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July 31

Cold fusion

Cold fusion is referred to as "bullshit" above, and the article seems to say as much. I thought that there were reputable scientists, like those at the the San Diego Navy SPAWAR lab, who continue to produce confirmatory results and have never wavered in their support for cold fusion (e.g. [1]) and some Italian outfit that claimed to have built commercial scale reactors in the past year.[2] What is the current status of that controversy? 99.39.4.220 (talk) 00:54, 31 July 2011 (UTC)[reply]

The 'Italian outfit' can be found here: Energy Catalyzer. Opinions on its validity are, shall we say, divided. AndyTheGrump (talk) 00:58, 31 July 2011 (UTC)[reply]

(edit conflict)Certainly not bullshit. Nothing approached with an open mind and a receptive scientific attitude should ever be labeled "bullshit". There is certainly no convincing evidence that it exists (nor known theoretical mechanism by which it COULD exist). But people trying weird science is how some of the greatest advances of mankind have been achieved. Anyone claiming to have "proof" of anything does not deserve your attention. There is only evidence in science, never "proof". And until there is more than occasional irreproduceable evidence of the phenomenon, I say that there is no reason to believe it exists.

That said, the SPAWAR results are intriguing, but I think that we need to think about horses before zebras, in that there is likely some other energy source for the miniscule amount of heat they are able to produce. And I can not find anything about an Italian commercial application, but I can tell you that scam artists have been selling cold fusion kits for years.-RunningOnBrains^(talk) 01:01, 31 July 2011 (UTC)[reply]

It is not bullshit to explore weird stuff in a scientific manner. Cold fusion should be researched as with anything else. It is bullshit to make definitive claims on the success of research when there are none. Let me reiterate: research = good, making up things = bullshit. Much of the problems with the cold fusion was too much bullshit, not enough good. --Jayron32 01:45, 31 July 2011 (UTC)[reply]

What he said ::points at Jayron:: -RunningOnBrains^(talk) 02:27, 31 July 2011 (UTC)[reply]

Strength of atomic bonds

Hello... I was trying to get an idea of the strength of bonds between atoms by imagining that a water molecule, for example, was blown up to a size where I could hold it in my hands and pull it apart. However, I'm not sure how I should scale the force. Should it scale with the cube of the molecule length? Anyway, doing the calculation properly, what would be the force required to pull apart a water molecule if it was, let's say, 10 cm long? What would be the force required to separate two water molecules of that size in liquid water (keeping the individual molecules intact)? (Ignore the question of whether you could actually grasp such a blown-up molecule, and other similar quibbles.) 86.179.2.163 (talk) 01:20, 31 July 2011 (UTC)[reply]

I can't think of a single principle that would allow you to chose the proper scaling factor and in the absence of a principle your question is not answerable. Dauto (talk) 01:45, 31 July 2011 (UTC)[reply]

(edit conflict)Well, unfortunately, I do have a quibble. If you scale the size you still have to figure out how to scale all the other properties, like mass and charge, and this may not be a trivial answer. The H-O bond strength is 460 Joules...note that this is an energy required to break the bond, not a force. This is because at the molecular level the force needed to break a bond is unimportant, it is the energy input needed to break the bond that is constant. For this reason the was that you scale mass and charge becomes important. I feel like someone else should be able to explain this better, but suffice to say it is not as simple an analogy to scale up bond-breaking as it might seem.-RunningOnBrains^(talk) 01:48, 31 July 2011 (UTC)[reply]

460 joules seems vastly too great a number? Did you miss off a factor of ten to the power minus something? Despite the objections, and despite not being able to work it out myself, I feel it ought to be possible to give a sensible answer. To put it another way, if I was shrunk down to the point where I could hold a water molecule in my hand, how hard would it be to pull it apart. It seems like a question that ought to have a sensible answer.... ! 86.179.2.163 (talk) —Preceding undated comment added 01:56, 31 July 2011 (UTC).[reply]

Actually, I was off in two ways. I was trying to say 460 kilojoules (110 kCal), but that's per mole, so take that number and divide by Avogadro's number to get the energy per bond. Sorry. Anyway, it is still a unit of energy, which is the crux of my argument. -RunningOnBrains^(talk) 02:19, 31 July 2011 (UTC)[reply]

