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July 1

Homogeneous Continuous-time Markov chains - analytical expressions for the finite time transition probabilities?

This looks like a question much better suited to the Wikipedia:Reference desk/Mathematics. Dolphin (t) 00:15, 1 July 2010 (UTC)

I have transferred this question to the Mathematics Reference Desk. See HERE. Dolphin (t) 00:30, 1 July 2010 (UTC)

Iron

Does the body use iron(II) or iron(III)? --75.25.103.109 (talk) 00:18, 1 July 2010 (UTC)

Both.._oops Fe2+ probably = See Hemoglobin - especiallly Hemoglobin#Iron's_oxidation_state_in_oxyhemoglobin . Sf5xeplus (talk) 00:21, 1 July 2010 (UTC)

Both. Most iron containing proteins are (nominally) iron(II), but iron is transported around the body (by transferrin) and stored (by ferritin) as iron(III). Physchim62 (talk) 10:24, 1 July 2010 (UTC)

Expiration of medicine

What does it mean for medicine to expire? Assuming we are dealing with a pharmaceutical that doesn't become harmful as it ages, how much of the active ingredients can be lost before it is considered expired? —Preceding unsigned comment added by 173.49.77.61 (talk) 00:22, 1 July 2010 (UTC)

Usually drugs just become less effective as they get older. Some could become harmful though - don't risk it! --Tango (talk) 00:27, 1 July 2010 (UTC)

Why do they get less effective? --RampantHomo (talk) 01:14, 1 July 2010 (UTC)

Various reasons. Oxidation, decay due to exposure to light (that's why they are often in dark brown bottles), etc.. --Tango (talk) 02:23, 1 July 2010 (UTC)

How about a reference here at the Reference Desk? The first google hit on why does medicine expire yields this Medscape article, in which a medical doctor concludes that almost all drugs are fine to take after the expiration date, which he says is merely the last date at which the pharmaceutical company asserts the drug is still effective. On the other hand, he cites one example in which a guy may have damaged his kidneys by taking expired tetracycline. Comet Tuttle (talk) 03:01, 1 July 2010 (UTC)

As an example, I recall from my chemistry class that Hydrogen peroxide molecules (H₂0₂) will drop the extra Oxygen atom at somepoint, and as this happens to collectively more Hydrogen Peroxide molecules, the contents of the bottle become more and more water, and less and less hydrogen peroxide, thereby significantly reducing the effectiveness over time. Falconus^{p t c} 03:34, 1 July 2010 (UTC)

Hydrogen peroxide is a special case - it's very unstable. Most (pretty much all) medicines are far far more stable. Concentrated hydrogen peroxide is unstable enough to be used as rocket fuel. Ariel. (talk) 10:32, 1 July 2010 (UTC)

Important distinction: it's important that a rocket fuel is "high energy" (thermodynamically unstable) not that it rapidly/spontaneously decomposes (somewhat unrelated kinetic issue, which is sometimes even a negative quality to have). Lots of excellent rocket fuels are easy to handle and store until they are made to react with a large release of energy. Some fairly low-energy molecules are labile enough that they can change slightly under mild conditions, leading to loss of intended activity. DMacks (talk) 14:17, 1 July 2010 (UTC)

I should have said "unstable enough to be used as a monopropellant rocket fuel". Ariel. (talk) 14:52, 1 July 2010 (UTC)

If the drug is not harmful after it expires, it may even still work due to the placebo effect. ~AH1^(TCU) 18:06, 2 July 2010 (UTC)

Help identifying this flower

I would like to know the species of this flower so I can add the image to the correct article, but I haven't got a clue! The background info for this photo is:

• In the UK in Oxfordshire
• Taken in March
• Found in open cut grass

Thanks in advance - Zephyris _Talk 00:47, 1 July 2010 (UTC)

Is it not speedwell (Veronica (plant) of some variety)? It looks like the thing I've always called speedwell, and if I had to guess from the articles I'd say it was Veronica persica. But hopefully a botanist will help you with the more specific species. 86.164.57.20 (talk) 01:40, 1 July 2010 (UTC)

Looking it up in Complete British Wild Flowers by Paul Sterry, which has photos of many kinds of Speedwell, then it could be Veronica persica, polita, or agrestis. V. persica - reddish stems, "white on the lower lip of the corolla". Cannot see a reddish stem, but the other part matches. The photo shown in that Wikipedia article is perhaps the wrong species therefore. 92.28.244.45 (talk) 09:57, 1 July 2010 (UTC)

Thanks guys, thats brilliant. From those descriptions it must be V. polita, I will replace the image on that page with this one. - Zephyris _Talk 13:13, 1 July 2010 (UTC)

I thought I had typed V. persica above, not polita. Anyway I have now corrected it, and the desrciption corresponds to V. persica, and the V. polita article photo needs to be changed back. 92.28.247.183 (talk) 19:15, 2 July 2010 (UTC)

Neodymium magnets

Do the neodymium and iron in neodymium magnets (Nd₂Fe₁₄B) oxidize over time? --75.25.103.109 (talk) 01:02, 1 July 2010 (UTC)

Not sure but they are almost always coated (ie Ni plated)...

" Neo magnets or rare earth magnets have poor resistance to corrosion and should have a coating or plating applied" [1] . Not sure of the details.Sf5xeplus (talk) 01:31, 1 July 2010 (UTC)

"Sintered Nd2Fe14B tends to be vulnerable to corrosion. In particular, corrosion along grain boundaries may cause deterioration of a sintered magnet. This problem is addressed in many commercial products by providing a protective coating. Nickel plating or two layered copper nickel plating is used as a standard method, although plating with other metals or polymer and lacquer protective coatings are also in use" from Neodymium magnet .. link from that article http://www.journalamme.org/papers_vol20/1369S.pdf Sf5xeplus (talk) 01:33, 1 July 2010 (UTC)

From first hand experience: Yes they rust! I have several of those magnets ripped out of hard disks. The glue they use to fasten it to the holder is very very good so I torn of the nickel layer and now the magnet is not protected at that point. After several weeks this area get redish coloured and if you touch it you have rust on your fingers. The material is sintered and has a large surface area making it vulnerable for corrosion along the former grains.--Stone (talk) 05:05, 1 July 2010 (UTC)

Weight loss

I was watching a comedy/sitcom TV series and there is a woman who is a bit...*ahem*... "un-thin" but she is not at all overweight or fat. She came up with this insane regimen to lose weight, eg eating nothing but 1 salad and coffee 3 days of the week and eating normally the other 4 but running 3 miles and biking 10. This made me think: how would a person who is of a somewhat higher but not immediately health-threatening weight lose weight? Like what kind of everyday changes should they make. This is on a hypothetical level of course, not for medical advice, since I noticed every diet seems to be geared only to the REALLY overweight people.—Preceding unsigned comment added by 68.248.225.254 (talk) 01:57, 1 July 2010

There is one and only one way of losing weight and it applies to everyone, regardless of their starting weight: you consume fewer calories than you burn. That is it (well, you could get liposuction, I guess). You can do that by eating less, exercising more or a bit of both. The important thing is to make sure you still get enough nutrition despite reduced calories - you still need plenty of protein, fibre, vitamins and minerals and a little fat. If you are interested in losing weight yourself then a nutritionist will be able to work out a diet for you that will have all the nutrients you need while having a certain number of calories based on the amount of exercise you will be doing. We can't give specific dietary advice here - there are too many factors involved. --Tango (talk) 02:31, 1 July 2010 (UTC)

The real trick is finding a method to ensure you actually do eat less than you expend. The diet mentioned above might work for that for that or another person, it might not. I've been fat since high school and tried several diets and exercise regimes and found the following: Exercise 3 times a week works a little for me, and at least makes me feel better. Low calorie diets don't work for me (I can't handle being always hungry). Low fat diets don't work for me (I can easily get or stay fat on bread alone!). Low carbohydrate diets (specifically strictly following the Atkins diet) worked (120kg to 98kg over 3 months) for me on my third attempt - it took me that long to learn that the diet doesn't work if I cheat or if I don't eat often enough. But everyone is different.

Every day changes that I've seen help people: Exercising more - cycling or walking to work or school; not snacking; eating 5 times a day rather than 3; eating only and exactly 3 times a day; not eating high fat food; not eating high calorie food; not eating processed food; cutting out a particular food (crisps, chocolate, bread, etc). --Psud (talk) 08:53, 1 July 2010 (UTC)

The general view is that diets don't work lifestyle changes do. There seem to be three main areas, actual nutrition, exercise and the psychology of nutrition and exercise.

The main thrust of most of the work I've seen has been that the most enduring weight loss is slow, over several months.

As with Tangos comments, it's a question of eating a balanced, healthy, diet and adding in some exercise. The exercise needn't be formal, but increasing exertion in daily life on a regular basis.

One of the main concerns about the psychology is that published diets are seen as some kind of magic bullet, rather than a stimulus to actually change behaviours. Hence the type of club based weight loss system where there is both mutual support, and tacit peer pressure.

As with Psud there are little changes, smaller more regular meals, although some of those are little more than formalised snacks. Snacking on fruit, nuts etc rather than processed products etc.

ALR (talk) 09:47, 1 July 2010 (UTC)

Ah, yes, thank you for making that point - I forgot to. Your new diet needs to be something you can happily stick to for the rest of your life, or you will almost certainly put the weight back on again. --Tango (talk) 15:59, 1 July 2010 (UTC)

I've recently started to think about a simple rule for health and weight loss: never eat processed foods. Only eat food in the same form it was alive, apart from cutting, and avoid meat also. Following this rule you would avoid the salt, fats, and sugars that routinely tempt us to eat too much. Something I do is to eat a lot of vegetables, as this fills you up, and fruits also. 92.28.244.45 (talk) 10:10, 1 July 2010 (UTC)

While there is merit to the above, you need to be careful that you are still getting B vitamins and protein if you are going to cut out meat completely. While I agree that vegie diets are easier to reduce calories, you do need to be aware of the nutrition provided by meat and ensure that you are including it. Googlemeister (talk) 13:32, 1 July 2010 (UTC)

You are probably thinking of Vit B12, which is abundant in fish and eggs. People also greatly overestimate the amount and quality of protein that adults need (according to these pages), and excess protein is now thought to be bad for you. Sardines are extremely rich in Vitamin B12, more than anything else you would eat I think. I suggested avoiding meat because of the saturated fat it contains, particularly with modern farming methods. The B vitamin folic acid is mostly obtained from vegetables I believe, as well as other B vitamins, although I am not an expert on individual B vitamins. If you are eating a diverse diet, then eating some red meat is going to be more harmful than good for you. 92.28.247.183 (talk) 19:22, 2 July 2010 (UTC)

I agree with ALR that diets don't work. I've read that this is because the logic of "eating less than you burn" does not apply. As long as you're not on a diet that severly restricts calory intake (say eating less than 1500 Kcal per day), the body will regulate its metabolic rate such that the fat cells stay filled to some fixed level. This also means that if you do lose weight as a result of sticking to a diet of, say, 1500 Kcal, you will eventually return to approximately your old weight if you start to eat a normal amount of kcals again, even if that is still less than what you were used to eating.

What does seem to work is eating healthier, sleeping better and getting more exercise. It could be that if you do this, the body will decide to keep the fat cells filled at a lower level. Count Iblis (talk) 15:39, 1 July 2010 (UTC)

You make a good point. I forgot to say that you need to consume only slightly fewer calories than your burn - your intake should stay at this new level for the rest of your life. Rapid weight loss doesn't work, you will almost certainly put the weight back on again. Your weight is a key factor in determining how many calories you burn, so as you lose weight the calories you are burning will reduce until you reach a new equilibrium and then you'll stay at your new weight. If you reduce your calories to 1500kcal/day then you won't reach an equilibrium before you starve, which is why you'll end up increasing your intake again. For a typical lifestyle, most people need around 2000-2500kcal/day. If you are overweight, chances are you are consuming more than that (and burning more than that due to your weight - most overweight people aren't constantly gaining weight). You shouldn't try to reduce your intake to less than 2000-2500, you should reduce it to 2000-2500. --Tango (talk) 15:59, 1 July 2010 (UTC)

See somatotype and constitutional psychology. For some people it will be more difficult to "lose weight". ~AH1^(TCU) 18:03, 2 July 2010 (UTC)

I'd recommend a look at [The Fat Nutritionist http://www.fatnutritionist.com/], particularly the Articles & Evidence on the right hand side. I especially recommend this article. In brief, if you've never dieted, have access to enough food, and have a fairly normal relationship with food, the weight you are is pretty much the weight you're always going to be. You can convert more of that fat to muscle, you can get fitter, but you're not going to be able to lose weight without eating less than you need, which is a bad thing. And once you've done that (by going on a diet) and failed, you will end up fatter than you were when you started. Human body shape varies, human weight varies (it follows a normal distribution), and there is no problem with being overweight as long as you're not morbidly obese. 86.164.57.20 (talk) 19:13, 3 July 2010 (UTC)

"There is no problem with being overweight as long as you're not morbidly obese"? The scientific consensus is that being overweight is very harmful - take a look. —Preceding unsigned comment added by 174.131.45.82 (talk) 19:29, 3 July 2010 (UTC)

No, the social consensus is that being overweight is very harmful. Whereas the paper I cited actually looked at 11,326 Canadians and whether they died, and found that being 'overweight' according to BMI was actually associated with lower mortality. Being 'obese' was associated with the same mortality as being 'normal weight'. Being 'morbidly obese' was associated with the same increased risk as being 'underweight'. And the way to end up morbidly obese when you're overweight is to start dieting and wait 10 years. 86.164.57.20 (talk) 20:10, 3 July 2010 (UTC)

"And the way to end up morbidly obese when you're overweight is to start dieting and wait 10 years." It is true that when some people try to lose weight, they manage to lose some weight but later gain more than they lost. This is more likely to happen if someone tries a diet that is very restricted in food choices or very low in calories. However, someone who is overweight should still try to lose that extra weight that he is carrying. Being overweight increases the risk of health problems, and people who are overweight can become obese if they do not change the bad habits that have made them overweight. Dieting sensibly enables a person to lose weight safely and gradually, with a lower risk of gaining it back. What we want to avoid is yo-yo dieting.174.131.40.152 (talk) 20:50, 3 July 2010 (UTC)

To help answer the original question, someone should ask his or her doctor before trying to lose weight to ensure that losing weight is necessary and safe. Some people can be told that they look fat or are overweight by people they know when, in fact, their weight is within the healthy range.174.131.40.152 (talk) 21:32, 3 July 2010 (UTC)

Suntan

I was playing tennis but I was wearing sandals and did not think to put sunscreen on my feet. Now they have some weird tanned stripes where the holes elt sun in. How long does it take to get "un-tan"? Is there any way I can speed up the process of getting un-tan, whih I assume is my body metabolizing melanin? THNX 68.248.225.254 (talk) 01:57, 1 July 2010

It's going to depend on your exposure to sun and varies from person to person, but it can easily be weeks. It would be quicker to just tan the rest of your feet, but I wouldn't recommend that - intentionally exposing yourself to sunlight is generally not to be recommended. You could try a fake tan, of course. --Tango (talk) 02:38, 1 July 2010 (UTC)

"Intentionally exposing yourself to sunlight is generally not to be recommended." Not recommend by who? Vampires? I am going to go out on a limb here and say that moderate exposure to the sun is generally not a major health concern. And I will base that statement off of the 6,000,000,000 or so people on this planet who encounter sunlight on a regular basis without keeling over. Granted if you are going to overdo it that results in a slight increase in your chance of skin cancer, but that doesn't mean you should not go out during the day. In fact, if you are not exposed to sunlight, you can have vitamin D deficiencies, so telling people not to go out in the sun can be detrimental to their health. Googlemeister (talk) 13:27, 1 July 2010 (UTC)

I said "intentionally". That means going outside for the purpose of exposing yourself to sunlight. Going outside for other reasons and just happening to be exposed to sunlight is a different matter entirely - you have to weigh up the benefit from whatever you are doing with the harm from the sunlight, and as long as you don't overdo it it is often worth going outside. The amount of sunlight required to get enough vitamin D is minimal - in fact, with a decent diet, it can be none at all, at most it's about 10 minutes with only your face and arms uncovered on an overcast day at temperate latitudes (for someone with reasonably fair skin). --Tango (talk) 16:06, 1 July 2010 (UTC)

"about 10 minutes with only your face and arms uncovered on an overcast day at temperate latitudes (for someone with reasonably fair skin)" at a time of day and year when the sun is high enough. Why would 'intentionally' exposing yourself to sunlight be any more dangerous than 'accidental' exposure: the sun doesn't check your intentions. For many people, intentionally exposing themselves to sunlight (because they otherwise live indoors, or in dark areas, or work night shifts, or...) is a useful and beneficial activity. And they don't keel over dead from it. Much advice in the UK (over the last 10 or so years) about sun exposure has been based on Australian practice, and I think we can all see the flaw in applying that to a country like Britain. 86.164.57.20 (talk) 19:03, 3 July 2010 (UTC)

The cells in the epidermis (the outer layer of skin) don't "untan" - they just die and eventually slough off. That process takes 27 days after the cell is first formed. In effect, you grow an entire new skin every four weeks! So if an epidermis cell was 'tanned' right after it was formed then the tan might not completely fade for 27 days. I think that's the worst case - but in practice, I doubt the suns' rays are able to tan those deepest layers of skin, so probably only the outer layer of 'older' cells got tanned - and they'll be around for less than 27 days. SteveBaker (talk) 03:50, 1 July 2010 (UTC)

Wash your feet a lot - it helps get rid of the surface of the skin. 92.28.244.45 (talk) 10:16, 1 July 2010 (UTC)

wth? How do you tan that quick? John Riemann Soong (talk) 16:02, 1 July 2010 (UTC)

A few hours of strong sunlight is more than enough for someone with reasonably fair skin to tan. Remember, the OP wasn't necessarily tanned enough for someone to notice them as being significantly tanned, just enough for there to be a noticeable difference between neighbouring tanned and untanned areas of skin. --Tango (talk) 16:08, 1 July 2010 (UTC)

Consistently wearing sandals in the summer will give you a "sandal tan". I find it usually fades by the wintertime. ~AH1^(TCU) 18:01, 2 July 2010 (UTC)

However, wearing sandals while playing tennis could have much more serious consequences than a "sandal tan"... 67.170.215.166 (talk) 23:03, 4 July 2010 (UTC)

Reptiles + amphibians

I currently have a 29 gallon fishtank that I'd like to transform into a aquaterrarium. I'm looking to drop the water level to about a fourth (and maintain the 7 fish I have now), add some newts and then put in 2-3 thick branches and have anoles in there as well. I'm under the impression that newts possess special toxic slime that they exude upon attack that will protect them from the anoles. Would that be a plausible set up? DRosenbach ^{(Talk | Contribs)} 02:31, 1 July 2010 (UTC)

Well many have poisonous skin eg Rough-skinned newt (also Newt#Toxicity), I suppose this would be easier to answer if you told us the type of newt.

Even so it's not clear that the anole would attack the newt (relative sizes?), or indeed that it would be aware of a poison, or that the poison actually acts as a deterrent (eg bitter) rather than species attrition.

Additionally I would wonder if the anoles and newts would be compatible in terms of their ecosystem - eg dry/damp??

87.102.17.114 (talk) 11:47, 1 July 2010 (UTC)

I would say fire-bellied newts, but I may get another type -- unsure. DRosenbach ^{(Talk | Contribs)} 03:35, 2 July 2010 (UTC)

Surface tension

On the surface of a liquid, there aren't any molecules above the surface molecules, hence they experience an attractive force towards the interior of the liquid. But what causes the increases force tangential to the surface? 70.52.45.181 (talk) 05:34, 1 July 2010 (UTC) Further questions: I'm looking at the following link, http://www3.interscience.wiley.com:8100/legacy/college/cutnell/0471713988/ste/ste.pdf. When considering how to define surface tension, the article makes reference to a C-shaped apparatus. Why is γ = F/2l and not F/l? It says something about there being two surfaces, but I only see one. Second question: In example one, the article considers the surface tension as applying a force outwards. But I thought surface tension was only inwards. I can see why compressing the liquid will produce some outward force against the needle, but why would that equal γL? It seems like those are two different phenomena. Thanks a lot guys! 70.52.45.181 (talk) 06:36, 1 July 2010 (UTC)

ok First question - there isn't really an attractive force towards the centre of the liquid (yes I know the pdf says that).. The molecules on the surface are in equilibrium, so the net force is 0. Obviously if you try to pull a molecule from the surface of the liquid then there will be a force resisting that. maybe this seems pedantic

Following on, still answering the first question: are you familiar with the Energy=force x distance relationship (or Force=dE/dx) - I think it makes surface tension a lot easier to understand... if so you can assume that the molecules at the surface are at a higher energy (since they are surrounded by less molecules .. and molecules gain energy by interacting with those next to them) .. Then assume that this 'surface energy' is proportional to area ie E=kA .. (k is a constant for air/water interface) - if you can work with that then it is fairly easy to show the answer to the second question without "hand waving" - it also proves that water surfaces generally form the shape of lowest surface area (ie a water droplet is spherical).

if not a different explanation will be necessary - but it gets a bit fiddly explaning just with text

Your third question - is this about a needle in water? (ie Example 1) - again - if you consider the surface area you will see that the surface tension acts in such a way to minimise the surface area (in this case it means flat) - so if the needle pushes the water down, the surface tension will try push the water back up into flatness.

Alternatively (3rd question) - try drawing the 'force arrows' in the water surface around the needle (as in diagram b page 2) - you will see that some of the force arrows are pointing upwards (more that 180 degrees of force arrows) - giving a net upwards force.83.100.252.42 (talk) 11:24, 1 July 2010 (UTC)

Sorry, if you don't mind I would like things to be kept to forces and such...grasping something in terms of energy and surface area-minimization is fine, but my understanding of this thing from a force perspective leads me askew, and that's something I have trouble living with. Sorry! 65.92.5.151 (talk) 03:26, 2 July 2010 (UTC) (PS my IP seems to be changing; don't worry it's OP, not an imposter :) ). 65.92.5.151 (talk) 03:27, 2 July 2010 (UTC)

Plaiting and fluffiness

When my hair has been plaited for more than a couple of hours, untwisting the plait and brushing my hair makes it go very fluffy. It eventually "calms down" hours later but can be pretty wild directly after this sequence.

