The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
I have every once in a blue moon read some intriguing news snippet about people prosecuted for using "an electronic device" to influence the odds of slot payouts. I assume that the organized crime syndicates that set up the casinos must have ways to make it as easy for someone to win big as it is for someone delivering a payment to arrange to lose everything; otherwise it wouldn't be a very effective way to tracelessly launder money. Though apparently lower-tech methods work pretty well.[1]Wnt (talk) 03:55, 25 September 2011 (UTC)[reply]
Why is it that when you defrost a bottle of frozen milk, that the liquid milk is initially richer in lactose, making it digustingly sweet? Only once the milk is completely defrosted, does the lactose concentration return to normal. Plasmic Physics (talk) 06:27, 25 September 2011 (UTC)[reply]
Oh. That reminds me, can you filter out the additives from denatured ethanol using fractional distallation? Add water to denatured ethanol, and partially freeze it so that it forms a crust of ethanol/water. Collect the crust, and repeat after adding more water. Melt all the crusts back down into a ethanol solution. Keep adding anhydrous magnesium sulfate untill anhydrous ethanol is obtained. Will the crusts be free of additives? Plasmic Physics (talk) 09:13, 25 September 2011 (UTC)[reply]
I don't think so - fractional freezing will enrich the liquid portion in the lower freezing part of the mixture -eg alcohol - but it will not remove methanol from the alchohol mixture - this remains in the alcohol part not the frozen water part.Imgaril (talk) 11:03, 25 September 2011 (UTC)[reply]
I don't have to worry about methanol - in modern times, methanol is banned in many countries, from being added as a denaturant. According to the article, the frozen component contains ethanol, that is what I am taking advantage of. Plasmic Physics (talk) 11:23, 25 September 2011 (UTC)[reply]
Do you have a reference for that? My understanding is that methanol is still the most common additive used to denature ethanol. That's what our article on denatured alcohol seems to say. The whole point of denaturing it is to make it undrinkable, so that methanol is highly toxic is hardly a reason not to use it. --Tango (talk) 15:31, 25 September 2011 (UTC)[reply]
Denatured alcohol may or may not contain methanol; what it does usually contain are isopropanol (gives you a tummyache like you wouldn't believe), denatonium and various ketones (taste awful) and also pyridine (makes you throw up). Note that the additives are purposely chosen so that they cannot be easily removed from the alcohol. 67.169.177.176 (talk) 02:25, 26 September 2011 (UTC)[reply]
The way fractional freezing/freeze distillation works is that when water (or any other substance) freezes, it wants to form a regular crystalline lattice, as that's the most energetically favorable way of forming a solid (especially for water, which has all those hydrogen bonds to form). Any impurity would disrupt that crystal lattice, so the substance tries to exclude it from the lattice when possible (but if you freeze something long enough or fast enough, the impurities will get trapped in the solid eventually). Therefore anything that's not what's crystallizing gets concentrated in the still-liquid portion. -- 174.24.217.108 (talk) 18:01, 25 September 2011 (UTC)[reply]
I had a second look at the references, they seem to contradict each other, so I retract the statement on methanol. What about the remaining additives? Can the alcohol be removed using the above technique? Plasmic Physics (talk) 22:29, 25 September 2011 (UTC)[reply]
As for freezing to make the solution denser, it works with artificial tear eye drops: DO NOT apply eye drops you have just taken out from the freezer, wait till it melts instead. – b_jonas20:54, 28 September 2011 (UTC)[reply]
Why do farms have soil?
So I've been reading the article on hydroponics, and it occurs to me that if it is a method which is cheaper due to various efficiencies savings, and reliable, it is in the main simply better. In particular, better for industrial growing because the disadvantages can be dealt with by using good precautions. But latifundia of farms don't use hydroponics, almost all fruit and veg is still grown in all countries using traditional fields. Why is this, surely a hydroponic farm would be cheaper and therefore 'better' for the farmers? Prokhorovka (talk) 11:29, 25 September 2011 (UTC)[reply]
"Using good precautions" certainly costs money. And soil is getting cheaper and cheaper. The same applies to the logistics, which enables you to trade products across the globe, and therefore, from places with even cheaper soil. Quest09 (talk) 12:02, 25 September 2011 (UTC)[reply]
Cheaper in the long run and in bulk amounts. It requires a substantial amount of capital and know-how to start up. Something inaccessible to most people.-- Obsidi♠nSoul12:34, 25 September 2011 (UTC)[reply]
Can you imagine the sheer complexity of having millions of hectares of hydroponics? How would you mechanically harvest a field of wheat or maize (let alone potatoes) without destroying the equipment? For veg it is already a reality (so long as 90 hectares is a latifundia) - check out Thanet Earth. SmartSE (talk) 22:20, 25 September 2011 (UTC)[reply]
The answer to this type of physics question always depends: how would the moon "escape" from its orbit? If you can specify that, we can follow through with the consequences by solving the equations of motion for the Earth-Moon system.
For example, if you hypothesize that a giant comet large and fast enough to change the moon's orbit were to impact the moon, .... well, we would need to calculate the effect that such a large comet has on the orbits of Earth and everything else in the solar system, too. We could solve that problem by setting up an n-body problem to model the solar system, including Earth, Moon, Sun, and other planets; and we would use perturbation theory to study how sensitively the system reacts when we add in a new comet on a course to impact the moon. The results are difficult to compute, but this can be done in a reasonable amount of time with a reasonable amount of effort.
Our article on tidal locking has some mathematics, including this equation, but it's lacking the context you would need to apply it in our hypothetical case. (The equation depends on some assumptions, e.g., that the moon is smaller than the earth, and so on - and the question is about Earth's day - so we need to reformulate that equation to solve for Earth as the object that will become tidally locked... which is just a little messy mathematical manipulation). Suffice to say: the timescales for tidal locking (i.e., engaging an orbital/rotational resonance) depend very strongly on the distance between the Earth and the Moon. By being farther away, the moon has a much smaller effect on Earth's rate of rotation. This effect (semimajor axis to the sixth power) almost always will overpower any other effect, including a change in the Moon's mass, angular momentum, etc., due to inelastic collision with a comet.
On the other hand, if you just want to make something up, "just imagine" that the moon magically changes its orbit, all bets are off. We can't meaningfully speculate what consequences follow when one law of physics breaks "because of magic." Anything could happen. Everything we know about the way the Moon's orbit couples into the Earth's rotation depends on the rules of physics as we currently understand them. Nimur (talk) 15:16, 25 September 2011 (UTC)[reply]
I'm afraid I was trying to just imagine that Moon has dropped suddenly from the 3 body or 2 body problem and I wasn't paying attention to other effects that will definitely take place. I was just interested in Earth's spin because I though it would be if affected the most significant thing we would realize (tides for example won't make things worse as I expect).--Almuhammedi (talk) 15:52, 25 September 2011 (UTC)[reply]
The tidal interaction between the Earth and Moon is causing the Moon to slowly move further away from the Earth and for the Earth to slow down in its rotation (lengthening the day). Removing the moon would stop that and the length of the day would become pretty much constant. There is a theory that the presence of the Moon has helped keep the orientation of the Earth's axis stable, so that orientation may start moving around more if the Moon disappeared. See What If the Moon Didn't Exist for a description of some work that was done trying to work out what the Earth would be like if the Moon had never existed (that's a little different to the moon disappearing now, but still interesting). --Tango (talk) 17:04, 25 September 2011 (UTC)[reply]
The implication that communication may travel backwards in time if neutrinos go faster than the speed of light is still nullified by the fact that neutrinos nor light can travel into the past by going more fast from point A to point B than in zero time. Going back in time requires time reversal and not just exceeding the speed of light. Try unburning a candle first. --DeeperQA (talk) 20:42, 25 September 2011 (UTC)[reply]
but since you mention it, I feel compelled to reply because what you said isn't correct. It would be correct if we lived in a Euclidean space. But Physicists know that this is not the case. That's been known for over a century now. This is a relativistic effect. Faster than light travel (FTLT) does indeed allow for messages to be sent backwards in time and that's why FTLT is believed by most physicists to be impossible. Dauto (talk) 23:16, 25 September 2011 (UTC)[reply]
Well I mentioned it because its usually the next thing to come. Lets say there is an electromagnetic wave that can travel in empty space but at slower speed due to a hiccup in its electromagnetic interaction. If you send info via that wave it will get there after the same info sent by light. You can send info by light versus by "slower" wave traveling back in time but it is not the same as time reversal except perhaps for using the speed of "slower" wave as your reference. My reference, however, is zero time in which it is not possible to travel at all. A candle will not burn in zero time and you can not unburn it if you started at a later time and tried to get time to go in reverse. --DeeperQA (talk) 01:14, 26 September 2011 (UTC)[reply]
See tachyonic antitelephone. If anything can travel faster than light, then it is always possible in principle to construct a system that allows information to be sent back in time. If your tachyon is only slightly faster than light (as per the neutrino claim), then one half of your system must be moving just slower than the speed of light in order for it to work, but such things are always possible in principle. Dragons flight (talk) 01:40, 26 September 2011 (UTC)[reply]
But either end could be the moving end, right? Maybe Wikimedia should build a stationary receiving end now, in case the Wikimedia foundation at some point in the future builds the 99.99+% speed-of-light spaceship needed to form the transmitting end. Imagine the advantages of letting editors from the future edit the encyclopedia. APL (talk) 02:00, 26 September 2011 (UTC)[reply]
Dauto and Dragons flight, the standard argument that FTL communication is always possible with tachyons makes very little sense, as I said in the #FTL neutrinos thread. It depends on the weird quasi-Newtonian assumption that the causal "future" for an emitted tachyon consists of later coordinate times in the rest frame of the emitter. This is all kinds of crazy. First, I can't imagine why anyone would expect it to be true in the first place, post-1905; it's comparable to emission theories of light. Second, both ends of a tachyon worldline, or neither, can be "the emitter" by this definition, so it isn't even logically coherent. Third, what exactly is "the rest frame of the emitter" in quantum field theory? Fourth, wouldn't "the emitter" here be a muon traveling toward the detector at much more than (1 − .0000228) c wrt the ground? Because if so, and you believe this silly rule, then the emitted tachyon would have to be going at much less than (1 + .0000228) c. -- BenRG (talk) 02:22, 26 September 2011 (UTC)[reply]
I like the idea of recording something now and preserving it for the future so that it can be edited later on. It is the sending the edit back in time that I do not think will work. --DeeperQA (talk) 02:27, 26 September 2011 (UTC)[reply]
Oddly enough, its true that the theory of relativity is not a Euclidean model of space, and yet still, it has resulted in a century's worth of complex and useful modeling. Nevertheless, that has not stopped me from trying to sort this mess out for my own peace-of-mind, thus I should be blogging my Euclidean toy model in the near future, as it might help us to incorporate a recently discovered quantum mechanics analog, as well the new neutrino data, if necessary... such that we can perhaps ditch a load of metrics. Modocc (talk) 04:35, 26 September 2011 (UTC)[reply]
entry and exit door lock logic
Some front door locks have a small chisel edge knob in the center of the door knob that turns to lock or unlock the door. For some doors the chisel knob position controls both exit and entry. On other doors the chisel knob position does not effect exit but only entry. Where can I find a table or diagram for all types of entry and exit door lock logic? --DeeperQA (talk) 18:53, 25 September 2011 (UTC)[reply]
Ok, let's try it another way. The top of the mountain is referred to as the death zone. What do you think might live in the death zone, where it is extremely cold and the air is so thin you have to bring your own oxygen? Beeblebrox (talk) 19:36, 25 September 2011 (UTC)[reply]
Well, come on, humans have to bring their own oxygen. Doesn't follow that everything does. I don't know whether there's any life that can survive in that environment (some sort of plant or cyanobacterium with some serious antifreeze in its tissues?) but the question is not absurd on its face. --Trovatore (talk) 19:47, 25 September 2011 (UTC)[reply]