OK, well, how would the force generated by my muscles decrease as I shrank? If I could pull 100 kg at full size, and then I was halved in size, could I pull 100/8 kg? 86.179.2.163 (talk) 02:07, 31 July 2011 (UTC)[reply]

Do an experiment. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 02:37, 31 July 2011 (UTC)[reply]

Silliness aside, your reasoning sounds correct. Various internet sources seem to claim that strength is ~~connected~~ proportional to muscle A) mass, B) volume, or C) cross-sectional area. Not sure which is correct, as no reliable source seems to weigh in on the matter as far as I can tell. -RunningOnBrains^(talk) 02:47, 31 July 2011 (UTC)[reply]

A and B seem pretty much the same, in this case, since the density of muscle is more or less a constant. C would also be the same, if we assume the length of the muscles being compared is the same. StuRat (talk) 18:29, 31 July 2011 (UTC)[reply]

OK, RunningOnBrains gave the OH bond energy, which I understand is the usual way of expressing bond strength, but what I would like to know is the force. For example, what force would be needed to pull a hydrogen atom off a water molecule? More specifically, I guess there would be a varying force as the atom was pulled away, so I would like to know the maximum force at any point during the separation (I'm not sure where that would occur). Does anyone have that information? Or just a ballpark figure that would be representative of a typical molecule would be great... 86.183.129.83 (talk) 21:55, 31 July 2011 (UTC)[reply]

As RunningOnBrains said, it doesn't make much sense to talk about force in this context. It's energy that matters. How that energy is delivered is irrelevant. --Tango (talk) 23:19, 31 July 2011 (UTC)[reply]

I don't understand why it doesn't make sense to talk about force. Could you not use force to pull two bonded atoms apart? Why is that force not (in principle) measurable? 86.183.129.83 (talk) 23:53, 31 July 2011 (UTC)[reply]

Energy is required to do work and work is required to break bonds. Force is important in that you may need a certain minimum amount of force to do work which may require that the minimum amount of force be applied over a minimum amount of time. --DeeperQA (talk) 00:11, 1 August 2011 (UTC)[reply]

I imagine two bonded atoms a bit like like ping-pong balls connected by a spring (albeit a spring that gets weaker at a great distance). If I could physically grasp the two nuclei and pull them apart, would it not require a (distance-dependent) force to do that, just like the ping-pong balls and spring? I do not understand why this force cannot be quantified, if that's what people are saying. Maybe I am fundamentally misunderstanding something. (To be clear, I am not looking for reasons why bond energy might be a more useful quantity, I just want to understand why the force that I am talking about is not, in principle at least, quantifiable, if that's the case). 86.183.129.83 (talk) 00:24, 1 August 2011 (UTC)[reply]

If one can measure molecular potential energy as a function of atomic separation, shown as a schematic here [3], then it is straightforward to measure the minimum required force as a function of the stretch applied. It would be proportional to the slope of that curve. Such curves could be modeled or measured in principle, but there is little practical value to doing so. In the subatomic world nearly all interactions involve transfers of discrete quanta of energy (e.g. quantum mechanics). So it is far more useful to consider the energy involved in various processes rather than than "forces" required, since in practice there is almost never any way to apply a discrete force at the molecular level. Dragons flight (talk) 19:38, 2 August 2011 (UTC)[reply]

Could you not estimate the force by getting a crystal such as an hourglass-shaped piece of ice, calculating the number of water molecules in its smallest cross section, and then measuring how much force was required to pull (not bend or snap) it apart? 2.101.14.124 (talk) 11:08, 1 August 2011 (UTC)[reply]

Here is a table of rupture forces. The atoms in water are held together by covalent bonds, which appear to be on the order of 1000 piconewtons, while the hydrogen bonds between water molecules rupture at a few piconewtons. --Heron (talk) 18:30, 1 August 2011 (UTC)[reply]

Thanks Heron, that is exactly what I wanted. So, let's take a ballpark figure of 10^-9 N. A water molecule is apparently about 3 x 10^-10 m long. If we scale force with the cube of dimension, then the force needed to break a 10 cm water molecule would be 10^-9 * (0.1 / (3 x 10^-10))^3, which is about 4 x 10^16 N. Did I do that calculation correctly? That seems like a silly answer, so maybe the question, and certainly the method, doesn't make sense. 86.176.211.251 (talk) 21:02, 1 August 2011 (UTC)[reply]