My hair is not normally fluffy and it doesn't "fluff" after brushing any other time.

It's very wavy in thick sections when it comes out of the plait and I'm guessing this has something to do with the subsequent fluffiness, but I'm missing the connection or reason. Does anyone know why plaiting produces the effect? I tried searching for an answer, but it seems plaiting is a common answer to fluffiness and I therefore get a lot of false positives, :) Maedin\^talk 06:57, 1 July 2010 (UTC)

Does a penis have other functions besides urination and copulation?

Does penis have any other functions besides urination and copulation? For example, helping maintain balance as a man ealks (nude), or being a primary way to get off access heat during a heat wave, or being a sensory organ regarding thr environment, like a nose, or - anything at all? 92.224.206.50 (talk) 07:21, 1 July 2010 (UTC)

Re "sensory organ": Two men were peeing off the end of a pier at night when one remarked, "Gee, the water is pretty cold." The other replied, "And deep, too." -- 60.49.38.251 (talk) 09:58, 1 July 2010 (UTC)

Yes, read this Small penis rule  Jon Ascton  (talk) 09:12, 5 July 2010 (UTC)

Writing one's name in the snow. Cuddlyable3 (talk) 10:36, 1 July 2010 (UTC)

Masturbation? Which is used by some as "stress reliever"? --Enric Naval (talk) 10:40, 1 July 2010 (UTC)

The penis provides the site for visible circumcision that is performed for ritualistic or religious purposes. Any usefulness for maintaining balance is disproven by the considerable numbers of women and men with tight trousers who do not fall over. Cuddlyable3 (talk) 10:43, 1 July 2010 (UTC)

As far as balance goes, if it's long enough, you can use it as a tripod. --Enric Naval (talk) 10:56, 1 July 2010 (UTC)

Its dysfunctions and complications can be useful indicators of health problems; hypertension, for example, or a hormonal imbalance. More indicative, of course, if the onset of said side effect is sudden and/or the age isn't elderly. Though clearly not its primary application, it's still a gauge that females don't necessarily have (or at least not as obviously). Maedin\^talk 12:33, 1 July 2010 (UTC)

It functions as a differential analyzer of the attributes "surprise" and "panic". Surprise is the first time it won't the second time and panic is the second time it won't the first time. Cuddlyable3 (talk) 13:13, 1 July 2010 (UTC)

It's also a good MacGuffin for troll-like questions. --Tagishsimon (talk) 09:45, 2 July 2010 (UTC)

Credibility of hypotheses

What is it about a particular scientific hypothesis that makes it likely to reflect reality, or unlikely to reflect reality?--220.253.100.166 (talk) 08:49, 1 July 2010 (UTC)

Experiment :) --Dr Dima (talk) 10:12, 1 July 2010 (UTC)

See the article Falsifiability. Cuddlyable3 (talk) 10:34, 1 July 2010 (UTC)

A hypothesis is just an idea - until it's tested in some way, we don't know whether it's true or false. However, some hypotheses are more likely to be true than others. For example: You might hypothesize that I'll type the letter "e" more than once in the next sentence. Sadly, that turns out to be wrong. It was a good hypothesis - but until it was tested, we didn't know whether it was true or false. In general, a hypothesis that requires a violation of the known laws of physics/chemistry/whatever is much lessmore likely to be false than one that doesn't. Overturning some piece of well-established science is much less likely than proving something that's already known to be possible. If you hypothesize that T.Rex was able to run at 20mph, that's not an impossible thing - and by measuring fossil bones and looking for T.Rex footprints - you might be able to prove it...but if you hypothesize that the jet of gasses from a pulsar shoot out at twice the speed of light - then we're very nearly certain that this is false - just from general physics principles. SteveBaker (talk) 19:09, 1 July 2010 (UTC)

Only an omniscient being could answer this question, since the rest of us have no sure knowledge of the underlying reality. The hypotheses are the only scientific guide we humans have to reality. Occam's razor is helpful in keeping things simple. William Avery (talk) 19:54, 1 July 2010 (UTC)

Career in nanotechnology?

[I'm asking this question on behalf of a friend] I'm almost done my undergraduate studies in physics and I have started, for the first time really, to consider my career options. I've always enjoyed physics, and I would absolutely love to continue doing research in it, but I have been cautioned about the dangers of entering a overly theoretical discipline of physics. But there are plenty of fields with "real-word applications" that are certainly very interesting. In particular, something like condensed matter physics or theoretical chemistry seem worthwhile to pursue. However, a few weeks ago an old teacher of mine suggested that I look into nanotech, nanoelectronics in particular. It seems fascinating, and the technology and lab equipment are very impressive. My sole caveat is this: will I lose the physics I enjoy if I go into this? I want to work with the equations and study the mechanics of things...will that be jeopordized in something like nanotech? Thanks. 70.52.45.181 (talk) 10:52, 1 July 2010 (UTC)

Nanoelectronics involves or can involve a lot of condensed matter physics, and some theoretical chemistry (of the right sort) is also applicable - so your friend can carry on as normal :)

83.100.252.42 (talk) 11:31, 1 July 2010 (UTC)

Nanotech is a wide range of different things - from the very practical stuff like improved water filtration for use in the third world - to the highly theoretical stuff like Molecular assemblers. See List of nanotechnology applications. These days, saying that you want to work in Nanotechnology is like saying you want to do Math...it's used all over the place in a vast range of industries. I think this means that you should be able to find a niche somewhere between the so-practical-it's-just-engineering to the so-unlikely-that-research-funding-is-unobtainable. Somewhere in that spectrum, you should be able to find a job that pays well and lets you do the science you love. In case you didn't already check it out, we have lots of articles on the subject, including Nanoelectronics itself - and also Molecular electronics, Molecular logic gate, Molecular wires, Nanocircuitry, Nanowires, Nanolithography. NEMS, Nanosensor, Nanoionics, Nanophotonics and Nanomechanics. SteveBaker (talk) 18:57, 1 July 2010 (UTC)

Thanks to both of you for the informative and detailed responses. 65.92.5.151 (talk) 03:23, 2 July 2010 (UTC)

Wasps

A wasp (either a yellowjacket or common wasp, I think) somehow got into my bed this morning, with predictably hilarious consequences. While I treated the stings with hydrocortisone cream, they were still extremely painful, even after the swelling had gone down. So, my question: Bees use apitoxin and wasps apparently use "a chemically different venom designed to paralyze prey, so it can be stored alive in the food chambers of their young". What is the mechanism which causes the prolonged pain in bee and wasp stings? Apitoxin says that bee venom contains proteins which cause inflammation, but that some of the components are anti-inflammatory agents. Why this apparent contradiction? --Kateshortforbob _talk 11:10, 1 July 2010 (UTC)

I'm no doctor but from the sound of the article, apitoxin is multi-faceted. Some ingredients are anti-inflammatory while others are inflammatory, but not all work at the same time or on the same tissue. I suspect this dichotomy is part of why it hurts so much, the venom does different things at different stages ensuring that the wound will continue to cause pain and elude the body's self defense mechanisms. --144.191.148.3 (talk) 18:52, 1 July 2010 (UTC)

A hydrophobic cup?

What if a drinking cup has its inner surface lined with a hydrophobic material and water is poured into it? 67.243.7.245 (talk) 12:34, 1 July 2010 (UTC)

You've never drunk out of a polystyrene cup? Physchim62 (talk) 12:41, 1 July 2010 (UTC)

Cup holds water then. This happens a lot - paper cups with water proof coatings for instance. Was there some particular detail you where interested in? Sf5xeplus (talk) 12:54, 1 July 2010 (UTC)

Paper cup describes all the hellish details in full measure.Sf5xeplus (talk) 12:55, 1 July 2010 (UTC)

But water remains in a PS cup. It's not completely hydrophobic then? 67.243.7.245 (talk) 20:53, 1 July 2010 (UTC)

It is completely hydrophobic. Hydrophobic materials do not counteract the force of gravity; they merely are not miscible with water. That's all hydrophobic means. It just means that the material does not dissolve in or mix with water. They do not repel each other with a force in the way that, say, two north poles of a magnet will. The simply don't mix. When oil (a hydrophobic substance) and water are shaken, the seperate, but not from any force more complex than gravity; the less dense oil floats to the top because it is less dense, and thus lighter per unit volume, than the water. They aren't "forced apart" by any force. Likewise, water placed in a hydrophobic cup will simply sit in the cup and not soak through it. If you made a cup of hydrophilic material, like say salt, the water would dissolve the cup, making it not very useful as a cup. --Jayron32 05:54, 2 July 2010 (UTC)

You will get a convex Meniscus. Conversely, in a hydrophilic cup the meniscus will be concave. Ariel. (talk) 14:50, 1 July 2010 (UTC)

the cup shivers uncontrollably and cowers away from the water, if you persist in pouring water onto it, it "shuts down", curling into the fetal position and crying silently. If it is a prolonged experience in which the cup cannot get away, it can be permanently deformed psychology, perhaps hardly speaking a word for the rest of its life (or until treated). medications can help, but "water therapy", the naive idea that you should fight fire with fire, and just immerse the cup completely until it overcomes its fear, has disastrous results and is totally discredited; don't even think about it. 92.230.234.237 (talk) 13:08, 1 July 2010 (UTC)(Nonsense shrunk. Cuddlyable3 (talk) 13:58, 3 July 2010 (UTC))

be VERY careful around hydrophobic cups, in case they bite you.... Physchim62 (talk) 16:15, 1 July 2010 (UTC)

Help identifying this fruit

Unknown fruit.

I took this picture of a fruit on a tree in the botanic gardens in Oxford in February and have absolutely no idea what it is! Can anyone help? The fruit itself was around 4-5 cm diameter. Unfortunately as it was in the botanic gardens I have no idea of where in the world it is native to and the tree itself was fairly nondescript; ~7-10 metres tall with a trunk diameter of around 40-50 cm. The time of the year also meant there were no leaves for reference although the fruit was happy to stay on the tree over the winter - it was absolutely covered in these fruits. - Zephyris _Talk 13:21, 1 July 2010 (UTC)

Umm... not to be glib but if you were at the Oxford Botanical Gardens then why did not you just ask a botanist? 76.199.154.122 (talk) 17:46, 1 July 2010 (UTC)

Well I took a picture of what I thought was the species label for the tree... Turned out to be the species label for the miniature daffodils under the tree! - Zephyris _Talk 19:30, 1 July 2010 (UTC)

Do you remember if the tree had thorns? If it did, this might be a Bael. Googlemeister (talk) 18:26, 1 July 2010 (UTC)

I don't think it did, see the additional image. - Zephyris _Talk 19:30, 1 July 2010 (UTC)

Here's the number: 01865 286 690 you can ask them what's the tree above the daffodils. 71.161.46.51 (talk) 21:03, 1 July 2010 (UTC)

You might also find their website useful. It includes an interactive map (which regrettably does not identify individual plants) and an e-mail address for queries. 87.81.230.195 (talk) 05:43, 2 July 2010 (UTC)

They haven't been much help have they? My money is on the Dove-Tree or Ghost-Tree, Davidia Involucrata[2]. "Fruit: 3x2.8cm, obovoid (ie egg-shaped with the big end at the opposite end to the stalk), much ribbed, deep green and slightly glossy, ripening dark purple. Pedicel (ie stalk) 10-14cm, much swollen at the fruit end." (A Field guide to the Trees of Britain & Northern Europe, Alan Mitchell, Collins 1974). The English names come from the spectacular large white oval bracts in late May. "Also regrettably known as the 'Handkerchief Tree'" says Mr Mitchell. Discovered in W. China in 1904; the remarkable story of it's introduction to Western science can be found here[3], pages 40-41. A photo of the Oxford Botanical Dove Tree in full foliage is here[4]. Alansplodge (talk) 18:48, 2 July 2010 (UTC)

The dove-tree looks very plausible! I think I would put my money on that... - Zephyris _Talk 22:20, 2 July 2010 (UTC)

Found another picture[5]; scroll down to page 14. Is this the one? Alansplodge (talk) 22:44, 2 July 2010 (UTC)

That is definately the one! Nice detective work, thanks! 95.147.73.134 (talk) 20:49, 3 July 2010 (UTC)

Could it be a variety of walnut? Walnut shells are hidden inside a covering that looks like a fruit when growing. 92.28.247.183 (talk) 20:03, 2 July 2010 (UTC)

I thought of that but walnut trees have big chunky twigs[6], the nuts have short stalks and no ribs. Alansplodge (talk) 22:39, 2 July 2010 (UTC)

Would the car explode?

If something happened and a car's intake valve didn't shut all the way (or the valve suddenly developed a serious fissure while the car was in use) and the compression stroke came back up and the spark plug went off, is there anything stopping the combustion flame from streaking back through the lines to the fuel tank and causing a catastrophe?20.137.18.50 (talk) 14:04, 1 July 2010 (UTC)

The fuel line is full of liquid fuel. You need air in order to support combustion, so it would be hard to propagate the flame back very far prior to where air is mixed with the fuel vapors. DMacks (talk) 14:09, 1 July 2010 (UTC)

A Carburetor if there is one, also makes a barrier since the fuel must go through a narrow nozzle.87.102.17.114 (talk) 14:26, 1 July 2010 (UTC)

If a valve gets stuck open, you don't get compression and the fuel doesn't ignite properly. Aside from the fact of there being no air in the fuel lines, there isn't likely to be any combustion either. What happens is that the car sputters along, sounding really rough and developing very little power, because it's only running on three cylinders (assuming it's a four cylinder car that is!). Gasoline really doesn't burn very easily unless it's under hot or under pressure and in vapor form mixed with air. SteveBaker (talk) 18:39, 1 July 2010 (UTC)

A backfire in the intake is not unheard of on poorly maintained cars for a variety of reasons (such as timing, like you mentioned), and while it can easily damage intake manifold parts like rubber seals or sensors it is not likely to cause the fuel lines to catch fire for the aforementioned reasons. --144.191.148.3 (talk) 18:42, 1 July 2010 (UTC)

Yep. Remember - the car is a four-stroke machine. On the first down-stroke, gas/air is sucked into the cylinder. On the up-stroke, it's compressed...only the valve isn't closed so this just pushes most of the gas/air mixture back out again. Then the spark goes off - but there's hardly any gas/air mixture to ignite - and it's not hot and under pressure like it should be - so it probably won't ignite. The lack of a fuel burn means that there is nothing to push the piston back down again - aside from the inertia of the car and flywheel...but it'll go back down - sucking in whatever fuel and air is left in the cylinder. Then the exhaust valve opens and the piston goes back up - pushing most of whatever was in there out to the exhaust. But not all of the exhaust gasses will go out - lots of them will be pushed back to the carb. SteveBaker (talk) 18:45, 1 July 2010 (UTC)

Some cars have a Two-stroke engine that typically has no intake valve. Cuddlyable3 (talk) 13:47, 3 July 2010 (UTC)

How much energy required to reach infinite speed?

I can't remember the equation for how much energy is required to reach a given speed. What is it again? And what do you get when you feed infinity through it? Presumably no answer, since reaching c takes infinite energy.--92.251.129.172 (talk) 15:04, 1 July 2010 (UTC)

E=1/2 mv² where E=kinetic energy, m = mass and v= velocity. So, yes, at v= infinity, E = infinity. This is the non-relativistic definition of energy and velocity, so I am sure there are adjustments to be made for speeds close to light speed, however these adjustments actually make it worse; you hit infinite energy at the speed of light (3.00 x 10⁸ meters per second). So, you cannot ever reach the speed of light. It's a pain in the ass for science fiction authors, but the speed of light is a hard limit. --Jayron32 15:13, 1 July 2010 (UTC)

I thought they usually invented some warp drive or hyperdrive that somehow let them past lightspeed?--92.251.129.172 (talk) 15:25, 1 July 2010 (UTC)

Such things are utterly impossible given the laws of physics as we know them. But science fiction is fiction (meaning "it ain't all true"!)...and they are free to invent anything they want in order to make the plot work...and hand-waving away the speed of light limit is something you pretty much have to start off doing if you want people to go to other stars. A few science fiction authors don't do that...but they are certainly in the minority. SteveBaker (talk) 18:27, 1 July 2010 (UTC)

Hence "invented".--92.251.158.103 (talk) 19:59, 1 July 2010 (UTC)

Yes. The one E.E. Doc Smith invented for Lensmen just got rid of the pesky m in the above equation, so you only have to overcome the resistance of the interstellar medium. Unfortunately, they answer still is infinite (and the original question is physically meaningless - you cannot go faster than the speed of light without violating causality, no matter what your trick is). --Stephan Schulz (talk) 15:29, 1 July 2010 (UTC)

Well, there's a hidden assumption there that your "trick" is frame-independent. It's imaginable that there is some as-yet-undiscovered physical process that is not frame-independent, and that defines a preferred frame of reference which we have not yet been able to detect. Then you might be able to use that process to send information at arbitrary speeds in that preferred frame, but unlike in the case of the tachyonic antitelephone, you might not be able to "close the loop" to send information back to where it started from (which would allow the grandfather paradox).

In this scenario (for which, certainly, I am not claiming there is any evidence at all; this is a thought experiment), there would still be some observer, in motion relative to the preferred frame, who could observe an effect occurring before its cause in his time coordinate. But so what? That's just a coordinate. As far as I can see, there is no danger of paradox unless you can create a causal loop.

Another possible way out is, keep frame invariance, but throw out isotropy — the magic FTL drive works only when you're headed towards cosmic right; the usual speed limit applies to cosmic left. --Trovatore (talk) 03:44, 2 July 2010 (UTC)

The relativistic formula for kinetic energy is:

$E_{k}={\frac {mc^{2}}{\sqrt {1-(v/c)^{2}}}}-mc^{2}$.

Formally, setting v to infinity gives you a negative kinetic energy of -mc² ! But this is meaningless, because E_k tends to infinity as v approaches c, so you can't go through the "light barrier". Gandalf61 (talk) 15:30, 1 July 2010 (UTC)

so we can create a perpetual motion machine! All we need to do is get something to infinite velocity, and it will generate tons of free energy! Or maybe we need to use negative energy to get something going that fast. Googlemeister (talk) 15:43, 1 July 2010 (UTC)

No - none of those things work. The answer isn't infinity - it's the square root of some negative quantity. If (v > c) then (v/c)² is greater than one - so 1-(v/c)² is negative. If you take the square root of a negative number on your pocket calculator, it probably says "E" or "Error". You get a complex number - and those can't exist as actual quantities in the physical world - whenever you find a complex number as the final answer to a calculation, you know you've screwed up somewhere. That's WHY this equation says you can't travel faster than light. If the answer was merely infinity, we could possibly imagine some kind of meaning to the answer...but a complex number means "your math is broken somewhere" - and the reason it's broken is because you put a disallowed number into the equation. SteveBaker (talk) 18:27, 1 July 2010 (UTC)

I understood the answer would be meaningless I was jsut curious about what it would be.--92.251.129.172 (talk) 18:41, 1 July 2010 (UTC)

As v approaches infinity, the imaginary part decreases without limit. Thus, the energy would be real if it's moving at infinite velocity. Of course, from any other point of reference, it's going at a finite, but faster than light, velocity, and it comes out imaginary, so it doesn't really help.

You can question why you would even want to move your rest mass. Arguably "you" are an algorithm that your brain is computing and that can be sent using photons from one machine to another machine at lightspeed. So, travel at lightspeed is possible. Count Iblis (talk) 15:47, 1 July 2010 (UTC)

Well what they mean is matter cannot move at lightspeed. Although you could claim that since we are moving away from a certain galaxy at 0.6 c, and that same galaxy is moving the other way at 0.6 c, we are moving apart faster than the speed of light. ALthough that still isn't going the speed of light.-92.251.129.172 (talk) 15:53, 1 July 2010 (UTC)

You could claim that, but you'd simply be wrong, because relativistic velocities (i.e. ones that are a substantial fraction of the speed of light) don't obey simple arithmetic where 0.6 + 0.6 = 1.2; they obey more complex arithmetic where the answer can never be greater than 1.0: this seems counter-intuitive to us because our everyday logic is based on everyday experience which never includes relativistic velocities, but it has been extensively proved by observations and experiments. 87.81.230.195 (talk) 18:09, 1 July 2010 (UTC)

As I understand it, though, there is a sense in which a galaxy beyond the event horizon can be said to be moving away from us faster than the speed of light. It gets tricky to state exactly what this means. That general sort of question belongs to general rather than special relativity, and the math is hairy, and there's no longer always as clear a choice of coordinate systems as you might expect. --Trovatore (talk) 01:41, 2 July 2010 (UTC)

I think you mean a galaxy outside of our light cone. The event horizon is something different entirely. --Jayron32 01:53, 2 July 2010 (UTC)

What I mean is outside the observable universe in a certain sense. I have never entirely gotten straight what is the accepted terminology on this, but our observable universe article appears to be talking about all objects that could potentially have been affected by an event in the past, by which we could also have been affected. I mean the time-reversed notion — all objects that could potentially influence a future event that we could also influence. The language at particle horizon suggests that event horizon is used in this sense. --Trovatore (talk) 01:57, 2 July 2010 (UTC)

There is a connection with light cones, but no, "outside our light cone" is not the point exactly. The point is that our forward-facing light cone, and the forward-facing light cone of the galaxy in question, have empty intersection. --Trovatore (talk) 01:59, 2 July 2010 (UTC)

They're not moving, or at least not that fast. It's just that the intervening space is expanding. — DanielLC 07:42, 2 July 2010 (UTC)

Well, I was careful to say "there is a sense" in which it's moving away from us faster than light. --Trovatore (talk) 07:49, 2 July 2010 (UTC)

question about new planet

Time is talking about a new planet that someone got a photo of that is 500ly away. This planet is described as orbiting it's star at 330 AU, and has a surface temp of 2700K. To me this sounds more like a red dwarf star then a planet. Could someone familiar with the methodolgy give me a rough idea of how they can differentiate between a large planet and a very small star in orbit around another star (like Proxima Centauri)? Googlemeister (talk) 15:41, 1 July 2010 (UTC)

Make life easy for us and give us a link to the article, please!

