Wikipedia:Reference desk/Archives/Science/2012 October 2

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October 2

Kinetic reclamation to your phone due to bodily movements.

I've heard of watches that recharge just by the movement of your wrists. It's piezo mechanics, right?

Why can't we put that in phones so that when they move with your body while in the pocket, your bodily movements help recharge them too?

If they're on their way, how far away is that technology, and why isn't it here already? Thanks. --129.130.237.27 (talk) 03:59, 2 October 2012 (UTC)

The power drain of a wrist watch is minuscule compared to a smartphone. Think about it: A little button battery can run a watch for 5 years, and that technology is like 50 years old. You can use your smartphone for what, like 8 hours before it needs a charge, and that's on supermodern batteries that didn't exist 5 years ago. The difference is like the difference between shooting a bullet with a gun (the smartphone) and throwing it (the watch). Not even a comparison. --Jayron32 04:15, 2 October 2012 (UTC)

Our article, orders of magnitude, claims a quartz watch uses about a microwatt, without a source. That's about one million times less power, on average, than a modern computerized mobile telephone (at least, while it is in use). While looking for sources, I found a fantastic detailed overview of the Seiko 7S26 mechanism; and Mike Murray's Everything You Ever Wanted To Know About Mainsprings (except, of course, a quantitative estimate of their potential energy content!) I seem to recall, though I'm unable to cite a source, that quartz digital watches consume more power than a mechanical spring-driven watch; but a quartz watch battery contains significantly more potential energy than a spring... which is why digital watches last longer than mechanical watches, between winding. Let the race begin to find a reliable reference! Nimur (talk) 04:24, 2 October 2012 (UTC)

OK, we've established that movements of the wrist won't be enough, but how about if we put a crank on the phone ? I have a portable radio that runs off a crank, and that uses a comparable amount of energy to a cell phone, I bet. Obviously cranking your phone to charge the battery isn't something you'd want to do often, but you might on occasion, like if your car breaks down and you then discover that your cell phone battery is dead. StuRat (talk) 04:52, 2 October 2012 (UTC)

They exist, though mostly they are advertised as "emergency" chargers: Crank like crazy for 5 minutes and you can get enough charge to call a tow truck. Maybe. Not enough to be practical to keep a constant charge, but enough to put a few minutes of call time on the phone. --Jayron32 04:57, 2 October 2012 (UTC)

Virtually all automatic watches have been mechanical - a swinging weight rewinds a spring, as with the Sieko cited by Nimur. I very much doubt that windup cellphones would catch on, for three reasons: Firstly, the typical radio frequency output of a hand held cell phone is 200 milliwatts (200 mW). The typical maximum electrical power into the speaker in the sort of transistor radio that you can get with a crank generator is 250 mW. That sounds about the same, but that's not the whole story. The cell phone's RF output is continous throughout each call. Assuming a typical call duration of 3 minutes, that's 36 Joules of energy, assuming the circuitry is 100% efficient at converted battery DC into radio frequency output (which it won't be). However, the ratio of average to peak power in voice and music is very low - a transistor radio may have a max output of 250 mW but the battery drain under typical programme conditions wil be more like 20 mW, or 1.2 Joules per minute. But there's more: The second reason: When you've had enough of listening to the news and the latest silly nonsense from the politicians, you can turn the radio off and energy consumption is then zero. But a cell phone must be kepton, in standby mode, so you can recieve calls. It's hard to find reliable data for phone standby drain, but 10 mW would not be unreasonable. That means a consumption, above the call consumption, of 0.6 joules per minute, 144 joules each 4 hours. Thirdly, a generator and crank of the necesary size would increase the volume and weight of a cell phone by about a factor of 10. Keit60.230.201.133 (talk) 05:14, 2 October 2012 (UTC)

Jayron's answer above includes pics that appear to about equal the size of the cell phone they charge, so not 10X. You might want to take one with you camping, for example (away from electrical outlets but still within cell tower range). Also, you don't need to leave your phone in standby mode. You can just wind it up, make a call, and turn it back off again. When you next wind it back up and turn it on, you should get any messages left for you. Many people in the older generation don't even like to leave them on, given the choice. StuRat (talk) 05:31, 2 October 2012 (UTC)

You must have magic fingers as when I tried Jayron's link, it didn't work. However Jayron makes it clear it's not very practical - he said "crank like crazy for 5 minutes...and make one (quick call) ...maybe". Yes, you can turn the phone off, but to catch on ie be a market success the phone has to work normally - stay on standby and let you make decent length calls. Orthwise only a few nutters will buy it. Don't forget that every time you turn a cellphone off, and everey time you turn it on, it enters into an automatic dialog with the network, so the network knows which base stations to route the call to, should it be worthwhile. This is so that the network can operate efficiently - each time a call (or text) comes in, the network doesn't try every base station in the country (and any other countries you can roam to) - the network only tries base stations at and adjacent to where they last found you, and doesn't try at all if it knows your phone is off. It means the sequence turn it on - make a quick call - turn it off is very inefficient energy-wise. Keit124.182.178.117 (talk) 13:12, 2 October 2012 (UTC)

