Wikipedia:Reference desk/Archives/Science/2006 September 23

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Fate of the Universe

Well, I just researched the Big Crunch/Freeze/etc... and I didn't find anything specifically answering my question... Which is: "Once the universe ends in the way that it does will it be reborn again like another Big Bang or just end completely without any life existing ever again?" It's a scary question for me to think about, but I'd like to know the answer anyway. Any theories present? -MF14

Our current understanding of physics does not allow us to extrapolate beyond the demise of the universe as predicted by current cosmological theories about the ultimate fate of the universe. It is hard to imagine how some possible observation of a physical event might lead to a new understanding of physics that would allow such an extrapolation. In some string-theoretical models there is an analytical continuation beyond what would be a singularity in Einstein-like models, but the current status there is that the theories "aren't even wrong". If you like scary theories, then the Big Rip is my favourite; a lot more chilling than the Big Freeze, for which one can imagine some scientific way out. --Lambiam^Talk 01:18, 23 September 2006 (UTC)

Interesting. The Bose-Einstein condensate addendum to the heat death of the universe (which is only hinted there) isn't mentioned in the ultimate fate of the universe article. Yet, here it is: Fermion-boson fate of universe theory. Fixed. -- Fuzzyeric 03:01, 23 September 2006 (UTC)

Either way dude, I wouldn't let it affect your long term plans. ;) --AstoVidatu 02:55, 25 September 2006 (UTC)

If you want something to worry about, consider the rogue asteroid problem ie Impact_event--Light current 02:59, 25 September 2006 (UTC)

Go read Isaac Asimov's The Last Question. B00P 19:38, 26 September 2006 (UTC)

I think Douglas Adams' idea was best - to paraphrase:

As soon as anyone figures out what the universe is all about, it will instantaneously disappear and be replaced by something even more bizarre and inexplicable. Another theory says this has already happened.

Mattopaedia 08:42, 28 September 2006 (UTC)

I like to interpret Robert Frost's poem "Fire and Ice" as being about the Big Crunch and Big Freeze respectively. And it even fits with the topic at hand; if one looks at the positive elements of humanity (and, via extrapolation, sentient life), then the Big Crunch might be preferable, as it seems entirely possible that a new universe could be born out of that. But if one focuses on humanity's negative aspects, a cynic might hope for the Big Freeze, where it seems unlikely that any subsequent universe would come of it. Chuck 19:06, 28 September 2006 (UTC)

Covering answers on a worksheet

I want to photograph a homework sheet and send the photo to my friend, but I don't want him to see the answers. Is there a quick and easy way to block the answers? I tried copying & pasting blank portions of the page onto the answers (using an image editor of course), but the photograph is not of uniform brightness. Also, this method takes too long. --Bowlhover 00:33, 23 September 2006 (UTC)

Snopake out the answers before you scan it ! 8-)--Light current 00:38, 23 September 2006 (UTC)

In case anyone is confused about this answer, like I was before doing a Google search, apparently the Snopake company made correction fluid before they became a stationery company. --Allen 00:41, 23 September 2006 (UTC)

Thats right, they used to move right along! I should have said Tippex--Light current 00:46, 23 September 2006 (UTC)

But erasing the answers and re-writing them will take even longer than copying & pasting blank pieces of the photo. --Bowlhover 00:56, 23 September 2006 (UTC)

Just wondering, why does it have to have a natural gradient of shade? Hyenaste ^(tell) 01:04, 23 September 2006 (UTC)

I don't have a scanner, so I used my camera to photograph the worksheets. The main source of light in these photos is the camera's flash, and since it's so close to the sheet, the distance to the centre of the sheet is less than the distance to its sides. Taking account of the inverse-square law, this represents a significant difference in brightness. --Bowlhover 01:36, 23 September 2006 (UTC)

You could turn the flash off and use a bright external light, instead. StuRat 05:49, 23 September 2006 (UTC)

Why not load it into MS Paint or something and use the spray paint tool to cover them. Anchoress 01:03, 23 September 2006 (UTC)

Which colour should I spray-paint the answers with? I'm pretty sure that no matter which colour I choose, it will end up looking like this. --Bowlhover 01:36, 23 September 2006 (UTC)

Do you want to send him the questions without the answers, but don't want him to see that you are hiding the answers from him? The best bet might be to tell him that your imaging techniques aren't working, and to dictate the problems to him. If you have photoshop, you could use the stamp tool to create a gradient similar to the background, maybe other paint programs have a similar tool...Tuckerekcut 01:16, 23 September 2006 (UTC)

No, I'm not trying to hide anything from him. (Some of these homework sheets were handed out days ago, and he won't believe that I didn't even start any of them.) My friend needs to print the homework out and do it, and I don't want it to look like this. Also, no, I don't have photoshop. I have the Gimp, though. --Bowlhover 01:36, 23 September 2006 (UTC)

If I understand you, you don't really care what your friend sees, as long as he does not see your answers. The problem, though, is that you don't want the teacher to see these weird light-colored squares on your friend's answer sheet, after he has turned it in, right?