Well, others have asked too. This astronaught chap went all the way to the summit in May 2009 to find out: "Searching for signs of life on Mount Everest could provide a window into the extreme environments that organisms might inhabit elsewhere in the universe. So, former astronaut Scott Parazynski will set up instruments to hunt down elusive evidence of life at the top of the world when he attempts to summit Everest Wednesday." Watch this space while I try to find out what happened. Alansplodge (talk) 20:18, 25 September 2011 (UTC)[reply]
According to our article, tardigrades have been found at altitudes of over 20,000 ft in the Himalayas, and they have survived exposure to the vacuum of space, so I imagine they could live at the top of Everest. Gandalf61 (talk) 20:44, 25 September 2011 (UTC)[reply]
This article states: "Pressure decreases with increasing altitude, such that at 10 kilometers (6.2 miles) above the Earth's surface, the pressure is only about one-fourth that at sea level. Organisms have been discovered growing on the top of Mount Everest, the highest point on the Earth's surface (more than 8.8 kilometers [5.4 miles])." Frustratingly there are no more details, but there is a list of sources at the end. Alansplodge (talk) 20:47, 25 September 2011 (UTC)[reply]
According to Alpine_Chough#Distribution_and_habitat, 'It has been observed following mountaineers ascending Mount Everest at an altitude of 8,200 m (26,900 ft)' (ref's from a book, so I can't check it myself). Not inconceivable that it might also make it up to summit or thereabouts for a scrat around sometimes. --Kurt Shaped Box (talk) 22:12, 25 September 2011 (UTC)[reply]
The thing about the oxygen is, from what I understand, most of us have lungs far too small for such high altitudes to deal with the scarcity of oxygen. So, with people living at higher altitudes, natural selection would favour larger lungs, ya? Couldn't successive generations gradually moving up say 20 ft each generation maybe be able to adapt to permenant life on top of the mountain (not saying that the larger lung adaptation is given)? Provided they have some sort of food and water source as well as some bit of warmth. Sir William Matthew Flinders Petrie | Say Shalom! 26 Elul 5771 22:47, 25 September 2011 (UTC)[reply]
You've hit upon the real issue, which is sustenance. If a tiny organism could survive the climate at the top of Everest, presumably it still has to eat something, unless it can somehow subsist on ice and snow. ←baseball BugsWhat's up, Doc?carrots→ 23:31, 25 September 2011 (UTC)[reply]
Yea, except in a few rare, extraordinary cases, animal life follows plant life. Animals won't migrate to a place without plants. APL (talk) 23:37, 25 September 2011 (UTC)[reply]
Maybe, That'd be about a 10,000 year experiment. However, 10,000 years later you may find that your entire experimental tribe had dwindled to nothing and gone extinct.APL (talk) 23:37, 25 September 2011 (UTC)[reply]
I'm not seeing it in the George Mallory article, so my memory may be faulty... but it seems to me I had read someplace that there was evidence of birds or other scavengers having fed on Mallory's body to some extent before he solidified. That wasn't at the summit, obviously. ←baseball BugsWhat's up, Doc?carrots→ 00:21, 26 September 2011 (UTC)[reply]
Mount Everest#Flora and fauna mentions some birds being seen at fairly high elevations, and a spider that lives at about 22,000 ft, but that is still well below the summit. Up here in AK we have the highly improbable ice worm, which scientists currently have a very poor understanding of, but I've never heard of them living in the Himalayas. Beeblebrox (talk) 00:35, 26 September 2011 (UTC)[reply]
You learn something new every day here on wikipedia. Worms that are so well-adapted to the frigid that if it gets above 40 they disintegrate. Amazing. Like a worm form of Dracula. And they feed on algae, and the algae's presence there on glaciers has to be a story in itself. It would be interesting to take some of these worms and algae to the top of Everest and see how they would do. ←baseball BugsWhat's up, Doc?carrots→ 00:43, 26 September 2011 (UTC)[reply]
Well plants only need CO2 and water and minerals they could get from under the ice on the surface of rock and light which they may need only 1/10th as much as full sunlight to grow. They produce oxygen and nutrients so animals could grow.--DeeperQA (talk) 01:38, 26 September 2011 (UTC)[reply]
I keep coffee mugs, as well as other cups and glasses in a cupboard that has solid shelves. I've noticed that some people like to set receptacles in there upside-down, so they won't collect dust on the inside while waiting for use. Others like to place them right side-up, so the lip of the vessel won't be resting on the potentially dusty surface of the shelf.
Is one method really more sanitary, or is this just down to personal preference? I apologize if this has been asked and answered before; my search of the archives didn't quickly reveal it. Thanks in advance for reasoned opinions. -GTBacchus(talk)23:08, 25 September 2011 (UTC)[reply]
For the two options you gave, it is personal preference. The completely sanitary methods I've witnessed there are two more choices. Once is to place a saucer on the shelf and the cup upside down on the saucer. So, the saucer is clean and the cup is upside down on the clean surface. The other is to have cupholders so the cups hang and don't touch anything (except the handle where they hang). -- kainaw™23:21, 25 September 2011 (UTC)[reply]
... and a further option that I thought Kainaw was going to suggest ... place the mugs right side up, but with a sheet of paper over the top so that dust doesn't fall in. Are we being paranoid about a bit of dust? -- we breathe it in all day! Dbfirs08:12, 26 September 2011 (UTC)[reply]
Paranoid? No. I don't care that I ingest dust. This question was asked out of academic interest. I'm not going to change the way I store cups, which seems to be about 50/50 between the two original methods asked about. -GTBacchus(talk)15:34, 26 September 2011 (UTC)[reply]
I was once grossed out at my grandparents house, since they stored glasses on a high shelf, with the opening down. Since the shelf was high, and they were short, they couldn't see that he shelf had dead bugs on it. Yuk !
Personally, I always rinse glasses after I remove them from the shelf, to get the dust and germs off them. Note that none of the methods listed so far prevent a spider from dragging it's pus-filled butt along the glass on the outside, where your lower lip goes when you drink. StuRat (talk) 12:25, 26 September 2011 (UTC)[reply]
I have taken to using the tops from Pringles tubes, and similar clear plastic tops from other similar food packaging (according to size requirements), to use as coasters on which to place upside-down drinking vessels, or as covers for ones displayed right-way-up (I collect branded beer glasses). {The poster formerly known as 87.91.230.195} 90.197.66.194 (talk) 15:21, 26 September 2011 (UTC)[reply]
Another variable: After washing, I have taken to stacking glasses and bowls irregularly such that they will be certain to dry out completely over a few hours. When stacking plastic bowls in the normal fashion, a seal can be formed that traps the moisture currently on the bowls. I have no proof that my method prevents explosive bacteria growth within the trapped water, but that's what I've been doing. Comet Tuttle (talk) 19:28, 26 September 2011 (UTC)[reply]
Another option is to have one cup and one bowl. You want to use it. You wash it. You know it is always clean when you use it. -- kainaw™19:32, 26 September 2011 (UTC)[reply]
Are there two separate questions here? Are you prepared for the Recoil associated by Newton's Law with the time derivative of the projectile momentum you describe? Here are ideal specifications: Railgun: Fits in pocket, uses ordinary AA batteries and is clearly marked "Warning: bullet comes out this end". Coilgun: Available in a range of pastel colours, fits in pocket, uses ordinary AA batteries and is clearly marked "Warning: bullet comes out this end". A page in Wikipedia gives many sources of information on railguns, among them is Count Zero by William Gibson that states: "You can rig a railgun to blow itself to plasma when it discharges." This danger should be addressed by an obligatory warning label.Cuddlyable3 (talk) 13:17, 26 September 2011 (UTC)[reply]
I mean, how can it be achieved using existing technology? Assume that recoil is accounted for, that the weapon operates in a vacuum, and that it is no longer than one kilometer in length. Plasmic Physics (talk) 13:34, 26 September 2011 (UTC)[reply]
You would need to be very specific on what you mean by ideal. Engineering is all about compromises. If you want it stronger, it will either be more expensive, or weigh more is a common trade off, so without knowing exactly what you want, specifications are quite impossible to provide. Also, I would be surprised if many here are deeply involved in the finer points to coil gun engineering, so even with a great list of specs you need, we probably can't help all that much. Googlemeister (talk) 14:00, 26 September 2011 (UTC)[reply]
OK, I'll see how far I can get. It is a naval artillery piece, opperating in a vacuum and zerogravity, it needs to be energy efficient, easy to repair, prefferably not self-destruct on firing. Cost and weight is not not an issue, assume infinite construction supplies, so the design can be as exotic. Plasmic Physics (talk) 14:14, 26 September 2011 (UTC)[reply]
I'm still baffled by what kind of answer you are seeking. It seems the specifications of this weapon are that if fires 600 metric ton or possibly .6 gram (big difference there between mg and Mg) projectiles at 30 km/sec, operates in a zero-g vacuum, and is around 1 km long. The caveat about existing technology makes this simple, however. Such a device probably cannot be built at all with existing technology. For one thing, we don't have the technology to lift a single projectile into orbit (the closest source of zero gravity and vacuum), much less the firing hardware. We can't even begin to imagine the other engineering problems and technological solutions. We are talking about a device 1 billion times more powerful than any yet constructed. Your 1 kilometer storage space would have to contain a large scale nuclear powerplant just to charge whatever you used to store energy to fire the device.
If you have any followups, I can try to help more, but you should specify if you are talking about a railgun or a coilgun (and why do you care which?) and if you meant metric tons (Mg) or milligrams (mg). And I'm still struggling with what kind of information would constitute an answer.gnfnrf (talk) 17:43, 27 September 2011 (UTC)[reply]
Coilgun specifically, as the rails in a railgun firing at these energies, would turn to plasma on the 1st firing. Eitherway, I've decided on a coilgun. And, megagrams nor miligrams. Assume the weapon is already in orbit, and it is powered by a nuclear fussion plant supplying all the energy it could need. Is it best to have multiple short coils, or few long coils? Would there be any advantage in using ferromagnetic liquids to better control the magnetic field evolution. How should the weapon be held together. This is the kind of information I need. Plasmic Physics (talk) 23:35, 27 September 2011 (UTC)[reply]
A gun that fires a 600-ton shell at a speed of 30 km/s in zero gravity will have a recoil that will certainly disturb the orbit of even the biggest battleship, very likely causing it to tumble out of control after the very first shot. I recommend you seriously consider reducing the caliber of your big cannon. 67.169.177.176 (talk) 23:47, 27 September 2011 (UTC)[reply]
Just to show what I mean: Suppose your battleship has a displacement of 45,000 tons (which is a pretty reasonable size), and you fire a 600-ton shell from one of the ship's main guns at a speed of 30 km/s. By Newton's third law, action = reaction, which (since your battleship is in orbit) translates to m1v1 = m2v2, or in this case m[shell]v[shell] = m[ship]v[ship]. So, when you fire that 600-ton shell with a speed of 30,000 m/s, your battleship will start going the other way at 600*30,000/45,000 = 400 m/s! And what's worse, this delta-v of 400 m/s will occur in a tiny fraction of a second, which will subject the ship (and its entire crew) to a sudden acceleration on the order of several hundred g's! Definitely a very bad idea. (Besides, there's no conceivable purpose for firing conventional munitions of this size -- any target requiring such massive firepower is better dealt with by using nuclear weapons.) 67.169.177.176 (talk) 05:56, 28 September 2011 (UTC)[reply]
Cruiser not battle ship. The cruiser is approximately 1.17 km long and has mass of over a 100 Gg. Is it possible for the cruiser to redirect the recoil? Have the gun recoil into a hydrolic chamber, combine this with explosive retrothrusters (a thruster that produces an instantaneous directed explosive force). Plasmic Physics (talk) 07:20, 28 September 2011 (UTC)[reply]
There is no way to disperse the momentum transfer of a shot like this. Shock absorbers disperse kinetic energy, but momentum is always conserved. Now, your ship is about twice as massive as the estimate above, but his point about the acceleration due to firing being dangerous is still valid. To reach 30 km/sec in 1 km, the projectile needs to accelerate at ~45,000 g's for 1/15th of a second. The ship would be accelerated to 180 m/sec in the same time, for a force of 275 gs. You could try to spread out the time by giving the weapon some recoil within the ship, but with only 170 extra meters to work with, you can't spread the recoil out even over a whole second. In the fist 1/15th of a second, the weapon recoils freely within the ship and moves 90 m. It is then slowed evenly over the next 80 m, which takes 8/9 of a second, and still causes over 20 gs of force on the ship. You can't just say "they'll use thrusters to counteract it", either. In the first case, the thrusters need to be exactly as powerful as the weapon. Even in the second case, they need to accelerate a 100000 metric ton object at 20 gs to counter the force of firing, which comes out to something like 20 billion newtons. This is the thrust of 588 Saturn V rockets.
