Wikipedia:Reference desk/Archives/Science/2009 August 28

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August 28

invisibility

if a suit emmits a spectrum of light which can't be seen by the human eye i.e. infra red or ultra violet will it be invisible or will it produce a wave like affect or distorted image.--Meloxicam (talk) 01:48, 28 August 2009 (UTC)

It won't make any difference at all. It would still block the visible light from behind it and would probably reflect visible light from in front of it as well. What is happening in parts of the spectrum we can't see is pretty much completely irrelevant for determining what we will see. --Tango (talk) 01:59, 28 August 2009 (UTC)

A suit emitting IR might, like a red hot stove, produce distortion of light waves passing near it, by heating of the air. A suit emitting UV migh cause other objects in the scene to fluoresce, just like a black light bulb. Parts of the spectrum we cannot see can have visible effects. Edison (talk) 02:32, 28 August 2009 (UTC)

"IR" and "heat" are not the same thing. All objects at any temperature emit EM radiation, objects are every day temperatures happen to emit IR. It isn't IR that distorts light near hot objects, it is the hot air caused by the actual heat of the object. Any frequency of EM radiation will heat up anything that absorbs it, what matters is just the total energy being emitted, not the frequency (although the frequency will affect how much it is absorbed). A suit will only cause heat distortions if it is very hot. I don't think the OP intended the suit to emit the radiation by being at the appropriate temperature (which, for UV, would be thousands of degrees C), it would just use bulbs (which may involve a filament at that temperature, of course). --Tango (talk) 02:50, 28 August 2009 (UTC)

The front of most remote controls emit IR when a button is pressed, they get neither hot, nor invisible. Admittedly those are only small amounts but I've seen high power IR spotlights for night time security cameras and to the naked eye they appear just as cold and visible. Vespine (talk) 05:11, 28 August 2009 (UTC)

What clothes look like in infra-red! [through an image intensifier that has nothing to do with IR - wrong type of night vision! --Tango (talk) 14:15, 28 August 2009 (UTC)] (Tango is incorrect: Modern NVG's see almost entirely in the near infra-red. SteveBaker (talk) 18:05, 28 August 2009 (UTC)) Tango is never incorrect, you should know better than to say such things! Look at the image description, it is from an image intensifier, not IR goggles. If it was IR you would be able to see that it was lit by an IR bulb somewhere near the camera - it clearly isn't passive IR - and it doesn't look like that to me. --Tango (talk) 18:32, 28 August 2009 (UTC) Near-IR looks like that - and modern image-intensifiers such as found in military NVG's (which is what was used to take this picture) see predominantly in the IR. Mid to Far IR correlates better with 'heat' and those look quite different (eg faces glow more than clothing). This is 'near' IR. SteveBaker (talk) 19:55, 29 August 2009 (UTC) Near-IR night vision is active, yes? It should be lit by an IR bulb near the camera, that images looks like it is lit by something off to the left. --Tango (talk) 15:37, 30 August 2009 (UTC) No - when used by the military, it's generally passive. The last thing you want to do at night is to be shining lights around the place (even if they are IR lights). There is enough near IR produced by the moon and stars for these light amplifiers to cope with. The directional lighting in that image is probably moonlight. SteveBaker (talk) 17:55, 30 August 2009 (UTC)

More than that - you and your clothes DO emit infra-red light. You are very visible on an infra-red "night vision" camera because of that. You certainly aren't invisible - and because human eyes can't see into the infra-red, we can't see that light without help from some high-tech gear. SteveBaker (talk) 13:26, 28 August 2009 (UTC)

(I think you have swapped visible/invisible there.) But yeah, this is a good example. When you press the button on your remote, you don't see anything coming out of that little light bulb on the front. That's what IR looks like. If you view it through something sensitive to that band of frequencies—like a cell-phone camera—you can see it as if it were regular light (which is a neat trick to pull on those who otherwise don't know about that). Nothing fancy to it, but you can't see it with your naked eyes. --68.50.54.144 (talk) 13:17, 28 August 2009 (UTC)

No, SB's usage was correct. You're visible on an infra-red camera not invisible. Nil Einne (talk) 18:36, 28 August 2009 (UTC)

What clothes may look like in infrared no dispute, hopefully (colouring is artificial of course, you can't visualise what humans look like in infrared naturally since we can't see infrared) Nil Einne (talk) 18:44, 28 August 2009 (UTC) [Thermal IR, yes. Near-IR is rather different and that is what I expect the OP was talking about. --Tango (talk) 18:50, 28 August 2009 (UTC)]

• When you see a suit, what you actually see is a reflection from light around it, usually this light has all colors, so you can see a wide spectrum of different colors. For a red suit, almost all the visible non-red light is absorbed by the suit, while only the red light is reflected back to your eye, and the suit looks red. If the light shining on the red suit was completely blue only, with no red in its spectrum, that same suit would look black. A suit which only reflects back IR and UV, and nothing in the visible spectrum, would simply look pitch black, but since it is opaque, it would not function like an invisibility cloak. Sjakkalle (Check!) 09:07, 28 August 2009 (UTC)

No one has addressed my comment about a suit that emitted UV causing fluorescence of objects. For the sake of argument, I would assume the suit emits the equivalent of a 20 watt "blacklight" compact fluorescent bulb. And the OP did not specify that the suit emits the limited IR energy of a remote control or the normal clothing on a non-cloaked individual. Perhaps it radiates all the heat generated by the wearer, which is perhaps 60 watts, in the IR and UV regions, while absorbing incident light in the visible spectrum. It would be black to normal vision, it would be bright when viewed in IR sensitive optics such as security cameras, and it might well cause dayglo paint, tonic water, white paper, cleaning products, clothing washed in certain detergents or a white shirt to glow due to the UV. Thus emitting UV would be more likely to make the wearer easily detectable to the unaided eye at night than just emitting IR. Note: many years ago my rock band would sometimes kill the lights except for a couple of UV fluorescent tubes. Teeth, shirts,and other parts of clothing would fluoresce brightly, as would some of the image on advertising signs or labels or beverages. A cloaked person emitting similar UV would have been easily detected making his way through the room. Edison (talk) 14:58, 28 August 2009 (UTC)

The UV light would attract small flying insects (but not mosquitoes) like a Bug zapper without the zap. The wearer would see interesting UV watermarks on some stamps, banknotes, passports, credit cards and other documents, as well as any stains of body fluids such as semen, blood, bile and urine. Ultraviolet#Human health-related effects of UV radiation gives information that suggests (s)he should wear both insect repellant and sunblock creams and not bite on anything that looks like a candy[1].Cuddlyable3 (talk) 15:46, 28 August 2009 (UTC)

If you want an invisible cloak, you would likely want to consider Broadband Exterior Cloaking. [2][3] They haven't got it scaled up to humans at visible light frequencies quite yet though. ;) Franamax (talk) 17:54, 28 August 2009 (UTC)

People who have had cataract surgery can see a little way into the ultra-violet. My mother noted that some flowers which had been plain before had blue spots or stripes visible on their petals after her surgery. It's believed that plants do this to attract bees (which can see into the ultra-violet). SteveBaker (talk) 18:25, 28 August 2009 (UTC)

v

Gravitational wave triangulation

LIGO says:

These sites are separated by 3,002 kilometers (1,865 miles). Since gravitational waves are expected to travel at the speed of light, this distance corresponds to a difference in gravitational wave arrival times of up to ten milliseconds. Through the use of triangulation, the difference in arrival times can determine the source of the wave in the sky.

