Science desk
< November 7	<< Oct \| November \| Dec >>	November 9 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

November 8

Convict Fish

In the BBC Life natural world documentary about fish there is a segment about a fish called the "convict fish", this is the clip [1] and I was trying to find out more about it and the like, however searches only ield the convict chichlids, which seem quite different and no where near as peculiar. Any idea what its taxonomical name is / any info on it? MedicRoo (talk) 01:09, 8 November 2009 (UTC)[reply]

See this page. We have an article at Pholidichthys leucotaenia, but it's not very informative. Deor (talk) 01:57, 8 November 2009 (UTC)[reply]

Nimh substitute for Nicad?

Can an Nimh battery of the same voltage and millampere-hour rating be used to replace a Nicad battery in a cordless phone? I thought not, because the charging arrangements can be different. Edison (talk) 03:59, 8 November 2009 (UTC)[reply]

Yes. I've done it many times, it works just fine. And you can (and should) use a larger milliamp rating too. Most (well, all the ones I looked at) cordless phone batteries are just AA or AAA batteries attached to a special connector. You can buy regular rechargeable AAA NiMh batteries and use them (it's a bit tough to solder - don't let the battery get hot). Ariel. (talk) 05:24, 9 November 2009 (UTC)[reply]

Kinematics Question

A rock is dropped from a seacliff and the sound of it striking the ocean is heard 3.0 s later. If the speed of sound is 340 m/s, how high is the cliff? The answer is 41 meters, but I don't know how to get it. —Preceding unsigned comment added by 174.6.144.211 (talk) 04:21, 8 November 2009 (UTC)[reply]

Call the height of the cliff h. Then work out how long it will take to fall that far (remember your constant acceleration formulae?) and how long it will take for the sound to travel that far. Then add them together and set it equal to 3 and solve for h. --Tango (talk) 04:35, 8 November 2009 (UTC)[reply]

The question is ambiguous and poorly framed because it does not specify where the listener is. The sound might well be heard by someone at the water level below the cliff, or 100 meters from the cliff in a boat, just as it might be heard by someone on the cliff. Edison (talk) 04:53, 8 November 2009 (UTC)[reply]

The question only makes sense if they are listening from the top of the cliff. If they are at the water level below the cliff it would be a trick question, since the sound thing wouldn't add anything. If they are in a boat elsewhere on the water then there isn't enough information in the question to answer it. While it would be good if the question were more precise we shouldn't over analyse things - if it is obvious what the question means then it is much better to answer it rather than pick holes in it. --Tango (talk) 05:15, 8 November 2009 (UTC)[reply]

(ec) The speed of sound is a red herring. Solve the problem assuming the speed of sound is instantaneous and see how it works out. Matt Deres (talk) 04:56, 8 November 2009 (UTC)[reply]

I assume the problem is that the correction for the error caused by the speed of sound is kind of nonlinear. If you simply subtract away the error caused and add it back to the time (say 3.12 seconds) that changes your answer and now you have to iteratively calculate your solution. John Riemann Soong (talk) 05:01, 8 November 2009 (UTC)[reply]

The additional term is absolutely not nonlinear (even though it is quadratic). Note the important difference between the meanings of linearity. Your suggestion to use an iterated approach indicates that you are confusing the definitions. This physics problem is described by a quadratic equation in time (t) (or in height h), which is (strangely) a linear equation. It is unfortunate that the word "linear" has two different meanings. The solution, though, is obtained by standard quadratic formula or as shown below by Gandalf and others. Nimur (talk) 17:14, 8 November 2009 (UTC)[reply]

That's not true. The speed of sound will make a difference. If the speed of sound were instantaneous you would get a larger value for the height of the cliff. --Tango (talk) 05:15, 8 November 2009 (UTC)[reply]

Wrong direction oops. But you might still see what I'm saying. The order of magnitude of the error looks to be around 10%, though the precise value of that error is unknown to me. (What is the error of the error?) John Riemann Soong (talk) 06:00, 8 November 2009 (UTC)[reply]

The speed of sound is definitely not a red herring, as sound moves slowly enough to make a significant impact on the solution. The speed of light, however, is fast enough to be assumed to be instantaneous in such a problem. StuRat (talk) 13:08, 8 November 2009 (UTC)[reply]