This other article (from National Geographic) has a bit more detail about the planet. Its mass has been calculated at roughly eight times the mass of Jupiter. That's certainly big, but it's not 'star' big. Our own article on stars discusses the minimum stellar mass, which is something like eighty times the mass of Jupiter. The boundary is governed by the minimum mass required to sustain stable nuclear fusion in the stellar core; the gravitational attraction holding the star together has to be high enough to balance the pressure generated by core hydrogen fusion. There is a gray area as you go to objects with masses lower than that, but bigger than gas giants: the brown dwarfs. The intro to that article discusses the definitions (and the challenges associated with assigning objects to one category or the other). In general, the boundary between large gas giants and small brown dwarfs is taken to be at around 13 Jupiter masses.

Finally, the high surface temperature of the planet in the story is a bit of an anomaly. It's not being heated (significantly) by internal fusion, nor is its heat drawn from the parent star. Instead, this is a very young planet (just a few million years), and its very high temperature comes from gravitational collapse. (The gravitational potential energy freed up as matter fell in to form the planet appears as heat.) The young Earth went through a similar period as it was forming billions of years ago, it took millions of years before the crust cooled and solidified. TenOfAllTrades(talk) 16:00, 1 July 2010 (UTC)

I do kind of have to agree, though, that this is not what I think of as a referent for the word planet. Really I think the IAU got the definition wrong in a number of ways — rather than distinguish between "planets" and "dwarf planets" on the basis of this fairly silly clearing the neighborhood concept, it would have been more revelatory to draw a line between the "real" planets (that is, the rocky planets), and the gas giants, which are just another sort of cat entirely. --Trovatore (talk) 02:09, 2 July 2010 (UTC)

Trovatore and his pet peeve. Dauto (talk) 04:07, 2 July 2010 (UTC)

Why don't we write to the IAU and tell them to replanetize Pluto? 67.170.215.166 (talk) 01:11, 4 July 2010 (UTC)

1RXS J160929.1-210524, btw. --Sean 16:20, 1 July 2010 (UTC)

Trench binocular

The german de:Scherenfernrohr has no interwikis - but I think this instrument is well known. See the picture on the right. Can you help? --Eingangskontrolle (talk) 16:56, 1 July 2010 (UTC)

If you are asking what the instrument is, it is a periscope.--Shantavira|^{feed me} 17:08, 1 July 2010 (UTC)

Periscope seems closest, or else binoculars. The first talks about use in trench warfare but does not mention the design that has separate periscope for each eye of the binocular. The second, in binoculars#Military, specifically mentions this design and application, but is only a small part of a larger and wider-ranging article mostly about binoculars not the key periscope idea. DMacks (talk) 17:11, 1 July 2010 (UTC)

...and it's not even mentioned in trench warfare. DMacks (talk) 17:17, 1 July 2010 (UTC)

There doesn't seem to be a specific article in the EN Wikipedia. They're mentioned in passing, as "trench binoculars", in the binoculars article. Searching Google, there doesn't seem to be a general English language term for them, beyond "trench binoculars" or "rabbit ear scope". -- Finlay McWalter • Talk 17:13, 1 July 2010 (UTC)

Can someone who can read German more fluently than I check if de:Scherenfernrohr and its cited ref seem notable enough to make a Trench binoculars article? Although File:British trench periscope Cape Helles 1915.jpg looks like a simple (monocular) design not binoculars...hard to see clearly, so maybe better to do trench periscope instead of specifically binocular. DMacks (talk) 17:15, 1 July 2010 (UTC)

The link is available in english http://home.arcor.de/thuernagel/sf14-e.htm same person has a whole set of articles on "military optics" (or 'tactical optics') III Taktische Optiken http://home.arcor.de/thuernagel/katalog.htm . There's a mixture of reliable info and some small amount personal reflection/speculation in their (which is all probably completely right) website.

Also known as "donkey ears" in the UK.

Maybe a redirect to a new extended section in periscope? or a complete new article on 'tactical optics'?

I can't see enough there as it is.. though the article would probably survive on it's own as a stub on the assumption that there is more info out there, and the obvious fact that such items are of social-historical significance.87.102.17.114 (talk) 18:03, 1 July 2010 (UTC)

It's my understanding that they're a bit more than just binoculars with a periscopic lightpath. I believe (but can't find a reliable source to adequately support, otherwise I'd have started on the article) that the rabbit-eared variety (like this) are intended for field artillery spotters. My understanding is that the two lenses independently targettable; the spotter picks his target and aligns both lenses on it - he can then read off the angle from a little gauge, look up the corresponding distance in a little trig table, and report that to the gunners. It's obviously not as accurate as a proper survey with two theodolites separated by a measured baseline, but it's going to be better (and quicker) than just guessing and having to walk trial shots up and down. -- Finlay McWalter • Talk 17:41, 1 July 2010 (UTC)

I think it's just for observers - the wide position gives better stereoscopic view (ie less foreshortening) (Is this correct?)

You're thinking it's a rangefinder - the person says Scherenfernrohre are the predecessors of the stereoscopical rangefinders - the same site has info on these - they can look very similar - but are more meaty - eg http://home.arcor.de/thuernagel/em61.htm .. they could use the thing as a rough rangefinder though,as the article says.87.102.17.114 (talk) 18:03, 1 July 2010 (UTC)

Some info with pictures through in english here http://www.paulstiger1.co.uk/WWII-Optics-Collection.htm and here http://www.fieldgear.org/optics.htm 87.102.17.114 (talk) 18:16, 1 July 2010 (UTC)

The internet eh http://www37.atwiki.jp/strike_witches/m/plugin/ref/?guid=on&serial=921&w=500 ... 87.102.17.114 (talk) 18:24, 1 July 2010 (UTC)

If anyone is interested in writing the article on these particular type of thing then de:Artillerietruppe_(Wehrmacht) is useful - it says that they were used by forward observers for artillery.

Also searching "scissors telescope" shows that they were also used attached to tanks (with reliable sources, not forums).87.102.17.114 (talk) 19:04, 1 July 2010 (UTC)

It's a http://en.wiktionary.org/wiki/telestereoscope too.87.102.17.114 (talk) 19:04, 1 July 2010 (UTC)

Do the positions of stars change much relative to each other?

Will Alpha Centauri always be "about" the same distance from Earth? Suppose one picked any two stars at opposite sides of the galaxy. Does their position change much relative to each other? —Preceding unsigned comment added by 92.251.129.172 (talk) 18:28, 1 July 2010 (UTC)

The Wikipedia articles for many stars, such as Alpha Centauri gives their proper motion in the infobox at the right. -- Finlay McWalter • Talk 18:32, 1 July 2010 (UTC)

So the sky will be very different in a million years time?--92.251.129.172 (talk) 18:34, 1 July 2010 (UTC)

Yes; this page has movies of the Big Dipper over 200,000 years, and it changes substantially in that time. -- Coneslayer (talk) 18:49, 1 July 2010 (UTC)

See radial velocity for a sense of how the distances of stars change over time. Stars have even swung very close to the solar system in the past, and will likely do so again in the future—see Gliese 710 (negative radial velocity) for example. ~AH1^(TCU) 17:42, 2 July 2010 (UTC)

Tori Vienneau

was the Tori Vienneau murder case a suicide  ?

http://www.youtube.com/watch?v=LM6bBVd0f7s —Preceding unsigned comment added by Alexsmith44 (talk • contribs) 18:30, 1 July 2010 (UTC)

Why are you asking us here? We're scientists, not Columbo. 87.102.17.114 (talk) 18:35, 1 July 2010 (UTC)

law —Preceding unsigned comment added by Alexsmith44 (talk • contribs) 21:08, 1 July 2010 (UTC)

World building.

I'm sure questions like this have been asked many many times here, but I think mine is a little different.

Let's say I'm making a fictional planet and I want to make it as scientifically realistic as possible. I think I can handle most of the biological and geological topics, but mostly what I wanted to know about was meteorological and climatological. Things like ocean currents, jet streams, air masses and weather patterns, etc. What would the reference desk suggest as guidelines for creating the weather patterns on a fictional planet?

Also, I think I can handle most of the geology, but what confuses me most is plate tectonics. I can grasp the basic concepts of rifting and subduction and so on, but I don't think I know enough to create a fictional tectonic system on a fictional planet. This is important because drawing the maps of my world requires being consistent with tectonics.

63.245.168.34 (talk) 19:42, 1 July 2010 (UTC)

Make it pretty similar to Earth and you should be fine. If you stray from that, then really it's anybody's guess. We've only studied one planet in any great detail, so we don't know what else is possible. --Tango (talk) 19:52, 1 July 2010 (UTC)

Are you sure about that? We know that weather patterns are defined by some pretty basic physical laws like the Coriolis Effect and so on, and we're familiar enough with these rules and how they effect the atmosphere and the oceans that we can predict the weather with reasonable certainty. Also, don't you think that making it too similar to Earth would be very unlikely? 63.245.168.34 (talk) 20:05, 1 July 2010 (UTC)

Not necessarily, perhaps scientifically, a planet can only be inhabitable if it is similar to earth. I mean how likely is it for life to develop on a planet with almost 0 tectonic activity? Of course if the book is not about humanss (or similar to human) colonizing a planet, but about a chlorine breathing insect race, then of course all bets are off. Googlemeister (talk) 20:57, 1 July 2010 (UTC)

There is a whole book on this, created just for you: World-Building: A writer's guide to constructing star systems and life-supporting planets. There are other books in the series with titles like Aliens and Alien Societies. (I haven't read World-Building but have read Aliens. The latter covers a lot of bases and made me consider aspects I hadn't thought about.) Comet Tuttle (talk) 21:16, 1 July 2010 (UTC)

This is actually very difficult to deal with in a comprehensive way -- just consider how much the Earth's climate has varied over the past billion years, mainly as a consequence of the continents moving around. It is widely believed, for example, that the Earth was in a snowball state from 800-600 million years ago, with the oceans frozen over nearly down to the equator, due to all of the continental landmass being concentrated near the equator. In short, you can get anything from rainforests at the poles to glaciers at sea level at the equator, just by shifting the land around. Looie496 (talk) 22:41, 1 July 2010 (UTC)

That's mostly what I'm curious about. How does the shape of the land affect the climate?

Also, as for the planet being inhabited or inhabitable, in the context of my fictional universe, it's a terraformed planet with a young, engineered biosphere. Only inhabited by species that were brought to it by humans. There are no aliens in my fiction.

63.245.168.34 (talk) 00:12, 2 July 2010 (UTC)

Land has profound effect on climate. Generally speaking, the direction of prevailing winds over landmasses determines things like where deserts and forests will appear. If prevailing winds have a long Fetch over water, then the land downwind tends to be much rainier than if the winds have a long fetch over land. Also, there are effects like rain shadow, which makes land upwind of a mountain range rainier than land downwind of the mountain range; which is why Seattle is much rainier than say Spokane. --Jayron32 01:06, 2 July 2010 (UTC)

Also, land tends to heat up (and cool down) far quicker than water: you're better off swimming in the Mediterranean in October than in April, even if the land temperatures are similar. Physchim62 (talk) 01:58, 2 July 2010 (UTC)

From observations on exoplanets, astronomers can deduce a thermal "cimate" map of some planets, as they are hotter in some areas and cooler in others. However land and ocean have not yet been confirmed on any exoplanets, not even Super-Earths, so the hypothetical planet should be like Earth in the sense that it has mountains and oceans. As for the effects of land patterns on the overall climate of regions, here is a question I posted here way back, when I was proposing a way to model the climate that is probably much less complicated than supercomputer simulations but do not involve mathematical formulae and calculations. As I never got an answer, I'm not sure how accurate my method is, but it should be a good general approximation of the climate of a planet given its land masses, but its functioning is reduced when less is known about the planet. Regular weather patterns can be inferred from the location of quasi-stationary highs and lows, but there will always be great variations. We know the general climate of Mars, yet we still don't have the ability to predict global dust storms that often occur on the Red Planet or what exactly triggers them. Also, if the climate changes on your planet, the effects of that change are likely going to be harder to predict than just the climate map in a static climate. ~AH1^(TCU) 17:24, 2 July 2010 (UTC)

Micro rockets

Suppose we want to send a miniscule payload from the surface of the Moon to the Space station. Then it seems to me that at least in theory, the rocket could be made very small. So, what is the minimum mass of a rocket that is able to let a payload of mass m (assume m < 1 microgram) escape from the Moon's gravity, as a function of m? Count Iblis (talk) 22:56, 1 July 2010 (UTC)

What kind of fuel do you plan to use? By mass, do you include everything involved or just the rocket (projectile) that goes up into the air? I ask because different fuels have different masses. Also, assisted blastoff reduces fuel required. Further, the length of thrust is important. Do you want an explosion or a controlled thrust? -- kainaw™ 23:11, 1 July 2010 (UTC)

For a payload that small, the payload is irrelevant. What matters is how light an engine you can make. The smallest rocket able to lift a payload of a microgram will be essentially the same size as one able to lift a payload of a few hundred grams. That said, I can't think of anything with a mass of less than a microgram that I would want to send from the Moon to a space station, certainly not on its own - just put it in with something else that is going. --Tango (talk) 23:15, 1 July 2010 (UTC)

I see, so it is not possible to make microscopic engines? Count Iblis (talk) 00:59, 2 July 2010 (UTC)

I wouldn't say it's impossible, but I don't know of anyone ever doing so. There just isn't any reason to try. --Tango (talk) 01:19, 2 July 2010 (UTC)

Just fire it from a rifle? (I don't know how the space station would catch the bullet!) Dbfirs 07:13, 2 July 2010 (UTC)

See Space gun. It's a lot harder to get things into space than people on here seem to realize. --Mr.98 (talk) 11:37, 2 July 2010 (UTC)

This question is about launching from the Moon into the Moon's orbit. Gravity is 5/6 that on Earth and orbital distance is about 1/4 that of Earth. So, there are two things making it much easier. Gravity is much less. Orbit is much closer. Further, there is almost no atmospheric drag. So, I do not see why shooting a small projectile into orbit would be a big deal. -- kainaw™ 12:24, 2 July 2010 (UTC)

I've read that there's a limit to how small you can make a rocket. Unfortunately, they didn't actually say what the limit was. That said, getting it off the moon is much less difficult then getting it off the Earth. — DanielLC 07:32, 2 July 2010 (UTC)

Bottle rockets are pretty small... Googlemeister (talk) 18:17, 2 July 2010 (UTC)

In The Andromeda Strain book and movie, they dealt with this briefly (much more so in the movie), where an extraterrestrial could live as this infectious microbe such that only one cell needs to be launched into space, then becoming a superorganism upon arrival and mass infection on a habitable planet. A suggestion was that there would be any number of evolutionary advantages for a cell-colonial intelligence to be successful, one of which was the ease of resources for going into space and colonizing. That's getting off-topic, but that's why I find this question interesting.

I think the best way to get into space for that, then, is to begin with high-altitude balloons filled with hydrogen (or rigid vacuum). An easy calculation: the density of air is 1kg/m^3, so a little over one cubic meter of balloon is required to lift 1kg of payload. For a single cell in a colonial organism (density about water, 1000kg/m^3), we can easily find out the kind of balloon we need, though considering that we could wrap our organism in a (10-micron)^3 natural foam bubble filled with hydrogen released naturally from alkali oxidation, it's easily imaginable that organisms would get up to the edge of space by such means without intelligence to help. Of course, the smart cell would the burn the hydrogen once reaching the edge of space for a friction-free ride wherever, though they still have to break escape velocity to travel outside their solar system.

At 1km up we orbit the Earth at about 8km/s, and we start hydrogen burning with specific impulse of about 1000s to try to reach escape velocity 11km/s. So for payload mass m, we need to burn about 0.4m in fuel hydrogen, which is a really good deal since we can get as much fuel as we want up there with these balloons. I feel I must be missing something. SamuelRiv (talk) 02:57, 4 July 2010 (UTC)

Why does the ocean appear so high?

I took this picture of the Atlantic Ocean from an elevated train.[7] Am I unfamiliar with nature or does the ocean appear elevated especially towards the horizon? 67.243.7.245 (talk) 23:31, 1 July 2010 (UTC)

We do not know how the camera was oriented. The technical term for this parameter is the elevation of the camera - the angle it makes with respect to the horizon - and not to be confused with its elevation above sea-level. Obviously, by angling a camera, you can make the horizon appear to be at any vertical position inside the image. In addition, there may be an optical illusion of distorted perspective if the train's window has a weird aspect ratio or orientation, or if the train was actually tilted when you photographed the scene. The train was high above the ground, (the other kind of elevation), then you may see a lot more land in the foreground than usual, contributing to the unusual perspective. Nimur (talk) 23:51, 1 July 2010 (UTC)

Well, that's all valid, but also it looks like the foreground may be sloping downward a bit, which is a frequent cause of illusions of this sort. Looie496 (talk) 01:22, 2 July 2010 (UTC)

Actually, we do know how the camera was oriented -- if you're looking toward a horizon where the sea seems to form a sharp boundary against the sky (rather than fading into haze), a line from you to that horizon must be horizontal. (Horizon. That's why we call it "horizontal".) But as Looie says, the foreground is sloping, so this may contribute to the shoreline looking lower than you expect. --Anonymous, 04:14 UTC, July 2, 2010.

I think the reason is rather perception (or the lack thereof) of depth (in the sense of distance). On the structured foreground one has a pretty good feel for how the distance increases as one looks from the bottom of the picture to the land-sea boundary. The unstructured sea does not offer any depth indicator (increasing haziness doesn't help on this fairly clear day) so one does not perceive any distance difference between the land-sea boundary and the horizon. That's why the sea appears like a vertical wall, the entire blue surface at the same distance. I hope I understood the question correctly. --Wrongfilter (talk) 15:25, 2 July 2010 (UTC)

I've noticed the same effect while standing in an airport close to sea level—the ocean in the distance appeared somewhat distant as well as a bit elevated. Other than perspective could some kind of mirage effect explain the phenomenon? ~AH1^(TCU) 17:10, 2 July 2010 (UTC)

Plane Crash

What happens actually when an airplane (say a Boeing 747) crashes. How and why do people die, by the intense heat produced or by kinetic energy ? Is death swift or slow and painful ?  Jon Ascton  (talk) 17:33, 1 September 2010 (UTC)

July 2

Phantom question (nothing to do with the paranormal)

Just a couple of days ago I've read a funny story on the Internet about an F-4 Phantom taking off from a combination civilian/military airfield. Supposedly, the Phantom pilot was in a hurry to take off but the tower told him to wait because of heavy civilian traffic; then, after some back-and-forth between the tower and the pilot, the controller said, "OK, if you can reach 14,000 feet within half the runway length, then you are cleared for takeoff; otherwise, continue to hold." Allegedly, what happened then was that the Phantom taxied into position at the approach end of the runway, engaged full afterburner, lifted off within half the runway length, and then climbed vertically to 14,000 feet, thus abiding by the tower's conditional clearance. (Here's the link to the site in question: http://www.businessballs.com/airtrafficcontrollersfunnyquotes.htm ) My question is, does the Phantom have a sufficient thrust/weight ratio to do that? I've never flown the Phantom, so I can't be sure that this is possible. Thanks in advance! 67.170.215.166 (talk) 01:48, 2 July 2010 (UTC)

Based on the numbers in our article, the plane could climb vertically if it was empty, but the thrust/weight ratio was only 0.86 in a fully loaded configuration. Looie496 (talk) 03:23, 2 July 2010 (UTC)

So in other words, the plane must've had no bombs, no missiles, and only a light load of fuel on board. (Which means that the pilot (1) wasn't flying far, and (2) must've been a real hotshot to climb in afterburner with such a light fuel load.) Thanks for the info, and clear skies to you! 67.170.215.166 (talk) 03:49, 2 July 2010 (UTC)

Although not, I think, literally possible, such a feat would be much closer to attainable by an English Electric Lightning. 87.81.230.195 (talk) 05:56, 2 July 2010 (UTC)

The F-15 Eagle could do it without ruffling a feather (I know, I've seen that at an airshow). 67.170.215.166 (talk) 01:13, 4 July 2010 (UTC)

Chatting back and forth with ATC, especially during heavy traffic would be seriously unprofessional and violate FAA regs. I hope this story is not based on fact. Googlemeister (talk) 18:16, 2 July 2010 (UTC)

Wormholes

Does a wormhole exert its own gravitational pull assuming they exist? Based off diagrams I have seen of wormholes they seem to have a gravitational well similar to that of a black hole, so wouldn't they have an event horizon around each of the mouths and trap any traveller going through them inside? Or does the negative energy density required inorder to keep a wormhole open cancel out the gravitational fields. I know that this is assuming that traversable wormholes exist and they might not, but thanks for the help anyway. --74.67.89.61 (talk) 01:56, 2 July 2010 (UTC)

Wormholes do not exist. They are hypothetical conceptualizations. Even within the realm of the hypothetical, they introduce irresolvable contradictions. Wormholes are an interesting, mind-bending game for geometrically inclined mathematicians. But we have never seen them; we have never seen any evidence of them; and our best efforts to consistently explain them would require bizarre and unrealistic physics. It would serve the world a lot better if the pop-science physics books would focus on actual, existing strange physics - like the solution to the generalized double pendulum. Have you ever seen one of these crazy contraptions? I can't comprehend the popular fascination with fictional physics, when there is so much unexplained in actual physics. Nimur (talk) 02:22, 2 July 2010 (UTC)

There is no currently known property that forbids the existence of wormholes (and lets be clear here, I am not talking about traversable wormholes, I am talking about any type of wormhole, stable or not, microscopic or macroscopic) although there are a lot of strange implications. We currently do not know enough to say whether they exist or not for certain. When black holes were found in general relativity, many scientists believe they were impossible even though there was no principle that made them impossible, and look where we are today, finding that black holes are actually common throughout the universe. Wormholes are in a similar state today of what black holes used to be in, being allowed by general relativity with nothing preventing them, but no evidence for them and strange impications. I am not saying that they exist or don't, I am simply saying we do not know enough to give a definite answer, which is why I said ASSUMING THAT THEY EXIST do they exert a gravitational pull, i am simply curious as to whether someone knows the answer to this. —Preceding unsigned comment added by 74.67.89.61 (talk) 13:59, 2 July 2010 (UTC)