How can a single piece of information (the difference between two times) determine the source of the wave? You need three pieces of information to determine a position in 3D space (hence the name). I can imagine two pieces of information being useful since the universe is fairly sparsely populated so there is likely to be very few potential sources along the line (or curve) that you can narrow the source down to, but there being only one source on an entire plane (or surface) seems unlikely to me. --Tango (talk) 01:57, 28 August 2009 (UTC)

From the papers I'm reading (original 1991 LIGO paper in Science and 1999 publication in Physics Today) the purpose for two distant laboratories 3000 km apart is to filter out local noise sources. However, the latter source also states that "one determines the direction and polarization of a gravitational wave by measuring arrival-time differences between geographically dispersed detectors." So, this may also have been a design goal. Two arrival-times define two spheres; the intersection of two spheres is a circle (I think? Somebody check me on that). If so, the source must lie on that circle of intersection. Furthermore, there is polarization information because the two sites also have two arms each - so that can further narrow the possible source locations on that locus. I have not, however, found any papers which specifically discuss techniques or examples of such source location using the LIGO. My experience with electromagnetic waves source-determination suggests that you would still need three geographically-dispersed receivers; but there are a lot of games that you can play (also, space is pretty sparse, so there's only a few black holes that would even be candidates as sources for detectable waves - that can help narrow down the source locations). Nimur (talk) 05:20, 28 August 2009 (UTC)

The intersection of two spheres is a circle (or a point or empty), but that's irrelevant. You would only get a sphere if you knew the emission time as well as the arrival time (so you could calculate travel time, which gives you distance). With just the arrival time you have no information at all, with two arrival times you can calculate the difference and that is useful information, but only one piece of it. --Tango (talk) 14:19, 28 August 2009 (UTC)

In N dimensions, you need N pieces of independent information in order to identify a point uniquely. Hence, in 3D space, you need three pieces of information to determine a position unambiguously. With just the difference between two times - you have only one piece of information. For example - if the pulse were to arrive at exactly the same time at both places, you'd know that the source was equidistant from the two receivers...but that's all you'd know. That means that the source could be any point on a vertical plane that's at right angles to the line between the two receivers. Assuming that the point lies in the plane of the galactic disk narrows it down to a line - and because the solar system is on one of the outer arms of the galaxy, one might guess that the direction lies towards the center of the galaxy - which narrows the search down to a single direction...but those are assumptions that might not be true.

If they had a third receiver, they could unambiguously narrow the source down to any point on a line - and only with a fourth receiver could they determine the position exactly in 3D space. That's why your GPS reciever can't tell you where you are unless it can see four satellites. SteveBaker (talk) 13:20, 28 August 2009 (UTC)

GPS receivers do not need to see 4 satellites if they have a very good clock as the satellites themselves broadcast timestamps for the signals they send out. For an 'average' GPS receiver 4 satellites are required though to determine (x,y,z) and t to an accuracy better than the internal clock can provide. JMiall₰ 17:35, 28 August 2009 (UTC)

First I want to point out that pinpointing a source in the sky in that context means only the two coordinates of a point in the celestial sphere, namely right ascension and declination. Second, the time difference can be suplemented, as pointed out above, by the polarization information which helps pinpoint the source. Dauto (talk) 14:24, 28 August 2009 (UTC)

But they're not looking for a dirac-function, they're looking for a periodic wave (emanating from black hole rotations, for example). This gives phase and timing information (assuming a sub-wavelength separation); so two locations + timing information + phase information may be sufficient to uniquely identify the source (meaning that the N pieces of information criteria is satisfied - as long as the phase assumptions hold). Needless to say, I can't find any documentation of successful identification by this technique in the above papers. Nimur (talk) 15:52, 28 August 2009 (UTC)

Also if they find a source that lasts more than a few hours then they can use the fact that the baseline is moving to triangulate from different orientations of the baseline and get a better directional fix. Gandalf61 (talk) 16:01, 28 August 2009 (UTC)

Relatively local area network

Can I join the wireless home network of my friend who lives on a nearby street about 800 meters away? (Uh, half a mile, isn't it.) I asked this question in an electronics shop and the man said no, not unless I built a tall tower to provide line of sight over the intervening houses for the radio signal. Is there no better way? 81.131.51.80 (talk) 02:31, 28 August 2009 (UTC)

I don't think you need a tower, just an aerial on the roof of each house with some appropriate kind of booster. The Computing desk might be a better place to find out what you need to boost the signal. --Tango (talk) 02:42, 28 August 2009 (UTC)

OK, but just to squeeze every drop of science out of the question while I'm here - would this work even if I can't see his roof from my roof due to other people's roofs being in the way? I think the signal would have to go through about six of the things.81.131.51.80 (talk) 02:45, 28 August 2009 (UTC)

Yes. The signal will spread out from his house (or yours for the return journey) and then diffract around the roofs. Those roofs may weaken the signal, but shouldn't block it entirely. --Tango (talk) 03:06, 28 August 2009 (UTC)

In fact, for that reason, an aerial on the roof isn't strictly necessary at all, the aerial could be inside, but you'll get better signal strength with it on the roof. --Tango (talk) 03:07, 28 August 2009 (UTC)

Thanks! That was what I was hoping, having seen diffraction mentioned further up the page, but I wasn't sure if it would work over such large obstacles so many times in a row. (If I imagine water waves instead of radio waves, it seems less extraordinary.) 81.131.51.80 (talk) 03:16, 28 August 2009 (UTC)

Diffraction happens most when the wavelength and the size of the object being diffracted around are comparable. The wavelength for wireless networks is (I believe) on the scale of millimetres, the roofs are on the scale of metres, that is close enough to get significant diffraction. (You won't, however, get much diffraction of visible light (which is on the scale of 100s of nanometres), which is why you can't see the sun when it goes beneath the skyline.) I rather suspect I am oversimplifying the situation here (optics isn't really my area), but I think the general idea is right and that is all you need. --Tango (talk) 03:30, 28 August 2009 (UTC)

Ha, I like the relaxed notion of "comparable". In most situations I encounter, millimeters are not comparable with meters, but I'll take your word for it. I just went and looked up radio waves and saw that amateur radio wavelengths can be exactly house-sized, so I thought maybe that would be good to aim for, but perhaps converting LAN data to ham radio is not a practical idea, I couldn't say, I know nothing. Leaving it at the same frequency is undoubted cheaper, anyway. 81.131.51.80 (talk) 03:58, 28 August 2009 (UTC)

You may also need a license to use those wavelengths. It would certainly work, though - ham radios definitely work over 800m! The notion of "comparable" can be made precise, but for this kind of thing anything within a few orders of magnitude is close enough. --Tango (talk) 04:15, 28 August 2009 (UTC)

Typical consumer wi-fi gear has a range of only about 100m outdoors, you can get amps and directional antennas and stuff, but I think they claim to improve the signal by 50%-200%, not by 800%. I don't think you'll get 800m out of consumer gear without repeaters in between. There are things like Long-range Wi-Fi but I don't know how cheap and easy that would be to set up. Vespine (talk) 04:22, 28 August 2009 (UTC)

That's an excellent article. "just placing standard USB WLAN hardware at the focal point of modified parabolic cookware" is an inspiring phrase. (Does seem to want line of sight, though.) 81.131.51.80 (talk) 04:42, 28 August 2009 (UTC)

Converting "LAN data" into "HAM data" is commonly done. See AX.25, packet radio, and this great web resource from Tucson Area Packet Radio. With UHF (and a HAM license - you need a powerful transmitter, and you should be licensed to operate it!), UHF-based IP networks have been established with wireless ranges over 25 miles. Nimur (talk) 05:23, 28 August 2009 (UTC)

