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May 16

Non-ethanol effects of alcoholic beverages

Can anyone find any information about why different alcoholic drinks have different effects independent of the volume of ethanol consumed? The effects of champagne, red wine, whisky, rum, gin and beer are all subtly different, yet I've never read anything to explain why. The volume of water has some effect with beer, but the hops have others; presumably the herbs used in gin make a difference too, as do the flavonoids in red wine. Are there are secondary metabolites that brewers yeast produces during fermentation that are bioactive in humans? SmartSE (talk) 01:00, 16 May 2012 (UTC)[reply]

Whilst we're at it, can anyone improve on the first sentence of Ethanol#Pharmacology? SmartSE (talk) 01:06, 16 May 2012 (UTC)[reply]

I'm guessing this is mostly observer bias. Someguy1221 (talk) 01:11, 16 May 2012 (UTC)[reply]

Someguy is probably close to the answer. There have been numerous studies which show that a large portion of drunkenness is psychological, and not directly related to the ethanol itself. I have seen countless studies and demonstrations which show the placebo effect on drunkenness; people who are given non-alcoholic beverages and told they are alcoholic show signs of intoxication. It is quite likely that different types of drinks make you feel certain ways, not for their chemical composition, but for the social implications of what they represent (champagne or fine wine feels sophisticated, big fruity drinks seem fun, etc. etc.) so your internal feelings likely represent something of that beyond the mere chemical effects of the drink. That's my guess, anyways. --Jayron32 01:47, 16 May 2012 (UTC)[reply]

There's a few other possible factors at play as well: Many people can have an adverse reaction to sulphites found in certain drinks, especially wine. Anecdotal evidence suggests that drinks high in sugar can result in a worse hangover, but I have yet to find a reliable source for this. Also, if you have had a bad experience with a certain liquor (as in, you end the evening worshiping the porcelain gods), it is more likely to make you feel ill in the future: see Taste aversion. -RunningOnBrains^(talk) 01:54, 16 May 2012 (UTC)[reply]

fusel oils are supposed to worsen hangovers. Staticd (talk) 07:15, 16 May 2012 (UTC)[reply]

I would personally expect drinks high in sugar to result in a worse hangover for the simple reason that eating and drinking a lot of sugar can give you many of the same symptoms as a hangover, in my experience. Presumably for similar reasons (dehydration). 86.161.213.137 (talk) 09:14, 17 May 2012 (UTC)[reply]

Ripping someone a new one

In among all the -ostomies and -plasties and whatnot, there's got to be a medical term describing the forcible creation of a new asshole. What would it be? --Carnildo (talk) 01:56, 16 May 2012 (UTC)[reply]

Giving someone a new asshole is actually a necessary medical procedure following certain types of rectal cancer. I see various articles calling this "rectal reconstruction", but I haven't seen anyone apply a fancy name to it. Someguy1221 (talk) 02:02, 16 May 2012 (UTC)[reply]

Ok, that actually made me laugh... I guess if we take the lead from tracheotomy, it would make it a rectumotomy, not sure if that sounds quite right, maybe rectotomy might need someone with some latin skills to figure out the correct spelling.. This reminds me of defenestration, the technical word for throwing someone (or something) out a window. Vespine (talk) 02:51, 16 May 2012 (UTC)[reply]

Maybe that medical procedure could help me care about things. DMacks (talk) 21:49, 19 May 2012 (UTC)[reply]

Newborns also can have an imperforate anus, which, as you can imagine, needs to be fixed fairly quickly. StuRat (talk) 03:18, 16 May 2012 (UTC)[reply]

What about colostomy which is quite a common procedure for people with advanced colonic cancer where removal of the colon is necessary. The exit is usually placed on the lower left or right front abdomen for the obvious reason that anastomosing the colon to the anus is difficult and likely to give poor control. Richard Avery (talk) 07:17, 16 May 2012 (UTC)[reply]

As proctology is the branch of medicine dealing with the rectum and anus, would it be proctostomy? As an aside, the proctologist at my local hospital is Mr Shatwell. Prime example of nominative determinism. --TammyMoet (talk) 08:44, 16 May 2012 (UTC)[reply]

Wouldn't a proctostomy be the removal of a proctologist impacted in your nether regions ? StuRat (talk) 16:15, 16 May 2012 (UTC) [reply]

The imperforate anus article mentions a "perineal anoplasty". --Sean 20:15, 16 May 2012 (UTC)[reply]

Some cats might or might not have had a Fistulotomy, which is the most common reason for the surgical procedure in question. Tevildo (talk) 21:06, 19 May 2012 (UTC)[reply]

List of Self Limiting diseases?