The problem is, we can't just assume they exist. We have to make some assumptions about how they exist - we have to change something in the laws of physics, and there are all kinds of ways we could do that (some more plausible than others). Depending on how we change physics, we'll get different answers to your question. If we talk about a specific theory of wormholes, for example the wormhole described by Matt Visser which is mentioned in Wormhole#Traversable wormholes, then the question is answerable. Unfortunately, I don't know the answer... I can't find the paper mentioned. --Tango (talk) 02:28, 2 July 2010 (UTC)

Forgive me for soap-boxing, but somebody needs to just come out and say it. Wormholes do not exist. Before anybody tries to contemplate the gravitational behavior of a hypothetical wormhole, they should be required to thoroughly, quantitatively, and correctly describe the behavior of this simple contraption, which is only under the influence of regular, earth gravity. Once you have mastered the mathematical techniques necessary for the generalized description of a system in ordinary geometries, you will have the foundations for the mathematical tools to play with general solutions in arbitrary geometries. Unless you have these techniques completely mastered, any description of wormholes is just gonna be a lot of handwaving and nothing more. That's why it is so frustrating to see so much meaningless and frivolous writing on the subject of wormholes, hidden behind the excuse of advanced science/magic and totally devoid of actual meaning. Nimur (talk) 02:34, 2 July 2010 (UTC)

In order to actually try to be encouraging, rather than discouraging, here is a list of topics you will need to learn thoroughly in order to understand geometries and topologies for wormholes:

• Algebraic geometry, specifically the representation of space as a set of abstract and mappable coordinates
• Multivariate calculus, especially as related to coordinate transforms and the ugly/beautiful details associated with the careful study the mathematics of continuity
• Geometric topology
• Advanced linear algebra, tensor algebra
• Any of the various reincarnations of mechanics in terms of representations of energy - like Lagrangian mechanics
• And of course, the relativistic theory of gravity

This list is not complete, but by the time you master those concepts, you'll already know what else you need to learn in order to study wormholes. Spend a lot of time on the mathematical foundations. These subjects are very difficult, but they are not inaccessible. They are requisite - these are the languages which are best suited to discussions of the geometry of space. "Simple english" just doesn't have the precision and unambiguity necessary to properly describe these sorts of systems. Nimur (talk) 02:45, 2 July 2010 (UTC)

I hold an MMath degree in which I studied (and passed) modules in all the fields you mention. I am qualified to discuss the subject of wormholes. We do not know if wormholes can exist or not, but from a mathematical point of view (which is the point of view I take, since I am a mathematician by training) the changes to our physical theories are minimal (basically, you just have to forget the weak energy condition). Just because we cannot demonstrate that the weak energy condition can be violated does not mean that we cannot investigate the consequences of such a violation. --Tango (talk) 03:07, 2 July 2010 (UTC)

I have found the paper! It is here. It describes a solution in which a traveller would feel no forces on travelling through. In the words of the paper, if you send a beam of light into the wormhole "the throat of the wormhole acts as a “perfect mirror”, except that the “reflected” light is shunted into the other universe." (The paper describes wormholes between two universes, but the idea should work for wormholes within one universe too - at least, I can't see any problems with it.) Away from the throat of the wormhole, the universes are just flat Minkowski space, so there would be no gravitational attraction (or repulsion, for that matter). --Tango (talk) 03:07, 2 July 2010 (UTC)

And that paper links to Wormholes in spacetime and their use for interstellar travel: A tool for teaching general relativity (1988). (PDF available). That paper lays out some conditions for wormholes: "It turns out that there are very simple, exact solutions of the Einstein field equations, which describe wormholes that have none of the above problems. ... The traversible wormhole solutions to Einstein's field equations are so simple that they can be used as a tool for teaching beginning relativity students how to solve the Einstein equations, how to interpret physically solutions they have obtained, and how to explore the properties of solutions." This is what the OP was really asking, I think. Thanks Tango for the reference above. Nimur (talk) 14:37, 2 July 2010 (UTC)

I see no reason to believe the OP was interested in the pedagogical properties of wormholes. Those properties are very important (my first rigorous exposure to wormholes was purely as a convenient example of a non-trivial solution to the Einstein equations, although that was the non-traversable Schwarschild wormhole, so not much use to the OP), but I don't think the OP cares about them. --Tango (talk) 15:11, 2 July 2010 (UTC)

To repeat what I've said in the past about these traversable wormhole "solutions": a solution of general relativity is a solution of the Einstein field equation G_μν/8πG = T_μν, and any spacetime geometry whatsoever "solves" that equation if you don't constrain T_μν. All you have to do is work out G_μν/8πG for your geometry and then declare T_μν to be equal to it by fiat. Thus to say that a traversable wormhole geometry is a "solution" of general relativity is to say nothing at all, unless you have some independent constraints on T_μν (i.e., some non-gravitational laws of physics). Even the simplest physical constraints on T_μν rule out all of these wormhole solutions. They are therefore not solutions in any meaningful sense. The authors of these papers do acknowledge this, but what they don't mention is how easy it is to write down wormhole and warp-drive and time-travel geometries with whatever crazy properties you want, in the absence of any physical constraints. There's nothing interesting or inventive about the particular geometries they wrote down. -- BenRG (talk) 22:51, 2 July 2010 (UTC)

I would say a "solution" of the Einstein field equation is an explicit description of both the spacetime geometry and the matter/energy distribution that produces it. Just saying that the stress-momentum-energy tensor equals the appropriate value doesn't really count. --Tango (talk) 23:07, 2 July 2010 (UTC)

But it has been pointed out that these solutions require "exotic matter" and other un-physical assumptions, in some distribution, to produce the wormhole geometry. The paper Tango linked at least did not require exotic matter to be present in the throat of the wormhole (hence the arbitrary label of "traversable"), but exotic/unphysical matter was required elsewhere in some distribution to produce the system and its geometry. Such exotic matter does not exist. (Rather, we have never observed it, nor do we have any reason to believe it exists, but it would be really neat if it existed and it is very hard to prove non-existence, so lots of people latch on to the idea). Nimur (talk) 00:05, 3 July 2010 (UTC)

Actually, the solution did involve exotic matter at the throat, just not covering the whole throat so a traveller could avoid it. Also, "traversable" usually refers to the wormhole existing long enough to travel through rather than to whether there is exotic matter in the way. While there is far from any conclusive evidence that exotic matter can exist, the Casimir effect does point in the right direction. --Tango (talk) 00:49, 3 July 2010 (UTC)

If I may attempt to answer OP directly (the linked paper does that, but it's at a rather high level), wormholes can be considered a topological defect in spacetime. When you have a gravitational well, like the attraction to a planet or whatever that causes curves in the rubber sheet diagrams that I assume you're looking at, those are geometrical properties, not topological. So geometrical = bumps and troughs and mountains, while topological = bottomless holes and tunnels and loops and various cartoon sight-gags.

Tango's first answer was that there have been different types of wormhole theories (mathematical equation sets) presented in physics, some of which show this rubber-sheet gravity well that ends up rubbering itself onto the other side of the universe. But others simply use a flat tunnel that you can walk through with no noticeable change in forces - you simply go in one end and come out the other. In both cases, we can make solutions in well-established General Relativity, but they require "exotic matter", which is basically matter with negative mass - i.e., it is gravitationally repulsive. Let me say this unqualified (I don't study GR) opinion as to your original question now: there are perfectly good reasons to imagine wormholes fitting either the gravity-well or the magic-tunnel description, but once we need any kind of matter, however unusual, to hold it up, there's probably going to be some (or lots of) gravitational discomfort when you walk through.

Finally, to touch on exotic matter, you are in agreement with most physicists if you find negative mass hard to swallow at this point. However, there are other places, like on a rotating black hole's event horizon, where exotic matter appears in some models (nice overview on all this). As sometimes matter and antimatter pop into existence and annihilate, this math shows sometimes mass and negative-mass pop into existence and annihilate.

Regardless, we've never seen any evidence of a wormhole existing naturally (it'd likely be visible, and because the universe is big, if it can exist then there'd probably be a lot of them), and it'd take more energy than currently conceivable by our civilization to mash up the topology of space ourselves were the cards of physics in our favor. But, as one researcher who works on this said, it's worth studying because it's simply fun and interesting. And indeed, all the cards have not yet been dealt. SamuelRiv (talk) 04:28, 3 July 2010 (UTC)

MP3 player (well, really amplifier) power consumption

I recently got a new iPod Nano (5G). It seems to have somewhat lower volume than my old (2G) one. It's fine for music, but very many spoken word programs (such as the excellent Open Yale Courses) are recorded with a much lower volume than music. As a consequence, I often listen to them at 100% volume, and even then some are quite quite for outdoor listening. I'm wondering what that does for battery life. Does the power consumption depend only or primarily on the volume level (on the principle that the signal is maximally amplified, even if there is not much of a signal), or does it depend primarily on the amount of actual sound energy coming out of the headphones? The first seems intuitive - if I crank up the volume, I use more energy. Physically, the second is more plausible - after all, where would the energy go otherwise? Does a amplifier amplifying a flat line get hot? If the second is true, how much of an effect is it? Will my mp3 player have a significantly longer battery life playing e.g. the Pastorale than playing Painkiller (and yes, I have both on my player ;-). --Stephan Schulz (talk) 09:30, 2 July 2010 (UTC)

As a general rule, the power consumption of a noramlly designed amplifier is approximately proportional to the volume it produces, rather than the setting of the volume control. --Phil Holmes (talk) 13:11, 2 July 2010 (UTC)

Thanks! So I should always use an endless loop of 4′33″ for battery benchmarking ;-). --Stephan Schulz (talk) 13:19, 2 July 2010 (UTC)

iPod Nano uses a Class D amplifier, and great pains are taken to reduce its quiescent current. (That means that "0 micro-amps" would be spent on audio power amplification if you played 4'33" through it - but of course, there are parasitics and losses, but generally very small). Realistically, you will benefit more in terms of battery life by turning the LCD screen / backlight off - this draws significantly more current than the audio system. Unfortunately the iPod nano internals are proprietary and their amplifier and audio system is integrated into the "anonymously marked" IC packages with Apple part-numbers only; it is impossible to get the exact specs. Nimur (talk) 15:04, 2 July 2010 (UTC)

Just as an aside, I believe you can set individual mp3 files to play with different volumes in iTunes — File > Get Info > Options > Volume Adjustment. You could boost the predictably quiet files by 100% before putting them on the iPod. (I'm reasonably sure those settings are transferrable to the iPod, but I haven't tried it.) --Mr.98 (talk) 19:11, 2 July 2010 (UTC)

is there such a thing as a "semi-permanent" or reversible tattoo?

so I really like the idea of getting a tattoo, but who knows if later I need to sell out to the man as most people in corporations do after a while. If I ever need to have Forbes and Fortune's cocks ramming down my throat, or, who knows, maybe even Wall Street's directly, they will not be moved by tattoos on me. So, my question is whether there exists a tattoo that is "permanent" in the sense that it doesn't "just" go away, it is a real tattoo and looks like one, but semipermanent or reversible in the sense that I can get rid of it for a fee later? For example, I can imagine that a certain class of dyes have a specific "antidote", a chemical that will cause them to break up and disappear into the body, despite the fact that they do not by themselves. Is there such a thing? Or any other way to get a semipermanent or reversible tattoo? Thank you. 92.230.232.169 (talk) 09:41, 2 July 2010 (UTC)

As far as I know, best-practice modern tattoos will respond well to laser treatment (laser light tuned to the dye is used to destroy the colour molecules, your immune systems takes care of the rest). There is a certain risk, though, both that the process will be incomplete, but also of allergic reactions. And it's expensive... --Stephan Schulz (talk) 09:46, 2 July 2010 (UTC)

so why dont they develop special dyes that respond to, say, a certain frequency of microwave radiation, or a certain "antidote" chemical, or anything else... —Preceding unsigned comment added by 92.229.13.177 (talk) 10:40, 2 July 2010 (UTC)

Our article Temporary tattoo#Temporary variants of permanent tattoos mentions InfinitInk which is supposed to be easier to remove with laser treatment. While are I'm sure inks which are easier to remove which could be developed, I'm not convinced your first idea would be much easier or cheaper then with lasers, particularly if you don't want the rough area you were tattoed to be cooked and also don't wont your tattoo to fade a lot over time due to natural exposure to EM radiation from the many forms of transmission common nowadays. Your second idea is more promosing but developing something safe and effective is probably not easier or cheap particularly since the antidote will need to penetrate the skin Nil Einne (talk) 14:16, 2 July 2010 (UTC)

There's also the far-off possibility of nano-tattoos, which could be reprogrammed as you saw fit. (Kind of goes against the point of a tattoo, if you ask me. They're not just an image, they're an image of commitment!) IMO, if you are worried about not looking corporate enough, get it in an area that can be easily hidden, like the shoulder or upper arm, which can be displayed if and when you want it pretty easily, but is also easily hidden under your suit. And please watch your language on here—it's rude to swear for no purpose, and makes you sound neither edgy nor interesting. --Mr.98 (talk) 14:33, 2 July 2010 (UTC)

I acknowledge your sentiment, though I wonder if it isn't prompted by your regularly ingesting more corporate c**k than is your preference. see? self-censored.92.224.207.197 (talk) 17:49, 2 July 2010 (UTC)

Henna? Woad? 92.28.247.183 (talk) 21:17, 2 July 2010 (UTC)

Copper(I) sulfate

I reduced copper(II) sulfate (blue) with ascorbic acid solution (colorless) to get a yellow solution. Is it copper(I) sulfate? It reacted with hydrogen peroxide to form a green solution again (it was green because of the additional acidity of the ascorbic acid). --Chemicali nterest (talk) 14:16, 2 July 2010 (UTC)

Sounds more like a copper ascorbate complex to me. Physchim62 (talk) 14:43, 2 July 2010 (UTC)

I disagree (that example is in anhydrous conditions) - in aqueous solution copper is reduced [8] [9]

Possibly you made nano-copper in the first step [10]. It's also possible that the some of the product at the first step was fine Copper(I) oxide (this paper [11])

Additionally 'Cuprous(I) hydroxide' is yellow and is obtained in alkaline conditions, though this seems very unlikely in acid conditions - see also Benedict's reagent and Fehling's reagent

This paper [12] is available online, and covers a lot of stuff relating to the first reaction. —Preceding unsigned comment added by 77.86.124.131 (talk) 15:56, 2 July 2010 (UTC)

The second step (green solution) sounds like it might be a copper(II) complex - which are often green with oxy-ligands - I'm not sure what happens to the ascorbic acid through all this - but I guess it will be a complex of whatever the ascorbic acid is oxidised to.Sf5xeplus (talk) 14:55, 2 July 2010 (UTC)

Well Cu(I) isn't very stable in water. I'm going with the idea of nanocopper -- you reduced Cu(II) to Cu(I) only to have it disproportionate into Cu(II) and Cu(0). (And the Cu(II) gets reduced again into Cu(I)...) And colloidal copper is much different than bulk copper....because of different surface energies and the Mie scattering it will take on a different colour...welcome to nanochemistry! John Riemann Soong (talk) 15:54, 2 July 2010 (UTC)

Also did you add Vitamin C in excess? Or? John Riemann Soong (talk) 15:57, 2 July 2010 (UTC)

Neither were in clear excess. When CuCl₂ is reacted with ascorbic acid, white CuCl is formed. I was wondering whether there is a similarity in the sulfate reaction. --Chemicalinterest (talk) 20:17, 2 July 2010 (UTC)

I will try tin(II) chloride as a reducing agent and see if there is a difference. --Chemicalinterest (talk) 20:21, 2 July 2010 (UTC)

The issue is that Cu(I) sulfate is probably very soluble in water whereas CuCl precipitates in excess and is thus most of the Cu(I) is shielded from disproportionation. Try reducing Cu(II) chloride in the presence of concentrated HCl.... also reduce Cu(II) sulfate with Vitamin C in clear excess, eliminate the chance of any remaining Cu(II).

Btw, do you have any sodium borohydride? John Riemann Soong (talk) 21:04, 2 July 2010 (UTC)

(I ask because NaBH4 is a little more environmentally friendly than Tin(II) and Tin(IV)...what do you do with your waste?) John Riemann Soong (talk) 21:08, 2 July 2010 (UTC)

Tin is ok unless you're a mollusc ... In the overall scheme of things using NaBH4 is just as damaging (production of) as a little solder+HCl, but I appreciate your point.77.86.124.131 (talk) 21:25, 2 July 2010 (UTC)

well the basic products of NaBH4 decomposition can be neutralised with acetic acid or vinegar... then dumped down the sink. You can't do that with Tin(IV). John Riemann Soong (talk) 21:27, 2 July 2010 (UTC)

Tin in the enviroment (in the quantities here) is unlikely to be a problem .. the main problem from tin is poisonous by products from its production ie smelting —Preceding unsigned comment added by 77.86.124.131 (talk) 21:34, 2 July 2010 (UTC)

Do you have a centrifuge? If you want to do the waste management yourself, you can recover the colloidal copper by centrifuge. Sodium ions, sulfate ions, phosphate ions and chloride ions don't stress the environment that much. John Riemann Soong (talk) 21:31, 2 July 2010 (UTC)

I would use tin(II) chloride in quantities in order of .2 grams; that shouldn't hurt the environment. --Chemicalinterest (talk) 00:21, 3 July 2010 (UTC)

More clues: When reacted with an excess of ascorbic acid, it forms a yellow solution. The yellow solution becomes darker as ammonia is slowly added. The color is similar to a dilute solution of iron(III) chloride in acidic conditions. Wisps of brown precipitate are formed. When much ammonia is added, it creates a very dark brown precipitate that lightens within 1 second to a yellow brown precipitate. The yellow brown precipitate reacts with small amounts of hydrochloric acid to form a colorless precipitate (CuCl), which dissolves in excess HCl to form a clear solution. The precipitate also reacts with hydrogen peroxide to form a greenish solution (Cu(H₂O)²⁺). Hope this helps. --Chemicalinterest (talk) 00:43, 3 July 2010 (UTC)

Concentration of NaOH

100 mL of 0.125 M NaOH was mixed with 64 mL of H₂O. Calculate the final concentration.

Is it incorrect to use the dilution equation in the following way?

(100 mL) (0.125 M) = (64 mL) (C)

Concentration = C = 0.20 M H₂O.

--478jjjz (talk) 17:04, 2 July 2010 (UTC)

Yes that's wrong.

The correct formula is :

  Initial volume
  --------------   x Initial concentration  = Final concentration
   Final volume  

The final volume is 164mL. 77.86.124.131 (talk) 17:16, 2 July 2010 (UTC)

So, then the final Concentration is

[(100mL)/(164 mL)] * 0.125 M = 0.0762 M

--478jjjz (talk) 17:44, 2 July 2010 (UTC)

HCl and NaOH

25.0 mL of 0.5 M NaOH was reacted with 45 mL of 0.12 M HCl. Calculate the final concentration of NaOH.

(0.0250 L of NaOH)*(0.5 mol NaOH) = 0.0125 mol NaOH.

(0.045 L) * (0.12 M HCl) = 0.0054 mol HCl

Since NaOH and HCl react in a 1:1 ratio, then HCl is the limiting reagent. The product is 0.0054 mol NaOH.

How do I proceed further?--478jjjz (talk) 18:03, 2 July 2010 (UTC)

You're starting materials indicate 0.0125 mol NaOH and 0.0054 mol HCl are availible to react. Since there is a 1:1 ratio, answer these questions:

1. If 0.0054 mol of HCl react, and the SAME NUMBER OF MOLES of NaOH react as well, then how many moles of NaOH actually react?
2. If you started with 0.0125 mol of NaOH, and you TAKE AWAY the number of moles of NaOH that reacted (#1 above) how many moles of NaOH are left over?
3. If you mixed two solutions, one of which contained 0.0250 L and the other which contained 0.045 L, what is the TOTAL VOLUME of the two solutions after mixing?
4. Now, you have some NaOH left (answer to #2) and it is in the combined volume (answer to #3). What is the final concentration, given that the concentration of the NaOH left is the LEFTOVER MOLES OF NaOH DIVIDED BY THE COMBINED LITERS OF SOLUTION.

Every problem you solve that looks very similar to this is solved roughly the same way. --Jayron32 18:14, 2 July 2010 (UTC)

1. 0.0054 mol NaOH react
2. 0.0071 mol NaOH left over
3. 0.070 L = total volume
4. (0.0071 mol NaOH)/(0.070 L) = 0.10 M NaOH.--478jjjz (talk) 18:29, 2 July 2010 (UTC)

Solution concentration equals amount of stuff divided by solution volume--say that like a hundred times or so. That is the formula that always answers any dilution question, because you will always have or can figure out any two of those variables and then use that formula to solve the third. The secret to avoiding mistakes is to always write units with your numbers so you know exactly what the value means (i.e., whether it's concentration, volume, etc.). So when you first say "(0.5 mol NaOH)" BZZT, that's a kind of careless mistake that can lead to all sorts of confusion and more serious understanding mistakes later ("L * mol" is...um...L*mol not mol--that's a pretty fundamental fact). It's not just nit-picky--if you are working through towards an answer and the units are not the right type for the question ("how many moles?" and you have an answer in liters) you immediately know you made a mistake. Say you have figured out "0.0054 mol NaOH", that's an amount of stuff. You can easily figure out the total volume in which it is dissolved (the volume of a mixure of same-solvent solutions is just the sum of the volume of each part). And now you have a formula you've said like a hundred times or so that uses those two to figure out concentration. DMacks (talk) 18:49, 2 July 2010 (UTC)

Titration

250.0 mL of an unknown HCl solution was titrated with 14.75 mL of 0.0762 M NaOH to the phenolphthalein end point. Calculate the molarity of the HCl solution.--478jjjz (talk) 18:41, 2 July 2010 (UTC)

The reaction is

HCl + NaOH ==> NaCl + H₂O

Total volume= 250.0 mL + 14.75 mL = 264.75 mL = 0.26475 L

Every concentration problem is the same, just different numbers. DMacks (talk) 18:49, 2 July 2010 (UTC)

We have 0.0012 mol NaOH --478jjjz (talk) 18:47, 2 July 2010 (UTC)

DMacks, I beg to differ with you. I don't know how many moles of HCl I started with. This is the source of my confusion.--478jjjz (talk) 19:00, 2 July 2010 (UTC)

If you have the number of moles of NaOH and wish to find the number of moles of HCl, look at the equation. It will tell you the ratio between the two. - Jarry1250 ^{[Humorous? Discuss.]} 19:05, 2 July 2010 (UTC)

0.0012 mol HCl reacted with 0.0012 mol NaOH. I don't know how many moles of HCl I started with; how do I find the moles of leftover HCl?

