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September 3[edit]

replacement of secondary alcohol (in benzyl position) with H[edit]

I suppose the use of hydride and the use acid catalyst don't go together?

Is there anyway of protonating an OH group, having it leave, making a secondary benzyl carbocation, and then have hydride take up the spot? I think the issue is that it's kind of difficult to have functioning acid and hydride in the same solution. (And I don't mean that acid and hydride, har har...) Would it be feasible to replace the OH with a halogen, extract the product out of acidic solution, and then put it in basic solution where hydride would perform a nucleophilic attack on the halogenated carbon?

John Riemann Soong (talk) 04:34, 3 September 2009 (UTC)

Try this on for size. Oxidize the alcohol to the ketone using PCC and then reduce the ketone to the alkane using the Wolff–Kishner reduction (aka Hydrazine reduction). Working with raw "hydrides" is kinda messy (like, they ignite in air), and reducing alcohols directly, even with "harsh" reducing agents like Lithium aluminum hydride, doesn't usually work so well.

--Jayron32 04:46, 3 September 2009 (UTC)

Are there any complications to oxidisation/reduction if the alcohol is in benzyl position? (Also my pen and paper problem set doesn't give me good marks for excellent yield lol ... I just need a theoretical mechanism...).

Delta Air Lines (DAL)

Carbocation formation as a mechanism first came to mind because of it. John Riemann Soong (talk) 05:21, 3 September 2009 (UTC)

I don't believe so. The benzylic position generally tends to improve reactivity of nearly ALL reactions at that site since the benzylic position will stablize cation/anion/radical formation via conjugation. I am not aware of any particular reaction being LESS favorable at the benzylic position.

I am pretty certain PCC will oxidize ANY secondary alcohol to a ketone, and Hydrazine/Hydroxide will reduce ANY ketone to the hydrocarbon. You may want to double check your textbook on these, but any introductory organic text should have both PCC oxidation and Wolff-Kishner reduction mechanisms in them. --Jayron32 05:28, 3 September 2009 (UTC)

Oh I was wondering about possible side reactions that would take place at the benzylic position, since it's as you say, more reactive.

I mean, if you convert the alcohol into a ketone, you're going to get a pi system interaction with the benzene ring. (But I guess this will make the carbonyl group more nucleophilic?) John Riemann Soong (talk) 05:38, 3 September 2009 (UTC)

"a hydride [...] to pull off the halogen, with another hydride coming in" doesn't make chemical sense: anion attacking halogen seems electrostatically unlikely, and I can't think of a way to make "BnCl + 2H– → BnH + ???" a good balanced reaction.

Once you have halide attached, you just displace it SN2 with hydride. Or else you would need something that attracts chloride to pull it off and so an SN1 (a cation not anion, since you want to pull off chloride in order to have a carbocation for hydride to attack).

Chloride is already a good leaving group, so you don't need to activate it further (unlike hydroxide, which you correctly observe needs to be protonated to become a good leaving group).

Simple alkali-metal hydrides aren't often used as H- nucleophiles for SNx reactions (more often Group-3 complexes like NaBH4). I think there are both solubility and mechanistic reasons, can't remember what specifically is important for this case.

There are actually ways to get hydride donors in a lewis-acidic environment.

Hypophosphorus acid is one such, and it (alone or with additional strong acid in solution) may be sufficient to allow you to protonate and then SNx your alcohol. There are other direct ways to accomplish this change too. The major chemicals for the direct transformation are regulated because this reaction is how to make speed (conversely, your local meth-dealer might know some clever ways to do it and/or suppliers for the chemicals).

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DMacks (talk) 04:55, 3 September 2009 (UTC)

Re: the hydride pulling off the halogen -- that was a total brain fart on my part (as you see in the history where my edit summary was 'remove silliness').
Hahahaha, I just realised meth synthesis relies on displacement of a benzyl secondary -OH too ...

lol what a coincidence. (I'm just trying to do a problem set.) John Riemann Soong (talk) 05:03, 3 September 2009 (UTC)

It sounds like you are thinking of converting the OH into a good leaving group eg
R-OH + CF3COCl >>> R-O-C(O)CF3
If not continue reading after the indent
The trifluoroacetic ester is a good leaving group, and can be displaced by nucleophiles - including potentially "H-" equiavlents, unfortunately the ester group is also susceptable to nucleophilic attack.

I think there is a synthetic method that involves a process like this for converting R-OH to R-H - but it's obscure, involving specific leaving group, and "H-" donor, and currently I can't remember it.

There's some examples here http://www.organic-chemistry.org/synthesis/C1H/deoxygenations.shtm
The first example using a thioester is actually a radical reaction, so doesn't count.
The chlorodiphenylsilane reduction is much nearer to a Sn2 reaction - probably via a cyclic intermediate.
Ah here it is - convert the alcohol to the toluylmethanesulphonate, then reduce with LiAlH4 [1] also [2]
See 4-Toluenesulfonyl chloride the reaction is also mentioned in that page too.
The benzyl group should just speed up the reaction (excluding steric effects) - for Benzyl alcohol I don't imagine any side reactions, for others such as 1phenyl butan-1-ol there's the possibility of elimination as a side reaction to make a styrene83.100.250.79 (talk) 11:48, 3 September 2009 (UTC)
The "via a halogen" method is also good - but usually done like this - convert the alcohol to the halogen, then metallate eg use lithium metal to make the Benzyl Lithium compound, the just add water , or another H+ source to get the alkane.
Ph-CH2-OH >>> Ph-CH2-Cl use thionyl chloride or something else Ph-CH2-Cl >>> Ph-CH2-Li use lithium Ph-CH2-Li >>> Ph-CH2-H add water (talk) 11:55, 3 September 2009 (UTC)
Note there are quite a few different ways to reduce halo compounds to alkanes, including direct 'hydride' reduction..

The link below gives many examples pages 1 onwards.

