Science desk
< January 3	<< Dec \| January \| Feb >>	January 5 >

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.

January 4

Observable universe

I don't understand this. If the universe is only 13.75 billion years old. Then how can we possible see something that further than 13.75 billion years light away up to 40 billion years light? Anything further than 14 billion light years would indicate light didn't have enough time to travel from that far to Earth since it would takes a lot more time for it to get to the Earth. 174.20.15.246 (talk) 03:16, 4 January 2013 (UTC)[reply]

(First), we can see in two directions, and (second) whatever we can see in one direction that is 13.75 billion years old has also moved some 6 million additional years away from us. So the visible universe is 13.75 plus 6-plus billion years times two across. 04:02, 4 January 2013 (UTC)

I already accounted for it, base from the article. The observation from any direction would be 90 billion light years from Earth. How is that possible?174.20.15.246 (talk) 04:15, 4 January 2013 (UTC)[reply]

It's been a while since I've looked at astronomy, but if I remember correctly, this is because the universe was smaller when the light was emitted, and as the light travelled, the universe expanded, increasing the distance between the source and you (the eventual viewer of the light), so by the time the light reaches you, it's travelled a much greater distance than when it first left; in fact, some sources may even have moved so far away that they are more than 14 billion ly away (meaning that if the universe stopped expanding at this moment, and the distance was kept constant, it would take whatever that amount of time to get here). To use an analogy: if you have a deflated balloon and point A and B, with an original distance as 5cm, and an ant (our "light particle/photon) starts travelling from A to B, but you then expand the balloon as the ant walks. The ant is still walking towards B, but the distance between the two is increasing, so that by the time the ant reaches B, the distance may no longer be 5 cm, but rather something like 100cm (it's a huge balloon, let's say). Now look at the amount of time the ant spent walking. If we stopped the balloon's expansion, and had another ant walking at the exact same speed from Point A towards Point B, after the same amount of time, that ant would not have reached point B like the first ant did. This is the same way that a light particle from a 40 Billion ly distant star could now not have reached us in the amount of time, but because the universe was smaller and the source closer when the light was emitted, the particle did reach us (although this travelling over stretching spacetime distorts lightwaves, if I remember correctly, which is what causes redshift). Bramble claw x 04:17, 4 January 2013 (UTC)[reply]

OK according to my understanding, let put it this way. If we see something 13 billions years light away, that means it was 13 billions years light from Earth 13 billion years ago or billions years ago right? So its "current location" right now is much further away due to the expansion of the universe. So right now it could be at 90 billion years light away but apparently we can't see its current location rather what we see is its past. I guess their definition of what is observation is different from mine. So let me ask a question: is it possible for us to see something that is like 20 billion lights years away at its current location right now?174.20.15.246 (talk) 04:55, 4 January 2013 (UTC)[reply]

I wouldn't say it was exactly 13B ly away when the light was emitted. You could probably accurately say it travelled for 13 billion years, but the distance would be relative; in most cases, you would need to look at the rate of expansion of the universe to know the exact distance it travelled. But the idea seems more or less right. For the second question, no. If it's 20 billion light years away, to see it in it's exact location right now, we'd have to perceive the photon emitted at this instant, which will not arrive here until at least 20 billion years (but longer because of the universe's expansion) because 1 lightyear is the distance light can travel in 1 year; to see it right now, the light would somehow have to cover a distance it normally takes 20 billion years to traverse in an instant. Bramble claw x 05:33, 4 January 2013 (UTC)[reply]

Well to keep things simple, I just make some assumption to say it is 13b ly when the light was emitted but I also mention other possibility when I said "billions years ago right?". I understand it now but I got to say that the definition of observable universe is pretty hard to understand especially for layperson or even someone with some knowledge of physic. This makes me think of how much "limited we are from the real universe", everything we are seeing right now are all the past! It's funny when I think about this so let say I see a person stands next to me but what I see is just an image that was carried by light in an instant so that image technically is not exactly the same at the moment when I see it but sometime in the past, very small though, like .00000...1 second ago. No matter where we look at the universe, we always look in the past but the difference the past here can ranging from few seconds ago to billion years ago.174.20.15.246 (talk) 06:00, 4 January 2013 (UTC)[reply]