I'm trying to find this list. It appears that wikipedia does not have such a category. Many definitions I've found will list a few, but so far I've got gastroenteritis, hepatits, the common cold, dyptheria, tonsillitis, llaryngitis. I imagine an exhaustive list would be huge, but can anyone find (or has seen) a list of the most common 20 or so? Vespine (talk) 02:43, 16 May 2012 (UTC)[reply]

I don't think many of those qualify as "self-limiting" as they are limited by our immune response. A truly self-limiting organism would control it's own population, say with waste which is toxic to itself in sufficient quantities. StuRat (talk) 03:13, 16 May 2012 (UTC)[reply]

Self limiting in medicine has a slightly diffent meaning to that of general biology. From the article: the term may imply that the condition would run its course without the need of external influence, especially any medical treatment. Vespine (talk) 04:16, 16 May 2012 (UTC)[reply]

Don't all diseases run their course with or without medical treatment? 112.215.36.183 (talk) 10:16, 16 May 2012 (UTC)[reply]

If you count "death" as an allowable outcome, then yes. I presume, however, the OP is asking about diseases that don't normally result in death if left untreated. If that is the case, then I'd remove diphtheria from that list. --TammyMoet (talk) 11:06, 16 May 2012 (UTC)[reply]

If you search wikipedia for articles containing self limiting, you get 2464 results. The description usually has enough info to know if it is a self-limiting disease. Examples: Scheuermann's disease, Scleredema, Vernal keratoconjunctivitis, Transient synovitis, Epiploic appendagitis, idiopathic scoliosis, Mondor's disease, Acute posterior multifocal placoid pigment epitheliopathy, Pityriasis alba, Cricopharyngeal spasm, Necrotizing sialometaplasia... For a list of the most common ones, I guess the best way would be taking a list of the say 50 most common diseases and check which ones are self-limiting. Problem is finding such a list, I can't even find two sources that agree on the nr 1 (gum disease according to one tabloid), even found a list that had sociopathic personality disorder in the top five. Ssscienccce (talk) 16:55, 16 May 2012 (UTC)[reply]

I thought there might have been a strict medical meaning but it seems there might not be.. Diptheria says it has a fatality rate of 5%-10%, lol, ok, I just searched for the source of the claim that diptheria is self-limiting back to Herbert_M._Shelton. Not exactly a reliable source.. But Ssscienccce has given me a great start. Thanks. Vespine (talk) 22:51, 16 May 2012 (UTC)[reply]

red giant's maximum diameter

How big is red giant at the maximum diameter. Can some red giant be as big as Mars orbit. How we know how big will our red giant be? Does some red giant get as big as 2 AU or as big as 3 AU? What is the size range of red giant from the smallest to the biggest?--69.228.133.188 (talk) 02:44, 16 May 2012 (UTC)[reply]

Does red giant answer any of this ? Note that the "size" is usually given as mass, which tends to be fairly constant, versus volume or radius, which can change dramatically. They list 0.5 Suns to 10 Suns as the range, with any stars more massive than that called red supergiants. (The illustration in that last article answers your Mars question.) As for how they know what the Sun will do, that's based on it's mass. StuRat (talk) 03:02, 16 May 2012 (UTC)[reply]

If your question was specifically "What is the biggest star?", see VY Canis Majoris. Its radius is estimated to be between 8.4 and 9.8 astronomical units, which is almost out to the orbit of Saturn. You may also be interested in our List of largest known stars. As far as our sun, it is estimated its radius will only be about 1.2 AU at its maximum, so the Earth will likely escape destruction due to its increased orbital radius by then (not that any of us will be around to care). -RunningOnBrains^(talk) 03:26, 16 May 2012 (UTC)[reply]

1.2AU isn't the maximum, it continues to say that it will eventually grow to 2AU. Vespine (talk) 04:14, 16 May 2012 (UTC)[reply]

My question was the biggest red giant in size, not the red supergiant or biggest star. I am asking for largest red giant only.--69.228.133.188 (talk) 05:34, 16 May 2012 (UTC)[reply]

Well then the answer is about 200 solar radii (2AU); the cutoff between red giant and red supergiant is quite fuzzy, if not arbitrary. For instance, Epsilon Aurigae at 135 solar radii is described as a red supergiant, while Rho Persei at 164 solar radii is considered merely a Asymptotic giant branch star (i.e., red giant).RunningOnBrains^(talk) 05:50, 16 May 2012 (UTC)[reply]

An inquiry into the nature of solenoids

The equation $B_{z}={\frac {\mu _{0}}{4\pi }}{\frac {2\pi IN}{L}}\left[{\frac {d+L/2}{\sqrt {(d+L/2)^{2}+R^{2}}}}-{\frac {d-L/2}{\sqrt {(d-L/2)^{2}+R^{2}}}}\right]$ gives the axial magnetic field strength $B_{z}$ in a solenoid. This is true for the case where every turn of conductor in the solenoid has the same radius $R$ . Would you expect the magnetic field strength measured in a coil consisting of many layers of turns, each with a slightly different radius, to be higher or lower in magnitude than that predicted by this equation using the inner radius of the solenoid? --130.56.84.118 (talk) 02:53, 16 May 2012 (UTC)[reply]

Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our policy here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. Anonymous.translator (talk) 03:23, 16 May 2012 (UTC)[reply]

GRRROOOAAAANANNNN. My line of reasoning was that the answer is higher because the many layers reinforce each other. Is this right? --150.203.114.37 (talk) 03:42, 16 May 2012 (UTC)[reply]

We actually can't answer your question because you didn't tell us enough about the multi-layered solenoid. Are you taking a solenoid with the same number of turns and simply altering the radii of the turns? Are you wrapping an additional solenoid around the original? Are you scrunching the solenoid up so that it's shorter and fatter? What is the distribution of radii with respect to the original radius? If you make the total circuit length longer, are you changing the total voltage to keep the current constant, or are you dropping the current to keep the total power output constant? Someguy1221 (talk) 04:08, 16 May 2012 (UTC)[reply]

I don't suspect this is a homework question, as it is more a test of algebra skills than magnetic understanding, and a generic math teacher is not likely to set such an applied question. On the other hand, electronics technicians and radio hams often ask this - they tend to only find the single layer formula in books etc.