The sound is going to reach you in about a tenth of a second; just barely within the significant figures provided. I agree that you could add that to the equation, but I don't see the point if you're not going to deal with much larger factors such as the placement of the dropper/listener(s) and the drag from the air, including up-drafts - not to mention the height of the waves below. Matt Deres (talk) 20:08, 8 November 2009 (UTC)[reply]

A nice problem. Expanding on Tango's approach: If the rock takes time t1 to fall the height h and the sound takes a further time t2 to reach the top of the cliff (making the assumption that the listener is at the top of the cliff) then t1 + t2 = 3 s. Substitute expressions for t1 and t2 in terms of h and you get an equation in h and sqrt(h). Rearrange and square both sides and you get a quadratic in h. Solve and throw away a spurious solutuion introduced by squaring both sides, and you are left with a value for h that is a little under 41 m. If you assume t2 is 0, then you get 44 m, which is a significant difference. Gandalf61 (talk) 13:30, 8 November 2009 (UTC)[reply]

Or you can do it in an iterative fashion, as John Riemann Soong suggested. That makes the math much simpler, especially if you only want a couple significant figures (and, really, any more than that is spurious accuracy anyway, when someone is timing how long it takes them to hear something, which can't be very accurate). In the case of a multiple choice test, you also have the option of plugging in the values provided to see which gives you the right total time. (You might say this is cheating, but knowing how to check the answers provided by others is also a valuable skill.) StuRat (talk) 17:53, 8 November 2009 (UTC)[reply]

I would strongly recommend that technique in multiple choice tests. Trying to work out the right answer is often a waste of time, it is much quicker to just work out which of the options is right. Even if you can't narrow it down to one answer all you need to do is rule out one or two options on each question to make random guessing of the other options likely to give you a pass mark (obviously it depends on the pass mark, how many options you are given and whether marks are deducted for incorrect answers). Maths isn't usually done using multiple choice, though, since the method is usually more important than the final answer. --Tango (talk) 19:03, 8 November 2009 (UTC)[reply]

are quantised values always rational?

The cardinality of the continuum provides much more than the cardinality of the natural numbers... if the wavelength of a photon is 1500/pi nm, and a HOMO-LUMO gap of a fluorescent molecule is 1500/(pi+1/e^99) nm, will excitation occur? John Riemann Soong (talk) 05:06, 8 November 2009 (UTC)[reply]

I think the difference in length there (I haven't actually calculated it) will be less than the Planck length, which means they can be assumed to be equal. Quoting lengths at greater precision than +/- the Planck length is pretty meaningless. --Tango (talk) 05:19, 8 November 2009 (UTC)[reply]

That said, the largest difference that still allows for excitation is probably larger than the Planck length, otherwise it would hardly ever occur. I know virtually nothing about HOMO-LUMO gaps, though, so I'll have to let someone else give a more specific answer to that problem. In general though, nothing is exact in physics, so we can always choose to use a countable subset of the real numbers (eg. truncate all decimals after a few hundred places, it won't make any difference to anything - a few dozen places would probably suffice for almost everything). --Tango (talk) 05:23, 8 November 2009 (UTC)[reply]

Basically I wonder how it is possible that when I set my photoreaction chamber for say, 380 nm, how it's possible that orbital excitation ever be possible. I understand that quite a sharp distribution can be produced, but how likely is that the photons generated will ever be of the 380.0501292201285028(...) nm wavelength required for excitation? I understand that the Boltzmann distribution and random heat may kick in and provide that little more or little less of energy required to make the transition (e.g. within a certain difference environmental fluctuations can "compensate"), but then this also asks the question whether the environmental fluctuations themselves are rational. I know that a lot of time the Boltzmann distribution of molecular velocities spans such a large range of possible values that we call it a continuum -- but are those values countable? John Riemann Soong (talk) 05:53, 8 November 2009 (UTC)[reply]