That's essentially right, but the chances of it being defined as actually "looking into the past" tend to decrease in proportion with the increasing knowledge of the person talking. Special relativity made it fashionable to eschew any notion of absolute time. (And please don't edit others' comments; "travelled" is UK spelling.) Evanh2008 ^{(talk|contribs)} 06:25, 4 January 2013 (UTC) [reply]

With that, I would like to thank StuRat for fixing the formatting error in my post above. Irony is many-splendored thing. :) Evanh2008 ^{(talk|contribs)} 06:44, 4 January 2013 (UTC)[reply]

This has been discussed before. Someone correct me if I'm wrong, but we can't see anything 40 billion light years away. The universe is bigger than 13.7 billion light years because the speed of light is with respect to space, and space has been expanding rapidly. It cah expand faster than the speed of light. Bubba73 ^{You talkin' to me?} 07:19, 4 January 2013 (UTC)[reply]

We can see things that "now" (if such a concept could be defined) are more than 16 billion light years away, but the light that we are now perceiving from those objects began its journey toward us when the objects were significantly closer (less than 16 billion light years away). Evanh2008 ^{(talk|contribs)} 07:27, 4 January 2013 (UTC)[reply]

Microscope image identification

The above images were photographed through the lens of an optical light microscope at 400x magnification. Both are water samples originating from a stormwater management facility (aka storm pond) in Richmond Hill, Ontario, Canada. The second sample was boiled in a glass beaker on a hot plate (reaching a temperature of 94 degress Celsius for 3 minutes). My questions are: is the first one, as I suspect, a Navicula diatom, or some other organism? And the second question is what are the crystals in the second image? It only appeared in water sample which were boiled (appearing as a thin, milky coloured film on the water surface), which led me to rule out beaker contamination, so I hypothesized that it was crystalized cellular debris from lysed cells; am I right, or are the crystals of another origin, such as dissolved minerals crystallized by heat? Your answers are greatly appreciated, Bramble claw x 03:56, 4 January 2013 (UTC)[reply]

Boiled at 94°C ? Are you sure you don't mean the water was evaporated at that temperature ? StuRat (talk) 06:39, 4 January 2013 (UTC)[reply]

Let me correct myself, I tried to boil the water, but it only reached 94°C and did not rise beyond that point. So more correctly phrased, the water was heated to 94°C for 3 minutes. Bramble claw x 17:23, 4 January 2013 (UTC)[reply]

Without knowing the chemical composition of the water, I'd guess that it's the dissolved minerals crystallising out of solution.--TammyMoet (talk) 10:21, 4 January 2013 (UTC)[reply]

So heat can cause minerals to crystallize? Bramble claw x 17:23, 4 January 2013 (UTC)[reply]

See Precipitation. Matt Deres (talk) 18:26, 4 January 2013 (UTC)[reply]

I know what precipitation is, thank you. The article says nothing about heat causing precipitation; it's a chemical reaction that causes precipitation. Are you implying that the heat catalyzed a reaction? Bramble claw x 20:19, 4 January 2013 (UTC)[reply]

Well, the heat causes the water to evaporate, increasing the concentration of the solutes to the point where they precipitate out of solution. StuRat (talk) 05:54, 5 January 2013 (UTC)[reply]

And likely precipitates from water are calcium carbonate or calcium sulfate. Graeme Bartlett (talk) 12:22, 5 January 2013 (UTC)[reply]

Thank you all. That still leaves the first image (which is much bigger than the 200px that I set it to): does anyone know if that is indeed a Navicula diatom, or some other organism? Bramble claw x 22:51, 5 January 2013 (UTC)[reply]

Electromagnetic wave

Can a proton emit photon? When I read about photon, it only mentions that photon is emit or absorb by electron.174.20.15.246 (talk) 04:02, 4 January 2013 (UTC)[reply]

According to proton decay, a decaying proton would create a positron and a pion; the pion would then decay into two gamma ray photons. So if proton decay were to occur, then, yes, protons could emit photons. However, that's the thing: proton decay is as of now only a theory. There is no experimental evidence of proton decay ever happening, so it's not clear whether proton decay does occur. So while a theory states it is possible, in real life, there is currently no evidence that protons emit photons ever. Bramble claw x 04:23, 4 January 2013 (UTC)[reply]