The 25-mile range isn't because of technical limitations, right? By transmitting LAN data at a shortwave wavelength and bouncing it off the ionosphere, I don't see why you can't transmit to the other side of the world. --99.237.234.104 (talk) 06:13, 28 August 2009 (UTC)

You will not be able to get anywhere near enough bandwidth to bounce off the ionosphere at high frequency. Also the ionosphere adds multipath distortion and is constantly changing. There are special HF modems that can do this job, but you will be lucky to get 9600 bps. You can certainly get multiple kilometers out of wifi. Both ends need a wire parabolic dish, you can get about 26 dBi from each antenna. You have to match polarization and line of site really is required. You could perhaps survive one roof in the way, but 10 or twenty will give you a total loss. Not only do you have to get your signal through, you also have to separate it from all the other WiFi signals around. Graeme Bartlett (talk) 09:21, 28 August 2009 (UTC)

Normal wifi gear operates at 2.4GHz or higher. This is strictly line-of-sight and will be attenuated by intervening walls: diffraction will not help. If you cannot see the neighbor's house, you are stuck. If you do have LOS, then you can get or make a pair of directional antennas that will work. The classical one is the "Pringle's can antenna." See [4] -Arch dude (talk) 13:12, 28 August 2009 (UTC)

Do directional antennas and diffraction mix, if you see what I mean? Can you have a situation where you're pointing a directional antenna at the edge of an object the waves are diffracting round, or is that just silly? 213.122.66.56 (talk) 14:08, 28 August 2009 (UTC)

You can get ranges as far as this from a pair of home WiFi units using the "Pringle can antenna" approach (just Google for that exact term). "Pringles" cans are foil-lined cardboard tubes - but for some (almost magical) reason, they work extremely well as highly-directional WiFi antennae. HOWEVER - you will definitely need a line-of-sight over the rooftops because you can only get that kind of a signal boost using very directional signals that would be disrupted by the intervening rooftops. SPECIFICALLY: You're going to need to be able to see his Pringle-can antenna from your Pringle-can antenna with nothing but air in-between - you'll need to tape a laser-pointer inside the can-antenna of one of them and have it shine onto the bottom of the can-antenna of the other. You'll need to mount them quite firmly (and obviously, waterproof them somehow) in order that the wind doesn't blow them out of alignmnent. If you manage that then you have a really good chance to make this work - even at 8 times the range the thing is designed to work at. However, if you don't make the antennae just right - and if they don't accurately line up - you definitely won't succeed. SteveBaker (talk) 13:04, 28 August 2009 (UTC)

As I understand it, Pringles cantennas aren't actually as good as people seem to think despite the hype and popularity. You're better off choosing a different can and doing things somewhat differently without really raising cost or difficult. This isn't something I've ever done myself or look into in depth, so I could easily be wrong but what I have seen before and re-reading thing things now, the theory and evidence seems fair enough [5] [6] [7] and in particular, I've never come across anyone claiming the opposite whether in theory or evidence/testing (although that doesn't always say much) Nil Einne (talk) 18:31, 29 August 2009 (UTC)

A more reliable alternative is an off-the-shelf microwave transmitter and receiver (this reply is being sent via one of these), but it still needs line-of-sight and is more expensive than the above solutions. I suppose wi-fi is microwave anyway, so the technology is the same, even if the name is different This range of frequencies uses a tiny dish as an aerial. Dbfirs 02:15, 3 September 2009 (UTC)

how to interpret

work done in a compression process in int(pdv).and shaft work in a compression process in int(vdp). how?how its possible? 220.225.98.251 (talk)

Consider some quantity of gas moving through the system and doing some mechanical work as it goes. The gas enters at volume V₁ and pressure P₁ and leaves at volume V₂, pressure P₂. The work it does is $\int _{V_{1}}^{V_{2}}{PdV}$ using the usual work forumla. Using integration by parts, $\int _{V_{1}}^{V_{2}}{PdV}=V_{2}P_{2}-V_{1}P_{1}-\int _{P_{1}}^{P_{2}}{VdP}$.

The $-\int _{P_{1}}^{P_{2}}{VdP}$ part is the shaft work and the $V_{2}P_{2}-V_{1}P_{1}$ part is the flow work. (Edit: I'm not sure this is right. I need to think about it.) Rckrone (talk) 06:48, 28 August 2009 (UTC)

This must relate to http://en.wiki.x.io/wiki/Wikipedia:Reference_desk/Archives/Science/2009_August_2#why_so.3F.3F.2F

(removed confused stuff)83.100.250.79 (talk) 12:16, 28 August 2009 (UTC)

Ok, I think I understand this better now. Enthalpy#Open_systems talks about it a little. The total work that the apparatus does can be divided into two types: flow work, which is the work done to force the gas through, and shaft work which is mechanical work done on an external object like a turbine. The energy that the apparatus gets to do its work comes from energy donated by the gas passing through and heat added to the system ΔQ. For a steady-state process (internal energy in the control volume isn't changing) that's adiabatic (ΔQ = 0), the total work done by the apparatus is equal to the work done by the gas, so W_shaft = W_gas - W_flow = $\int _{V_{1}}^{V_{2}}{PdV}-(V_{2}P_{2}-V_{1}P_{1})=-\int _{P_{1}}^{P_{2}}{VdP}$. This is only true for a steady state adiabatic process. Rckrone (talk) 17:19, 28 August 2009 (UTC)

But the PV terms are not equal to the energy of the gas - the energy of the gas is kPV where k is a dimensionless constant relating to the number of degrees of freedom of the gas (I think)

ie δW = d(poutVout) − d(pinVin) + δWshaft

does not integrate to Work = change(PV) + shaft energy

The questioner has already asked a question based on a false supposition (linked above) - I would not be surprised that they are doing the same here.

I wouldn't waste any time on attempting to reverse engineer unfounded statements.

The PV terms aren't supposed to represent the internal energy of the gas, but the energy required to move the gas out of the control volume (and the energy gained when gas forces itself in). To push a volume of V out against a pressure of P requires work PV. This is called flow work or pV work. Here's a source I found on it [8]. The change in internal energy of the gas is measured by the work that the gas does (since in an adiabatic process the two are equal). Some of the internal energy that the gas is donating goes toward pushing itself out of the control volume (flow work) and the remainder is shaft work. δW = d(poutVout) − d(pinVin) + δWshaft does integrate to Work = change(PV) + shaft energy. You might be right about the OP trying to waste everyone's time though (but maybe not). Rckrone (talk) 18:01, 28 August 2009 (UTC)

To the OP:

Can you give a single example (such as a book reference or web page) where this formulation int(VdP) is used?

You've already asked one question that assumed something was true when it wasn't ie your question linked to above.

Please don't expect others to do your work for you.

I have suspicions that you are deliberately wasting other peoples time - Can you please prove me wrong and give a source from where you are getting these statements?

If you do not I will treat any further posts as vandalism - if you feel this is unfair then feel free to complain about it.

