Wikipedia:Reference desk/Archives/Science/2010 May 10
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May 10
[edit]rate for humans
[edit]What is the maximum number of calories a human could burn an hour without setting themselves on fire? (by burning I mean through exercise and by setting themselves on fire I mean with a match like some crazy protester.) 71.100.0.29 (talk) 02:15, 10 May 2010 (UTC)
- I seem to recall that the highest scientifically measured metabolic rate was during tests of the Gossamer Condor, a human-powered aircraft. (Gossamer Condor and Albatross: A Case Study in Aircraft Design, AIAA, 1980). I vaguely recall numbers on the order of feeding the cyclist a 10,000-calorie-per-day diet, and working all those calories out of him by having him pedal at top efficiency for many hours each day. [1]. This was about the maximum; feeding the cyclist more sugar, or training him harder each day, did not effectively get any more energy out of him. This thesis, Human Powered Helicopter (1991, Naval Postgraduate School), has a nice plot on page 12, indicating maximum power output plotted vs. time that power can be sustained. It is feasible to sustain approximately 1.5 horsepower (or about 1000 calories-per-hour) for about one minute. It is feasible to sustain half that rate for a much longer period of time. After a few hours at ~500 calories-per-hour, though, a human's entire daily food intake would be used up; hence the research into intense super-nutrition, sugar-syrups, and so forth, for the human-powered flight. Per the OP's question, these metabolic rates are well below the autoignition temperature of a human, so through exercise alone, it is impossible for a human to ignite. Nimur (talk) 02:31, 10 May 2010 (UTC)
- I've personally burned 1000 calories in an hour running on a treadmill, if the numbers were accurate, and I was nowhere close to a top-level athlete. Note however that the calories the body burns do not equal the power output of the person -- well over half of it goes into waste heat. My understanding is that somebody like Lance Armstrong during a race will burn over 1500 calories per hour for at least a couple of hours. Looie496 (talk) 03:29, 10 May 2010 (UTC)
- I would not believe what the treadmill says unless you have an oxygen mask on. There is a factor of ten in unknowns which they are guessing. Ranulph Fiennes has been quoted as saying that racing in the artic you lose weight on 8000 calories a day but you cannot metabolise more than this. He quotes 15000 calories output but beware trying to match the performance of someone who cut off his own finger-tips... --BozMo talk 11:41, 10 May 2010 (UTC)
- What do you mean by 'There is a factor of ten in unknowns which they are guessing.'? What unknowns, and why a factor of ten? --22:07, 10 May 2010 (UTC)87.114.95.229 (talk) signed too late—Preceding unsigned comment added by 87.114.95.229 (talk) 22:06, 10 May 2010 (UTC)
- I would not believe what the treadmill says unless you have an oxygen mask on. There is a factor of ten in unknowns which they are guessing. Ranulph Fiennes has been quoted as saying that racing in the artic you lose weight on 8000 calories a day but you cannot metabolise more than this. He quotes 15000 calories output but beware trying to match the performance of someone who cut off his own finger-tips... --BozMo talk 11:41, 10 May 2010 (UTC)
Mystery Plant
[edit]Hello! I am hoping there are some botanists here who can shed some light on this plant for me. (pun not intended.) This plant has recently come into my possession, and the prior owner was just as clueless as I am. I am hoping to find out what species this plant is, or at least it's family, so I can best take care of it. Any help is much appreciated. (A few more images are available if needed.) Avicennasis @ 04:25, 10 May 2010 (UTC)
- It is definately a type of succulent plant, likely a member of the Crassulaceae family. My best guess is this is a Crassula, likely a Jade plant or some similar. You might want to search through various genera and species of the Crassulaceae family to find a good match, if the Jade plant isn't it... --Jayron32 04:41, 10 May 2010 (UTC)
- You might, in the absence of an answer here, like to try Flikr. Accounts are free and they have a group especially for unidentified plants (craftily called "What plant is that")[2]which I can confirm provides an excellent and quick service. Whatever it is it needs much more light than it is (or has been) receiving. It definitely looks etiolated. Caesar's Daddy (talk) 06:17, 10 May 2010 (UTC)
- This is some variety of Echeveria, possibly E. elegans. Here [3]is a photo of several varieties of which the one at front centre look like yours but is in a normal form and not drawn up from lack of light. Richard Avery (talk) 19:59, 10 May 2010 (UTC)
russia
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This question inspired an article to be created or enhanced: |
why are many (not all) russia drunks? —Preceding unsigned comment added by Tom12350 (talk • contribs) 04:36, 10 May 2010 (UTC)
- [citation needed]. Do you have any evidence that Russians are more prone to alcholism than people of other cultures? There are stereotypes of Russians as such, but there's lots of bullshit stereotypes out there, many of which have zero basis in fact. Russians drink, but so don't people from other countries. --Jayron32 04:46, 10 May 2010 (UTC)
- Alcoholism in Russia happens to be a redirect to the "Russia" section of the Binge drinking article, and that section states that Russian binge drinking is basically on par with most other European countries. It doesn't specifically discuss alcoholism in Russia. Comet Tuttle (talk) 05:27, 10 May 2010 (UTC)
- The Russians seem to think alcoholism is a "national calamity" (Pravda, 2006) and a "national disaster" (RIA Novosti, 2010). This article says "Consuming on average 32 pints of pure alcohol per person per year, Russians are among the heaviest drinkers in the industrialized world." According to this academic commentary, "Alcohol, as a central component of life in Russia, has been commented on, by Russians and by travellers from other countries, since at least the tenth century AD." Fee fi fo fum, I smell the blood of a new article, but I don't have time today. Clarityfiend (talk) 06:04, 10 May 2010 (UTC)
- Aren't the Irish the biggest drunks of all? :-) 67.170.215.166 (talk) 08:43, 10 May 2010 (UTC)
- I discussed this in a conversation with one Irish and one Russian friend of mine. I think the conclusion was that Irish drink more, but Russians drink harder. Vimescarrot (talk) 09:45, 10 May 2010 (UTC)
- I'll drink to that. Cuddlyable3 (talk) 18:29, 10 May 2010 (UTC)
- I discussed this in a conversation with one Irish and one Russian friend of mine. I think the conclusion was that Irish drink more, but Russians drink harder. Vimescarrot (talk) 09:45, 10 May 2010 (UTC)
- Aren't the Irish the biggest drunks of all? :-) 67.170.215.166 (talk) 08:43, 10 May 2010 (UTC)
but why —Preceding unsigned comment added by Tom12350 (talk • contribs) 19:15, 10 May 2010 (UTC)
- My unsourced, inexpert take on it: Drinking has been deeply ingrained in Russian culture for centuries and therefore more or less accepted. In modern times, Russia, and the Soviet Union before it, were dreary places for the masses, so they drank as an escape. Clarityfiend (talk) 19:57, 10 May 2010 (UTC)
- Another possible reason would be the extremely cold climate, which would encourage some people to drink hard liquor in order to "keep warm" (a potentially dangerous practice that can in some cases lead to hypothermia or even death). FWiW 67.170.215.166 (talk) 02:03, 13 May 2010 (UTC)
Where can I find paid gigs to be a guinea pig for a medical experiment within 150 miles of Manhattan, KS? (With room & board included?)