There is an energy-time uncertainty analogous to the position-momentum uncertainty (Heisenberg's uncertainty principle), so a energy difference of ΔE = c*h/(1500*e^99) = 1.34E-71 J means that the "uncertainty in time" - that is, the time it takes for the transition to occur, is Δt = ℏ/(2 ΔE) = 3.94E+36 seconds. So unless your typical observing times are of that order of magnitude, you don't need to worry about such a deviation. Icek (talk) 11:47, 8 November 2009 (UTC)[reply]

And you can think of it in a different way: If you have your light source producing the light of this specific frequency for some relatively short time period (like a second, an hour or a year). The actual frequencies that the light source produces are determined by a Fourier transformation of the amplitude of the electrical field; for your frequency f, the electrical field is ideally sin(2πf t), but the light source is only turned on for a finite amount of time and has some intensity profile when being turned on and of (usually not going from zero to full intensity and back down to zero abruptly, but even if it does, my argument still applies), and therefore some function g(t) which is zero for the time the light source is turned off has to be multiplied with sin(2πf t) in order to get the real amplitude of the electrical field. Now Fourier transform the resulting function - you will get a frequency spectrum with a sharp peak at your frequency f, but the width is still far larger than the deviation in wavelength/frequency you mentioned. Icek (talk) 12:06, 8 November 2009 (UTC)[reply]

And of course, the relevant time is not the time the light source is turned on, but its coherence time, if it is shorter. Icek (talk) 12:11, 8 November 2009 (UTC)[reply]

I still have the issue of cardinality. The frequency spectrum may be treated as continuous, but are the actual elements of the spectrum rational? The number of elements must be finite: if I have a 100W bulb releasing 380 nm (ish) photons, there can only be 1.9 * 10^20 photons that are released every second. But the number of possible values near 380 nm is uncountably infinite. Why should any of those photons match my energy gap? John Riemann Soong (talk) 14:49, 8 November 2009 (UTC)[reply]

Countably vs. uncountably infinite is irrelevant here. Photons do not have a precise frequency. It's only possible to measure the frequency of a photon for some amount of time that isn't arbitrarily large. So from the Heisenberg uncertainty principle, it's only meaningful to talk about the frequency of a photon to a precision that isn't arbitrarily small. It's not like the difference between two rational numbers, which can be arbitrarily small. Put another way, the number of distinguishably different frequencies that a photon can have isn't even countably infinite. There exists a very large but finite set of frequencies, such that any photon will always be measured to match a nonempty subset of those frequencies, accurate to the uncertainty of the frequency of the photon. Red Act (talk) 15:43, 8 November 2009 (UTC)[reply]

The key point is that they don't need to exactly match your energy gap since nothing is exact in physics. You can work out a degree of precision that is high enough to make absolutely no difference to your experiment (not just negligible difference, but no difference, since the error will be less than the inherent error caused by the laws of physics). That means you can always model the frequencies as rational numbers (by truncating the decimal expression, say). Alternatively, you could model them as irrational numbers of the form pi*q (or any number of other forms) where q is rational, in which case there are still only countably many. So, the physical quantities aren't inherently rational numbers, they are just imprecise enough that we can model them using a countably subset of the reals - we get to choose which subset, within reason. --Tango (talk) 19:11, 8 November 2009 (UTC)[reply]

Interactions rarely happen in isolation. In the case of a photon striking a molecule, for example, in addition to electronic excitations, you can also transfer kinetic energy, and excite rotational or vibrational states. The energy in the transition can be fuzzy in part because it doesn't all have to come from or go to the same place. Dragons flight (talk) 20:41, 8 November 2009 (UTC)[reply]

pharmaceutical questions

Q 1.what are the application of two and three component system in formulation of liquid dosage form? Q 2.what are the application of HLB in the formulation of W/O and O/W emulsions?

thank you for your help,

your sincerly

getye —Preceding unsigned comment added by Getisha (talk • contribs) 11:54, 8 November 2009 (UTC)[reply]

This sounds a bit like a homework question - if that is the case, then as suggested at the top of this page please show where you've gotten stuck after exerting some effort, and we'll try to get you un-stuck. In any case, we do have a very informative article on emulsions, which includes medical/pharmaceutical applications. -- Scray (talk) 15:13, 8 November 2009 (UTC)[reply]

aminoacyl-tRNA bond

How in the world is this a "high energy bond"? A sugar alkoxide doesn't look like a very good leaving group to me -- and IIRC the amino acid is not attached via the anomeric carbon. John Riemann Soong (talk) 12:11, 8 November 2009 (UTC)[reply]