83.100.250.79 (talk) 17:35, 28 August 2009 (UTC)

Firefighting with seaplanes

Recently, in the fires that desolated Greece, I got the impression that the fire fighting seaplanes were picking up water directly from the sea. Do they really use seawater against fire? Quest09 (talk) 10:21, 28 August 2009 (UTC)

Sure. But they prefer fresh water, if that's handy. Our article on aerial firefighting doesn't really address this, but salt water is more likely to lead to corrosion and other conditions that require extra maintenance hours, so sea water isn't the preferred source of water. Still, if sea water's what they've got, that's what they'll use; an emergency is an emergency, after all. -- Captain Disdain (talk) 11:01, 28 August 2009 (UTC)

However, would they use this water on forest fires too? And, wouldn't that ruin the ground for many years? It seems less damaging to let it burn... Quest09 (talk) 11:34, 28 August 2009 (UTC)

The salting the earth question is interesting: but how long would it take to leach out the salt added in the firefighting operation? And for practical purposes, is the amount of salt deposited of real moment, in comparison with, for instance, wind-blown spray over the millennia, or the salts deposited by the conflagration? --Tagishsimon (talk) 11:49, 28 August 2009 (UTC)

This page says salt contamination is taken into consideration for sensitive vegetation and water supply areas. I guess several things need to be considered: how much damage the fire would cause, how far away a fresh water supply is, how much water (and thus salt) would be needed, and what the effects of that amount of salt will be on the land type in question. 88.114.222.252 (talk) 11:56, 28 August 2009 (UTC)

one way salb

i looked evry where , without any use . now designing a one way slab we put the main reinforcement in the short direction but when you take astripe in each direction you will find that the one in the long direction is carrieng more load and resisting more moment . so shouldnt we be putting the main reinforcement in the long direction.please dont use math i need avery clear exponation even other people beside engineers can figure it out ....? —Preceding unsigned comment added by Mjaafreh2008 (talk • contribs) 12:09, 28 August 2009 (UTC)

Usually reinforcement is put in any part that is too weak as it is to carry, support, or withstand an expected load.

Also what is the "one way slab" for, what is it for?83.100.250.79 (talk) 12:21, 28 August 2009 (UTC)

Without further details it's impossible to say where the reinforcement should go in your example.83.100.250.79 (talk) 12:19, 28 August 2009 (UTC)

will the way i visualized it like this ... assuming a two beams crossing each other at midspan now to carry a specified load applied at that point of intersection then which one of the two beams need the biggest reinforce steel ratio to support that load ... —Preceding unsigned comment added by Mjaafreh2008 (talk • contribs) 12:47, 28 August 2009 (UTC)

It's still not clear, please supply a diagram, or actual dimensions.83.100.250.79 (talk) 12:50, 28 August 2009 (UTC)

This sounds like a civil engineering matter where some mathematics is unavoidable and viability of the resulting structure could have legal consequences. I don't think we can provide the kind of consultation that requires. Cuddlyable3 (talk) 12:53, 28 August 2009 (UTC)

If the slab is long but not very wide, then obviously the main tendancy to bend will be along the long direction - so the main reinforcement beams need to go along the length to resist that bending. However, I presume you need some reinforcement across the width too - resulting in a "waffle-slab" approach. The trouble is that the spacing and depth of the beams and the degree of post-tensioning that goes on does indeed require some math - and it depends on the results of a proper soil survey too. I would have thought that you'd either have a book or table of some kind that specified what you need for different shapes and sizes and under different soil conditions - or that you'd employ a structural engineer to design the slab properly using math and stuff. The consequences of a slab cracking years after the building is completed are pretty severe for the owners (I know - it's happened to me!) and I'd be horrified if someone was just guessing how much reinforcement the thing needed on the basis of asking a question to a bunch of unknown people on the Internet! SteveBaker (talk) 12:53, 28 August 2009 (UTC)

But a one-way slab is by definition supported across the short way. What you're thinking of is a case where you can only support at the distant ends. Or in the case of a two-way slab (W > 0.5L), you would need to consider the long direction as well as the short one. Franamax (talk) 18:55, 28 August 2009 (UTC)

And isn't there another option: diagonal reinforcement ? Those would have the advantages of providing more rigidity and stopping cracks parallel to any of the walls, but some would need to be somewhat longer than a straight reinforcement, in order to reach from one corner to another, and others would be shorter. If the reinforcement is free-floating, this might work fairly well. However, if attached to the frame at the ends, this would be a bit trickier (if the components don't meet at a right angle).

If we're still discussing length-wise or cross-wise reinforcements, the site will make a big difference. Is this on a hill ? If so, I'd put the reinforcement in the direction going up and down the hill. But, better yet, put in reinforcement in both directions, so you're covered no matter what the stresses are. StuRat (talk) 14:52, 28 August 2009 (UTC)

will .... first of all in engineering they call it one way slab and the wright way is to provide steel reinforcement along the short dimension , so its known where to put the steel . what iam asking for is why ,,, i think we should provide the steel along the long direction . and we cannt provide steel reinforcement along both directions because it wont be economical at all .

How do you know one way is the right way - give a reference for your statements or stop wasting peoples time. ok? Why do you think the short reinforcement is supposed to be the right way to do it -does it say so somewhere? Where?

Stop wasting peoples time with you unfounded statements ok?.83.100.250.79 (talk) 17:24, 28 August 2009 (UTC)

You need to provide some reference for your statements that

the wright way is to provide steel reinforcement along the short dimension

This seems wrong - please provide evidence showing from where you got this statement. Otherwise I will assume that you, like others are making up false statements in an attempt to waste other peoples time. I view that sort of behaviour as vandalism and will treat it accordingly.

You should realise that disruptive behaviour is vandalism.

If you feel have been falsely accused of disruptive behaviour by me please provide evidence that your statements have a legitimate basis, if you feel I am treating you unfairly then please feel free to complain.

83.100.250.79 (talk) 17:41, 28 August 2009 (UTC)

Iam acivil engineer ... thats how i know ... because we design it ... but i need a full understanding about it theres alot of terms that we use all the time but we didnt have afull understanding for it ...?? and i think you should cool alittle bit ... iam not intersted in wasting your time ... ????????????????????????????????/ —Preceding unsigned comment added by Mjaafreh2008 (talk • contribs) 17:52, 28 August 2009 (UTC)

You know what - I don't believe you. Why is a civil engineer asking simple questions on the internet - you should already know the answer. You should also know as an engineer that you need to communicate the problem clearly to other people - something you have failed to do - You have not supplied any information about dimensions, material, points of attatchment, etc despite being asked. You have failed to provide any source for your statement that "reinforcement should be across the short length".

Why should people waste their time (which is what you are doing) trying to justify statements that you have not provided a source for, and may not even be true.83.100.250.79 (talk) 18:04, 28 August 2009 (UTC)

83.100, you seem to enjoy hectoring posters here more than actually answering their questions. If you wanted to know what a one way slab is, you could have tried that handy new thing called Google. If you want a source that short way is the right way, you could have looked that up pretty easily too, since the answer is "all of them". But if it will make you feel any better, here's a nice Powerpoint with lots of pictures.[9] Now please try to provide answers in a polite and helpful way, and if you're worried about people's time being wasted, well, the only time you control is your own. I didn't have any trouble understanding the question or giving a simple non-numeric answer. Franamax (talk) 18:45, 28 August 2009 (UTC)

Also 83.100.250.79, please consider that an OP's first language might not be English. Not all the other language wikis have reference desks. If you don't understand what the person is asking, then tell them so politely and ask them to clarify. Also there is no requirement for questioners to produce sources for their statements, you may ask for them politely, but they don't have to give you any. Asking or answering questions on the ref desk isn't like editing an article. Also please assume good faith. 152.16.15.144 (talk) 06:15, 30 August 2009 (UTC)

Your problem is that the slab is supported all around, so your assumption that there is more stress in the long direction is incorrect. The load is borne primarily in the short direction. Think of it as the load wants to find something to hold it up as quickly as possible, so it "looks" for whatever support is closest. Franamax (talk) 18:08, 28 August 2009 (UTC)