Force probably doesn't vary as the cube of the linear scale. What is it as the square of the linear scale? 64.134.228.55 (talk) 00:50, 2 August 2011 (UTC)[reply]

Well, as the square of the linear scale, it would be 10^-9 * (0.1 / (3 x 10^-10))^2, which is about 10^9 N. Going with the square of the linear scale may make some kind of sense. Force = Power x Time / Distance. If I was shrunk down by a linear factor of k, then power would surely scale with k^3? Time is unchanged, and distance scales with k, so overall force should scale with k^2. Still, 10^9 N seems very large. Could anyone check that calculation for silly mistakes? 86.181.202.137 (talk) 11:48, 2 August 2011 (UTC)[reply]

I agree. Here's another way of looking at it. Imagine your molecule is not a molecule but an iron bar. When you scale up the bar, the force required to break it (by pulling) is its tensile strength times its cross-sectional area. Tensile strength is independent of size (in the macroscopic world, anyway), so the force would scale as the area, i.e. the square of the linear dimension. It doesn't surprise me that the tensile strength of a water molecule is huge. After all, it takes a lot of energy to electrolyse water. --Heron (talk) 18:46, 2 August 2011 (UTC)[reply]

And here is some more data: the Young's modulus (stretchiness) of atomic bonds. Compare these to the moduli of more common materials in our Young's modulus article. The upper and lower limits of the YM of a covalent bond, 200 to 1000 GPa, are the same as those of steel and carbon nanotubes respectively. --Heron (talk) 21:10, 2 August 2011 (UTC)[reply]

I think your iron bar analogy is a good one. Assuming the bond strength in a water molecule is roughly the same sort of order of magnitude as in a tough solid substance such as iron or steel, then it's plausible that pulling apart a blown-up water molecule say 10 cm across would be similar to trying to pull apart an iron bar of that diameter, which obviously would take a great deal of force. I don't know why, but somehow I was visualising that I would be able to pull it apart with my hands... 86.179.5.247 (talk) 23:36, 2 August 2011 (UTC)[reply]

Severed hands

If one's hand is sliced off, is he likely to bleed to death? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 02:46, 31 July 2011 (UTC)[reply]

Why not? Cutting you wrists seem to be an effective mode of suicide. Plasmic Physics (talk) 03:07, 31 July 2011 (UTC)[reply]

Unreliable sources have claimed otherwise. Also i remember reading a statement (admittedly online) about cutting wrists hardly ever resulting in anything more than a scar (paraphrased near quote). Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:11, 31 July 2011 (UTC)[reply]

Cutting ones wrists often does result in a scar for a few reasons. Someone might find the person, call the local emergency services, and save the person. The person may just be trying to get attention and do it in front of someone who then calls. The person may not do it well enough, pass out, it clots, and they're found. Often the person doing the cutting doesn't do it well enough to actually exsanguinate themselves, passes out, wakes up later, cries, and goes on with their life. What do you expect, they're amateurs and probably don't know that much about what they're actually attempting. Dismas|^(talk) 03:37, 31 July 2011 (UTC)[reply]

So, considering all that, what would completely severing the hand do? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:57, 31 July 2011 (UTC)[reply]

If no medical attention were given, and no attempt was made to staunch the flow of blood? Kill them. If drastic lifesaving measures were used, they may survive. --Jayron32 04:22, 31 July 2011 (UTC)[reply]

Agreed. The most immediate difference would be made by applying a tourniquet. StuRat (talk) 04:23, 31 July 2011 (UTC)[reply]

Really? What to Expect the Toddler Years claims that applying pressure to the wound is usually sufficient to staunch the flow of blood even in such extreme circumstances. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 04:32, 31 July 2011 (UTC)[reply]

That's what a tourniquet is, a means of applying pressure to an extremity, for the purpose of reducing the flow of blood. StuRat (talk) 06:04, 31 July 2011 (UTC)[reply]

As I understand it, there's a big distinction in practice. Usually a tourniquet doesn't apply pressure to the wound itself, but to the blood vessels upstream of the wound. Secondly, a tourniquet typically cuts off almost all blood flow to the extremity, as opposed to direct pressure to the wound itself which still allows blood flow to vessels not compressed (e.g. on the other side of the limb). "Use a tourniquet" is a much more drastic intervention than "apply pressure to the wound", and the two should not be treated as synonyms. -- 174.24.213.112 (talk) 18:03, 31 July 2011 (UTC)[reply]