You should have access to your own reference desk for all these scary questions with words of which I've never heard! DRosenbach ^{(Talk | Contribs)} 15:02, 8 November 2009 (UTC)[reply]

I don't know the answer to the question, but Aminoacyl tRNA synthetase and Transfer RNA say the amino acid is attached as an ester linkage on 3'OH terminus of tRNA chain, which is an adenosine ribose ring. Man, our Transfer RNA and Aminoacyl-tRNA articles need help--so full of vague biochemistry descriptors of the actual amino-acyl key! DMacks (talk) 06:38, 9 November 2009 (UTC)[reply]

A sugar alkoxide ... I can't see how RO- is a good leaving group simply because it's on a sugar. John Riemann Soong (talk) 10:53, 9 November 2009 (UTC)[reply]

As far as a leaving group goes, the enzyme probably protonates it prior to/simultaneously with expulsion (a very common tecnique in enzyme catalysis). So it's not really the alkoxide leaving, but an alcohol. I'm not sure who classified it as a "high energy bond", but it's certainly on par with, if not higher energy than, the final peptide bond, and certainly higher energy than the free amine plus free carboxylic acid. For what it's worth, the Aminoacyl-tRNA article mentions "the net reaction is only energetically favourable because the pyrophosphate is hydrolysed" -- 128.104.112.237 (talk) 00:16, 11 November 2009 (UTC)[reply]

Occam's razor + envenomation immunity

How would envenomation immunity be preferentially beneficial for only the vast minutia of the species, such as mongooses and opossums? I'm not attacking evolutionary theory per se...I was just wondering what someone from a non-theistic perspective would say. Pardon me if this is merely another form of an 'inexplainable intermediate form' question.

Ah, yes -- second question -- is this merely another form of the same question that begs to know how inexplainable intermediate forms can lead to otherwise explainable end forms? DRosenbach ^{(Talk | Contribs)} 15:18, 8 November 2009 (UTC)[reply]

The key phrase from the article is "feeds on snakes". Only a minority of species feed on snakes. Looie496 (talk) 15:38, 8 November 2009 (UTC)[reply]

In case it's not obvious to all readers, Looie496 is referring to the section on immunity among other animals in the Snake venom article. I agree that this is pretty straightforward. One could surmise that some animals got into fights with venomous snakes, and those that had genetic variations that favored survival in such encounters lived to reproduce, and some of those animals gained a food source (venomous snakes). It's clear that frequent and intense selective pressures strongly drive evolution. -- Scray (talk) 15:57, 8 November 2009 (UTC)[reply]

So it's because of the overwhelming environmental pressure on snake-eaters to be able to profitably and safely consume their dinners that such mutations would be favored -- whereas non snake-eaters, albeit standing to benefit at large from envenomation immunity, have no such pressure (in the form of removal by of individuals by envenomation death from the population who do not possess such a mutation).

Would it be begging a question to ask why any other organisms don't develop protective entities against their main damage source -- like to ask why all pinnipeds don't possess defensive tusks to defend against shark attacks? Basically, the concept of natural selection allows us to explain the world -- but we do not use it to ask why not? DRosenbach ^{(Talk | Contribs)} 16:04, 8 November 2009 (UTC)[reply]

For something to be selected for, it needs:

a) to be present in the population (via mutation, for example)

b) things with it to produce more surviving offspring than those without it, in the form it currently is (not the form that will be arrived at after 100 generations of selection)