[edit]Now you see, I've heard about willing members of society becoming the subjects of medical experiments where a medical school tests prototype drugs and/or experimentally new medical procedures on any willing participant.
I think nowadays would be a good time to start becoming a medical test subject. However, it must be within reasonable distance (~150 miles) of Manhattan, KS, with local lodging also paid for by the medical school or whatever medical research lab is doing these experiments.
So could someone help me find these gigs? In whatever links you give, be sure it'll also give the amounts each test subject will be paid for each experiment (or time-period of these sets of experiments).
For those of you who are about to ask me to just Google it, I already did, and the results were not at all helpful.
--Let Us Update Wikipedia: Dusty Articles 06:39, 10 May 2010 (UTC)
- It seems highly unlikely you could support yourself with such a "career", most medical test subjects receive a very small honorarium for participating in the test; usually enough to cover travel expenses and maybe give you enough cash for a night on the town (say, $50-$100). Many will also pay for medical expenses should any complications arise. I can't think of a single lab that would cover room and board, nor would it likely be possible to participate in enough tests to live off of. Many tests may be mutually exclusive, and most require you to have a medical condition first, and already be under the treatment of a doctor. For example, if they were testing a radical new cancer treatment, it would do you no good if you don't have cancer. Also, some tests you participate in may preclude you from participating in other tests. There are some paid testing where they do take "anyone off the street" who volunteers for the test, but these are so unreliable, rare, and poorly paid that I can't imagine you supporting yourself on such a gig. --Jayron32 06:48, 10 May 2010 (UTC)
- There is the NASA bed rest study, where they do provide room and board, except you have to stay in a bed without getting up. If you are interested: https://bedreststudy.jsc.nasa.gov/ Ariel. (talk) 07:24, 10 May 2010 (UTC)
- I'm a unit assistant for a clinical trials unit at my University hospital. We board test subjects for a week, and offer taxis/rides into town. We pay them 300 dollars / wk and they get 15/dollars a day for meals via hospital vouchers. But yeah, there are conditions or predispositions we target. John Riemann Soong (talk) 07:25, 10 May 2010 (UTC)
- If you can stretch your travel radius to Wales and don't mind getting the sniffles, consider signing up at the Common Cold Research Centre. Cuddlyable3 (talk) 10:02, 10 May 2010 (UTC)
Flexing axe heads
[edit]I split a fair bit of wood (not commercial, just for home), often with a standard axe and a sledge hammer. The complete book of self sufficiency says one thing you should not do is hit the back of a stuck axehead with a sledge hammer, using it as a splitting wedge. The book says doing this will break the axe handle but I find the tempation to do so irresistable when the axe sticks in a tough log. And the book is right, it breaks the axe handle (not at once but over a few hours misuse), AFAICT not because the sledge hammer ever hits the wood but apparently because the heavy steel axe head deforms enough under the strike of an eight pound sledge hammer to damage the wooden handle in the middle of it (??is that possible). The bit of the handle inside the axe head pulverises, leaving the rest of the handle intact (so I can plane off and put the axe head back on a couple of inches lower. So what's the deflection mode of the axe head and can I do anything to stop the damage to the handle? This particular bad habit is a great time saver. What if a cut the handle so it had a top and bottom gap and was held by the sides? Or put in a piece of rubber or something? Any ideas? --BozMo talk 10:26, 10 May 2010 (UTC)
- As an aside, you can get splitting wedges (basically a wide axe-head without a shaft). If the axe gets stuck in the log-end, then tap the wedge into the wood in the crack you've made, free the axe, tap the wedge in a bit more to secure it, then whack it with the sledge hammer. CS Miller (talk) 10:45, 10 May 2010 (UTC)
- Thanks I have a splitting wedge but I need to get through logs in a matter of a few seconds and collecting a wedge from where it landed, putting it into a log, tapping with a smaller hammer, wriggling out the axe and then moving the wedge centrally and then hitting it with a big hammer takes ages compared to 50% one strike 50% two strike with an axe and hammer --BozMo talk 10:49, 10 May 2010 (UTC)
- Based on what I know about axes and hammers (which is not that much), the sledge puts the back side of the ax blade under compression, causing the hollow channel in the blade that holds the handle in place to flex ever so slightly in the direction of the blow, and to put a bending stress on the handle, which eventually cracks from material fatigue. But hey, I could be wrong about this, no? As far as what you can do about it, cutting the handle so that it's only held by the sides is a very bad idea -- it will just let the ax-head wobble and maybe lead to a major accident. One possible solution would be to get an ax with a handle made of some ductile material (not rubber, but some kind of hard plastic could work), or machining a broad groove in the handle where it passes through the ax-head and putting a sleeve of hard rubber on it, then putting the ax-head back on so that it rests on that sleeve. Another alternative would be an all-metal ax, like a tomahawk or something. FWiW 67.170.215.166 (talk) 10:57, 10 May 2010 (UTC)
- (ec) I don't think it's because the axe-head is deforming and breaking the handle - I think it's because of the inertia of the handle. When you hit the head, it moves very abruptly - the unyielding metal of the hammer and the head mean that when the fast-moving hammer hits the stationary head, the axehead has to accelerate from zero to something like half the speed of the hammer head in the very short amount of time over which the two can deform. The acceleration is likely to be hundreds of g's - which is then (more gradually) reversed as the axe is slowed down again by its passage through the wood that you're chopping and is again stationary. Meanwhile - that LONG axe-handle has to follow along. The end that's attached to the axe-head has to accelerate just as the head does - but the center of mass of the handle is roughly halfway along it - so the inertia of the handle acts a long way away - so it has lots of leverage - and the result will be all sorts of really nasty forces propagating along the handle...enough (evidently) to weaken, and eventually break it.
- What surprises me is that some sledge hammers also have long handles - how come the sledge-hammer's handle doesn't break for the same reasons?
- I understand that using a splitting wedge slows you down - but not as much as a broken axe handle. This job is simply going to take longer than you thought.
