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July 17

How do people sleep during daylight conditions?

Ny-Ålesund at 79°N claims the world's northernmost hotel, which should be able to locate a sleep mask for their guest. If not, try the airport. DreadRed (talk) 11:13, 17 July 2013 (UTC)[reply]

I brought dark curtains with me, otherwise I would have gone mad! 83.109.151.51 (talk) 09:54, 17 July 2013 (UTC) (Count Iblis posting from his vacation address)[reply]

Maybe this requires more explanation. Are you being forced to sleep during the day and be up at night, or are you maybe posting from very high northern latitudes with a marked lack of darkness? HiLo48 (talk) 10:00, 17 July 2013 (UTC)[reply]

The latter, I'm now at the Northernmost hotel on Earth and they don't have decent curtains. 83.109.151.51 (talk) 10:08, 17 July 2013 (UTC)[reply]

Everybody's different. Whether it's noon or midnight when I'm sleepy, I just close my eyes and fall asleep. But then, I may have already gone around the bend. Clarityfiend (talk) 10:13, 17 July 2013 (UTC)[reply]

After working all night, I find it rather easy. Dismas|^(talk) 10:17, 17 July 2013 (UTC)[reply]

Yes, genuine physical tiredness does it for me. HiLo48 (talk) 10:35, 17 July 2013 (UTC)[reply]

You can get eye masks or maybe improvise a blindfold. --TammyMoet (talk) 11:04, 17 July 2013 (UTC)[reply]

I wonder if the OP was expecting a scientific answer? I sometimes plonk another pillow over my head and if angled correctly can block out most of the light. This also helps to block out barking dogs, screaming children and nagging wives. Sandman1142 (talk) 11:30, 17 July 2013 (UTC)[reply]

Yeah, a pillow works fine for me, too. I used to angle it to avoid near suffocation, but now I'm used to breathing straight through it. Actually handy for drowsiness. And yes, works great for muffling sound, too, especially feather pillows (I plug wet toilet paper in my ears, just to be sure). A bit hot some days, but beats laying awake all night (or day). InedibleHulk (talk) 11:36, 17 July 2013 (UTC)[reply]

A sleep mask combined with high-quality earplugs will shut out most of the potential interferences. ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:10, 17 July 2013 (UTC)[reply]

The earplug article warns of health risks which is why I prefer a pillow. As it is, I use headphones far too often, which has a related set of prolonged use dangers such as tinnitus and higher risk of infection. At the risk of asking a medical question, does anyone here have similar concerns? Sandman1142 (talk) 12:53, 17 July 2013 (UTC)[reply]

I wouldn't recommend earplugs on a routine basis, no. ←Baseball Bugs ^{What's up, Doc?} carrots→ 13:38, 17 July 2013 (UTC)[reply]

Every body's different, but in my ten years of regular nightly earplugging, I've had just three occasions where a tiny piece went in too far and some minor swelling, deafness and water-in-the-ear dizziness happened. Lasted about a day and a half each time. I say it's a fair trade. The trick (if you can call it that) is to just get it damp, not soaked. Fold it into a square, and stick it in flat side first. Never a round or bullet shape. I've tried regular foam earplugs, too, but find they're not near as effective. InedibleHulk (talk) 03:04, 18 July 2013 (UTC)[reply]

You might want to consider getting some custom-made earplugs. (eg [1]) I know some people in the audio business (who greatly value their hearing) who had their ears scanned and custom-made earplugs produced. The very precise fit they get from that makes the plugs vastly more effective and much more comfortable. But it's not cheap - you're looking at $150 per pair. SteveBaker (talk) 13:48, 18 July 2013 (UTC)[reply]

Thanks, but no thanks. Affording the toilet paper is enough of a challenge for me. And at least one audio designer likes that, too. InedibleHulk (talk) 20:21, July 18, 2013 (UTC)

I can certainly understand the expense issue - but none of the earplugs in that review are of the kind that are custom-molded to your ears - so there is no grounds here for saying that they're no better than wadded up TP. The problem with TP is that it doesn't allow for venting (which - as they point out in a couple of those reviews - is important for hearing your own voice and a couple of other reasons), they can also trickle water into your ear and trap it there for extended periods, potentially resulting in swimmers'-ear kinds of infection, and if you use them for long enough, they start to dry out which dramatically reduces their effectiveness at attenuating high frequencies - which is not a good thing if you're trying to protect your hearing. SteveBaker (talk) 02:40, 19 July 2013 (UTC)[reply]