Yes, a proton emits a photon whenever it moves - and even when it isn't moving! In modern physics, we actually describe the proton as continuously emitting a sea of virtual photons; some of these convert to real photons when they interact with other particles. Nimur (talk) 04:33, 4 January 2013 (UTC)[reply]

If it is just a theory so far then I just go with it can't emit photon. How about can photon absorb photon? In other word, what would happen if a photon hits a proton? Would it just pass right through the proton? Or would it get absorbed by proton?174.20.15.246 (talk) 04:34, 4 January 2013 (UTC)[reply]

According to the article http://wiki.riteme.site/wiki/%C4%8Cerenkov_radiation, emmision or absorption of a photon can be done by any charged particle (including therefore protons) that is accelerated or decellerated, respectively - the energy has to go/come from somewhere. Wickwack 60.230.207.240 (talk) 04:40, 4 January 2013 (UTC)[reply]

The energy does not have to go/come from somewhere if it is virtual photons we are speaking of (which is the case when like charges repel - protons exchange virtual photons which carry the Coulomb force), it just can't be around for too long. Heisenberg's uncertainty principle states that the conservation of energy may be violated to at most ΔE in a short enough time interval Δt such that ΔEΔt≤ħ/2. (Δ technically means uncertainty - we are assuming that our real observations of both are 0) 72.128.82.131 (talk) 20:26, 4 January 2013 (UTC)[reply]

So would it require certain situation that proton would absorb and photon when photon hits proton? In other words, do proton "always" absorb photon when being hit?174.20.15.246 (talk) 04:58, 4 January 2013 (UTC)[reply]

Inelastic Compton scattering only changes the photon's energy so although it imparts momentum to the proton the photon is not absorbed. --Trillianthcircuit (talk) 07:25, 4 January 2013 (UTC)[reply]

Seas of virtual photons being continuously emitting seems really vague. Are these carriers actually enumerated or are they just an assumed part of the wave function that maps the near field? --Trillianthcircuit (talk) 07:25, 4 January 2013 (UTC)[reply]

If you'd like less vagueness, consider reading the mathematical article I linked earlier this week. Nimur (talk) 07:36, 4 January 2013 (UTC)[reply]

In classical physics, anytime a charged particle is accelerated, it will radiate electromagnetic energy (photons). Protons entering a gravitational field will radiate. The physics for protons is the same with the difference being charge and mass. Incidentally this is also why the electrons circularly orbiting the protons to create atoms - fails as a model.--DHeyward (talk) 13:09, 5 January 2013 (UTC)[reply]

Validating the Big Bang or is it unfalsifiable?

Correct me if I'm wrong, but the size of the observable universe appears to be at the limit of current telescopes is it not? But as the scopes get better, we should stop seeing more distant galaxies at some point, thus if this observable myopic cliff has not been definitively found as of yet, then I wouldn't think the Big Bang model has been validated. My question is this, if the newer scopes show us a bigger universe, will the Big Bang parameters simply get updated again and again such that its unfalsifiable or is there a breaking point? --Trillianthcircuit (talk) 06:50, 4 January 2013 (UTC)[reply]

There is no distance limit to a telescope. The reason we can't see much further is because there isn't all that much more to see. With better telescopes we'll see more detail and better what we see already. If there were galaxies extending out further away we'd have seen some. Parameters haven't been updated, if anything the estimate for the age of the universe has been reduced a bit since the original estimates from Hubble but not recently. What is seen at the far limits is different from what is around now and the big bang theory is quite well established. If more was seen it would mean a major change in science not just tweaking some parameters. Much as scientists would like a major upset like that along with the new money it would bring it does not seem at all likely. Dmcq (talk) 12:39, 4 January 2013 (UTC)[reply]

Well, there may not be a distance limit to a telescope, per se, but there's certainly a distance limit to light. If the light coming from faraway galaxies is too faint then there's a chance any instrument we come up with won't be able to detect it. No matter how good telescopes get, no one will ever be able to tell me how many finger's I'm holding up from an observation point in the Andromeda galaxy. Evanh2008 ^{(talk|contribs)} 12:52, 4 January 2013 (UTC)[reply]