In general, yes, but in the case of a severed hand, I can't see how you could "apply pressure to the wound". Does that mean you'd clamp a hand tightly over the end of the stub ? I can't see that working. Perhaps you could squeeze the arm tightly further up, though, which is precisely what a tourniquet would do. StuRat (talk) 18:24, 31 July 2011 (UTC)[reply]

I recently took a first aid course and they no longer teach tourniquets. They don't event teach the use of pressure points. Apparently, people tended to do more harm than good with them. Direct pressure and elevation are usually sufficient even in extreme cases. It's not difficult to apply direct pressure to the stump of an arm. If your hand is all you had available, then use it, but ideally you would put a large sterile dressing over it (perhaps using your hand to apply more pressure, since it would be difficult to tie a tight bandage in that location). If you don't have a dressing, you could take your shirt off and use that. In addition, you elevate the wound above the heart. That should be plenty to stop the bleeding. I have read that in cases of total traumatic amputation, the bleeding is actually quite small. If an artery is completely severed, it tends to contract, which slows the bleeding. --Tango (talk) 18:43, 31 July 2011 (UTC)[reply]

I even more recently took a first aid course as part of Scouting/Guiding. We talked about how tourniquets are no longer taught in more first aid courses, but also talked about the possibility (if you're letting the children take any sort of controlled risk, as we are supposed to) that you actually might need to, and that obviously we wouldn't use it just because someone was bleeding, but that if someone has lost a limb and/or is gushing blood so that a significant proportion is on the ground, and the ambulance is not here, your choice is basically tourniquet or dead child. If you let children use axes and saws, you can try to teach them safety and control the situation, but you also need to plan for the worst. It's all very well talking about the danger of cutting off circulation to flesh, but is someone is bleeding out and you can't stop it, they will die. So you try another way, and if it doesn't work and they are still bleeding catastrophically, you do what can save their life. We also talked about how the way we were taught to make a tourniquet as children ('as tight as possible') is the real problem, and that you should only tighten until the bleeding has just become slow enough to control, or just stopped if there's no other way to stop it. Catastrophic bleeding actually comes at the top of list of priorities for first aid, these days. You can't circulate oxygen around the body if there is no blood. 86.164.73.187 (talk) 20:35, 31 July 2011 (UTC)[reply]

Certainly, in a matter of life and death you do whatever you need to do to save the life. It's very unlikely that a tourniquet will ever actually be necessary, though. It only helps for bleeding limbs and try tend not to bleed that much. When people bleed out, it is usually from abdominal wounds. On my course they mentioned that when they had taught tourniquets, they taught people to release it every ten minutes for a few seconds to allow the limb to re-oxygenate. For anyone considering using a tourniquet, be aware that you could face legal action if you do it wrong and cause more harm than good, since it isn't a recommended part of first aid (so may not be covered by good Samaritan laws). --Tango (talk) 21:46, 31 July 2011 (UTC)[reply]

We also talked about litigation. We were told that it basically never happens, and that if it does we would rather be in the position of a live child with some disability than a dead child, and that the organisation would stand behind us if we'd taken reasonable action based on what we'd learnt. We were told that the bogeyman of 'I'll be sued for helping' is basically a myth in this country, and far more likely to lead to injury and death than us doing reasonable things. Again, these tourniquets were only if you'd tried everything else and it hadn't worked, and were just to reduce the pressure enough that you could use other techniques effectively. Tourniquets are still used in medicine, but much more carefully than in the past. If you're just trying to keep the blood inside until a paramedic arrives and takes over, we were told that tourniquets could indeed be a vapid part of our arsenal (just not used indiscriminately or over-tightly). 86.163.1.126 (talk) 15:56, 2 August 2011 (UTC)[reply]

Why on earth would I let my children use this?] Whoop whoop pull up ^{Bitching Betty | Averted crashes} 21:40, 31 July 2011 (UTC)[reply]

I'm removed the giant image you posted here of an antique horseman's axe because it was distracting and completely irrelevant.

It's perfectly normals for children to be allowed to use modern wood-cutting axes and hatchets with varying degrees of supervision. The Boy Scouts do it all the time. That's ages 10 through 18(In USA). And I'm sure kids who grow up on farms or out in the woods also learn to use those tools at an early age. But that doesn't really have anything to do with the question. APL (talk) 01:32, 1 August 2011 (UTC) [reply]

Don't forget to warn them to be careful with That Axe, Eugene...