...which is because modern mobile telephones are cellular phones, not general-purpose radio telephones. Cellular mobile telephones require an elaborate digital communication protocol, and must remain in contact with a cellular base-station, in order to operate. Contrast this to, for example, a conventional civilian aviation handset, whose RF transmitter is on exactly only when transmitting voice signal; and whose receiver amplifier requires very little power. Nimur (talk) 16:23, 2 October 2012 (UTC)

I'm certainly not suggesting that the crank would be used normally, you'd go with the usual battery operation and wall outlet recharging, then. And the crank charging unit would be separate, so you don't have to carry it around with you (although you could, in a pocket or purse, say). StuRat (talk) 18:02, 2 October 2012 (UTC)

Infant mortality rate and birth rate

Is it true that if the Infant mortality rate goes down in an area (i.e. child survivability goes up), the birth rate usually goes down more so that the population declines? Bubba73 ^{You talkin' to me?} 05:07, 2 October 2012 (UTC)

Damn skippy. See Demographic-economic paradox for a fuller treatment. --Jayron32 05:35, 2 October 2012 (UTC)

I don't know what "damn skippy" means, but in any case the statement is not true. The literature on the relationship between infant mortality and fertility struggles to find any consistent effect at all, and certainly not a large enough effect to counteract changes in infant mortality. It's hard to even imagine how that could come about. Looie496 (talk) 05:39, 2 October 2012 (UTC)

You could read the article I provided a link to. Or you could provide your own links or references. See [1] for your other question. --Jayron32 05:45, 2 October 2012 (UTC)

Well, as you point out, there is a lot of evidence that increases in prosperity can reduce birth rates to the extent of causing a population decline. But that article doesn't say that changes in infant mortality alone can do it. A Google Scholar search for articles on the topic finds a number of them, none of them supporting such a result as far as I can see, but none that is recent and authoritative enough to be worth citing; see for example http://www.nber.org/papers/w1528. Looie496 (talk) 06:13, 2 October 2012 (UTC)

I suppose part of the problem is teasing out a specific correlation. There's a melange of factors which leads to modern development, and the things which lead to decreased infant mortality tend to get all wrapped up in the same sorts of things (education, industrialization, improved access to health care) that leads to "development" generally speaking. What you'd need is some sort of society where somehow there was access to fully-modern prenatal care, but with no other development at all. It's hard to imagine such a place existing, so there isn't really a great way to run the experiment. What we're left with is the broad trends, that show that as a population becomes more "developed" in an economic sense, the birth rate goes down (as does infant mortality), but I'm not sure there's a causative or directly correlative effect which could be isolated for those two AND ONLY those two variables. --Jayron32 06:34, 2 October 2012 (UTC)

It's false. See Qatar, UAE, and Bahrain for counter-examples. A8875 (talk) 07:54, 2 October 2012 (UTC)

See outlier. Come back when you have a question. --Jayron32 13:58, 2 October 2012 (UTC)

They certainly look correlated, but that doesn't mean that one is the cause of the other. Like Jayron pointed out, the two are probably both just small parts of overall development. Here is a graph showing the two values plotted against each other: [[2]] 209.131.76.183 (talk) 11:48, 2 October 2012 (UTC)

The (loose) association is known as the demographic transition. Itsmejudith (talk) 13:20, 2 October 2012 (UTC)

Those graphs are quite interesting, and the type of information I was looking for. Bubba73 ^{You talkin' to me?} 14:00, 2 October 2012 (UTC)

I'm in favor of helping children in underdeveloped countries live, but is that making the problem worse in the future? Bubba73 ^{You talkin' to me?} 17:12, 2 October 2012 (UTC)

Good question. like it could explain the genocide that occurs in Africa. those that were "supposed" to be dead a long time ago are eventually killed by the militias fighting for food, etc.165.212.189.187 (talk) 17:33, 2 October 2012 (UTC)

It's a complex problem. One the one hand, compassion dictates that we don't let people starve. On the other hand, long-term continuous foreign aid in the form of basic necessities may also be what is keeping these countries from developing their own native food and clothing industries. Here is just one article on the detrimental effect of clothing donations on the domestic clothing industry in Nigeria. here is a similar paper which poses the same sorts of problems related to food aid and farming. However, issues noted in the D-E paradox and Demographic transition articles noted above point to the solution likely coming from development and education in general. I've seen some studies which show a marked improvement in living conditions in areas with properly applied Microcredit systems that allow development of native industries instead of blanket food aid. Which is not to say that food aid isn't needed in some dire cases, but it isn't the end of the solution. --Jayron32 17:42, 2 October 2012 (UTC)