Part of (b) is not only that there is an advantage to this thing, but that the advantages outweigh the disadvantages including use of resources. It is perfectly valid to ask 'why not?', but very often the answer is that the cost (in terms of food/energy, time, etc) is greater than the advantage. Organisms do develop protective measures against their main damage source if this reduces the number of surviving offspring, but some forms of defence interfere with others. For example, the Leopard Seal. Leopard seals are very fast swimmers, and they have all sorts of adaptations to allow them to swim faster. They also have sharp teeth, which do not interfere with their ability to swim fast. Let us suppose a leopard seal were born with the beginnings of defensive tusks: larger teeth which did not fit completely inside its closed mouth, but protruded. This leopard seal would be less streamlined than other leopard seals, and thus slower at swimming away from sharks. It would be more likely to be caught by an attacking shark, and those little semi-tusks would be very little good at stopping the shark from eating it. So, this seal with proto-tusks would be less likely to survive to produce offspring, because the proto-tusks would interfere with the existing defence method (swimming very fast). On top of this, there may be other risks associated with proto-tusks: they might be more likely to break than normal teeth, putting the seal at greater risk of infection; they might interfere with normal eating; they take more energy to produce than normal teeth, energy that could be put to other uses.

So, it is fine to ask 'why not?': a lot of progress is made by considering 'why not?'. 86.142.224.71 (talk) 16:45, 8 November 2009 (UTC)[reply]

There are two answers here - both have an influence:

The problem is that things like venom protection requires energy to grow and maintain - in times of famine, animals that have only what they need to survive will out-compete animals that have all of these rarely-useful but energy-consuming adaptations.

Evolved features don't just stay there when they aren't being actively used. If some species evolves venom protection - but isn't gaining much of a benefit from it, then random mutations that result in that venom protection becoming weaker would not be selected against...the amount of protection would then gradually decrease just from random changes. This reversal of adaptation is actually a surprisingly fast process because mutations that destroy some adaptation are vastly more likely than adaptations that produce benefit. It's like if you took a large hammer and started smashing bits in the engine compartment of your car...what are the odds that a particular random hammer blow will improve the performance of your car? Almost (but not quite) zero. Almost all hammer blows will make the car worse. However, if you never drove your car at night, hammer blows that smashed headlamps or broke the wiring to the headlamps or the switches that turned them on and off would be neither beneficial nor any immediate loss to you. Hence, suppose some species of mouse had venom immunity - it would pretty soon get random mutations in some individuals that would destroy or reduce that ability. In an environment where poisonous snakes were everywhere - the mice without the genes for venom immunity - or with weakened immunity - would be eliminated from the gene pool pretty quickly - so the survivors would all retain that protection - and their offspring would remain venom-protected. But if there are no venomous snakes around, then there is no such corrective mechanism. A mouse with a genetic mutation that wipes our (or reduces) their immunity suffers no disadvantage and can spread it's weakened protection into the next generation. Because destructive mutations so vastly outnumber benefitial ones - any currently useless features can be knocked out without anything preventing the resulting 'defective' animals from reproducing.

For the pinnipeds - you can't just evolve tusks without evolving a stronger skull to absorb the impact - also a more muscular neck perhaps? So this is not necessarily a single, simple adaptation. Perhaps the tougher (larger) skulls prevent efficient live birth out on the ice? Can they still efficiently catch and consume fish with those big tusks out there? Will the lack of streamlining result in them failing to out-swim sharks and then suffering even worse predation? This is a really complex question - and it's by no means obvious that tusks would be an advantage. The answer is very likely that tusks are not a winning option for these animals - or they'd probably have them by now.

SteveBaker (talk) 16:54, 8 November 2009 (UTC)[reply]

Being resistant to snake venom is only one way to avoid harm from snakes. Another is to avoid them or kill them. Being resistant isn't a perfect strategy, as you're still getting attacked and bitten by snakes.

It might help to look at how it is that mongooses are resistant to snake venom. Most snake venoms, three-finger alpha-neurotoxins, act on the nicotinic acetylcholine receptors (nAChRs) to block them, which messes with the prey's neurotransmission. Mongooses and other resistant species have an altered binding site in their nAChRs that doesn't allow the toxin to bind but still allows acetylcholine to bind.[2][3]

If a species doesn't regularly get bitten by snakes due to their absence or having other means to combat snakes, there won't be a selection pressure on the nAChRs to develop resistance to snake venom binding. Mongooses eat snakes and hedgehogs rely on hiding and rolling up rather than running away, so will face snake bites. Snakes also have acetylcholine receptors that are resistant to snake venom, as they also get attacked by snakes! Because acetylcholine receptors have things to do other than avoid binding snake venom, like neurotransmission, there is a balance between doing their normal job and becoming resistant to snake venom.