Assuming you want to know what happens if you have the same total number of turns in a multilayer solenoid as in a single layer solenoid, with the current, wire gauge, and everything else the same, for comparison, the asnwer is simple: Apply the formula to each layer separately. The total axial field strength is the sum of the fields contributed by each layer. As the layer radius is lower (denominator) terms in the calculation, the layers are progressively less effective moving out from the inner layer. Therefore the total axial field strength in a multi-layer solenoid is less than that for a single layer coil of the same turns, and less than that estimated by taking the inner layer radius. However, if wire gauge is reduced so that the radial distance occupied by the turns is the same, you'll get near enough the same field. This is because the improved contribution of the inner layers is balanced by the reduced contribution of the outer layers. Keit120.145.9.168 (talk) —Preceding undated comment added 04:40, 16 May 2012 (UTC).[reply]

There seems to be some laziness in presenting an equation which contains 2 pi in the numerator and 4 pi in the denominator. Please simplify and ask again. Is it correct to infer that mu nought is the permeability of free space, I is current in amperes, and L is inductance in ohms? Is R in meters or centimeters? (My education included physics textbooks using both units of distance). Edison (talk) 05:34, 16 May 2012 (UTC)[reply]

You made me look at his formula again. It doesn't look right - for a start the Ln (natural log) symbol is missing. However, L in this case is not inductance (& inductance is not measured in ohms), but the length of the coil. Keit124.178.152.203 (talk) 07:18, 16 May 2012 (UTC)[reply]

Looks right to me, L being the length, and d the distance from the center (between -L/2 and L/2). If I understand the question correctly, it's whether using this formula for a multilayer solenoid, with R equal to the inner radius, would underestimate or overestimate the field strength? Assuming we're still talking about a solenoid of length L with N turns, I'd say the measured field strength would be lower, since the real radius is larger. Ssscienccce (talk) 17:44, 16 May 2012 (UTC)[reply]

Orbital Chainsaw

The Orbital Chainsaw would consist of thousands of small "teeth" satellites in Low Earth orbit. Each such tooth would have a reasonable sized solid state laser that was powered by ultracaps that were then refilled by solar power during the 99% of the orbit when that tooth wasn't over a useful target.

Each tooth is only over a target area for a short period of time, so only needs enough power storage to cover lasing for that short period of time.

The Orbital Chainsaw would have a thousand and one uses. From laser propulsion for laser powered aircraft and satellite launches (to fill in more teeth satellites), to weapons uses such as shooting down hostile ICBMs or satellite launches and dealing with gatherings of terrorists or other dissidents.

So why hasn't this been implemented yet? (What exactly is the flaw in this brilliant (pun intended) scheme of global domination?) Hcobb (talk) 10:58, 16 May 2012 (UTC)[reply]

You would need literally millions of satellites. Not very practical. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 11:09, 16 May 2012 (UTC)[reply]

Expensive. Plasmic Physics (talk) 11:10, 16 May 2012 (UTC)[reply]

Very—it would require more money than the entire U.S. gross national product. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 11:12, 16 May 2012 (UTC)[reply]

I don't know about the number but I agree cost seems to be a big factor. Consider say the cost of Iridium satellite constellation (which only involves 66 active satellites) or Globalstar or Orbcomm involving communications satellites. For example if we take Iridium, scale up the cost for your fancy laser satellites, times it by 50 for your proposal involving thousands of satellites (with some discount for the large number) and you probably end up with something more then the entire Military budget of the United States. Also while probably not banned by the Outer Space Treaty, I think it's clear such an extreme Militarisation of space is unlikely to be popular. The Strategic Defense Initiative was rather unpopular, and considering the existing ability of the US to nuke the entire planet many times over, the advantages seem slim considering the cost and unpopularity of such a move, mutually assured destruction is generally considered to remain a powerful deterence to any large enemy. In terms of 'rogue states', the US already have their Missile Defense Agency allegedly for that purpose. And smart bombs and UAVs to deal with gatherings of people they don't like. I think many would question how achievable most of your stated goals are anyway (using them from satellite launches sounds like wishful thinking to me). Nil Einne (talk) 11:37, 16 May 2012 (UTC)[reply]

Just compare the cost of the system against what is spent on commercial jet airliner fuel every year and it pays for itself in no time.