Relatedly, I've seen analysis of why current Somaliland (formerly British Somaliland) is fairly highly functioning and stable, compared to the basketcase that is the rest of Somalia (formerly Italian Somaliland), claiming that it is at leastly partly that Somaliland has received hardly any foreign aid, and hence politicians who wanted money had to negotiate with local groups to get funding (leading to a level of democracy), and local markets were not undercut. 86.159.77.170 (talk) 19:55, 7 October 2012 (UTC)

There is also breastfeeding infertility, the tendency of women breastfeeding live children for a few years not to ovulate during that period. μηδείς (talk) 00:44, 3 October 2012 (UTC)

derivation of lensmaker's equation for a thick lens

What is the derivation of the lensmaker's equation for a thick lens. It seems hard to find on the internet. (Why isn't it on Wikipedia?) 137.54.11.202 (talk) 05:44, 2 October 2012 (UTC)

Does Lens_(optics)#Lensmaker.27s_equation answer your question? --Jayron32 05:46, 2 October 2012 (UTC)

By googling "derivation lensmaker's equation" I found the following two derivations[3][4]. I personally don't think derivations belong on Wikipedia(unless the derivation warrants its own article), since they add unnecessary bulk to the articles. People looking for an equation and people looking for a derivation of the same equation are on completely different skill levels. A8875 (talk) 07:20, 2 October 2012 (UTC)

I had always presumed that the lensmaker's equation is strictly an empirical first-order approximation. This is why it doesn't hold up very well to modern, complicated materials like compound glasses. The incredibly over-priced, but totally-without-equal, textbook Applied Photographic Optics, has no substitute: it thoroughly runs through the physics and the actual engineering specifications for many common lens glasses and compound lens groups. Nimur (talk) 16:18, 2 October 2012 (UTC)

The reason why I need a derivation is that I am solving a problem for an exotic lens, so I'm trying to use the same technique to derive the thick lens equation to create an equation for my exotic lens. 199.111.224.96 (talk) 18:21, 2 October 2012 (UTC)

If your lens is exotic enough that the lensmaker's equation won't work, you would probably have to analyze it by tracing rays through it rather than trying to derive a new lensmaker's equation. Commercial lens design software does this, but it is quite expensive.--Srleffler (talk) 04:27, 3 October 2012 (UTC)

Nimur, the lensmaker's equation is not just an empirical formula. It's based on a simple, geometric model for ray propagation. As you guessed, it is a first-order approximation; specifically the paraxial approximation. Geometrical optics in the paraxial limit is sometimes called Gaussian optics.--Srleffler (talk) 04:27, 3 October 2012 (UTC)

But ray-propagation refraction is based on a linear fit to an experimentally-measured index of refraction. There are more elaborate methods to model index of refraction. The text I linked above outlines a 3-parameter dispersion model for different types of common lens glass materials; using that model, the effects of apochromaticity, and chromatic aberration, are all accounted for; but the lens-maker's equation assumes that n (index of refraction) is constant across the visible spectrum. In fact, I believe I discussed this extensively in December of last year in response to a question about modeling index of refraction. Everything in physics is always experimentally derived; even if you design a ray-tracing algorithm where each photon interacting with each iota of matter in the glass lens is calculated from quantum-mechanical first-principles, you still need an experimental value for the fundamental physical constants. Nimur (talk) 17:02, 3 October 2012 (UTC)

I think we are using the same terms differently. An empirical equation is one which is obtained purely by finding an expression that fits experimental data, without any theoretical model of the process. The lensmaker's equation is based on a simple model of light propagation. It is therefore not an "empirical equation". Almost everything in physics is experimentally derived, but not all formulas are empirical.

The lensmaker's equation does not assume that the index of refraction is constant across the visible spectrum. It assumes that you are using the correct value of n for the wavelength that you are interested in.--Srleffler (talk) 17:27, 3 October 2012 (UTC)

Fair enough. I was probably abusing the terminology, "empirical equation," in a way that wasn't clear. Nimur (talk) 17:43, 3 October 2012 (UTC)

what is the resistor for?