There's also the issue of evolutionary arms races - snakes that don't have their venom work on their prey won't be as successful as snakes whose venom still works, so snake venom will evolve to avoid resistance and still be effective. The selection pressure on snakes to have effective venom will be greater than on most of their prey to be resistant, as every snake needs to feed but not every prey animal in most species needs to be resistant to snake venom, as they rarely get bitten. Fences&Windows 18:52, 8 November 2009 (UTC)[reply]

Two questions: swine flu in rodents, and ladybugs in November

Hi. Can the pandemic H1N1 virus infect rodents (esp. squirrels)? Also, how common is it to see a ladybug in November at my location? I'm from Southern Ontario. Thanks. ~A H 1^(TCU) 15:44, 8 November 2009 (UTC)[reply]

You might want to post such different questions separately. As for the ladybug part, I'd say it's uncommon, but we have had some rather warm weather this November, and I've even seen mosquitos this November, right across the Detroit River from you, here in Michigan. StuRat (talk) 17:44, 8 November 2009 (UTC)[reply]

Rodents are used in some lab experiments on influenza, e.g. [4], so many flu strains can infect rodents, but the majority of wild mice are resistant to influenza:[5]. There has been one report of a wild mustelid, a stone marten, being infected with H5N1:[6] It's birds, especially aquatic birds, and pigs that are the main animal reservoirs for influenza:[7]. Rodents and squirrels may carry plague, rabies and Hanta virus among other infections. Fences&Windows 18:20, 8 November 2009 (UTC)[reply]

The reason I ask the question about the squirrel is because I recently saw a dead squirrel on the grass, with no signs of trauma or injury (although its jaws were open). Although I had seen a squirrel as roadkill (that wasn't it) and also a different dead squirrel years before, I was curious to what disease, if any, the squirrel may have had that led to its death. My first thought was H1N1, but that would only be possible if they can contract the virus.

About the ladybugs, I've already seen two today, and perhaps one last evening. It has been a very warm day today considering it's November. Do ladybugs usually move somewhere else when it's cold, and does prolonged warmth in late autumn have an effect on its wintertime distribution? ~A H 1^(TCU) 19:34, 8 November 2009 (UTC)[reply]

So the question is really, "What can a squirrel have died of if there is no obvious external trauma?". Not sure. Poisoning is possible.

Ladybirds/Coccinellidae will find somewhere warm to overwinter, like roof spaces, where they go into diapause. Fences&Windows 23:18, 8 November 2009 (UTC)[reply]

I'm in Toronto and I've seen ladybugs in my house a few times in the last couple of weeks, which is unusual. --Anonymous, 05:08, November 9, 2009.

The ladybugs are not the normal ladybugs, but rather Harmonia axyridis, the Asian Lady Beetle. They're slightly larger than our native ladybugs and are considered a pest. Not only do they like over-wintering indoors, the little buggers bite. Squish 'em! Matt Deres (talk) 01:19, 13 November 2009 (UTC)[reply]

Proteus Syndrome or Brittle Bone disease

This is just my science fiction idea, not a real medical question. Why not replace the entire skeleton of individuals with these disfiguring bone conditions with titanium? Trevor Loughlin 80.0.101.110 (talk) 15:59, 8 November 2009 (UTC)[reply]

Ah -- titanium is great, but it is nonresponsive and non-sensory. Bones may appear to be nothing but "framework," but they are dynamic centers of growth and regulation. Not to mention a tremendous (and perhaps the only) store for calcium when the body needs it for things like nerve and muscle electrical conduction and growth. Titanium cannot replace that. DRosenbach ^{(Talk | Contribs)} 16:08, 8 November 2009 (UTC)[reply]

Also, the interior of the bones is the place where blood cells are manufactured. Consider the chemical environment that requires - how blood gets into and out of the marrow. SteveBaker (talk) 16:20, 8 November 2009 (UTC)[reply]

Who needs blood cells? I'll have a rack of titanium ribs with a tall glass of hemolymph? DRosenbach ^{(Talk | Contribs)} 17:55, 8 November 2009 (UTC)[reply]