As for the number needed, assuming a reach of 200 km, fewer than 5000 satellites are needed to cover the entire Earth. Hcobb (talk) 12:05, 16 May 2012 (UTC)[reply]

Unlikely. Assuming this implausible amount, you would require 6332.573979 satellites. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 12:13, 16 May 2012 (UTC)[reply]

Exactly who is going to pay to get them up there? Plasmic Physics (talk) 12:19, 16 May 2012 (UTC)[reply]

Yeah. It is much easier to get citizens to pay for things to move them places than it is to get anyone to pay for orbital weapons already racked with other problems. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 12:46, 16 May 2012 (UTC)[reply]

And capacitors to store power collected in one orbit (1 hour) would have to store megawatts of energy. These would be big and heavy. Graeme Bartlett (talk) 12:39, 16 May 2012 (UTC)[reply]

And that would give complete 100% global coverage at all times (die penguin scum!). A few dozen would be sufficient for satellite launches. (At which point the launch costs go way down.) Hcobb (talk) 12:22, 16 May 2012 (UTC)[reply]

"And that would give complete 100% global coverage at all times." Wrong. Because circles do not tesselate, you would need at least twice that number. And practically, you would need twice that, to provide redundancy and allow for failures. Which comes out to over 25,000 satellites. And a number of satellites this large would be extremely difficult to keep from occulting each other, colliding with each other, dragging each other out of orbit etc. Dyson sphere#Dyson swarm lists the problems with such a large swarm of satellites. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 12:33, 16 May 2012 (UTC)[reply]

They only have to worry about occulting each other if they block a significant fraction of the sunshine striking the Earth. Given a a few meter diameter satellite every hundred km, this isn't much of a problem. As for energy storage the satellite would only be in a position to fire for an average of 200 km (not the full diameter of 400 km, because passing directly overhead is rare) and since it is traveling at 7 km/s it only fires for half a minute. Given a solid state laser of 100 kw, this is only three megajoules of power storage or 30 kg of ultracaps. Hcobb (talk) 12:59, 16 May 2012 (UTC)[reply]

Were the satellites light enough to be significantly helped to orbit by each others' lasers, then whenever one of the satellites fired its laser, the recoil would be enough to throw it into a higher orbit, possibly causing it to tumble as well, and forcing EITHER the launch of a replacement satellite, in which case the satellites would be literally one-use-only, necessitating replacement whenever fired, which would VASTLY increase the costs, OR the expenditure of large amounts of fuel to bring it back down to its rightful orbit.

Also, for attacks on ground-based targets, atmospheric attenuation might be a problem for objects near the satellite's horizon.

Firing the laser against property of another nation would be an act of war. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 13:04, 16 May 2012 (UTC)[reply]

It is also doubtful whether another space-faring nation would allow the development of such a system without deploying its own laser-armed killer satellites. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 13:11, 16 May 2012 (UTC)[reply]

Because of the energy conversion issues, the Solar sail effect of the solar cells would be much greater than the laser recoil and almost, but not quite, vanishing small. It still makes a tiny bit of sense to be able to tilt the solar cell wings to (very very slowly) bleed off excess rotational momentum from the stabilization reaction wheels. One imagines that any country able to achieve Prompt Global Strike against any target on or near the Earth within one second of locating that target wouldn't care very much about what other countries thought about it. Hcobb (talk) 13:21, 16 May 2012 (UTC)[reply]

See my post above about enemy killer satellites. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 13:24, 16 May 2012 (UTC)[reply]

Also, the force exerted by the laser on another satellite would be exactly equal to the force exerted on the satellite firing its laser. Have you any knowledge of Newton's third law of motion or the law of conservation of momentum? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 13:30, 16 May 2012 (UTC)[reply]

See Space-based solar power. I'm afraid you're golfing pretty far into the woods this time. There are limits on how precisely you can focus a laser from orbit. And using light defeats much of the purpose of space-based power since it gets absorbed in the atmosphere anyway. And a geosynchronous array loses only 75 minutes of sunlight to the Earth's shadow twice a year. And the satellites can be useful for power generation without being military assets. Wnt (talk) 14:54, 16 May 2012 (UTC)[reply]

Whether it is practical or not, I can't see why it would be useful. You can cover the whole Earth's surface with just a handful of geosynchronous satellites, so why bother building thousands of LEO ones? --Tango (talk) 23:06, 16 May 2012 (UTC)[reply]

In LEO the laser beam would have a much smaller spot on the ground. So it could be more concentrated, causing less damage outside its target area. Also from your geostationary orbit fixed location you may get a building in the way of your target. Graeme Bartlett (talk) 21:40, 17 May 2012 (UTC)[reply]

I'm totally out of my depth on this one, so this is probably moronic, but ... is it conceivable to make a "virtual" satellite to focus the signal from the geosynchronous microwave satellite tightly, by etching some sort of perfect lens into the ionosphere with high-frequency radiation to establish more and less conductive regions? Wnt (talk) 19:03, 18 May 2012 (UTC)[reply]

Due to air currents, I don't think so. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 02:55, 19 May 2012 (UTC)[reply]

Compatibility (mechanics)

Can someone explain to me in ver simple terms what compatibility in mechanics is? I've read the article on Wikipedia & many books but they all confuse me. Thanks.Clover345 (talk) 13:50, 16 May 2012 (UTC)[reply]

It means that a certain kind of parts can be used with ("are compatible with") a certain device. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 13:54, 16 May 2012 (UTC)[reply]

Well, I've read the Wikipedia article http://en.wikipedia.org/wiki/Compatibility_(mechanics), and I'm none the wiser. But it has nothing whatever to do with what Whoop whoop said. The word "compatibility" is apparently used in the sense that certain conditions must be satisfied for the math theory of deforming/bending a solid by distributed force to be valid, such as not tearing. The article needs to be re-written, or at least a preamble added, in ordinary language, so us ordinary mortals can figure out what possible use it might be. The wiki article was written by some math nut who likes the sound of his own gibberish, and thinks that links to other math nut stuff means something. Wickwack120.145.133.138 (talk) 15:30, 16 May 2012 (UTC)[reply]

It's unusual to see an article entirely written by one editor, but no, it's not total gibberish. You might like to read our article on Finite strain theory first, but unless you are "into" tensors, that might not make much sense either. I agree that the lead needs to be written for non-specialist readers. Dbfirs 16:04, 16 May 2012 (UTC)[reply]

I think it has to do with: 1) the deformation must be continuous, ie a curve or surface must still be a curve or surface after the deformation, and 2) the formulas or matrix representing the distortion have more unknowns than there are independent variables, so you'll have additional constraints, which are the compatibility equations. Edit: just found the Continuum mechanics article. Ssscienccce (talk) 19:05, 16 May 2012 (UTC)[reply]

Whether I saw the moon.