--478jjjz (talk) 19:11, 2 July 2010 (UTC)

What does the "phenolphthalein end point" mean? DMacks (talk) 19:26, 2 July 2010 (UTC)

It means to the point where all the acid (HCl) and base (NaOH) have completely reacted.--478jjjz (talk) 19:30, 2 July 2010 (UTC)

With none left over? DMacks (talk) 19:32, 2 July 2010 (UTC)

Since I don't know the starting moles of HCl, I don't know how much of it is left over. I want to know if this problem from one of my chemistry quizzes is even solvable.--478jjjz (talk) 19:34, 2 July 2010 (UTC)

Maybe the context of my question wasn't clear. Again, for the definition of the phenolphthalein endpoint of a titration, you say the acid and base "have completely reacted". How many moles of base do you need in order that X moles of acid gets "all reacted"? Does the definition here mean you have added just an arbitrary small amount, exactly as much of one as the other, or a huge excess amount? How much un-neutralized acid (or base) is present according to the definition of your endpoint? DMacks (talk) 19:42, 2 July 2010 (UTC)

• 0.0012 mol NaOH has reacted with X amount of HCl. I don't know the initial concentration of HCl which I can multiply by the volume to get the # of moles of HCl. If I have, let's say, 0.000005 moles of HCl at the start of the reaction, then HCl would be the limiting reagent & none would be left over.--478jjjz (talk) 19:51, 2 July 2010 (UTC)

If you have 0.000005 moles of HCl at the start of the reaction, how many moles of NaOH would you add to get that amount of HCl all reacted? DMacks (talk) 19:53, 2 July 2010 (UTC)

They would both be in a beaker. 0.0012 mol NaOH reacts with with all HCl and (0.0012 -0.000005) mol NaOH is left over in the basic solution in the beaker due to the excess NaOH.--478jjjz (talk) 19:56, 2 July 2010 (UTC)

There's your problem: the very definition of your endpoint is when the solution reaches a very specific pH (the color changes), which tells you exactly how much H+ is present at that point. The casual wording of the reaction description might be throwing you: if you have >0 of compound X present, it is not "all reacted", it's excess unreacted. All reacted means all of it is reacted, not just "as much as can". If I have 10 H+ and add 5 OH-, my OH- is all reacted, but my H+ is not all reacted. The first drop of base you add to the acid completely reacts, but you (hopefully) recognize that this is not the endpoint of the titration (would be silly because every titration would be "1 drop"). You keep adding base until (for example) neutrality (until the H+ is "all reacted"). DMacks (talk) 20:05, 2 July 2010 (UTC)

• The purple words above have been copied verbatim. I have concluded that this problem isn't solvable. I have to speak to the instructor and let him know that it is preposterous to put unsolvable problems on quizzes to make me get a low grade. (If anyone can somehow solve this problem, then please do!)--478jjjz (talk) 20:11, 2 July 2010 (UTC)

The problem is perfectly solvable. In fact, it is a routine calculation in analytical chemistry! Just just need to figure out the amount of HCl in the initial solution, then divide by the volume of the initial solution to find the concentration. Physchim62 (talk) 20:16, 2 July 2010 (UTC)

(ec) To be honest, if I were the teacher, I would tell you to go back and relearn what "endpoint" means in the context of a titration experiment. Again, per the technical definition of phenolphthalien endpoint, you know exactly the concentration of unreacted acid in the solution at that point (i.e., excess, compared to the amount of base you added). DMacks (talk) 20:17, 2 July 2010 (UTC)

This problem was in a lab quiz. The #s in the above purple problem have nothing to do with the experiment that I had performed during the prior class. As it stands, I re-affirm that the purple problem is unsolvable.--478jjjz (talk) 20:22, 2 July 2010 (UTC)

The first step in pretty much any titration problem is the calculate the amount of reagent you have added from the burette. Now here, you've added 14.75 mL of a 0.0762 M solution: amount is concentration times volume, so you have added n(NaOH) = 0.01475*0.0762 = 0.001124 mol. You know the equation, so you know that that hydroxide has reacted with 0.001124 mol of HCl. So how much HCl is left over? Take a quick look at our article on phenolphthalein, and you will see that it changes colour at pH 8.2… so when the phenolphthalein changes colour, there's no hydrochloric acid left! So your initial solution of HCl contained 0.001124 mol HCl in 250 mL, in other words it was a 0.001124/0.25 = 0.00450 M solution. Physchim62 (talk) 20:41, 2 July 2010 (UTC)

Let's start again

14.75 mL of 0.0762 M NaOH was used - that's 14.75 x 0.0762 = 1.12395mmol of NaOH (A)

It was titrated to an end point with an acid-base indicator in the reaction NaOH + HCl >>> NaCl + H₂O , that's as 1:1 reaction so 1.12395mmol of HCl must have been used (B)

Molar concentration is related to number of moles and volume by the equation:

                      Number of moles
   Concentration =  ------------------  (C)
                         Volume

You've been asked to find the molality (concentration) of the HCl solution, so that's 1.12395mmol ÷ 250ml (D)

Be careful with units, since both the quanities in D are 'milli' in this case the answer is is mol/litre , which is the same of molality, no further work needed.

If you didn't get any part please ask about it. I've labelled each step A,B,C etc .77.86.124.131 (talk) 20:33, 2 July 2010 (UTC)

Your methodology differs from the section above this problem. Therein, the leftover moles that hadn't reacted were used to find the concentration. On the other hand, in this problem, the number of moles that have reacted have been used to find the concentration.--478jjjz (talk) 20:38, 2 July 2010 (UTC)

• The problem should have said " Calculate the initial molarity of the HCl solution." User:Physchim62's explanation has made me believe that 0.001124/0.25 = 0.00450 M HCl is indeed correct for the initial concentration.

The point is that you start out with 250 mL of an HCl solution: the implication is that it is a sample of a larger volume of solution that you have sitting in a bottle somewhere. Physchim62 (talk) 20:58, 2 July 2010 (UTC)

I have finally reconciles the approach of this problem and the one above it. In both cases, we have used the unreacted substance to find concentration. In this one, we found the unreacted amount prior to the reaction to get the concentration. In the previous problem, we found the unreacted amount after the reaction to get the concentration.--478jjjz (talk) 20:50, 2 July 2010 (UTC)

The other point with a titration is that you use an indicator (here, phenolphthalein) so that you know exactly when the last of the reactant has been used up. In effect, you already know the final concentrations – they're both zero. Physchim62 (talk) 20:58, 2 July 2010 (UTC)

I thank you and others who helped me with this problem.--478jjjz (talk) 21:12, 2 July 2010 (UTC)

The "endpoint" in this case is mentioned in the main titration article. See Equivalence point. ~AH1^(TCU) 16:22, 3 July 2010 (UTC)

Temperature of the ocean

What IS the temperature of the water 1000s of meters down in the worlds oceans? —Preceding unsigned comment added by 86.182.34.28 (talk) 19:20, 2 July 2010 (UTC)

Pretty close to freezing. 35-40 deg F. Volcanic vents the exception. Googlemeister (talk) 19:25, 2 July 2010 (UTC)

"Pretty close to freezing" may be a bit misleading. Those temperatures are close to freezing at sea level. However, due to the great pressures experienced at that depth (not to mention the salinity), the water isn't really close to becoming ice 1000s of meters deep. -- 174.24.195.56 (talk) 20:56, 4 July 2010 (UTC)

Abyssal plain#Terrain features has lots of interesting information on this subject. --Tango (talk) 20:05, 2 July 2010 (UTC)

Four degrees celsius, because that is the temperature at which water is most dense. See below, and the article - sorry, my error. Thermocline has good information for you. Because of salinity-based and thermal-based upwelling, there may be local variations, but in the very deep ocean, there is a reasonably static temperature profile (even if there is mass transfer of the water). Nimur (talk) 20:30, 2 July 2010 (UTC)

As I learned after giving the same answer here some time ago, the four degree value holds true for fresh water but not for salt water. The link you pointed to explains that the freezing point of seawater is about minus two Celsius, and the density increases right down to the freezing point. Looie496 (talk) 00:20, 3 July 2010 (UTC)

Take a look at this for global ocean temperatures at the depth of 1000 metres (select the "1000 m" option; here is a permanent link for June 30, 2010). ~AH1^(TCU) 16:13, 3 July 2010 (UTC)

Audio amplifier power output

Why is the output of a domestic audio amplifier not inversely proportional to the loudspeaker impedance. I was taught that P=V^2/R but amplifiers dont seem to obey this law when you look at their specifications. Why not?--Bellwelder (talk) 20:12, 2 July 2010 (UTC)

Mostly because the amplifier itself has impedance as well - this makes your equation:

   P=V2/(Rspeaker+Ramplifier)

The other additional factor in old fashioned amplifiers is the power supply capacity of the power supply ie the transformer - there's a limit to how much power a transformer can supply related to its inductance, but I don't know the equation.

Yes but the output impedance of a feedback amplifier is very low (approaching zero)--88.104.90.10 (talk) 10:51, 5 July 2010 (UTC)

77.86.124.131 (talk) 20:41, 2 July 2010 (UTC)

You will get maximum power transfer from the amplifier to the loadspeaker when their impedences are matched, not when the speaker impedence is minimum: this is known as the maximum power theorem. Physchim62 (talk) 21:47, 2 July 2010 (UTC)

natural skin oil

What oils most closely mimic natural skin oil: for example, olive oil, peanut oil, canola oil, mineral oil, vaseline, coconut oil, palm oil, etc.? —Preceding unsigned comment added by Alexsmith44 (talk • contribs) 20:34, 2 July 2010 (UTC)

Human skin oil is called Sebum , there's a simplfied analysis at Sebaceous_gland#Composition - of those components most are not found in vegetable oils, so a comparison between them isn't that viable. (Mineral oil and Vaseline are pure petroleum product and not like skin oil at all).77.86.124.131 (talk) 20:47, 2 July 2010 (UTC)

Very roughly it's similar to 40% vegetable oil + 20% beeswax + 10% vaseline + 10% soap (though it's not salty like soap) .. hope that helps.77.86.124.131 (talk) 20:59, 2 July 2010 (UTC)

Jojoba, which is strictly speaking not an oil but a wax, is said to have the nearest composition to human sebum. --TammyMoet (talk) 07:35, 3 July 2010 (UTC)

It sounds quite similar to 25% of the make up of sebum - I used 'beeswax' as an equivalent above, but johoba is a better match.87.102.21.49 (talk) 11:22, 3 July 2010 (UTC)

Name these two plants?

In a park in southern England. http://img186.imagevenue.com/img.php?image=04455_DSCF0003_122_354lo.JPG The one on the right foreground had grape-like berries earlier in the year, on the now dried fish-bone-shaped parts. Thanks 92.28.247.183 (talk) 21:04, 2 July 2010 (UTC)

Just to give the same warning as before, the image link gives a popup to a NSFW site77.86.124.131 (talk) 21:18, 2 July 2010 (UTC)

Back left is I think some kind of fig (ficus), but there's an awful lot of them to choose from. Perhaps the one in front is mahonia? 213.122.27.137 (talk) 22:42, 2 July 2010 (UTC)

The plant at the back is the False Castor Oil plant, (fatsia japonica), see here [13], I agree the plant at the front is a mahonia of some sort. Richard Avery (talk) 05:13, 3 July 2010 (UTC)

Thanks 92.15.0.208 (talk) 11:24, 3 July 2010 (UTC)

Annual architectural plants

I bought some annual seeds to sow in my garden in southern England and was dispointed to discover that they were only 2 or 3 inches high when flowering. Are there any big garden plants that will grow as an annual, particularly in semi-shade? 'Architectural plant' means a large plant at least three or four feet high. Thanks. 92.28.247.183 (talk) 21:12, 2 July 2010 (UTC)

Rhododendron? I think you mean a small shrub of which there are many - was there anything else you wanted in the plant that could help narrow the search?77.86.124.131 (talk) 21:20, 2 July 2010 (UTC)

Some varieties of Hollyhock (Althaea aka Alcea) would certainly qualify: they are often used as the 'backdrop' plant (in front of which shorter flowers are placed) in the traditional English Cottage garden. Foxgloves (Digitalis) would fit the size criterion, but are perennial or biennial.

Leafing quickly through Dr D. G. Hessayon's The Bedding Plant Expert, other candidate annuals with some varieties in the 3-4 feet range include: Love-lies-bleeding (Amaranthus); African Daisy (Arctotis); Spider Flower (Cleome); Datura; Sunflower (Helianthus); Burning Bush (Kochia); Larkspur (Delphinium); Annual Mallow (Lavatera); Nasturtium (Tropaeolum); Sweet Pea (Lathyrus); Ornamental Maize (Zea). 87.81.230.195 (talk) 22:41, 2 July 2010 (UTC)

Thanks, although I thought all Hollyhocks were at least biennial. Are there any that are annual? 92.15.0.208 (talk) 11:26, 3 July 2010 (UTC)

Well, we have this gigantic encyclopedia here someplace...um, let me see...aha! Hollyhock says "They are biennial or short-lived perennial", so no. SteveBaker (talk) 14:45, 3 July 2010 (UTC)

Dr Hessayon says otherwise: "Hollyhocks are sometimes grown as perennials in the border, but rust disease generally weakens the plant and it soon becomes a sorry sight. It is better to treat Hollyhock as a biennial or to grow one of the annual varieties." [My italics]. 87.81.230.195 (talk) 18:25, 3 July 2010 (UTC)

The USDA] disagree. Their website (which is the reference for our hollyhocks article) clearly states perennial/biennial. SteveBaker (talk) 03:38, 4 July 2010 (UTC)

I venture to suggest that the USDA are referring to the wild-growing parent plant rather than all of the domestic cultivars created for the horticultural market to be grown in gardens. Many plants that are primarily perennial or biennial in the wild have had annual varieties bred by commercial nurseries. Note also that the OP asked for plants "that can be grown as an annual", which horticulturally one can do with a biological perennial that reaches maturity in the first year simply by removing it after its first flowering - our own Annual plant article states "Some perennials and biennials are grown in gardens as annuals for convenience . . . ." That said, I find it difficult to believe that such an eminent gardening authority as Hessayon would explicitly refer to "annual varieties" if none such existed, but this is I suppose also possible. Perhaps we need a tie-breaker. 87.81.230.195 (talk) 23:05, 5 July 2010 (UTC)

Aaaand . . . try this. The very first google hit from "annual hollyhock seed" gives this site page which includes the paragraph "The small annual hollyhock 'Majorette Mixed' which has large semi-double blooms is ideal to grow in a flower border with other annuals and perrenials [sic]." I rest my case. 87.81.230.195 (talk) 23:23, 5 July 2010 (UTC)

I recommend Sunflowers. They are annuals and some varieties grow to be 6' tall. However, the more natural varieties grow to three or four feet and produce gigantic, long-lasting orange blossoms. They grow spectacularly quickly. SteveBaker (talk) 14:45, 3 July 2010 (UTC)

FWIW this year we have both of sweet peas and sunflowers grown from seed a month or so ago next to each others. They are both currently about 175 cm tall, both growing fast but the sweet peas are much bulkier and will go further if supported. If you only want four foot and want quick growth Californian poppies are not bad (but a bit vulgar to look at, hence the name I guess). Of the above list Nasturtium is fun because the buds are edible (they taste like capers) but will only grow high if supported. --BozMo talk 21:44, 3 July 2010 (UTC)

July 3

Enthalpy-entropy chart

Hi. On this enthalpy-entropy chart, can someone please tell me what 'x' represents (ie x = 10%, x = 20%, x = 30%...).

The person who actually created the graph seems to have 'retired' from Wikipedia, and another user asked for a caption to be added to it.

Thanks in anticipation,  Chzz  ►  01:32, 3 July 2010 (UTC)

It's either Percent humidity, i.e. the ratio of actual vapor pressure to vapor pressure at the dew point, or some sort of meaure of proximity to the critical point. --Jayron32 01:38, 3 July 2010 (UTC)

I found a few descriptions that say that the x lines are "constant moisture or quality" lines. I don't know what that means, though. Looie496 (talk) 04:45, 3 July 2010 (UTC)

Constant moisture implies relative humidity, which is the link I speculated on first... --Jayron32 04:55, 3 July 2010 (UTC)

This is a Mollier Diagram for water-steam. It is named after Richard Mollier a German physicist. His water-steam diagram depicts the properties of steam and is used to estimate the performance of steam engines and steam turbines. X represents the quality of the steam. X=90% refers to steam which consists of 90% (by mass) of dry steam and 10% colloidal liquid water. (Cloud is is an example of colloidal liquid water mixed with dry gas. In a cloud, the dry gas is air and water vapor.) X is not referring to relative humidity which implies the presence of air. The diagram shows the critical point which is a property of water, not humid air. Outside the boundary labeled as 100% the steam is superheated - it contains no colloidal liquid water, and the steam is hotter than the boiling point at the prevailing pressure.

X is sometimes called the dryness fraction. At Mollier diagram it says Lines of constant dryness fraction are drawn in the wet region ...

Steam with X less than 100% is called wet steam. Steam with X=100% is called saturated steam or dry steam. When dry steam is heated it is called superheated. See Steam and Superheated steam. Dolphin (t) 06:48, 3 July 2010 (UTC)

Medroxyprogesterone versus medroxyprogesterone acetate

Hola,

I'm trying to figure out the difference between medroxyprogesterone versus medroxyprogesterone acetate. Anyone know? Anyone have any sources that can be used to distinguish the two? WLU (t) (c) Wikipedia's rules:^simple/_complex 01:57, 3 July 2010 (UTC)

Other than the obvious difference in chemical structure, what are you looking for? It appears that the 17-acetate is the only one used medically - both of the references in the MP article actually relate to MPA. The Merck Manual entry indicates that the former is "supplied as the acetate". The article in MedlinePlus uses the terms interchangeably. Many references discuss plasma levels of MPA, so my initial impression that the acetate is readily hydrolyzed is probably false. -- Scray (talk) 02:59, 3 July 2010 (UTC)

the following 4 posts were copied from WikiProject Medicine, as discussed there

(Original research alert) First, pregnanes and progestins are not mutually exclusive. Pregnane refers to the structure of the molecule, while progestin refers to its biological activity. Medroxyprogesterone is the active molecule. In order to stabilize medroxyprogesterone in a form that can be administered orally or intravenously, it's acetylated (see the final step in File:Medroxyprogesterone_acetate.png. In the body, the acetate residue is degraded and you get the active molecule back.

If you're talking about administering medroxyprogesterone as a pharmaceutical, then you're technically talking about MP acetate. If you're talking about steroid biochemistry in vivo, then it's more correct to refer to medroxyprogesterone, period. I'm not sure what implications this has for our article structure (and this is just me talking - it's been awhile since I took pharmacology, and I don't have a supporting source at my fingertips, so take it with a grain of salt because I have been known to be wrong). MastCell ^Talk 03:58, 3 July 2010 (UTC)

MastCell, thanks for correcting me (pending further corroboration/refs). That was my initial sense (see the RD/S history for my original comments if curious) but I couldn't quickly find anything reliable that supported me. I'll look again tomorrow if no one weighs in further. -- Scray (talk) 04:02, 3 July 2010 (UTC)

I wish I had some refs handy to back me up. BTW, I saw your comment at the Ref Desk about MPA levels ([14]). Are you sure those are actually looking at levels of medroxyprogesterone acetate? I'm not familiar with measuring levels of it. On the other hand, there's an extensive literature on monitoring plasma levels of mycophenolic acid (also abbreviated MPA), which is the active metabolite of mycophenolate mofetil (MMF). Like most immunosuppressants, MMF pretty finicky and has a narrow therapeutic window, so a lot of work has been done on MPA pharmacokinetics (mycophenolic acid) in the organ transplantation literature. You've probably already looked into this, but is it possible that the "MPA" levels you saw were actually mycophenolic acid, rather than medroxyprogesterone acetate? MastCell ^Talk 04:30, 3 July 2010 (UTC)

I'm very familiar with mycophenolate; the particular paper that made me doubt myself was this one, which led me to PMID 6457936, which is entitled "Medroxyprogesterone acetate in human serum". On re-reading the abstract, this may simply be sloppy terminology, since the RIA they're using might not distinguish medroxyprogesterone from the acetate. Clarity remains elusive, but I think you're probably right. -- Scray (talk) 05:01, 3 July 2010 (UTC)

end of material copied from WikiProject Medicine -- Scray (talk) 05:14, 3 July 2010 (UTC)

Can you clatify how or what property you want to distinguish, or what sort of test (chemical,biological,physical) you want to use to distinguish them.87.102.21.49 (talk) 11:27, 3 July 2010 (UTC)

I'm mostly interested in being able to find sources that distinguish between the two. Medroxyprogesterone and medroxyprogesterone acetate are both stubs and so far I haven't been able to find sources that are clear on the difference.

