Wikipedia:Reference desk/Archives/Science/2006 June 8

Humanities Science Mathematics Computing/IT Language Miscellaneous 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 at one of the pages linked to above.

< June 7 Science desk archive June 9 >


Linux and SATA drives

edit

Further my question on Sata Drives, windows will not accept them as a boot drive, if I wanted to have duel boot, would Linux be able to use them.-User talk:erniehatt Ernie

I've moved this question from the original location at #Windows as it's largely unrelated to the original question, and it will get more attention at the bottom of the page.-gadfium 01:24, 8 June 2006 (UTC)[reply]
My current system boots both Windows XP and Kubuntu amd64 from the SATA drive. I did have problems installing both operating systems, because the machine also has an IDE drive, and both OSs put their boot loader on the IDE drive even though the system gives the SATA drive priority. I had to disconnect the IDE drive to install them correctly. In the case of Windows XP, this may be because I was installing from an original XP CD, not one which had Service Pack 2 already applied. XP was unable to see more than the first 140 (approx) GB of the disk until I added SP2. I note that Kubuntu calls the SATA disk /dev/sda, which is a name I would normally associate with a SCSI disk under Linux. Kubuntu calls my IDE disk /dev/hda as expected.
In short, I don't see why both Windows and Linux can't boot from a SATA drive, unless there is another drive in the system confusing them, or your BIOS is set up to not include the SATA drive as a boot device.-gadfium 01:41, 8 June 2006 (UTC)[reply]
Linux and Windows will boot off almost ALL SATA chipsets with no problem. It's the chipset, not the drive. --mboverload@ 04:18, 8 June 2006 (UTC)[reply]

When sata drives first came out, I was caught in a situation with my new Linux machine, in that you couldn't directly install to sata. But you could always boot from it, once you figured out how to install. --Zeizmic 12:02, 8 June 2006 (UTC)[reply]

THanks for that, but what I would like to be able to do is leave Windows on the IDE, and put Linux on the sata, is that possible.-Ernie
As far as I know, yes. Once you're in GRUB you can boot from different hard drives. To get to GRUB, the BIOS needs to boot from the hard drive which has GRUB in its MBR. Setting up a dual boot like this depends on which hard drive the BIOS boots from first -- you can check that in the BIOS. Say you have the BIOS booting the SATA hard drive first, then you would put GRUB on the MBR of the SATA hard drive. But the lines in GRUB which boot to Windows would have to refer to, say, hd1 instead of hd0, to boot from the other IDE drive. Your numbers will probably be different because I don't know the exact setup of your drives, but it's definitely possible to do. -- Daverocks (talk) 08:05, 10 June 2006 (UTC)[reply]
Thankyou very much.-[[user:erniehatt|Ernie

medical question on horses

edit

Hi thank you for taking the time to help me with a question its a test question that i find very tricky on my test and just cant make up my mind on which one it would be, i have been all over the internet trying to determain it, so here it goes; a long tubelike isterument called a rhionoscope is used to examine the nasal cavity of a horse with nasal discharge and obstructed airflow, the instrument may pass through "spaces in the nasal cavity called? 1. conchae 2. nasal septum well thats what i have it narrowed down to but im just not sure :( thank you again for your time and your help would be appreciated...

Greek for nose is rhino, for look is scopos, so you are of course describing a rhinoscope. It goes into a nostril so the doc can look around in the nose. A septum is always a dividing wall between anatomic cavities, and does not refer to a cavity itself. Conchae are shell-shaped anatomic structures, including partially enclosed spaces. I have no idea what a horse nose is like inside but suspect your answer is conchae. A sinus is an anatomic term for a more fully enclosed space. There are usually sinuses connecting into the upper reaches of mammalial noses, but you didn't list that as a choice and perhaps even horse sinuses do not have passages large enough to stick a rhinoscope into. Your teacher can flunk me if I'm wrong. alteripse 02:43, 8 June 2006 (UTC)[reply]

Water absorption in loose bowel movements

edit

I have been wondering about this for a while. I know that most water absorption from a fecal bolus takes place in the large intestine, and that a good amount of that water is absorbed in the decending colon. I am also aware that when one is suffering from diarrhea (and at other times), most of the loose stool collects in the sigmoid colon before release. However, is it possible (perhaps with coniderable concentration and sphincter strength) to "hold in" diarrhea so long that it solidifies? That is, can one still absorb water from a fecal bolus that is on the verge of "coming out"?Tuckerekcut 03:03, 8 June 2006 (UTC)[reply]

So the question is along the lines of "Does significant water resorption take place in the sigmoid colon"?--inksT 03:11, 8 June 2006 (UTC)[reply]
Really, the question is: "Can one hold in a loose stool so long that it becomes a normal one," and all the rest of this babbling is me thinking out-loud.Tuckerekcut 03:19, 8 June 2006 (UTC)[reply]
IIRC, water absorption (sp?) in the large intestine is driven by osmosis and accounts for roughly 10% of total water absorption in the digestive tract (the small intestine accounts for the remaining 90%). Since diarrhea is generally caused by increased motility and decreased time spent in the intestines (and vice versa for constipation), I would venture that the answer to your question is yes (assuming the characterisitics of the sigmoid colon are roughly comparable to the large intestine in general).--inksT 03:29, 8 June 2006 (UTC)[reply]

I guess this begs a new question, then: Are the sigmoid colon and rectum comparable to the rest of the intestines with respect to water absorption? The answer to the first question, too, does not necessarily follow from the phenomenon of constipation with bowel retention because whereas an already firm stool might have enough volume and bulk to remain partially in an absorptive section of the colon, a watery stool may slip completely into the rectum.Tuckerekcut 04:06, 8 June 2006 (UTC)[reply]

C'mon, he just wants to know if you could hold in diarrhea long enough for it to solidify. Try it yourself and find out! —   The Mac Davis] ⌇☢ ญƛ. 07:25, 8 June 2006 (UTC)[reply]
Really, this is Wikipedia. What we need is a peer-reviewed study, published in a reputable, scholarly journal where people had to hold it in for, say, a few hours to a day...(shudder)--inksT 08:19, 8 June 2006 (UTC)[reply]
What we need is some common sense. Experience (ahem) says that it's not possible to "hold" diarrhea, so it's a moot question. To support my observation, I quote from Diarrhea: "...[T]he [excessive] water dilutes toxins as well as triggers contractions of the intestine due to increase in intestinal distension. These contractions push the contents of the lower GI tract towards and out of the anal canal." So physiologically, one's gut is working to prevent one from "holding it." --Ginkgo100 talk · contribs 03:41, 9 June 2006 (UTC)[reply]

Physics problem

edit

A cyclist intends to cycle up a 14.0 degree hilll whose vertical height is 120 m. Assume the mass of bicycle plus person is 75.0 kg. If each conmplete revolution of the pedals moves the bikes 5.10m along its path, calculate the average force that must be exerted on the pedals tangent to the circular path. Neglect work done by friction and other forces. The pedals turn in a circle of diameter 36.0 cm. Answer: 802 N

I'm just overwhelmed with all the information given here. How would I start off with this problem and then proceed to the next steps? It gives way too much numerical data for me to make sense of.....Any help would be appreciated. Thanks. C-c-c-c 04:12, 8 June 2006 (UTC)[reply]