Survival probably wouldn't depend on "drastic lifesaving measures". Some basic first aid would suffice. You're probably right, though, that left untreated it would be fatal.--Srleffler (talk) 04:47, 31 July 2011 (UTC)[reply]

Perhaps "immediate" would be a better term than "drastic". StuRat (talk) 06:05, 31 July 2011 (UTC)[reply]

Don't forget about cauterisation. Plasmic Physics (talk) 05:37, 31 July 2011 (UTC)[reply]

Now that really does seem "drastic", especially when done in the field with whatever hot object is handy and without any anesthetic. StuRat (talk) 15:45, 31 July 2011 (UTC)[reply]

OWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW! Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:57, 31 July 2011 (UTC)[reply]

Yes, do forget about cauterisation! It almost always does more harm than good. You would eliminate any chance of reattaching the hand (which is entirely possible with modern medical techniques - see microsurgery). Even if reattaching the hand wasn't going to be possible anyway, you would eliminate the chance of getting a clean stump without amputating more arm, which could make use of a prosthetic more difficult. There are also all the usual risks associated with severe burns, which can be life threatening. --Tango (talk) 18:43, 31 July 2011 (UTC)[reply]

It is extremely possible for a severed hand to develop some kind of infection which left untreated or the person has a weak immune system could possibly die from that. Bleeding to death is possible but its more likely to happen if both hands got severed which is hard for one person to do. --86.45.132.106 (talk) 20:41, 31 July 2011 (UTC)[reply]

I'm not sure I'd be that worried if the severed hand developed an infection - I'd be more worried about the parts still attached to the body... ;-) AndyTheGrump (talk) 02:26, 1 August 2011 (UTC)[reply]

Read about Aron Ralston. Bus stop (talk) 01:49, 1 August 2011 (UTC)[reply]

I never did understand why cauterization was presented as a first aid solution - infection in ancient times was so often a death sentence, and in modern times people know better. Hippocrates regarded it as a last resort: "What drugs will not cure, the knife will; what the knife will not cure, the cautery will; what the cautery will not cure must be considered incurable". My feeling is that only European civilizations from a few centuries ago could have been so ignorant of medicine as to rely on it to stop bleeding... Wnt (talk) 21:15, 2 August 2011 (UTC)[reply]

IR tech

Can an infrared or thermal imaging sensor detect a person buried under snow? 67.169.177.176 (talk) 03:59, 31 July 2011 (UTC)[reply]

It depends on the density of the snow and the sensitivity of the sensor. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 04:01, 31 July 2011 (UTC)[reply]

So a top-of-the-line IR sensor like those used by the military could do it easily, while a cheap $200 IR camera would be useless? Makes sense to me. BTW, am I right that you have an interest in aviation? 67.169.177.176 (talk) 04:11, 31 July 2011 (UTC)[reply]

Yup. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 04:30, 31 July 2011 (UTC)[reply]

Do you have an actual reference for this? Or are you completely making it up off the top of your head? APL (talk) 13:09, 31 July 2011 (UTC)[reply]

No need to cite the obvious... Whoop whoop pull up ^{Bitching Betty | Averted crashes} 16:33, 31 July 2011 (UTC)[reply]

If you said that about the specific claim on the encyclopaedia proper you would still find yourself challenged and the claim legitimately removed, and rightly so. Nil Einne (talk) 17:10, 31 July 2011 (UTC)[reply]

It would also depend on the depth of the snow and how long it has been in place. Right after an avalanche, the surface of the snow should be about the same temp everywhere, while, after some time, the snow near the body should be slightly warmer. StuRat (talk) 04:19, 31 July 2011 (UTC)[reply]

Unfortunately, by that time most of the victims would be dead. See Avalanche#Human survival and avalanche rescue. 67.169.177.176 (talk) 05:19, 31 July 2011 (UTC)[reply]

True, that's why I said "the body". So IR would likely only be useful for recovery of bodies, not for rescue. StuRat (talk) 05:59, 31 July 2011 (UTC)[reply]