In that case, you can use MSPAINT, and use the cut-paste commands... capture a square of the blank background page, copy it, then paste it back in and move it so it covers up one of the blanked-out regions. Repeat until you have all of the answers blanked out. It will not be perfect, but it'll have the correct background color instead of being white. 192.168.1.1 7:51 22 September 2006 (PST)

This is actually what I did, except with Gimp instead of MSPAINT. However, the "blank" background does not have a constant brightness. So if I copy a square and move it to another location, the contrast will be obvious. --Bowlhover 03:41, 23 September 2006 (UTC)

Another suggestion: Go to a copy center, like Kinko's, make a straight copy, use liquid paper on the copy, let it dry, then copy that. Copiers don't distinguish between various whites as much as a digital camera will. The total cost should be less than a dollar. StuRat 03:02, 23 September 2006 (UTC)

Don't use cut and paste, use the clone tool. Gimp has it. The fact that there's a gradient makes it a bit more difficult, but as long as you clone parts of paper along the same gradient line it's very difficult to notice the difference. It takes a little practice to make it look perfect, but I've been fiddling with pictures like that for years, and as long as I have a 1-inch section of relatively clear base texture I can blank out anything. There are a lot of techniques to make it easier, one might be to unify the brightness of the entire photo using an inverted mask or something like that, but you don't seem to have the time to do that.

A useful tool for this purpose is the GIMP Resynthesizer plugin, which provides a "smart remove selection" tool. However, I agree that for this purpose it's overkill — just adjust the levels until the black is truly black and the white truly white, and then you can just cut out the parts you don't want. You'll have to do the adjustment anyway, if you want the resulting image to look like the original after printing. —Ilmari Karonen (talk) 19:38, 23 September 2006 (UTC)

Of course, no matter how well you blank the answers out, there are always ways for your friend to find out what you did.  freshofftheufoΓΛĿЌ  04:13, 23 September 2006 (UTC)

What ways ? StuRat 05:49, 23 September 2006 (UTC)

That's what I'd like to know. Police always claim they have fail-safe ways of finding out (possibly analyzing patterns in the code segments?), although I'm not sure how much that applies to non-compressed media.  freshofftheufoΓΛĿЌ  07:55, 23 September 2006 (UTC)

Whatever you do to blank the spaces, first clip the white to make it even (and very white!). Basically, you shift the white point that way. Use the histogram or 'levels' or whatever it is called in Gimp and move the right slider to the left (if it works the same as in Photoshop). Another way to do this might be to increase the contrast. DirkvdM 06:03, 23 September 2006 (UTC)

Wouldn't it be easier to call up your friend and read out the answers which you want him to know ? Wikicheng 07:38, 23 September 2006 (UTC)

Here's some general advice on using a digital camera as a copier:

• Don't take the picture with the camera too close to the document to be photographed. Keep a distance of about 5 ft (1.5 m) between the camera and the document. Optically zoom in the document if necessary. This reduces geometry distortion and will result in more uniform illumination.
• Make sure that the document to be photograph lies flat.
• When you shoot the picture, leave some space around all sides of the document to make sure that the edges and the corners of the document are visible. This will be helpful if you need to correct the orientation or perspective distortion of the image.
• If you use autoexposure, point the camera at something a little darker (e.g. a grey sheet of paper placed next to the document to be photographed) and lock the exposure. If you point your camera at a mostly white sheet of paper, your picture will be under-exposed for your purpose. As an alternative, manually adjust the exposure compensation until you get the desired exposure.
• If you're photographing a black-and-white document, convert the picture taken into a grey-scale image to reduce the file size.
• Adjust the color curve so that most of the pixels will be clipped to white. You may also need to adjust the curve at the other end so that printed text will look more black.
• You can compensate for non-uniform illumination using a gradient mask (radial or linear, as appropriate) and selectively brighten only a portion of the image, if you're familiar with how it's done in photo editing.

Just do whats easiest for you.Evildoctorcow 05:38, 30 September 2006 (UTC)

If you were using a scanner, I would recommend you convert it to one bit. This works very well especially with a high res (600 dpi - 1200 dpi is enough) and you don't have any photos/complex images. It gives excellent quality printouts and you don't have gradient problems and it gives tiny files too with the right compression (any CCITT group 3/4 compress works well, e.g. PDF). However with some quick calculations, you probably could only achieve about 150-300 DPI with most digital cameras unless you took multiple photos and combined them (and then you might have macroing problems. However you should still get good quality with 1 bit although I would recommend you double the res if under 250DPI or so (you can usually do this when converting to 1 bit). BTW, I've never heard of snopake and only vaguely of Tipp-Ex. I usually call it Liquid Paper or just correction fluid. Also there appears to possibly be some confusion above. Unless you do it very very well, you can usually detect modifications in photos. However it will not be possible to know what the answers on the paper were. Of course, if you used 1 bit, and did it carefully, it would easily be nearly impossible to tell the answers were removed. Nil Einne 08:01, 30 September 2006 (UTC)

Science Question

How can i get more information about salamanders from your site,like their habitat, what they eat things i can ues to do a report"""""""""

You should sign your comments with four of these: ~. The key is usually on the top-right top-right of your keyboard.