It's not that bad. First find the force of gravity the cyclist has to overcome. Then figure out the mechanical advantage of the whole system, including the gears and the inclined plane, in other words, the ratio between the distance the pedals move and the vertical distance the bike travels. You can split that into multiple steps: remember, mechanical advantage is multiplicative, so the advantage of the whole system is the advantage of the inclined plane (= the sine of the angle) times the advantage of the gear system (= the distance the pedals move over the distance the bike moves). —Keenan Pepper 04:26, 8 June 2006 (UTC)[reply]
Oh, and usually when there are suspiciously many pieces of information, you should look for red herring that don't have any effect on the answer, but in this case there are none. —Keenan Pepper 04:28, 8 June 2006 (UTC)[reply]
1) Don't let yourself become overwhelmed! I bet you can answer this problem if you can see it as a few smaller questions: First, find out how much force will be needed to move the biker (you can think of him and the bike together as a frictionless block on an inclined plane) up the plane. Since no time is involved, we can assume that the biker is not going to be accelerating horizontally, and thus we only need to take into account the force of gravity. Thus, enough force must me used to counteract the force of gravity. So draw a free diagram of the box on the inclined plane, and find out what gravity's component force is parallel to the downward slope of the inclined plane. The biker must create a force equal to this component force in order to avoid deceleration. ( You will use the slope and biker's mass in this section
2) Now that you know how much force will be needed to keep the biker from slipping down the inclined plane, find out how much force is needed on the pedals to create this forward force. First, use torque = r x F (with r = the radius of the circle with the circumference 5.1m) to find the torque needed by the wheels to produce the forward force, then use the same equation in reverse (this time with the radius of the pedals) to find the linear force needed to create the same amount of torque on the pedals. (This section takes into account the distance traveled over one rotation, the radius of the pedals, and the answer from part one.)
That's it! notice that the height of the hill is irrelevant, and is just thrown in there to confuse you., Good Luck!Tuckerekcut 05:05, 8 June 2006 (UTC)[reply]
Right... wonder why I didn't notice that... there is a red herring after all. —Keenan Pepper 05:43, 8 June 2006 (UTC)[reply]
Notice that this is a question about work and energy, so thinking in terms of forces and accelerations is an unnecessary complication. During one revolution of the pedals the rider exerts an average tangential force F on the pedals over a distance πd (where d is the diameter of the pedals) so the work done by the rider is Fπd. Make the simplifying assumption that the speed of the bike is constant and the wheels do not slip, so the kinetic energy of the bike and rider is constant. No work is wasted overcoming friction, so all of the work done by the rider goes into increasing the potential energy of the bike and rider. The increase in potential energy is mgh where m is the combined mass of bike and rider, g is acc. due to gravity, and h is the vertical distance travelled during one revolution of the pedals. Trigonometry gives you h, then you can find F from F = (mgh) / (πd). Gandalf61 09:56, 8 June 2006 (UTC)[reply]

Theoretical question on time

edit

Hi, my girlfriend is writing a story and she asked me the following question, but while I can make guesses, I was wondering if anyone could answer it with reference to actual theories. Her question is: What would happen if time collided with a differently moving time?

Is there any way in which this would be possible? Or if not, could you point me in the direction of the reason why? Thanks! Phileas 04:38, 8 June 2006 (UTC)[reply]

How does time move? Does it have 'velocity', changing distance with respect to, um, time.... Peter Grey 05:23, 8 June 2006 (UTC)[reply]
The question is syntactically valid but semantically meaningless. It's like "Do colorless green ideas sleep furiously?". In order for my reply not to be a complete disappointment I'll direct you to this suprisingly readable paper that discusses the possibility of multiple dimensions of time. —Keenan Pepper 05:48, 8 June 2006 (UTC)[reply]
She is obviously thinking of time as something like a piece of matter—a physical, concrete, thing; an object. This is not the case as far as we know. Time is a dimension, a measurement of that dimension, a property of the system. Can you say "what if width hit another width" or "what if one piece of direction hit another direction?" —   The Mac Davis] ⌇☢ ญƛ. 07:28, 8 June 2006 (UTC)[reply]
But that's quite right, because, as we all know, time flies like an arrow, and an arrow could collide with another arrow flying in a different direction even if such a collision is quite rare. – b_jonas 09:05, 8 June 2006 (UTC)[reply]
Anyway, you can make nice sci-fi writing about this. One that comes into my mind is a universe colliding to another one, which is described in the StickManStickMan webcomics. – b_jonas 09:11, 8 June 2006 (UTC)[reply]
Thanks everyone. Very informative. And yes, I think she is planning to write some sci-fi about it! Phileas 00:27, 9 June 2006 (UTC)[reply]

Nuclear Engineering

edit

Is it possible to construct a bomb in such a way so as to use some sort of filler(as only 3%-5% of the material gets fissioned)to make an efficient device without unfissioned material after explosion?

It's hard to say, as nuclear weapon design details are secret and much of the public domain information is merely intelligent guesswork. According to our article on nuclear weapon design, fusion-boosted weapons can have an efficiency of about 40%, so 60% of the material remains unfissioned.
Why do you want to know about maximising efficiency? If you want to minimise fallout, what you need to do is maximise the proportion of the total yield which comes from fusion rather than fission, as fission is essentially fallout-free. If your point is "bang for buck", use a fission-fusion-fission weapon with an outer case of cheap and readily obtainable natural or depleted uranium. --Robert Merkel 05:29, 8 June 2006 (UTC)[reply]
Is it possible to speculate that if it is possible to compute the portion of the supercritical mass that undergoes fission,then let only that portion be fissionable,the rest be some sort of filler material,so as to make the entire fissionable material to undergo fission.
By 'fissioned' and 'unfissioned' I assume you DO NOT mean 'fissile' and 'unfissile'. Fissile and unfissile mean they can be 'fissioned' and 'unfissioned,' but not necessarily are. This question is incoherent—the amount and kind of fissile material, the material that gets fissioned, is what matters, not what percent of the material in the bomb, because that percent IS the bomb. —   The Mac Davis] ⌇☢ ญƛ. 07:32, 8 June 2006 (UTC)[reply]
I think the questioner is asking is it possible to arrange things so that all, or nearly all the fissile material in the bomb (ie the U-235 or Pu-239) actually fissions, by careful bomb design. The answer to their question appears, from the open literature, to be "no". --Robert Merkel 12:27, 8 June 2006 (UTC)[reply]
Though you can probably get the efficiency pretty high if that is your primary concern. --Fastfission 13:38, 8 June 2006 (UTC)[reply]

Why isnt it possible to make a smaller fissile core(Pu or Ur) and suround it with another metal that produces neutrons(say lithium)and expect the entire solid fissile core to undergo fission with full efficiency and no unfissioned material?


The quick answer is no. The reason you need the other 95-97% of the fissile material is because of the low neutron cross-section of it. i.e. Most of the neutrons go straight through undeflected and unabsorbed. If you have a 1000g core, of which only 40g of material are fissioned, why can't you only start with 40g instead of wasting a kilo of enriched uranium? The answer is because the 40g doesn't capture 100% of the incoming neutrons, in fact, it hardly captures any. So you need all the fissile material to capture just a fraction of the neutrons. -- Eh-Steve 09:24, 9 June 2006 (UTC)[reply]

Lightning Rod

edit

What happens to be the concept of using a lower amperage fuse for a large current for a short duration , Is there any mathematical explaination about it?

I'm not sure I understand the question, but I'll take a shot at it. A fuse is designed that if a large amount of current attempts to pass through it (typically the result of a short, like a toddler sticking a fork in a socket, for example), it will burn out, breaking the circuit.
The mathematical explination is simple enough - Ohm's law, Voltage = Current * Resistance. For the time scales we are considering, the voltage is effectively constant (120 volts in the US), and if someone shorts the line, the resistance is very close to 0. Therefore, the current will be very high - suffecient to kill someone. Raul654 05:09, 8 June 2006 (UTC)[reply]

Actually I was wondering how does a lightning rod survive a lightning strike..