The point of all this, and I do have one, is that this is science fiction. Nothing like the device you describe can possibly exist given present or plausibly extrapolated technology. So there is no point in asking for how the engineering details would be worked out, because they can't.
Most of the more extravagant problems of this device are in the mass of the projectile. You have a muzzle energy of 2.7e14 joules; this is about the same as a 60 kiloton nuclear bomb. So, if you used an actual nuclear bomb, you could delivery many many times the energy, even if it would be more widely dispersed. But lets say for some reason you are stuck on an inert kinetic impact weapon. Reduce the shell weight by a factor of 100, make it 6 metric tons. To keep the impact energy, it has to go only 10 times faster, at 300 km/sec. The shell is and the weapon are subject to more accelerating stress, but if we can imagine handling 45,000 gs, why not imagine handing 450,000? The good news is that the momentum transfer falls way off. You are down to a 10th of the gs, which still requires some recoil space to dissipate, but becomes survivable.
Now, here's the key. Make the engineering details up. Do ferromagnetic liquids sound cool to you? Then go for it. There is no right answer, because nobody knows how to do it. gnfnrf (talk) 04:12, 29 September 2011 (UTC)[reply]
I need a weapon can compensate for very large target distance. Nuclear weapons in this case, is only lethal on direct contact, so is too expensive compared. Plasmic Physics (talk) 07:48, 28 September 2011 (UTC)[reply]
It is not really the kind of weapon to get into a firefight with, it is a rock to kill an ant, a final resolution. It is targeted by an A.I., requiring confirmation by the person on command. You tend to use it only if you're certain that it will find the target. Plasmic Physics (talk) 21:45, 28 September 2011 (UTC)[reply]
Also, the "hydrolic" (which BTW is not a word) chambers on ordinary (non-space) artillery pieces only work because the gun is set into the ground, or the ship on which it is mounted is floating in the water; i.e. they depend on the resistance of the ground or water to actually absorb and dissipate the recoil. Since your jumbo-cruiser is in space, the only resistance it offers to the recoil is that of its own weight, which is completely insufficient even with "hydrolic" chambers. As for explosive retrothrusters, they will have to fire AT THE EXACT SAME INSTANT as the main guns AND also create nearly the same amount of thrust as the gun. Even if such a massive explosive force could be assembled AND detonated instantaneously at precisely the right instant (which is practically impossible), such a tremendous explosion close aboard would most likely blow a big hole in the main armor, or worse. 67.169.177.176 (talk) 01:55, 29 September 2011 (UTC)[reply]
And that, is why I'm asking for solutions to problems such as these. This ship is classed as a light cruiser. According to the system used, a carrier is typically on the order of 3000 meters long. With 2 meter thick hull armour, a thermal pulse will only do superficial damage. I'd be more worried about the gamma pulse. Plasmic Physics (talk) 02:54, 29 September 2011 (UTC)[reply]
In this case, here's my answer to your question: A coilgun with the specified caliber and muzzle velocity is WAY too powerful for the ship's size, and will be a danger to the ship and crew. Nor do I see any need for an artillery piece of this size. You'd be much better off having the ship armed with much smaller artillery pieces (perhaps firing only 1-ton shells instead of 600-ton -- which would still be A LOT of firepower by anyone's standards), possibly with nuclear shells for the largest, most hardened targets as well as conventional ones for most other targets. 67.169.177.176 (talk) 05:02, 29 September 2011 (UTC)[reply]
Those studies are just about recording visual images you are currently thinking about. Next they need to be able to scan visual images stored in your memory, then non-visual images, then they need to read your personality, then they need to put it all together to form a consciousness. Each of those steps is a massive leap. So, there's a long way to go, yet. StuRat (talk) 12:06, 26 September 2011 (UTC)[reply]
They aren't even quite that, yet. What they are matching bloodflow patterns in your visual cortex to a library of bloodflow patterns from other visual cortexes. Which is pretty damn cool and just the tip of the iceberg, to be sure. But they aren't really about recording visual images — they're looking at patterns and finding ways to correlate those to other patterns, which through very clever work actually corresponds in a recognizable way to images actually seen. But they aren't really able to show what the visual cortex "sees" in its raw form, and it's not clear that it even works for things you are "currently thinking about," as opposed to "stuff your eyes actually are seeing right then." --Mr.98 (talk) 14:20, 26 September 2011 (UTC)[reply]
Only if you consider that in both cases something complex is broken down into small, manageable units, but this is true of many tasks. StuRat (talk) 20:35, 27 September 2011 (UTC)[reply]
It was meant to be a wry comment on the value of predictions about future technology. Just because something is technologically feasible does not mean it actually will be adopted. --Mr.98 (talk) 14:17, 26 September 2011 (UTC)[reply]
For something less ambitious, how about recording video clips containing your thoughts on a variety of topics, and indexing them on your computer, so your family can access those clips whenever they wish ? StuRat (talk) 12:12, 26 September 2011 (UTC)[reply]
A single copy of your software for backup purpose is allowed but you may not distribute copies yourself of yourself. Before you go to immortal dataspace please install Skype so that one can still converse with you posthumously. Will Mr. Magill in Manchester grant an exclusive interview to this uplifting magazine ? Cuddlyable3 (talk) 12:43, 26 September 2011 (UTC)[reply]
Sadly, even if the software did exist, (and it doesn't), your desktop PC likely lacks sufficient memory to store your consciousness with full fidelity. Also, I suspect most humans last longer then most desktop PCs, and most hard drives contained therein. Googlemeister (talk) 13:56, 26 September 2011 (UTC)[reply]
The human brain has 1011 neurons in it — which is 1000GB worth if each neuron were just a byte. But what's really important are the connections between neurons — there are about 1014 to 1015 of those (or, if every connection was 1 byte, between 100TB and 1PB). (In these byte estimates, I am using true GB, TB, and PB definitions, not GiB, TiB, or PiB). So I don't know where 20TB would come from, as a number. All of this makes a lot of arbitrary and no doubt incorrect assumptions though about how much data this technology would require. Assuming these were just stored as fairly simple numbers, you could probably achieve extremely high levels of compression... but anyway, these numbers are kind of meaningless except as an order of magnitude estimate of how complex the human brain is as an organ. (Numbers from Neuron#Connectivity.)--Mr.98 (talk) 20:49, 26 September 2011 (UTC)[reply]
In case it was, has the author thought about how unhappy such an existence would be? It would effectively be like living as a quadraplegic, able to see the world in front of you, but not actually go into it. Of course by the time the tech is available, advances in robotics might make my arguement null. Sir William Matthew Flinders Petrie | Say Shalom! 27 Elul 5771 15:45, 26 September 2011 (UTC)[reply]
Well, the robotics needed is far beyond the current state of the art, but the technology needed for mind uploading is quite a bit farther beyond the current state of the art. Looie496 (talk) 17:37, 26 September 2011 (UTC)[reply]
Believe it or not you can be simulated by a variety of methods of which neural networking is the current favorite used by IBM's Watson. Just restrict the input data to all of your past and present thoughts (rules) and Watson or its equivalent will use neural networking to simulate you. My personal favorite in terms of strictly logical thinking. --DeeperQA (talk) 19:04, 26 September 2011 (UTC)[reply]
IBM Watson does not use a neural network. IBM Watson uses a large, eventually consistent database and advanced natural language processing, as well as massively parallel computing. A neural network is a programming-paradigm for solving specific types of numerical math problems. It's actually a very poor algorithm, in my opinion, compared to formal numerical optimization. Neural networks have an amazing tendency toward system-instability and are not very resilient at escaping from local minima. The term "neural network" has been coopted by various science-fiction writers who think it "sounds cool;" in science fiction, it is used to describe "any type of smart computer software." In fact, a neural network is just a specific technique for solving estimating solutions for matrix algebra. User:DeeperQA, most of what you wrote above is incorrect. You might want to read the Watson FAQ. Watson was never intended to "simulate" a human. It is intended to "greatly improve information seeking tasks" and "help make computers more effective at communicating in human terms." Nimur (talk) 21:43, 26 September 2011 (UTC)[reply]
Where did you get the idea that IBM's Watson used neural networking? Or even that Neural Networks were the "Current favorite"? I thought they mostly passed out of fashion in the late eighties. I think nowadays they're mostly the "Current favorite" for High School level AI projects. (You know, science fair projects titled "Can a Computer be Taught to Recognize Faces?".)
Also, that article you linked is horrible. I wish I knew who wrote it so I could never read anything by him or her again. APL (talk) 21:53, 26 September 2011 (UTC)[reply]
This only serves to further my theory that there are only about twelve actual people on the internet; four of them are reference-desk trolls. Nimur (talk) 22:32, 26 September 2011 (UTC)[reply]
The question looks like an LC item. In any case, he's alluding to what Jor-El did in the first Superman film. Kryptonian technology was rather more advanced than ours (although their common sense apparently wasn't). Another approach could be the way Spock copied the essence of his being to Bones in Star Trek II. ←Baseball BugsWhat's up, Doc?carrots→ 01:56, 27 September 2011 (UTC)[reply]
I wouldn't say 'years absent'. They've been coming back to the RD with accounts and IPs every so often usually getting blocked when they went too far. The most recent account should be obvious from their promotion of their fairly obscure (i.e. if you see someone promoting it you have to wonder who that person is) work both above and on their talk page (which I noticed a few weeks ago) as well as the similarity of intererests and views expressed on the RD. Nil Einne (talk) 02:55, 27 September 2011 (UTC)[reply]
Hang on, read the Consciousness article on this very site. Capturing a human conscience is not simply a matter of recording their memory - you would have to capture and replicate all of the operational aspects of their brain AND the input and output routes that helped to form and will be needed in order to continue to form the structure and experience of the concsiousness i.e. connecting the replicated consciousness to the outside world via an accurate simulation of a human with a body, eyes, ears etc.. otherwise the 'person' wouldn't respond or react to the world or experience it in a manner equal to the way they did in their original body - crucial to get right if you want an accurate simulation of a person and all their characteristics. I guess you could modify the programme to take account of the lack of a real body, but you'd be massively limiting the realism of the simulation. In short to accurately reproduce a human mind, I suspect that you would need to accurately reproduce an entire human being. — Preceding unsigned comment added by 213.120.209.210 (talk) 15:34, 27 September 2011 (UTC)[reply]
Maximum number of options from which a human can make a rational decision on optimal choice
What does research indicate is the largest number of applicants a single hiring manager could rationally consider and deliberately pick a single optimal final choice from without resorting to something arbitrary yet decisive such as pulling a resume out of the stack of equally acceptable applicants and being done with it, thereby reducing the winner's attribution of success to pure luck? 20.137.18.50 (talk) 14:31, 26 September 2011 (UTC)[reply]
There's the well-known paper, "The Magical Number Seven, Plus or Minus Two", which claims 7±2 as the number of items the average person can usefully consider simultaneously. That said, I see no inherent reason that a hiring manager couldn't break an arbitrarily large applicant pool into some sort of bracket-style process. That might not get the "best" candidate at the end, for certain values of "best", but it's certainly not success by "pure luck". — Lomn15:04, 26 September 2011 (UTC)[reply]
I would suspect that a hiring manager not subject to any Workplace stress in his/her position in the real would would be like unto a Spherical cow. There are many things about life that inhibit ideal behavior. I wanted to know the usual result of how successful the optimization effort is after taking the real world into account (which has such pressures and more pretty consistently). 20.137.18.50 (talk) 15:36, 26 September 2011 (UTC)[reply]
375 resumes and a week? No problem, at least for your statement. Divide and conquer. Interviews of finalists the last couple of days. I'll repeat: "might not get the 'best' candidate at the end, for certain values of 'best', but it's certainly not success by 'pure luck'". — Lomn15:40, 26 September 2011 (UTC)[reply]
Would your grouping not be simply mechanical "pulling out of the stack," making it a matter of luck whether someone was in the group that got picked all the way to the finals, making the finalists at least lucky that they got there? To elaborate, imagine the question: Why did you initially put applicant #253 in group number two? Did you have a rational reason for doing so? What did he/she have in common with the other 53 people in group number two and not have in common with everyone not put in group number two, and what made you eliminate group two? Was it the aspect that all of them had in common (if indeed there was one)? In other words, what's the difference between your grouping game and just picking seven resumes off the top or bottom or middle? If one of the applicants that was eliminated for being in the wrong 53-person group upon your first elimination was clearly better upon hearing him/her in an actual interview than any of the other finalists, would not that be very lucky for the finalists that he/she wasn't there? This could be the case for any of the 368 humans you didn't see that got eliminated along the way. 20.137.18.50 (talk) 15:48, 26 September 2011 (UTC)[reply]
You seem to have confused my proposal with GTBacchus'. His appears to arbitrarily eliminate groupings. I'm saying "grab 5 resumes, set aside the best, discard the others. Proceed thusly through all 358 resumes. Now return to your set-aside stack of 70 and repeat. Interview the appropriately small subset of finalists." This is necessarily a rough grading (thus "for certain values of 'best'"), but it is not random selection nor significantly luck-driven. — Lomn16:32, 26 September 2011 (UTC)[reply]
What kind of terrible HR person would do anything arbitrary? Allow me to specify more detail: I would separate them into groups according to some criterion (e.g., type of experience), and then make eliminations based on my professional judgment. Then I would choose a finer criterion, etc. Sheesh.