The following may be of interest : http://skeptics.stackexchange.com/questions/2335/what-evidence-of-the-big-bang-theory-does-exist Sfan00 IMG (talk) 12:33, 4 January 2013 (UTC)[reply]

We can "see" (i.e. detect and map in considerable detail) the cosmic microwave background radiation, which consists (roughly - some hand waving here) of photons emitted at the end of recombination in the early universe. Beyond/before this point the universe was a hot dense plasma which was opaque to electromagnetic radiation. So in a sense we can already "see" as far as it will ever be possible to "see". Gandalf61 (talk) 12:58, 4 January 2013 (UTC)[reply]

The current estimate is based on the distance to galaxies in the deep field of Hubble in accordance with the Big Bang and the red shift. At one time, we could not see that far and did not know about expansion, but with larger and better scopes we have been able to see further. Thus to make theory fit the data, they added cosmic inflation and then cosmic acceleration. Hence, the model fits this data for now, and Hubble can't resolve fainter galaxies that might emerge in abundance with a larger telescope. Thus my question is not trivial because if we see even more faint galaxies, then based on previous model changes, I would expect that astronomers would simply increase their inflation parameter because the CMB data does not appear to stop them from indefinitely increasing this parameter. Another thing that bothers me, is the mantra that the inflation rate is somehow meaningless, and in a terribly ironic way they are right, for they would repeatedly increase the inflation rate indefinitely to accommodate more galaxies, thus supposing Eric Lerner is correct that the Big Bang is wrong, I am asking is there any other part of the model such that a repetitious parameter change could eventually break it, or could it be just an unending shell game? In addition, Dmcq and the articles mention that the nearby oldest galaxies differ from the more distant younger galaxies. That would certainly mean that our neighboring galaxy formations have a similar history, but this important information would be more informative if a summary of this data is given or displayed in an article and I would appreciate accessible references too. --Trillianthcircuit (talk) 14:56, 4 January 2013 (UTC)[reply]

Eric Lerner? Well, his version of Plasma cosmology has an alternative explanation of microwave background radiation, but other astronomers have found lot of problems with his theory (at the end of Eric_Lerner#The_Big_Bang_Never_Happened). Even if the Big Bang theory was proven to be wrong, Lerner's theory wouldn't become more accepted until those problems were solved. I think that Lerner hasn't made any successful prediction using his theory, which is another point against it. --Enric Naval (talk) 18:41, 4 January 2013 (UTC)[reply]

The OP's question is interesting; let me put it more succinctly:

Is the Big Bang Theory falsifiable? Specifically, what could we hypothetically observe that would be inconsistent with the Big Bang having occurred?

I've seen lots of discussion in various places about what observations that we have made are consistent with the theory; but how would we falsify it? I sort of remember that back in the 1990s there was substantial consternation about the fact that cosmologists dated the universe's beginning at around 13 billion years ago, but that the oldest stars appeared to be more like 19 billion years old. I recall that the paradox was resolved when people found better ways of estimating the ages of stars. Would it contradict the Big Bang theory if, hypothetically, we found some stars that are solidly dated at well over 14 billion years old? Duoduoduo (talk) 17:33, 4 January 2013 (UTC)[reply]

This would contradict the currently accepted standard cosmological model, and will falsify it. But it will not be sufficient to falsify the Big Bang because different models are possible. Ruslik_Zero 18:33, 4 January 2013 (UTC)[reply]

Of course big bang cosmology could have failed many tests, and not long ago it was thought to be on shaky ground, as you said. It dominates the field now because it continues to be consistent with large bodies of new experimental data that have ruled out every proposed alternative. In other words, it has succeeded in the same way as any other scientific theory. I encourage you to read Ned Wright's excellent cosmology site, in particular the FAQ which lists various independent sources of evidence for the big bang, evidence against some alternatives including Eric Lerner's model, and the tutorial which is the best explanation of big bang cosmology I've seen. -- BenRG (talk) 18:54, 4 January 2013 (UTC)[reply]