Ahh! Here is exactly what you where asking for! (in addition to the one already given by DMacks)[3] Page 13, scheme 15, bottom example -- note here that NaBH4 is used becuase even though it can act as a hydride donor, it is quite stable to mild acid (?) (well more than most - I'm slight dubious about that without more info., but the other examples in the text are good).

There's a lot more examples in the text. (talk) 12:26, 3 September 2009 (UTC)

how is -COOH formation oxidisation?[edit]

This aspect of COOH formation always seems to be glossed over ... but I don't really get how it's "oxidisation". I can totally see how increasing unsaturation (via elimination or oxidising an alcohol into an aldehyde) is oxidisation, because the organic compound actually sees electrons exit its system.

But I don't see how RCOOH is any more electron deficient compared to R(C=O)H.

Plus, when I look at the thermodynamics of alcohol metabolism in the body, I see that aldehyde formation is moderately endothermic (e.g. ethanol to acetaldehyde), but the conversion of aldehyde into -COOH is quite exothermic .

John Riemann Soong (talk) 05:31, 3 September 2009 (UTC)

It's just the "formal oxidation state" that increases - obvious the carbon stays as a 4 bonded atom.
If you consider RCHO + 'O' >>> RCOOH
It's clear that oxygen has been added - hence oxidation, though in fact the O has just been inserted into the C-H bond.

The molecule has been oxidised, but the carbon atom hasn't really changed it's oxidation state. (talk) 09:05, 3 September 2009 (UTC)
Ah, wikipedia has an article - see Oxidation number versus Oxidation state (note that Hydrogens at considered to have oxidation +1 , not -1 ; when in organic compounds) (talk) 14:17, 3 September 2009 (UTC)
Organic redox reaction does explain it quite well.

--Jayron32 20:23, 3 September 2009 (UTC)

pressure distribution , how amount and area of effect works[edit]

Hi guys

Ok I have been puzzling on this for a while so can someone answer me this. I know from watching a show called mythbusters that even the most high powered rifles cannot knock a person backwards by force alone.

The question I have therefore is this. If you add up the amount of kinetic energy, or simply the amount of pressure I can produce with my hands when I push someone, this would not be very much compared to a bullet, put simply if you could focus all the kinetic energy of my body to one point the size of a bullet, it would not penetrate steel or shatter concrete.

A bullet however can do this. My question is how come if less kinetic energy is transferred to the object that I push does it move further backwards than when It is struck with a bullet. I.E I can push someone over but a high powered bullet cannot. Does the area of my hands, much wider than the surace of a bullet have anything to do with it.

I know weight comes into it as well, Im talking pure pressure, I could produce about 50 psi at best, while bullets can produce thousands of psi. How does this work?

Thanks guys —Preceding unsigned comment added by (talk) 09:37, 3 September 2009 (UTC)

Force and energy are NOT the same thing. Someone owes me a dollar - because it's a Wikipedia Ref Desk rule the I get a dollar every time someone confuses the two!

Kinetic energy is calculated as energy = mass x velocity2/ 2. Velocity squared means that when you double the speed that something is moving, you quadruple the amount of energy it can deliver. When you gently poke someone in the chest with your finger - it's moving at maybe 0.1 meters per second. Even a really crappy pistol can propel a bullet at maybe 400 meters per second - 4000 times as fast - and if the bullet weighs about the same as your finger, that's 16 million times as much kinetic energy...OUCH!!

However, the force of the recoil for the shooter and the force applied to the victim is calculated by force=mass x acceleration. Which means that the forces involved don't depend on the speed - but the rate at which the bullet speeds up or slows down. For the victim, that means that the distance the bullet travels through their body - which determines the time it takes to come to a complete stop - and that determines the force it applies while it's slowing down.

Of course with a really high powered weapon, the bullet may go in one side of the body and out the other without stopping or even slowing down by very much - which reduces the force it applies still more!

So that's why the bullet produces immense amounts of kinetic energy (which can do you a whole lot of no good) - but not enough force to knock you backwards 10 feet like in the movies.

Newton's laws tell us that every action has an equal and opposite reaction - so the force of the recoil felt by the shooter is equal to the force felt by the victim (minus a bit due to air resistance). However, in a rifle, that force is spread out over many square inches by the big, fat stock of the rifle - where the victim feels the same force - but has all of it applied at the teeny-tiny tip of the bullet.

Hence the pressure that the victim feels is much more than the shooter - which is why the bullet ends up embedded in some vital organ while the shooter doesn't have the recoil blow his shoulder off. SteveBaker (talk) 11:55, 3 September 2009 (UTC)

So would they get knocked back 10 feet if they had a bulletproof vest on, which quickly decelerated the bullet? Googlemeister (talk) 13:13, 3 September 2009 (UTC)
We should be considering momentum here to determine the speed with which someone is knocked back.

If the bullet is stopped by the vest then mass times velocity of the bullet will be equal to the mass of the person times the speed at which they start to move backwards. Because the person is many times the mass of the bullet, their speed will be low, but if the person is caught off-balance, they may well stagger backwards. A marginally greater momentum is given to the gun (and the person firing it), but this is only a problem if one is holding the gun lightly, or balancing on one leg to fire.

Dbfirs 15:21, 3 September 2009 (UTC)

Also, if the person was running for cover and shot near the center of mass while his feet are in the air, it might cause his path to veer slightly.

Sagittarian Milky Way (talk) 17:11, 3 September 2009 (UTC)

bus bar[edit]

what is the use of bus bar atGSS?Satyendra singh sihoria (talk) 10:39, 3 September 2009 (UTC)

Busbar may help you; you'll need to explain what you mean by GSS if you need a more specific answer.

-- Finlay McWalter • Talk 10:41, 3 September 2009 (UTC)

Attachment of dental crowns[edit]

Dental crowns are attached to the root of the tooth by a metal post being put into a hole drilled into the root. How is this post kept firmly attached to the root, particularly when newly made, and for upper teeth?

Cabot Oil and Gas (COG)

What stops it falling out? Thanks. (talk) 10:43, 3 September 2009 (UTC)

According to post and core, the post is either smooth-sided and held in place only by cement, or is serrated or threaded to engage with the remaining tooth root.