Nice 'correction'. :) DirkvdM 07:36, 24 September 2006 (UTC)

Don't you mean the top left? DirkvdM 06:05, 23 September 2006 (UTC)

No its in the middle right on my KB--Light current 06:06, 23 September 2006 (UTC)

Well it's bottom right on mine, between the alt-gr and the ctrl. No wonder folk have difficulty finding it.--Shantavira 06:55, 23 September 2006 (UTC)

I meant top-left, even though they're on the top-right of my Japanese keyboard haha. I'm all messed up in the head!  freshofftheufoΓΛĿЌ  07:52, 23 September 2006 (UTC)

As for your question about salamanders, remember that there are many many different kinds of salamanders, so there's no easy answer to something like "where do salamanders live?". About the only thing you can assume is that most of them live close to, or in, water. Check out the articles for individual species/suborders for more specific information about each type of salamander: Cryptobranchidae, Hynobiidae, Ambystomatidae, Amphiumidae, Dicamptodontidae, Plethodontidae, Proteidae, Rhyacotritonidae, Salamandridae, and Sirenidae.  freshofftheufoΓΛĿЌ  04:20, 23 September 2006 (UTC)

Sliding...erm, thing (physics)

An object travels toward a 25-degree ramp at 18m/s. It travels 16.2m along the ramp. Find the force of friction. Is there enough information in this question to get the answer? BenC7 05:58, 23 September 2006 (UTC)

P.S. Not homework.

TRavels up the ramp and then stops at 16.2m? I think youre gonna need the weigt of the object for friction calculations.--Light current 06:04, 23 September 2006 (UTC)

Ah wait a mo. Some the kinetic energy of the object has been dissipated as heat caused by friction. THe remainded has gone into the potential energy of the object now at rest. You can work out how much energy is lost due to friction by subtracting the PE from the KE. THis however does not give you the force of friction.--Light current 06:11, 23 September 2006 (UTC)

You do know the vertical displacement, which is what matters for the potential energy at the point where the block stops... you can then assume that if all the potential energy became kinetic energy, you could solve for terms of your velocity in terms of the mass... then you could do some kinematics to solve for the acceleration, when given the displacement, and rewrite that acceleration as a force in terms of Newton's Second Law. The only change of velocity there would be due to the deceleration caused by the force of friction, no? Titoxd^(?!?) 06:23, 23 September 2006 (UTC)

No: you also have to remember the component of gravity acting down the slope that will add to the frictional force in decelerating the object. THis is a very interesting question! Is the ramp sloping up or down? I assumed up.--Light current 06:32, 23 September 2006 (UTC)

It's not that interesting. The slope can be either down with very strong friction, or up with less friction. To know the force, you need to know the mass (it's proportional to it) - if you only want to know the friction coefficient, you're good to go (let m be the mass, it will cancel). For each case, you know the changes in kinetic energy and potential energy. The difference (or sum, depending on how you look at it) is the work done by friction. Divide by the distance travelled, and you have the force. -- Meni Rosenfeld (talk) 06:42, 23 September 2006 (UTC)

Of course Work = Force x distance moved. So the answer is that there does appear to be enough information to slove the problem! So solve it!--Light current 06:49, 23 September 2006 (UTC)

Again, that depends. The mass is not given. So you can express the force as a function of m, and you can find the friction coefficient as a number, but you cannot express the force as a number. -- Meni Rosenfeld (talk) 07:09, 23 September 2006 (UTC)

The question also didn't explicitly state that it hits the base of the ramp at 18 m/s, so it could lose some speed due to friction, before it arrives. If we allow for this possibility, then we can't answer the question in any way. StuRat 09:44, 23 September 2006 (UTC)

Yes yes. Well all previous respondents assumed that we had frictionless travel before it hit the ramp other wise as you say the solution could not be obtained.

In fact this can be deduced from the question in that since it travels toward the ramp at a given speed, we must assume that speed is constant: An object travels toward a 25-degree ramp at 18m/s.(constant velocity is implied) So, as long as it is traveling towards the ramp, its doing 18m/s. THis eliminates any possibility of energy loss before it hits the ramp!. I hope that explains our position! --Light current 11:21, 23 September 2006 (UTC)

I'm not sure you can make that assumption. If that were the case, why wouldn't they state it simply as: "An object starts up a 25-degree ramp at 18m/s" ? StuRat 13:52, 23 September 2006 (UTC)

Its badly phrased by the physics /mechanics teacher. An object hits a ramp at 18m/s would be best.--Light current 13:55, 23 September 2006 (UTC)

I prefer my way. Your way might involve some transfer of energy to the ramp from the impact of "hitting". StuRat 16:15, 23 September 2006 (UTC)

I think there is not enough information. The vertical reaction force F = mg with which the ramp pushes back to your sled can be decomposed into two orthogonal vectors of, respectively, magnitude F·cos α (directed upwards normal to the ramp) and F·sin α (directed backwards along the ramp), where α stands for the slope angle of the ramp. If μ denotes the coefficient of friction, the force of friction X is then X = μF·cos α. The total backwards-directed force G has then magnitude G = μF·cos α + F·sin α = mg(μ·cos α + sin α). The deceleration is then a = G/m = g(μ·cos α + sin α). The distance s travelled until the object reaches zero speed equals s = v²/(2a) = v²/(g(μ·cos α + sin α)), where v is the initial velocity. We may assume that g is known, and s and v are given (assuming that all forward speed gets translated into speed along the ramp!), and we have to find the value of X = μF·cos α = μmg·cos α. We have two equations (s = v²/(g(μ·cos α + sin α)) and X = μmg·cos α) with four unknowns (μ, α, X, and m). --Lambiam^Talk 14:21, 23 September 2006 (UTC)