OK, so basically, if we do not reach a deep field resolution where we stop seeing more galaxies, then because the Big Bang appears correct for various reasons, this is not a problem because theorists would have no problem with increasing the expansion factor. Of course, that would be frustrating to not find the edge of the sea (unless you are an ancient mariner sailing one), but hardly newsworthy. --Modocc (talk) 20:29, 4 January 2013 (UTC)[reply]

The most distant thing we can see is the cosmic microwave background; it is older, hence more distant, than any galaxy we might see (according to big bang cosmology). Its uniformity is evidence that the universe is homogeneous out to that distance. So we already know that we can't see the edge of the "sea", in that sense. Since more distant visible objects are older, and galaxies didn't form until some time after the big bang, there should be a gap between the most distant visible galaxies and the distance of the CMBR. If we could determine that the galaxies ended closer or farther than that it would at least contradict current theories of structure formation.

The Hubble Deep Field images, if that's what you and the OP are talking about, are not a major piece of evidence for the big bang. Currently the most important evidence is the CMBR spectrum: it is far too uniform to be starlight, it has small inhomogeneities of the right size to explain later structure formation, and its angular power spectrum matches the rather complicated prediction of inflationary ΛCDM over a wide range of scales. It was the COBE data that first gave astronomers a high level of confidence in big bang cosmology. -- BenRG (talk) 22:05, 4 January 2013 (UTC)[reply]

The Hubble constant gives a rate of accelerating expansion of about 74 k/s/Mp, and back-of-the-envelope calculations suggest that that could easily allow for a recessional velocity of two objects relative to each other that is well in excess of the speed of light. This could hypothetically allow for there to be a cut-off point beyond which we could not see (since the objects beyond that point would be receding from us too quickly for their light to ever reach us), but the CMB observations pretty much rule that out entirely. Evanh2008 ^{(talk|contribs)} 01:34, 5 January 2013 (UTC)[reply]

There is nothing special about a recession velocity of c. We have seen galaxies with recession velocities larger than c, and the recession velocity of the CMBR is more than 3c. The distance at which the recession velocity is c, i.e. c/H₀, is sometimes called the Hubble distance and is currently about 14 billion light years (the similarity to the age of the universe is coincidence - in most eras they are not similar). In Lambda-CDM the Hubble distance eventually approaches a constant value of 18 billion light years, and does mark the boundary between what you can and can't see in that epoch, if I'm remembering right. -- BenRG (talk) 05:09, 6 January 2013 (UTC)[reply]

Oops! Thinking about that now, you're right. The only noticeable effect of excessive recessional velocity would be excessive redshift, I think. In that case, I would say the only thing that's likely to result in galaxies being "too far away" to be imaged is shot noise and the amount of light actually reaching a telescope from great distances. In any case, CMB is very good evidence for the Big Bang. Evanh2008 ^{(talk|contribs)} 03:05, 7 January 2013 (UTC)[reply]

Frame of reference

Is it possible to change a non-inertial frame to an inertial frame or an inertial frame to a non-inertial frame ?
Is it possible for the same frame to appear as inertial as well as non-inertial for two different observers or for only one observer ?
In which year it was claimed that carbon form more number of compounds than any other element ?
In which standard (class) calculus, space-time, and theory of relativity is taught to the students of USA ? Want to be Einstein (talk) 12:11, 4 January 2013 (UTC)[reply]

For #4, calculus is often taught in several classes over a year or two (depending on the depth), starting often in a student's last year of high school or their first year of college. Space-time and relativity are often covered in very brief overview in introductory (high school or first year of college) physics classes, but generally aren't covered in detail until later years of college. --Jayron32 13:04, 4 January 2013 (UTC)[reply]

I asked question #4 because I am a student of standard 9 and I have read much more about these topics from books. Nowadays I am reading on these topics from Wikipedia. There is enormous information about these topics on Wikipedia. Want to be Einstein (talk) 13:18, 4 January 2013 (UTC)[reply]