- As far as I know, the primary reason for not striking the back of an axe with the hammer is avoiding dangerous splinters. Never hit steel on steel. I think you might do well to invest into a heavy rubber mallet - that should be easier on your tools and easier on your eyes, and it can be used with your current work flow (although you may need a few more blows). --Stephan Schulz (talk) 11:50, 10 May 2010 (UTC)
The OP needs a Log splitter. Cuddlyable3 (talk) 18:27, 10 May 2010 (UTC)
- Curiously I did look into it but actually the better ones used by farmers are screw head ones which our article does not even mention, run off a tractor engine and they cost a couple of thousand dollars. For my 6 or 7 metre tonnes of logs a year thats an overkill. The cheaper hydraulic ones like in the WP article seem to be okay for dealing with small diameter plantation wood or a coppice but hopeless when you have big hardwood logs which comes from falling trees in an old garden, and the very big hydraulic ones are hopelessly expensive for light use. So axe it is and it looks like I will go with Steve's advice. All aerobic anyway. --BozMo talk 19:38, 10 May 2010 (UTC)
Another way to split wide logs is to aim your axe so that the near end of axe's blade lands at the near end of log. Then on your second swing, aim the axe so that the near end of the blade hits just after the crack you've made. Hopefully the two cracks will join and the axe will go all the way through. CS Miller (talk) 20:07, 10 May 2010 (UTC)
- You can get axes with toughened heads suitable for being whacked with a hammer. I'll have a sniff about and if I can find a link to such will post it here. DuncanHill (talk) 22:17, 10 May 2010 (UTC)
- OK, what you need is a splitting maul. The blade is shaped for efficient splitting (a felling axe has a different profile, and will be less effective for splitting). The poll is toughened, so it can be used as a sledge or should be OK to use a sledge on it. Try a good manufacturer such as Gränsfors Bruks [4]. DuncanHill (talk) 22:23, 10 May 2010 (UTC)
- The technique I was taught was that when the axe gets stuck, rotate the axe handle 180 degrees, lift the whole thing up and strike the base block with the axe head lowest. If you are strong this often splits the wood. Polypipe Wrangler (talk) 10:57, 12 May 2010 (UTC)
Calculating the velocity of a cannon ball and cannon
[edit]Hi all,
I was trying to work out the velocity that a cannon ball and cannon would have after the cannon is fired in my question above (How do wheels on a cannon affect the force on the cannon ball?), but I think I failed. Can someone help?
Suppose we have a 1kg cannon ball and a 100 kg cannon. Suppose the explosion of the gunpowder provided 100 Joules, and that this was transformed perfectly into the KE of the cannon and cannon ball. How fast would the cannon ball move?
Thanks a lot, — Sam 76.24.222.22 (talk) 11:39, 10 May 2010 (UTC)
Conservation of energy: 1/2*(1kg)*(vb)^2+1/2*(100kg)*(vc)^2=100J
Conservation of momentum: 1*vb+100*vc=0
vc=-vb/100
1/2*(1kg)*(vb)^2+1/2*(100kg)*(-vb/100)^2=100J
1/2*(1kg)*(vb)^2+1/2*(100kg)*(-vb)^2/100^2=100J
(1/2*(1kg)+1/2*(100kg)/100^2)*(vb)^2=100J
vb=sqrt(100J/(1/2*(1kg)+1/2*(100kg)/100^2))=14.0719509m/s
vc=-0.140719509m/s 157.193.175.207 (talk) 13:46, 10 May 2010 (UTC)
- edited to correct sign error 157.193.175.207 (talk) 13:48, 10 May 2010 (UTC)
- (ec) I won't do your homework for you, but the key to this problem is conservation of energy and momentum. Since you can probably assume that both quantities are initially zero, you know the final energy is 100 J, and the final momentum is zero. You can work from there to find the velocities. anonymous6494 13:51, 10 May 2010 (UTC)
Interesting. So it seems from my calculations that, while it certainly makes a difference if the cannon is big and heavy, it doesn't make a huge difference
Using a cannon weight of 10kg, we get vb=sqrt(100J/(1/2*(1kg)+1/2*(10kg)/10^2)) = 13.48 m/s
Using a cannon weight of 100,000kg, we get vb=sqrt(100J/(1/2*(1kg)+1/2*(100000kg)/100000^2))=14.14m/s
Is that right? So in answer to my earlier question, bolting the cannon directly to the ship, so that the weight of the cannon would be equal to the entire weight of the ship, would make the cannon balls travel slightly faster, but not a lot. So putting wheels on the cannons (for which there are a bunch of other reasons) doesn't rob very much energy from the cannon balls. Is that right? — Sam 63.138.152.189 (talk) 14:31, 10 May 2010 (UTC)
- You could just put blocks behind the wheels and move them when you want to reload and have the best of both worlds. Googlemeister (talk) 15:41, 10 May 2010 (UTC)
- I didn't check the math but the conclusion seems right. The ratio of kinetic energies each of the two objects gets is the inverse of the ratio of their masses. So for a 10kg cannon, the cannon ball is already getting most of the total energy (10/11 of it). Rckrone (talk) 17:16, 12 May 2010 (UTC)
Solubility of nickel chromate
[edit]Does nickel chromate react with hydrochloric acid when it dissolves, or does it just dissolve? Another related question is: Is chromic or hydrochloric acid stronger? You only have to answer one of these. --Chemicalinterest (talk) 11:40, 10 May 2010 (UTC)
- Question 1) Depends on how you define "react" and "dissolve". If nickle and chromium ions cannot coexist in solution together, then it seems unlikely that the presence of excess hydronium or chloride would effect the situation. If you are trying to make an insoluble salt dissolve, then something sort of complexation is probably needed, much like how at VERY high pH's, soluble copper hydroxide complexes form which can cause Cu(OH)2 to redissolve as the Cu(OH)42- ion. The only lewis base in your proposed system would be the chloride ion, so I would suspect that, if it DOES dissolve, you are creating NiCl42-, which is a complex ion attested to in the Nickel(II) chloride article. For Question 2) Chromic acid is not a compound which readily exists either isolated or in water. Stable "chromic acid" generally only works as a mixture of chromate (or dichromate) salts and a strong acid such as hydrochloric or sulfuric; and the strength of these chromic acid solutions probably depends on the strength of the co-acid present. From the point of view of Arrhenius theory, there is no acid which can be "stronger" than HCl anyways; it dissociates extensively in water; since acid strength is a measure of % dissociation, no acid may be stronger than HCl, though several "tie" it in terms of acid strength. --Jayron32 13:48, 10 May 2010 (UTC)
- There is: No acid dissociates 100% in water. But does it react with HCl, i.e. to form Cl2, or does it just dissociate and dissolve? According to my anionic activity theory: Ni2+ is higher up on a standard reduction potential chart than H+. If chromic acid is stronger, than the Ni2+ will stay with the chromate, causing no reaction. If hydrochloric acid is stronger, the H+ ion will take the chromate, forming chromic acid. I know some people don't like my idea of ions in solution "sticking" together, but that is what my opinion of anions is. --Chemicalinterest (talk) 14:29, 10 May 2010 (UTC)
- (edit conflict with below) My Table of standard reduction potentials shows that Cl2 is a (slightly) better oxidizer than chromate (or dichromate, whatever), which, if we believe that, means that the Cl- will not be oxidized to Cl2. As to your first question specifically, look at Nickel Chromate, where it says it dissolves in an HCl solution, forming a "yellow solution". This is, as Jayron says, likely to be Ni(Cl)42-: see here. Finally, Chromic acid is weird (as Jayron points out), so it doesn't really make sense to talk about it in the sense of a normal acid. Usually it's used as a strong oxidizing agent, rather than as an acid. The article makes is unclear as to what species is actually in solution, (Chromium trioxide vs. H2CrO4), but it really doesn't matter. Buddy431 (talk) 15:21, 10 May 2010 (UTC)