What's to survive? It doesn't have any biological or electronic parts to be damaged, it can't catch fire. If it's low enough resistance it won't melt. I guess the worst thing that could happen is along the way to ground connections could short out.--Anchoress 06:31, 8 June 2006 (UTC)[reply]
Who would want to put a fuse into a lightning rod ? Well, turns out such devices are actually used for testing lightning rods - see here. Gandalf61 09:26, 8 June 2006 (UTC)[reply]

Combo Drive/Computer

edit

Why cant a cd-burner/DVD reader burn dvd's?

See Compact disc and DVD. The discs and the read/write methods are similar but not exactly the same. Dismas|(talk) 05:44, 8 June 2006 (UTC)[reply]
Right. If they made it able to burn DVDs, it would have to be more expensive, because it would need a more powerful and accurate laser, among other things. —Keenan Pepper 05:51, 8 June 2006 (UTC)[reply]
You can always buy an external burner. --Proficient 03:07, 10 June 2006 (UTC)[reply]

That's weird, my post to this page disappeared

edit

And I looked thru the diffs but I can't figure out when or how. I posted this and didn't notice until almost an hour and half later that it was gone. Looked thru every diff, didn't see it go. Reposted it (paraphrased) here.--Anchoress 06:24, 8 June 2006 (UTC)[reply]

See this edit (two edits after you posted your original comment. I can't explain the edit though. — Knowledge Seeker 07:23, 8 June 2006 (UTC)[reply]
Possible edit conflicts failing to update properly. Next time try editing on notepad first then cut and paste.--Jondel 07:36, 8 June 2006 (UTC)[reply]
To whom is your comment directed? — Knowledge Seeker 07:38, 8 June 2006 (UTC)[reply]
  • Thanks for the replies. Knowledge Seeker, thanks for spotting that. Jondel, I know about edit conflicts, and this wasn't one. It saved fine without any errors, and I saw my post on the page after saving. Then the next time I looked it was gone. This had nothing to do with edit conflicts or problems saving, the post saved and was clearly compiled to the page. But thanx for the comment anyways, if I'm making a large addition I *do* edit elsewhere and add, it's a good idea.--Anchoress 08:01, 8 June 2006 (UTC)[reply]
Anchoress, before an edit conflict can occur, the first edit must be saved to the page. It will show up on the page. Then the second user gets the conflict, and accidentally kills the first edit. What you've described is consistent with what happens in an edit conflict. -lethe talk + 08:38, 8 June 2006 (UTC)[reply]
Well, aren't 'cha gonna put it back?
Lethe, can you explain that more clearly? How would that explain what happened to my edit (as per Knowledge Seeker's link above)? I've seen lots of edit conflict windows, and they never resulted in lost data, either mine or that of the editor before me.--Anchoress 09:24, 8 June 2006 (UTC)[reply]
Well first of all, my point was that it was not you, but rather Tuckereckut who had the edit conflict. So the fact that you yourself never saw the edit conflict window means nothing at all. If anyone saw the edit conflict dialogue, it would have been him. As for the fact that you've never seen an edit conflict result in loss of data, let me suggest that you revise your position: never before today have you seen it occur. Today you did see it occur. I have seen it occur in the past, so I don't know what else to tell you except "welcome to the suckiness that is MediaWiki". Looking at the edits, it appears that Tuckerekcut made two edits in rapid succession, the second revising the first. Probably he used the back button after submitting. This often allows to bypass the edit conflict mechanism. This sort of thing happens a lot on high traffic pages. Is it clear now? -lethe talk + 09:56, 8 June 2006 (UTC)[reply]
Sorry Anchoress, reacted too fast and replied. I wanted to say the same thing that Lethe said.--Jondel 10:15, 8 June 2006 (UTC)[reply]
Thanks for the info you guys.--Anchoress 10:21, 8 June 2006 (UTC)[reply]

Oops. No harm intended, it was 2 AM here.Tuckerekcut 19:32, 8 June 2006 (UTC)[reply]

NP.--Anchoress 09:22, 11 June 2006 (UTC)[reply]

Does mental calculation improve concentration, memory and other mental capabilities unrelated to calculation? I need conclusive , authoritve proofs from Doctors, psychologist and am searching the web. I believe it does, it's just that a cite source is demanded at the mental calculation page. --Jondel 07:37, 8 June 2006 (UTC)[reply]

You're may be using Google Scholar, which is ok. You could also try PubMed ([1]) with some keywords. I did have a quick try on PubMed but didn't find anything immediately relevant.--inksT 08:02, 8 June 2006 (UTC)[reply]
Thank you very much for the PubMed. Yes I 've been using google.--Jondel 09:12, 8 June 2006 (UTC)[reply]
Could you define 'mental calculation'? Do you mean doing math in your head, thinking four moves ahead in chess, doing the word puzzles in the paper without writing anything down, or...?--Anchoress 08:04, 8 June 2006 (UTC)[reply]
Math in your head. No chess and puzzles just things like 37 x 48 in your head , square roots , etc.--Jondel 09:12, 8 June 2006 (UTC)[reply]
4 moves ahead in chess and word puzzles also improves concentration, memory and other mental capabilities, I believe. But that's aside the topic of mental calculation.--Jondel 09:21, 8 June 2006 (UTC)[reply]
  • Yes it does, but if you want authoritive proof, asking a doctor or psychologist directly is going to help you much more than searching the web. Mental excersize (calculation, puzzles, riddles and the like) improve the neuron connections in your brain which is good when it comes to memory and other things you use your brain for. It's been proven that people who do puzzles on a regular basis are less likely to suffer dementia and other brain problems in the long run. As for concentration, the more you practice it, the easier it gets. - Mgm|(talk) 08:52, 8 June 2006 (UTC)[reply]
I know. But we need authoritive sources.--Jondel 09:17, 8 June 2006 (UTC)[reply]
Mgm, are you sure about this statement: "It's been proven that people who do puzzles on a regular basis are less likely to suffer dementia and other brain problems in the long run."? How have the studies you refer to been able to separate cause and effect? People with a latent or beginning dementia might be less likely to want to do puzzles on a regular basis than those who have no latent brain disease.
I searched PubMed for (cognitive[All Fields] AND (("aptitude"[TIAB] NOT Medline[SB]) OR "aptitude"[MeSH Terms] OR abilities[Text Word])) AND ("training"[All Fields] OR "excercise"[All Fields]), and for (cognitive[All Fields] AND (("aptitude"[TIAB] NOT Medline[SB]) OR "aptitude"[MeSH Terms] OR abilities[Text Word])) AND mental[All Fields] AND (calculation[All Fields] OR computation[All Fields] OR arithmetic[All Fields]), with very few relevant hits.
  • A 1992 review paper, Nolan KA, Blass JP, Preventing cognitive decline. Clin Geriatr Med. 1992 Feb;8(1):19-34. (Abstract) states, rather disappointingly that "Although they are rapidly becoming more numerous, the efficacy of cognitive training programs in preventing or slowing cognitive decline has not yet been demonstrated."
  • This paper: Martini L, Domahs F, Benke T, Delazer M. Everyday numerical abilities in Alzheimer's disease. J Int Neuropsychol Soc. 2003 Sep;9(6):871-8. (Abstract) did not appear to be relevant in itself, but if you could get a copy from the library, it might have some relevant references.
  • The abstract of this paper: ACTIVE: a cognitive intervention trial to promote independence in older adults. Jobe JB, Smith DM, Ball K, Tennstedt SL, Marsiske M, Willis SL, Rebok GW, Morris JN, Helmers KF, Leveck MD, Kleinman K. Control Clin Trials. 2001 Aug;22(4):453-79. (Abstract) states that cognitive training interventions (memory, reasoning, and speed of information processing), have previously been found to be successful at improving mental abilities under laboratory or small-scale field conditions. Maybe there is something in the references of that paper which may be relevant.
  • This paper, Dehaene S. Varieties of numerical abilities. Cognition. 1992 Aug;44(1-2):1-42. (Abstract) might also have relevant references.
If any reliable data exists, I would expect it to be either from training regimens for the elderly, or perhaps some early work in experimental psychology. The results of the search suggest that there simply is no experimental data to support the claim. It may, of course, still be true. My guess is that this assumption has never been tested rigorously. When you think about the study design that would be needed, and the amount of work that would be involved, that is perhaps not so surprising. --vibo56 talk 17:49, 8 June 2006 (UTC)[reply]
MGM's statement is correct, I believe, and it doesn't make any assumptions as to cause-and-effect. I had to read numerous articles on this in neuroscience classes back in college, and it's been shown both for people who do crosswords on a regular basis, and for people who were more verbose and descriptive in their writings during young adulthood (I could possibly dig up the references). What has been shown, as you rightly note, is a correlation, not a causation. However, cognitive reserve has been possibly shown in a number of cases, which seems to show that people who use their brains more show fewer signs of dementia, even after onset of pathologies such as Alzheimer's. So the reserve doesn't prevent the neural damage, but it does seem to prevent, or at least delay, their effects. A moment of searching finds this as a pretty good recent review article. — Asbestos | Talk (RFC) 18:27, 8 June 2006 (UTC)[reply]
In the context that MGM presented it, I read this as an argument of a causal relationship, but you are right in pointing out that the statement, in itself, does not imply causality. As for the mental reserve argument, the attraction towards intellectually stimulating activities and the slower progression of dementia may have a common cause (the mental reserve), or the mental training may lead to a greater mental reserve, and therefore be the cause of slower onset of dementia. The original question was whether data exists to support the hypothesis that doing mental calculation has other, positive effects on cognitive function. I doubt that the such data exists, but would be delighted in being proven wrong. --vibo56 talk 19:25, 8 June 2006 (UTC)[reply]