Or for another way to think about it, why does the slab care how long it is? If it 1 meter wide and 10 meters long, why is that any different than if it was 10 kilometers long? It would only be important if you were supporting the slab only at the widely separated ends, in which case you would definitely need to reinforce it the long way (and if it's 10km long, you will need an awful lot of rebar;). Franamax (talk) 18:12, 28 August 2009 (UTC)

you could look at it like this ... at midspan where the moment is maximum and assuming that the slab is consist of two strips crossing each other at midspan now ,,, deflection will be the same amount at midspan because as we know both strips are attached at that point . and using this info that elastic strain in ex. concrete is 0.003 then the short side need more steel to prevent strain from exceed this value where cracks could start to form . so if you provide enough steel in the short direction to prevent cracks cracks surly wont form along the long direction. mabey this is the answer. and mr 83.100.250.79 please dont call people lair ... unless ofcoarse if its make you feel good .--Mjaafreh2008 (talk) 18:43, 28 August 2009 (UTC)

Or here's yet another way of looking at it, related to the first thought experiment of two crossing beams: pretend you have two beams of equal cross-section, one (beam A) 3 times longer than the other (beam B), with no load, just their own weight. We know for sure that A will have a higher deflection than B. But we also know that if the beams are joined at the centre, they must have the same deflection, so in effect, beam B is supporting A, and A is actually two beams of length A/2 (actually it's a compound beam with two spans, but whatever). Now stick in a few more "B" beams across the width. Now you have A as a compound beam with four spans of length A/4. Since deflection varies as L^3 and stress varies as L^2, pretty soon you don't have to worry at all about A. Everything is carried by the "B" beams, except right at the very ends of the slab. Don't know if that makes it clearer or not... Franamax (talk) 19:12, 28 August 2009 (UTC)

thank you very mush ... will as you said i could google it , but its better to disscus it with someone , soory for wasting your time .--Mjaafreh2008 (talk) 20:11, 28 August 2009 (UTC)

Not a problem, and I don't think you wasted anyone's time. We all choose how to use our time here. It's easy enough to search for things on Google or look them up in books, and you can always find the equations. But when you get stuck on why the equations are the way they are and need help in understanding the conceptual basis, well, that's exactly what we're here for. Hopefully at least some of us helped you, and feel free to ask more questions. (And I see that our concrete slab could use a bit of work, I'll put that on my list) Franamax (talk) 22:56, 28 August 2009 (UTC)

Comparing male/female weight loss on The Biggest Loser

Sorry about this banal question, but I'm going to be competing with some female friends in a weight loss race, and I want to make it fair so they cant complain when I win. I see in The Biggest Loser that it's proportional - they use percentage of body weight lost - is that how they deal with the fact that males tend to lose weight faster than females?

thanks Adambrowne666 (talk) 12:20, 28 August 2009 (UTC)

I suggest you agree with your friends about how your weight changes will be compared before starting the race. Then there need be no complaints afterwards. Cuddlyable3 (talk) 12:48, 28 August 2009 (UTC)

If that's all they do then that's all they do, so there's your answer. One of the reasons why men and women tend to lose weight differently is that, when working out, women often gain a fair portion of muscle mass. That dulls the weight loss they might otherwise notice. So if you wanted to be completely correct about it, percentage wouldn't inherently be correct, but it should be more than sufficient for your purposes. ~ Amory (user • talk • contribs) 13:11, 28 August 2009 (UTC)

Exercise increase the basal metabolic rate, what fat you burn at rest on a fasting stomach. According to our dieting article, this increase accounts for a lot of the weight loss from exercising. With equal heights the BMR is roughly BMR = 10*weight + s where s is a constant. Depending on whether you'r male or female s is +5 or -161. This means that the weight-term dominates the equation. Given two people who exercise equally, the fattest will loose the most weight. So as a measure of who's exercising most during a diet, percentage loss is a good measure and gender can be ignored. EverGreg (talk) 13:13, 28 August 2009 (UTC)

You have neglected to say what units you are measuring weight in... --Tango (talk) 14:28, 28 August 2009 (UTC)

I think on the biggest loser, it's not done that way just to compare male and females, but also people that are different amounts over weight. It's much more feasable to lose 100 lbs if you're 350lbs than if you're 250lbs to start with, for example. The % loss is probably a fairly fair way to go about it (although as pointed out it ignores muscle gains). Perhaps you could also compare BMIs (which will suffer from the same problem). TastyCakes (talk) 14:50, 28 August 2009 (UTC)

Doesn't this question request medical advice? You should consult your doctor before participating in a "weight loss race." Wikipedia can't specifically advise you on what a safe or desirable weight should be; nor what weight-loss regimen is healthy. Nimur (talk) 15:54, 28 August 2009 (UTC)

But we can speculate on what makes a fair competition, no? An egg eating contest might well have health implications, but we could still pontificate on the best way to judge such a competition (clearly most eggs as a percentage of body weight would be fairest). TastyCakes (talk) 16:00, 28 August 2009 (UTC)

This isn't medical advice, it is advice about judging a competition. There is no diagnosis, prognosis or treatment suggestion involved. --Tango (talk) 16:02, 28 August 2009 (UTC)

A lot of the time, it's the skinny ones who cram down the most eggs in absolute terms, so judging based on % body wt. might just skew things even farther in their favour. And I agree this is not a mediq per se, but a warning that rapid weight loss can be quite dangerous is certainly appropriate. Franamax (talk) 19:19, 28 August 2009 (UTC)

Yeah, look how many skinny people are champion gurgitators. Body mass has little to do with it, if anything it should be handicapped by height but even that's hard to correlate. --66.195.232.121 (talk) 20:28, 28 August 2009 (UTC)

I think those people are skinny precisely because they have an overactive metabolism (which also leads to their increased appetite). I should know, because I'm thin as a stick but I absolutely devour buffets and am constantly hungry. (Perhaps I have an enzyme deficiency or something...) John Riemann Soong (talk) 21:56, 28 August 2009 (UTC)

Possibly true, but when you're ingesting eggs at a rate of one a second, your metabolic rate isn't really going to help you fit more in, since the contest is over basically before the digestive process has even begun. I would rather speculate that a relative lack of fatty tissue around the stomach allows it to expand more in a short period of time. (And if you are able to enjoy eating lots of food while avoiding weight gain, I'd be thinking enjoy it while it lasts! Good food is one of the true pleasures of life. :) Franamax (talk) 22:42, 28 August 2009 (UTC)

Thanks, all, for your informative answers. Adambrowne666 (talk) 08:01, 31 August 2009 (UTC)

Recharging Electronics in Different Countries

If I take my iPhone to different countries around the world and plug it in to recharge from empty will the charge time vary based on the local electric power standards? TheFutureAwaits (talk) 16:35, 28 August 2009 (UTC)

Probably not, but make sure you have an adapter that can cope with all the different voltages you may encounter (ie 110V and 230V). SpinningSpark 17:26, 28 August 2009 (UTC)

And in addition to the voltage difference, be aware that plug configuration varies a lot around the world, beyond just the standard North American and European shapes. Franamax (talk) 20:53, 28 August 2009 (UTC)

Assuming everything works correctly, the wall voltage goes through a transformer which changes it to the proper voltage for the iPhone to use (and converts it to DC). At that point there's no difference in what the phone is receiving. Rckrone (talk) 17:39, 28 August 2009 (UTC)

Ah, thanks this makes sense. Though I was hoping higher voltage meant faster charge time... TheFutureAwaits (talk) 21:34, 28 August 2009 (UTC)

I hate to pass on rumours[10] but they say there is a problem of iphones exploding. Cuddlyable3 (talk) 13:37, 29 August 2009 (UTC)

what is the counterion for DNA?