- There is: No acid dissociates 100% in water. But does it react with HCl, i.e. to form Cl2, or does it just dissociate and dissolve? According to my anionic activity theory: Ni2+ is higher up on a standard reduction potential chart than H+. If chromic acid is stronger, than the Ni2+ will stay with the chromate, causing no reaction. If hydrochloric acid is stronger, the H+ ion will take the chromate, forming chromic acid. I know some people don't like my idea of ions in solution "sticking" together, but that is what my opinion of anions is. --Chemicalinterest (talk) 14:29, 10 May 2010 (UTC)
- Look, dude, you can't just invent your own acid-base theory. Science doesn't work like that. What inadequate hole in the existing theory are you trying to plug? Not the least of which is that your theory doesn't work, as it presupposes the existance of things which do not exist. You keep insisting that substances which exist in water simply do not; like a week ago when you kept trying to find discrete NaCl particles floating in water. Lets see if we can dispense with some of the multitude of misconceptions in the above proposal you have. Regarding HCl's dissociation in water: Insofar as there is no perfection in the universe, yes, nothing is 100%. However, insofar as we must use measuring devices to detect things, there is no way to measure the amount of discrete HCl particles in water, so it is as functionally close to 100% dissociated as we need it to be. One can make comparitive gas-phase measurements of acid strength between acids, but any acid which has a higher aqueous-phase pKa than hydronium is effectively 100% dissociated in water, so it is completely moot to decide on strength between, say, HCl and H2SO4. Compare the pKa of water (14) with the pKa of Hydrochloric Acid (-7). That's a difference of 21 pKa units, or a difference in equilibrium constant of 1021. That means that at a 1 molar concentration of HCl, we have something on the order of SQRT (6 x 1023/1021) or about 25 molecules of HCl per liter. There is absolutely no way that this is not as close to 100% dissociated as you need it to be. Secondly, the reduction potential chart has NOTHING TO DO WITH ACID-BASE STRENGTH. There's just no way that the oxidation/reduction relationship between hydrogen and nickel and chromate and chloride has anything to do with how they will react in an acid-base sense. You are completely ignoring things like coordination chemistry, complex ion formation, lewis theory, stuff like that, all well established and functioning parts of chemistry. You can't just have an opinion about anions "sticking" together, with no experimental proof, and then act like this is somehow a "valid" theory. You're just making shit up, and that isn't really what science is all about!!! --Jayron32 15:06, 10 May 2010 (UTC)
- (ec) Make sure you don't base any further predictions on a model that is your opinion which contradicts reality. Even in college, we make all sorts of approximations and "known to be not quite correct" explanations, but we know when we are "not correct". Things might work well in a few limited cases, so we are very careful not to apply those things outside of their correct realm. In normal (not many-molar concentration) solutions, strong electrolytes like HCl are so dissociated there's no reason to consider the few molecules that are bound. In a water solution of two strong acids, neither takes the H+...the H+ is solvated by the water (as H3O+) because water is a stronger base than the anion from any strong acid and there is like 55 molar concentration of water in glass of water (vs only a few molar at most concentration of the counter-anions). DMacks (talk) 15:08, 10 May 2010 (UTC)
- With react I mean a redox reaction. I can demonstrate the anionic activity series by analyzing the reaction of barium chloride and copper sulfate. Since sulfuric acid is a stronger acid than hydrochloric acid, the cation of a more reactive metal will take the stronger acid, forming barium sulfate. Sodium acetate is reacted with hydrochloric acid. Since sodium is the cation of a more reactive "metal" according to the standard electrode potential page, the hydrogen will bond with the acetate, forming acetic acid and sodium chloride. Calcium hydroxide and sodium carbonate react to form sodium hydroxide and calcium carbonate. Because carbonic acid is stronger acid than water, then calcium must be expected to bond with carbonate because it is higher up on the standard electrode potential chart (not the activity series). This anionic theory is just a helpful way to predict whether certain salts will react or not. It really concerns precipitates though, not reactions in which all four of the salts are soluble, the reactants and the products. --Chemicalinterest (talk) 19:37, 10 May 2010 (UTC)
- What I mean by sticking together is that, like in the reaction of barium chloride and copper sulfate, the barium kicks the chloride out and takes sulfate, leaving copper chloride. Essentially, the copper chloride exists since the Ba2+ and SO42- ions are out of solution. --Chemicalinterest (talk) 19:40, 10 May 2010 (UTC)
- The copper chloride doesn't exist. What exists is discrete copper ions and discrete chloride ions. You are confusing a notational convenience (like writing CuCl2 (aq)) with actual reality. Let's take your theory, and show where it doesn't work: If I mix Lead(II) acetate and sodium perchlorate, I would get absolutely no precipitate. This is despite the fact that a) acetic acid is weaker than perchloric acid and b) lead is higher up on the standard reduction potential table. Using either your "which acid is stronger" theory OR your "standard reduction potential" theory, that one doesn't fit. I can find any random number of mixtures which will not work; that you have found some random ones that coincidentally do means nothing. --Jayron32 20:16, 10 May 2010 (UTC)
- What I mean by sticking together is that, like in the reaction of barium chloride and copper sulfate, the barium kicks the chloride out and takes sulfate, leaving copper chloride. Essentially, the copper chloride exists since the Ba2+ and SO42- ions are out of solution. --Chemicalinterest (talk) 19:40, 10 May 2010 (UTC)
- With react I mean a redox reaction. I can demonstrate the anionic activity series by analyzing the reaction of barium chloride and copper sulfate. Since sulfuric acid is a stronger acid than hydrochloric acid, the cation of a more reactive metal will take the stronger acid, forming barium sulfate. Sodium acetate is reacted with hydrochloric acid. Since sodium is the cation of a more reactive "metal" according to the standard electrode potential page, the hydrogen will bond with the acetate, forming acetic acid and sodium chloride. Calcium hydroxide and sodium carbonate react to form sodium hydroxide and calcium carbonate. Because carbonic acid is stronger acid than water, then calcium must be expected to bond with carbonate because it is higher up on the standard electrode potential chart (not the activity series). This anionic theory is just a helpful way to predict whether certain salts will react or not. It really concerns precipitates though, not reactions in which all four of the salts are soluble, the reactants and the products. --Chemicalinterest (talk) 19:37, 10 May 2010 (UTC)