Hello, in this circuit, what is the 1MOhm resistor for? Also, shouldn't there be a series resistance in series with the potentiometer? Thanks in advance! Asmrulz (talk) 14:09, 2 October 2012 (UTC)

The 1M resistor sets the time constant (and the scaling factor) for the differentiator. An extra series resistance in line with the potentiometer wouldn't help or hurt the measurement of the time-derivative of current, because it is constant; it might be a good idea to prevent accidental shorting of the battery. Many potentiometers don't span all the way to zero ohms, so it's not necessary in practice. To fully analyze the role of the resistor, you should write the circuit equations in terms of a complex impedance, which will allow you to solve the circuit in the Laplace domain, permitting a straightfoward accounting for the time/frequency effects as well as the differentiation. Nimur (talk) 14:25, 2 October 2012 (UTC)

I haven't thought of this as an RC element *slaps himself on the forehead.* Thanks again. Asmrulz (talk) 15:01, 2 October 2012 (UTC)

Out of curiosity, how would I go about writing the circuit equations? Asmrulz (talk) 15:59, 2 October 2012 (UTC)

Kirchoff's circuit laws, appropriately using complex impedance instead of simple resistance. A typical first course in circuit analysis goes over the common techniques to write out equations for each node and then uses the techniques of linear algebra to solve simultaneously, giving the voltage at each node, and the current in each branch. Nimur (talk) 16:14, 2 October 2012 (UTC)

Like this, just in complex numbers? Asmrulz (talk) 18:10, 2 October 2012 (UTC) not quite, probably... Asmrulz (talk) 23:39, 2 October 2012 (UTC)

Yes, network analysis is the basic technique. The page you linked is specifically about DC network analysis: you notice that every example contains only resistors and batteries, never a capacitor or inductor or active circuit element. That's because those components require AC analysis, or complex impedance, which is a subject in mathematics that some introductory circuit texts try to avoid because it's a little bit harder than basic arithmetic. But, there's no magic; there's just a few rules for the algebra when complex numbers are involved; and then a handful of common techniques, like multipole expansion or separation to partial fractions... these are just common mathematical recipes that help you solve the parallel-and series- circuit equations that commonly show up in circuit analysis with capacitors and inductors mixed in. Nimur (talk) 14:58, 3 October 2012 (UTC)

Morphine/Dilaudid conversion

Does anyone know what 20 mg of Kadian converts to in terms of hydromorphone? This is simply a factual question; I am not seeking any kind of medical advice. Joefromrandb (talk) 14:50, 2 October 2012 (UTC)

Anyone?... Nevermind...too late.165.212.189.187 (talk) 17:29, 2 October 2012 (UTC)

Some kind of Wikispeak? Joefromrandb (talk) 17:50, 2 October 2012 (UTC)

I think the OP is suggesting this is a homework question. Nil Einne (talk) 20:27, 2 October 2012 (UTC)

Our Hydromorphone article, while fairly poor, ('and it can be said that hydromorphone is to morphine as hydrocodone is to codeine and, therefore, a semi-synthetic drug' - is it a copyvio or copied from a public domain source without talk page attribution or something?) says this:

Hydromorphone's oral-to-intravenous effectiveness ratio is 5:1 and equianalgesia conversion ratio (hydromorphone HCl to anhydrous morphine sulfate, IV, SC, or IM) is 8:1. The oral equianalgesic conversion rate (hydromorphone HCl to morphine SO4) can vary between 5:1 to 8:1. Therefore, 30 mg of immediate-release morphine by mouth is similar in analgesic effect to about 4–6 mg of hydromorphone by mouth (requiring extra care during conversion & titration), 10 mg of morphine by injection, and 1.5 mg of hydromorphone by injection.

Given the state of the article, I can't vouch for these figures. But it does indicate an unsurprising issue, this isn't a simple factual question and there's no simple conversion. These are related but different drugs so don't have any perfect correlation in effect. (Our article also says other things which indicate this.)

Also your question is fairly unclear. In the subject you mentioned 'Dilaudid', in the question you mentioned 'Kadian'. The later is apparently an extended release form of morphine sulfate in capsules and the former a immediate-release form of hydromorphone hydrochloride in tablets or liquid. From this it sounds like you're referring to oral ingestion (obviously an important consideration) but the lack of consistency and generally limited information makes this unclear. Notably, if you're referring to an extended release form of morphine and an immediate release form of hydromorphine, this wasn't clearly specified and is unlikely to be obvious to anyone unfamiliar with the specific brands you mentioned in the subject or in the question.

Nil Einne (talk) 20:15, 2 October 2012 (UTC)

BTW I've added one link above from the article as again while not a great article, it'll give you an idea of how to begin to look for answers (plural intentional) Nil Einne (talk) 20:24, 2 October 2012 (UTC)

Thank you very much for the help! And although I am perpetually learning, I can assure you my "homework" days are decades behind me. Joefromrandb (talk) 20:33, 2 October 2012 (UTC)

And perhaps I was more vague than I needed to be. Specificallly, I was comparing an 8mg immediate-release Dilaudid to a 20mg extended-release Kadian. Joefromrandb (talk) 20:40, 2 October 2012 (UTC)

Bowel flora

What happens to the bowel flora if a person only receives IV nutrition? Does it "starve" to death? If not, how does it obtain sufficient nutrients? Thanks in advance.--Leptictidium (mt) 14:56, 2 October 2012 (UTC)

Do people only receive IV nutrition? I'm pretty sure that for any nutrition is dealt with via Feeding tube. Intravenous therapy is used for fluid or electrolyte replacement, but not generally for caloric needs, at least on a long term basis. --Jayron32 16:34, 2 October 2012 (UTC)

Yes, check out "Total peripheral nutrition", for example.--Leptictidium (mt) 16:55, 2 October 2012 (UTC)

Google is your friend. This list came from Google Scholar. Use it wisely. --Jayron32 17:05, 2 October 2012 (UTC)

what is the formula for the angular magnification for a single lens?