α is known. And I have already discussed what happens with m. And doing this using energy considerations is much easier. -- Meni Rosenfeld (talk) 14:31, 23 September 2006 (UTC)

It has been my experience that those who are good at math, such as physics teachers, often have poor language skills. This question is yet more confirmation. Edison 15:17, 23 September 2006 (UTC)

That's why I take pride in being equally poor at both. :-) StuRat 16:15, 23 September 2006 (UTC)

The question is (was) for a grade 11 physics student who I am tutoring. He recounted the question to me from an exam he had just taken, saying that he was confused because there did not appear to be enough information. After trying to work it out myself, and seeing you guys have a go, I think it is safe to deduce that you cannot work out the force of friction from the information given. So it is possible that either there was a mistake in the original question, or it was not remembered properly. I probably would have felt a bit silly as a tutor if you could work it out! BenC7 01:23, 25 September 2006 (UTC)

I would say it was likely not remembered properly. Specifically, if a person dismisses a piece of info as not relevant, they aren't likely to recall that info later, even though it may, in fact, be critical to solving the problem. StuRat 19:57, 25 September 2006 (UTC)

Potential energy and mass

Hi everyone.

When two Deuterium nuclei merge into a Helium nucleus, energy (and lots of it) is released.

It is often explained by saying that the meass of a Helium nucleus is less than twice the mass of a Deuterium nucleus, and the difference in mass is converted to energy according to the famous E=mc².

However, why is this explanation necessary? There is a strong nuclear force acting between the protons and neutrons. When these get closer together, an energy equal to the difference in potential energy - that is, the integral of the force between the two points - is converted to other forms. This is the same as the calculations of potential gravitational energy in classical mechanics (of course, the nature of the force is different), and should be enough to explain the released energy.

What does it mean? Does it mean that the energy released is the change in mass plus the change in potential energy? Or does it mean that the additional mass is itself the potential energy? Does this apply to other forces as well? For example, when two distant, massive objects are moving away from each other, do their masses increase because of the additional gravitational potential energy? What kind of mass is this? Inertial, gravitational, both, or something else? If this is so, is there even such a thing as "potential energy", or are there only changes in mass?

Taking this into account, and the idea that the relativistic mass of an object increases when it is in motion, would it be correct to say that every form of energy is actually mass? From which it can be concluded that mass and energy are the same thing?

I will be grateful if anyone can clear that up, or direct me to references which will. Thanks. -- Meni Rosenfeld (talk) 06:31, 23 September 2006 (UTC)

Conservation of mass-energy--Light current 06:36, 23 September 2006 (UTC)

Wrong link, but I've managed to find the right ones, and they're helpful, thanks. Let's see if I got this straight:

• The additional mass is itself the potential energy (I'm not too clear about how that might be, but whatever).
• It applies to other forces as well.
• This mass is both inertial and gravitational.
• There is no "abstract" potential energy which appears only in calculations, but only observable differences in mass.
• Mass and energy are the same thing.

Right? By the way, does this apply only to special relativity or to general relativity as well? -- Meni Rosenfeld (talk) 06:55, 23 September 2006 (UTC)

So potential energy is always manifested as mass?--Light current 07:03, 23 September 2006 (UTC)

Yes. All forms of energy contribute to the mass of the system by the infamous formula m = Ec⁻². There is a well-known quip that when you climb a mountain, your mass increases. In doing so, you however convert energy stored in your body as fat or ATP to "height" energy (E = mgh) as well as to heat that is lost to the environment. So, theoretically, there should actually be a (miniscule) mass loss. --Lambiam^Talk 11:28, 23 September 2006 (UTC)

And I believe that the Sun loses something like 4.25 million tonnes per second in radiated energy. Richard B 15:21, 23 September 2006 (UTC)

I think this is why it becomes harder and harder to accelerate when you approach the speed of light - your kinetic energy starts making a significant contribution to your mass. Clarityfiend 18:21, 23 September 2006 (UTC)

O, now I understand! When I try to run really fast it's always as if I hit a wall. --Lambiam^Talk 22:16, 23 September 2006 (UTC)

Okay, thanks everyone. -- Meni Rosenfeld (talk) 15:39, 23 September 2006 (UTC)

Well actually you answered most of it yourself! So thanks to you!--Light current 16:11, 23 September 2006 (UTC)

Say not: "when you climb a mountain, your mass increases." Say instead: "when you climb a mountain, the mass of the planet-person system increases." --GangofOne 00:33, 24 September 2006 (UTC)

Oh? What energy was added to the planet-person system in that process? --Tardis 15:09, 25 September 2006 (UTC)

Potential energy due to mgh (where h is the height of the mountain)--Light current 15:15, 25 September 2006 (UTC)

My point was that that energy was already there -- in the form of the chemical (food) energy the person used to climb the mountain (or even in the batteries of the person's autogyro, or...). If talking about the planet-person system, energy can only be added from the outside. Such internal workings as the rearrangement of the person with respect to the planet are entirely irrelevant, just as the different distributions of particle velocities in the person's blood. Of course, Earth is constantly exchanging (mass-)energy with the outside: sunlight, meteors, blackbody radiation from the surface, radio broadcasts, cosmic rays, etc. But none of those are non-negligibly involved in the process of climbing the mountain. --Tardis 16:04, 25 September 2006 (UTC)