- OK, your example had several fallacies: None of these (lead(II) acetate, sodium perchlorate, sodium acetate, and lead(II) perchlorate) is insoluble. Second, sodium is higher up on the standard reduction potential than lead. So sodium would keep perchlorate. It doesn't really apply when all of the products are soluble though. As you said, compounds don't exist as discrete ions. --Chemicalinterest (talk) 00:29, 11 May 2010 (UTC)
- "I can demonstrate the anionic activity series by analyzing the reaction of barium chloride and copper sulfate. Since sulfuric acid is a stronger acid than hydrochloric acid". You're not going to get very far in validating your hypothesis if aren't even using correct data for your fundamental example: hydrochloric acid pKa=−8 vs sulfuric acid pKa=−3. DMacks (talk) 20:47, 10 May 2010 (UTC)
Here is a question: Do stronger acids displace weaker acids from weaker acid salts? Does hydrochloric acid displace acetic acid from sodium acetate? Does hydrochloric acid displace boric acid from sodium borate? Does sulfuric acid displace hydrofluoric acid from calcium fluoride? Does sulfuric acid displace hydrochloric acid from sodium chloride? If it does, then hydrochloric acid is weaker than sulfuric acid. --Chemicalinterest (talk) 00:29, 11 May 2010 (UTC)
- The answer to your first three examples is "yes": hydrochloric acid will create acetic, hydrofluoric, and boric acid from the appropriate salts. This is because those three are "Weak acids", and do exist as single species in water. However, the answer to your last question (hydrochloric and sulfuric acid in water) is NO. Both of those are "Strong acids" that will completely dissociate in water: there will not exist either HCl molecules, nor H2SO4 molecules, in the water. Instead, there will exist H3O+, Cl-, and HSO4-. I think your confused about "acid strength" and the table of standard reduction potentials. An acid-base reaction is not an oxidation-reduction reaction. The trouble is, many acids are also oxidizing agents: so called "oxidizing acids" (that should be an article. Edit: and now it is!). That is, the anion acts as an oxidizing agent stronger than the H+. These include nitric acid, chloric acid, Chromic acid, and to a lesser extent sulfuric acid, among others. While they may seem similar, the "strength" of an acid refers to it's ability to dissociate, not it's strength as an oxidizing agent. Many highly oxidizing acids are also strong acids: nitric, perchloric, and sulfuric, for example, so it's easy to confuse the two different properties. Buddy431 (talk) 04:27, 11 May 2010 (UTC)
- See Hydrogen chloride#Laboratory methods: they produced hydrogen chloride from sulfuric acid by its reaction with sodium chloride. (But can sulfuric acid be produced from hydrochloric acid?) --Chemicalinterest (talk) 10:49, 11 May 2010 (UTC)
- I see. What they're doing is taking dry sodium chloride (i.e. salt) and adding concentrated sulfuric acid to it. The HCl is then released as a gas (where it's usually called "hydrogen chloride" rather than "hydrochloric acid"). You couldn't do the same with sulfuric acid only because sulfuric acid isn't volatile: it won't bubble out as a gas. Additionally, sulfuric acid can be prepared as a nearly pure liquid, while hydrochloric acid, when a liquid, always exists as a solution. So if you tried adding even the most concentrated hydrochloric acid to a sulfate, you would just dissolve the sulfate and get an acidic solution. Buddy431 (talk) 14:55, 11 May 2010 (UTC)
- See Hydrogen chloride#Laboratory methods: they produced hydrogen chloride from sulfuric acid by its reaction with sodium chloride. (But can sulfuric acid be produced from hydrochloric acid?) --Chemicalinterest (talk) 10:49, 11 May 2010 (UTC)
@ChemicalInterest: I don't want to sound whiney here, but why don't you and John Riemann Soong just exchange email addresses and be done with it? I for one have used up all my Good Faith, and am getting quite fed up with what appears to be constant use of W:RD instead of your local classroom resources and library and learning how to do basic research on your own... DaHorsesMouth (talk) 01:00, 12 May 2010 (UTC) What is the ref desk for then? Asking questions about things that puzzle you. --Chemicalinterest (talk) 22:21, 12 May 2010 (UTC)
I'm searching for the name of a possible "fallacy"
[edit]I've seen an article here on Wikipedia, but cannot remember the name. I've just searched the list of fallacies, but could not find it there. It is about a scenario when the situation is getting worse, nearly everyone knows about it and about possibilities to avoid it, but they don't do anything against it because that would lead to short-term loss of comfort or competitiveness... so they doom themselves on the long term. There was a name for this, and even scientific game theory -related explanations, if I remember correctly. --131.188.3.21 (talk) 12:10, 10 May 2010 (UTC)
- I think you may somewhat misremember the tragedy of the commons. The problem there is not a loss of comfort or competitiveness (albeit that might be a result of cooperative action), but that at any time the optimal individual behavior is to stress the commons as much as possible, even if that leads to a degradation that's bad for everyone. The two standard ways out of this are to privatize the commons (hard to do with oceans or the air, for example), or to enforce rational behavior from all participants via regulation with punishment. --Stephan Schulz (talk) 12:44, 10 May 2010 (UTC)
- Thanks, but I know this dilemma, and think this is not the one I'm looking for. This is about acting to max out one's own benefit and disregarding the common goals. What I'm searching for is when, for example, decision makers could solve a situation (or at least keep it from worsening) but refuse to act out of fear that the temporary loss of comfort might make them lose power or not win the next election. --131.188.3.20 (talk) 13:03, 10 May 2010 (UTC)
- Sounds a lot like Prisoners dilemma. Googlemeister (talk) 13:16, 10 May 2010 (UTC)
- Not at all. I'm sorry if that sentence was not clear enough, but by "This is about acting to max out one's own benefit and disregarding the common goals" I meant the example given above (tragedy of the commons) was about it, but that's not the one I'm looking for. --131.188.3.21 (talk) 13:30, 10 May 2010 (UTC)
- Sounds a lot like Prisoners dilemma. Googlemeister (talk) 13:16, 10 May 2010 (UTC)
- Thanks, but I know this dilemma, and think this is not the one I'm looking for. This is about acting to max out one's own benefit and disregarding the common goals. What I'm searching for is when, for example, decision makers could solve a situation (or at least keep it from worsening) but refuse to act out of fear that the temporary loss of comfort might make them lose power or not win the next election. --131.188.3.20 (talk) 13:03, 10 May 2010 (UTC)
- I think the kind of situation our OP is thinking about is like if you had an oil leak on your car. You can either spend $10 to top up the oil once a week - or $300 to fix the problem properly. On any given week, topping up the oil is easiest and cheapest - but over a year, you've spent $520 and a dozen visits to the store to buy oil and 52 x 10 minutes = 8 hours of your time dealing with it...when you could have spent $300 and 6 hours and fixed it properly. Sadly, I don't know of a convenient name for this behavior. It's very similar to the tragedy of the commons...but not exactly that. How about "short term thinking" - or "horizon thinking"? SteveBaker (talk) 15:20, 10 May 2010 (UTC)
- Drifting off topic, but I'm not sure those terms best describe your scenario, Steve. You might be well aware of the long term disadvantage of repeatedly topping up, but simply not have $300 available for the better solution, a sort of individual variation on the Poverty trap.