The X-ray images on that page show a variety of post designs. Gandalf61 (talk) 11:39, 3 September 2009 (UTC)

Prosthetic crowns are cemented to teeth that exhibit decay or are fractured to an extent that does not allow a filling (intracoronal restoration) to be placed -- thus, the restoration must be placed around the remaining tooth structure, and is referred to as an extracoronal restoration.

Crowns are routinely cemented to the underlying tooth structure with a number of various types of dental cements. When there is not enough remaining tooth structure to provide adequate physical retention for a prosthetic crown (although cement is used, a crown placed onto a tooth that does not provide adequate physical retention, e.g. overtaper, will not remain there predictably), a core build-up is necessary to compensate for the missing internal tooth structure.

If there is not enough tooth structure to retain a core build-up, the core build-up can be built around a post that has been cemented into the coronal portion of of a root canal that has had a portion of it's root canal filling (after root canal therapy has been completed) removed.

While the core build-up is often a resin-based composite material which bonds to the tooth structure, the post and crowns are affixed with cements like zinc oxide phosphate (ZOP), flowable resin-based cements and other similar materials.

Upper teeth have root canals for housing posts just the same as lower teeth, and without proper cementation, a lower tooth will not remain in position any more predictably long-term because of chewing forces, sticky foods and the tongue -- gravity will have very little effect on displacing upper crowns and allowing lower crowns to remain.

DRosenbach(Talk | Contribs) 03:42, 4 September 2009 (UTC)

Why do ruins end up buried under soil?[edit]

It's strange, should be simple, but I just can't work it out. Where does the soil come from that buries anything lying around for long enough? Are the continents getting thicker?

Adambrowne666 (talk) 10:56, 3 September 2009 (UTC)

(EC with Steve below) Mostly it's just localised redistribution of material.

I think the four main mechanisms involved are:

  1. Dust and other small soil particles are blown on to the ruins by the wind, and build up significantly over long periods. In areas subject to sandstorms and similar phenomena, the periods may not be so long. Think how often one has to sweep a path clean.
  2. Plants (which are mostly made of molecules extracted from the atmosphere) grow on, up through, and over the ruins, trapping more windborne dust/soil, and themselves die and decay on top of them forming a mulchy component - this can build up surprisingly quickly (you should see my neglected patio!).
  3. Earthworms churn up the underlying and surrounding soil causing more solid objects to sink into it, as Charles Darwin famously observed.

    Other larger burrowing creatures can contribute to this effect.

  4. People may build on top of the ruins using less durable materials and later abandon the site, allowing those overlying structures themselves to decay into soil components.
Hopefully a trained archaeologist will quantify, correct and expand on this initial stab. (talk) 11:46, 3 September 2009 (UTC)
Simplistically: When there is erosion due to wind, rain, rivers, ice-sheets & glaciers, dirt gets eroded from one place - and deposited someplace else.

Hence, by the law of averages, you might expect about half of ancient ruins to be eroded to nothing and the other half to be buried in the resulting dirt - which protects them from erosion.

Obviously the ones that were eroded away are never found - so pretty much the only ruins we ever find are the ones that were buried.

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However, it's likely that the nature of erosion processes is to transport material from high peaks and dump them into lower valleys (well, mostly) - and because people like to build their homes close to water sources - that may well result in there being more ancient habitation in valleys than there on mountain-tops...so I guess, statistically, more ruins get buried than are eroded away.

The continents don't get thicker - just smoother - the mountains get lower and the valleys higher. But on geological timescales (too long to concern the preservation of human-constructed ruins), mountain ranges are pushed up by the buckling and colliding of continental plates and are also raised by volcanoes - and the ancient, smoothed-off, eroded land is subducted away to be recycled in the earth's core - or gouged out by rivers and glaciers.

So there are two sets of processes - the plate techtonics and vulcanism make the world rougher and erosion mostly makes it smoother. SteveBaker (talk) 11:28, 3 September 2009 (UTC)

The Straight Dope did an interesting article about this : How Come Archaeological Ruins are Always Underground. APL (talk) 13:18, 3 September 2009 (UTC)
We also have an article on subsidence, which is the geological term for changes of the ground surface.

In many cases, the entire ground level is actually lowering, and new material is being replaced on top. In some cases, the rate of subsidence can be measured in inches or feet per century; this effect alone may account for some archaeological sites being buried. Nimur (talk) 16:15, 3 September 2009 (UTC)

Thanks, all, for your thorough and informative answers.

Adambrowne666 (talk) 07:45, 6 September 2009 (UTC)

There was a show on BBC that showed that all human ttraces would be gone from the earth except maybe the Hoover dam in about 100,000 years due to the building material used - also it showed that Nature (trees plants ect) reclaim the cities in poretty quick time say 70 - 80 years. would be good to watch if you really wanted to.Chromagnum (talk) 08:02, 8 September 2009 (UTC)

What do stock traders do?[edit]

Is it just a gamble, no matter what, or are there people able to predict the movements of the stock market?--Quest09 (talk) 11:44, 3 September 2009 (UTC)

See fundamental analysis, and the slightly pseudoscientific technical analysis.

--Mark PEA (talk) 12:15, 3 September 2009 (UTC)

I'll take a loot at these two links above. However, I consider the technical analysis not better than astrology and the fundamental analysis very suspicious.

-- (talk) 17:50, 3 September 2009 (UTC)

Fundamental analysis isn't suspicious. You analyse the company and work out how much it is worth, divide that by the number of shares and you get what the share price should be. If that is higher than the current price, you buy, if it is lower, you sell.

It is perfectly logical. The problem is that everyone else can do the same analysis and reach the same conclusions so there won't generally be much of a difference between the current price and the calculated price. This is known as the efficient market hypothesis. The only way to reliably make money on the stock market is by breaking one of the assumptions in that hypothesis - the most obvious way to do that is to get hold of information other traders don't have (either by having better contacts or by interpreting the same data better).

