands before eating.
120.145.145.144 (talk) 01:07, 16 October 2013 (UTC)[reply]
From a guitar forum, questions e-mailed to RoSH website: Question #2. Do I have to take any special disposal measures if I remove old lead solder from any equipment? Answer: No. Ssscienccce (talk) 14:30, 19 October 2013 (UTC)[reply]

Human mortality question edit

Based on known age-specific mortality rates, what is the expected time between successive deaths of the world's oldest inhabitant?→31.54.112.70 (talk) 22:42, 15 October 2013 (UTC)[reply]

A little over one year. You don't have to estimate it, since we know the true answer: World's oldest person#Chronological list of the verified oldest living person since 1955. Someguy1221 (talk) 22:48, 15 October 2013 (UTC)[reply]
You would have good luck asking this question at the Math desk. It's a basic stats question and the Oldest person page is a large enough dataset to get a good estimate. I don't know how to do the math for you offhand, but they will. I'm guessing a poisson distribution would be a good start. Shadowjams (talk) 02:29, 16 October 2013 (UTC)[reply]