And my apologies for cross-posting this to the extent I did. To make things worse, I'm going to compile the posts at Talk:Medroxyprogesterone 17-acetate. WLU (t) (c) Wikipedia's rules:^simple/_complex 13:34, 3 July 2010 (UTC)

Can you expand on "distinguish between the two" - they are different compounds (the strcuture, CAS number, formula, name, molar mass etc are all different) : one is the acetate ester of the other. How do you want to distinguish between them? Chemically, physically, by effects?? 87.102.21.49 (talk) 14:03, 3 July 2010 (UTC)

jail

can you have a electric space heater in jail if a relative brings it to you? —Preceding unsigned comment added by Alexsmith44 (talk • contribs) 08:15, 3 July 2010 (UTC)

It depends on the category of the jail, some low category jails may allow it, but not very likely due largely to the cost implications on the jail, I mean, just imagine all the prisoners having space heaters. In normal medium and high category prisons the possibility is zero for a range of security reasons. Richard Avery (talk) 10:11, 3 July 2010 (UTC)

what are the security reasons? —Preceding unsigned comment added by Alexsmith44 (talk • contribs) 10:17, 3 July 2010 (UTC)

The heater could have parts that could be used as a weapon.--Mr.K. (talk) 10:52, 3 July 2010 (UTC)

Or the heater could be used to start a fire. Really, I doubt any prisons would allow it. Maybe a halfway house, but that's about it. — Lomn 12:22, 3 July 2010 (UTC)

I doubt they would allow it - there are strict limits of the luxuries you are allowed in prison, and space heaters are a serious fire risk. They'll probably let your relatives bring you an extra blanket (in fact, the prison may well bring an extra blanket if you ask). A prison should be heated to a reasonable temperature anyway. If you have a medical condition that means you need a higher ambient temperature, they may well make allowances for you. (I'm assuming a prison in a developed country with a good human rights record - if either of those things don't hold, then there is no guessing what horrible conditions they might keep prisoners in.) --Tango (talk) 15:28, 3 July 2010 (UTC)

No jail in North America, at least, would permit a heater, due to fire risk, load on the electrical system, weaponization, and as a useful place to store contraband. Acroterion _(talk) 16:08, 3 July 2010 (UTC)

I imagine that's correct, but since this is the Reference Desk, perhaps someone would like to find a reference on the subject? --Anonymous, 17:15 UTC, July 3, 2010.

Do jail cells even have power sockets? This would seem unlikely to me Nil Einne (talk) 17:54, 3 July 2010 (UTC)

They might. Prisoners are sometimes allowed TV's or radios in their cells or reading lights. There might be somewhere to plug the vacuum cleaner. --Tango (talk) 18:28, 3 July 2010 (UTC)

I had a friend in a low security prison in the US. He was permitted to have some electronic devices (Specifically, I bought him a box fan). I don't know if he would have been allowed a space heater. It seems though that the heat was more of a problem than the cold, which is the case in many densely populated buildings. Buddy431 (talk) 21:08, 3 July 2010 (UTC)

It depends on the jail. In some, such as that described here, just about anythings goes. Zoonoses (talk) 12:35, 4 July 2010 (UTC)

Trapped in a fusion reactor?

Supposing a man got accidentally locked in a fusion reactor such as JET and it was switched on. Is the plasma dense enough and does it last long enough to do serious harm? We will assume he has a spacesuit so that the effects of vacuum can be ignored. —Preceding unsigned comment added by 80.1.80.3 (talk) 08:21, 3 July 2010 (UTC)

The temperatures are higher than the surface of the sun. The whole problem with fusion reactors of this kind is that the plasma has to be contained in a 'magnetic confinement' to avoid it vaporizing the walls of the reactor. So our poor test subject would likely be vaporized in a small fraction of a second. Still, that will prevent the huge amounts of radiation and energetic particles from being a problem. SteveBaker (talk) 14:39, 3 July 2010 (UTC)

We are talking about the same nuclear reaction that drives the sun, if a person could not survive on the sun, how could they survive in the nuclear reactor? The interior of the reactor has to be hot enough to keep the plasma ionized and it has to be compressed to undergo fusion, and considering people are made of matter, they would be vaporized just like the hydrogen and/or helium gases that power the reaction. In addition, some methods for heating the gases use high intensity microwaves, essentially turning the divice into a giant microwave oven. In other words, between the temperatures required for fusion, the radiation and energy produced by the reaction, and some of the ways the gases are heated, there is no way the unlucky person in the reactor would survive. —Preceding unsigned comment added by 74.67.89.61 (talk) 14:56, 3 July 2010 (UTC)

I apologise for my colleagues being so dismissive of this question. You are absolutely right that just because a plasma is very hot doesn't mean it would actually burn someone. Parts of the interstellar medium is plasma at 10,000K but obviously that wouldn't burn anyone because it is so sparse. Our article on the Joint European Torus says the plasma lasts 20-60s, so that would be long enough to burn if it is dense enough. So the question, as you say, is what is the density of the plasma? As the anonymous response says, the density has to be at least above a certain level in order for fusion to take place. Unfortunately, I can't find a description of what that density actually is. The JET website is full of interesting information, but apparently not that piece of information. The pressure inside the Sun is enormous, but the temperature there is more than 100 times the maximum temperature achieved at JET, so clearly the two situations are not comparable. --Tango (talk) 16:01, 3 July 2010 (UTC)

I believe exposure to neutrons would be a major problem, ignoring thermal energy flux for the moment, since the reaction liberates large quantities of fast neutrons (whose energy is harvested from the lining of the tokamak). Acroterion _(talk) 16:06, 3 July 2010 (UTC)

The neutron radiation could easily be a serious risk, you are right. Does anyone know the neutron flux for JET? --Tango (talk) 16:43, 3 July 2010 (UTC)

The original poster mentioned density. Is the pressure / amount of plasma sufficient? You can have very high temperature (= velocity) atomic particles, but if there are not too many of them, their combined punch will not be able to heat a cup of tea, much less vaporize anyone. I guess it's not so much about temperature than about total thermal energy. Also perhaps the type of particles matters; some kinds, while quite hot, might pass through our intrepid adventurer without sharing their kinetic energy? 88.112.56.9 (talk) 16:16, 3 July 2010 (UTC)

You can calculate the density from the pressure and amount of plasma, yes, but I can't find those numbers either. Total thermal energy would be a good way to estimate the harm, but the thermal conductivity of the plasma would be a factor too - a hot metal bar burns you more than a plastic bar with the same temperature and thermal energy, since more of the energy actually gets to you. --Tango (talk) 16:43, 3 July 2010 (UTC)

My intuition (= no sourceable fact content in this comment) suggests that gas/plasma uses a different mechanism to impart heat energy than a solid does. Gas goes: atom-skin. Solid goes: atom-atom-atom-atom-atom-atom*100000-skin. Atoms in tenuous gas/plasma do not conduct, they just hit you. 88.112.56.9 (talk) 19:54, 3 July 2010 (UTC)

Plus inside the plasma, the person would be crushed because the plasma is compressed.--dance (talk) 20:41, 3 July 2010 (UTC)

How much is it compressed by? I can't find any figures for the pressures they use. --Tango (talk) 21:15, 3 July 2010 (UTC)

The first thing they do before establishing the plasma is remove all the air with very good vacuum pumps. He will simply die of the lack of air. The person will ruin the vacuum, because the water in the body of a person takes ages to evaporate after some time hours or days they will open the chamber and find the person.--Stone (talk) 22:00, 3 July 2010 (UTC)

N.B.: per the OP: "We will assume he has a spacesuit so that the effects of vacuum can be ignored" -- Scray (talk) 22:09, 3 July 2010 (UTC)

They're going to let a multi-billion-dollar fusion reactor sit around for days with no vacuum before they bother finding out why? Boy, that fusion-workers' union — you don't mess with them, huh? --Trovatore (talk) 22:05, 3 July 2010 (UTC)

It won't be no vacuum, just a very slightly poorer vacuum. --Tango (talk) 01:07, 4 July 2010 (UTC)

For the purposes at hand, I imagine that's equivalent to no vacuum. --Trovatore (talk) 01:21, 4 July 2010 (UTC)

Thank you! I went to try and look that up (and didn't really succeed) and found just the information we need. This page says: "The densities of the hydrogen plasma that can be confined by magnetic fields are very low, about one million times lower than the density of air." I think that's low enough to be pretty harmless, even at temperatures of 100 million Kelvin. --Tango (talk) 01:33, 4 July 2010 (UTC)

No way! Just stop and think about what you're saying. The ENTIRE point of going to all that trouble with magnetic confinement is to prevent the plasma from vaporizing the walls of the reactor! If spacesuit material could comfortably withstand the heat/pressure then making a fusion reactor would be a pretty trivial problem - you'd make a simple spherical container and line it with 'spacesuit material' and you'd be done...so much for 60 years and billions of dollars of research! Our Tokamak article says "Magnetic fields are used for confinement since no solid material could withstand the extremely high temperature of the plasma."...so our intrepid experimenter is very, very dead from heat alone. You - and the others who say that the pressure is too low - need to explain why you'd need magnetic confinement if heat was not a major problem at the pressures the reactor is designed to run at. SteveBaker (talk) 03:26, 4 July 2010 (UTC)

That argument doesn't necessarily add up. I read the statement you referenced as "the magnets are needed to keep the plasma pure enough and at a high enough temperature" rather than "the magnets are needed to protect the structural integrity of the device". "Plasma vaporizing the walls of the container" isn't the same thing as "completely destroying the walls". From Magnetic confinement fusion, it appears that the magnets are to prevent "sputtering", where a portion of the wall is vaporized, thus lowering the temperature of the plasma. It's unclear how much of the container wall is actually vaporized, though. It's conceivable that a very small vaporization of the container (or man in a spacesuit) could ruin the experiment (by lowering the temperature and adding heavier element contamination), but still not compromise the structural integrity of the container (or health of the man in his spacesuit). Additionally, it appears that the plasma is only contained, heated, and compressed for a couple of seconds at a time. I don't know how much material would be vaporized in a couple seconds, and whether an ordinary spacesuit would stand up to it. Buddy431 (talk) 04:21, 4 July 2010 (UTC)

Right, that jibes with a vague memory of something I read somewhere, that the reason it is so essential to keep strict magnetic confinement is to keep the plasma hot; any failure of confinement, and consequent contact with a solid object, means your temperature instantly drops out of the working range for fusion. What happens to the solid object might also be a concern, of course, but keeping the plasma hot is a sufficient reason for confinement in itself, even if the damage to the torus were tolerable. --Trovatore (talk) 05:48, 4 July 2010 (UTC)

That's correct. If there were contact between the plasma and the walls, heat would be conducted from one to the other. A few wisps of plasma (which is all there is) will only remove a very thin layer from the walls. See the answer to the question "Q: I recently heard a sort-of scientific urbanmyth that at some point during its operation the plasma inside thetorus touched the top and the whole thing jumped up in the air. Isthere any truth to this story?" on the FAQ on the JET website (here). --Tango (talk) 06:25, 4 July 2010 (UTC)

Metric system

What's the metric prefix for 10²⁷? --75.149.70.22 (talk) 14:28, 3 July 2010 (UTC)

There isn't one. The official list is at SI prefixes, and stops at 10²⁴. Physchim62 (talk) 14:34, 3 July 2010 (UTC)

Our article on SI prefixes mentions hella as an option (not, it seems, used outside the USA). --88.117.82.138 (talk) 14:36, 3 July 2010 (UTC)

"Hella" is an idea at the moment – it isn't used anywhere, and doesn't seem likely to be adopted anytime soon (if at all). If they chose "hella" for 10²⁷, what would be the prefix for 10³⁰???? Physchim62 (talk) 15:11, 3 July 2010 (UTC)

hellalotta- you asked for it 87.102.21.49 (talk) 15:19, 3 July 2010 (UTC)

I was thinking more along the lines of "f*ckloadsa", but we can always save that for 10^33... Physchim62 (talk) 15:38, 3 July 2010 (UTC)

I wouldn't say that it isn't used anywhere. Dismas|^(talk) 00:37, 4 July 2010 (UTC)

Isn't "Hella" the Norse name for the underworld? 67.170.215.166 (talk) 01:20, 4 July 2010 (UTC)

_{try the search box?} —Preceding unsigned comment added by 77.86.10.42 (talk) 12:19, 4 July 2010 (UTC)

the sun

approximately how many fusion reactions occur within the sun every second? And how much light from it hits the moon?80.47.187.29 (talk) 17:22, 3 July 2010 (UTC)

The energy released by the creation of one Helium-4 nucleus by the Proton–proton chain reaction is 23.4MeV. The luminosity of the sun is about 4x10²⁶W. If we divide one by the other, we get about 10³⁸ Helium-4 nuclei being created every second. The amount of light hitting the moon is the solar constant multiplied by the cross-sectional area of the Moon. That's 1.4kW/m²*(pi*1700km^2)=7.5x10¹² W=7.5 TW. --Tango (talk) 17:39, 3 July 2010 (UTC)

nucleus diameter

How scientists measure the diameter of a nucleus? —Preceding unsigned comment added by Lukyshubham (talk • contribs) 17:58, 3 July 2010 (UTC)

You can get an estimate from the Geiger–Marsden experiment. Physchim62 (talk) 18:04, 3 July 2010 (UTC)

As Rutherford himself might have illustrated, imagine you are Jesse Ventura in Diamonds Are Forever (film). Jesse "The Body" has to find a single diamond hidden in a soft-brick wall in 10 minutes, with only his minigun in hand. He fires millions of bullets spread randomly on that wall, slicing cleanly in-and-out of the clay for 8 minutes until finally he hears a loud "ping" as one of the bullet literally bounces back. "Bullseye", he says with a smirk, then spits his tobacco on the floor... before being skinned alive by The Predator.

The moral of the story is that Rutherford fired alpha particle bullets into a metal foil wall to find the size of that diamond, the nucleus, as a ratio of the clay brick wall, the electron cloud. He fires millions of alphas, and millions go clean through the foil, but occasionally one or two bounce back. The cross-sectional area of the nuclei is then the ratio of #particles bounced back divided by # particles fired, multiplied by the area of the foil wall. Jesse Ventura fired a million bullets, one bounced back, onto a 10x10-meter wall, so the length of the diamond is about sqrt(100m^2 / 1000000) meters = 1cm. SamuelRiv (talk) 03:18, 4 July 2010 (UTC)

Which Diamonds Are Forever (film) are you talking about? Jesse Ventura never was in that movie -- it was Sean Connery as James Bond, and there wasn't any scene with a Gatling gun or a diamond hidden in a brick wall (I know, I've seen it on DVD)! 67.170.215.166 (talk) 04:19, 5 July 2010 (UTC)

It's in the "deleted scenes". Actually, I thought I made it quite clear that this was a remake of Diamonds Are Forever starring Jesse The Body and replacing Kidd and Wyndt with The Predator and requiring Bond to find a diamond in a clay wall using only his minigun for some reason. AND YET IT'S STILL MORE REALISTIC THAN THE ORIGINAL! SamuelRiv (talk) 08:19, 5 July 2010 (UTC)

In case the OP wasn't talking about an atomic nucleus and instead wanted to know about a cell nucleus, you can use a microscope. Smartse (talk) 21:12, 4 July 2010 (UTC)

Decreasing metabolism?

Right now, there is great interest in increasing metabolism to help combat obesity (including interest in leptin and uncoupling proteins). I was wondering whether there is any research being done about how to decrease metabolism. Why is it that when we talk about solutions to global starvation, we never talk about metabolism – we always talk about growing more food? I should be particularly interested in any effect that would help the body burn several hundred calories less each day.174.131.45.82 (talk) 18:09, 3 July 2010 (UTC)

That's easy: do less. That's not usually an option for people facing starvation, though. --Tango (talk) 18:26, 3 July 2010 (UTC)

Doing less would burn fewer calories and prevent metabolism from increasing as a result of exercise. I was looking for a way to decrease metabolism other than inactivity or caloric restriction, however.174.131.45.82 (talk) 18:59, 3 July 2010 (UTC)

Along the lines of medication that decrease metabolism? As far as I know, that's usually considered a side effect rather than a benefit... Ks0stm ^(T•C•G) 20:08, 3 July 2010 (UTC)

I doubt that using medications would be safe, and I do not know of any medications that decrease metabolism as a side effect (many drugs cause unwanted weight gain - insulin can cause some people to gain weight, but it actually increases the metabolic rate).174.131.40.152 (talk) 20:58, 3 July 2010 (UTC)

I think the human body already has pretty well developed famine modes which it goes into when near starvation. I doubt tinkering with a few chemicals would improve what happens naturally. --BozMo talk 21:38, 3 July 2010 (UTC)

We do have drugs that can decrease metabolism, but if your metabolic range is within normal range, it is not a good idea to use these drugs. They would cause someone to develop hypothyroidism which does have have very desirable effects. —Preceding unsigned comment added by 76.91.30.156 (talk) 02:03, 4 July 2010 (UTC)

I already knew about ATD's (antithyroid drugs). What I am wondering is this: why do some people have slow metabolisms (excluding people with hypothyroidism and Cushing's syndrome)? How are they different from us? What could we learn from them? One answerer also talked about the body adapting to starvation. It is true that metabolism does slow down a little during starvation, but this occurs, of course, only when a person does not consume enough calories. Would it be possible to decrease metabolism so that a person would need fewer calories in the first place, without ever actually reaching a negative energy balance.174.131.68.138 (talk) 02:20, 4 July 2010 (UTC)

Wouldn't that just make them fat? 67.170.215.166 (talk) 07:57, 4 July 2010 (UTC)

[Geography] Locate or name the lake by picture

I've seen some different wallpapers that picture one lake.

These are links to 3 different wallpapers with this lake: (Warning: links spend traffic, these are large pictures (1920x1080 or larger))

http://i8.fastpic.ru/big/2010/0704/2a/56aa011b5871afdeece9e24644d4682a.jpg

http://www.goodfon.ru/download.html?id=39203&rash=1920x1080

http://www.goodfon.ru/download.html?id=30598&rash=2048x1536

Where's this lake, does anybody know? And does it have a name?

--Grue12 (talk) 22:13, 3 July 2010 (UTC)

Moraine Lake in Banff National Park, Canada. Deor (talk) 22:16, 3 July 2010 (UTC)

Thanks a lot! --Grue12 (talk) 22:35, 3 July 2010 (UTC)

Superball dynamics

Hi, I was reading an article on Superball dynamics, and they said that during a collision with a table or something, the normal component of the velocity will change sign without justification. Why would this be true? 74.15.137.192 (talk) 23:01, 3 July 2010 (UTC)

It's basically reflection. The kinetic energy gets converted into elastic potential energy as the ball compresses and then that elastic potential energy is converted back into kinetic energy in the other direction as the ball rebounds. --Tango (talk) 23:54, 3 July 2010 (UTC)

But why wouldn't this potential energy get converted to rotational kinetic energy or tangential kinetic energy? 74.15.137.192 (talk) 00:12, 4 July 2010 (UTC)

It just doesn't. That's the way physics works. You can't get rotation with a torque and you can't get tangential movement without a tangential force. The force between the ball and the table will always be normal to the table. --Tango (talk) 00:18, 4 July 2010 (UTC)

But if the ball is spinning and has an initial component of its velocity parallel to the table, then surely there can be some torques and frictional forces, no? 74.15.137.192 (talk) 00:31, 4 July 2010 (UTC)

Ah, if it's spinning that's a different matter entirely. A non-spinning ball with an initial parallel component to its velocity will end up with a slightly slower parallel component due to friction - that's completely independent from the normal component, though, which is why I didn't mention it. --Tango (talk) 01:05, 4 July 2010 (UTC)

Sorry for not being clear. Is your explanation still valid? 74.15.137.192 (talk) 01:36, 4 July 2010 (UTC)

Wait a minute, Tango - I think IP74 is also asking why all the force is normal, even without a rotational component but with a component parallel to the table, since that parallel part gets a bit "stuck" on the frictional table surface and thus decreases the angle of reflection. That would lower the final velocity but conserve both energy and momentum since the table is pushed in the process.