The difference between a non-inertial frame and an inertial one is that the former is accelerating where the latter is not. You can't "change" one into another just by changing your point of view or using some mathematical trickery. They really are different. If you're in a rocketship and you feel yourself being pushed back into your seat (or whatever) then you're in a non-inertial frame. When you cut the rocket motors and coast, then you're no longer accelerating so you're in an inertial frame of reference - and you can immediately tell that this is the case by the fact that you're now floating free of your seat. Because such a simple experiment allows you to know which kind of frame of reference you're in - there is no way to "change" one into the other...except by turning off your engine. SteveBaker (talk) 17:38, 4 January 2013 (UTC)[reply]

What about the third question ? Want to be Einstein (talk) 09:28, 8 January 2013 (UTC)[reply]

bitcoin

I read the bitcoin article and think I have understood proof of work concept. A bit coin is a really long number which produces a hash which has quite a number of zeros. Ok, so bitcoins are really hard to find. But once I find the really long number, what is there to prevent me from transferring it to two people? How is a bit coin uncopiable? How is the transfer of ownerhsip ensured? Does it depend on the asumption of good faith of the bit coin clients? Cplusplusboy (talk) 15:14, 4 January 2013 (UTC)[reply]

The bitcoin approach requires a public consensus about a database; all transactions are recorded publically in a block cipher chain. If you make a transaction, and fail to announce it to the network, the transaction doesn't validate against the history of all prior bitcoin transactions; so phony transactions are "tamper-evident." This still requires some non-trivial technical effort: each transaction needs to be published and validated, and a community consensus is required to arbitrate which database stores the canonical record of transactions when there are conflicts. Nimur (talk) 16:25, 4 January 2013 (UTC)[reply]

And a bitcoin itself is not a hash or anything else. It is simply the result of transactions between bitcoin wallets. Transactions are from one wallet to another, and the miners do the work of putting transactions into the block chain and verifying things. This is computationally intensive, so the miners are rewarded with transactions that create new bitcoins, or eventually with transaction fees. 209.131.76.183 (talk) 17:51, 4 January 2013 (UTC)[reply]

Ah, it seems I had not understood the block cipher chain thing and 10 minute delay in txactions.. I'm currently going through their wiki, which seems to have more details [[1]]. — Preceding unsigned comment added by Cplusplusboy (talk • contribs) 09:08, 5 January 2013 (UTC)[reply]

Coastal armoring thoughts

What is causing more coastal erosion, climate change induced sea-level rise or coastal armoring? Does anyone have any thoughts on this question? Thanks. This is an honors thesis research question-not a homework question. — Preceding unsigned comment added by 99.146.124.35 (talk) 18:39, 4 January 2013 (UTC)[reply]

"Coastal armoring induced sea-level rise" ? Please explain. Are you talking about the sea level rising due to dropping concrete blocks and such into the ocean along the coast ? If so, that rise would be absurdly small, far less than a millimeter. Now, coastal armoring can cause erosion, but not by that mechanism. Basically it can redirect the force of waves onto a different area, and cause that area to erode more quickly. StuRat (talk) 20:02, 4 January 2013 (UTC)[reply]

I meant to say "climate change induced sea level rise or coastal armoring" -anon

I see you changed your original post. Coastal armoring presumably helps more than it hurts, or they wouldn't do it. So, in that respect it does less harm, although, in one particular spot, coastal armoring may cause a worse problem than global climate change. StuRat (talk) 05:47, 5 January 2013 (UTC)[reply]

what is the pH of borate-carbonate-bicarbonate solution? why is no formula given for the pH calculation of a hetereogeneous buffer solution? is the pH of an amino acid solution its pI?

What is the general solution for a heterogeneous buffer like ammonium acetate? Somehow, the folks who wrote buffer solution seem to think that all the interesting buffers worthy of discussion only consist of weak acids and their conjugate bases. Buffer solution doesn't discuss heterogeneous solutions at all. What is the general solution for buffering pH with multiple polyprotic substances?

The reason is that my friend and I are trying to reverse-engineer a proprietary aquarium reef buffer sold by Seachem which has a buffering pH of 8.3. It cannot contain any ammonium/ammonia or phosphates (this would be detrimental to the aquarium ecology). I was thinking it was a mixture of sodium borate and bicarbonate/carbonate, but I don't know how to determine whether the buffering pH would be 8.3.