[Unindenting] Well I guess it probably wouldn't be any use, but a couple of years ago I saw a piece on Animal Planet or something where geriatric beagles who were put through mazes and given memory exercises and taught new tricks retained their faculties better and longer than the 'control beagle' group.--Anchoress 19:31, 8 June 2006 (UTC)[reply]

Animal studies could certainly be the way to go to prove causality, if the study was well designed. If mental training can be shown to delay dementia in animal studies, I would find it reasonable to assume that the association observed in humans is indeed causal. --vibo56 talk 20:01, 8 June 2006 (UTC)[reply]
Well if you think it's any use, here is an article on the beagle study.--Anchoress 20:23, 8 June 2006 (UTC) ADDENDUM Actually it looks like there may be some interesting data on humans on the same page. Check it out.--Anchoress 20:25, 8 June 2006 (UTC)[reply]

If you want a citation, I read this in A User's Guide to the Brain by Dr. John Ratey, a psychiatrist. It's a very approachable book for laypeople. --Ginkgo100 talk · contribs 20:10, 8 June 2006 (UTC)[reply]

Thank you very much for your discussion and inputs. This is a big help. I'll try to research the above to add sources on the article.--Jondel 01:46, 9 June 2006 (UTC)[reply]

low vs. high pressure

edit
IN our physics book we learned an equation P=F/A,

where p is pressure , f for force and A for area. it means if force is constant than pressure is inversly proportional to area. but in fluid mechanics book i read that when water at high pressure passes thrigh a nozzle or a notch(which obviously have area reducing), the pressure reduces and velocity increases, isn't it against the above mentioned equation, according to which by reducing area pressure should increase! kindly solve my problem!

Your physics book is correct, the equation is  , meaning that the pressure on an object will be equal to the force being applied to that object, divided by its area. This explains why it is more painful when your foot is trodden on by a high-heeled shoe as opposed to a trainer (sneaker): the high-heeled shoe has a lower surface area, and thus applies more pressure with the same amount of force. I suppose water will come out of a nozzle at high velocity because it was originally under high pressure, but is moving into a low pressure environment. An analogy could be when you shake a can of Coke and then open it; the cola will spill everywhere at quite a high speed, as it goes from a high pressure environment inside the can to the low pressure environment outside. Andrew 10:14, 8 June 2006 (UTC)[reply]
(1) There actually is less force because the force is not being applied only against the cross-section of the fluid, but also against the nozzle. (2) By definition a fluid has no resistance to shear forces. A fluid does not transmit a linear force - generally you want energy density Energy/Volumne = Force/Area = Pressure. The increase in velocity leads to a decrease in pressure. Peter Grey 15:21, 8 June 2006 (UTC)[reply]

Games where the rules change

edit

I've read a number of studies that show that playing "games where the rules keep changing" is good for the brain, warding off dementia and generally keeping you in tune. My question: Are there any pencil-and-paper one-player games that would be something like that? I know of games like nomic, and thhe card-based versions like fluxx, but is there any thing remotely similar in a one-player game? — Asbestos | Talk (RFC) 15:18, 8 June 2006 (UTC)[reply]

Mathematics? Theoretical physics? Conscious 16:18, 8 June 2006 (UTC)[reply]
Huh? Just pencil and paper. If you know of a pencil-and-paper self-modifying game that involves theoretical physics, that's fine. — Asbestos | Talk (RFC) 18:30, 8 June 2006 (UTC)[reply]
I was mostly kidding. But if you come up with a new theory, you will be effectively changing the rules of the game. Conscious 10:51, 9 June 2006 (UTC)[reply]
Play with cards : Eleusis. Each one puts a card in turn and one of the players says if it is OK or not with the rule he devised at the beginning. The one who guesses that rule takes h[er|is] place and a new game starts with a new rule. --DLL 21:24, 8 June 2006 (UTC)[reply]
P.S. Alone, it's harder, like trying to tickle oneself. --DLL 21:25, 8 June 2006 (UTC)[reply]
At university we would occasionally play Calvinchess, a variation of Calvinball using a chessboard and chessmen. Very entertaining, but still generally requiring at least two players. --Ginkgo100 talk · contribs 03:35, 9 June 2006 (UTC)[reply]

Hmmm... Doesn't sound like we're finding any one-player games. I feel like it must be possible, but I can't quite see what it would be like. One thing I'm thinking is that the rule-changing can happen in one of two (or more?) ways: Either the player consciously makes new rules, trying to put them together in some way that guarantees his victory (obviously with some limitations on what kind of rules can be made), or the rules change automatically as a result of his previous move — e.g. if the board lookis like this, then only these kinds of moves are allowed. Hmmm, I'm seeing potential here. — Asbestos | Talk (RFC) 11:59, 9 June 2006 (UTC)[reply]

Much of the fun of games where the rules change is pitting one intelligence against another, and the entertainment afforded by watching another person's creativity at work. If you were the kind of person that was entertained by your own brilliance, you would know it by now. Other than that, you could adapt the idea of one of the card games to your own purposes: Create a deck of cards with rules (or aspects of rules) on them, then at intervals during a normal one-player game, draw one (or a set) and follow its (or their) instructions. In the case of a set-based game, you could have a relatively small number of cards and, through the magic of combination, quite a few rule changes. And, you can simulate the challenge of playing against someone (somewhat) by including "goal" cards, with changes in the conditions by which you win the game. Draw them at more-or-less random, preset intervals, and follow the instructions. It would be just like a turn-based game where the opponent keeps setting you back in some way, in that anything you'd just done would probably wind up being counterproductive. Hey, this has some serious potential. Black Carrot 22:54, 9 June 2006 (UTC)[reply]
Correction: Since you say pencil-and-paper, you could write up a small handbook with the same info in it, then use dice (or a random number calculator if you know any easy ones) to do the random drawing for you. Black Carrot 22:56, 9 June 2006 (UTC)[reply]

Another one is nomic. Emmett5 02:30, 12 June 2006 (UTC)[reply]

NeuroNetworks

edit

Has anyone being able to develop a complete theory for the cognitive neurosciece and neurons (linking,learning,etc...)? If so, I'd like to know where to find it.