The article bone has quite a list of uses for bone - protection and structure aren't the only ones. Vimescarrot (talk) 16:54, 8 November 2009 (UTC)[reply]

Bones are "alive", not dead, and they play an important role in a number of bodily processes. You can no more just replace your bone with metal than you could your eyes or stomach. --Mr.98 (talk) 18:42, 8 November 2009 (UTC)[reply]

I was thinking about it, and returned to post another comment. Titanium replacement parts are not entirely fantasty ala X-Men. I cannot speak of joint replacement (for that is not my field, yet it certainly does occur), but as this idea relates to titanium dental implants, they do not work just like teeth do. Teeth are attached to and supported by the alveolar bone via the periodontal ligament fibers, -- basically modified Sharpey's fibers. These allow for a number of things, including proprioception and cushioning of occlusal (biting) forces, things that are absent with implants, which are osseointegrated, essentially ankylosed to the bone into which it is placed (bone fuses to titanium directly). Ask anyone with enough missing teeth that were restored with implants so that they are able to differentiate the feel of biting on the implants from biting on their natural teeth and they will tell you it doesn't feel the same. That's why dental gurus like Gordon Christensen say 'Implants don't replace teeth -- they replace no teeth." DRosenbach ^{(Talk | Contribs)} 19:12, 8 November 2009 (UTC)[reply]

I do think that full-skeleton replacements probably are via X-Men, though. You do need your bones, even if some of the smaller bits can be replaced. --Mr.98 (talk) 00:03, 9 November 2009 (UTC)[reply]

I saw a film of a Proteus Syndrome woman with enormous legs, and a Proteus Syndrome small girl with a deformed face, but the rest of their body was unaffected, so there would be room for bone marrow elsewhere. For example, could a complete skull be replaced, bearing in mind that it has only one moving part, the jaw?80.2.205.110 (talk) 10:03, 9 November 2009 (UTC)Trevor Loughlin[reply]

Can alpha-carbon protons (in a carbonyl/ketone) hydrogen bond?

Do they undergo deuterium exchange? Do they form hydrogen bonds? How would you explain why dibenzyl ketone has a much lower MP than benzil? John Riemann Soong (talk) 16:50, 8 November 2009 (UTC)[reply]

This is just a guess, but maybe benzil's planarity allows much closer packing and so more van der Waals interactions. Dibenzyl ketone has a more awkward 3D shape that probably doesn't allow many VDW forces to form. --Mark PEA (talk) 18:34, 8 November 2009 (UTC)[reply]

Furthermore, I would speculate that the full conjugation in benzil allows the pi-systems to participate in intermolecular bonding to a much greater extent than in dibenzyl ketone, where there are two sp³-hybridized carbons which break the conjugation, and do not allow intermolecular pi-molecular-orbital interactions as fully. Come to think of it, this is pretty much the exact same thing as the planar/non-planar difference noted by Mark PEA, but from a different perspective. --Jayron32 04:33, 9 November 2009 (UTC)[reply]

Would be interesting to find X-ray crystal structures of the two solids to see exactly how they pack. DMacks (talk) 06:31, 9 November 2009 (UTC)[reply]

The crystal structure of benzil has been determined several times, the most recent one I could find was this: Acta Cryst. (1987). B43, 398-405. Benzil is not planar in the solid, the OCCO dihedral angle is about 108°. I couldn't find a structure determination for dibenzyl ketone.

Ben (talk) 16:04, 10 November 2009 (UTC)[reply]

Calculated Error query

Okay so I have a slight problem, I'm trying to calculate extension error of a Hounsfield Tensometer calibrated too 0.1mm

So i know to calculate extension I'll use the equation 0.01*100 then i have to devide the answer of that by a value, But i've forgotten what value i should use, Should it be the number of measurements up to the elastic limit or the total number of measurements or what? Help please! —Preceding unsigned comment added by 92.18.81.39 (talk) 21:11, 8 November 2009 (UTC)[reply]

You can buy the article Push-pull fatigue modifications to a Hounsfield tensometer which costs $30. Cuddlyable3 (talk) 22:04, 8 November 2009 (UTC)[reply]