The disadvantage of your method is that the first 5 you grab might be the 5 best. The next 5 you grab might be the 5 worst. Why eliminate one of the best and keep one of the worst in the very first step? -GTBacchus(talk)17:09, 26 September 2011 (UTC)[reply]
There is a huge qualitative difference that is being overlooked here. One issue is the problem of selecting the optimal member from a large number of items in a group. A second issue is the problem of defining "optimal" by combining a large number of variables - some that are weakly defined - and trying to distill this into a single value suitable for strict "better-or-worse" comparison. The former is a problem of scale, and is fairly trivial to deal with. The latter is a problem of dimensionality. Even with a powerful computer, a high-dimensional optimization problem is difficult to solve; a human may sometimes outperform a digital computer on such problems by using heuristics to sparsify the search-space.
Formally, if the set is comparable, and follows a strict total ordering rule, or at least partial ordering, it is trivial for a human to optimize the selection. The most appropriate technique would be the manual application of a sorting algorithm; the best algorithm to choose depends on the size of the set and the expected distribution. There will be a time-vs.-space tradeoff for each algorithm, and a tradeoff between "best-case," "average-case" and "worst-case" performance. The use of a sorting machine or a digital computer will invariably speed up this process, but is not required. A human can run quicksort on a pile of paper resumes, if they know how to do it. (Use selection sort if you have a small desk; use quicksort if you only have ten minutes until your next meeting).
If the set is highly dimensional and is not strictly orderable, the human will be trying to solve a high-dimensional nonlinear optimization problem; human brains and modern computers are poorly equipped to solve these problems in the general case. Resume-sorting is a good example: there's not a "quality number" printed on top of each resume; the human must estimate whether Resume #3,041 is "better" than #3,042 by applying heuristic interpretations of the resume content. The assumption that there is an optimal choice is formally an assumption that some total-ordering metric can distill a one-dimensional objective function out of the problem, and then we can try to find its "best value."Nimur (talk) 17:31, 26 September 2011 (UTC)[reply]
That is pretty much how I do a lot of things. For this, I'd make two piles: Above average and below average. Then, I'd toss the below average in the garbage. I'd repeat again with Just Average, and Better than Average on the previous Above Average stack. I'd toss the lower group in the garbage and repeat. Eventually, my good stack will have a few resumes that I can look at in detail - even though I've already scanned them multiple times and know them pretty well. It probably make me comfortable to sort this way because it is just a binary tree sort, which is something I've worked with for about 30 years. -- kainaw™17:45, 26 September 2011 (UTC)[reply]
Honestly what is going to happen is that of the 375 applications, probably 200 will be at least qualified, and the top 50 will have only a negligible difference in quality. Thus mix the pile of applications and take the first 50 from you new mixed pile. Toss the rest. This will have the added advantage of separating the lucky from the unlucky applicants, since you don't need unlucky people in your company. Divide that group into qualified vs unqualified. Toss the unqualified. You should be down to around 30 qualified applicants. Now go through in detail in groups of 10 and keep very highly qualified, and toss those who are barely qualified. Interview these remaining applicants and have a nice week. Googlemeister (talk) 18:22, 26 September 2011 (UTC)[reply]
Simply use a search engine to find all of the applicants that included all of the attributes you desire. If the results are greater than one add another quality until you find "The One". --DeeperQA (talk) 19:21, 26 September 2011 (UTC)[reply]
There is, of course, no guarantee that any particular combination of search criteria will result in precisely one hit. There's also the practical matter of what kind of search engine to use, and the question of applying judgment to individual cases. Can search engines screen for everything that we expect HR professionals to understand? Why not replace them all with robots, then? -GTBacchus(talk)00:59, 27 September 2011 (UTC)[reply]
Some writers start by trying to memorize every word in the dictionary and to that end playing crossword puzzles and scrabble. The psychologist's goal as a scientist is to form words and phrases that describe the most esoteric of conditions. All that is required to automate the process of identification by search engine is to use these words as keywords and to use them in the personnel database(s) to be searched. The key is to use all of them by either stating that they define or do not define the entry. The creating of new words, phrases and phrase combinations may one day be the privy of computers but for now it is still pretty much limited to the privy of human beings. See Optimal Classification--DeeperQA (talk) 03:34, 28 September 2011 (UTC)[reply]
And what guarantees that some combination of criteria gives you exactly one hit? What if your first 7 criteria narrow the field to 5 candidates, and then every other criterion you try next cuts it to zero. Then what?
I believe that last time, you were refered to the Bristol Stool Scale and informed that feces is far too variable a substance to come up with a single numerical answer for the density of stool. It will vary wildly from person to person, and from sample to sample for even the same person. --Jayron3200:23, 27 September 2011 (UTC)[reply]
Jayron is correct (as usual). If all faeces stools were the same density we would not have developed the "sinker" and "floater" terms! These two words validate what Jayron is saying. Richard Avery (talk) 06:19, 27 September 2011 (UTC)[reply]
Crowbar in a blender...
Is there any blender currently on the market that can successfully blend a crowbar without breaking? This is related to the Will It Blend? YouTube vids. People keep requesting that he do a crowbar, but he keeps finding ways to avoid doing a crowbar.
Also, would it actually be possible to blend diamonds? I know that he blended cubic zirconia in one vid when people were requesting diamonds (due to the cost), but those are less hard. --Kurt Shaped Box (talk) 01:05, 27 September 2011 (UTC)[reply]
The blades of a blender cannot slice or cut a diamond, but if it strong enough, it can smash the diamonds into dust. While diamonds are hard to scratch or cut, they are quite fragile, they can be smashed using a hammer. Plasmic Physics (talk) 03:16, 27 September 2011 (UTC)[reply]
"Will it Blend" is an advertisement, I think it's safe to say that if that brand of blender were capable of it, he would do it.
But seriously, a crowbar would be tough to get through quickly with anything short of a plasma torch. (And if a blender's blades can't make it through the thing in one go, it'll just jam.) APL (talk) 07:14, 27 September 2011 (UTC)[reply]
I thought that there might actually be a video on YouTube of someone trying to blend a crowbar using a BlendTec blender (and failing), just to diss Tom. Not so. Yet. --Kurt Shaped Box (talk) 11:59, 27 September 2011 (UTC)[reply]
Hypothetically-speaking, how would one go about designing a blender capable of slicing up a crowbar? I'm just picturing the infomercial now - the pitchman being all like "Look! It will even blend a crowbar and still make guacamole!". Billy Mays would have approved. --Kurt Shaped Box (talk) 23:32, 27 September 2011 (UTC)[reply]
Well, it would need to be huge, with blades far bigger than a crowbar (imagine a battleship's prop), and the blades would need to be coated with something hard enough to cut through the steel in the crowbar (corundum would be fine, no need for diamond), and the blades would need to spin at a hellacious speed, which would require a huge amount of energy. StuRat (talk) 00:37, 28 September 2011 (UTC)[reply]
Machines which grind off pavement have steel teeth with carbide cutting tips welded on. The same trick is used on carbide circular saw blades. It is a very old trick. Archeologists found an axe from (if I recall correctly) the dark ages which had a hard but brittle steel welded (by heating and hammering) onto a softer iron base, so that it could maintain a sharp cutting edge but would not shatter as easily as the harder and more brittle edge material. Industrial metal shredders are close to what is desired, shredding thick metal castings, but no indication they would handle a crowbar: [2]. I have seen shredders at scrap metal recycling companies which could handle an automobile, including the frame and axles, and I expect a crowbar wouldn't even be noticed. The cutting part shown at 2:20 has a rotating cylinder with long teeth sticking out which basically shears the metal into little pieces. That video is a toy sized operation compared to some. Others have a pair if meshing cutter arrays:[3]. Here they shred steel rebar ans structural steel such as angle. Seems like such a company could build a "blender" with a vertical hopper and a spinning vertical shaft with teeth projecting from it. From captions on some of the videos, it would likely need couple hundred kw motor. Edison (talk) 15:26, 28 September 2011 (UTC)[reply]
Those videos are awesome. I'm sooo going to be watching those 'Shred of the Month' vids tonight. It's like a document shredder - but on steroids (anyone else here love feeding the document shredder?). Also, car shredder redirect created... --Kurt Shaped Box (talk) 18:24, 28 September 2011 (UTC)[reply]
Homopolar generator
If a conductor were buried at ether the North or the South pole at the rotational axis of the Earth and another at the equator and an insulated wire laid so that a meter could measure any voltage or current between them what would the voltage and current be? --DeeperQA (talk) 01:21, 27 September 2011 (UTC)[reply]
The question is how much power would be generated if such a scheme did in fact work. Could enough power be generated to solve the energy crisis? --DeeperQA (talk) 03:17, 27 September 2011 (UTC)[reply]
It (voltage and current) depends on how much work (watts) you try to extract from it - which in turn is related to the (reaction) force being exerted on the wire due to the motion of the earth - the general equation is given at the top of Lorentz force. You need the magnetic field at the earths surface, and the angle it makes with the surface of the earth (Magnetic dip ?)Imgaril (talk) 11:00, 27 September 2011 (UTC)[reply]
To a first order of approximation, the Earth's magnetic field is fixed in magnitude and rotates along with the circuit, so there's no time-varying change in magnetic flux through the circuit, which means no current is induced. However, there can be geomagnetically induced current due to geomagnetic storms. However, GIC isn't steady or reliable, and hence would make a poor energy supply. Red Act (talk) 16:41, 27 September 2011 (UTC)[reply]
What if you forget about the equator and build some superconducting hexagons over the north magnetic pole? As it moves you sap the energy from one hexagon, dismantle it, and move it around in the direction the pole is going. How much energy would that produce? Is it possible to trap the pole and keep it from moving with enough superconducting hexagons cutting across the field lines - or even deliberately give it freedom to move only in a direction you want? (Note: this is theory only - real superconductors have limits to the current they can carry, which I think in practice would severely limit the amount of power extracted) Wnt (talk) 18:51, 27 September 2011 (UTC)[reply]