Reading about the Perceptron is a good place to start. ...IMHO (Talk) 15:41, 8 June 2006 (UTC)[reply]
What else? Anything more complete? If no, I'll have a go at it for myself. Seriously.
There's a whole Category:Neural networks. Conscious 16:15, 8 June 2006 (UTC)[reply]
What exactly do you mean by a "complete theory"? What kind of answer were you hoping for? The mechanics of learning in the brain fills several very large textbooks currently on my shelf. How memory works fills several more textbooks. Neuroanatomy, alongside neurogenesis and cell death, is another huge topic. What would a "complete theory" look like? The category you're looking for, by the way, is probably Category:Neuroscience and maybeCategory:Cognitive science. It's much easier to have a complete theory on something with a fixed set of rules, like a neural network. — Asbestos | Talk (RFC) 18:05, 8 June 2006 (UTC)[reply]
I agree with Asbestos. Aside from the question being a bit ill-defined, there's a huge mass of information that you'd need to know before you could go about trying to make a "complete theory." Parts of such a theory do exist though. Temporal Difference Learning is a fair approximation to operant conditioning (augmenting the more abstract and problematic Rescorla-Wagner model of classical conditioning). Kohonen's SOM can broadly reproduce aspects of early sensory areas. More abstract models also exist for, e.g., planning behavior. All of these approaches are, however, abstract and generally quite divorced from the reality of the biological neural networks that support such computation in the brain. This is all to say: the brain is complicated and such a theory as you're after is probably a long way's off. I'm a computational neuroscientist and I have done a fair bit of modelling many parts of the brain and aspects of cognition, and from that it's clear that: 1. we currently lack enough computing power to make detailed, quantitative models/theories that are widely applicable in the brain, and 2. the tiny details of neuronal functioning have a significant effect on the population/network activity and those details vary as you move around the brain, so any theory that attempts to subsume everything will likely be a very complicated one. Of course, if you can prove me wrong, I'd be very happy to see such a theory. 128.197.81.181 23:08, 8 June 2006 (UTC)[reply]
Maybe the article engram is relevant. --Username132 (talk) 22:54, 8 June 2006 (UTC)[reply]

kilograms = liters of h2o and vice versa

edit

Why is a kilogram no longer a part of the difinition of a liter? I use it all of the time to measure the volume of a container simply by filling the container (liquid) with water and then subtracting the empty container's weight and just like magic I get the volume in liters. What's goin' one here? ...IMHO (Talk) 15:38, 8 June 2006 (UTC)[reply]

You have it the wrong way round. The liter was defined as a volume, based on a pre-existing measure, the meter. The gram was then defined as the weight of 1 cm3 of water at 4 deg, and from the gram we get the kilogram. However, the weight of 1 cubic cm of water is not standard, and so a standard kilogram measure was created using independant units. The last article linked has the explanations. — Asbestos | Talk (RFC) 15:51, 8 June 2006 (UTC)[reply]
Actually, the kilogram isn't defined using independent units; it's defined to be the mass of a lump of platinum-iridium kept in a vault in Paris. Somewhat embarrasingly, the lump has lost 50 billionths of its mass since it was made, which means that anything that had a mass of 1 kilogram 100 years ago, and hasn't changed since, now has a mass of 1.000000050 kg. They're trying to sort this problem out by redefining it as the mass of a particular number of atoms, but haven't worked out the details yet. EdC 04:51, 9 June 2006 (UTC)[reply]
The liter is a derived unit, equal to one cubic decimeter (one one-thousandth of a cubic meter). Currently, the meter is defined as the length of the path travelled by light in absolute vacuum during a time interval of 1/299,792,458 of a second. This is an extraordinarily precise defined value that we can measure to an arbitrarily high degree of precision, at least as long as you have good enough instruments.
Practically speaking, nobody outside a lab actually measures out meters (or decimeters, or liters) with a light and stopwatch. Rather, we use various approximations to go about our business. The density of liquid water is fairly close to 1 kilogram per liter, so in practice we can use a liter of water as a stand-in for a kilogram mass, and vice versa. Note that since water does change a bit in volume (and therefore density) depending on its temperature, you shouldn't rely on this volume to mass conversion for more than rough approximation. TenOfAllTrades(talk) 16:18, 8 June 2006 (UTC)[reply]
Great this was the answer I was hoping to find since in many instances it is sufficient to have only a rough approximation such as when you want to know if the scale at the grocery store is off by a whole lot. With a rough kilogram/liter equivalent you can quickly check the store scale with a 2 liter bottle of pop which equals about 4.4 lbs. Scale says 5 lbs shop somewhere else. ...IMHO (Talk) 17:02, 8 June 2006 (UTC)[reply]
Well sure. I thought you were asking why the kg was no longer part of the definition of liter. Obviously the weight is almost exactly the same as it's always been, and certainly neither the weight nor the definition has changed in any of our lifetimes. — Asbestos | Talk (RFC) 17:58, 8 June 2006 (UTC)[reply]

Electrical resistance

edit

Hi, physics AS tomorrow and im stuck on this. State the effect on the ammeter reading and the voltmeter reading (of a circuit) as the temp of resistor is increased. Obviously with increased enthalpy the resistance of the circuit will increase, thus the current will decrease. My problem is that what happens to the voltage if the resistance increases and the current decreases (due to IR=V)?

Thanks for any help chemaddict 15:56, 8 June 2006 (UTC)[reply]

If the voltage source has nonzero internal resistance, then a decrease in current will lead to the decrease of voltage on internal resistance. Guess how this will affect the voltmeter reading if the electromotive force isn't changing. Conscious 16:13, 8 June 2006 (UTC)[reply]
Hi, you're right to say that the resistance will increase as its temperature increases. But what happens to the voltage across it will depend on the circuit that it's in. For example, if it's connected directly across a battery (that is of comparatively low internal resistance) then the voltage across it isn't going to change significantly, but if it's connected across a battery in series with another resistor, then the voltage across it will increase. But in any case, in any simple circuit you'd be reasonably likely to be considering, the current through it would decrease. If you get a chance, then please clarify the question by describing the circuit you have. If not -- well, all the best in your exam! Arbitrary username 16:17, 8 June 2006 (UTC)[reply]

Hi, yeh circuit with emf of 3v (negligable internal resistance) attached to a 'bundle' of thin insulated copper wire, an ammeter in series with this and a voltmetre in parallel around the copper wire. thanks chemaddict 16:27, 8 June 2006 (UTC)[reply]

Okay, so yes in this case the voltmeter reading isn't going to change as the internal resistance of the voltage source is negligible. Arbitrary username 17:35, 8 June 2006 (UTC)[reply]

Active X controls and other internet security considerations

edit

I have several questions

1. I am soliciting an opinion about Javacool's Spyware Blaster which prevents the installation of ActiveX spyware. Spybot recommended it, I don't know much about it. Does it eat up a lot of memory to run? Would it significantly benefit someone like me whose surfing habits wouldn't upset a parental control module set to neurotic?

2. Anyone know any useful security extensions for Firefox? I have the NoScripts, but that's it.

3. Should I use OSCP to validate certificates in Mozilla Firefox? I really have no idea what that is.

4. How do you prevent CDs from automatically running programs upon insertion? I have heard all this junk about music CDs running and installing stuff surreptitiously and frankly I don't want to wonder when I play a CD or DVD. Can you disable it so that it won't run its contents upon insertion without first letting me see what is on the blasted thing?