I saw a thin white cresent, perhaps waxing, not far above the horizon, roughly in the east, from London at 9:00 this morning. Was it the moon? 82.31.133.165 (talk) 14:01, 16 May 2012 (UTC)[reply]

Very likely yes. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 14:06, 16 May 2012 (UTC)[reply]

The Moon is in its waning crescent phase right now, and would appear as a thin crescent near the eastern horizon (and not too far from the sun) in the mornings. Also, there just isn't anything else that looks like a crescent moon in the sky, really, ever. TenOfAllTrades(talk) 14:10, 16 May 2012 (UTC)[reply]

Actually, the crescent was indeed waning. I confused waxing and waning. Thanks. 82.31.133.165 (talk) 19:20, 16 May 2012 (UTC)[reply]

Definitely waning; it needs to be a new moon soon so we can have this weekend's eclipse. You can always tell waxing vs waning by if the moon is visible at sunset or sunrise: visible at sunset means it's waxing, visible at sunrise means it's waning. -RunningOnBrains^(talk) 20:37, 16 May 2012 (UTC)[reply]

Or remember the word DOC, D for waxing, O for obviously full moon, C for waning, note the curve of the letters. Richard Avery (talk) 22:06, 16 May 2012 (UTC)[reply]

Maybe I'm missing something, but that doesn't seem like a very helpful mnemonic. Won't it depend on what direction you're facing? Also how do you keep track of which is waxing and which is waning? -RunningOnBrains^(talk) 22:13, 16 May 2012 (UTC)[reply]

Won't it depend on what direction you're facing? - Technically yes, but only if you can get yourself to somewhere beyond the far side of the Moon to be able to observe it from that side. From the Earth, which is where most of us live most of the time, it appears the same everywhere, at least in terms of general shape and orientation. -- ♬ Jack of Oz ♬ ^{[your turn]} 22:29, 16 May 2012 (UTC)[reply]

I agree DOC does not seem useful, DOC seems to refer to crescent or gibbous but the moon goes throug each phase both waxing and waning: Lunar phase. Waxing or waning indicates whether the moon is going towards full or away from full (towards new), which you can not easily tell from casual observation, except to note whether it is morning or evening, as Running on Brains pointed out. Vespine (talk) 22:41, 16 May 2012 (UTC)[reply]

Doesn't a moon with a crescent on the right side appear to have the crescent on the left side when the viewer is upside down ? And, being an Aussie, you must be aware that you are upside down from the civilized world. :-) StuRat (talk) 22:41, 16 May 2012 (UTC)[reply]

(edit conflict)If I'm facing south and the moon is curved like a "C", then when I turn and face north the moon will look like a "D". I know I'm not mistaken... -RunningOnBrains^(talk) 22:40, 16 May 2012 (UTC)[reply]

Thank you guys, I'm really glad I'm not crazy :) -RunningOnBrains^(talk) 23:03, 16 May 2012 (UTC)[reply]

If you're facing south and looking at the moon and then turn and face north, then you can't see the moon any more because it's behind you... --Tango (talk) 23:13, 16 May 2012 (UTC)[reply]

And, if what you say is true, the HOLLYWOOD sign would read DOOWYLLOH from the front, depending on which way you're facing the sign. Is that your actual experience? If so, I suggest a good optician or even a psychiatrist. -- ♬ Jack of Oz ♬ ^{[your turn]} 23:33, 16 May 2012 (UTC)[reply]

Okay, but does the Hollywood sign move across the sky and rotate over the course of the night? When the moon is near the horizon, it is neither a C or a D, it is a U (the spines of a crescent always point up from the horizon). When it is directly above, it could be any of these, depending on what direction your body is facing as you look up. I'm not trolling, I'm legitimately trying to point out that this method just doesn't work. -RunningOnBrains^(talk) 00:45, 17 May 2012 (UTC)[reply]

And when I say "always" points up, I'm talking seeing the moon at night. I assume you can't see a small crescent while the sun's up (although I've admittedly never tried to look for it). -RunningOnBrains^(talk) 00:51, 17 May 2012 (UTC)[reply]

You can but it's more like hide the Sun behind an object and look x degrees along the ecliptic for it. Sagittarian Milky Way (talk) 18:26, 18 May 2012 (UTC)[reply]

What is the connection between the letter D and the word waxing? Or the letter C and the word waning? Even if you are restricting yourself to one hemisphere (as StuRat says, the moon is the other way around in the opposite hemisphere), I don't see how the word DOC helps you remember anything. --Tango (talk) 23:13, 16 May 2012 (UTC)[reply]