If you can do that, you can make money. Technical analysis, on the other hand, is complete nonsense - it only works because of self-fulfilling prophecies (all the technical analysts see a certain pattern and think the price will go up, they then buy and that pushes the price up).

--Tango (talk) 18:45, 3 September 2009 (UTC)

Apparently, if the efficient market hypothesis is true (what seems plausible), and you are not breaking any law (like using insider information), then you'll never make money with the fundamental analysis alone.

Les meilleurs plateforme de trading crypto

Great surprises come only from the fact that people start to speculate (=no fundamental analysis) about the influence of the weather/terror/politics in a company. Therefore, any money that you invest, would simply be a bet, wouldn't it?--Quest09 (talk) 10:46, 4 September 2009 (UTC)

That's right - if all publicly available information is already incorporated into the stock price, you can't make money by analysing the publicly available information.

However, research indicates (I don't have the refs off-hand) that financial markets are not strong-form efficient which means that 1)not all publicly available information feeds investor's decisions and 2)different investors perhaps interpret the information differently.

So your last comment is probably not accurate - evidence would suggest that money can be made by being better at fundamental analysis than others. Remember, equity analysts face a daily battery of information (market/trading reports, interviews with management, economic data, company financials, analyst reports) and it takes a fair amount of skill to sift out the useful bits.

Zain Ebrahim (talk) 11:02, 4 September 2009 (UTC)

Well that's if you want to speculate -- you basically are driven to look at undervalued shares and buy them. After all, to sell them to someone, there must be some value in the shares. In an IPO, shares are sold, giving capital to the company in return for some cut of future profit (voting rights, whatever). The company uses the IPO money, expands, and pays investors back (e.g.

in the form of dividends). As the company grows the share price rises, because the company's profits will probably increase. If you're a speculative trader, you prolly don't want to hold on to the stock for too long, simply because when you buy stock you're tying up capital in that stock (think opportunity cost).

Theoretically, the idea is to find rough diamonds, and stocks create an incentive to fund their expansion.

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(But such funding only occurs during an IPO of course -- the rest of the trading afterwards is what the investors do with their "reward".)

The game is also slightly different for private stocks or funds, etc. There's this firm that looks like it has a promising business plan, and it seeks investors to give it capital. In return, it promises you a cut of future profits.

Except in an IPO, this aspect of the game doesn't exist for public stock. John Riemann Soong (talk) 19:09, 3 September 2009 (UTC)

"Mass" of colour[edit]

As a partially red-green colourblind person reading Color blindness, I was surprised to see the following line in that article's section "Design implications of color blindness":

Thirdly, for some color blind people, color can only be distinguished if there is a sufficient "mass" of color: thin lines might appear black while a thicker line of the same color can be perceived as having color.

I've often experienced this, finding it harder to distinguish the colour of a line one pixel wide than a line ten pixels wide. For you who can see normally — is it generally equally easy to distinguish the colours of small and large amounts of colour?

I've always assumed that it would be harder for smaller amounts for anyone, simply because less colour is present. Nyttend (talk) 13:24, 3 September 2009 (UTC)

Pretty much just as easy. If something is relatively far away, and the sizes are out of proportion (TONS of blue and a little green) then maybe, but not really.

The only time it actually ever sort of happens (for me, anyway) is when there are very similar colors - light gray text on a large, off-white background from far away can look fuzzy. I think a bigger issue is that when there are lots of colors together it can be hard for some to focus on individual colors rather than an amalgam, but it is still quite discernable. ~ Amory(user • talk • contribs) 14:01, 3 September 2009 (UTC)

In my experience, if an object is in the center of my vision, (ie: I'm looking right at it) I can distinguish the color down to very small sizes.

Almost, but not quite, down to the visual threshold. Certainly identifying the color of a single bad pixel on a monitor at a reasonable desk-sitting distance is no problem at all. APL (talk) 14:40, 3 September 2009 (UTC)

I think distinguishing between similar colours is easier for everyone where there is a large expanse of each, but small areas will be more of a problem for those whose receptor cones are less efficient at distinguishing.

Dbfirs 14:52, 3 September 2009 (UTC)

Same color illusion is worth a look for how even 'normal colour sight' (?) people's eyes get confused. (talk) 15:37, 3 September 2009 (UTC)

That's a different phenomenon, though -- it depends on higher level processing.

The basic and universal fact is that the color pathway has lower resolution than the contrast pathway -- the brain combines information from the two to make "guesses" about color. You can see this in action in the "Just GREY" illusion here, for example. Looie496 (talk) 16:36, 3 September 2009 (UTC)

The density of rods and cones in the eye is the issue here. There are about 90 million rods and 6 million cones in a "normal" eye.

The rods are denser around the edges of the eye - but even in the center of our visual field there are about 6 times as many rods as cones. Since rods are responsible for monochromatic 'brightness' detection and the cones do color - you can readily see that very small features lose their color because there isn't sufficient spatial resolution of cone cells.

Hence the observation that our article makes is true of normal people as well as colorblind people. In SOME kinds of colorblindness there are even fewer cones than usual - because (perhaps) all of the green-detecting cones are missing - that would make the effect more pronounced - but that's not true of all kinds of colorblindness.

SteveBaker (talk) 18:12, 3 September 2009 (UTC)

Your input is quite helpful. Part of the reason I was surprised is the contrast between blue and black in my sight — while I see blue (and yellow) without problem, if I go into Microsoft Word and change the colour of a comma or a single letter from black to blue, I don't generally notice it at first glance; I need to pay attention or change the colour of other letters to notice it.

Nyttend (talk) 21:58, 3 September 2009 (UTC)

I think I would have difficulty noticing a blue comma in black text depending on the shade of blue and I don't have any form of color blindness. For example the wikipedia link color is very noticable, but the visited link color is less so. For an entire word it stands out reasonably well, but if it were a single letter I might miss it if I weren't specifically looking out for it.