Some responders have been too quick to dismiss the question. When a solar flare reaches the Earth, utilities sometimes experience hundreds of volts and hundreds of amperes through neutral conductors when the transformer neutral is grounded at each end of the line. This is seen in lines under a hundred kilometers, at various orientations. There are also earth currents of electricity of varying magnitude, due to different points of the Earth';'s surface being at varying potentials. They were studied by Sir Humphrey Davy, Becquerel and others from the beginning of the 19th century. W.H. Barlow in the 1840's found them strongest in a N-S orientation. They vary during the day and during the year, and are influenced by electrical storms, and relate to the auroras, and to sunspots.. They were sometimes large enough to interfere with telegraph operations. They generally prevented telegraphy from August 29-September 4, 1859. It produced 800 volts on a 600 km line in France. It produced an effect near Boston equal to 200 Grove cells, or about 380 volts. Telegraph lines could in fact be worked without batteries, powered only by the earth currents. In Europe that particular storm had its strongest effect in a NE-SW direction, but would have likely had a huge effect from pole to equator. In 1881 a similar storm produced 1.1 volt per km in England. You could take 1.1 volts/km times 10,000 km from the pole to the equator as a representative value when conditions are favorable, and expect 11,000 volts between the ends of the line. If the ionospheric currents which induce the earth currents have more localized circulation (on the scale of 1000 km,) then over such a long line there could be some cancellation. As a result, a somewhat longer or shorter line might have a higher end-to end voltage. A telegraph line in the 19th century had appreciable resistance, which held the current to .28 amperes in one report from the last work cited, in a 200 mile line from New York to Providence, corresponding to 644 volts. The ground resistance at the terminals and the resistance per unit length of the conductor would be the limiting factors holding down the current.Today researchers use the term "Telluric current"in place of the older "earth current," to indicate that current in seawater is included in the topic. The latter source, from the 1980's indicated that a 2650 volt potential had been measured along a transatlantic cable in the 1950's. It also notes (p245) that in 1921 a geomagnetic storm produced 1kv over a 100 to 200 km line, with 2.5 amperes of current available (probably one or the other conditions would apply: high voltage at zero current, or high current in a short circuit condition.) The source discusses telluric problems on long power lines, including 100 amps of telluric current in 1972, sufficient to damage power transformers. Edison (talk) 19:12, 27 September 2011 (UTC)[reply]
Red Act's answer above is incorrect. It's such a common misconception that it has a name--the Faraday paradox--that is also discussed in the homopolar generator article. The first error is saying that the Earth's magnetic field 'rotates with the planet'. Since the field is circularly symmetrical, there is no way of knowing whether it's rotating or not; in fact, 'rotates' has no physical meaning in this case. You can't paint a mark on the field so that you can watch it going round. The error may be caused by people imagining that the field is a bunch of field lines rotating, when in fact field lines are a drawing aid and don't actually exist. The field is really a continuum. The second error is the belief that the generator works by the relative motion of the rotor and the field: it does not. What matters is the relative motion of the rotor and the pickup wire, both of which are within a constant (unipolar) magnetic field. I'll stop here because half of you probably don't believe me, but I leave you to read the relevant articles. --Heron (talk) 19:37, 27 September 2011 (UTC)[reply]
Are you saying that the difference in potential between the center of the copper disk and the edges of the copper disk is such that a wire attached to the center of the disk will "drain" that potential at each point on the edge of the disk that it touches and that continuous flow then depends upon the wire touching new points on the edge of the copper disk? --DeeperQA (talk) 19:57, 27 September 2011 (UTC)[reply]
Not quite. The potential (I think it could be called electromotive force or EMF) does not exist until the disk rotates. It is created by the rotation of the disk relative to the collector. Imagine that you cut away almost all of the disk leaving just a narrow wedge between the axle and the collector. Now you rotate that wedge through say 1°. From the start to the end of that rotation, the geometry of the electric circuit has changed slightly. Before the rotation, let's say that the axle, wedge, collector and external wiring were all in a single plane. After the rotation, the circuit is no longer in a single plane, because the wedge is now out of the plane, and the current has to return through a tiny arc of the collector and back into the original plane. This is equivalent to tilting the plane of the original circuit slightly. The tilted plane is now not quite parallel to the magnetic field lines, so it has gone from having none of the flux passing through it to having a tiny bit passing through it, and that is how it picks up an EMF by electromagnetic induction. Now, if the collector circuit were rotating with the disk, none of this plane-tilting would happen and there would be no EMF. This is all hand-waving stuff and probably sounds oversimplified to a proper physicist (I'm not one). The real answer is due to the Lorentz force on electrons, but I can't remember how that works so I've gone for the geometrical explanation instead. I believe you need to use special relativity to explain the Lorentz force properly. --Heron (talk) 20:44, 27 September 2011 (UTC)[reply]
I follow you up to the point of the plane. In my imagination this plane is horizontal and perpendicular to the axis of the disk which is vertical. To follow what you are saying though it seems I am using the wrong orientation for the plane. The plane you are referring to seems to be in line with and parallel to the vertical axis on which the disk is rotating such as is a knife used to cut a piece or wedge of pie. Is the plane you are referring to in line with the axis of the disk or in line with the disk? --DeeperQA (talk) 01:03, 28 September 2011 (UTC)[reply]
The plane is in line with the axis and perpendicular to the disk. If the axis is vertical then so is the plane. Therefore the magnetic field vector is parallel to the plane and does not pass through the circuit. If the plane tilts slightly then a small amount of flux will pass through it. This change in flux through a loop, according to classical electromagnetics, is what induces the current. --Heron (talk) 13:20, 28 September 2011 (UTC)[reply]
So you are saying that because the disk is turning for a moment in time the circuit moves with it and becomes slightly tilted to the plane such that opportunity is presented to the magnetic flux to pass through it? --DeeperQA (talk) 19:54, 28 September 2011 (UTC)[reply]
So I can leave the disk and the magnet stationary and just move the wire around the perimeter of the disk to generate electricity? Homopolar Generator--DeeperQA (talk) 22:30, 28 September 2011 (UTC)[reply]
What's the best way (if any) to store and/or transmit sensitive scientific data in such a manner that said data would be immune to tampering (or nearly so) during storage/transmission? If such a method exists, would it be possible to transmit the data in real time from an aircraft in flight over water? 67.169.177.176 (talk) 03:47, 27 September 2011 (UTC)[reply]
One more thing: If the plane's RNAV system is integrated with the scientific instruments and the recording/transmitting computer, should the navigator/scientist enter his private key before or after aligning the RNAV? (I'm guessing before, but I need to be sure.) 67.169.177.176 (talk) 04:52, 27 September 2011 (UTC)[reply]
An aircraft's flight-critical or flight-essential systems (such as RNAV) should not be integrated with a mission payload. While they might reside on the same computing system, their operation should be independent such that no misconfiguration of mission equipment impacts the safety of the aircraft. — Lomn13:08, 27 September 2011 (UTC)[reply]
Assuming, though, that we're talking appropriate levels of independence, and that the mission payload is pulling some read-only data from RNAV -- I'd have to say, in the very general case, that you always pull data after initialization. It's not at all clear, though, what RNAV data you'd be pulling (a flight plan?) or what "aligning" RNAV would mean, particularly in the context of cryptography. — Lomn14:01, 27 September 2011 (UTC)[reply]
This is correct -- the RNAV feeds position data (including groundspeed/drift angle) to the mission computer, which then uses it to correlate the scientific data (temperature, humidity, solar radiation, levels of CO2 and trace gases, etc.) to the plane's position. So in case of a computer crash, the scientific data is wiped out, but the RNAV system is still operational. As for "aligning" the RNAV system, this involves spinning up the gyros, torquing the accelerometers, and then feeding the plane's current position into the system; this is done before each flight, and is the navigator's responsibility. 67.169.177.176 (talk) 00:07, 28 September 2011 (UTC)[reply]
Inertial nav systems -- gyros, accelerometers, and the like -- are not permissible for primary RNAV inputs, as they cannot maintain appropriate NSE. Note also that you're talking positional data, not navigational data. It appears that you ought to just pull from a GPS box. — Lomn12:33, 28 September 2011 (UTC)[reply]
The advantage of the inertial navigation system (supplemented by a Doppler system), in this context, is that it provides a continuous readout of the plane's acceleration, which in turn allows the computer to calculate the direction and strength of air currents -- a valuable piece of information in the context of studying climate change. GPS cannot give this information in the raw form like inertial and Doppler can. So even if inertial won't be needed for navigation (most of the time, anyway), it will still be used as a scientific instrument. 67.169.177.176 (talk) 05:16, 29 September 2011 (UTC)[reply]
To detect deliberate tampering (i.e., tell legitimate data from bad) you need either a message authentication code or a digital signature. Digital signatures are more expensive to calculate but have some advantages that may or may not matter in your application. To protect against tampering (i.e., ensure that all legitimate data gets through) is impossible in general because the attacker can always just block the signal completely. Error correcting codes might help in certain situations. I can't answer your followup question because I know zilch about aircraft control systems. -- BenRG (talk) 08:00, 27 September 2011 (UTC)[reply]
In this case, I am not concerned about the signal getting completely blocked; my main concern is how to prevent the data from being fraudulently altered during storage or transmission. FYI, the aircraft in question is performing a flight around the equator as part of a mission to study climate change (as well as being a memorial flight for Amelia Earhart) -- so the main concern in this case (as far the the data is concerned) is to prevent either the crew or the scientists from altering the data at will for political purposes, just in case they have a mind to. 67.169.177.176 (talk) 00:07, 28 September 2011 (UTC)[reply]
It is possible to prevent tampering, rather than just detect it, by using narrow beam transmission and directional antennae at both ends, say to and from a series of satellites. The only way to block such a transmission would be to physically place the jamming signal generator between the plane and satellites, or send a massive EMP to overwhelm the signal. StuRat (talk) 00:22, 28 September 2011 (UTC)[reply]
Thanks for the ideas, everyone. Now, how big is a typical narrow-beam directional antenna? Please understand that it would have to fit into an Electra 10-E, which is not only pretty small to start with, but already crammed full of extra fuel tanks (which take up the forward 1/3 of the cabin), navigational equipment, scientific instruments, etc., etc... 67.169.177.176 (talk) 06:13, 28 September 2011 (UTC)[reply]
Question about the Galileo Statement
Galileo theorized that in the absence of air, all things would truly fall with the same acceleration and 300 years later demonstrated this by the crew of Apollo-15 on the Moon (which has gravity but lacks air) by dropping a hammer and a feather.
As moon was seen from both the feather as well as hammer with two different gravitational fields [g] therefore just wondering what was the falling acceleration of moon towards aforementioned feather and hammer from the following possibilities [if i'm not wrong]?
1- Moon had the gravitational acceleration of hammer i.e "g" of hammer
2- Moon had the gravitational acceleration of feather i.e "g" of feather
3- Moon had the net "g" of feather and hammer
So, is Galileo's statement correct [Theoretically] if the senario is considered in the absence of all other gravitational attraction.?68.147.43.159 (talk) 03:52, 27 September 2011 (UTC)Eccentric Khattak#1[reply]
The force on the moon will be the gravitational attraction from the hammer plus the gravitational attraction from the feather. The moon will accelerate towards the centre of gravity of the (hammer+feather) system, a point which will be much closer to the hammer than the feather: so the moon will move very slightly in the direction of the hammer. (The hammer and feather will also accelerate very slightly towards each other, but this tiny movement won't significantly affect the moon's trajectory.)