Center pivot irrigation

Why is Center pivot irrigation argriculture laid out on a square grid? Surely there would be more efficient ground usage on a hexagonal grid. -- SGBailey 07:05, 23 September 2006 (UTC)

Thought I saw some pics of this using hex grid. Or was that oil tanks?--Light current 07:24, 23 September 2006 (UTC)

I agree that it would be more efficient in a very large field, but not in small rectangular fields, because the hex grid would tend to leave half a circle out at each edge. Also, the non-irrigated areas might be useful for storage, farm equipment, crops that require less moisture, livestock, offices, housing, etc., so don't assume it's all wasted. StuRat 09:04, 23 September 2006 (UTC)

In addition, the length of the piping leading to the watering stations might be significantly longer per irrigated area, since the pipes would either need to be doubled or placed at an angle to reach all the points in a hexagonal grid:

                                                 WATER
 *---*---*----+-*-+-*-+-*------*---*---*-------- SUPPLY
 |   |   |    | | | | | |     / \   \            LINE
 *   *   *    * | * | * |    *   *   *
 |   |   |      |   |   |     \   \   \
 *   *   *      *   *   *      *   *   *
 RECT GRID     HEX GRID        HEX GRID

Then again, my diagram seems to indicate that the total length is the same is they are placed at an angle. StuRat 09:23, 23 September 2006 (UTC)

Here are pics (also) showing hex grids: [1][2]. There may be a relationship with the local price of (arable) land. If land is dirt-cheap, the simplicity of working with a square grid may prevail. If land is at a premium, you want to use it to the maximum extent. --Lambiam^Talk 11:13, 23 September 2006 (UTC)

The simple reason, at least in Nebraska, is that country roads are laid out on a one-mile grid. It's much easier to lay out land in rectangular patterns, and the roads were laid out before center-pivot irrigation came along. If the fields were hexagonal and the roads had to zigzag around the fields, then driving around in the country would be very annoying. ;-) —Bkell (talk) 19:04, 23 September 2006 (UTC)

Nostrils--how long and why?

I ve had difficulty for some time with my left nostril. (Been to the doctors a number of times- so no need to advise I go see one). Any way I was wondering how long each nostril is before it (preumably) joins up with the other one. Why do we have two nostrils anyway? (Oh, and please dont say: In case one gets blocked 8-) )--Light current 13:16, 23 September 2006 (UTC)

Why not say the correct answer ? I frequently have one sinus and nostril blocked, and the other is clear. StuRat 13:46, 23 September 2006 (UTC)

Same as my problem. But mines always the left sinus/nostril. Snap! 8-)--Light current 13:48, 23 September 2006 (UTC)

That is the answer basically. I think I heard them say it on Beakman's World once. Just think of the trouble's you'd be having now if you only had one nostril.  freshofftheufoΓΛĿЌ  14:34, 23 September 2006 (UTC)

OK but how far do they go before joining up into the airway?--Light current 14:35, 23 September 2006 (UTC)

I can only guess from this picture, but they appear to converge quite close to your pharynx, right in the middle of your ears. I had no idea there were so many holes in the front of my head.  freshofftheufoΓΛĿЌ  14:40, 23 September 2006 (UTC)

Yeah picture is unclear. I belive tho' that you can get ito the sinuses by probing round the labarynth of passages in there and there is a procedure for cleaning out the sinuses via the nose. Somone who had it done said it felt a bit like having your brain sucked out 8-)--Light current 14:49, 23 September 2006 (UTC)

If somebody dressed like an ancient Egyptian offers to clean out your sinuses with a large hook, I'd look elsewhere. :-) StuRat 16:10, 23 September 2006 (UTC)

Id be shouting 'Mummy!, Mummy!'.. please dont make me a mummy'--Light current 16:14, 23 September 2006 (UTC)

Tube shoved up one nostril. Water turned on. Dish held underneath the other. Been there, seen it, had it done when I was 10. Not pleasant. --G N Frykman 17:29, 23 September 2006 (UTC)

What sort of water do they use? Saline?--Light current 00:37, 24 September 2006 (UTC)

Who says we all have two nostrils? Don't do cocaine, kids. Rockpocket 04:40, 24 September 2006 (UTC)

Seriously, though. It is highly likely there is a lot more going on with the human nasal septum than just a barrier to divide the nasal cavity into two. In rodents for example, there is a vomeronasal organ (VNO), a septal organ of Masera [3] and a septal organ of Gruneberg [4], all of which are associated with the nasal septum. In humans, there is disagreement over whether there is an active VNO on the septum, but there certainly is a small, regressed pit where one used to be. There is still much to be learned about human olfaction, but recent work from Linda Buck and others are beginning to suggest there there are unusual types of olfactory receptor neurons, and the smart money is that they are, at least in part, associated with these unusual organs associated with the septum. So, i would say that there may well be a very important reason we have two nostrils - because the seperator is functional. We just don't quite know in what way yet. Rockpocket 04:56, 24 September 2006 (UTC)

You've found the missing link between Humans and Dolphins! --192.168.1.1 10:14, 23 September 2006 (PST)

How long? Run your thumb along the roof of your mouth until you feel the edge of the hard palate and the beginning of the soft palate. At that point the tip of your thumb is about 5-10mm (~1/3") past the posterior margin of the nasal septum. So, its nearly as long as your thumb. Why? Yes, 2 nostrils are better than one for snot management, but it also increases the surface area of the nasal mucosa (along with the nasal turbinates) which help to warm and humidify the air you breathe, making it nicer for your lungs to handle (think of the difference between mouth and nose breathing when you're in the snow). Happy olfaction! Mattopaedia 08:59, 28 September 2006 (UTC)

BTW, the reason it's hard to figure out from the pisture is because the nasal septum is not shown in that illustration. It only shows one of the lateral walls of the nasal cavity - those things are the turbinates, and under those are some entrances to some of the sinuses, particularly the maxillary sinus and frontal sinus. Mattopaedia 09:03, 28 September 2006 (UTC)

Why does some hair stop growing at a certain length?