Unless by some miracle someone is buried at a shallow depth (which means the snow warms faster) and/or has some kind of physiological traits that allow him/her to survive being buried without suffocating or freezing to death. (It seems there's always a case or two like that after any major building collapse {earthquake, 9/11, etc.}, and I've even read of one case where someone was dug up alive from an avalanche after being buried all day long.) Could be a good plot twist. 67.169.177.176 (talk) 08:25, 31 July 2011 (UTC)[reply]

You would need a passive microwave receiver to detect heat radiation from the body, at radio frequencies this travels through snow easily. These exist on some satellites, but I don't know if yo can get a hand held model. Graeme Bartlett (talk) 12:07, 31 July 2011 (UTC)[reply]

Thanks, but I'm looking for a helo-mounted model right now. (In any case, if you're looking for survivors on foot, then rescue dogs make a much better sensor.) 67.169.177.176 (talk) 01:32, 1 August 2011 (UTC)[reply]

Some burglar alarms contain them, so I don't see why not. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 12:49, 31 July 2011 (UTC)[reply]

Link please?

For me, Google is only turning up a couple of patents. But that's no help, since completely impossible things are often patented. APL (talk) 13:12, 31 July 2011 (UTC)[reply]

Here's a fun thread where this technology is discussed, one person, who claims to be a scientist studying exactly this says this about the equipment to find people with microwaves : "After decades of development, you currently need 2 skidoos, a sled, and 3 people to operate the proprietary equipment." APL (talk) 13:44, 31 July 2011 (UTC)[reply]

Perhaps we need to wait for the patents to expire so competition can kick in and provide a cheap, reliable, easy to use, and effective product. StuRat (talk) 15:19, 31 July 2011 (UTC)[reply]

Better yet, pay someone in the Patent office to revoke the patents. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 20:45, 31 July 2011 (UTC)[reply]

Oh, my, the things I would do with a portable terahertz imager and a powerful tunable terahertz maser. Find people under snow, look at the neighbors naked in their beds, reprogram their skin to make fetal keratin and blister up all over their bodies... so many possibilities, so little time. Someone has this equipment... Wnt (talk) 01:25, 1 August 2011 (UTC)[reply]

fart. Erm, flatulence

~~Exactly what is a fart composed of? Oh, that's obvious, it depends on what you eat/drink/do/say/whatever, but w~~ What is the general base for a fart? Is it true that one can fart next to a lighter and get the fart fire cloud thingy? An editor since 10.28.2010. 05:37, 31 July 2011 (UTC)[reply]

a) It varies, of course, but Hydrogen, Methane, and 'air' (Nitrogen, Carbon dioxide, Oxygen)

b) Methane and Hydrogen burn explosively, so yes; it can also lead to burns, and a trip to hospital.

Wikipedia covers it, in a) Flatulence#Composition of flatus, b) Fart lighting. Chzz ► 05:50, 31 July 2011 (UTC)[reply]

Thank you. Chzz saves the day again --Since 10.28.2010 05:53, 31 July 2011 (UTC)

YouTube used to have videos of people lighting their farts. Don't know if there are still any around. Dismas|^(talk) 13:05, 31 July 2011 (UTC)[reply]

Adam Savage did it on Mythbusters (Episode 48 outtakes). — Michael J 06:45, 1 August 2011 (UTC)[reply]

Cymbal-eared aliens

If a species of aliens had big ears shaped like cymbals, at the sides of their heads, what would their hearing be like? Megaconworlder (talk) 08:40, 31 July 2011 (UTC)[reply]

Out of this world?--78.148.142.14 (talk) 10:54, 31 July 2011 (UTC)[reply]

Not very good, cymbals are mostly flat except at the centre. Sound wouldn't be focused very well by cymbal shaped ears. Plasmic Physics (talk) 11:55, 31 July 2011 (UTC)[reply]

See pinna (anatomy).--Shantavira|^{feed me} 12:50, 31 July 2011 (UTC)[reply]

Cymbals are not mostly flat, by the way. They may lok that way from afar or on TV but they are actually almost never flat. It's much more obvious if you see from beneath. Beeblebrox (talk) 22:04, 31 July 2011 (UTC)[reply]

True, but probably still not a good shape for an ear. There's no focal point. APL (talk) 02:10, 1 August 2011 (UTC)[reply]

OK, conical. A cone is still flat along a generatrix, meaning there is still no focal point for the conical section. The central buldge however, does have a focal point. Essentially, the large conical outer section of the cymbal has no effect on sound focussing, and is superfluous. They are incorporated for a different purpose in the musical instrument, to improve tintinnabulation qualities. Plasmic Physics (talk) 14:07, 1 August 2011 (UTC)[reply]

NATO notation for color specification?