Since F = (G m1 m2)/r, and a=F/m the total acceleration on the moon will be G (mfeather+mhammer)/r, or with a 1kg hammer and 10g feather about 4x10-17 m s-2 (by my calculations). If you drop them from one metre up, the moon will move something of the order of the diameter of a proton. The moon's surface will be much bumpier than this (plus there will be stray gas and dust molecules on the way down), and its gravitational field is non-uniform (because it's not a perfectly homogenous sphere) so in theory the moon will hit the hammer first, but in practice the difference in gravitational attraction will be unimportant to the outcome. --Colapeninsula (talk) 10:40, 27 September 2011 (UTC)[reply]
Is it correct to say that even if force experienced by hammer is more due to its more mass, acceleration gained by the hammer is same as that of feather because hammer has more inertia due to its more mass? - 61.16.182.2 (talk) 04:03, 28 September 2011 (UTC)[reply]
Gender-specific athletic performance
I just noticed that the ratio for men's and women's world records in the marathon is 1:1.0953 and that of the 100 meter sprint is 1:1.0959. I find it fascinating that they are currently equivalent to the thousandth decimal. Is this a coincidence or is this performance ratio present is other events? The Masked Booby (talk) 03:55, 27 September 2011 (UTC)[reply]
Just crunching numbers for a couple other events, I got a ratio of 1.1215 for the 800m, 1.1319 for the mile, and a ratio of 1.1893 for the high jump, so it looks like a coincidence. Googlemeister (talk) 13:46, 27 September 2011 (UTC)[reply]
Bringing back an old topic. In hyper valent molecules, is it correct to say that the hypervalent atom only contributes to some of the orbitals, meaning that the remaining ligands effectively borrow the otherwise LUMO orbitals to complete their valencies? Take pentachloro-λ5-phosphane or pentachloridophosphorus as an example. According to my theory, the phosphorus only has a total bond order of 4, despite having 5 ligands. The three equatorial bonds have a bond order of one each, the orthogonal bonds have a bond order of a half each. To make this arrangement work, the two half order bonds will have to be spin paired. Essentially, phosphorus does not contribute to the HOMO. This would should result in a triagonal bipyramidal molecule with a 1+ charge on two chlorines each, and a 2- charge on the phosphorus. The two positively charged chlorines, will move to opposite locations to minimise electrostatic repulsion. The remaining chlorines will move to equatorial positions. Because half bond order bonds are lower in energy than 1 bond order bonds, they are longer. As a result, the orthogonal bonds should be longer than the equatorial bonds, which is what is observed. This is my own idea, the three-centre-four-electron article doesn't make much sense to me, so I synthesised this one. I call it orbital borrowing, since the ligands borrow an orbital(s) from the host, without the host contributing. Another thing to note, is that this orbital borrowing should only be a stable system for molecules where the gap energy between the HOMO and LUMO orbitals of the host is small. So orbital borrowing shouldn't be happening for 1st period elements, except under extreme conditions. is orbital borrowing correct, and is it a useful description? Plasmic Physics (talk) 11:56, 27 September 2011 (UTC)[reply]
Orbital "borrowing" is a useful heuristic to describe what happens in either hybridization theory or molecular orbital theory. However, it is important to remember that in MO theory, the idea is to describe all orbitals of the molecule as belonging to the molecule as a whole; without giving actual care as to where they "came" from. When a molecular orbital has a character which resembles what would be expected from the mathematical combination of two atomic orbitals, we can say it "formed" from those orbitals; likewise if a molecular orbital is largely identical to the atomic orbital in an unbonded atom, we can say that the orbital "did not participate in the bonding" of the molecule; but these are heuristic approximations. Strict MO theory actually treats all electrons (and thus all orbitals) as belonging to the molecule as a whole and describes the orbital space around the entire molecule, and does not pretend to assign electrons or orbitals as "originating" at any particular atom. MO diagrams, in this way, present a good heuristic for predicting what the MO structure is going to look like; but to be scrupulously correct, actual molecular orbitals are described purely by the wavefunction of the molecule in exactly the same manner as the atomic orbitals are described purely by the wavefunction of the atom. And you've also gotten way past the level of mathematics I have experience with, so I'm not sure I can get more detailed than that. The concept of "hypervalency" is a way to jibe the mathematical results of the wavefunctions with the heuristic predictions of the simpler models, like the "octet rule" and Lewis/Valence bond theory. In actuality, MO theory doesn't treat "hypervalent" molecules as a special case; they just are what they are. --Jayron3216:55, 27 September 2011 (UTC)[reply]
Love hurts
Our articles on limerence and being lovestruck mention chest pain as a "symptom". Having experienced this on more than one occasion, I know that it's not made up. My question is, what's the physiological basis/mechanism for this pain? What's making my chest hurt when I'm in love/lovesick? ElMa-sa (talk) 12:53, 27 September 2011 (UTC)[reply]
This is OR, but my impression is that the ache actually arises from the solar plexus, not the heart. In fact, when people have actual damage to the heart, the pain sensation is frequently referred to other body parts, such as a shoulder. Looie496 (talk) 14:27, 27 September 2011 (UTC)[reply]
That article focuses on the worst cases, but in general, stress weakens the heart, allowing for some ballooning to take place, which causes chest pain. There are many kinds of stress and this particular kind tends to follow deeply emotional events - which is why it is referred to as broken heart syndrome. -- kainaw™14:45, 27 September 2011 (UTC)[reply]
I scanned over the news-website you linked; noticed strong claims; and so as I am typically inclined to do, I searched for the original paper. The article didn't mention where it was published, so I turned to Google Scholar search. All I turned up is ... nothing. It appears User:Edison also searched for an original publication back in April and independently reached the same conclusion: who is "Peter Braun"? Where does he work? Where did he publish his research? Absent such clarifications, we can't really evaluate the claims. This is a case of serious shortcoming of internet "journalism." The original "newspaper report" does not actually answer any of these questions. I think we should probably call "citation needed" on that Telegraph article. Nimur (talk) 16:55, 27 September 2011 (UTC)[reply]
Have you looked at the amount of mercury involved? Yes, it is technically correct to say that a CFL contains mercury, but the amount involved is minuscule. For comparison, the amount of mercury released into the environment by smashing a CFL is far less than that released by burning coal to power an incandescent bulb for the 10000 hours or so that a CFL will last. --Carnildo (talk) 01:11, 28 September 2011 (UTC)[reply]
Yes, because of the soldering. That mercury and lead and tin and zinc are all considered hazardous materials these days is just a ploy to make you get rid of those for free so they can recycle them. The official reasoning is usually just some chemical compounds containing those metals that are actually harmful on your health, but you can find harmful compounds of any element; or allergic reactions, but you can find allergy and skin irritation for any material as well. I wonder which classical metal will be turned dangerous next: iron, copper, silver, or gold? – b_jonas19:40, 28 September 2011 (UTC)[reply]
If it did, it would have reduced the nitrogen atom from the +5 oxidation state to the +4 oxidation state, which would mean that something else would have had to have been likewise oxidized. Look at the oxidation states of all of the elements in your mixture, and see if any is likely to be oxidized. --Jayron3218:23, 27 September 2011 (UTC)[reply]
From personal experience, no. Potassium nitrate and excess sulfuric acid is reacted to produce nitric acid, which can be disilled off as an azeotrope. No brown gas is evolved, the nitrate ion is preserved in the substitution reaction. Plasmic Physics (talk) 01:28, 28 September 2011 (UTC)[reply]
Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. --Sean17:55, 27 September 2011 (UTC)[reply]
It does not sound like it is necessarily a homework problem. A sensible homework problem would be more likely to ask how the color changes when the compound reacts with something. It is not a violation of the rules to just answer a question here with a referenced answer. I see lots of sites with discussion of similar sounding compounds; maybe there are various nomenclature conventions. Is "iron(II) thiocyanate", also discussed here as an indicator, different from "iron (II) isothiocyanate?" The latter site discusses Fe(SCN)x compounds; are these different from the OP's "CNS" compounds? Edison (talk) 18:40, 27 September 2011 (UTC)[reply]
Thiocyanate is an ambidentate ligand (I took freshman chemistry with the guy who discovered this general property in organometallic chemistry). Thus, SCN is different from NCS in that SCN bonds to the iron at the sulfur, while NCS bonds at the nitrogen. I just noticed that the OP is asking about CNS, which I am unfamiliar with, but doing a quick search indicates that CNS is an alternate way of notating the thiocyanate ion (I always use SCN), apparently used in applications where the actual bonding is irrelevent or unknown. --Jayron3218:48, 27 September 2011 (UTC)[reply]
I remember a ton of similar problems when I took inorganic chemistry, based on ligand field theory. Different geometries and numbers of ligands, and different orbital energies of the ligands and different oxidation states of the metal all play a role in predicting the energies of the transitions among electronic states. That's the same as saying "what are the UV/vis spectral characteristics?", which is really just "what color is it?". One could even analyze this pair based on HSAB (different iron charge and effect of added "S" have on the nitrile ligand)? That latter is sort of like Edison's idea, except the "reaction" is the removal/addition/electronic-tuning of the ligands on one to get to the other:) DMacks (talk) 02:18, 28 September 2011 (UTC)[reply]
Anecdotally, I know several people who have adopted small dogs from shelters, and many such small "rescued" dogs have a pathological fear and hate of strange men, maybe due to their life experiences, or maybe because they're nuts. I've known several dogs whose owners got tranquilizers prescribed for the animals to calm them down so they don't bounce off the walls. I have known several nutty cats which go berserk and start scratching or biting for reasons known only to themselves. Maybe behavior which seems nutty to us makes perfect sense to the animal. Edison (talk) 19:51, 27 September 2011 (UTC)[reply]
Most of the cases of 'animal mental illness' I've come across inevitably have the animal kept under the close auspices of, you guessed it, humans. Cf. dogs, zoo animals. Nietzsche wrote a little about how he thought animals would regard humans as the 'sick animal, the laughing animal'. Vranak (talk) 00:56, 28 September 2011 (UTC)[reply]
Here's the full quote:
I fear animals regard man as a creature of their own kind which has in a highly dangerous fashion lost its healthy animal reason – as the mad animal, as the laughing animal, as the weeping animal, as the unhappy animal. The Gay Science. Vranak (talk)
Are you kidding? The only dogs I have known that did not go into kill mode when hearing or seeing a person riding a bicycle are those who have been trained and conditioned by a loving owner to accept such persons as friends. Dogs come into this world as completely neurotic killing machines until guided by their owner with love through the most normal of human activities and situations. --DeeperQA (talk) 03:19, 28 September 2011 (UTC)[reply]
What you call neurotic Cesare Millan might called 'prey drive' -- which would be very useful if a dog had to go feral and catch its own dinner. Vranak (talk) 03:55, 28 September 2011 (UTC)[reply]
In terms of prey most domesticated dogs (family pets) are too well fed for food to be the driving force behind their neurosis. Rather their neurosis is to establish dominance over the territory they inhabit in the mind of anything and everything that enters it, right up to the point of who shares their food bowl. What love and conditioning and training and food do is merely raise the trigger point to a higher level. For a feral dog or cat that level may be only slightly above that of a wild animal. Check news articles for the past month and you will find one in which a family pet mauled and killed a new born because the parents were expecting the dog to automatically accept it as a family member without introducing the dog to the child. For the dog the infant was something that had entered its territory without introduction and anyone's demonstrated approval and in need of being killed and eaten like a person cuddling a rabbit they are about to set down in front of a pet snake for its next meal. --DeeperQA (talk) 05:51, 28 September 2011 (UTC)[reply]
How does that mean neurosis? Human beings usually automatically kill any snake or spider they find in hysteria that it might be venomous. Does that make humans neurotic as well? You are making the classic anthropocentric mistake of classifying animal behavior by how they relate to humans. -- Obsidi♠nSoul06:37, 28 September 2011 (UTC)[reply]
So you are saying that puppies are completely neurotic killing machines. Seems harsh. :) For interest, in my area there are several packs of feral dogs and my experience of their behavior doesn't really match your descriptions. They are effectively wild. They live, move, hunt/scavenge as packs and they are rather good at finding snakes it seems. They invest a lot of time establishing their dominance hierarchy or playing (...hard to tell the difference) and sleeping of course (..it's hot) to the extent that they are almost nocturnal. They don't really interact with people or seem very interested in people at all. They seem to reserve barking for ritualized, quite rare interactions with neighboring packs and encounters with potentially dangerous snakes. They are oddly quiet. Their territories seem to be rather fluid and change quite rapidly over time. The area is rural so there isn't really any kind of population pressure. I cross various pack's territories on foot quite often and if anything, I think the packs tend to react with indifference or moderate fear when they see a large primate crossing their territory. They certainly don't go into fight or flight mode or act aggressively. Sean.hoyland - talk06:50, 28 September 2011 (UTC)[reply]
Trigger Point Ally, Trigger point... I am saying is that all that changes is the threshold. For puppies it is not until the first really painful bite on the nose by a litter mate that reveals puppy love may not be the only thing a puppy is made of. Dogs that have not been conditioned, trained, and reinforced with food and shelter to tolerate others and even strangers in their environment may have a very low Trigger Point of becoming neurotic over the presence of undesired guests. Even those which have been conditioned may "loose it" if their mind is in control and not their owners when faced with the need for action. Dogs which are left on their own forced to rely upon their own decisions may have no threshold of tolerance at all. Hence dead baby brought to you by an otherwise loving pet. --DeeperQA (talk) 07:52, 28 September 2011 (UTC)[reply]
Read Animals in Translation the works of Monty Roberts if you want to read about animal neurosis and do not include me or my family in any so called humanity which automatically kills spiders or snakes. My reaction is to photograph them, or capture them and release them a day later. μηδείς (talk) 00:31, 29 September 2011 (UTC)[reply]
How many fields?