Sifaka talk 16:48, 8 June 2006 (UTC)[reply]

Sorry that I can't answer all your questions, but to address number four, the default key in Windows for disabling run-on-load software on removable media like CDs or DVDs is to hold down the shift key when putting in the media. --Kuzaar-T-C- 18:48, 8 June 2006 (UTC)[reply]
In Windows XP, for a more permanent solution: go to My Computer, right-click on your CD drive, select Properties, then select the AutoPlay tab at the top. For each "content type", you can choose from a list of predefined actions, or select Prompt me each time to choose an action. --LarryMac 20:44, 8 June 2006 (UTC)[reply]
Question number 1: If you're using Firefox, no ActiveX can run, so you wouldn't need the "Spyware Blaster" program. (unless you're using a special ActiveX-enabled version of Firefox or have a Firefox ActiveX extension, but don't use those.) Question number 2: Firefox is inherently quite secure, and you shouldn't need too many extensions to make it more secure. Still, there are still are some good security extensions which you can search for here. Question number 3: I don't know much about validating certificates, but from my experience Firefox is quite picky about validating certificates, so you should be fine. Question number 4: A good solution was suggested above, but I'll add to it. If for some reason, you keep AutoPlay enabled, you can still hold the Shift key while you're putting the CD in the drive to stop AutoPlay. I hope this has helped. :) -- Daverocks (talk) 08:16, 10 June 2006 (UTC)[reply]

humming in my head

edit

When I am very quiet I can hear humming in my head. I don't think it's Tinnitus, and it is high rather than low pitched (which is how The Hum is described). I've wondered if it could be caused by the electrical activity in the brain. The lower limit of hearing in humans is cited as about 20 Hz, which is only slightly above the that of Alpha waves. However, if that were the case one would expect a low-pitched sound. Any ideas? --Halcatalyst 17:57, 8 June 2006 (UTC)[reply]

Do you have any dental work ? People do sometimes pick up radio signals there which are transformed into an audio signal. They aren't very loud, but seem so since they are inside your own head. In rare cases, people can even make out words or music being broadcast. Another possibility is that it's just a malfunctioning nerve in your ear. This happens all the time, but would be temporary and in only one ear. StuRat 18:21, 8 June 2006 (UTC)[reply]
I strongly suspect it's tinnitus. I get the same thing (in fact, I have it right now) and it seems to match the description in the article. Nice theory for The Hum though. Skittle 18:23, 8 June 2006 (UTC)[reply]
Why do you (the questioner) think it is not tinnitus? Are you immune to tinnitus for some reason? --Kainaw (talk) 18:27, 8 June 2006 (UTC)[reply]
You can't hear your brain. That's just silly. Even if you could, you would go through adaptation to accept it as something to be ignored. —   The Mac Davis] ⌇☢ ญƛ. 02:47, 9 June 2006 (UTC)[reply]
FWIW< I get tinnitus, and what I hear is a very high pitched hum, a bit like the "carrier wave" of an old TV. Grutness...wha? 02:58, 9 June 2006 (UTC)[reply]
Assuming that Halcatalyst is correct in thinking that it's not tinnitus, then I think I know what it is.... I have the same thing.
IANAMD, I don't have the knowledge of biology or physics to define it properly, and I'm only sharing what I've heard, from someone who really should know what he's talking about (I asked an ear specialist.... I saw him about an unrelated matter, but asked about this because I take my hearing very seriously) and this was about twenty years ago, so please don't jump all over me if it's totally wrong.
It's like a reference tone.... a sort of continuously generated frequency against which the brain evaluates other sounds. Most people are not conciously aware of it, (precisely because of adaption, I should think) and it's not clear why any of us are able to hear it. TheMadBaron 03:45, 9 June 2006 (UTC)[reply]
In an oft-quoted passage of John Cage's book Silence he describes entering a soundproof room, and he hears

... two sounds, one high and one low. When I described them to the engineer in charge he informed me that the high one was my nervous system in operation and the low one was my blood in circulation.

I have no opinion on the authenticity of this claim. —Blotwell 11:47, 9 June 2006 (UTC)[reply]
Well, the articles on Cage and the work those sounds inspired (4'33") suggest that it's probably not true. Interestingly enough there may be another explanation - your ears are generating the sound. The ear actually has a feedback mechanism that as well as hearing sounds will, under the right conditions, produce them (it's supposed to help you hear very quiet noises), and while these sounds are usually not independently audible, there have been cases where other people have heard the noises someone's ears make. I may be able to WP:CITE a source or two in a few days' time. Confusing Manifestation 12:54, 9 June 2006 (UTC)[reply]
What you describe sounds like tinnitus (no pun intended). - Cybergoth 14:22, 9 June 2006 (UTC)[reply]

This happens to me too. Sometimes when I'm in bed I swallow and all of a sudden there a high-pitched noise in my ear. I always thought it was just background noise but I guess it's more than that. But it only happens in the middle of the night for me (could that mean something?) --Jonathan talk 21:45, 9 June 2006 (UTC)[reply]

Are you sure it's not tinnitus, because it seems like it would be that, despite you not thinking it is?--Proficient 03:13, 10 June 2006 (UTC)[reply]

Where is the best place to live on Mars?

edit

If Mars is terraformed, where would the best place to live be? Considering climate, mountains, water supply and others I might not be thinking of. It would be a great help if someone could help me with this.

If you go here there are some pictures of Mars:

These are the two I’m basing it on: http://www.lns.cornell.edu/~seb/celestia/marsc-1k.jpg http://www.lns.cornell.edu/~seb/celestia/mars-mola-2k.jpg

Others: http://www.lns.cornell.edu/~seb/celestia/gallery-002.html

~Cathy~

It would probably be difficult to predict the climate and water supply of a particular place on Mars before it is terraformed. Terraforming is a rather intensive undertaking, and it would change a whole lot about the planet's surface. —Bkell (talk) 18:17, 8 June 2006 (UTC)[reply]

But it wouldn't change how the planet is shaped (mountains), would it?

~Cathy~

Here are some pictures I've modified with water... sorry that you have to kinda download it... I just uploaded them real quick. http://www.esnips.com/web/harry-potter-here ~Cathy~