Okay, I should prolly know this, but they don't ever mention where the positive counterion for the phosphate in DNA goes! I mean, if you have millions or billions of basepairs of this stuff in a single chromosome, it seems to me you have some major charge separation ... is it like sodium and potassium ions around the DNA? And when you're during agarose gel electrophoresis and you do DNA extraction, I assume these counterions are still hanging around in solution (or you'd have a major test tube of negativity, lol...), and it is these ions that flow in opposite direction of the DNA? But I'm really amazed, because chromatin gets wound into these megadense structures and I would think there would be some major electrostatic repulsion. John Riemann Soong (talk) 17:32, 28 August 2009 (UTC)

At first I was thinking that, the phophates may be protonated at physiologic pH, which would mean that counterions would be unneccessary. Upon actual research, that turns out not to be the case. However, This 2004 study implies that the DNA is actually in a diffuse sea of sodium and potassium cations, which provides the charge balance; however the counterions don't actually bond to the phosphates directly. This google search turns up LOTS of material on this topic. --Jayron32 18:44, 28 August 2009 (UTC)

Hmm, I'm just perplexed at why everyone seems to neglect to mention where the counterions go (or how they affect things) at the elementary level. The paper you cite gives short-length DNA chains -- but let's take say, a chromosome. Why don't the negative phosphates generate some major repulsion? I'm sure that the K+ and Na+ will have hard time reaching the phosphate groups deep within the chromosome. For that matter, I'm kind of perplexed at why DNA is called an "acid", when the NH2 groups and phosphate groups would seem to make it at least weakly basic. If DNA is already negatively charged, I'm sure that the pKa of DNA (or of one base pair unit) must make it a weaker acid (and a stronger base) than water. John Riemann Soong (talk) 21:46, 28 August 2009 (UTC)

Have a look at the histone article. There is a discussion of exactly what you are asking under the sub-heading "Structure". From the article: "Helix-dipoles from alpha-helices in H2B, H3, and H4 cause a net positive charge to accumulate at the point of interaction with negatively charged phosphate groups on DNA". --- Medical geneticist (talk) 00:54, 29 August 2009 (UTC)

I don't really know the answer to that question. I have always assumed that there would be enough hydrogen atoms around to balance the charges. Isn't that why DNA is considered an acid? Dauto (talk) 03:30, 29 August 2009 (UTC)

Well shouldn't it be called deoxyribonucleic conjugate base then? Are Na+ / K+ really the primary counterions, or is it H+? John Riemann Soong (talk) 20:04, 29 August 2009 (UTC)

If you get full dissociation in aqueous solution, there isn't really any "possession" of counter ions - that is, you can't say "these ions are associated with DNA, these with these acidic metabolites, these with acidic protein sidechains, etc." The neutralization of charge is a bulk effect - all positively charged species contribute to countering all negatively charged species, and vice versa. The closest you come to is when histones and other proteins bind to the DNA. The remaining charges are neutralized by whatever else is around. In large part this depends on the pH and what else is dissolved. At pH 7-8, there isn't a lot of H⁺ around, and other ions are needed (indeed - if the other ions haven't been added, you wouldn't *be* at pH 7-8). Inside the cell there are some positively charged amines floating around, but I imagine that most of the cations would be potassium rather than sodium (cytosol concentrations of ~200 mM vs. 5 mM - for E. coli, from Bionumbers). In an agarose gel, it will be countered by whatever counterion you've placed in the buffer. For TAE buffer/TBE buffer, that would be Tris. -- 128.104.112.102 (talk) 20:01, 31 August 2009 (UTC)

t--

Is K+ (as well as Ca++, Mg++ and whatever cations happen to be there) at a higher concentration in the nucleus? Basically the issue is that DNA would seem to be a pretty concentrated mass of negative charges, and a neutral histone could probably form polar bonds with DNA, helping to delocalise some of the charge, but this means negative charge areas would be created on "another" side (inside the histone?). DNA must create a pretty strong electric field inside the cell that would probably pull cations towards it! So these cations could be "associated" with DNA, roughly speaking. John Riemann Soong (talk) 00:49, 1 September 2009 (UTC)

Also, bacteria don't have histones, but I imagine that with a circular chromosome you're still going to get quite the localisation of negative charge .... wouldn't it be in quite a high energy conformation? John Riemann Soong (talk) 00:56, 1 September 2009 (UTC)

Histones aren't neutral. They usually contain a large number of lysine and arginine residues, which are positively charged at physiological pH. In fact, this is generally true of all nucleic acid binding proteins - overall, they have a net positive charge. The coulombic attraction between the positively charged proteins and the negatively charged nucleic acid adds to the binding affinity, and also serves to neutralize the negative charge on the phosphate backbone. I don't know about the difference in ion concentrations in the nucleus and cytoplasm, but it could be slightly different, although there is a bunch of other negatively charged items in the cytoplasm (ribosomes, mRNA, ATP, various small molecule and protein phosphates, various organic acid metabolites) which may boost the amount of cations in the cytoplasm. Remember also that water has a pretty strong dipole moment, and as such can shield/solvate charged particles quite well, so strict neutrality isn't needed. (Indeed, there is usually an electrostatic membrane potential that is maintained by the cell.) Also, while bacteria don't have histones per se, they do have histone like DNA binding proteins, as well as a host of other DNA binding proteins. -- 128.104.112.102 (talk) 15:30, 1 September 2009 (UTC)

Brain Development

Am I right in thinking that the brain stops developing at age 18 and the between the ages of 18 and 24, the brain only changes in the way it works in order to adapt to the adult social environment. In other words the brain of an 18 year old brain is fully developed structurally and has the same capabilities as a 25 year old brain but has just not yet adjusted to the adult environment. Thanks in advance for any answers. 86.139.54.213 (talk) 20:44, 28 August 2009 (UTC)

Neuroplasticity drops off sharply before puberty (and this seems related to the linguistic critical window). Neural stem cells continue to be active for most of the adult life (though neural stem cells are more active in children). Are you in an argument about maturity of 18-year-olds or something? "Structure" and "social environment" are all pretty vague terms. If you amputated an adult arm, the part of the brain that used to control that arm will be restructured to do something else. (though not perfectly -- sometimes you get phantom limbs.) John Riemann Soong (talk) 21:26, 28 August 2009 (UTC)

I've been reading around this area and alot of research suggests that the brainc ontinues to develop after age 18 up to about 25. However I don't quite understand how. The assumption I made above was that by age 18 the structuring process of the brain which occurs during adolescence has finished and the brain has its full capabilities and that after that the change is more of a change like you described with phantom limbs to adapt to the adult environment. 86.139.54.213 (talk) 21:35, 28 August 2009 (UTC)

As far as I understand, the brain was thought to stop developing at around 18, but more recent thinking is that it can continue to change or develop at any age. One example is registered London taxi drivers who have to memorise a lot of streets - known as "The Knowledge" - and the part of their brain that remembers things grows larger. 89.240.194.145 (talk) 22:54, 28 August 2009 (UTC)

Brain development has many aspects, and some of them continue throughout life, although most of the structure is in place by ages 5-6. Most of the brain's nerve cells are created before birth, and there is actually a drop in numbers after birth as redundant cells are eliminated, but there are a couple of special areas where new neurons are apparently created throughout life. One is the dentate gyrus of the hippocampus, and there is evidence that the newly created neurons play a role in storing new memories. This is just one aspect; a complete answer to the question would be very long. Looie496 (talk) 23:23, 28 August 2009 (UTC)

There was a mid-decade newsmagazine article that said the body pretty much reaches adultivity at 18, but brain development isn't really complete until about 22, and somewhat even until way into the 20s, so the traditional 21 age of complete majority would really be 22 psychologically. (though some things about wisdom teeth and bones aren't until 22) Does anyone know the paper(s) this was based on? Sagittarian Milky Way (talk) 00:07, 29 August 2009 (UTC)