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Rckrone (talk) 00:20, 4 September 2009 (UTC)

earth without moon[edit]

If the moon were to magically disappear, what would the impact to the earth be? Would there be a mass extinction? Googlemeister (talk) 14:53, 3 September 2009 (UTC)

See Wikipedia:Reference desk/Archives/Science/2009 January 29#Earth without a moon.

There's some really good talk there, and not just the obvious stuff about tides. ~ Amory(user • talk • contribs) 15:16, 3 September 2009 (UTC)

Interesting. So it would seem unlikely that there would be a mass extinction? Googlemeister (talk) 15:40, 3 September 2009 (UTC)


You are asking for scientific predictions of outcomes in a fundamentally unscientific scenario. Maybe the magical disappearance of the moon creates huge flocks of flying pigs which blot out the sun and trigger a new ice age.

Who can tell ? Once you postulate magic then all bets are off. Gandalf61 (talk) 15:57, 3 September 2009 (UTC)

"Once you postulate magic then all bets are off." Coming from an account named after Gandalf. ~ Amory(user • talk • contribs) 17:27, 3 September 2009 (UTC)
Hmmm....are all bets off for Maxwell's demon? (talk) 17:46, 3 September 2009 (UTC)

A mass extinction of what? People? I don't really see why species not directly influenced by the moon (that depend on the tides, for example) would be affected, especially not immediately.

Whether life would have evolved in a similar manner had the moon never existed is another matter... TastyCakes (talk) 16:02, 3 September 2009 (UTC) TastyCakes (talk) 16:00, 3 September 2009 (UTC)

You may be interested in the following documentary: The Universe - The Day the Moon Was Gone.

A Quest For Knowledge (talk) 16:58, 3 September 2009 (UTC)

The complete elimination of lunacy? Edison (talk) 19:06, 3 September 2009 (UTC)
Werewolf movies would make a lot less sense to future generations. Googlemeister (talk) 19:49, 3 September 2009 (UTC)
According to Tides#Forces, the Sun has 46% of the effect on tides as the Moon does.

As such, the tides would always be about as strong as they usually are at their lowest. Also, the poles are always at low tide, so weaker tide would mean more water there, and less everywhere else.

I suspect all this would have drastic consequences on plants that live on the coast. I may be wrong, though. — DanielLC 00:28, 4 September 2009 (UTC)

Tin Pan Alley would be stumped to come up with romantic songs rhyming "croon," "June", and ????? Businessmen currently seek to tear down the physical block of New York City which was "Tin Pan Alley" for comercial development.

Seems a common theme.Edison (talk) 04:01, 4 September 2009 (UTC)

  • See this essay[4] and a documentary narrated by Patrick Stewart called "If We Had No Moon". I watched it last week, by complete coincidence. The major difference would be a destabilisation of the axis of rotation of the Earth; it rotates at a near constant axial tilt of 23 degrees now, but without the Moon it would oscillate much more, like happens with Mars, and this would play havoc with the climate.

    See Planetary habitability#Orbit and rotation and Axial tilt#Long period variations. Fences&Windows 01:16, 9 September 2009 (UTC)

CO2e emissions[edit]

Hi. What are the carbon emissions produced by a 12-hour commercial flight? Assume a Boeing 777, travelling at typical speed for a continental flight, using regular fuel.

Calculate the CO2 emissions produced, in addition to the equivalent GHG warming potential of any other greenhouse gases resulting from fuel combustion and on-air energy usage. Preferable units include pounds, kilograms, and metric tonnes. Total emissions for the flight are preferred to emissions per passenger, although that's fine too if calculateable.

This is not homework. Also, how much net CO2 might one square metre of tropical rainforest remove in its lifetime, assuming current climate conditions and current CO2 concentrations for the entire lifetime of the square metre, and assuming a lifetime of 40 years? Thanks. ~AH1(TCU) 15:54, 3 September 2009 (UTC)

This looks a lot like a homework problem.

We're not supposed to solve those. Looie496 (talk) 16:23, 3 September 2009 (UTC)

(ec) Just thought I would point out that fuel consumption on an aircraft is extremely non-linear. So, if anybody is planning to calculate "average fuel consumption" and multiply by flight duration... reconsider?

Navigation menu

It would be better to find an actual end-to-end fuel consumption estimate for a 12-hour flight on a 777 and use that as a baseline for estimation.

Nimur (talk) 16:24, 3 September 2009 (UTC)

I highly doubt it is a HW question because of the level of complexity. To be honest there is absolutely no way to give a good answer with the supplied information. first, there is something like 6 different variants of the 777, each would have a different fuel consumption profile.

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Second, weather conditions will be a huge factor for this flight, especially whether the flight will be against the wind or with the wind, and what speed of wind do we have. Also, we need to know how loaded the plane is. Do we have a full load of fuel and passengers, or is the flight half full? Even this information was known, the resulting calculations would be quite substantial. Probably not something you could spit out in 5 minutes.

As to the rain forest part, there is still some question as to whether rainforest scrubs CO2 at all since the carbon just renters the atmosphere when the plants die and decay. Googlemeister (talk) 16:31, 3 September 2009 (UTC)

Thanks, but I already indicated that "This is not homework".

I will now bold the relavent sentence in my question. Also, is any calculus required to solve such a problem? ~AH1(TCU) 17:22, 3 September 2009 (UTC)

Such claims are ignored because homework-askers have been known to lie.


Comet Tuttle (talk) 19:42, 3 September 2009 (UTC)

I didn't disbelieve it, I just failed to see it -- sorry. Looie496 (talk) 20:57, 3 September 2009 (UTC)
Google is your friend.

A five second search turned up a bunch of links. This says saving one kilo of fuel saves 3.16 kilos of CO2.

This says a 747 uses about 36,000 gallons of fuel on a 10-hour flight (approx. 1 gallon per second or 5 per mile) so by simple calculation using the density found at Jet fuel, that flight would produce around 383,000kg of CO2.

Also, here's a a link to calculate the emissions for any flight. ~ Amory(user • talk • contribs) 17:40, 3 September 2009 (UTC)

Personally I'm more concerned about the massive consumption of fuel than the emission of CO2.