The Wikipedia page on the Union Flag has a section on its color specification. One of the specifications is in some “NATO” notation (e.g. “8305.99.130.4580” for the blue color of the flag). What color notation is that? --98.114.98.196 (talk) 13:07, 31 July 2011 (UTC)[reply]

NATO Stock Number - see the link. The wikilink in the article to the North Atlantic Treaty Organization was misleading; I'll change it. Chzz ► 16:04, 31 July 2011 (UTC)[reply]

HRT self-medication

I want to learn about HRT self-medication. Specifically, do people who do it buy the medication online or do they extract it from shop bought products? Please, spare me the "it's dangerous go see a doctor" speech; I'm not a transsexual and I've no plans to self-medicate. I'm just curious about how it's done. — Preceding unsigned comment added by 78.126.136.91 (talk) 14:55, 31 July 2011 (UTC)[reply]

Perseids, August 2011

Hi! I've been watching the Perseids for a few years now. The peak of this year's meteor shower is 13 August. However, full Moon also comes that night and will ruin the show. My question is: what night should be perfect for watching the Perseides if we consider both the peak of the shower and the moon phase (but not the weather)? 92.36.179.99 (talk) 16:17, 31 July 2011 (UTC)[reply]

Well, according to our Perseids article, the best viewing is in the pre-dawn hours, and to get darkness during that period you should watch a couple of days before the full moon. But if you don't want to be watching for meteors at 4 AM, you might prefer to watch a couple of days after the full moon, when it will be dark for a while after the sun sets. (I hope I got that right; I always have trouble with moon phases.) Looie496 (talk) 17:43, 31 July 2011 (UTC)[reply]

The very best time is still August 12 / 13, but after the moon sets. Because the moon is full there is only a small window when it will still be dark after the moon sets. At my location in the US, that gives a couple hours immediately before dawn. Dragons flight (talk) 17:51, 31 July 2011 (UTC)[reply]

Ciências - Química

Scientific American: A reportagem refere-se ao Sequestro de Carbono presente na atmosfera por meio de tubulações especiais, sendo devolvido ao subsolo em alta profundidade para ser reabsorvido, tornando-se útil novamente em um futuro distante, em forma de gás ou até mesmo de Petróleo(?). Qual seria a certeza científico-tecnológica da eficácia de tal procedimento? — Preceding unsigned comment added by 200.103.171.184 (talk • contribs)

The question refers to carbon sequestration. But we aren't going to answer in Spanish, and I don't see that point in an answer the OP won't be able to read. Looie496 (talk) 17:25, 31 July 2011 (UTC)[reply]

That's not Spanish. It's Portuguese! Portuguese happens to be my mother tongue, so you can take my word for it. For a second I didn't even realize that the question was not in English since I can read either of them effortlessly. He is asking if it is true that carbon dioxide re-inject underground might turn back into natural gas or oil in a distant future. Dauto (talk) 18:38, 31 July 2011 (UTC)[reply]

Well then, no. There's nothing to reduce or hydrogenate the carbon dioxide before it diffuses into rock fractures and sedimentation. 208.54.5.187 (talk) 20:36, 31 July 2011 (UTC)[reply]

Translation to Portuguese: Não porque não há como reduzir ou hidrogenar o dioxido de carbono que simplesmente se difundirá pelas rochas e sedimentos. Translated by Dauto (talk) 21:23, 31 July 2011 (UTC)[reply]

Agreed. Translation to Portuguese: Concordo. Dauto (talk) 21:23, 31 July 2011 (UTC)[reply]

Kepler’s 3rd law

Assuming the Kepler’s 3rd law is valid at the atomic level do the electron shells follow Kepler’s 3rd law? --DeeperQA (talk) 23:59, 31 July 2011 (UTC)[reply]

It sounds like you are using the old Dalton model of the electrons as little spheres which orbit the nucleus just like planets orbit the Sun. The modern conception is more that electrons exist as probability clouds, so classical orbital mechanics don't really apply, no. StuRat (talk) 00:06, 1 August 2011 (UTC)[reply]