Also, how do I calculate the pI of a solution containing multiple amino acids (how does ProtParam do it?) as well as the buffering pH? 72.229.155.79 (talk) 18:58, 4 January 2013 (UTC)[reply]

The calculation for a heterogeneous buffer solution will simply be the simultaneous solution of the equilibrium equations for the two systems. If you're using the technique in the Buffer solution article:

ICE table for a two monoprotic buffers
	[HA]	[A^-]	[H⁺]	[HB]	[B^-]
I	C₀	0	y	D₀	0
C	-x	x	x+z	-z	z
E	C₀-x	x	y+x+z	D₀-z	z

Working the equilibrium constant equations as before will give you two equations, one for Ka-A and one for Ka-B. You can then solve one for the change variable (e.g. solve the Ka-B equation for z), and then substitute into the other, and then you can solve for the other change variable. Extend to however many species you have. Analytical techniques may be a bit tricky, so you may need to resort to numerical calculations. Similar approaches can be used for multiple polyprotic buffers.

At some point, though, it gets easier to work forward from the Henderson–Hasselbalch equations. Assume a change in pH from A to A±x, and then calculate what the change in species concentration (and thus the added H⁺) would have to have been to make the change. If all you're concerned about is how well a solution will hold a pH of 8.3, that's probably the better approach, anyway.

What you'll find is that unless the desired pH is within about one pH unit of one of thee pKa's of one of the buffers (ignoring the Bicarbonate buffering system gas-exchange issue), you'll not get appreciable buffering - if A doesn't buffer by itself, and B doesn't buffer by itself, the A+B combination won't do anything special unless they interact somehow. (And amino acids aren't special - treat them just like any other polyprotic acid) -- 205.175.124.30 (talk) 04:36, 5 January 2013 (UTC)[reply]

Syrian missiles

Do Syrian war planes have laser guided missiles/bombs? If not, how accurate are unguided missiles/bombs? Is it any better than the unguided missiles/bombs of WW2--Jonharley667 (talk) 19:51, 4 January 2013 (UTC)[reply]

This site Xairforces.net lists; "• Air-to-ground-missiles (ASM):

AS-7 Kerry (Kh-23 general purpose), AS-9 Kyle ( Kh-28 confusional), AS-10 Karen (H-25ML/MR model - a general purpose / Kh-25MP antiradar), AS-11 Kilter (Kh-58 antiradar), AS-12 Kegler (Kh-25PS confusional), AS-14 Kedge (Kh-29 general purpose), AT-10 Stabber, (AT-2B Swatter B (9M17M Skorpion-M MCLOS 9K8 (Falanga-M)); ATGM helicopter Mi-24/25), HOT (ATGM for Gazelle helicopters)" .

Of those the AS-10 Karen, AS-12 Kegler AS-14 Kedge are laser guided. So they do have them, but if you're trying to get a few people hiding in a ruined house, I would think that you would go for a cheaper option, like unguided rockets or free-fall bombs. Are they more accurate than the WWII versions? I suspect not very much, but I couldn't find any hard data. Alansplodge (talk) 21:09, 4 January 2013 (UTC)[reply]

(I'm not sure how this wasn't marked as an edit conflict, but the answer above is probably going to be more helpful.) The Syrian Air Force article has a list of their aircraft; look for the "Combat aircraft" and "Attack helicopter" sections for attack aircraft. Of those craft, the Mikoyan MiG-29, the Sukhoi Su-24, and the Sukhoi Su-17 are compatible with laser-guided weapons, according to their articles. From my quick search of Wikipedia on the subject, I'm only finding specific references to more advanced air forces, such as those of the US, Israel, and India, using laser-guided missiles, but I'm not too familiar with the subject myself. This could reflect the encyclopedia's systemic bias. --BDD (talk) 21:11, 4 January 2013 (UTC)[reply]

Are they able to acquire laser guided bombs though, even if the plane accepts them? In Libya their planes could accept them but it didn't seem they used them as war footage of Libyan jet runs showed high inaccuracy. Also is a laser guided missile the same thing as a laser guided bomb? --Jonharley667 (talk) 01:21, 5 January 2013 (UTC)[reply]