By the way, Tango, not sure if you teach undergrad or high school physics labs ever, but if you ever get to use those collision carts on the low-friction airtrack, have your students model the curve that the computer outputs at the half-second during collision - it's a nice spring-storage SHM, illustrating that springs are everywhere, even in supposedly-rigid metal carts. SamuelRiv (talk) 03:05, 4 July 2010 (UTC)

Well, the OP was asking about the normal component of velocity, which is effected by the normal component of the force between the ball and table. It's always difficult to answer "why?" questions in physics - there always comes a point where you just have to say "That's just the way it is.". --Tango (talk) 06:15, 4 July 2010 (UTC)

I was trying to answer OP's question after your first explanation, about rotation components. Regarding the great "why" of the Normal Force, I defer at that point to "this is where we introduce the Conservation of Momentum". Indeed, the Normal Force can be explained in terms of smallball-bigball for most purposes where the Normal is actually useful. When it's not useful, it's a convention issue (we can use Lagrangian mech instead - that gave one class some relief that there was some arbitrariness in this grandiose of subjects). SamuelRiv (talk) 06:31, 4 July 2010 (UTC)

What I understood from this was, the normal force doesn't depend on any tangential or spin velocity (that just adds frictional forces), and because the normal velocity reverses in a no-spin/no-tang. velocity scenario, it must always do so. Is that the right reasoning? 74.15.137.192 (talk) 07:01, 4 July 2010 (UTC)

Time Dilation

The faster something moves, the slower time passes for it, but doesn't this prove something is moving which indicates absolute space? Assume that there is a train passing a train station and bob is on the train and alice is observing the train pass on the station. If alice watches bob's actions through a window she will see what bob is doing in slow motion. Likewise if bob watches alice's actions, it will seem everything is happening super fast (assuming the train was traveling at a signifigant proportion of the speed of light). According to relativity, as long as neither person is accelerating, both alice and bob could argue that they aren't moving and the other person is, but if time changes based off speed and time was passing slower for bob, wouldn't that prove that he is the one moving because if he was trying to argue that alice was the one moving, then she would seem to be acting in slow motion. Because it is bob that time is passing slower for doesn't that prove he is the one moving which violates relativity? —Preceding unsigned comment added by 74.67.89.61 (talk) 23:53, 3 July 2010 (UTC)

Actually, both Alice and Bob see the other moving in slow motion. This seems like a contradiction, but because of the finite speed of light it does actually work. If you do the maths, it all works out. --Tango (talk) 23:56, 3 July 2010 (UTC)

It's only if they come back to compare there clocks that there's a problem. But that's hard to do. Either it takes light from one clock a finite amount of time to reach the other person (and everything works out, as 74 asserts), or one (or both) of them have to accelerate to meet the other... Which means he (or she, the one who accelerates) is at some point in a non-inertial reference frame, where special relativity is no longer sufficient, and general relativity is needed to explain the time dilation effects. Buddy431 (talk) 02:51, 4 July 2010 (UTC)

No, general relativity is not necessary to understand the correct answer to that question. There is no gravity in that question and no gravity means no need for GR. Just because you used the word acceleration does not mean that GR is required for a correct answer. What Tango said is essentially correct. The solution to the apparent paradox passes by the definition of proper time. More on that later (it's late and I'm tired). Dauto (talk) 03:49, 4 July 2010 (UTC)

One of the central principles of GR is that the effects of acceleration are indistinguishable from the effects of an equivalent gravitational field. So I don't see why you say that a problem that involves acceleration but no gravity does not need GR, since we can rephrase the problem in terms of a gravitational field with no acceleration, and the results must be the same. Surely acceleration and/or gravity => non-inertial reference frame => GR, not SR ? Gandalf61 (talk) 13:16, 4 July 2010 (UTC)

There is no acceleration involved, and no gravity. The question refers to special relativity. --Wrongfilter (talk) 15:07, 4 July 2010 (UTC)

The original question is just SR, yes. But if you read the thread you will see that Buddy431's extension, where Alice and Bob meet up again to compare clocks in the same reference frame, introduces acceleration and hence GR - it is a version of the so-called twin paradox. And this was what Dauto's reply and my own reply to Dauto were referring to. Gandalf61 (talk) 17:22, 4 July 2010 (UTC)

You don't need GR to deal with nongravitational acceleration. SR is fine. You can, if you like, use general relativity and say that in an accelerating reference frame there is a gravitational field, but that's just words. The GR predictions are the same as the SR predictions, and the mathematical framework in which they're derived is the same. There are no GR corrections. It's true that you can't accelerate an inertial reference frame, but that's an intrinsic property of inertial reference frames, not an inadequacy of the theory describing them. You can accelerate objects of finite extent, like rocket ships, and SR correctly describes the acceleration. -- BenRG (talk) 19:17, 4 July 2010 (UTC)

But if that were true, then we wouldn't need GR at all, as we could convert any problem involving gravitational fields into an equivalent problem with accelerations and no gravity, and then solve it using SR alone. Is that what you are saying ? Gandalf61 (talk) 22:51, 4 July 2010 (UTC)

No, that's not what he is saying. The only problems that can be solved that way are the ones where there is no source of gravity. GR is a theory of gravity. When there are no sources of gravity GR reduces trivially to SR. In GR the presence of Energy (or momentum) density (or flux) leads to the warping of space-time. In the absence of any (large) energy-momentum density-flux GR predicts a flat spacetime AKA Minkowski space-time. The theory that describes the motion of objects within that flat space-time is SR. The objects that are moving within that flat space-time may be accelerating but the space-time is still flat and therefore SR is enough to describe that kind of situation. Just because I used the word acceleration in that last phrase doesn't mean that GR is now required since the space-time remains flat. I hope that helps clear that point. Dauto (talk) 01:58, 5 July 2010 (UTC)

Hmmm. So an observer who notices that spacetime in the frame of reference in which they are at rest is not flat can either (a) assume they are in a gravitational field, and use GR in their own frame of reference to handle this non-flat spacetime or (b) assume they are accelerating, and use SR in an inertial frame of reference (which is accelerating relative to them) in which spacetime is flat ? And they will get the same results in either case ? Gandalf61 (talk) 09:19, 5 July 2010 (UTC)

Not quite. A curved space will be curved in any frame of reference (No matter what system of coordinates you chose to use the surface of the earth will never be a plane). But there is always a flat reference frame that is tangential to any given point of the curved spaca-time which is good enough an approximation for small enough a region around that point. The better the approximation you need the smaller the region will have to be. Within that region gravity can be replaced by an accelerated motion in that tangential flat space. That's why the equivalence principle is said to be correct only locally, but not globally. Globally the non-flatness becomes evident and shows up as tidal forces that cannot be removed by a change in reference frame. The equivalence principle is a consequence of the fact that the gravitational forces really are pseudo-forces that come about because of the choice of a non-inertial reference frame. In that sense gravity is a force similar to centrifuge force or to Coriolis force. And that's also why gravitational forces are proportional to the mass of the object in which it acts (Centrifuge and Coriolis forces also are proportional to the mass). Gravity actually belongs on the right side of the equation F=ma. The difference is that because of the curvature of space there isn't a single inertial reference frame capable to cover the whole space-time. different inertial reference frames have to be used at different points in space-time. Dauto (talk) 13:28, 5 July 2010 (UTC)

The misunderstanding is at the very beginning of the question: For the thing that is moving there is no change in "how fast" time passes. Your wrist watch almost advances at the same pace, whether you're lying in bed, riding on a train or on an interstellar spacecraft or even sliding through a wormhole (hypothetically). Your wrist watch (alternatively, all the physiological processes in your body) runs according to your personal time, so called "proper time" (as hinted at by Dauto). If Bob rides on the train, then Alice, standing on the platform, will see Bob's wrist watch run slow compared to her own watch. Conversely, Bob will see Alice's wrist watch run slow compared to his own. The situation is symmetric, which means that there is no absolute time. Sounds weird but that's the way it is and there are no logical contradictions. --Wrongfilter (talk) 15:07, 4 July 2010 (UTC)

Ah, I think this is actually where I meant to put this comment, rather than at the "neutron star" thread above.

It is important to distinguish what Alice and Bob see from what happens in Alice's and Bob's coordinate systems; these are not necessarily the same thing at all. What they see is affected by the passage of light between them, and is subject to the Doppler shift and to various distortions that are not purely relativistic (I'm sure we have an article somewhere, but the title is too long to remember; something like visual appearance of fast-moving objects?).

The safe, if slightly laborious, way to keep track of it all is to imagine that Alice and Bob have both set up a network of space probes that extend as far as necessary in all directions. The probes make sure they are at rest with respect to one another by shining laser beams at one another and making sure there's no Doppler shift, and they keep their internal clocks synched up by the laser beams as well (they transmit the reading on their clock; the neighboring probe makes sure that it's correct after subtracting the time the light takes in transit). Alice's probes are at rest with respect to Alice, and Bob's with respect to Bob. Then Alice's probes take pictures of Bob's watch as Bob passes them, and Bob's probes take pictures of Alice's watch. The "slowdown" is computed by comparing the pictures with their timestamps.

That's all in flat spacetime, pure special relativity. When you have gravitational effects it all gets more complicated, and there are more assumptions that need to be specified. Near the event horizon of a black hole these all become especially bad. --Trovatore (talk) 03:19, 5 July 2010 (UTC)

July 4

Forward-swept wings

How stealthy are forward-swept wings? And would a forward-wing fighter make a good carrier aircraft? And how efficient are they at high speeds? --The High Fin Sperm Whale 00:39, 4 July 2010 (UTC)

(1) Not very, they tend to focus radar waves when viewed from head-on. (2) Forward-swept wings alleviate the boundary layer separation problems associated with swept-back wings, which allows a higher critical angle of attack and thus a lower landing speed. (3) About as efficient as swept-back wings, the reduction in wave drag is a function only of the absolute value of sweep angle without regard as to which way the wing is swept. FWiW 67.170.215.166 (talk) 01:35, 4 July 2010 (UTC)

Stealth: I would have thought that stealth wise it would be much the same as a back sweep, but not as good as 'current' steath aircraft (thinking of the B2 here) where blending the wing/body reduces reflective surfaces. However this site enemyforces.net says of the Sukhoi Su-47 'Berkut', "The forward-swept wing has a lower radar signature from the front hemisphere"

Carrier: What makes a "good carrier aircraft"?. Low take off speed would be one factor, which is why some (F-14 Tomcat) had variable-sweep wings, (though back-swept and which also seem out of fashion now, see [15]) Swept wings have better stall charateristics at high angle of attack, which is relevant for all fighter aircraft. Supposedly, no forward-swept wing(FSW) planes are in production, so I don't think anyone are contemplating Variable forward sweep, yet. Then again, see this at dreamlandresort.com which purports to show a Grumman Northrop design for a FSW with variable sweep! Here too [16] more recently. And Northrop Switchblade on Wikipedia!

This site century-of-flight.net says "Aircraft with forward-swept wings are highly manoeuvrable at transonic speeds". --220.101 (talk) ^\Contribs 01:54, 4 July 2010 (UTC)

Variable FORWARD sweep?! Sounds like a perfect recipe for structural failure! 67.170.215.166 (talk) 05:08, 4 July 2010 (UTC)

Why? I can't see why it would be any more of a problem than variable backward sweep. --Tango (talk) 06:18, 4 July 2010 (UTC)

More twisting loads on the wing pivots, for one thing -- what do you think is the reason why swept-back wings can be made from ordinary aluminum but swept-forward wings require fancy-schmancy composites? 67.170.215.166 (talk) 08:00, 4 July 2010 (UTC)

"More twisting loads on the wing pivots" - I can't see any justification for that statement.77.86.10.42 (talk) 12:18, 4 July 2010 (UTC)

If you experiment with a piece of paper, then you'll see why... 67.170.215.166 (talk) 23:48, 4 July 2010 (UTC)

You know I have absolutely no idea what that means, please read Wikipedia:Reference_desk/Guidelines#Guidelines_for_responding_to_questions - it's quite counterproductive to make unverifyable claims.87.102.23.18 (talk) 01:37, 5 July 2010 (UTC)

What I mean is, take a long, fairly narrow strip of paper and a portable fan, and hold it at an oblique angle to the fan's airflow at the end nearest the fan (as if it were a swept-back wing with the wing root facing into the airflow); the strip of paper should flutter a little about the horizontal, but come back to the horizontal after each oscillation. Got that? Good, now reverse the strip of paper so that you're holding it at the end farthest from the fan (as if it were a swept-forward wing with the wing tip facing into the airflow); now the strip of paper should twist one way (either up or down) as far as it will go, and stay that way. This is such a basic experiment for illustrating the aeroelastic properties of swept-back vs. swept-forward wings that I'm perfectly shocked that the three of you (Tango, 77.86 and 87.102) have no idea about it or about the implications of its results. Seriously, if the three of you had at least a little basic visualization skills between you, then you would've got my point about twisting loads without me having to digest the very basics for you. 67.170.215.166 (talk) 04:38, 5 July 2010 (UTC)

Yes but a wing isn't a completely unrigid structure - in the experiment you describe the strip of paper just bends back completely and flaps about in both cases - there is an article aeroelasticity and I can't see a mention of what you are describing it's not flutter. I originally thought you were talking about up/down (flapping) torsion, but I'm not sure if you mean torsion force that attempts to force the wing back.. (The link to 'responding to questions' was to explain the usefulness of supplying interwiki links or external references to help confirm or explain what you are saying)'.87.102.23.18 (talk) 13:30, 5 July 2010 (UTC)

Doesn't matter- I think Acroteroin has explained it below.87.102.23.18 (talk) 13:35, 5 July 2010 (UTC)

sorry for the inconveinience - it just wasn't clear what you were saying87.102.23.18 (talk) 13:49, 5 July 2010 (UTC)

Yeah, the Goa'uld really loved their death gliders. nerd joke. --mboverload@ 07:12, 4 July 2010 (UTC)

The WP article on FSW makes only passing mention, but FSWs exhibit divergent aeroelastic flutter characteristics which amplify wing twist in certain flight regimes, leading to failure. It can be controlled using composites, but the structure necessary to deal with it increases weight unacceptably in metal construction. Digital flight controls are also helpful. In general, in stealth design, re-entrant angles are avoided, as they focus reflections (think of a headlight's shape) rather than diffusing them. Acroterion _(talk) 16:35, 4 July 2010 (UTC)

Tin allotrope conversion

Why didn't the tin in my pewter spoon change from the beta form to the alpha form when it was in a freezer for a couple weeks? --Chemicalinterest (talk) 12:20, 4 July 2010 (UTC)

Firstly the change doesn't always happen immediately - ie the reaction can be delayed like crystallisation from a supersaturated solution.

Secondly Pewter contains metals that inhibit the allotrope change - see also Tin_pest#Modern_tin_pest_since_adoption_of_RoHS 77.86.10.42 (talk) 12:40, 4 July 2010 (UTC)

Yes it contained antimony and that is one of the substances used to prevent the conversion. --Chemicalinterest (talk) 18:24, 4 July 2010 (UTC)

Only pure tin is subject to tin pest -- that's why these days you always find it alloyed with other metals. 67.170.215.166 (talk) 23:33, 4 July 2010 (UTC)

No. Please stop giving incorrect answers. (Some forms of) Alloyed tin is also subject to tin pest. eg [17] p17/64. 87.102.23.18 (talk) 01:38, 5 July 2010 (UTC)

Fine, from now on I will not answer questions about metallurgy, since my knowledge of this subject is admittedly not up to date. As for questions about aerodynamics, I would recommend that you refrain from answering, considering the obvious lack of knowledge about the subject that you demonstrated during the discussion about swept-forward wings. 67.170.215.166 (talk) 10:23, 5 July 2010 (UTC)

Most tin alloys are not subject to tin pest. --Chemicalinterest (talk) 11:56, 5 July 2010 (UTC)

So the tin buttons on the French tunics which caused Napoleon to retreat from the Russian winter (tall story from chemistry teacher that one) were supposed to be pure? Does anyone know if its actually true? --BozMo talk 11:38, 5 July 2010 (UTC)

It would be hard to tell whether that story was true. --Chemicalinterest (talk) 11:58, 5 July 2010 (UTC)

Piano Lid Prop angle

Most modern grand pianos' lid props appear to form a 90º angle where they meet the underside of the piano's lid. It seems logical to me that the lid prop is less likely to slip at that angle because there would be a direct load transfer of the weight of the piano's lid to the support stick. That is, grand piano manufacturers intentionally use a 90º angle for safety reasons. Could someone show me the mathematics, perhaps using vector analysis, to prove my hypothesis? The reader may want to visit http://en.wikipedia.org/wiki/Grand_Piano to see a couple of pianos that do not appear to use the 90º angle. Note the Louis Bas grand piano of 1781 and Walter and Sohn piano of 1805.Don don (talk) 16:52, 21 June 2010 (UTC)

here with an angle less than 90 degrees downward slippage is impossible without raising the mass of the lid

You already asked, [18] as Dmcq might have hinted(?) - your hypothesis is wrong - at 90 degrees the only thing that is stopping slippage is friction. Whereas at more acute angles slippage is impossible since the mass of the lid needs to be raised for the support to be able to fold into the resting position.Sf5xeplus (talk) 13:59, 4 July 2010 (UTC)

Yes, after I posted my question on the mathematics page I thought that it should have been placed in the physics section. How can the original entry be deleted?Don don (talk) 18:25, 5 July 2010 (UTC)

I assume you didn't mean upward slippage of the support which should be prevented by a block.Sf5xeplus (talk) 13:59, 4 July 2010 (UTC)

Perhaps the most sensible solution would be to have slippage prevention blocks on both sides of support? Sf5xeplus (talk) 14:00, 4 July 2010 (UTC)

Also aren't some fixed and hinged at the lid ? Sf5xeplus (talk) 14:05, 4 July 2010 (UTC)

Perhaps I should clarify the angle to which I am referring in my original question. Looking at the picture of a grand piano found at the right, let point A be the center of the piano lid hinge pin to the left of the pianist; point B is the center of the support stick’s hinge pin; and point C is where the support stick contacts the piano lid. Currently, I have in my possession data (segment lengths AB, BC, and CA) from more than one hundred pianos. Using trigonometry, I have found that the average for angle C is approximately 90 degrees, plus or minus one degree. I know of no piano that has a hinge at point C.Don don (talk) 18:45, 5 July 2010 (UTC)

The premise of your question is wrong; most concert grand pianos don't have a fixed angle for the lid. Generally the lid angle can be adjusted either by elongating a telescopic prop (like this one) or by using a replacement prop (this article on recording gives 38° for the long prop and 10° for the short one). The piano lid functions as a reflector for mid- and high-frequency sound; Music, physics, and Engineering by Harry F. Olson shows a 15dB differential of 4kHz volumes horizontally rightward (outward) of the lid vs. the hingeward side. In performance, concern grands have to function well in a variety of settings - on the level with the audience, above them (e.g. on the stage in a multi-use hall where the audience is seated on a flat floor like a basketball court) or below them (e.g. on the stage in a concert hall where the audience on raked seating). So the musician will want the lid adjusted to best present the piano's sound to the audience (further diagrams in Olson suggest this effect is marked for at least the top half of the piano's range). So they'll set the lid angle to suit the performance space. It's not going to slip because it's not relying on flat friction to hold it up (there's a hole, or several). -- Finlay McWalter • Talk 15:34, 4 July 2010 (UTC)

I have been unable to find the "telescopic prop" on the website to which you refer above (beethovenpianos.com). However, I thank you for your comments on angle A. (See my clarifying paragraph above.) Using the data that I have collected so far, an angle of approximately 32 degrees, plus or minus 2 degrees, seems to be standard when the longest lid prop is used on a grand piano. There are only three sizes available commercially for a lid prop, that I am aware of, 31", 30" and 20".Don don (talk) 19:01, 5 July 2010 (UTC)

The prop resists the moment of the lid tending to turn on its hinge. The value of the moment is MGR.cosØ where M=mass of lid, G = gravitational constant, R = distance from hinge to center of gravity of lid, Ø = angle of elevation of lid. The force P exerted by the prop on the lid may have two components: P.sine@ and P.cos@ where @ is the angle between the prop and the lid. P.sine@ is a force parallel to the underside of the lid and, if present, it will cause the top of the prop to slide against the lid unless this is prevented by a block, notch or hole. P.cos@ counteracts the lid turning moment which implies that MGR.cosØ = PS.cos@ where S = distance from hinge to prop contact. The situation @ = 90 degrees is interesting because P.sin(90 degrees) = 0 meaning the prop has no tendency to slide against the lid. Nothing here requires R = S so the lid angle Ø can be chosen by the prop length and the OP's hypothesis is not unreasonable. Cuddlyable3 (talk) 16:19, 4 July 2010 (UTC)

Yes angle=90 is a Saddle point Local extrema, but saddle points local extrema aren't really safe at all.77.86.10.42 (talk) 16:50, 4 July 2010 (UTC)

No. It's just a local extremum. Cuddlyable3 (talk) 13:31, 5 July 2010 (UTC)

thanks corrected 87.102.23.18 (talk) 13:38, 5 July 2010 (UTC)

Surely the most logical reason for choosing the angle they do is to best project the sound from the piano towards the audience as it reflects off of the heavy lid. I doubt it has anything whatever to do with the angle of the prop. SteveBaker (talk) 04:46, 5 July 2010 (UTC)

SI prefixes (redux)

The earlier question about 10²⁷ got me wondering... Among the SI prefixes, why was deca- given the two-letter prefix da? I realize it couldn't be d because that is for deci-, but why couldn't it have been D, just as we have m for milli- and M for mega-? Thank you kindly. — Michael J 16:41, 4 July 2010 (UTC)

Did you spot SI_prefixes#Proposed_changes - it seems they are aware of the inconsistences, but haven't done anything yet.77.86.10.42 (talk) 17:32, 4 July 2010 (UTC)

I saw that. I just wondered why it wasn't done in the first place. — Michael J 17:43, 4 July 2010 (UTC)

I see what you mean. The page Kilo- has the answer - the odd ones out hecto, deca, kilo (and centi, deci, milli) are original Metric system prefixes (year 1795) so they were already introduced and in use when the SI-system was started. I suppose it would have been confusing or impossible to get people to change from lower case prefixes which they had been using for over 100 years

Clearly the original metrix prefixes don't follow the 'same word root/ upper or lower case pattern' as do the later SI type.

In fact the metrix prefixes use a greek derived word for 10,100,1000 and a latin derived word for 0.1,0.01,0.001 .... Metric_system#Prefixes 77.86.10.42 (talk) 18:27, 4 July 2010 (UTC)

Adidas Jabulani

Please see http://commons.wikimedia.org/wiki/Category:Adidas_Jabulani - as far as I can tell all official 'jabulani' balls are the '8 sided' truncated tetrahedron design, yet there are other designs in commons

File:Adidas Jabulani 2010 World Cup Replique 1.JPG

eg fake?

Can anyone confirm that this is a fake? Thanks.77.86.10.42 (talk) 16:46, 4 July 2010 (UTC)

I can't directly confirm, but I would note the presence of the word "Replique" in the name of the file (replica?), and also the fact that the match balls seem to be smooth, whereas there is distinct stitching on the one shown in the file. --TammyMoet (talk) 16:49, 4 July 2010 (UTC)

The Replique's are the cheaper version for sale to the mass market. See http ://hubpages.com/hub/Which-adidas-Jabulani-World-Cup-2010-official-soccer-ball-is-right-for-you (url split as for some reason that site is on the blacklist) for a truncated list of the different available Jabulani's. Nanonic (talk) 17:35, 4 July 2010 (UTC)

Thanks, the replicas seem to be more expensive than the ones with the new design.. odd

Resolved

Diopters

magnifying lens

magnifying and a diminishing lens

In corrective lenses, what does the unit diopter determine? Is this to do with how curved the lens is, how thick it is? Clover345 (talk) 18:39, 4 July 2010 (UTC)

Please read the article Dioptre. Cuddlyable3 (talk) 18:45, 4 July 2010 (UTC)

Thank you. I read that before I asked but its too scientific for me to understand. Clover345 (talk) 21:26, 4 July 2010 (UTC)

It's a measure of the strength of the lens - a higher value means a more powerful (either magnifying or diminishing lens)

A higher power lens is more curved, and being more curved means it is thicker. For a magnifying lens it's thicker in the middle, for a diminishing lens it's thicker at the edges.

So a higher dioptre lens will be thicker, and more curved than a lower dioptre lens.