I use earplugs every day, there is no way I could sleep during daytime with all the noise here. The only alternative would be to live in some quiet place and sleep from e.g. 10 pm till 7 am. If you have to sleep from 6 am till 3 pm and you live in busy place, you can forget about sleeping well without earplugs. I know that there are many people who say that they sleep through noise and are not bothered by daylight, but then I've noted that they are not exactly examples of people who are physically very fit who would need at least 8 hours of good quality sleep to maintain a top level fitness. Count Iblis (talk) 09:20, 20 July 2013 (UTC)[reply]

It helps to block out blue light from your field of vision for one hour prior to attempting to sleep to increase your serotonin levels. Plasmic Physics (talk) 13:10, 17 July 2013 (UTC)[reply]

Around here some people cover the window with black plastic garbage bags or aluminium foil. Or if you are like me you just go to sleep. CambridgeBayWeather (talk) 14:20, 17 July 2013 (UTC)[reply]

Stuff like washing machines, people passing by the street, cars, or singing birds doesn't annoy me. For me, it's White noise or random noise, and I can block it out. But I can't sleep when I hear a TV or a prolonged conversation. My brain keeps trying to keep up with the argument, or with the conversation. --Enric Naval (talk) 19:02, 17 July 2013 (UTC)[reply]

I'm back. Sleep masks have the problem of causing you to sweat at the parts covered by the sleep mask, that sweat will move into your eyes, causing you to wake up. I know that many people can't sleep during daylight conditions, and yet the hotels (the expensive ones) above the polar circle don't have decent curtains. In my own home I have curtains in the sleeping rooms that block out all the daylight.

During daylingt conditions when I fall asleep, I immediately wake up again. This seems to be caused by rapid eye movement; the light in the room is not uniform, rapid eye movement will cause a strong signal which will wake me up. However, many people can sleep during daylight conditions without problems, so I was wondering why they are not bothered by this effect. Count Iblis (talk) 22:09, 19 July 2013 (UTC)[reply]

Measuring rate of rotation by the Doppler effect

Follow-up from this question about the cricket revometer, how can the rate of rotation of an object be detected by the Doppler effect? Apparently this has been done for Venus, does it work for smaller objects - for instance the ball in a sports game? ManyQuestionsFewAnswers (talk) 13:31, 17 July 2013 (UTC)[reply]

If your view of the rotating object is somewhat equatorial, one edge is moving toward toward you and the opposite edge away from you (once you've subtracted the net motion of the ball as a whole). But if your aspect is polar then this won't help. Given the small diameter and low rate of rotation of e.g. cricket balls, the Doppler shift differential between the opposite edges is tiny; for galaxies (which have huge diameters and spin at high speeds) it's another matter entirely. -- Finlay McWalterჷTalk 14:49, 17 July 2013 (UTC)[reply]

Using optical doppler (ie color shifts) is unlikely for something as small and slowly spinning as a cricket ball - and utterly impossible when using a TV camera to capture the motion as in a cricket revometer (because TV cameras only capture three colors - and not a complete spectrogram). I suppose that if you aimed a focussed beam of sound at the object then acoustic doppler might pick up a different sound reflection from one side of the ball versus the other. SteveBaker (talk) 13:43, 18 July 2013 (UTC)[reply]

That's a misunderstanding of how TV cameras work. While TV cameras encode light into three primary colours, red, green, and blue, they detect and resolve any pure spectral colour within a range that approximately corresponds to the range of the human eye. In concept, if a scene contains a wavelength that is half way between red primary wavelength and green primary wavelength, it will be encoded as 50% red and 50% green. If the scene contains a wavelength much closer to green it might (say) be encoded as 25% red and 75% green. If TV cameras only detected and encoded three colours they obviously would be useless - all scenes contain colours ranging from very narrow bandwidths to very wide bandwidths. (The actual coding is a little more complex due to technical reasons.) TV cameras cannot tell the difference between a pure monochromatic colour and a mix of colours occupying a narrow band centred on the same wavelength. But that does not matter, as here we would be interested in the group wavelenth shift. However, Steve is still right to believe that using a TV camera to detect ball rotation speed by doppler effect is quite unlikely as the doppler effect on the red/green/blue ratios is too small to be practial, of the order of some fraction of the speed of the ball surface divided by the speed of light, i.e around 0.001%. And if the axis of rotation points to the camera, there will be no doppler shift at all. Analog electronics is good at resolving to only 0.1% at best. 60.230.221.33 (talk) 15:40, 18 July 2013 (UTC)[reply]