Like say, a magnifying glass? Everywhere I look on the internet, the formula is always given for microscopes or telescopes with an "eyepiece", which is totally inappropriate to my problem. 199.111.224.96 (talk) 18:22, 2 October 2012 (UTC)

It sounds like you want a textbook introduction to geometric optics! Our article does have a formula for magnification for a simple lens, which is suitable for a high-school-level approximation. (This equation, which is commonly used in simple lenses, expresses magnification as a magnitude, in terms of focal-length and object distance). Whether that formula applies to your lens, or not, is entirely what makes optics a not-very-easy subject. If you've never studied optics formally, a decent introduction is found in Tipler's Physics for Scientists and Engineers, (in the second volume, Chapter 30-something in the 2nd edition). If you really really want to study optics, you should start with formal analysis of geometric optics and then study generalizations of electromagnetic wave propagation; so that you can appropriately model the lens or optical path you care about. Nimur (talk) 18:30, 2 October 2012 (UTC)

I think the problem is that a single lens can give you magnification that is so distorted and dim as to be useless. So, the question becomes "what is the useful magnification", which depends on the application and is subject to opinion. StuRat (talk) 18:33, 2 October 2012 (UTC)

A simple lens doesn't have a single angular magnification. The magnification depends on how far away the object and/or image planes are. Magnifying glasses are a special case: these are commonly quoted as having a particular angular magnification. You can find the formula for this at Magnifying glass#Magnification (inline within the text). There are two formulas, both of which presume a "typical" human eye. One formula presumes that the magnifying glass will be used to bring an image to the near point of the eye. This gives the highest magnification, and is the value typically quoted when magnifying glasses are sold. The second formula presumes that the lens will be placed about 1 focal length from the object to be viewed. This gives slightly lower magnification, but is often more convenient. The actual magnification a person experiences depends on the ability of his or her eye to accommodate. A young person experiences much less magnfication when using a magnifying glass, compared to an old person with presbyopia.--Srleffler (talk) 17:20, 8 October 2012 (UTC)

Animal eyesight and the electromagnetic spectrum

I was thinking about how awesome it would be to be able to see radio waves directly and not needing them to be interpreted by television sets, wireless receivers, etc. (watching TV and browsing the Internet directly with your eyes just by looking at open air!), which led me to the following questions:

• Why did animals evolve eyesight focused toward the infrared, visible, and/or ultraviolet parts of the electromagnetic spectrum, instead of evolving sensitivity toward other parts of the spectrum such as radio waves, microwaves, x-rays, or gamma radiation?
• Under what sorts of environmental conditions might we expect animals to have instead evolved sensitivity toward those other regions of the spectrum?
• How would such hypothetical animals perceive the world? For instance, radio waves can penetrate right through walls, with distortion being minimal enough that wireless receivers read them just fine. Does that mean a theoretical animal that sees only radio waves would be unable to see walls?

—SeekingAnswers (reply) 18:59, 2 October 2012 (UTC)

A picture is worth a thousand words. I'll let this one do the speaking. --Jayron32 19:22, 2 October 2012 (UTC)

Yes, and that's the problem. All they would see is the source of the radio emissions, so things like stars and lightning. Animals need to see what's right around them, so visible light, ultraviolet, and infrared are ideal, as they reflect off nearby objects, and some are actually produced by certain organisms (like mammals producing IR). Sound and vibrations, while not part of the EM spectrum, similarly react with, and are produced by, the local environment, so are useful. Similarly, the ability to "see" electrical fields is useful. StuRat (talk) 19:10, 2 October 2012 (UTC)

There are environments where there is no light, and blind organisms evolve there. (While some organisms can give off their own visible light, most do not, and sufficiently murky water would make even that approach unusable.) An organism living in space might not be able to use sound and vibrations, and, if inside a dark cloud of gas, might not be able to see, either. I'm not sure if being able to detect those other wavelengths of EM would be of much benefit, though. StuRat (talk) 19:30, 2 October 2012 (UTC)

(ec with Jayron) The solar spectrum and the opacity of the atmosphere to electromagnetic radiation have a lot to do with the wavelengths to which the eyes are sensitive. The Sun's output peaks at ~550nm and so does the sensitivity of the eyes. The articles on eyes has some general information on this, and Color vision has quite a lot of information. Astronaut (talk) 19:28, 2 October 2012 (UTC)