I found some useful images at Eyeglass prescription. Note that a negative value is used to denote a diminishing lens.77.86.10.42 (talk) 21:35, 4 July 2010 (UTC)

If it has a positive diopter value you can use a lens as a Burning glass, as some boys discover to the discomfort of a few ants. Cuddlyable3 (talk) 22:15, 4 July 2010 (UTC)

If the sun is bright enough, you can even use a lens to light a campfire (like in The Mysterious Island). 67.170.215.166 (talk) 23:30, 4 July 2010 (UTC)

Though note that in Lord of the Flies, William Golding got it completely wrong by describing the short-sighted Piggy's spectacles being used in this way, even though lenses to correct myopia (such as my own) are diverging, somewhat diminishing my respect for the author when I read the book as a child. I thought I also remembered him describing a crescent Moon rising at sunset, but have subsequently failed to find such a passage - anyone else recall this? 87.81.230.195 (talk) 23:59, 4 July 2010 (UTC)

I've read the Lord of the Flies in high school, but it was a long time ago and I don't remember the details; the one thing I do remember is the conflict between Ralph and Jack, and the gradual transformation of all those civilized schoolkids into savages. 67.170.215.166 (talk) 04:44, 5 July 2010 (UTC)

You (87.81) are right about the crescent moon mistake - I remember it being mentioned in one of Martin Gardner's books. I'll try to find a reference later. AndrewWTaylor (talk) 08:14, 5 July 2010 (UTC)

Reptile's sense of time

Would a reptile, being cold-blooded, subjectively experience a cold day as passing very quickly, but a hot day as going on for much longer? 92.15.12.165 (talk) 19:24, 4 July 2010 (UTC)

There is no reason why it would. Cold-blooded does not mean it has cold blood; it only mean that it is poikilothermic. --Chemicalinterest (talk) 20:07, 4 July 2010 (UTC)

Do I really need to point out to you that "cold-blooded" is a common synonym for poikilothermic and that I and other people using that phrase do not mean that it literally has cold blood?

I'm wondering if, since its body would be hotter on a hot day, therefore its neurones should be faster, therefore its gets more thinking done within a constant time period compared with a cold day. Hence subjectively objective time seems to go more slowly for it on a hot day. 92.15.12.165 (talk) 21:49, 4 July 2010 (UTC)

This is an interesting question. We don't fully understand how humans keep time neurally on a minute-to-minute basis, much less reptiles. We do have a good understanding of how circadian rhythms are implemented in the brain, and I believe there is evidence that the day-clocks in cold-blooded animals are temperature-compensated to some degree, but do run a bit slower when body temperature drops. In short, the answer is not known. (And I haven't even addressed whether "subjective" actually means anything for a reptile.) Looie496 (talk) 21:55, 4 July 2010 (UTC)

Right, the question as phrased is about reptiles' qualia, which in principle are not amenable to objective analysis. The only way to know them for sure is to be the reptile in question. --Trovatore (talk) 22:12, 4 July 2010 (UTC)

Yes, but sleep passes very quickly for people (?) and there must be some objective difference in brain activity between the two.. Surely someone must have done a study on 'nerve activity in hot and cold reptiles' for us to be able to draw some sort of conclusion about the 'level of awakeness' or 'quality of experience' in reptiles as they get colder, if not whether times passes quick for them.77.86.10.42 (talk) 22:26, 4 July 2010 (UTC)

Nerve activity certainly slows down as body temperature drops. But there is no guarantee that timekeeping has a simple relationship to neural activity. Circadian timekeeping, for example, doesn't depend on neural activity -- the timekeeping process is driven by gene transcription. As I said, we really don't know at this point what mechanism determines subjective time, even in humans. It is probably neural activity at some level, but there are many types of neural activity, with different temperature dependencies, some of them pretty shallow. Human time estimation is definitely affected by body temperature, but since we are warm-blooded it is possible that there is less evolutionary pressure to counteract this in mammals than in reptiles. Looie496 (talk) 22:52, 4 July 2010 (UTC)

Discussion moved to Wikipedia_talk:Reference_desk#Grammitical_edits_to_question_header_on_science_desk_causing_offence please discuss there, and do not edit the questioneer's section heading further. Thank you.

Is there a behavioral explanation?

Is there a behavioral explanation for why the toilet-trained cat in this video is putting toilet paper in the toilet bowl? Bus stop (talk) 20:03, 4 July 2010 (UTC)

Yes, the cat has probably seen the human "parents" using paper after doing shit. So it has recorded the scene and trained itself to do so though of course it cannot understand why they do this. For the cat it is a ritual that its "parents" (who are humans of course) but to cat they are just big cats who are leaders of the pack who provide the cat food and protection, and what they do must be followed as a sign of respect  Jon Ascton  (talk) 21:57, 4 July 2010 (UTC)

No. It is instinctive behaviour of a cat to scratch earth over its stools. Tame cats try to do this even when there is nothing to scratch, as when the one in the video scratches at the plastic seat. At some point it discovers that a paper roll gives endless scratching satisfaction, that's all. Training a cat this way does not involve the owner performing for the cat to imitate. Cuddlyable3 (talk) 22:08, 4 July 2010 (UTC)

No, I have seen dogs and cats trying to involve the acts they have seen humans doing, and trying to reach out for electric switchs etc for no reason  Jon Ascton  (talk) 23:04, 4 July 2010 (UTC)

I could understand cats and dogs hitting switches for things because there's an immediate "reward" in that some action occurs that they desired. I find it harder, nigh impossible, to believe that a cat would paw at toilet paper just because they had seen a human do it. There's no "reward" for the cat. If the paper is close enough, as in the video, they may paw at that. Notice also that the cat in the video paws all around the seat. It's not unusual for a cat to paw at the walls and other vertical surfaces around a litter box either. (WP:OR warning) Around one of our litter boxes, we have a piece of hard plastic because we found the cats were scratching at the wall and causing damage. They still scratch at the plastic sheet but now it doesn't damage the walls. PS That's a good sized deuce for a cat! Dismas|^(talk) 00:19, 5 July 2010 (UTC)

Maybe pet psychology is different in India than in the West (!) Perhaps the evolutionary relation between human and cat/dog followed a different route to evolve in India, you know due to we-are-spiritual-you-are-materialistic factor  Jon Ascton  (talk) 03:50, 5 July 2010 (UTC)

Strange beetle

I took a picture of a beetle and uploaded it to Flickr. Can anyone help me identify it?

Americanfreedom (talk) 22:03, 4 July 2010 (UTC)

Reminds me of a Cockchafer, (which is probably isn't) - maybe you could look at the family Scarabaeidae while you wait.. or http://commons.wikimedia.org/wiki/Category:Species_of_Scarabaeidae (there's only 287 to go through..)

I think you need to tell us where you are to aid the indentification, eg region of the country you're in, not your address :) 77.86.10.42 (talk) 22:07, 4 July 2010 (UTC)

And how big is it, _{and what is that thing in the background - a tray with a picture on or something?}77.86.10.42 (talk) 22:10, 4 July 2010 (UTC)

This is definitely not a cockchafer, look at the antennae. Kinda hard to tell from that angle, but my guess would be that this is a fairly large longhorn beetle, maybe Cerambyx sp. or Neocerambyx sp., family Cerambycidae. Some of those guys, although of course not the largest of the beetles in general, tend to get impressively large. Let me know where you took the picture and maybe I can narrow it down to the species. --Dr Dima (talk) 23:08, 4 July 2010 (UTC)

I can't help directly, but here is a site [19] that may help you now or in the future. Caesar's Daddy (talk) 07:27, 5 July 2010 (UTC)

I took a closer look at the longhorn beetle picture you've got, it may be Prionus sp.. --Dr Dima (talk) 08:04, 5 July 2010 (UTC)

abbe or v number

I haven't been able to find a thing on this - I was wondering what would be a typical abbe number (or some other comparative figure at other wavelengths) (whatever measure of chromatic aberration is relevant) for a camera lens assembly - ie is it higher than 59 as per crown or CR39 glass. I was thinking about a mid priced lens rather than super expensive, or super cheap..77.86.10.42 (talk) 23:02, 4 July 2010 (UTC)

Mig-29

Does the Mig-29 require a starter cart to start its engines? Or does it have self-start capability? Thanks in advance! 67.170.215.166 (talk) 23:37, 4 July 2010 (UTC)

This article says it has a "GTDE-117 gas-turbine starter-APU". Moreover this article has a "RD-33 Engine Start" section which says that stored air (from engine bleed) is normally used to start the engines, with the APU or battery-only as fallbacks. I dread to think how quickly you'd flatten your battery starting (presumably just one) engine. -- Finlay McWalter • Talk 00:04, 5 July 2010 (UTC)

Don't know about this Mig thing, but think I heard very interersting thing about the starting-of-engine process of a large propellor (probably a pre WW2 model) aircraft. They put a 12 gauge shotgun cartridge and fire it to start the engine ! Is that true...? Jon Ascton  (talk) 03:20, 5 July 2010 (UTC)

Yep - they really exist. See Coffman engine starter. SteveBaker (talk) 04:37, 5 July 2010 (UTC)

I knew that the Canberra bomber used a cartridge starter; didn't know that piston-engine aircraft also used this system... 67.170.215.166 (talk) 04:48, 5 July 2010 (UTC)

Well, talking of novel uses of 12 gauge shotgun cartridge, is there any machine which exploits the rush of its blast to dig a hole in ground when you have no time for a shovel ? Jon Ascton  (talk) 05:56, 5 July 2010 (UTC)

Submarine aircraft carriers of Japan employed pre-heaters for aircraft engine oil so that minimum time would be spent while vulnerable on the surface for starting and launching floatplanes. Cuddlyable3 (talk) 13:18, 5 July 2010 (UTC)

July 5

Giant isopod-- survival outside of water

Can a giant isopod survive on land? If so, for how long? (I ask because I've seen photos of them on land and apparently alive, such as [this], unless it's fake). 68.123.238.146 (talk) 03:22, 5 July 2010 (UTC)

That photo is definitely a fake. The trilobites vanished in the Permian extinction about 250 million years ago. The ICanHazCheezburger.com site is all about faked photos of one kind or another. Modern isopods are things like wood lice - but there is the Giant isopod Bathynomus - that can grow to a half meter or more in length. They live in very deep ocean areas though - I don't think they'd do well out of water. SteveBaker (talk) 04:35, 5 July 2010 (UTC)

So I can buy realistic plastic Trilobites?

...Excellent... --mboverload@ 04:39, 5 July 2010 (UTC)

Ancient Machines

This site seems to be deliberately laid about yarn. They mention a "University of Chandigarh" which does not exist. The guy who made this up knew that there is a town called Chandigarh in India allright so thought that must have uni too ! There may be other loopholes too that need to be exposed  Jon Ascton  (talk) 03:42, 5 July 2010 (UTC)

Sorry, don't understand your question. Laid about yarn? --mboverload@ 04:13, 5 July 2010 (UTC)

I think he's stringing you along:) Myles325a (talk) 08:10, 5 July 2010 (UTC)

(EC) My guess is that the university is a self-made non-accredited diploma mill, as pretty much no credible scholar would entertain the so-called ancient astronauts theories. Not even the champion of the hypothesis in Chariots of the Gods? has much of anything in the way of credentials. In other words, academia is immune to this nonsense, but the public isn't, especially when this kind of trash appears on National Geographic channel's Is It Real? see critique. Alas, pseudoscience runs strong still, but we can take comfort that cocaine/opium or worse brain tonics aren't sold every which-way in every CVS nowadays, as would be in the late 19th century. SamuelRiv (talk) 04:16, 5 July 2010 (UTC)

Excuse me but there is no need to begin to bad mouth a university or an individual because a 'new-ager' has mentioned it/them in one of their 'new-age' articles. Please do some research before jumping to conclusions, both SamuelRiv and JonAscton. Thank you.87.102.23.18 (talk) 15:28, 5 July 2010 (UTC)

Just to clarify the scholar was Dr. Ruth Reyna who obtained a PhD from University of Poona 1961 , [20] (also written books eg amazon link). The Chandigarh university appears to be Panjab University, and the documents are supposedly held there. I don't know where the documents come from or anything about their authenticity. 87.102.23.18 (talk) 15:23, 5 July 2010 (UTC)

I assume these are the documents [21] ?

Hypothetical Example I have a crackpot theory that the observable universe is explained by tiny silver men inside my eyeballs casting shadows on my retina, and by invisble golden mice jumping up and down on my head to simulate gravity.. In my article I reference Isaac Newton's work on optics and gravity .. Science is not advanced by attempting to discredit the work of Newton, or Cambridge University simply because I mentioned them in my blog..87.102.23.18 (talk) 15:44, 5 July 2010 (UTC)

I don't understand the question here. The "strange artefacts" site is obviously just another conspiracy theory site, not to be read when sober. anything else? Physchim62 (talk) 16:01, 5 July 2010 (UTC)

I think the aim is to show some flaw - I would guess that dating the documents would be one obvious route to disproving the 'ancient astronaut' theory. If the documents prove to be old then that would be a suprise too.87.102.23.18 (talk) 16:08, 5 July 2010 (UTC)

To IP87: fair enough, the fault is mine. Regarding the original question, I hope I answered that with the links in my first response which discuss the whole history of the ancient astronauts kookery. Regarding the "documents" linked, they are actually modern drawings of what are assumed to be descriptions of flying machines in one of the Vedas, if I recall from a badastronomy.com debunking. It's like if I were to draw a picture of Noah's Ark, except I assume beforehand that it's actually a spaceship. SamuelRiv (talk) 16:12, 5 July 2010 (UTC)

Additionally there's no shortage of flying machines/people/creatures in many other ancient texts (along with gold growing rings, cloaks of invisibility etc etc) . Many people require more than just some text describing space travel to make them believe that it's literally true. The usual requirement for a historical document to be taken seriously as containing elements of fact is supporting evidence, usually archaeological remains, or secondary independant documents corroborating (eg such as matches in historical events/people in the bible and in eqyptian historical sources). 87.102.23.18 (talk) 16:47, 5 July 2010 (UTC)

Could you clone a human female from male cells?

As a male cell contains both an X (female) chromosome and a Y (male) chromosome, would it be possible to pair up two of the X chromosomes and create a human female from the male's genome? And if a boy was cloned (XY) as well as a girl (XX), could the two of them theoretically breed and start a population that derives from a single parent? I am supposing that human females can't do this because they do not have a Y chromosome. We are used to thinking that only females can create life, but the whole cloning idea is reminiscent of the Genesis story in which Eve is created from Adam's body, and not vice versa.Myles325a (talk) 08:28, 5 July 2010 (UTC)

I am not used to thinking that only females can create life. Cuddlyable3 (talk) 13:11, 5 July 2010 (UTC)

I suppose it might be possible, in theory, to remove the Y chromosome and duplicate the X chromosome to create a female "clone" from a male cell. But that cell now has two identical X chromosomes, so any recessive genes on the X chromosome have been duplicated, which will reduce the chances of the "clone" developing normally. Similarly, mating clones will result in a population with very low genetic diversity, which leads to low fertility rates. Gandalf61 (talk) 13:27, 5 July 2010 (UTC)

Strange hydrostatic aroud Australia

Hello dear thinkers. Please excuse my uneasy English, I'm French. For those who prefer the French version, here it is [22].

On this page Australie : nouvel état des lieux du réchauffement climatique they are explainations concerning the rise of water of the oceans linked to the global warming. My question is about a single strange sentence: It's written that "This rise of water is not uniform (all aroud Australia ) : on the south, the sea has risen at a speed of 3 mm per year during the recent years, but has risen at a speed of (from 7 to 10) mm per year on the north coast."

Question : How can the sea rise at a different speed on 2 shores opposite the same island ? (even if this island is a near-continent). With the difference between the 2 speeds (7-3)=4 mmm per year for 10 years long it makes 4 cm (more than 1 inch).

When I asked this question on the Reference desk in French for the 1rst time some months ago I got answers refering to earth gravity changes, but I think it's not conclusive. I also got answers sending me to Geoid but I also think that this geoid doesn't change enough to explain this strange difference between 2 "sea rise speeds".

Sorry for being so long asking my question, thank you very much for your "cogitations". Joël DESHAIES- Rheims-France---90.18.59.162 (talk) 14:07, 5 July 2010 (UTC)

Surprising as it may sound to you, the answers you were given are the correct ones. Dauto (talk) 14:38, 5 July 2010 (UTC)

The mistake you seem to be making is to look at the issue as a problem in hydrostatics, when the oceans are a dynamic system with continual energy input from the Sun. The height of a column of water depends on other factors apart from gravity, for example the temperature and the salinity. Just to give one well known example, the mean sea level (MSL) of the Pacific Ocean at that end of the Panama Canal is 20 cm (8 inches) higher than the MSL of the Atlantic Ocean just 77 km (48 miles) away. Physchim62 (talk) 15:42, 5 July 2010 (UTC)

Yes, there are 'anomalies' in sea level height, but how to explain the anomaly in the rate of sea level rise ?? 87.102.23.18 (talk) 16:28, 5 July 2010 (UTC)

Well, I think the question has a point. It is possible to have different sea level changes at different points, but it clearly isn't possible for the differences to build up continuously at a steady rate for very long. And the explanation wouldn't be in terms of earth gravity changes, but more likely in terms of changes in wind patterns and ocean currents, which push the water from place to place. As Psychim62 pointed out, there is a difference of 20 cm between the Atlantic and Pacific sea levels at Panama, caused by winds and currents. Looie496 (talk) 16:31, 5 July 2010 (UTC)

Perhaps the answer is related to average changes in wind/current direction then , happening at the same time as (and probably related to) the overal sea rise/warming effect. I don't know how to search for such data though.87.102.23.18 (talk) 16:40, 5 July 2010 (UTC)

The Australian tectonic plate[23] moves North (causing seismic activity Java - Indonesia). Might this induce a N-S tilt in Australia? Cuddlyable3 (talk) 18:44, 5 July 2010 (UTC)

Difference between mud and loam

What is the difference between mud and loam, please? I am interested mainly in their use as a construction material.

Wikipedia says that loam consists of sand, silt and clay and that mud consists of soil, silt and clay, but can also contain sand. Thus, to me not being a native speaker both terms seem to have the same meaning. --146.107.3.4 (talk) 14:11, 5 July 2010 (UTC)

In British English, "loam" is used to describe a type of soil (a type that often becomes "muddy" in the British climate): "mud" is a more general term. Physchim62 (talk) 14:31, 5 July 2010 (UTC)

(ec)In agriculture Mud generally refers to ground matter/earth/soil that is clay rich - (ie not really peat, topsoil, humus though they will be described as mud when very wet). it is generally clay like in nature, but may be more liquid. Loam is generally more particulate, and so has better drainage.. though when wet it can be described as mud. It is not as slicky as clay.

In argricultural terms they are different. In building terms they are less distinct see http://www.thefreedictionary.com/loam : "A mixture of moist clay and sand" which is very similar to mud as used in Mudbrick.87.102.23.18 (talk) 14:38, 5 July 2010 (UTC)

You might find this useful http://www.tbe-euro.com/downloads/SustainableBuildingConference-Austria1.pdf : in construction terms I haven't found an example where the two can't be used fairly interchangably.87.102.23.18 (talk) 14:42, 5 July 2010 (UTC)

More Were you thinking of mudbricks and walls or more in terms of groundwork/subgrades? In terms of foundations etc this book makes some useful distinctions [24] 87.102.23.18 (talk) 15:04, 5 July 2010 (UTC)

The difference is a matter of water content mainly. Mud always means soil that is saturated with water enough to be sticky and squishy. If it dries out, it isn't mud any more. Loam is simply a type of soil with no particular implication about water content. If you saturate loam with water, it turns to mud. Looie496 (talk) 16:20, 5 July 2010 (UTC)

Is it Gun cotton?

This is what we found inside, besides shot itself, when we open a of 12 gauge shotgun cartridge in India. Is it so in the West ? I mean what exactly you find in yours in real life (as opposed to Wikipedia article)? What exactly is the stuff I have shown ? Is it gun cotton, or cordite what ever ? (the scale in centimeter) Jon Ascton  (talk) 17:24, 5 July 2010 (UTC)

I think it is either cordite, or, more likely, Smokeless powder. Guncotton isn't black. --The High Fin Sperm Whale 17:36, 5 July 2010 (UTC)

No, it ain't black either (only seems in pic).It is in fact dark green  Jon Ascton  (talk) 17:42, 5 July 2010 (UTC)

It's not pure gun-cotton nitrocellulose which is white like cotton. Nor cordite (image in article). It's almost certainly a form of smokeless powder. You might find this useful [25] - the powder is similar but slightly lighter. If the cartidge was more full than those in the links with powder then they could be using real gunpowder which is less powerful.87.102.23.18 (talk) 17:53, 5 July 2010 (UTC)

You might find the cutaway shells here [26] [27] interesting to look at.87.102.23.18 (talk) 18:05, 5 July 2010 (UTC)

(edit conflict)In that case, I'd guess smokeless gunpowder (smokeless gunpowder has so many different formulas and ingredients, probably one of them is dark green, or it could even be coloured like that on purpose). --The High Fin Sperm Whale 17:56, 5 July 2010 (UTC)

Yes, this is exactly what I found. The link seems to say that such shot is only meant for bird game, i.e. not in serious ammunition ? What about the standard shotgun ammo in the west ? And what about pistol and revolver etc. ?  Jon Ascton  (talk) 18:26, 5 July 2010 (UTC)

And, the other question, actually I already asked it in another post - talking of novel uses of 12 gauge shotgun cartridge, is there any machine which exploits the rush of its blast to dig a hole in ground when you have no time for a shovel ?

shotgun shell with slug

Erm - shotguns are mostly for bird game, (or firing at rats etc) - the small amount of powder is due to it's high power - shotgun cartridges were originally made for gunpowder - see Shotgun_shell#Construction_of_a_typical_shotshell " Modern smokeless powders are far more efficient than the original black powder used in shotgun shells, so very little space is actually taken by powder" - even the ones for bigger game have a lot of wadding eg see Shotgun slug (image right).

Were you thinking about military use? 94.72.242.84 (talk) 18:40, 5 July 2010 (UTC)

For rifle or pistol cartridges I can't do better than this [28] which has numerous cutaway diagrams of the 'shells'.94.72.242.84 (talk) 18:44, 5 July 2010 (UTC)