I'm not sure I get what Jack is saying, the moon DOES look upside down when viewed from the other hemisphere. You'd have to flip Hollywood along the horizontal axis too to get the correct picture, but since a C and a D are symmetrical along that axis, you don't notice it as much with the moon. That's why Orion and most of the other constellations are upside down when viewed from Australia, because northerners made them up. Australian Aboriginal constellations like the emu will look upside down from the north. Vespine (talk) 23:52, 16 May 2012 (UTC)[reply]

Actually, I might be wrong about the emu because I think even us southerners look south at emu, you'd probably have to be on the south pole to look north at it. Vespine (talk) 23:58, 16 May 2012 (UTC)[reply]

Forget this D and C stuff. "Waxing" means "getting bigger" and "waning" means "getting smaller" - as in the size of the illuminated portion of the moon as seen from earth. Next time, class, I'll explain what "gibbous" means. :) ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:59, 16 May 2012 (UTC)[reply]

Resembling a gibbon ? :-) But seriously, why do we use such an archaic term as "waxing" instead of "growing" ? What would it be like if we used "waxing" for anything else ? "Hey mom, can you measure me ? I want to know how much I waxed today !" :-) StuRat (talk) 00:06, 17 May 2012 (UTC)[reply]

My interest in this topic is waning. ;) Vespine (talk) 00:25, 17 May 2012 (UTC)[reply]

Except for this bit "it appears the same everywhere, at least in terms of general shape and orientation." That's definitely wrong, even the Moon phases article points out that the moon looks different from the northern and southern hemispheres. Vespine (talk) 00:27, 17 May 2012 (UTC)[reply]

OK, I overstated my argument. I'm trying to get to the bottom of what Runningonbrains said: If I'm facing south and the moon is curved like a "C", then when I turn and face north the moon will look like a "D". If there are 30 people standing in a circle out in a moon-lit field, and they all look up at the Moon, ALL of them will report seeing a C-shaped object, or ALL of them will report seeing a reverse C-shaped object, or ALL of them will report seeing a (roughly) circular object. Is this not true? I cannot get my head around the claim that it will appear C-shaped to some but the reverse orientation to others in exactly the same place at exactly the same time, depending on which why they happen to be standing on the ground. That would be true of an object that is MUCH, MUCH closer, such as a photo of the crescent Moon stuck on your bedroom ceiling. But the actual, real Moon is WAY further away than that and it simply does not change its orientation in a matter of seconds as the observer changes their attitude to it. -- ♬ Jack of Oz ♬ ^{[your turn]} 22:06, 17 May 2012 (UTC)[reply]

So if the dean is waxing wroth, he's getting bigger, Bugs? That will certainly make it harder for Roth to wax the dean when it's his turn. Deor (talk) 00:37, 17 May 2012 (UTC)[reply]

EO is way ahead of you on that one.[1] :) "Waxing wroth" is an obscure way of saying "growing angry". As to the verb form of "wax", which appears to have nothing to do with the noun form, my old Webster's indicates the word is cognate with Latin and Greek terms that are also the basis for the term "augment". So instead of a "waxing crescent", we could have an "augmenting crescent". But that doesn't alliterate with "waning" very well. :( As regards the moon looking "upside down", the "tilt" of the crescent is going to vary depending on your latitude, just like the sun's angle will vary - however, the "midpoint" of the crescent moon is always going to be "aimed" at the rising or setting sun. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:35, 17 May 2012 (UTC)[reply]

If you google [waxing] you'll see other relatively common uses of that old-fashioned term. "Waxing poetic" seems the most obvious, along with alliterative pair, "waxing and waning", which can refer to most anything, like "ebb and flow". ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:41, 17 May 2012 (UTC)[reply]

One other thing: On the night of May 5th we had that "super" full moon, so it stands to reason that it would be in waning crescent phase 11 days later. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:48, 17 May 2012 (UTC)[reply]

There's actually heaps of sites that explain this if you google "does the moon look upside down", but I can't find any that have a clear illustration showing why. I might try to make one later. Vespine (talk) 00:40, 17 May 2012 (UTC)[reply]

Indeed, just thinking outer-spatially about it, the moon is going to appear 180 degrees different from the north pole to the south pole. -RunningOnBrains^(talk) 00:49, 17 May 2012 (UTC)[reply]

Controversy on Gravitational Singularities

If the universe had begun from a single point of zero volume and infinite density and mass, then why is the present universe so imperfect? — Preceding unsigned comment added by 205.178.233.168 (talk) 14:04, 16 May 2012 (UTC)[reply]

"Imperfect" in what way? If you mean the uneven (yet still highly homogeneous and isotropic) distribution of matter, our article on cosmological inflation gives credit to quantum fluctuations magnified to large scales during that period, and further references the galaxy formation and evolution and structure formation articles. — Lomn 14:15, 16 May 2012 (UTC)[reply]