I am guessing : electromagnetic, gravitational , weak , strong, possibly higgs.
Every particle is made up of fields and the relationship is not one to one. Most of those fields have components that may or may not be counted as separate fields depending on how you choose to count them so counting fields is not a trivial matter. For instance, quarks come in three different colors and each color possibility should be counted separately for some purposes such as statistical mechanics where the number of independent states of an ensemble must be accurately calculated. Often times though identical quarks with different colors are not counted separately since they are so similar. Dauto (talk) 21:31, 27 September 2011 (UTC)[reply]
The second question has an even more subtle answer. Most of those fields are believed to be massless at high enough energies, but at low energies, after electroweak symmetry breaks, many of them acquire mass through the Higgs mechanism. Only massless particles propagate at the speed of light. Dauto (talk) 21:35, 27 September 2011 (UTC)[reply]
The real answer is that fields are a concept which act as a model that allow us to probe the details of how the universe works in a mathematically consistent way, and which best approximates the actual behavior of "reality". (It should be noted that, in this way, fields are not any different than "particles" or any other physical concept). Fields can be said to exist for any force or property which propagates through space (or space-time) and decays. We use fields because they work. --Jayron3221:41, 27 September 2011 (UTC)[reply]
In modern physics "field" has a pretty clear definition and it is reasonable to ask how many there are (though, as Dauto said, there are different ways of counting them). -- BenRG (talk) 07:16, 28 September 2011 (UTC)[reply]
The number may increase dramatically if supersymmetry is discovered. And since supersymmetry is broken, there then likely exists hidden sector fields corresponding to particles that unlike the supersymmetric particles don't interact with Standard Model particles at all (except via gravity). Count Iblis (talk) 22:18, 27 September 2011 (UTC)[reply]
Yeah, LHC has been pretty hard on some of the more creative ideas, but supersymmetry is still alive and well despite the rumors of its demise. Dauto (talk) 13:18, 28 September 2011 (UTC)[reply]
Although, if the Higgs is not found soon, that is then evidence for new physics. E.g. there could exist many different types of dark matter particles to which the Higgs can decay. That then hugely broadens the resonance signal for the Higgs, making it much more difficult to extract from the background. Count Iblis (talk) 15:52, 28 September 2011 (UTC)[reply]
The LHC has closed many windows in the mass spectrum for the Higgs, but the range from 115 to 130 GeVs is still open because of the large amount of background. There seems to be a peek forming in that region but it is still too early to tell. That peek seems broader then expected which goes along with your remark about peek broadening. That broadening might be spurious though. - Heck, the whole peek might be spurious. We may have to wait at least another year before things become more clear. Dauto (talk) 18:31, 28 September 2011 (UTC)[reply]
Suppose I was to wind a coil of wire around the equator lots and lots of times. In this configuration with the Earths magnetic field the coil takes on the same (albeit linear rather than "U" shaped) voice coil or solenoid. Could I do anything useful with this coil such as listening to the Earth's magnetic field or sending dance tunes to aliens in the far reaches of space? --DeeperQA (talk) 20:57, 27 September 2011 (UTC)[reply]
Not that I'm aware, although you could use up the world's copper output for a while. Electricity is generated when a wire loop moves with respect to a magnetic field. Yours would be pretty static. Regards, RJH (talk) 22:50, 27 September 2011 (UTC)[reply]
You could probably pick up fluctuations in the field, such as from geomagnetic storms, though. Of course, you could probably do almost as well with a much smaller coil... --Carnildo (talk) 01:14, 28 September 2011 (UTC)[reply]
Silver mining
My crew of dwarves that I'm running a dnd quest for is about to liberate a silver mine from a band of marauding orcs who took it over from the human kingdom recently. They've mentioned desire to actually take it over and manage it during down time from adventuring so I thought I'd investigate what would be involved. I've taken a look at the silver mining page and have decided that the mine is actually a lead mine and the silver is extracted by melting the lead with zinc. My question is this: About how much lead could one expect to need in order to get a measure of silver? Are we talking about a tonne of lead for a pound of silver type ratios or would it be much greater than that? Thanks for any insight :) 142.244.35.91 (talk) 21:21, 27 September 2011 (UTC)[reply]
If it helps any, an example on pg. 41 in Principles of Mining by a certain former president gives 20 ounces of silver per ton, of which 15 ounces is recovered. You could just use 2d20 ounces/ton. Regards, RJH (talk) 22:08, 27 September 2011 (UTC)[reply]
Make sure you ask your DM for the maximum rainfall in the local climate so you can take precautions against flooding (e.g. dig out lower levels under your main shafts so your miners have time to escape.) Also you should be able to hire some low level henchmen magic users or clerics to help locate the mineral seams and maybe set up some kind of a magic smelting furnace to cut down on operating costs. 69.171.160.139 (talk) 22:59, 27 September 2011 (UTC)[reply]
I'm able to track down information on the past and current crystallization of the Earth's outer core along the boundary with the inner core, but I can't find anything about future projections for inner core expansion and the freezing out of the magnetic dynamo process. If memory serves, there was a story about that in the press a while back (perhaps in a science magazine), but I can't find it anywhere. Does anybody have a reference I could use, or recall where that story appeared? Thank you. Regards, RJH (talk) 22:01, 27 September 2011 (UTC)[reply]
Yes I think that might be it. Thank you. Unfortunately, in skimming through it, that doesn't appear to answer the question I'm trying to resolve. I.e. how long it will take for the core to freeze solid. I guess then that may be just too difficult to model at the moment. Regards, RJH (talk) 22:43, 27 September 2011 (UTC)[reply]
Given the current rate of 0.5 mm/yr and assuming that rate stays constant (which is probably not a good assumption as the proportion of iron drops and the radius increases), I get (3,480 − 1,220 km) × (1,000,000 mm/km) / 0.5 mm/yr. = 4.5 gyr. But either way it's not something I can reliably cite, alas. Regards, RJH (talk) 21:47, 28 September 2011 (UTC)[reply]
It's an energy loss problem, so the scaling factor ought to be change in volume rather than change in linear scale, but yeah I don't see any thing one can cite either. Dragons flight (talk) 22:05, 28 September 2011 (UTC)[reply]
It's an energy flux problem. The exact composition of fissile material in the core is not known, and ignoring such heating factors will thorw off any estimate. μηδείς (talk) 00:27, 29 September 2011 (UTC)[reply]
I took the wrapping off several sets of batteries and now I can't tell which ones are rechargeable (and some of them are) and which ones are not. Not one of them says "rechargeable". I think the Energizer Lithium AA 03 2023 batteries are the rechargeable ones, but I can't seem to find anything on Google that says they are or they aren't, just lots of specs about their uses in cameras. Anybody out there know? Thanks Bielle (talk) 03:04, 28 September 2011 (UTC)[reply]
Having kept looking at batteries in general, I have concluded that if they don't say "rechargeable" writ large, then they are not. Thus, the Energizer Lithiums are not rechargeable. Comments appreciated. Bielle (talk) 03:15, 28 September 2011 (UTC)[reply]
The chemistry in rechargable batteries is very different than the chemistry of other batteries. Attempting to recharge a non-rechargable battery involves unfun events like "leakage" and "overheating" and "explosion". Don't do it. Every rechargable battery is always clearly and unambiguously labeled as such, so if it doesn't say "rechargable" in a giant freindly font on the side of the battery, don't stick it in a recharger. --Jayron3203:37, 28 September 2011 (UTC)[reply]
Lithium is a violently excitable substance if it is abused. A lithium fire would probably ruin your day. It cannot be put out by common extinguising substances such as water or carbon dioxide. Roger (talk) 09:26, 28 September 2011 (UTC)[reply]
In other words, say hello to the new battery-sized hole in your table. And to the huge blackened areas in your floor and ceiling. And I bet that any smoke entering your lungs will be completely healthy and won't leave you coughing for weeks. --Enric Naval (talk) 10:31, 28 September 2011 (UTC)[reply]
As a further note, the only common rechargable AA batteries are NiMH, NiCd and NiZn. Of these, NiMH are what you're most likely to encounter, NiCd are rare nowadays for various reasons including them containing cadmium, their general low capacity, the memory effect etc. NiZn are still relatively new and should be used with care since their nominal voltage is fairly high, may be too high for some devices. NiMH and NiCd batteries can usually be charged in the same chargers but NiZn need their own chargers. You can get rechargable Lithium-ion batteries (there are actually several kinds of rechargable lithium ion batteries) in AA size equivalent, more commonly called 14500 but these are not intended to be used with most devices accepting AA batteries because of the much higher voltage. These need special chargers and aren't really intended for the consumer market (although are popular in some circles) and instead are generally used for packs and in devices where they aren't intended to be removed, and you will not find Duracell ones. Because of the risks, I wouldn't recommend them if you don't know what you're doing. As has been noted, lithium batteries should be treated with care, even primary ones. Nil Einne (talk) 12:41, 28 September 2011 (UTC)[reply]
Normally if batteries are not rechargeable they will say "do not recharge" on them in small print somewhere. I have plenty of batteries which are rechargeable but don't explicitly say "you can recharge this battery" on them 82.43.90.142 (talk) 09:54, 28 September 2011 (UTC)[reply]
Thanks for all your help. I did some checking in with electronics stores today and they all, without exception, agree with Jayron: if it is rechargeable, it will say so; if it does not say so, do not try recharging for all the reasons shown above. I am concerned that 82.43.90.142's response is backwards, and may be dangerous, though perhaps this varies from country to country. Bielle (talk) 02:37, 29 September 2011 (UTC)[reply]
solar wind movement and direction
solar wind is the current of ejected particles from sun corona , those particles have 250~750 km/s velocity , which is further than sun gravity field escape velocity (180km/s).In addition the particles obey electromagnetic field rules which make them to have spiral rotation , suppose any particle (for example proton)how will it move at its way in interstellar space?akbarmohammadzadeIRAN--78.38.28.3 (talk) 03:36, 28 September 2011 (UTC)[reply]
I don't think so. They are only related in the sense that plasma can sometimes behave like a type of liquid. The current, widely-accepted stellar model holds that the cores of hydrogen fusing stars are entirely composed of plasma matter. At the millions of Kelvin needed for nuclear fusion of hydrogen, how can it be otherwise? Ergo, the properties of liquid matter don't appear directly relevant. Regards, RJH (talk) 21:55, 28 September 2011 (UTC)[reply]
Remember, there is pressure also to contend with. Plasma is not a state of matter, at least it is not distinct from solids, liquids, or gases. After all, you do get solid plasmas, liquid plasmas, and the more familiar gaseous plasmas. Plasmic Physics (talk) 22:06, 28 September 2011 (UTC)[reply]
When the fighter aircraft pass over our heads they make such terrible sound, but why does the big Jumbojet(or Airbus) don't make so much noise though it has more and far bigger engines, when it is landing or taking off near us at airport etc, though it is quite close. 124.253.129.113 (talk) —Preceding undated comment added 04:15, 28 September 2011 (UTC).[reply]
The planes don't always break sound barrier. Breaking sound barrier is a momentary phenomenon that last a fraction of a second...most of the time they are traveling below speed of sound 124.253.129.113 (talk) —Preceding undated comment added 05:02, 28 September 2011 (UTC).[reply]
Jumbo jets have high-bypass turbofan engines, which derive most of their thrust from pumping cold air through the bypass ducts and out the exhaust nozzle; this is for better efficiency at medium-high speeds (500-600 knots or so), but it also muffles the roar of the hot gases from the combustion section of the engine. Jet fighters, on the other hand, have straight turbojet engines that derive most if not all of their thrust from the hot gases; these produce more power at very high speeds (> Mach 1), but obviously don't have the same muffling effect from the cold air. 67.169.177.176 (talk) 05:32, 28 September 2011 (UTC)[reply]
The above is totally correct. If it's worth mentioning, this is not an accident, noise consideration is a considerable factor in commercial jet engine design. If anything, apart from absolute performance, fighter jets are, in complete contrast, designed to shock and awe. Vespine (talk) 06:15, 28 September 2011 (UTC)[reply]
Not as a general statement; it depends on what kind of stealthing the aircraft is concerned with. For instance, a helicopter operates a low altitudes where being noticed by foot soldiers can be a threat. Thus, stealthing a helicopter reasonably includes sound-deadening measures. Stealth aircraft operating at high altitudes, on the other hand, won't care about sound output to nearly that extent. If they're quiet, it's more likely as the result of a happy design accident (perhaps aforementioned high-bypass turbofans are good for stealth purposes due to heat of exhaust) than intentional design methodology. — Lomn18:50, 28 September 2011 (UTC)[reply]
I was going to say a simiar thing that noise is not generally a factor on "stealth aircraft" but our article on the state of the art F-22_Raptor#Stealth seems to disagree. It says The aircraft was designed to be less visible to the naked eye; radio, heat and noise emissions are equally controlled. Vespine (talk) 23:31, 28 September 2011 (UTC)[reply]
I'll provide a "yes, but" to that. Reading the source document, there is but a token generic statement ("To make a stealthy aircraft, designers had to consider... muffling noise...."). That's it. The rest of that article discusses relevant stealth features in detail -- radar signature, visual signature, emcon, and heat signature. There is zero mention of noise reduction as an actual meaningful objective of F-22 design. I note also this image, from this F-35 acoustics test report, notes that the F-22 is the noisiest fighter in the modern US arsenal at both minimum and military power. The F-35, itself a stealth design, is also relatively loud. Finally, note that the acoustics study is for purposes of noise exposure to maintenance personnel and the general public around military airfields. It's not a combat consideration at all. — Lomn03:18, 29 September 2011 (UTC)[reply]
Although the noisiest aircraft in the world seems to be a propeller plane, the Tupolev Tu-95:
Its blades, which rotate faster than the speed of sound, according to one media source, make it arguably the noisiest military aircraft on earth,[2] with only the experimental 1950s era Republic Thunderscreech turboprop powered American fighter design as a likely rival.