Terraforming probably wouldn't change where mountain ranges are (although maybe it would, if it were done by creating volcanoes or something). But there are many factors besides mountains that make somewhere a good place to live. You will want to know about the climate, nearby water, natural hazards (like storms), vegetation, and so on. Think of what the Earth would be like if it were "unterraformed". As an example, the coast of Norway is about the same latitude as northern continental Canada, but their temperatures in January differ by about 30°C because of the Gulf Stream. This effect would be very difficult to predict if Earth had no oceans and very little atmosphere. How can you even predict what sea level would be on a terraformed Mars? —Bkell (talk) 18:54, 8 June 2006 (UTC)[reply]
While I agree with your point about the Gulf Stream, a Martian sea level is easy to predict -- just pick a contour line and color everything below it blue. — Lomn Talk 19:01, 8 June 2006 (UTC)[reply]
At second thought, you're probably also pointing out the uncertainty of the quantity of water, not the shape an arbitrary quantity would take. Anyway, to go back a bit: the original poster might want to check out Kim Stanley Robinson's Mars trilogy, which (among other topics) discusses terraforming. As a quick guideline, though, you would probably aim for an equatorial low-elevation region to maximize temperature and air pressure. If a society is capable of terraforming a planet, they're also presumably capable of building water pipelines, so I wouldn't worry overmuch about proximity to potential oceans. — Lomn Talk 19:08, 8 June 2006 (UTC)[reply]
Written before your last comment: That would be a Martian coastline you're predicting, not the Martian sea level. How do you figure out which contour line to choose? You would choose the one corresponding to the predicted sea level. Now how do you predict the sea level? You will have to know how much water is currently on Mars, how much would be in liquid form after terraforming, how much would be above ground, whether water would be imported or created during the process of terraforming, and so on. —Bkell (talk) 19:10, 8 June 2006 (UTC)[reply]
I was interpreting the question as asking for the most aesthetically pleasing or the most comfortable place to live on Mars, assuming that the planet had been "Earthified", hence my comment about nearby bodies of water. Your recommendation of an equatorial low-elevation region seems to be mostly an engineering recommendation, since it would be easiest to convert such a location to a habitable environment. Of course, one can actually make reasonable engineering recommendations. ;-) —Bkell (talk) 19:16, 8 June 2006 (UTC)[reply]
You can't just color everything blue below a certain contour line, can you? Because Mars is spinning, sea level will not remain exactly at the same level around the globe. Isn't that why the sea level on one side of the Panama canal is different than the sea level on the other side? --Kainaw (talk) 20:31, 8 June 2006 (UTC)[reply]
Some of my first response was also based on the aforementioned sci-fi stuff, which postulates that terraformers could exercise some control over the amount of water released, thus predicting an approximate coastline. As for the Panama Canal question, that's not an issue of centripedal force bur rather of lunar tides and America impeding the free flow of water. Mars doesn't have moons capable of causing significant tides. However, centripedal force does cause planetary deformation (sea level at the equator on earth is a few miles farther from the center than sea level at the poles). Depending on whether/how a contour line is drawn on a martian map, this might/might not matter -- I presume it would be readily predictable in advance, though, and it would still provide a good rough approximation — Lomn Talk 20:35, 8 June 2006 (UTC)[reply]
This is off the subject, but what do you mean by "America impeding the free flow of water"? Surely the sea level was different on either side of Panama before the canal was built, or even before the United States existed. —Bkell (talk) 22:07, 8 June 2006 (UTC)[reply]
America the continents, not the country. Landmasses delay the tides and so the coastlines of the Atlantic and Pacific, though close as the crow flies, are incredibly distant with respect to sea level fluxuations. — Lomn Talk 03:22, 9 June 2006 (UTC)[reply]

I just need a small section around 220,000 km² big with water nearby, not a lot of mountains (but some), and good climate. Just make a good guess... you'd be closer than what I would guess. Plus, I have no idea how much space 220,000 km² would take up on Mars! ~Cathy~

Well, obviously, unless huge mirrors were used to provide additional illumination to polar and mid-latitude regions the tropics will, in general, be the warmest place. One possibility that occurs to me is the bottom of Valles Marineris. ---Robert Merkel 23:54, 8 June 2006 (UTC)[reply]

Why the bottom? (BTW, I've just started to learn about Mars (about two weeks or so) ~Cathy~

Generally, you'd want the bottom to maximize air pressure. As is, the lowest elevations on Mars are (with respect to air pressure) comparable to or lower than that found atop Everest. — Lomn Talk 03:23, 9 June 2006 (UTC)[reply]
Not only is the air going to be thicker, it's going to be warmer. Additionally, water flows downhill, so you'd expect there to be water at the bottom of the largest canyon in the solar system if liquid water was indeed flowing on your terraformed Mars. --Robert Merkel 07:13, 9 June 2006 (UTC)[reply]

Acetyl-CoA synthesis

edit

How is acetyl-CoA synthesized? The article on CoA says that "Coenzyme A (CoA, CoASH, or HSCoA) is adapted from β-mercaptoethylamine, panthothenate and adenosine triphosphate" and also "The conversion of pyruvate into Acetyl-CoA..." Not sure if these mean that CoA can be synthesized from either. Does it? Jack Daw 18:17, 8 June 2006 (UTC)[reply]

Coenzyme A is the universal carrier of acyl groups, a special case of these are acetyl groups. The acetyl groups may be derived from glycolysis, such as in your example (pyruvate), and they may also be derived from the breakdown of fatty acids and certain amino acids. The statement "The conversion of pyruvate into Acetyl-CoA..." is a slightly misleading, the reaction is
pyruvate+CoA+NAD+ → Acetyl-CoA + CO2 + NADH,
according to my 1975 copy of Lubert Stryer, Biochemistry (yes, I'm considering buying a newer copy :-) ).
CoA is recycled, it is neither synthesised nor consumed in this reaction.
According to the same source, the biosynthesis of CoA starts with pantothenate (vitamin B5), which is first is linked to cysteine, a carboxyl group is lost, AMP is added (from ATP), and finally the product is phosphorylated. I'm not sure about where the β-mercaptoethylamine enters, maybe it's an alternative pathway for synthesising CoA. --vibo56 talk 18:55, 8 June 2006 (UTC)[reply]
The word enzyme is reserved for biological catalysts, and coenzymes are simply the active sections of binary enzymes. Thus "coenzyme" is mostly a fancy word for a type of catalyst. Since catalysts by definition are not altered by the chemical reactions they catalyze, Coenzyme A is not used up in acylation. (This is less of an answer and more clarification / support for the addition immediately preceding this one).Tuckerekcut 20:09, 8 June 2006 (UTC)[reply]
  • No. You're confusing two things, one is the synthesis of brand new (Acetyl-)CoA, which occurs from Adenosine, etc. The other is the synthesis of Acetyl-CoA from CoA which occurs in the Citric acid cycle. Once synthesised, Coenzyme A serves to pick up an acetyl group (forming Acetyl-CoA) which it later loses, forming CoA again, but the base CoA stays the same. --BluePlatypus 16:31, 9 June 2006 (UTC)[reply]

Virtual Machine Screensaver

edit

On Mepis Live (linux) there is a screensaver called 'virtual machine' that appears to display a load of random coloured letters and numbers across the screen, kind of matrix stylee. Are these randomly calculated or are they actually displaying some sort of information like what the processor is up to? --Username132 (talk) 19:55, 8 June 2006 (UTC)[reply]

From the program's description: "a virtual machine that can be used to make a simulation of automata construction under randomly changing conditions." I don't really know exactly what that means. --Kainaw (talk) 20:29, 8 June 2006 (UTC)[reply]
It means nothing. Well technically it means "a fake machine that can be used to simulate an automatically interacting device with random input", but simplified to non-bullshit English that basically means "a program that runs on random numbers", which answers the first user's question.  freshofftheufoΓΛĿЌ  03:42, 9 June 2006 (UTC)[reply]
Or... it might be referring to cellular automatons. I don't have access to the program, so its hard to say whether its using a specific type of algorithm, or just spouting bullshit. --Monguin61 19:58, 14 June 2006 (UTC)[reply]

How does this edit conflict thing work?

edit

So I just added a (small) contribution to the Acetyl CoA question, and ran into an editing conflict. I simply accepted without changing (as I seem to have screwed this up earlier today...) and the contribution is not recorded. Is the newer edit always just erased in these situations? The edit conflict wasn't even in the same section as the other additions (they did not overlap). Needless to say this is very frustrating, as it is the second time in 14 hours.Tuckerekcut 20:07, 8 June 2006 (UTC)[reply]

I have redrafted my contribution to the CoA question, but the frustration remains, why would the newer article be automatically removed in an editing conflict?Tuckerekcut 20:13, 8 June 2006 (UTC)[reply]