- Re the OP's question, Endowment effect isn't quite what is described but is perhaps related. 87.81.230.195 (talk) 15:53, 10 May 2010 (UTC)
- It might be that my memory is failing me, but I still think I've seen an article about it here on Wikipedia. I think the examples were more politically oriented, for example decision makers avoiding to do something just because they fear their ratings will drop, or that society itself became to value its comfort too much and let things get worse over time. --131.188.3.20 (talk) 16:13, 10 May 2010 (UTC)
- The situation you described reminds me of groupthink. Vranak (talk) 16:28, 10 May 2010 (UTC)
- Is there a name for the fallacy of believing that all fallacies have names? --Tagishsimon (talk) 16:33, 10 May 2010 (UTC)
- Or the fallacy that all forms of wrong-mindedness are indeed fallacies. Vranak (talk) 17:17, 10 May 2010 (UTC)
- I'm just searching for an article I've once seen. It might not even have "fallacy" in its name, because I already searched that list. --131.188.3.21 (talk) 16:50, 10 May 2010 (UTC)
- You are describing a situation like a Nash equilibrium in mathematical game theory. This is where any change in strategy by any player (competitor) will produce (short-term) losses. In such a situation, people tend to do nothing, even though this may be bad for all parties. Staecker (talk) 17:16, 10 May 2010 (UTC)
- I'm surprised we don't have an article on the monkey trap. The monkey puts its hand into a transparent jar to grab the bait, but can't withdraw its clenched fist from the jar; it would have to let go of the food in order to free itself, but refuses to do so. It "prefers" to remain trapped.--Shantavira|feed me 08:15, 11 May 2010 (UTC)
- I think you might be looking for dynamic inconsistency. Maedin\talk 11:36, 11 May 2010 (UTC)
Physical fitness, health, and running
[edit]I've tried looking online for advice on general physical health and fitness, but I find pretty much nothing that I could consider to be a reliable source - they're all either:
- People trying to sell something, or
- People who have learned everything from people trying to sell them something.
So I've turned to the Wikipedians for some fitness advice. I'm considering going running, twice a week (my current fitness regime consists of sitting in front of my computer every day). Now, most websites seem to recommend doing it more than that, but they mention doing it specifically to increase things like muscle development, endurance, etc. Thing is, I'm not interested in any of that - I just want to live a little longer (and maybe get some free endorphins); I want to be healthier for no more reason than to be healthier. A friend told me that, if that was the case, I may as well just walk the same distance instead.
So; is running twice a week enough to achieve anything? Is walking really just as good (and what's the difference)? And why is it so hard to find objective sources for this kind of info? Vimescarrot (talk) 12:47, 10 May 2010 (UTC)
- There is no way to predict what your optimal training regime should be. There are some important guidelines that you should stick to, though. If you are older than 35 years old and you have not been physically active since a very long time, then you should not start any intensive physical activity like running straight away. You should then start with walking and then very gradually build up your fitness over the course of several months by gradually increase walking pace and duration, before you start intensive physical training like running. Even then, it is advisable to check with your doctor before you switch to running.
- Even if you are younger than 35, it is still advisable to start with walking. This is because the frequency and duration of trainings are important. If you are not fit, there is no way you could run for half an hour, five times per week. But it is likely that you can start with brisk walking every other day for 30 minutes. As your fitness level improves, you can replace one of these brisk walking trainings by a jogging exercise. Some time later you can replace all the brisk walking excercises by jogging exercises. At that time you can also replace one of the jogging exercises with running.
- This gradual increase in intensity has the advantage that when exercising you get can interpret the feedback your body gives you better. If you were to start running straight away, then you may get out of breath after 2 minutes. You don't have yet developed any feeling to adjust your pace to the right level. Most people in this situation tend to run way too fast.
- Depending on your age, you may be able to get into a 5 times per week running routine within perhaps half a year or one year. So, the short term goal should actually be to get fit enough to be able to train intensively. I have been excercising at a five times per week half an hour fast running routine for many years now, and all I can say is that the results are very good. Count Iblis (talk) 14:21, 10 May 2010 (UTC)
- I agree with that advice. Walking for 20 minutes every evening is better than running for 40 minutes twice a week. Get a dog...they won't let you forget or duck out of your training schedule! SteveBaker (talk) 15:14, 10 May 2010 (UTC)
- If you want to improve your general fitness then you need to do something that gets your heart rate up. It doesn't need to be very strenuous, as long as it increases your heart rate a bit. If you don't do much exercise at all now then a brisk walk would probably do it. I think the usual advice is that you should do it for about 30 minutes three times a week as the minimum to be effective. If you want more detailed advice then you could join a gym - they will usually have trainers that will help you put together a training regime (or you can hire a freelance trainer for a session or two). --Tango (talk) 15:25, 10 May 2010 (UTC)
- If you want an exercise programme that wasn't designed to extract money from you, consider the Royal Canadian Air Force plans 5BX (for men) or XBX (for women), which were published in an inexpensive paperback by Penguin Books titled Physical Fitness from 1964. Note that, as our articles say, some of the exercises involved are now thought not to be ideal if performed unsupervised, but the programmes' general outline might be useful to you. 87.81.230.195 (talk) 15:44, 10 May 2010 (UTC)
- Actually, when we had this conversation, I meant walk once a day or something, because we had been talking about certain inconveniences Vimes has of running every day and there only really being the option to do it twice a week. --KägeTorä - (影虎) (TALK) 17:45, 10 May 2010 (UTC)
- Having fluctuated all my life between periods of intense physical activity and periods of torpor, I disagree about walking being better than running. Running even once a week is far more useful to your heart, lungs, energy level, and weight than any amount of walking (unless you walk up steep hills), especially if you can work up to covering five miles at a moderately brisk pace. Looie496 (talk) 22:24, 10 May 2010 (UTC)
Thanks for the responses, everyone. Vimescarrot (talk) 05:45, 11 May 2010 (UTC)
methane hydrate ignition
[edit]Would heating methane hydrates to 25C cause them to ignite or detonate if they are at 150 bara pressure and surrounded by water, or would they just melt? Googlemeister (talk) 15:28, 10 May 2010 (UTC)
- No. They can't ignite/detonate without a source of oxygen...and there isn't one because you're underwater. Since you're obviously thinking about the gulf oil spill - I should point out that the oil that's coming out of the well-head is up at 150C - so if heat was enough, it would already have exploded! They might melt with the heat of oil collected beneath the dome - but that doesn't seem to be happening. The water down there is extremely cold! SteveBaker (talk) 15:35, 10 May 2010 (UTC)
- Yeah, I was wondering why they couldn't rig their collector unit with electric heating or something since it looks like the methane hydrates are forming and blocking their pipe. I mean it's not like putting that in would be cheap, but its got to be cheaper then their clean up costs. Googlemeister (talk) 15:39, 10 May 2010 (UTC)