Cellular Renewal

I have heard that every seven years, all the cells in a human body will be replaced. I am quite skeptical of this claim, so I wanted to know about it's factual accuracy. Is it true? Also, aren't there some cells, especially neurons in the brain, that are not replaced?128.227.85.113 (talk) 19:57, 2 October 2012 (UTC)

No, that is not true. But it does seem to have a grain of truth, or was based on a true claim. According to the website for Stanford's Institute for Stem Cell Biology [5], "Every single cell in our skeleton is replaced every 7 years." (emphasis mine). Granted, that web page does not link to a specific reference for that claim, but I'm willing to trust them on this one. SemanticMantis (talk) 20:24, 2 October 2012 (UTC)

Also, some cells are replaced much faster, like red blood cells, which only last a few months. StuRat (talk) 20:27, 2 October 2012 (UTC)

This is a decent link on the subject: [6]. --NorwegianBlue ^talk 22:28, 2 October 2012 (UTC)

The way I heard this one a long time ago was that every seven years the atoms would be replaced because of how the (electrons? protons?) swap places with each other so much. Is that totally wrong? Would there be a viable equation or something to determine the atoms version? ~ R.T.G 00:17, 3 October 2012 (UTC)

No. Electrons can bop around, but protons and neutrons stay with the atom (barring nuclear fusion, nuclear fission, and radioactive decay). StuRat (talk) 01:55, 3 October 2012 (UTC)

Entire atoms are exchanged to. Much of the molecules in your body are constantly being repaired and regenerated on a molecule-by-molecule basis; a carbon atom that was part of a fat cell yesterday could be part of hemoglobin next week, and be breathed out as CO2 in a few weeks. I have no idea on the time scales involved, but it isn't just the electrons that shuffle around. The "renewal of the body" thing, taken on an "atom-by-atom" basis is a classic example of the Ship of Theseus/George Washington's Axe paradox... --Jayron32 03:08, 3 October 2012 (UTC)

Jayron is right. That's why cells have nuclei, to produce new protein molecules by genetic transcription, ultimately from chromosomal DNA. Red blood cells die within about a month because they cannot transcribe from DNA, lacking cell nuclei. One can't give a seven year expiration date, but all living cells obviously regenerate their molecular structure or die. μηδείς (talk) 03:20, 3 October 2012 (UTC)

A more ignorant question than usual: Do red blood cells have DNA? More to the point, is the premise of Jurassic Park, of mosquitoes trapped in amber and full of dinosaur blood whose DNA could be harvested, theoretically possible? ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:07, 4 October 2012 (UTC)

Red blood cells don't have a proper nucleus, so no, they don't have the same sort of DNA that other cells do. However, there are many nucleated cells in blood, including white blood cells, so you could get a full DNA sample from those. As far as the rest of the Jurassic Park scenario; which would involve using that DNA to make a T-Rex... no comment. --Jayron32 05:27, 5 October 2012 (UTC)

DNA is a stable molecule, but it still decomposes over time. There's no way there would be usable DNA in an amber sample.128.227.214.249 (talk) 20:19, 10 October 2012 (UTC)

The fossil part of Fossil fuels

How can we know that fossil fuels are originally from fossils? Before life appeared on Earth, there should have been plenty of carbon around, so, couldn't the fuel been formed be it? OsmanRF34 (talk) 22:09, 2 October 2012 (UTC)

While we wait for an answer to your question, I give you Abiogenic petroleum origin for the nay-sayers ;) --Tagishsimon (talk) 22:14, 2 October 2012 (UTC)

I know that there is such a kind of fringe theory about oil, but it doesn't explain the other side of the equation. Why is it regarded as common wisdom that oil's origin is from fossils? OsmanRF34 (talk) 22:19, 2 October 2012 (UTC)

Well, from direct observation we can see that plants and animals produce oils and methane (hopefully not too directly on that one). Methane is also produced by natural processes, but we don't know of any inorganic process which produces oil or coal. StuRat (talk) 22:27, 2 October 2012 (UTC)

In the case of anthracite (hard) coal, you sometimes find the fossils right in the coal: [7]. We also have intermediate steps, like peat bogs, around today. StuRat (talk) 22:20, 2 October 2012 (UTC)

"fossil fuel" does not mean "fuel made of fossils", it means "fuel that is a fossil", where "fossil" means "of the ancient past". That's all it means - fuel from long ago. -- Finlay McWalterჷTalk 22:21, 2 October 2012 (UTC)