John Riemann Soong (talk) 18:47, 3 September 2009 (UTC)

That, of course, is the full plane, which isn't THAT bad when you consider that it's moving 250 people at 5 times the speed of highway cars at 0.2 mpg. --antilivedT | C | G 04:27, 4 September 2009 (UTC)
That is also not the specified type of airplane.


What would happen if an ant crawled up my nose or into my ear while I was sleeping?

Would it find conditions favorable to stay awhile? Would it get lost? Would it be able to reach my brain or lungs? —Preceding unsigned comment added by (talk) 16:29, 3 September 2009 (UTC)

Unless it was going to make its own tunnel, it could not get to your brain. It would also have a hard time getting to your lungs because it would trigger the coughing reflex.

Googlemeister (talk) 16:33, 3 September 2009 (UTC)

While following up on a dubious claim in an article a while ago, I came across this very real and very reputable report from the Royal Society of Medicine, Doctor, there are maggots in my nose. Gross. The article also cites some case studies of other insect species infestations.

The most common are fly larvae; and most of the medical techniques known to human-doctors are attributed to prior research from agriculture, livestock, and veterinary procedures.

Imax B6 CONNECTION BREAK with venom batterys with universal connectors RESOLVED

Nimur (talk) 16:49, 3 September 2009 (UTC)

And let's not forget about the earwig. Baseball BugsWhat's up, Doc?carrots 17:49, 3 September 2009 (UTC)
But, earwigs don't really do that.

APL (talk) 18:33, 3 September 2009 (UTC)

Involuntary Ingestion of an Airborne Diptera Specimen by the Geriatric Female: Incidental Report and Treatment MethodologyCuddlyable3 (talk) 18:46, 3 September 2009 (UTC)
Without looking, I'm guessing that's a report about an old lady who then swallowed an arachnid that wiggled and wriggled and tickled inside her.

And so on. :) Baseball BugsWhat's up, Doc?carrots 20:59, 3 September 2009 (UTC)

Technically, she swallowed the spider to catch the fly. I presume she will die.Quietmarc (talk) 02:11, 4 September 2009 (UTC)
In addition, According to the report, she took the further absurd step of consuming live avian fauna in an attempt to eliminate the aforementioned arachnid. It is predicted that this will be fatal. APL (talk) 02:42, 4 September 2009 (UTC)

~ Amory(user • talk • contribs) 02:50, 4 September 2009 (UTC)

As an emergency measure, non-morbid feline ingestion may be indicated. (talk) 02:55, 4 September 2009 (UTC)
Ants can't survive for long outside of a colony. Vimescarrot (talk) 07:46, 4 September 2009 (UTC)
If they exude toxins while inside one, what is the 'antidote?

Edison (talk) 15:27, 4 September 2009 (UTC)

  • Other invertebrates can act as internal parasites, but I don't know of any reports of ants doing so. However, Pheidole ants have been known to cause hair loss.[5]Fences&Windows 01:38, 9 September 2009 (UTC)
p.s. Re: Nimur's comment, with maggots this is called myiasis or flystrike, and I can gross everyone out with this paper: Furuncular myiasis of the breast caused by the larvae of the Tumbu fly.

It even has videos. Fences&Windows 01:43, 9 September 2009 (UTC)

nucleus volume[edit]

Does the volume of the nucleus of an atom grown in proportion to the volume of multiple cubes or in proportion to the volume of an expanded sphere? In other words does the volume increase in proportion to adding marbles to a bag or in proportion to blowing up a balloon? --Taxa (talk) 16:38, 3 September 2009 (UTC)


The characteristic radius of a nucleus can be defined in many ways, but the collisional radius is a very complicated parameter. See nuclear cross section for a quick introduction. In general, a quantum mechanical treatment is necessary to define this parameter.

Nimur (talk) 16:43, 3 September 2009 (UTC)

Where would the actual radius plot on this diagram? --Taxa (talk) 17:03, 3 September 2009 (UTC)

Our Nuclear size article uses an expanding-sphere model. DMacks (talk) 17:32, 3 September 2009 (UTC)
Okay I see that - density remains the same for all atoms - hence a solid and not a homogeneous sphere.

--Taxa (talk) 21:32, 3 September 2009 (UTC)

Sorry my earlier post was so brief; I didn't explain thoroughly enough. Nucleons are sufficiently small that they behave quantum mechanically. This has many important implications for things like the definition of their size.

Because we can't actually measure the nucleus directly, we instead estimate its size by bombarding it with a beam of particles, and statistically determining how many of those particles "bounced off" - thus determinining a collision cross section. However, and this is very critical - the collision cross section of an atomic nucleus depends on how we set up the experiment. That is, changing the incident energy, or the spin, or polarity, or any other quantum characteristic of the incident beam, changes the size of the nucleus that is measured. See here for an example of nucleus-particle resonance.

So, it's not really possible to tell you the "volume" or the "length" of the atomic nucleus, let alone how it scales with proton-number; because it doesn't scale in any classical sense. Instead, every nucleus has its own characteristic cross-section profile and interacts uniquely with various incident particles.

One of the simplest models, as documented in Atomic nucleus#Nuclear models, approximates the trend as a power law, r ≈ Z1/3 (Z=atomic number), or the similar r ≈ A1/3 (A=atomic mass); but these are very bad approximations. This entire chapter from the source I linked earlier, may be helpful for conceptual understanding (though it is written for a reasonably high level physics background, you can probably parse through it anyway - it's surprisingly light reading!): Nuclear Cross Sections and Neutron Flux, from the Department of Energy handbook on Reactor Theory.

(Also available at the DOE Standards website. Nimur (talk) 18:00, 3 September 2009 (UTC)

Note that for a cube, a sphere, or any other 3D shape scaled proportionally, any length measurement scales in proportion to the cube root of the volume.