(FWIW I'm not sure that Doppler effect is what is used to measure the spin on a cricket ball, it was just one answer in my earlier unresolved question. I was curious how the Doppler effect might be used to measure rotation though - presumably the idea is to use electromagnetic waves rather than sound, but what wavelength would you pick, how would you detect it, and what accuracy would be available? ManyQuestionsFewAnswers (talk) 19:03, 18 July 2013 (UTC))[reply]

Well, if you know the size of the object you're measuring (eg for measuring the rotation speed of a planet) then you use doppler shift for the side of the planet that's moving away from us and for the site that's moving towards us - and from the difference between the two, you can figure out the rotational speed. These techniques are fairly precise - astronomers have used them (for example) to measure the wobble induced into a star by a planet orbiting it. Our Doppler spectroscopy article covers much of this. SteveBaker (talk) 19:36, 18 July 2013 (UTC)[reply]

When do we call a celestial body a moon?

How are rocks classified as moons? I mean, there are many big and small rocks orbitting planets, but we don't call all of them moons. Yashowardhani (talk) 13:47, 17 July 2013 (UTC)[reply]

Natural satellite#The definition of a moon which says there is not an established lower limit on what is considered a "moon" -- Finlay McWalterჷTalk 13:49, 17 July 2013 (UTC)[reply]

Presumably it would have to be a naturally-occurring moon or "satellite". What we commonly call a "satellite" is short for "artificial satellite". The countless objects in the rings of Saturn, Jupiter, etc., would qualify as "moons" under a broad definition, but because they are bits of a larger entity they are called "rings" rather than "moons". ←Baseball Bugs ^{What's up, Doc?} carrots→ 13:51, 17 July 2013 (UTC)[reply]

So there's a size limit or something? And what do you mean by "naturally occuring"? --Yashowardhani (talk) 15:21, 17 July 2013 (UTC)[reply]

"Naturally occuring" means not a man-made satellite, like Mars Reconnaissance Orbiter for example. Gandalf61 (talk) 15:38, 17 July 2013 (UTC)[reply]

"Naturally occuring" means the Death Star doesn't qualify. That's no moon. --Onorem (talk) 15:44, 17 July 2013 (UTC)[reply]

My go-to reference for such things, de Pater and Lissauer's Planetary Science book, uses the term "moon" and "satellite" mostly interchangeably when describing planet-moon systems within our solar system or discussing the general case of planet-moon system formation. The IAU (who are a sort of well-respected authority on such terminology) made a lot of noise when they re-categorized Pluto in year 2006. IAU and associated researchers released numerous technical papers and news releases for public consumption intended to explain how the word "planet" is defined.

The definition for "planet" needed changing, because over the last ten or twenty years, we've discovered many new objects - extrasolar planets - and the scientific community ought to use consistent terminology to describe new discoveries. The status-quo was changing at an accelerating pace, and the definition needed to be made much more clear than it had ever been before.

Ultimately, the definition for "moon" isn't a problem," because there are really a very small number of known moons; it is very infrequent to discover a new moon-like object; so most planetary scientists who specialize in such things don't get bothered by the semantics. But the recent announced discovery of a previously unknown moon at Neptune is a good example where the terminology is stretched. Here's the long and short of it: such "moons" are so tiny that we know very little about them. Very few people study them and write about them; and so there isn't a large enough community for a standardized terminology to become very strongly entrenched. One could debate whether a 20 kilometer rock is properly termed a moon or a satellite or a dust-speck, but first one would have to find somebody to take an opposing position. Nimur (talk) 18:25, 17 July 2013 (UTC)[reply]

But are all the grains of sand in a gas-giant's rings moons? CS Miller (talk) 21:15, 17 July 2013 (UTC)[reply]