Sure, but the question is why the biogenic origin theory is favoured over the abiogenic in the cases of liquid and gaseous fuel. Oddly, the best we have on this seems to be at Abiogenic_petroleum_origin#State_of_current_research which summarises arguments in favour of the biogenic origin theory. --Tagishsimon (talk) 22:25, 2 October 2012 (UTC)

One of the things to consider is that we've got examples of every stage along the mechanism from living things to coal. Consider things like Peat which is just very young coal, Lignite, Bituminous coal, anthracite, etc. You can pretty much find all of the steps along the mechanism right now. What we don't find is large quantities of all of the steps of so-called "abiogenic coal". --Jayron32 22:32, 2 October 2012 (UTC)

Fossil, according to Wikipedia, means something which is fossus, dug up from the ground. Fossil fuels are made of stuff that plants produce when they decompose and are found in places where plants seem to have been decomposing. Scientists can not only theorise how oil occurs, they can/could (it's all gone remember) could predict whereabouts it would be and how much would be there, even under the ocean, so they know something. Household garbage can be compressed down some way to squeeze petrol out of it sort of like coal can be squeezed into diamonds in a pressure machine and those things are probably about as direct as any evidence we could ever get without time travel. Sap from some trees, for instance of doubt, drips and pours in large amounts. It hardens and becomes amber with little million year old insects in it. Maybe it's not tree sap after all, But it's made of tree sappy stuff and has little tree insects stuck in it. But don't worry about that because the oil is all gone away to a better place now with the copper and other useful stuff. ~ R.T.G 00:00, 3 October 2012 (UTC)

Have I missed where we have an article on abiogenic coal? Thomas Gold seems to have suggested that some bituminous coal may have a non-biological origin. I am not aware of any serious, detailed, broadsweeping theory that all coal can be explained without reference to decaying plant matter. μηδείς (talk) 03:13, 3 October 2012 (UTC)

Why would you think there would be plenty of carbo around before the advent of life (photosynthetic life, in particular)? Given that it would be sitting in an atmosphere full of oxygen, for billions of years? Gzuckier (talk) 06:31, 3 October 2012 (UTC)

There was carbon in the form of calcite in limestone from the Archaean, the oldest hydrocarbon source rocks are of Paleoproterozoic age page 46. Mikenorton (talk) 09:57, 3 October 2012 (UTC)

oh, OK, sure. I was thinking of elemental carbon, like big lumps of coal lying around on the ground. Gzuckier (talk) 16:46, 3 October 2012 (UTC)

This paper describes what is known about the early composition of the Earth's atmosphere and specifically discusses this in relation to the origins of life (and therefore the availability of carbon). Mikenorton (talk) 12:39, 3 October 2012 (UTC)

Rating rechargeable battery input and output

I think most people get cheaper electricity at night and I wanted to work out if it would be viable to buy large batteries, such as those used in solar energy systems, charge them at night and then run some home electrical devices from them. Okay so working out, I understand wattage. KWh just means kilowatt per hour. My heater here beside me is 2kWh and that means while switched on it uses 2 per hour chared to me at 32cent each. Now, try to figure out how many kWh it takes to charge a battery and how many kWh it will return is not so straightforward and there isn't much explaining it on the internet from what I see. It's just a pie in the sky but if you could get back much more than 50% of the energy expended charging etc., no reason not to work that out.. There is some discussion board stuff on the net and I probably work it out after a while but might take me a few hours reading and frowning... Anybody on here just kind of know? ~ R.T.G 23:27, 2 October 2012 (UTC)

I can't see that working out. There's the inefficiency in charging and discharging the batteries, and the initial cost of batteries, plus maintenance costs, since they don't last long. And, from an environmental POV, there's all those old batteries to dispose of. A better approach would be to heat an insulated tank of water at night, then use that hot water to heat the home during the day. StuRat (talk) 23:36, 2 October 2012 (UTC)

It's more for the electricty itself. Water heating and conversion would be impractical for me (Ireland climate, small apartment), but if you could charge one of these extremely efficient new cells and discharge enough for the PC, the kettle, or even oven etc.. I know probably unlikely but also kind of frustrating not to do the equations very easily. I have a torch powered by 2 CREE batteries about mid size between AA and D cell, light as softwood, but it's bright enough to dazzle and goes for a couple hours, I've no idea however how much wattage it takes to charge and how much it releases, only that it seems powerful and that large size batteries can be made of the same stuff. I know if it could be done people would probably be doing it already but how close we are I can't tell. ~ R.T.G 00:14, 3 October 2012 (UTC)

Cree makes LEDs not batteries. It sounds like you have some random lithium ion cells, probably 18650 ones branded by some random manufacuturer in China with some random brand they knew was associated with torches. Nil Einne (talk) 14:46, 3 October 2012 (UTC)