So far as I know - correct me if I'm wrong - this "point" never really existed, not even in theory. To quote our article, "This singularity signals the breakdown of general relativity. How closely we can extrapolate towards the singularity is debated—certainly no closer than the end of the Planck epoch." Given how much stuff is said to have happened in the first second after the Big Bang - see Chronology of the universe - it is hard to think of any part of the process as "actually" being "instantaneous" in any meaningful sense. It is simply absurdly fast and absurdly small by our yardsticks - by the vibrations of cesium atoms, for example. But in the philosophical sense - the sense abused by some as described in Religious interpretations of the Big Bang theory - I think it would make more sense to view our universe as infinitely "old" (in the sense of a logarithmic time), but "receding to the horizon" as we look backward to ages where different properties of matter predominated. So there is no need to explain this contradiction - indeed, if we extrapolate from any time we can actually observe, the hypothetical first point, if it existed, would be an imperfect point. Wnt (talk) 14:44, 16 May 2012 (UTC)[reply]

Yes, since things tend to happen much more quickly, and in a much smaller area, at higher energies, it's reasonable to take this logarithmic view of the early universe. Still, the laws of physics presumably bottom out somewhere, so there would still be a beginning in traditional big bang cosmology, though it wouldn't be a singularity.

The other issue is that inflationary cosmology explicitly replaces the early universe, including the supposed singularity, with something quite different, which has no obvious beginning. So in a sense we're back where we used to be, with a universe that has existed for an unknown time, perhaps forever, with the details seemingly beyond the reach of present-day experiment. -- BenRG (talk) 20:17, 16 May 2012 (UTC)[reply]

Spontaneous symmetry breaking Hcobb (talk) 16:52, 16 May 2012 (UTC)[reply]

Do not complain of an imperfect universe. It is written that the actual world is the Best of all possible worlds. 84.209.89.214 (talk) 14:51, 18 May 2012 (UTC)[reply]

Prove it. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 02:57, 19 May 2012 (UTC)[reply]

Beryllium barrier

What would be happened if beryllium-8 were a stable isotope? --84.61.181.19 (talk) 17:09, 16 May 2012 (UTC)[reply]

The entire universe would scream in horror. Looie496 (talk) 17:32, 16 May 2012 (UTC)[reply]

Big bang nucleosynthesis#Heavy elements briefly discusses one consequence. I'm not sure how much sense it makes to discuss this happening in isolation though - would it be possible to change physics in such a way that this isotope becomes stable, but nothing else is affected? 81.98.43.107 (talk) 18:00, 16 May 2012 (UTC)[reply]

If beryllium-8 were 10 eV lighter, it would be stable. --84.61.181.19 (talk) 19:20, 16 May 2012 (UTC)[reply]

Yes, but in order to achieve this you would have to change fundamental laws of physics, which would likely have dire consequences for life as we know it. -RunningOnBrains^(talk) 19:59, 16 May 2012 (UTC)[reply]

10 eV or 100 keV? --84.61.181.19 (talk) 20:28, 16 May 2012 (UTC)[reply]

Irish Great whites

If global warming is going to make the sea levels rise and warm up a bit does this mean that sharks like the great white might end up appearing in irish coastal waters. — Preceding unsigned comment added by 86.41.81.98 (talk) 21:20, 16 May 2012 (UTC)[reply]

Actually, quite possibly the opposite effect would happen in Ireland. The Gulf Stream, which brings warm tropical water to the coast of Ireland, is driven by a complex circulation in the North Atlantic. Melting northern polar ice is, in many models, predicted to disrupt this process, actually interrupting the flow of the Gulf Stream, causing the water (and likely the climate) of Ireland and other parts of Western Europe, to actually get colder. That's because, while the term is "GLOBAL" warming (that is the average temperatures of the entire world, averaged together) are getting warmer, the effects of those changes are complex, and can result in some local areas getting colder. That's why the preferred term is "climate change" as it captures the more complicated nature of what melting ice and overall warming will be. See Shutdown of thermohaline circulation which explains in some more detail how global warming can have a cooling effect on Ireland, which is named specifically. --Jayron32 21:30, 16 May 2012 (UTC)[reply]

I prefer the term "global weirding". StuRat (talk) 21:50, 16 May 2012 (UTC)[reply]

I have always had a problem with the "Gulf Stream Shutdown" theory, mainly because it's not purely a thermohaline circulation; it is also a dynamically-driven western boundary current. Seems like an incredibly unlikely scenario to me. Computer models seem to agree with me too, from the lede of the "shutdown" article: "In coupled Atmosphere-Ocean General Circulation Models the THC tends to weaken somewhat rather than stop, and the warming effects outweigh the cooling, even over Europe" -RunningOnBrains^(talk) 23:08, 16 May 2012 (UTC)[reply]

Indeed, the Gulf Stream itself will continue as long as the Earth continues to rotate at roughly its present rate and the Atlantic Ocean is not ice covered. Short Brigade Harvester Boris (talk) 00:26, 17 May 2012 (UTC)[reply]

Probably. The question becomes whether the nature of the Gulf Stream, such as the specific direction it travels, the temperature and salinity of the water it caries, and the way it affects the climate of Europe, which is in question. It is a complex situation, and does not have a simple answer, or even necessarily a known one. That's part of the major issue with climate change: If its effects were well understood, they could be prepared for. --Jayron32 00:29, 17 May 2012 (UTC)[reply]

There's no "probably" about the continuation of the Gulf Stream, or its direction, just basic fluid dynamics. See western boundary current. You're right about questions of thermal transport, salinity, and thus effects on climate of Europe (though in strict terms the latter are attributable to extensions of the Gulf Stream such as the North Atlantic Drift, rather than the Gulf Stream proper). Short Brigade Harvester Boris (talk) 01:45, 17 May 2012 (UTC)[reply]