Here's a question i've always been curious about: why does some hair stop growing at a certain length, while other hair will continue growing "indefinitely". For example, human head hair can be grown down to the toes (and beyond? Note that indefinitely is in quotes above because its possible a maximum length for human hair does exist, but I'm not sure), but if i cut my cats' fur, it will always grow back to the same length it was originally. Is this a function of the hair follicle itself "sensing" the length (via the weight perhaps??), or is it simply because the hair is shed at a fixed interval, which means the individual hairs cannot grow beyond a certain length before they are removed? The latter seems like a good theory, but human eyebrows are fixed length and dont seem to "shed".

Note that I know that the commonly-held belief of hair growing back thicker or longer every time it is shaved is false...no matter how many times you shave an eyebrow, it will always grow back to the same length and shape.

--65.105.3.194 16:22, 23 September 2006 (UTC)

You may know about one myth, but you've been fooled by another! I'm sure it's detailed somewhere along the lines in the hair and hair follicle articles, but hair doesn't stop growing at a certain length, it falls out at a certain lenth. There's a continuous cycle of hair growth, and when the time comes (signalled by a chemical message, apparently), it just falls out. Notice that short hairs almost never fall out! Most human hair has a max length, but only in a few rare cases can humans grow hair "indefinitely". Some kind of mutation I guess.  freshofftheufoΓΛĿЌ  16:30, 23 September 2006 (UTC)

How was I "fooled" by another? In my question I asked whether falling out was the reason. Also this answer is self-contradictory; why would a short hair never fall out...if falling out is the only way that hair length is regulated, wouldn't short hairs fall out more often than long ones? I'm willing to accept that a certain type of hair will grow at a constant speed and fall out after a constant period of time, thus regulating the length (this seems to be what happens with my cats hair for example), but it seems like in some instances this isn't the case (such as an eyebrow, which will grow back at a rate similar to the rest of the hair on the body when shaved, but then seems to stop without falling out). --17:33, 23 September 2006 (UTC)

Sorry, I might not have been clear. The question you ask about hair is actually a pretty common one. It's almost as if there's an urban myth that hair stops growing at a certain length. Short hair never falls out, because hair doesn't usually fall out until it gets longer. The hairs that fall out naturally are usually the longest ones.

Once again, with eyebrows it only appears that the hair doesn't grow. You lose a fair percentage of your eyebrows every day when you sleep (the long ones) and it is constantly regenerating. Apparently there is a relatively long phase at the end of an eyebrow hair's existance in where it doesn't grow much, but it's not that much different than any of the other hairs on your body.  freshofftheufoΓΛĿЌ  17:45, 23 September 2006 (UTC)

This is much clearer now, thanks. --24.159.108.105 19:44, 23 September 2006 (UTC)

It does seem odd that head hair would ever grow long enough to be in the way, particularly by becoming a tripping hazard. It's hard to see what type of evolutionary advantage such hair could possible offer. StuRat 17:11, 23 September 2006 (UTC)

Perhaps it was a mutation which occured after humans had developed the ability to cut their hair. It may have spread as a result of being attached to another beneficial mutation and because the hair could be cut, it was not a detriment to survivability. --65.105.3.194 17:41, 23 September 2006 (UTC)

An old roommate of mine hair that would "stop" growing after it reached her shoulders; she's never been able to grow it longer than that. This might seem freaky, but freshofftheufo apparantly has more than two eyebrows. Considering that temperature can affect the rate at which hair grows (that's the function of goosebumps), it and possibly other factors may also affect the rate at which other hairs grow so that perhaps body hairs grow a bit faster after you've just shaved because it's colder (or hotter?) but once the hair gets to a functional length the sensory demand for faster growth diminishes and the hair's growth rate stays slow until it falls out. AEuSoes1 20:17, 23 September 2006 (UTC)

I can assure you I have two very well formed eyebrows thank you very much! I was only pulling my information from the hair follicle article, where it states that eyebrows have a 9 months telogen phase at the end of the growth cycle in which the hair doesn't grow (as much?). Maybe I was interpreting the information incorrectly.  freshofftheufoΓΛĿЌ  05:21, 24 September 2006 (UTC)

I haven't had a haircut for about 15 years and my hair doesn't grow beyond my shoulders. Also about 15 years ago I stopped using shampoo to stop the fallout (which worked). A friend once told me my hair has split ends (figures, I love splitting hair :) ). So maybe split ends are nature's way of preventing hair from getting too long and shampoo undoes that. Did your rommate use shampoo and if she did, what kind? Maybe something 'light' like baby shampoo? DirkvdM 07:51, 24 September 2006 (UTC)