Actually, that was a topic I started writing about, then deleted, and eventually edit-conflicted with myself. No, and the thing is, there's usually a very clear difference in size between moons and ring particulate matter. But some ring systems have really big particles! So, just like planets and dwarf planets, the boundary is a bit unclear. The key is clearing an orbit - moons almost definitionally do clear their neighborhood, and ring-particulate matter does not - but there aren't enough edge-cases where the terminology is unclear enough to really matter. Again, my favorite planetary science book calls the planet-moon-system relationship a sort of microcosm with a lot of similarities to the sun-planet-system. Nimur (talk) 01:10, 18 July 2013 (UTC)[reply]

Energy of Earths Spin

I've noticed a lot of these science questions sound more like science fiction; which bypasses the whole no predictions disclaimer. So I've read that the earths spin creates more energy than several thousand nuclear bombs; most of which I know is just from the mass. But still, why don't we see more spinning half molten iron spheres providing power on a smaller scale if that shape is so effective. You can tell I've never been inside of a power plant before, I would love to know what the name for that part is if it is already in common use. CensoredScribe (talk) 15:14, 17 July 2013 (UTC)[reply]

You're referring to dynamo theory - the blanket term describing the geophysical model of semi-molten iron spinning inside Earth and producing a magnetic field. Power plants also have a dynamo, but this term is a little archaic; nowadays, you'll probably hear "generator" or "turbine" or "unit" more commonly spoken in day-to-day operations. Of course, more specific terminology exists for various items in a modern power station. At the "core," all the Earth-Dynamo theory says is that molten iron makes a magnet, and that magnet is spinning around (as a semi-fluid). No new energy is being created. The Earth is just transforming energy from heat and motion into a magnetic field. There are massive quantities of heat and motion inside the Earth, still left over from planetary formation billions of years ago. But no new energy is made by the dynamo.

Spinning molten iron - or any other magnet - doesn't produce energy; it provides a way to transform thermal and kinetic energy into electromagnetic energy, and we can recapture that energy using a coil of wire. In a power station, we already have permanent magnets (or powerful electromagnets made with another coil of wire - a detail that at first seems like a head-scratching perpetual motion machine; but in fact, the energy powering these secondary magnets also comes out of the same thermal or kinetic energy that the plant produces). Making a molten iron core to replace the already very good magnets would introduce new hazards, engineering complexity, and inefficiency.

Even if we consider the giant magnet that we get for free - the Earth - we can't really use it for electric power generation. Very little usable electric energy can be extracted, because Earth's magnetic field is weak - something like a thousand times weaker than a crummy disposable refrigerator magnet. ...And perhaps a hundred thousand times weaker than a refrigerator magnet strong enough to be useful. You'd need an absolutely enormous machine to extract meaningful amounts of energy. Instead, we see that Earth's magnetic field interacts very slowly with very large processes, like the solar wind. A lot of energy is involved, but that energy is not in a usable form for human activities. Nimur (talk) 16:17, 17 July 2013 (UTC)[reply]

There are a lot of misapprehensions about this kind of thing. The earth doesn't "produce" energy - it "contains" energy. It might be possible to extract energy from the earth's rotation (by slowing the earth down by a teeny-tiny amount) - but it's not renewable - it's like mining coal: When it's gone, it's gone.

The largest nuclear weapon ever tested released 50Mtons - or about 2x10¹⁷ Joules. That's about the same as the total amount of sunlight striking the surface of the earth in one second...or about 1% of the US's annual electricity consumption. The total rotational energy of the earth is 2x10²⁹ Joules - so about the same as a trillion of our largest nuclear weapons.

The amount of energy contained in the earth's revolution is immense. But it's not because of spinning iron spheres and magnetic fields or weird and wonderful like that...it's just very large, very heavy, and spinning very quickly.