There's a lot of discussion out there about storing energy from off-peak times. This is a very reasonable idea, and can potentially aid in the efficiency of the entire network, as well as save money for users. The cost/benefit analysis is tricky, but you may be interested in the idea of using a flywheel to store energy. I can't sort through them right now, but /home flywheel energy storage/ presents several interesting results on google, and some are commercial products for home use: [8]. See also our article on flywheel energy storage. SemanticMantis (talk) 01:40, 3 October 2012 (UTC)

BTW, you said your heater is 2 KwH. I think you mean it's 2 Kw, which means, if you use it for an hour, that makes 2 KwH. The main problem with batteries is that they are expensive and don't last, with inefficiency being a minor concern. The flywheel suggestion may work, because, unlike batteries, it shouldn't need to be replaced every few years. If you want electricity, rather than just heat, another option is to pump water into a water tower at night, and use that gravitational potential energy to run an electricity generator as the water flows back down to a lower tank, during the day. StuRat (talk) 01:43, 3 October 2012 (UTC)

The most common way that energy is stored from off-peak hours, at least residentially, is a Storage heater, which basically heats up some bricks at night, and allows the heat to escape into the room during the day. Storing heat is nice because it is, pretty much by definition, 100% efficient - any electricity you use will be converted into heat, and the design is such that most of it can be directed when and where you want it. If you want to get energy back in a usable form (to power your computer, or whatever), it's a little harder. On a municipal level, it's probably most common to pump water up, and then let it fall back down to reclaim the energy: Pumped-storage hydroelectricity. This is what is done at the Robert Moses Niagara Hydroelectric Power Station: They generate more power at night (because they don't have pushy tourists who want to look at the falls), but they have higher demand during the day. They therefore pump water up into a man-made reservoir at night, and let it come down during the day to provide supplemental electricity. Not practical on a residential level, to say the least. One of the ways being looked at to store power during cheaper times (again, mostly on a larger scale, though possibly adaptable to a household) is a flow battery [9]. The general term for this sort of thing, by the way, is Load balancing. Sorry I'm not providing much specific help, but I thought I'd point out some of the things that are done. Buddy431 (talk) 04:21, 3 October 2012 (UTC)

From what I read, it looks like a lead-acid or lithium-ion will cost you more than the energy they can store in total, i.e. even if you could charge them for free, you'd still be losing money. And when you compare with the market price, the price at which companies are selling to one another, the whole thing looks even worse: at the moment it's quite high, 60€/MWh or 0.06€/kWh for peak hours. Storing the electricity will cost you maybe 3 times as much, even if you get the power for free. See link. Water to a reservoir: assuming 100% efficiency, pumping 36000 liters 10 meter higher will store 1kWh. Don't expect a small pump to be very efficient, not in pumping nor in generating electricity. And you need two reservoirs, don't forget. I cn't think of much to use the cheaper energy to your advantage. Maybe unplugging the fridge and freezer during the day, if the highest temperature is still acceptable. (You can fill all free space with water filled containers to increase thermal mass.) Won't make that much difference, but at least you'll be paying a bit less, not a lot more. Ssscienccce (talk) 17:37, 3 October 2012 (UTC)

They (some sort of battery) can't cost more if you charge them for free because that would render renewable energy useless, and I don't know about the safety and cost of keeping a flywheel in the home powerful and precisely engineered enough to run cookers and heaters off, but basically I have the impression that the idea isn't often considered, which happens. Been looking at renewable energies for years myself and not had that idea or given it any consideration. Oh well I will just have to do the sums but thank you for all the energy info. ~ R.T.G 09:25, 4 October 2012 (UTC)

The advantage of batteries is that they are portable. So, while you end up paying more for energy delivered by battery, having an extension cord running to your flashlight isn't very practical, and this is where batteries shine. StuRat (talk) 16:56, 4 October 2012 (UTC)

That's (one of the) the problem(s) with renewable energy at the moment, solar and wind energy are unreliable since they depend on the weather, and storing the energy is still more expensive than burning coal. It's not the cost of the energy needed to charge them, it's the cost of buying them and the limited number of charge-discharge cycles you get. If they lasted forever there wouldn't be a problem, but lead-acid for example only last maybe 600 cycles, so after a 1kWh battery has stored a total of 600kWh you have to buy a new one for 90€. That's 0.154€ per kWh, without the electricity cost. Unless you pay only 0.16€ for electricity at night, you're gonna be paying more than 0.32€ for the electricity you've stored (actual cost will be higher because the efficiency of charging and discharging is only 75-85%). Electricity suppliers give you the lower night rate because they can't store the energy themselves and many power plants have to keep running at night (coal plant takes 12 hours to start up again, nuclear may take weeks, gas turbines only minutes). If it was cheaper to store the energy than to sell at the price they do, it would make no economic sense for them not to store it themselves, and they could use bigger, more efficient storage options than you can. Ssscienccce (talk) 13:47, 5 October 2012 (UTC)