- I believe they considered pumping hot water into it to do that - but the problem is that the thing filled up so quickly that now it's bouyant! This thing is the size of a three storey house and it's made of steel and concrete...it takes a LOT of ice to make something that big and heavy (200 tons!) bouyant! SteveBaker (talk) 16:23, 10 May 2010 (UTC)
- The different in density between oil and water over 1.5 km height all the way up the pipe has got a lot of buoyancy... but they should have thought of that. --BozMo talk 17:22, 10 May 2010 (UTC)
- I'm sure they knew that - that's the reason the thing had to weigh 200 tons! But the clathrate buildup was evidently either not expected or not expected to be so serious. SteveBaker (talk) 22:39, 10 May 2010 (UTC)
- The different in density between oil and water over 1.5 km height all the way up the pipe has got a lot of buoyancy... but they should have thought of that. --BozMo talk 17:22, 10 May 2010 (UTC)
- I believe they considered pumping hot water into it to do that - but the problem is that the thing filled up so quickly that now it's bouyant! This thing is the size of a three storey house and it's made of steel and concrete...it takes a LOT of ice to make something that big and heavy (200 tons!) bouyant! SteveBaker (talk) 16:23, 10 May 2010 (UTC)
- Yeah, I was wondering why they couldn't rig their collector unit with electric heating or something since it looks like the methane hydrates are forming and blocking their pipe. I mean it's not like putting that in would be cheap, but its got to be cheaper then their clean up costs. Googlemeister (talk) 15:39, 10 May 2010 (UTC)
Evian bottle ring
[edit]whats the ring at the top after you take off the cap off made out of on Evian water bottles made of of ? it dosent feel like plastic it feels like acrylic or something. it gets like sticky and leaves a gross residue on your hands after its been open for a day or 2 or if it gets wet. just that ring thou not the whole bottle. heres a pic of a bottle. im taking about the part circled in red. (make sure you click to maximize the pic size)
http://img96.imageshack.us/img96/5116/water1i.jpg —Preceding unsigned comment added by Tom12350 (talk • contribs) 20:45, 10 May 2010 (UTC)
- It's just a part of the cap - when you get an unopened bottle of water, it's attached to the cap - as you unscrew the cap, you break the connection between ring and cap. This is done so that you can tell that the water bottle hasn't been tampered with. I don't understand why it would leave a residue or anything though. SteveBaker (talk) 22:34, 10 May 2010 (UTC)
- It's made of polypropylene. And water should not affect it at all. So I don't know what the residue you mention is. Ariel. (talk)
how do u know its made from that? the ring feels like a differnt material than the cap —Preceding unsigned comment added by Tom12350 (talk • contribs) 02:58, 11 May 2010 (UTC)
- It'll feel weaker and smoother different because it's not as thick and it doesn't have the cap's structure keeping it rigid, or the grooves for grip. If the cap was as smooth as the ring, you'd never be able to grip it to open it. Vimescarrot (talk) 05:47, 11 May 2010 (UTC)
- If you examine one, you'll see that it's moulded as an integral part of the cap, and you break the link between them when you open the bottle. It's therefore made from the same material as the cap. If it gets sticky and disgusting over time, you may like to think about whether you ever drink from the bottle directly, and leave a residue of your mouth on th ring? --Phil Holmes (talk) 09:06, 11 May 2010 (UTC)
Can Crusher
[edit]For my science class I have to build an aluminum can crusher. It has to crush cans consistently, into a thin piece of metal. It has to be based on a simple machine. I'm debating between dropping a piece of wood onto the can from a height, and using a pulley to keep it up until I need it to drop, and using a lever to crush it. Which one would work better? I'm planning to build it using wood, but if anyone has a better idea, I'd be open to it, and the same with another idea for the concept of the machine.
tl;dr version: What's the best way to build a can crushing machine?
RefDeskAnon (talk) 21:11, 10 May 2010 (UTC)
- Well, I would expect a lever would be faster to reset after you crush a can. Googlemeister (talk) 21:13, 10 May 2010 (UTC)
- Very true. RefDeskAnon (talk) 21:51, 10 May 2010 (UTC)
- "Which one would work better" depends on a jillion variables (weights, distances, acceptible cost, etc.). It's a science class, you say? Do the experiment! Pick an approach, build it an test it. Change it a little to see if it gets better or worse. Each test doesn't have to be the fully built machine, just the key parts. How high a drop does it take to crush? How well can you aim from that distance? How long a lever does it take to crush? How well can your materials tolerate that much force? etc. DMacks (talk) 21:15, 10 May 2010 (UTC)
- The problem is that the project is due pretty soon, and I still need to get the wood. But I hadn't thought of just doing the key parts. Thanks, both of you. Any more ideas would be appreciated. RefDeskAnon (talk) 21:51, 10 May 2010 (UTC)
- I just stomped the can flat with my boot, very fast, no special tools needed. Nowadays the recyclers don't care. Graeme Bartlett (talk) 22:02, 10 May 2010 (UTC)
- The problem is that the project is due pretty soon, and I still need to get the wood. But I hadn't thought of just doing the key parts. Thanks, both of you. Any more ideas would be appreciated. RefDeskAnon (talk) 21:51, 10 May 2010 (UTC)
- I actually have a can crusher - it bolts to the wall, has a hopper for the cans and a large vertical lever that pushes on a plunger to crush the can end-to-end. Of course the classy "science fair" way to do it would be to "pinch" them like this! SteveBaker (talk) 22:31, 10 May 2010 (UTC)
- How are you planning on building your lever? The strength of that could offer some creative options. Dropping the wood seems the easiest option but basically that's it, you drop the wood and hope it's heavy/fast enough to crush the can. —Preceding unsigned comment added by 87.114.95.229 (talk) 22:43, 10 May 2010 (UTC)
- Measuring the least amount of weight you need to reliably crush a can would be a part of the science in this. Do experiments with a variety of cans. Do some math - figure out the most weight you needed to crush each can so you know which kind are toughest. Then plot graphs of the amount by which the can gets shorter for a given amount of weight. That will allow you to decide the point at which adding more weight doesn't make much difference to how much the can gets crushed. Machines that my son and I built for science fairs that required dropping something in a controlled way benefitted from using a few feet of 2" or 3" PVC pipe (buy it from any DIY store). This guides the weight precisely onto the target and helps you to always drop it from the same height in a repeatable way that gets you more reliable data. Once you have the data - you can think how to build the machine. (How about a weight on the end of a pendulum arm that swings down and crushes a can laid horizontally in front of the bottom-most part of the swing? You could raise the weight by pulling on a lever mounted to the top of the pendulum.) SteveBaker (talk) 01:01, 11 May 2010 (UTC)
- In case you didn't already Google for "Can Crusher", you could look at this web site has a bunch of different can crushers for sale. (The one I have is this one). SteveBaker (talk) 01:04, 11 May 2010 (UTC)
- A lever will work a LOT better than dropping a weight. An average person with a lever can easily impart 1000 pounds of force on something. You are unlikely to get anywhere near that by dropping something. (You might have a bit of an impact force, but that will only dent the can, you need longer duration force to crush it totally.) Ariel. (talk) 03:52, 11 May 2010 (UTC)
- ... and avoid using soft wood for the part in contact with the can. If it has to be wood, use a hardwood, but metal would be better. Dbfirs 07:52, 11 May 2010 (UTC)
Use a car? Polypipe Wrangler (talk) 11:02, 12 May 2010 (UTC)
Do wormholes violate the 2nd 1st law of thermodynamics?