She still shampoos as far as I know. She's also black and repeatedly straightens her hair as well as trims it periodically to keep the split ends out. I'm trying to recall what she uses but I can't remember. She stopped sleeping with me and I stopped talking to her so I don't think I'll ask. AEuSoes1 08:36, 24 September 2006 (UTC)

Nuclear fusion in stars - iron

I've been told that the reason why elements heavier than iron are not produced in stars is because the fusion reaction becomes exothermic for iron nuclei. Can anyone explain why this is, and why such elements can be created in a supernova? Thanks 80.169.64.22 17:12, 23 September 2006 (UTC)

Isn't the most common hydrogen to helium fusion reaction also exothermic ? That's what keeps the Sun hot, isn't it ? StuRat 17:30, 23 September 2006 (UTC)

Iron is such a stable element (pays its bills, doesn't drink, is always on time for work) that there aren't any(?) ways to convert it to heavier elements that don't require adding energy. If you examine the diagram in Island of stability, you can see iron securely located in the "Stable Mountains". However, during a supernova, there's plenty of energy to go around! Clarityfiend 18:01, 23 September 2006 (UTC)

If I understand the articles about iron and nuclear fusion correctly, that's actually not true. It can be converted into nickel, but I don't really get what they mean by disintegration in the article. - Dammit 18:35, 23 September 2006 (UTC)

since it's the most stable element, eventually everything in the universe will end up as iron. THere was a nice popular science book about this called The Iron Sun.--Light current 18:46, 23 September 2006 (UTC)

That's weird. If ⁶²Ni is more tightly bound than ⁵⁸Fe (the iron isotope previously thought to be the most stable), why would it disintegrate (break down into lighter elements) so much faster? Clarityfiend 19:12, 23 September 2006 (UTC)

Heger and Woosley simulated various stellar collapse phenomena to estimate population production. Fig. 2 there has a region labelled "nickel photodisintegration" for very massive star collapse. Fig. 3 there shows suppressed formation of heavy elements. Figs. 4 and 5 show that this suppression is reduced for smaller initial masses. It is known that the silicon burning process can more or less allow jumping straight from ²⁸Si to ⁵⁶Ni. ⁵⁶Ni converts to ⁵⁶Fe in a couple of days. Note that ⁶²Ni is not accessible from any of the ⁵⁶Ni precursors by the alpha process and so would have to be results of synthesis of heavier elements which decay via beta radiation. Since silicon burning occurs so late in the process (last weeks before implosion) there's not enough time for enthalpically unfavored production of heavy isotopes and multiple decay to produce significant quantities of ⁶²Ni. Additionally, at such late times, the temperature in the star is high, ~ 3 GK. The island of stability shifts to slightly lower masses because heavy nucleii keep boiling off nucleons. -- Fuzzyeric 21:40, 23 September 2006 (UTC)

The Supernova article talks about what we're discussing. It confirms what I read somewhere: during the last stages, you end up with shells of increasingly heavy elements, an outer layer of hydrogen over a layer of helium over a layer of carbon...all the way down to a core of iron. Always liked that image. Clarityfiend 00:13, 24 September 2006 (UTC)

It goes like this:

If you compute the mass of a neutron and the mass of a proton individually, they come out to certain values. If you multiply each of these by two and add them together, though, this value is larger than the actual observed mass of a Helium nucleus (two protons and two neutrons). As the subatomic particles have fused, some of the mass has been converted into energy. This release of energy is what makes the reaction take place, and is what powers the very sun itself. This missing mass, or packing fraction, is most noticable between Hydrogen and Helium, but is still noticable as one moves up the atomic scale. That is... until Iron-56 (gasp)

At Iron-56, fusing nuclei together no longer releases energy, but necessitates it. Thus, when a star has gotten this far, it finds itself out of fuel and explodes (usually) in a nova or supernova. So much energy is present at this time that many atoms are driven "uphill" past Iron-56 to form heavier elements such as Zinc, Uranium, etc. Hope this explains things. See also Helium fusion, Fusion power, and best of all, read Atom: Journey Across the Subatomic Cosmos by Isaac Asimov where he explains this quite skillfully and in great detail. An excellent read overall, as well. Cheers, Dar-Ape (talk) 18:54, 23 September 2006 (UTC)

Also, check out this graph: Image:AvgBindingEnergyPerNucleon.jpg —Keenan Pepper 19:00, 23 September 2006 (UTC)

So from the graph, iron is bound to be the most stable and populous element in the universe?--Light current 21:44, 23 September 2006 (UTC)

Most stable doesn't imply most populous. There is more hydrogen-1 in the universe than any other nuclide, even though it's one of the least stable, simply because enough stars haven't been burning long enough. —Keenan Pepper 04:03, 24 September 2006 (UTC)

No I meant it will be. THats why I said 'bound' 8-)--Light current 01:48, 25 September 2006 (UTC)

And also, more complete data indicates that ⁶²Ni should win based on binding energy but won't due to its not being on the alpha ladder from ²⁸Si through ⁵⁶Fe. This article adapts Fewell's plot of data from The 1983 Atomic Mass Table. -- Fuzzyeric 05:13, 24 September 2006 (UTC)

So it's like the Bates Motel of fusion. Clarityfiend 17:59, 24 September 2006 (UTC)