The Wikipedia:Help desk is a better place for these kinds of questions, but yes, basically the newer version is discarded if you let it do so. The edit conflict screen shows you two text boxes and clearly states that the one on the top is the one that will be saved. Your edits are shown in the one on the bottom. What you need to do is to copy-and-paste your edits into the one on the top, making any adjustments as necessary to account for the fact that the article you meant to change has now already been changed. This is necessary, as how else would wikipedia understand what to do with two potentially conflicting edits? It also alerts you to the fact that something new has happened between the time you started editing and when you pressed "save". — Asbestos | Talk (RFC) 20:16, 8 June 2006 (UTC)[reply]
I never realised how complicated the edit conflict issue was, cuz I just thought everyone did what I do, which is select and copy my new unsaved text, exit the edit window without saving, go into a fresh edit window, and add my text to the newest version. I also try to edit small sections when they exist rather than editing from the top edit tab all the time. You know I didn't even know until just now that it was possible to deliberately overwrite previous changes from the edit conflict window; that's why I didn't understand lethe's explanation above. LOL There was an edit conflict while I was replying.--Anchoress 20:16, 8 June 2006 (UTC)[reply]
I know I'm pointing out something you most probably know very well, but just to be on the safe side: When you want to edit a section, you do click on the small [edit] link on the right hand side just above the horizontal line that defines the section, and not the "edit this page" tab at the top of the page, right? --vibo56 talk 20:29, 8 June 2006 (UTC)[reply]

Yeah, I have just been editing with the little link accross from the section header, but that has apparently not eliminated editing conflicts when other sections are changed. It does seem like the software could just add in all information contributed after a conflict, rather than discarding the most recent. I suppose the "edit conflict" warning window is still neccesary, to allow users to reformat, but I don't see any reason why wikipedia shouldn't be able to add all the edits together. Judging from the comparison section of the conflict screen, it "knows" where to put the new information, and simply "chooses" to make the user do the copying and pasting.Tuckerekcut 20:48, 8 June 2006 (UTC)[reply]

That's largely a function of the page in question. For a large talk page like this, where users are editing only their own paragraphs, sure, automated conflict resolution is viable. What about an article edit, though, where two people are altering the same paragraph at once? Computer systems aren't capable of intelligently resolving that situation on a consistent basis, so merges are left to the human users in question. — Lomn Talk 21:26, 8 June 2006 (UTC)[reply]

There do appear to be three problems that could be addressed, however:

1) There is no way to edit the top "introduction" section of an article without editing the entire article. This makes edit conflicts more likely.

2) When two people do edit the same section at once, rather than giving a section-level edit conflict, this causes an article-level edit conflict. This means that anyone saving from that point is saving the entire article, and more likely to encounter additional conflicts with unrelated edits in other sections.

3) Self-edit conflicts are possible, usually as a result of errors and/or using the back button on the browser. If both edits were made by the same user, it should just ignore the earlier one and accept the final edit. StuRat 21:51, 9 June 2006 (UTC)[reply]

  • Unless two editors are trying to edit the same line (or append text after a given line), MediaWiki does resolve edit conflicts automatically. That includes introduction edit conflicts. Titoxd(?!?) 04:56, 12 June 2006 (UTC)[reply]
    • That's not true. Try finding a random section in this page, clicking edit, then waiting ten minutes or so before saving. You'll get an edit conflict, whether or not the conflict was on the line you were editing. I just proved myself wrong. Um, ok, my mistake. — Asbestos | Talk (RFC) 13:47, 12 June 2006 (UTC)[reply]
Answer to 1) - not true, you can edit the introduction with a small URL hack. Copy the URL for the "edit section" link (for instance, http://en.wikipedia.org/w/index.php?title=Wikipedia:Reference_desk/Science&action=edit&section=79), and replace the section=79 with section=0. Enter the new URL into your browser's address bar and voilà! — QuantumEleven 10:47, 13 June 2006 (UTC)[reply]

Distance Light Travels

edit

Last night I took my flashlight and some dog food to my backyard to feed my dog. While walking back inside I was throwing my flashlight (while turned on) up in the air and catching it. It occured to me that the light has tyo go somewhere as there aren't any trees above it or anything. How far does the light go? If an airplane were to fly above, would the light shine in the passengers eyes? Just FYI I have

 
My kind of flashlight

this kind of flashlight. Thanks schyler 21:36, 8 June 2006 (UTC)[reply]

It'll keep going until it hits something, but it'll spread out so much that the airplane passengers, for instance, probably won't be able to make it out as brighter than the background. Some of the individual photons, though, will surely make it far into space to be absorbed by interstellar hydrogen atoms. —Keenan Pepper 21:58, 8 June 2006 (UTC)[reply]
The thing is, it goes in a cone rather than a cylinder, which you can verify because the spot gets larger as you shine it on more distant surfaces. As the same amount of light spreads over a larger area, it gets dimmer. —Keenan Pepper 22:03, 8 June 2006 (UTC)[reply]
Light which "stays together" more (say, in a laser) keeps on going pretty high (that is, is identifiable as coming from a single source and is very visible). Hence there have been a number of people who have gone to jail for shining laser pointers at airplanes and things like that and actually making it into the cockpit with them. (see this story about it, for example). --Fastfission 00:19, 9 June 2006 (UTC)[reply]
Wow, with that kind of flashlight, be careful about shining it out into space unless you want to be accused of the second "most reckless act in the history of mankind". alteripse 01:47, 9 June 2006 (UTC)[reply]
I don't mean to detract too much from the discussion at hand, but I couldn't help but wonder - why do you consider the Pioneer plaque to be foolish, alterelipse? — QuantumEleven 10:51, 13 June 2006 (UTC)[reply]
He doesn't - he was quoting someone else (whose name escapes me at the moment) who said that about the Pioneer plaque. The reason is that any hostile aliens could use the pioneer plaque to find mankind and destroy/enslave us (ala Battlefield Earth). Raul654 10:53, 13 June 2006 (UTC)[reply]

An interesting thing to try is going out on a country road at night. Point the flashlight at a reflective roadsign some distance away. Even a small household torch will cause a reflective sign of good highway quality to reflect from a distance of up to 1/4 of a mile (400m). Grutness...wha? 03:01, 9 June 2006 (UTC)[reply]

Also, "hits something" also applies to particles in the air, most of which are probably water, which would deflect or possibly absorb much of the light. On a super clear day it may be possible to see a speck of light from a rather powerful flashlight, but on a foggy/rainy day you might not even be able to see the light from a giant light house, many hundreds of times stronger than your average flashlight.  freshofftheufoΓΛĿЌ  03:37, 9 June 2006 (UTC)[reply]

Note that the spotlights used for advertising new stores (and sometimes used to spot airplanes at night during wars) can light up objects thousands of feet up, even though they aren't lasers. They make up for their higher light scattering rate (than lasers) by using more photons to start with. You can tell if a lot of scattering happens by whether the beam itself is visible. On a foggy night, it will be quite visible, but won't go very far, while on a clear night it will be almost invisible and will go quite far. Stars can even send light through billions of light years of almost completely empty space to reach us. StuRat 21:32, 9 June 2006 (UTC)[reply]

Compare it to a 1mW laser pointer. If we knew at what distance a laser pointer is barely visible at night, then we'd know that a flashlight would only be barely visible at a much closer distance. Most google hits only mention at what distance the spot is visible on a surface. One says that a 95mW green laser beam is visible at 45 miles distance. --Wjbeaty 05:26, 11 June 2006 (UTC)[reply]

The most interesting answer is that in all likelihood some, maybe most, of the photons you are sending out will travel unimpeded through intergalactic space for many billions of years, toward the end of the observable universe, in the same way as we now are observing light from early galaxies that has travelled that far through space before "hitting something" - a telescope... Space is pretty empty. - Mglg 03:25, 15 June 2006 (UTC)[reply]