Asking whether something scales more like a cube or like a sphere isn't really meaningful since they both scale the same. Rckrone (talk) 19:12, 3 September 2009 (UTC)

My reference to the cube root of a cube and the radius of a sphere may be confusing because of the difference in the reference I am making. The reason for referencing a cube is that a cube can approximate the space between the spheres (I lost my equation for this. It was in my bag of marbles) Anyway if you stack a bunch of blocks (small cubes) so that they make a large cube and take the cube root of the large cube and divide by 2 then you have something to compare with the radius of a sphere formed by a bunch of marbles in a bag of marbles where there are spaces between the marbles.

One would think this sphere would have a larger radius that a sphere based on the absence of spaces between marbles (nucleons) which should be the smaller spheres that would compose the bigger one - the atom. I'm just wondering whether the neutrons and protons retain their spherical shape when part of a nucleus like a bag of marbles or if they meld and loose their shape and "liquefy" so to speak and become indistinguishable inside the nucleus as individual nucleons.

Is it a bag of marbles or a water balloon? --Taxa (talk) 21:09, 3 September 2009 (UTC)

To summarise, if you mean: a) what is the exact size of a nucleus and how does it grow with the addition of a nucleon, your question is unasnwerable per Nimur.

Nucleon sizes cant be measured as they are defined by probability densitys and so depends on the critical probability you define as 'the edge' of the nucleus. b) how does it scale with the addition of a nucleon your question is pretty much meaningless per Rckrone. In that with all n dimensional shapes a 1-dimensional measurement scales with respect to the nth root of the nD-volume. Elocute (talk) 19:26, 3 September 2009 (UTC)

Finally don't forget binding energy. Adding nucleons will add volume (let's take a theoretical definition like a 95% probability density cutoff with a bombardment by a particle with a characteristic energy, used for all measurements), but I suspect how volume increases or decreases also depends on whether binding energy per nucleon increases/decreases or not.

John Riemann Soong (talk) 19:38, 3 September 2009 (UTC)

Sounds like you favor the marble model.

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--Taxa (talk) 21:35, 3 September 2009 (UTC)

Instead of speculating about marbles, or guessing about which model is favored, you might consider reading the sources I linked, or the relevant articles at atomic nucleus. Unfortunately, modern physics is not very intuitive; it really needs to be learned by studying the basic formulations of quantum mechanics. These maths weren't invented for the fun of it - they're the accepted models because the observations we have of nuclear size contradict intuition, and require a more complicated branch of physics to model their behavior.

You can replicate these observations, but the machinery and equipment necessary is sort of expensive, making it hard to do in your backyard laboratory - but they've been independently confirmed hundreds of times.

Quantum mechanics is hard, and weird, but you can understand it if you just try carefully. But speculation about incorrect intuitive ideas is not a substitute for studying prior experimental evidence and theory. This is not meant to be harsh or mean - I'm just trying to help you get on the right track to really understanding the problem - it sounds like you're stuck on formulating a visual model of the size of the nucleus, but that will not help you correctly understand the real physical behavior.

Nimur (talk) 22:22, 3 September 2009 (UTC)

Okay, then you seem to favor the solid or balloon model. I'm not trying to be mean or harsh either but only to understand which concept is correct. While I do not have a backyard lab I do have intuition and the ability to conduct a thought experiment and have basic calculator skills to explore the idea I find most intriguing, which is the idea that the nucleus is like, if not identical, to the electron quantum shells in which an applicable equation such volume of a sphere is based upon integer values (see graph above) similar, if not identical, to integer wavelengths for the orbital shells of the electron around the nucleus.

--Taxa (talk) 00:20, 4 September 2009 (UTC)

No, it's not a solid or a balloon. It's an atomic nucleus, and it behaves like atomic nuclei. The best model is the quantum mechanical description of the cross-sectional radius, which is not a measure of a volume. Nimur (talk) 03:56, 4 September 2009 (UTC)
I think you are missing the point. Let me repeat. A cube and a sphere of the same volume have different root dimensions. The cube root for a cube is greater than the radius for a sphere of the same volume as the cube.

If the nucleons remain as individual spheres inside the nucleus then the volume of the nucleus will also be greater than if those same nucleons meld together as one single larger sphere and in effect expand the volume of the single spherical nucleus. Quantum mechanics is not thing more than requiring integer values based on very intuitive phenomenon like integer values for number of waves in an electron shell but even though integer values are required as individual nucleons suggest only the volume remains quantum because it is composed of fixed size units called protons and neutrons.

Thus to find the radius of the nucleus for each atom you simply use the atomic number (an integer value) as the volume of a sphere and then find its radius unless the nucleons remain as marbles in a bag. --Taxa (talk) 06:23, 4 September 2009 (UTC)

Taxa I think it is you that has missed the point, your two models are identical apart from a multiplacative constant of .

Notice on your graph you have that your models predict different sizes for a nucleus of 1 nucleon (which in your intepetation of your "thought experiment" means that putting a ball in a bag makes its volume different from putting it in a balloon). All your two models do is postulate a different size of the nucleons of which your atoms are made. And since the size of the nucleon has been shown to be an irrelavant concept, particularly is the classical 'hardball' sense (please not that quantum mechanics implications strech far further than simply the posit of the quantisation of values), I don't think your question really means anything.

—Preceding unsigned comment added by Elocute (talk) 09:54, 4 September 2009 (UTC)

You still misunderstand... a sphere made of a bag of marbles has space between the marbles and therefore a larger diameter than a bag with an orange in it of the same volume as the total volume of the marbles excluding the spaces between them.

Its common on the Wikipedia reference desk to find that when someone does not understand a question they blame the question and not themselves. --Taxa (talk) 13:39, 4 September 2009 (UTC)

TAXA, you are the one that is missunderstanding it.

Your bag of marbles is indded bigger than the Orange but the ratio between the two is a constant that does not depend on how many marbles you have which means it has no effect at all on the scaling which was you original question.

Wikipedia:Reference desk/Archives/Science/2009 September 3

Dauto (talk) 15:39, 4 September 2009 (UTC)

Thank you Duato. The ratio is constant as dauto says, and the tessellation question is irrelevant as we are not dealing with classical particles. Nucleons do not have edges. Elocute (talk) 19:30, 4 September 2009 (UTC)
If I interpret the most recent clarifications correctly, Taxa, you are more concerned about space-filling tesselations, which are a purely classical description, than you are about the actual mechanics of the atomic nucleus.