Order of magnitude values

A probably silly question from a math dummy: on scales that use an order of magnitude between whole numbers (e.g., the Richter scale) what are the values for each decimal? With metallicity, a scale is used where +1 equals x10 solar and -1 equals 1/10th solar metallicity. If O.O = solar, does 0.1 = double solar, 0.2 = triple, etc.? What doesn't make sense to me is you reach x9 at 0.8 and the next order of magnitude at 0.9. Or, on scales like this, is there some other formula for increasing/decreasing values? Marskell 17:41, 23 September 2006 (UTC)

These scales are called Logarithmic scales, the article will probably make it a lot clearer to you. - Dammit 17:55, 23 September 2006 (UTC)

Or not. Basically, 0.1 translates to an increase of 10^0.1, 0.2 to 10^0.2...1.0 to 10¹. In other words, it's not a linear increase, it's an exponential one. So the difference between a 3.3 earthquake and a 3.4 one is smaller than that between a 3.4 and a 3.5. Clarityfiend 19:27, 23 September 2006 (UTC)

Or maybe either one depending on the particular scale? Clarity (sorry if this is really stupid) but 10^0.1 is simply equal to 10 yes? Thus any #.1 is redundant? Marskell 21:50, 23 September 2006 (UTC)

According to my calculator 10^0.1 is about 1.26 - Dammit 22:02, 23 September 2006 (UTC)

If you are using Windows XP, start the calculator (Start -> All programs -> Accessories -> Calculator). If the calculator has 27 keys, click on View, then click Scientific, so that you have 58 keys. To find a linear number that corresponds to a logarithmic number, type 10, then the x^y key, then the logarithmic number, then the enter key. So a metalicity of 0.1 would be about 1.26 times Sol, and a metalicity of 0.2 would be 1.58 times Sol. On other calculators, the key to use may be labeled x^y or just ^. --Gerry Ashton 22:55, 23 September 2006 (UTC)

I assure you WinXP isn't the only OS with a calculator. :) DirkvdM 07:54, 24 September 2006 (UTC)

10^0.1 = 1.26, 10^0.2 = 1.58, 10^0.3 = 2.00, 10^0.4 = 2.51, 10^0.5 = 3.16, 10^0.6 = 3.98, 10^0.7 = 5.01, 10^0.8 = 6.31, 10^0.9 = 7.94, 10¹ = 10. Notice that it goes up 0.26 in the first interval, but a whopping 2.06 in the last. Clarityfiend 23:55, 23 September 2006 (UTC)

Thx all. Marskell 05:59, 24 September 2006 (UTC)

One quibble: it's not that meaningful to use subtraction to say that it "goes up 0.26 in the first interval, but a whopping 2.06 in the last". It's a multiplicative scale, so the right way to think of it is that when it goes from 0.1 to 0.2 it's being multiplied by 1.259 (1.259 × 1.259 = 1.585), and when it goes from 0.2 to 0.3 it's being multiplied by 1.259 (1.585 × 1.259 = 1.995), and when it goes from 0.3 to 0.4 it's being multiplied by 1.259 (1.995 × 1.259 = 2.512), and so on, up to 0.8→0.9 = 6.310 × 1.259 = 7.943, and 0.9→1.0 = 7.943 × 1.259 = 10.000. (In other words, it's just a tenth-scale version of the way when we go from 1.0 to 2.0 it's a factor of 10, and from 2.0 to 3.0 it's another factor of 10, meaning that from 1.0 to 3.0 was a factor of 100. But this doesn't mean that 2.0 to 3.0 was "a whopping 90".)

Where does that magic number 1.259 come from? It's the tenth root of 10, or 10^0.1, or in other words, the number which, when you multiply it by itself 10 times, you get 10. —Steve Summit (talk) 04:05, 25 September 2006 (UTC)

It's meaningful to illustrate exponential behavior in fairly concrete terms to a self-confessed "math dummy", rather than overwhelm him with unnecessary details. Clarityfiend 07:44, 25 September 2006 (UTC)

User:Molecular Hamiltonian/monobook.js

Could someone please delete this? this is the 2nd time I've created a monobook that has kept me from editing. Since this monobook doesn't do anything other than crash my browser, could someone please delete it for me? since I can't edit it while logged in, I'm kind of at a loss--172.148.167.123 20:30, 23 September 2006 (UTC)

• In the mean time, I'll be using this one--Laplacian Operator 20:35, 23 September 2006 (UTC)

Tried deleting but page has been protected from editing for some reason. Youll have to contact an admin.--Light current 20:39, 23 September 2006 (UTC)

Okay, I deleted it. Can you log in and post here just to confirm that you're that user? -- Consumed Crustacean (talk) 20:49, 23 September 2006 (UTC)

Thanks, same person, no problems, from now on I'll be more careful with my monobook (; Molecular Hamiltonian 20:55, 23 September 2006 (UTC)

maps

where in the world can you find a map of where peacocks are living today?—Preceding unsigned comment added by 74.107.171.198 (talk • contribs)

'Bout time somebody asked this question! - R_Lee_E (talk, contribs) 22:53, 23 September 2006 (UTC)

Well I guess it's about time somebody answered it then. First off, peacocks don't survive long without their friends, the peahens. So what you want to search for is Indian peafowl. This and the peafowl article have some information about their range. If you search Google and Google images for "indian peafowl map" it comes up with several results that will answer your question.--Shantavira 13:53, 24 September 2006 (UTC)