No, I am aware that an electron cloud might even be a field but nevertheless what I am asking is whether the shells, in any regard, are related by two factors such as semimajor axis and period of orbit. --DeeperQA (talk) 00:21, 1 August 2011 (UTC)[reply]

The answer is no, electron "shells" have nothing at all to do with orbiting, even though the term "orbital" is used. Electrons don't "orbit" in the same way that a planet orbits a star; distance from the nucleus is described by a variation of the Rydberg formula, but this "distance" isn't the same thing as the "distance" a planet orbits from the sun. There is no correlation because its a completely different concept. If you want to get beyond a grade-school understanding of electrons, you need to let go, in every way, the idea that the electron is a little ball racing little circles around the nucleus. It works nothing like that. --Jayron32 01:19, 1 August 2011 (UTC)[reply]

I wouldn't say that. All the QM formulas started by working with classical concepts. The electron has an angular momentum, which matches where it is and "how fast it is orbiting"; it can't exceed the speed of light while orbiting as we discussed with end of the periodic table. It just isn't really a nice neat little ball, but a sort of wave of probability. Now looking up Kepler's third law, "The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." Now the infamous Bohr model gives orbital radius (they're circular) proportional to the square of n; the velocity is proportional to the inverse of the square root of the radius (and thus 1/n); so the period (circumference / velocity) should be proportional to n cubed. That means that the radius : period is indeed square : cube, just like in Kepler's law. Bohr's model gives real results for hydrogen atoms, so it's not irrelevant. Now quantum mechanics and the Schroedinger equation involve more complex terms for interactions between charges, and end up resorting to numerical methods for solution, but the three-body problem isn't really so tractable even in orbital mechanics. Also note: Rydberg atoms (i.e. atoms that are almost but not quite ionized, with electrons orbiting a nucleus at a great distance) actually do have electrons moving in "Keplerian elliptical orbits", according to the article. Wnt (talk) 02:15, 1 August 2011 (UTC)[reply]

In most cases Kepler's equations cannot be used to make predictions about the orbits of electrons. Excepting some special cases as mentioned above.

Electrons orbiting a nucleus...

Cannot be described as solid particles.
Do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves.
Are never in a single point location, although the probability of interacting with the electron at a single point can be found from the wave function of the electron.
Jump between orbitals in a particle-like fashion. For example, if a single photon strikes the electrons, only a single electron changes states in response to the photon.
Retain particle like-properties such as: each wave state has the same electrical charge as the electron particle. Each wave state has a single discrete spin (spin up or spin down).

"Despite the obvious analogy to planets revolving around the Sun, electrons cannot be described as solid particles. In addition, atomic orbitals do not closely resemble a planet's elliptical path in ordinary atoms. A more accurate analogy might be that of a large and often oddly-shaped "atmosphere" (the electron), distributed around a relatively tiny planet (the atomic nucleus). One difference is that some of an atom's electrons have zero angular momentum, so they cannot in any sense be thought of as moving "around" the nucleus, as a planet does. Other electrons do have varying amounts of angular momentum."

see Atomic orbital.

Electrons are odd ducks that are not ducks. - ArtifexMayhem (talk) 03:26, 1 August 2011 (UTC)[reply]

The atomic orbitals derive from the same Hamiltonian that classically gives you the Keplerian orbits, so some probabilistic variant of Kepler's laws must apply to them, but I'm not sure exactly how it works. The classical orbits and quantum orbitals are closely related.

That d-orbital image is terrible. The five individual orbitals are badly hand-drawn (compare this), and the "combination" orbital is complete nonsense. A combination of all orbitals of a given type should look like a sphere, if it looks like anything. I can't believe that image is being used in major mainspace articles after all these years. Surely there's something better on Commons. -- BenRG (talk) 08:00, 1 August 2011 (UTC)[reply]

Clearly orbitals involve quantum weirdness - one way to think of this is to note that the orbital structure is not actually linked to the atom, but to the observer. As I understand it, if you determine an electron is in one particular p orbital, and you change your viewpoint by 45 degrees along one of the axes you've defined, it now has equal odds of being in two different p orbitals according to your new coordinate system. Even so, I wouldn't go so far as to say that an electron in an s orbital has no angular momentum - after all, it has spin angular momentum, which inevitably involves the electron's cloud-like essence going in a circle of some sort around the nucleus. Wnt (talk) 21:03, 1 August 2011 (UTC)[reply]