I'm not quite sure what everybody means when they say the brain "stops developing". The brain continues to have the ability to learn throughout your life, which requires the formation of new connections between neurons. StuRat (talk) 14:30, 29 August 2009 (UTC)

I'd say your assessment is quite accurate. The brain stops 'expanding horizons' at about 18, and after that it's filling in the gaps left by that expansion, so to speak. Vranak (talk) 16:42, 29 August 2009 (UTC)

(Wrote this while ago but got distracted and never posted) As most of the discussions above indicate, this is a rather complicated and in some ways I would say unanswerable question since it depends by what you mean by structural brain development, stops etc. Generally speaking your brain is going to be changing until you die. Precisely how will depend of course on a variety of factors particularly how you train it. Size wise (well weight), the brain is already about 25% of the adult brain at birth [11] and broadly speaking is structurally very similar. Of course human brain development is much more then just getting larger and taking on the right structure. As said earlier, the brain is always changing and I'm pretty sure there are some quite significant changes past 18. For example, one thing that often comes up particularly when it comes to driving is that the parts of the brain involved in risk taking continue to change significantly until about 25 [12] [13] [14] [15]. It's sometimes said that the brain reaches maturity or adulthood at 25, but as I've indicated this is really overtly simplistic Nil Einne (talk) 20:23, 30 August 2009 (UTC)

Kirchhoff current law validity

It was quiet interesting when I had a discussion with some friends about Kirchhoff current law. Someone told me that there is some current lost in the case of a lamp/bulb due to electrons-ions effect. I'd like to verify this, and how can I restate Kirchhoff current law?--Email4mobile (talk) 21:44, 28 August 2009 (UTC)

It sounds like somebody is playing games with the definition of a closed system. Kirchoff's current law always holds, because conservation of current is equivalent to conservation of particles (# electrons in = # electrons out). There are no relevant electronic processes which create or destroy electrons; photon radiation by incandescent heating certainly does not create or destroy electrons. It might be *remotely* possible that some electrons thermally escape from the hot filament, and result in a static charge buildup on the exterior of the lightbulb; this is a very minor effect, if it is even measurably present in a lightbulb. Anyway, current is still conserved, if you count the flow of electrons via thermionic emission as part of the wire-in, wire-out system. A more general way of stating Kirchoff's law is in the form of the displacement continuity equation, $\nabla J=-{\frac {\partial \rho }{\partial t}}$; most people consider the charge buildup to constitute current (by definition). Nimur (talk) 22:16, 28 August 2009 (UTC)

In fact, Kirchoff's current law holds even if electrons are created or annihilated since electrical charge is always conserved. Rckrone (talk) 01:13, 29 August 2009 (UTC)

General fomula for electric fusing

Given a material's characteristics (type, length, cross sectional area) is there a formula to relate time-current fusion assuming normal conditions?--Email4mobile (talk) 21:49, 28 August 2009 (UTC)

Yes there is, the resistance of the material is a major factor, you need to make assumptions about how fast it will radiate heat, (and whether or not it's local surroundings form insulate the heat produced.

You also need a formula for Thermal_radiation - when the heat (due to resistance, dependent on current) generated is greater than the heat that can be dissapated at the fuse melting point the fuse fails. Note that the resistance depends on temperature too.83.100.250.79 (talk) 22:01, 28 August 2009 (UTC)

By "fusing" do you mean "melting?" Model the energy dissipated in a unit length of the material versus the heat radiated or conducted, along with the ambient temperature and tension on it. A bare conductor would carry more current without melting than an insulated conductor, and a conduit filled with conductors would lead to melting at lower current than the same conduit with one pair of conductors. Conductors at a very high ambient temperature would melt at lower current than conductors at low ambient temperature. Even the surface color or emmisivity of a conductor would affect the current it could carry before melting. In some household plug fuses, the spring tension on a fuse element causes it to open at a lower current than if it were not under tension. Edison (talk) 00:34, 30 August 2009 (UTC)

Yes I do mean "melting". The reason I raise this question is to estimate an approximate but general formula for the fuses melting curve given in data sheets, thanks..--Email4mobile (talk) 21:55, 31 August 2009 (UTC)

why is ammonia more acidic than water?

Is it because it has more protons to donate? It's kind of curious to me, because shouldn't oxygen tolerate the negative charge better than the ammonia? Or does having more hydrogens to delocalise the negative charge improve ammonia's acidity? If NH2- is a stronger base than OH-, why is NH3 more acidic? John Riemann Soong (talk) 21:52, 28 August 2009 (UTC)

Ammonia is a weak base. It is not more acidic than water. You should read that article, which describes its ionization process, and Ammonium for some electrochemical properties of the ion that forms. This MSDS from Texas A&M Chemistry says Ammonia has a pH of 11.6 at 1M and describes it as a dangerously corrosive alkaline. Ammonia also has interesting properties when mixed in low concentrations with chemical solvents at other net pH 's - it can become dramatically more dangerous and can create hazardous fumes. Nimur (talk) 22:06, 28 August 2009 (UTC)

I just did. I was actually curious about its pKa, and I note its value (9.75) is even lower than that of phenol (9.95). But NH2- has no resonance stabilisation! Why is it so acidic? I know it's a weak base, but ammonia seems to have some significant (though not symmetric) amphoterism. Water is usually known as an amphoteric substance, but ammonia seems to give water a run for its money. John Riemann Soong (talk) 22:11, 28 August 2009 (UTC)

What do you mean ammonia is not more acidic than water? Water's pKa is 15.74 ... a drastically weaker acid than ammonia. In fact, I'm wondering why hydrogen bonding doesn't form between ammonia molecules as it does for HOH molecules, since ammonia has a lone pair to donate, and it also has protic hydrogens to donate as well. Is it the lack of symmetry (3 protic hydrogens and 1 lone pair versus 2 protic hydrogens and 2 lone pairs)? Here's my working hypothesis: ammonia's 3 protic hydrogens give it an advantage over water, which only has two, resulting in a higher pKa. But nitrogen isn't as good at handling negative charge, so I wonder why those two effects don't cancel out. (I note that water is only an order of magnitude or two more acidic than alcohols, when it has twice the number of acidic hydrogens, so surely ammonia having 50% more acidic hydrogens can't lower the pKa that much??) John Riemann Soong (talk) 22:18, 28 August 2009 (UTC)

There are multiple definitions of acid and base. I think you're using an uncommon definition, and incorrectly applying it to compare Ammonia's dissociation constant to that of water. See also this detailed description of Ammonia and its pKa. You might also want to read Amphoterism about chemicals which have both acidic- and alkaline- properties, simultaneously. Nimur (talk) 22:19, 28 August 2009 (UTC)

It's uncommon to define acidity as a function of pKa? I suppose I don't know what the solvent context for the reported pKa value in question. If the 9.75 value is the self-ionization value, then the pKb of ammonia (in ammonia) should also be 9.75, shouldn't it? (Since there would always be equal amounts of NH2- and NH4+ with pure ammonia). Plus, I'm not sure how you would measure the self-dissociation constant of a gas...so I'm guessing the 9.75 value is that of ammonia in water? I don't see how I'm using pKa incorrectly here. If ammonia's pKa in water is 9.75, and water's pKa in water is 15.76, surely ammonia must be more acidic than water? John Riemann Soong (talk) 22:37, 28 August 2009 (UTC)

Also, what is the mathematical relationship between the number of acidic hydrogens (assuming that they are chemically equivalent) and pKa? I note that ethylene glycol has a pKa of 14.22 whereas ethanol has a pKa of 15.9. I actually suspect that the (straight-chain) alkyl group stabilises the negative charge on a deprotonated alcohol via hyperconjugation, so perhaps a 50% decrease in the amount of acidic hydrogens by itself should actually result in a much greater pKa difference than the pKa of alcohols would suggest? (Having a tert alcohol increases the partial negative charge on the carbon next to the alkoxy oxygen, but would probably only negate a minority of the stabilising effect, so perhaps it's possible that having more or less protic hydrogens can affect pKa that drastically, an effect which the alkyl group mostly masks?) And is it true that NH2- enjoys some hyperconjugative stabilisation? John Riemann Soong (talk) 22:37, 28 August 2009 (UTC)

Sorry for the 3 posts in a row, but can I double check to ensure that the 9.75 value is correct? A google search is telling me that some people think ammonia has a pKa of 9.25, which makes it even more acidic than I thought it was. John Riemann Soong (talk) 22:39, 28 August 2009 (UTC)

(see below) Basically google is giving you the wrong figure. It pKa ammonia =34

The OP (J R S) is talking about this acidity

NH3 >>> NH2- + H+

not the acidity/basicity in water -

I think there must be some wrong figures somewhere - Sodium amide definately deprotonates water.