See our article Laser-guided bomb which explains how it works. I'm not sure if the Syrians have them, or use them, but it looks as though some of their aicraft are capable of using them. But as I said before, cost is going to be a critical factor for a poor country in a long war. Laser Guided Bombs (LGB) are cheaper than Air to Surface Missiles (ASM) but unguided bombs are cheaper than either. Alansplodge (talk) 02:00, 5 January 2013 (UTC)[reply]

Unguided bombs are not cheap for use against specific targets, only against large areas. You need to use a lot of them to have any chance of hitting a target. Dmcq (talk) 04:46, 5 January 2013 (UTC)[reply]

Modern aircraft with Continuously Computed Impact Point bombsights (such as the F-15 Eagle) can deliver unguided bombs with almost as much precision as guided bombs (using a variation of the dive bombing tactic), but this requires a highly skilled pilot -- which is why you never see the Syrians achieve this kind of precision. 24.23.196.85 (talk) 05:05, 5 January 2013 (UTC)[reply]

Constantly computed impact point seems to get there. I'm surprised they don't have the system autopilot the plane, perhaps they do now for all I know. That would reduce the skill requirements. I'd have thought they would still need to get fairly close to have a good chance of hitting a building but I'm no expert. Dmcq (talk) 14:18, 5 January 2013 (UTC)[reply]

It is not unusual for export version of warcraft to have "different" capabilities than those used by the designing nations. Rmhermen (talk) 15:35, 5 January 2013 (UTC)[reply]

(ec-wtf?)

The superpowers with sophisticated systems like CCIP have highly trained personnel. Autopilot would be redundant to them, and it would open a window of vulnerability if the plane was shot at with the autopilot on. Basically, a pilot employing CCIP would try to go into a zero-g dive for a moment, to make the impact point stabilize on the target, and then drop an unguided bomb, or in a high-threat environment, not dive at all and drop a volley ("ripple") of bombs "across" the target, to nail it at least once in the process.

And yes, exported versions are usually limited when it comes to weapon compatibility. - ¡Ouch! (^{hurt me} / _{more pain}) 18:18, 5 January 2013 (UTC)[reply]

So what kind of laser guided bombs do the Syrians have? --Jonharley667 (talk) 07:07, 6 January 2013 (UTC)[reply]

Given Assad's evil genius demeanor, presumably something like this. μηδείς (talk) 17:12, 6 January 2013 (UTC)[reply]

Railroad switch question

The railroad switch article contains this:

Joints are used where the moving points meet the fixed rails of the switch. They allow the points to hinge easily between their positions. Originally the movable switch blades were connected to the fixed closure rails with loose joints, but since steel rails are somewhat flexible it is possible to make this join by thinning a short section of the rail itself.

Points/switches on model rail layouts have, I believe, pinned hinges at the place in question, and the wording above implies that this used to be the case in full-size operation but that now the steel rail used will flex each time the points are changed. Am I right in assuming that any failure due to metal fatigue would take far longer to occur than the normal lifetime of the rail, brought to an end by the wear caused by train movement over it? — Preceding unsigned comment added by 31.54.246.112 (talk) 20:47, 4 January 2013 (UTC)[reply]

As long as the stress level stays well below the yield point, the fatigue life of a typical metal part is measured in millions of cycles. For steel parts, if the stress level is below the fatigue limit, the fatigue life is infinite.

(If the stress level is above the yield point, the fatigue life can be as low as tens of cycles, as demonstrated in the Grayrigg derailment, where the spreader bars for a switch went from "passes visual inspection" to "complete failure" in less than twelve days.) --Carnildo (talk) 02:50, 5 January 2013 (UTC)[reply]

That could just mean the visual inspection was insufficient to detect the micro-cracks. StuRat (talk) 05:43, 5 January 2013 (UTC)[reply]

If they were "micro-cracks", then by definition they were too small to see with the naked eye. 24.23.196.85 (talk) 20:42, 5 January 2013 (UTC)[reply]

Yes, and that's what makes a visual inspection alone insufficient. StuRat (talk) 21:40, 5 January 2013 (UTC)[reply]