As Dauto and others have pointed out, space-filling tesselations of spheres grow at the same volumetric rate as space-filling tesselations of cubes, except for a multiplicative constant.

This means that your two models, (cubes or marbles - neither of which apply to a nucleus), both grow at the same rate. If you want some help visualizing this, we have an article on sphere packing. I can't stress enough, though, that these classical descriptions completely and utterly fail to apply to an atomic nucleus "volume", which has a scale-length defined by its quantum mechanical interactions.

Nimur (talk) 16:05, 4 September 2009 (UTC)

He(or she) seems to be asking whether (or not) the protons and neutrons fuse to become one "super nucleotide" , spherical in itself. See the "bag of 1cm marbles vs 10cm orange" analogy above. (talk) 19:08, 4 September 2009 (UTC)
eg for a 13 item nucleus - is it 13 times the "volume" of a proton (ie a sphere 13 times the volumetric size), OR , the volume of a filled dodecahedron including the gaps between (ie the volume of 13 proton sized spheres packed together) I can't answer this. (talk) 19:12, 4 September 2009 (UTC)
Obviously if you look at the curves in the above diagram you see that they are proportional to each other.

What I am asking is to see a plot of the actual curve produced by empirical measurement of the radius of actual nucleus.

Perhaps it does not even follow a curve but a jagged line. I want merely to compare it to the curves above. --Taxa (talk) 16:27, 4 September 2009 (UTC)

Oh my days Taxa, you are quite trying at times!

Do trust that I am trying to help you here, but as we have established the nuclear radius is roughly proportional to the third root of the nucleon number. As for actual measurements, they can't be done! Nucleons and nuclei do not have edges they are defined by probability fields which are in turn defined by quantum interactions which are not analogous to anything in classical physics (In fact the very tenet which Dirac uses to derive his requirement for a quantum theory Principles of Quantum Mechanics is that scale is not relative, it is infact absolute).

If you look through our various articles you will probably come across the phrase "probability cloud" this is the closest analogy I can give you, clouds don't have edges and neither do nuclei, you can't measure the physical size of either (without subjectively defining a cut off point), you can only measure whats in them. Elocute (talk) 19:42, 4 September 2009 (UTC)

I just want this for the nucleus.

--Taxa (talk) 17:49, 4 September 2009 (UTC)

There's a measure of cross section using thermal neutrons here [6] not a graph unfortunately - I don't know if that is of any use, I expect a neutral particle's cross section is a better measure than that of a charged a particle. I believe the situation is too complex to draw a direct parallel between nuclear size and neutron cross section.
You seemed to be asking above as to whether the indiviual protons and neutrons remain "discrete" or become a "throbbing glue" in the nucleus - as such you might be interested in the quark model - which makes the elementary particles up of three(?) quarks, - it's possible therefor to postulate whether or not there is interchange between these quarks (cf.

resonance in chemistry if you recognise that).

In the resonating model - there should be some mathematical connection between the number of ways the quarks can interchange and the nuclear stability - as I remember, attempts to get an equation that relates number of quarks in the the quark model of the nucleus, and the nuclear stabilty failed (though others may have succeeded).

This could be taken as tentative evidence that the structure remains marble like. However it could also mean the quark model is total nonsense, or that many scientists are incompetent.. however the same methodology does give a clear explanation of why nuclei with equal numbers of n and p are the most stable - assuming quark exchange between individual p and n ignoring quark colour whatever that is..

this might suggest the "big ball" model - but there's nothing we known about quarks to suggest they wouldn't do the same in the 'bag o' marbles model

Anyway why not take a look at "quark" (which is probably horribly out of date by now) (talk) 18:45, 4 September 2009 (UTC)
I checked out the table, which is the same place the chart above got it's values.

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Naturally one would expect higher values for neutron moderators but the cross section data seems to go way beyond that and appears to be grossly flawed.

Hence, I looked up absorption cross section which seems to portray a similar curve as the curves above. Absorption cross section I don't know yet how to interpret this. Perhaps the difference with the atomic radius differences being attributed entirely to electron shells.

I was hoping to find nucleus shells as stated before. --Taxa (talk) 20:06, 4 September 2009 (UTC)

I'm not a nuclear physicist -but for the absorbtion cross section I think you would need the data at higher than x-ray energy - anything below that will be heavily influenced by the electrons, which will mask any interaction with the nucleus - the graph does go to high energys, but is fairly bland. (talk) 23:35, 4 September 2009 (UTC)
The line marked K in the diagram marks the upper limit of electron interations with the ray.

There should be some absorbancies above this energy due to the nucleus - but I don't know exactly how high in energy one needs to go to see that. (talk) 23:40, 4 September 2009 (UTC)

(Outdent) I was re-reading the above, trying to find how we've had such a difficult time explaining the issue, and I think I came across a critical gap in communication. Taxa's statement, "Quantum mechanics is not thing more than requiring integer values", was never corrected.

Quantum mechanics much more than that - it begins with the postulate that certain values are quantized - that's just the "quantum" part. Then, there's also the "mechanics" part - which is the entire reformulation of physical interaction congruent to the implications of quantized energy, space, etc. This reformulation of physics is what makes up the bulk of quantum theory, (including nuclear physics).

Until you recognize this, there's no hope - it isn't simply a matter of having discrete numbers of nucleons - every physical description of those nucleons will deviate from the classical formulas.

This includes "volume", which does not strictly scale with nucleon count. Nimur (talk) 17:23, 5 September 2009 (UTC)

(Post deleted by author)
We are not dealing with classical mechanics done with integer values.

There really is no analogy between classical and quantum. Your original question really doesn't make any sense still, this is more complicated than your original question implies you had anticipated. Elocute (talk) 15:04, 6 September 2009 (UTC)

(Post deleted by author)