SteveBaker (talk) 19:46, 17 July 2013 (UTC)[reply]

... and, of course, we do extract energy from the earth's rotation (and the moon's) when we use tides to generate electricity. If we did this on a much larger scale worldwide, we would possibly notice a minuscule slowing of the earth's rotation (resulting in extra leap seconds) and a minuscule increase in the rate at which the moon is receding from the earth (currently about an inch and a half per year). Other factors such as glacial rebound have a bigger effect. Dbfirs 22:43, 17 July 2013 (UTC)[reply]

I'm not sure about us slowing down the earth by using tidal power. We're not actually extracting any extra energy, we're merely directing energy that gets extracted anyway into useful work. 202.155.85.18 (talk) 07:08, 19 July 2013 (UTC)[reply]

Yes, most of the energy we capture would otherwise have been dissipated as heat as the tides ebb and flow. I'm not sure what the effects of widespread restriction of tidal flows would be, but I suspect it would be to reduce the slowing. In the back of my mind I had the common practice of pumping water up at high tide and using it to generate extra electricity at low tide (releasing much more energy than the original pumping required). This technique does extract extra energy from the earth-moon system, but it might well be more than offset by the reduction in tidal flow caused by other power-generating techniques. In any case, any effect is swamped by other factors over which we have no control. Dbfirs 07:39, 19 July 2013 (UTC)[reply]

Note the dynamo is powered by radioactive decay in the case of the Earth, and perhaps tides in the case of the Moon, or could briefly be started by energy from a collision; but in any case, the energy in the dynamo is being refilled by some other source, just as in a power station. [2] Wnt (talk) 19:28, 19 July 2013 (UTC)[reply]

reaction of amino acids with formaldehyde...a simple way to decarboxylate to form a hetero-diene?

Start at say pH < 4, using a non-acidic or non-basic amino acid (i.e. tyrosine or alanine something). The amino group first becomes an imine group (standard rxn), eliminating water. When the carboxylic group and the amino group are both protonated, what prevents enolization from occuring and attacking the formaldehyd? The enol is stabilized by conjugation with the imine, and after the attack on a second formaldehyde, a hemiacetal is created-- the alkoxide is readily stabilised by the COOH group next to it. When the COOH group is reprotonated, 6-center concerted decarboxylation readily occurs because the hydroxl group is a readily available base, the alpha carbon -COOH bond moves between the alpha-carbon and the formaldehyde-carbon, i.e. water and CO2 are eliminated as a concerted step.

The result is R-(C=CH2)-N=CH2), i.e. a diene. This could be used as a starting point for all sorts of useful Diels-Alder reactions. But I don't see this mentioned very often... it's just a one-step rxn with formaldehyde and some acidity.

This also seems like a good way to turn a carboxylic acid group into a group that can be readily reduced to a methyl group, something that seems sought after.

(I would draw but I don't have access to a scanner or Chemdraw atm.) 76.23.197.50 (talk) 23:11, 17 July 2013 (UTC)[reply]

At pH < 4 the alkoxide is already protonated, not likely to be a strong base. If you're under acidic conditions, it would be the (as you say) enol (not enolate) that is present to attack the formaldehyde. And the formaldehyde carbonyl itself may become protonated first (becoming a better electrophile) in order to induce the neutral enol (a not-very-good nucleophile in general, and now also with additional stabilization?) to attack it. I can't envision where you get a hemiacetal structure (especially if it's part of the [carboxylic acid]→[enol]→[β-hydroxy acid (aldol addition product)]→[6-membered-ring decarboxylation].

The overall idea of [enolizeable carboxyl]+[carbonyl]→[alkene]+CO₂ is actually well-known in a variety of malonic ester and related syntheses (aldol condensation#Scope is a similar example). The novel idea in yours is the use of the imine type of reactant (rather than β-carbonyl) to get to the heterodiene structure.

The [amino acid]+H₂C=O→[imine] under acidic conditions is well known, but appears not to proceed further. A few aldol addition reactions of that imine product are popping up in my SciFinder searches to give the Β-hydroxy product but I'm not finding any that give decarboxylation. DMacks (talk) 02:35, 19 July 2013 (UTC)[reply]

Sorry hemiacetal is the wrong word to say. Gosh. I meant to say beta-hydroxy.

If the beta-hydroxy product results, why wouldn't decarboxylation result? The imine is not sabotaging the electronics of the decarboxylation or the enolization is it? (Also I'm not sure if I'm reading correctly ... has the imine-beta-hydroxy-COOH species been observed at least?) The six-membered ring decarboxylation would seem to be facile. Anyway thanks for searching for me. Maybe it's not a viable route. I had a few syntheses in mind where using the amino acid backbone would simplify many syntheses. Is there a more potent catalyst than a leaving group-cum-weak-base to force decarboxylation? What if I were to use a Lewis acid? 64.134.102.213 (talk) 00:47, 20 July 2013 (UTC)[reply]