[edit]I know wormholes are theoretical, but lets assume they are real for the purposes of a thought experiment.
Lets take two mouths of a wormhole. Mouth A and Mouth B. Mouth B is stationary, but you accelerate Mouth A to 99% the speed of light. Now any object you send into Mouth B will come out of Mouth A and should now be moving at 99% the speed of light as well right? Unless Mouth A slows down as a result of accelerating the object, wouldn't this be a violation of the 2nd 1st law of thermodynamics since the object would have been accelerated without energy being conserved. ScienceApe (talk) 22:00, 10 May 2010 (UTC)
- Are you sure you're talking about the second law? You don't seem to mention entropy anywhere in your thought experiment, so I don't really see how the second law comes into play. --Trovatore (talk) 22:03, 10 May 2010 (UTC)
- Indeed, I think ScienceApe probably means the first law. The second law is also a problem with wormholes, though (they mess with causality, since one end can be time dilated relative to the other, allowing time travel). --Tango (talk) 22:05, 10 May 2010 (UTC)
- These are the kinds of reason why very few people think that wormholes are 'real'...and even if they are, what their properties might be. One rather likely possibility is that you can't actually move them...or perhaps one end cannot move with respect to the other (so they both have to move together). If those kinds of restrictions are imposed then there is there still a problem with either thermodynamics or time travel? SteveBaker (talk) 22:23, 10 May 2010 (UTC)
- That might fix it, I'm not sure how it would work, though. You can't say the ends can't move, since that would require everything in the universe to stay still since motion is relative. Not move relative to each other might be manageable, but I'm not sure how. Wormholes simply not lasting long enough for anything to pass through them is quite likely, if you can't get rid of them all together. --Tango (talk) 22:48, 10 May 2010 (UTC)
- I was thinking that you might not be able to move them...because there is no way to grab a hold of them - or that there is no way to apply forces to them maybe. Since they are "connected" it seems somewhat plausible that you might not be able to move one end without also moving the other. Dunno - I think it's vastly more likely that the either don't exist - or are too narrow to allow things to pass through them - or (as you say) that they have such a short life that it's irrelevent...I dunno...they just seem like so much wishful thinking with zero evidence behind them. SteveBaker (talk) 01:09, 11 May 2010 (UTC)
- That might fix it, I'm not sure how it would work, though. You can't say the ends can't move, since that would require everything in the universe to stay still since motion is relative. Not move relative to each other might be manageable, but I'm not sure how. Wormholes simply not lasting long enough for anything to pass through them is quite likely, if you can't get rid of them all together. --Tango (talk) 22:48, 10 May 2010 (UTC)
- These are the kinds of reason why very few people think that wormholes are 'real'...and even if they are, what their properties might be. One rather likely possibility is that you can't actually move them...or perhaps one end cannot move with respect to the other (so they both have to move together). If those kinds of restrictions are imposed then there is there still a problem with either thermodynamics or time travel? SteveBaker (talk) 22:23, 10 May 2010 (UTC)
- Indeed, I think ScienceApe probably means the first law. The second law is also a problem with wormholes, though (they mess with causality, since one end can be time dilated relative to the other, allowing time travel). --Tango (talk) 22:05, 10 May 2010 (UTC)
- I never understood why people (ScienceApe: you are not the only one) assume that if one end of the wormhole is moving and the other stationary the object sent inside would take on the properties of the exit hole. If the exit hole is moving, then the exit location would constantly change, but the exit velocity of the object traveling inside it wouldn't. Unless you assume the object is bouncing off the walls of the wormhole (like water in a tube), but then by traveling in it you are slowing down the wormhole. Also, a two dimensional wormhole entrance (like the opening of a pipe) has problems with conservation of momentum. You can cause an object to change direction, which violates that law. I think if a worm hole exists, it would have to be three dimensional - meaning you can enter it from any direction, and you leave in the same direction, with the same velocity you entered. Ariel. (talk) 01:38, 11 May 2010 (UTC)
- The equivalence principle implies that you can't put these sorts of restrictions on ends of a wormhole. Any object, including a wormhole mouth, can be manipulated by gravitational tugging, and there's no way for the wormhole to "compensate" for that. Given both ends of a wormhole, I think you can always create a closed causal loop by luring them along worldlines of different lengths. Pipe-like wormholes like the one in Star Trek: Deep Space 9 are unrealistic even by wormhole standards. I think some people are misled by illustrations like this into thinking that black holes are "flat". It's actually the accretion disc that's flat. -- BenRG (talk) 09:28, 11 May 2010 (UTC)
- I never understood why people (ScienceApe: you are not the only one) assume that if one end of the wormhole is moving and the other stationary the object sent inside would take on the properties of the exit hole. If the exit hole is moving, then the exit location would constantly change, but the exit velocity of the object traveling inside it wouldn't. Unless you assume the object is bouncing off the walls of the wormhole (like water in a tube), but then by traveling in it you are slowing down the wormhole. Also, a two dimensional wormhole entrance (like the opening of a pipe) has problems with conservation of momentum. You can cause an object to change direction, which violates that law. I think if a worm hole exists, it would have to be three dimensional - meaning you can enter it from any direction, and you leave in the same direction, with the same velocity you entered. Ariel. (talk) 01:38, 11 May 2010 (UTC)
- Well Ariel if what you are saying is true, then time travel should also be impossible with wormholes. In any case, if the object that leaves the mouth isn't traveling at the same speed as the mouth that would be like stepping off of a car moving at 200 mph, as if you were stepping off of a stationary car. I don't really think that's true. ScienceApe (talk) 01:21, 12 May 2010 (UTC)
- Yes. No time travel. I don't believe it's possible in principle at all, but more specifically not with wormholes either. I had a thought. Say you poke a long stick through a wormhole. So now you have one end of the stick stationary, and the other moving. Now grab the moving stick and try to stop it. What happens to the wormhole? Do you slow down the wormhole, because you are slowing down the stick? If so, then their's your answer: the energy comes from the energy containing in the motion of the wormhole. If the wormhole does not slow down, then what happens to the stick? Ariel. (talk) 07:52, 12 May 2010 (UTC)
Yes I meant 1st law, my bad. I'll change the topic name. ScienceApe (talk) 22:32, 10 May 2010 (UTC)
- Energy, as we classically understand it, is not globally conserved in general relativity. Instead a compound entity, the stress-energy-momentum pseudotensor is conserved. In essence this says that you can change the total energy in the universe if and only if you change the configuration of space-time as you do it. So, yes, if you accelerate matter through your wormhole then there must be a counter-reaction in the wormhole itself, presumably causing it to slow down. Dragons flight (talk) 01:26, 11 May 2010 (UTC)
ScienceApe, by changing what you've written previously you're making some of the responses to your question appear nonsensical. I've changed it so that your old version appears, crossed out, next to your new version - hope you don't mind - it just makes the contributors who responded look slightly less foolish. Vimescarrot (talk) 05:40, 11 May 2010 (UTC)