I think you may have got the pKa for ammonium, not ammonia - that's an error on the part of the people who write the books/web pages - they say the pKa of ammonia when they are describing the pKa of NH4+

This is better [16] pKa ammonia/amide is 34 , (ammonium is 9.24) I assume this is near enough to 9.75 to not cause any problems

The 3 H's on N should add slightly to the acidity (compared to a hypothetical NH2, or an OH2 were the N and O have the same electronegativities), but this won't be a major factor. Less than 1 pKa unit.

83.100.250.79 (talk) 22:48, 28 August 2009 (UTC)

Wow, thank you .... I've gone back and fixed the article. It makes me so mad when people are sloppy and end up describing the pka's or pkb's of the ions instead. I was guessing solvent acidity because it's kind of hard to self-dissociate as a gas. How do you measure the self-dissociation constant of a pure gas anyway? One more question though ... I assume a 50% increase in the number of protic hydrogens results in a pKa decrease of around (or less than 1), but a 50% decrease (ceteris paribus) seems to affect the pKa way more than this (in alcohols the alkyl group's hyperconjugation probably masks this somewhat). Am I right in thinking that the relationship of the number of protic hydrogens to (pKa1) acidity isn't linear? John Riemann Soong (talk) 22:53, 28 August 2009 (UTC)

Wait a minute - alcohol hyperconjugation - you mean like in ethylene glycol? There's a much simpler thing going on here which is the stabilisation of -O^- by an 'adjacent' OH group

     H
    /
   O   O-
   |   |
 2HC---CH2

You can see how the H can be 'shared' by the two O groups - think resonant hybrids - is this what you meant - I don't think that is usually termed hyperconjugation. (I might have forgotten)

The effect of having more H's is complex - you can view it theorectically as having N times more hydrogen (ie equivalent to the concentration being N times more concentrated). As a first approximation this would change the pKa by log(N). (does this make sense to you?)

It's possible to draw up more complex models - but they usually fall apart outside a narrow range of compounds - because of all the other things that can alter acidity.83.100.250.79 (talk) 23:05, 28 August 2009 (UTC)

Wow, thanks for addressing ethylene glycol. I guess I don't have to make a new question after all! Is the inductive effect also at play in ethylene glycol? (Or is the single oxygen too far away?) I'm thinking something analogous to what happens in trichloroacetic acid. Would making the OH groups further apart basically weaken this "H+ delocalisation" significantly? If I had OH groups substituted on opposite ends of n-octane, for instance, would approximately see a pKa decrease of log 2 compared to n-octanol? John Riemann Soong (talk) 23:15, 28 August 2009 (UTC)

Also, by "hyperconjugative stabilisation" I was referring to the possibility that the alkyl group on ethanol for instance, helps delocalise some of the negative charge of the EtO- anion. Theoretically the difference in pKa between water and ethanol should be 0.30, right? But we observe a pKa difference of 0.14. The alkyl group seems to mitigate half of the lost acidity. John Riemann Soong (talk) 23:18, 28 August 2009 (UTC)

And does a tert alkyl group actually destabilise the conjugate base because the tert-carbon is likely to have more partial negative charge than say, a methylene carbon? What would explain say, isopropyl alcohol's increased acidity compared to n-propanol? Interestingly, methanol has a pKa of 15.5 -- it's more acidic than water! So the hyperconjugative stabilisation seems to have some observable effects (pKa stabilisation of 0.56? theoretically methanol's pKa should be 16.06 compared to water if the alkyl group didn't do anything...) John Riemann Soong (talk) 23:28, 28 August 2009 (UTC)

The inductive effect always works a bit, even 4 atoms away - but it gets weaker an weaker.. So 3chloro propanol will be ever so slightly more acidic than plain propanol. etc. Same goes for OH groups..

The log 2 effect I think would appply to any di-alcohol, that's just the effect of having more OH groups (don't quote me on that because I haven't got a reference - but I'm sure the reality isn't far off)

The H+ delocalisation should work best when the compound can make 5 and 6 membered rings (eg 1,2 and 1,3 diols) - (5 and 6 membered rings are the best - anything over get's a bit 'wobbly')

The difference in pKa between water and alcohol - the 0.3 figure is a very rough estimate (ie the value should be the same order of magnitude) - but maybe yes.

t-butanol is less acidic than s-propanol - this could be the electronic effect of the alkyl group - BUT - the larger the size of the alkyl group the more sterically hindered the O- is - which affects solvation - in other words in t-butanol the O- anion of the base is more difficult to solvate - I would expect this to have quite a significant effect in this case (the relative electron withdrawing abilities of the different plain alkyl groups being about the same.) In the absence of the solvation effect t-butanol might be expected to be as acidic as ethanol - because the larger molecule can very weakly stabilise the anion very slightly, and the other factors (excluding solvation) are nearly the same.

I think the difficult of solvation effect must be greater than the stabilisation due to hyperconjugation in t-butanol - as you mentioned above - measuring gas phase acidities would cut out the solvation effect - I don't know much about gas phase methods - if you want to know I suggest a separate question for it.

Yes methanol is a good example of hyperconjugation .

There's actually one book that covers all this (and super acids) very well - it's called "Physical and mechanistic organic chemistry" by R.A.Y Jones . Don't know if it's still in print - but you can get it from amazoncom from $1.40 (used)! (academic books are always expensive - typically about £30 for this new = ~50US $.) It should be library gettable - it's quite a common text for higher level chemistry. It's the size of a paperback novel, It may or may not be too complex - I haven't got a copy so this is from memory. But it definately covers all these topics.83.100.250.79 (talk) 00:14, 29 August 2009 (UTC)

I've given a link [17] - it seems to be out of print which may explain why some people are asking $150 for one on amazon. 83.100.250.79 (talk) 00:18, 29 August 2009 (UTC)

There's a link here [ http://books.google.co.uk/books?id=z3jXKOYuqQAC&pg=PA245&lpg=PA245&dq=methanol+acidity&source=bl&ots=dKaix0e3Or&sig=6jNkscLbVbfbwPrr1jPbfMXI6VU&hl=en&ei=Q3qYSs_ICpPajQet1IW3BQ&sa=X&oi=book_result&ct=result&resnum=8#v=onepage&q=methanol%20acidity&f=false] for the gas phase acidities of alcohols - page 245. It doesn't use the term hyperconjugation - but the data can be explained in terms of hyperconjugation quite well. It also covers solvation effects on page 246. It should go some way to answering most of your alchohol questions.83.100.250.79 (talk) 00:50, 29 August 2009 (UTC)