I wouldn't call it "basic" fluid dynamics! That'd imply that it's the sort of thing we could use a simple equation for. Or, that it's the sort of topic that gets covered in an introductory textbook on fluid motion! The hydraulic equation is basic fluid mechanics. Bernoulli's law is basic fluid dynamics. The statistical dynamics of an ideal gas flowing in a perfectly thermally isolated closed box is fairly intermediate or advanced fluid dynamics. But climate and oceanic currents? That's some serious world's-largest-supercomputer kind of cutting-edge theoretical physics research. We have fairly sparse coverage in Wikipedia. One day I intend to work on improving our articles on global circulation models, but it's really dense stuff. Nimur (talk) 03:42, 17 May 2012 (UTC)[reply]

The equations which lead to a western boundary current are simple... or as simple as you get on the scale of fluid dynamics, anyway. They are laid out here quite nicely. Western boundary currents are unavoidable even in extremely simplified models of large bodies of water on a rotating sphere. -RunningOnBrains^(talk) 03:51, 17 May 2012 (UTC)[reply]

Urgent Antihistamine Question

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.

If you have a question about the effects, uses, or dosage of medication, consult a medical professional. --Jayron32 23:14, 16 May 2012 (UTC)[reply]

Fair does. However, since "the deed is done" and I am genuinely not asking about my situation any more I will repost what I believe to be a compliant version just in case it interests people:

Antihistamine and general drug dosage question

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.

If you have a question about the effects, uses, or dosage of medication, consult a medical professional. --Jayron32 23:14, 16 May 2012 (UTC)[reply]

Since you made your intent known in the previous version, people here cannot unremember the past. It is best to just drop this line of questioning, and ask a doctor if it is of concern to you. Or use the Wikipedia or Google or WebMD search function where no one will try to stop you. --Jayron32 23:35, 16 May 2012 (UTC)[reply]

I appreciate that but I'm not at all asking about the medicine as it relates to me, or tbh that medicine at all. I'm genuinely interested in the theoretical question. Although as the actual medicine is starting to kick in now, I will just ask tomorrow at some time when it is obvious it's just for interest, then we can all rest easy

Egg Centri c 23:40, 16 May 2012 (UTC)[reply]

Talk to someone who's qualified to answer, such as your doctor or maybe your pharmacist. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:50, 16 May 2012 (UTC)[reply]

One suspects that the complainants here must be associated with the medical profession, as once they have branded the defendant as "sick" and permanently denied him the rights other people would have on this basis, their work is done! Wnt (talk) 04:35, 17 May 2012 (UTC)[reply]

We can't answer any questions on specific medical matters, it doesn't figure into that how hypothetical it might be. It's a moral and a legal issue, if someone reads the archives or the thread here and makes an assumption they could become seriously hurt. It's not just because of the original question-asker, but anyone who might read the advice and be tempted to follow it rather than the guidance of a certified medical professional. HominidMachinae (talk) 04:39, 17 May 2012 (UTC)[reply]

No, I think you've misunderstood. We can't offer medical advice. We can indeed answer questions about medicine. --Trovatore (talk) 05:29, 17 May 2012 (UTC)[reply]

Physicians can bluster about the importance of following medical advice, but unless the person can pay them their shakedown, even the most medically necessary medicine will be banned from the patient by law, even though it is otherwise affordable - just as their advice on this would be limited to paying customers. I am wearied of their ethics. Wnt (talk) 04:53, 17 May 2012 (UTC)[reply]

Some of us are fortunate enough to live in a country where medical advice is free at the point of use. Here in the UK, if I need medical advice of the sort that's often asked for here on WP, I can go and ask my GP, or a pharmacist, and pay nothing for the privilege. (I've already paid for it in my taxes, but I'll complain about tax spending on aircraft carriers or unimplemented national databases before I complain about tax spending on keeping me and everyone else healthy.) AlexTiefling (talk) 08:15, 17 May 2012 (UTC)[reply]

Wnt apparently would stake his very life on the free (and worth the price) advice from strangers on the internet, rather than pay a trained professional. That's up to him. And if he suddenly stops editing one day, we'll know what happened. But wikipedia has rules against giving medical advice, and if Wnt doesn't like that rule? Too bad, too sad. ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:27, 17 May 2012 (UTC)[reply]

Yeah, well, here in the U.S. they just closed down the local hospital, one which had stood for decades, in a sensible business decision reflecting that there is just not that much demand for medical care in the Third World. The prisons for people who sell allopurinol without a prescription, of course, well, those received a perky plus with all the money the state saved cutting the education budget. Wnt (talk) 21:32, 17 May 2012 (UTC)[reply]

What has that got to do with wikipedia's rules against giving medical advice? ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:04, 17 May 2012 (UTC)[reply]

Nobody here ever pretended to give medical advice; then someone called it a "moral issue". Wnt (talk) 03:10, 18 May 2012 (UTC)[reply]

Advising someone NOT to see a doctor qualifies as medical advice. ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:43, 18 May 2012 (UTC)[reply]

So make an appointment with your doctor so you can ask him if you should make an appointment with your doctor. :-) StuRat (talk) 05:52, 18 May 2012 (UTC) [reply]