See Republic XF-84H: "Unlike standard propellers that turn at subsonic speeds, the outer 24–30 inches of the blades on the XF-84H's propeller traveled faster than the speed of sound even at idle thrust, producing a continuous visible sonic boom that radiated laterally from the propellers for hundreds of yards. The shock wave was actually powerful enough to knock a man down."Alansplodge (talk) 00:58, 29 September 2011 (UTC)[reply]
Galileo was first to demonstrate that all objects fall at the same rate in the absence of an atmosphere. As it is said that the earth and the apple fall toward each other but apple looks a lot to falls to the earth as compared to the falling of earth toward the apple which is so tiny to be detected.
Let's imagine earth is a homogeneous sphere therefore in the following cases, what would be the direction of direction [falling] of earth in the absence of all other gravitational attraction including atmosphere?
1- If two apples start falling simultaneously from ANTIPODEAN trees from same altitude.
2- If two different masses [say one apple and other big asteroid] start falling simultaneously from same ANTIPODEAN altitude.
I think there is some very fundamental goof here. He says apple and earth fall towards each other. The apple is OK, it falls i.e. it moves in a straight line towards earth, but does earth move towards apple ? I think no, as it is already "falling" - it is moving around the sun - that is technically falling, no ?. Now, how can something move towards two different directions at same time ? 124.253.137.182 (talk) —Preceding undated comment added 05:49, 28 September 2011 (UTC).[reply]
Different vectors. Any object in space is moving along a path that's the "sum" or "net effect" of its vectors (there may be a different word for that, but I can't think of it just now). Earth and moon orbit around a common point. That point orbits around the sun. The sun presumably orbits around the center of the Milky Way Galaxy. And the Milky Way is presumably heading along an approximately straight line away from wherever the Big Bang occurred. So the earth, viewed in isolation and relative to all these different forces, would be taking a very interesting path. As noted in the earlier section, the falling apple has a theoretical effect, but so small as to be negligible. ←Baseball BugsWhat's up, Doc?carrots→ 05:52, 28 September 2011 (UTC)[reply]
Fair answer except for the moment where you talk about a place where Big Bang happened since Big Bang did not happen in a place. Big Bang is an expansion of the whole universe, not an event within the universe with specific time and space coordinates. Dauto (talk) 14:03, 28 September 2011 (UTC)[reply]
Again you are essentially setting up a Three body problem. What don't you understand? It's fairly simple. In scenario 1, given "perfect conditions, the earth will stand perfectly still, pulled equally by the two apples. For 2, assuming the asteroid is more massive then the apple, then it will have more of a gravitational effect on the earth then the apple, however unless the mass of the asteroid is a meaningful fraction of the mass of the earth, the effect will still be insignificant. I don't actually understand what you mean by Also, would gravity "g" of two equal planets cancel each other if placed on each other? No, if your adding their masses, their gravity would also add up, not cancel out. Note, the gravity would NOT double with a doubling of the mass, it's one of those inverse square law situations. Vespine (talk) 06:09, 28 September 2011 (UTC)[reply]
The force on an object due to gravity does double when the mass is doubled. This means that the acceleration due to gravity remains the same. Depends exactly what the OP meant, though, which isn't clear. Grandiose(me, talk, contribs) 09:42, 28 September 2011 (UTC)[reply]
With both planet's particles being present, the various forces these generate will be present too, but they can create a static equilibrium with forces opposed to each other. If you lived in the place where the two equal planets are joined together, the net force due to gravity is zero. See for instance Lagrangian point #L1. In a similar situation, gravitational acceleration decreases in a tunnel or well with increasing depth, see Gravity of Earth. Living between the two planets should be interesting, being nearly weightless and being able to easily move around but with a significantly greater rate of fall as you climbed out. The building materials on the surface would need to be progressively stronger too. The L1 position is unstable, so apples will fall towards the planet that is nearest, but one might become adapt at calculating the inertia needed to miss the surface in a fall (see Douglas Adams' novels, :-), for more detail). Once you reached either "end of the world" the gravity due to both masses will not be doubled on account of the one planet being further away. So locally, there is no doubling, but as you leave the system, the difference in the distances to each planet can become so insignificant that the gravitational acceleration due to both can be practically twice that of only one planet. --Modocc (talk) 17:54, 28 September 2011 (UTC)[reply]
faster than light question
Just to clarify the predominant understanding of traveling faster than the speed of light actually means I take it then that if I am traveling on the back of a light beam and a faster than light is coming up behind me that from my perspective it is in my past and when it passes me it is in the future while I remain in the present? --DeeperQA (talk) 07:37, 28 September 2011 (UTC)[reply]
In many ways, FTL travel introduces many real paradoxes which cannot be resolved by a simple explanation in the English language (or indeed any language). --Jayron3213:20, 28 September 2011 (UTC)[reply]
You keep asking the same question. Do you expect to get different answers? Speed is relative - that is: different observers can see objects moving differently. For someone in a train, their luggage is just sitting their on the floor, not moving at all, but that same piece of luggage is observed to be moving by someone at the train station. It turns out that the rules for transforming speed between different observers is more complex than believed by 19th century physicists and an object seen moving faster than light by one observer will be observed moving backwards in time by another observer leading to severe paradoxes. The simplest and most logical way to avoid those paradoxes is to assume that faster than light travel is not possible. Dauto (talk) 13:32, 28 September 2011 (UTC)[reply]
Also, if you are traveling on the back of a light beam, you have no past and no future since your time stops due to an infinite time dilation factor. From the point of view of a photon, the universe is flattened into a 2-dimensional pancake and the photon is created and absorbed in the same spot simultaneously. Dauto (talk) 13:57, 28 September 2011 (UTC)[reply]
I can't see anything there that justifies answering this question with a yes. Nutrition does matter, but to my knowledge there is no evidence that muscular exercise affects height. Looie496 (talk) 16:27, 28 September 2011 (UTC)[reply]
I did explicitly say bone mass. While it does not (or at least only very slightly) affect bone length, surely you can agree that being a bit more healthy can make the difference? -- Obsidi♠nSoul16:53, 28 September 2011 (UTC)[reply]
From my knowledge of developmental biology, increase in bone length is not caused by increase in muscle mass, and the diversion of resources to the muscles and premature bone closure by increased hormone levels might instead lead to stunting. If this is a serious question ask a specialized pediatrician. There are far to many factors and risks to take comments here as above the level of nonsense. μηδείς (talk) 00:24, 29 September 2011 (UTC)[reply]
Ok now you're both putting words into my mouth. Read my original post again. First I never said anything about bone length. I said bone mass - i.e. bone strength, bone health. And I, again, explicitly said with proper nutrition and that it all depends on genetics in the end. And no, muscular mass is tied intimately (literally) to the bones they're attached on. You can not exercise any major muscle without affecting bone as well. And to the contrary, just as exercise has no effect on increasing bone length, neither does it stunt it. Unless we're talking about one of those crazy parents pushing ten-year olds into bodybuilding with steroids or malnourished kids working high-intensity jobs in sweatshops. I mean, jeez, why do you think we had PE classes? To stunt our growths? Nevertheless, I agree, talk to a pediatrician. Overenthusiastic and incorrect exercise can result in injuries that can stunt growth.-- Obsidi♠nSoul03:04, 29 September 2011 (UTC)[reply]
What manner of substance is this?
So, I was watching videos of shredders shredding things and found myself on this one, which shows large chunks of metal igniting and burning and popping and sparking whilst going through the machine. The video just calls it 'speciality metals' and there are a few guesses in the comments section - but does anyone here know for sure what the hell that stuff is? --Kurt Shaped Box (talk) 19:19, 28 September 2011 (UTC)[reply]
Looks a lot like magnesium to me; for liability purposes they may not want to name the metal, which is why they use the term "specialty metals". Magnesium is a fairly reactive metal, and it is flammable, even in pure CO2. (google "magnesium and dry ice" for some fun videos). The bright flashes look a LOT like magnesium when it burns, and grinding it up like that could easily produce enough heat via friction to ignite it. --Jayron3219:28, 28 September 2011 (UTC)[reply]
Mass M is required to be levitated (stationary) in a gravitational field of strength g by pumping out a fluid of density rho over a cross-sectional area A (e.g. a ducted fan blowing out air). Assuming no losses, how much power is required?
Clarification: In case unclear, I mean that the fan is attached to (part of) the mass M and is also hovering (as opposed to blowing upwards from the ground). 86.179.118.99 (talk) 22:16, 28 September 2011 (UTC)[reply]
If it helps, this is called a gravity burn. "Assume no losses"? The entire thing is a loss! No work is done on Mass M - it isn't moving - which means that 100% of the work is done on the exhaust-gas! Nimur (talk) 22:38, 28 September 2011 (UTC)[reply]
I mean no losses to heat/noise/etc (other than after the fluid has escaped and done its business, obviously). For example, a real fan would not convert all of the energy it consumes into moving the air in the required direction, I assume. Thanks for the link to the article, but it doesn't really help to me verify that exact formula, as far as I can tell. 86.179.118.99 (talk) 23:21, 28 September 2011 (UTC)[reply]
I concur with your formula. It is a measure of the minimum rate at which downward kinetic energy must be imparted to the air in order to stay stationary in the limit that air is incompressible. In practice, there will be corrections (potentially large ones) from the fact that air is a compressible fluid and the fact that the amount of energy used to operate any real fan will always be significantly larger than the amount of downward kinetic energy acquired by the air. Dragons flight (talk) 01:38, 29 September 2011 (UTC)[reply]
Unless I'm making some gross error of intuition, wouldn't it be more? Think of what a propeller does in incompressible water vs what it does in air. Vespine (talk) 04:19, 29 September 2011 (UTC)[reply]
Normally, cantennas DIY projects require that you have a USB-WiFi stick + cable or some other way of connecting the cantenna to your WiFi circuity. However, would it be possible just to do a cantenna without that? Just imagine a normal catenna + some other improvised devise forwarding the signal to your laptop (without messing with the laptop). 22:55, 28 September 2011 (UTC)
I wanted actually an inexpensive solution for connecting a laptop too far away from the router. A directional antenna (cantenna wa my first thought). I'll try to find a DIY repeater. Quest09 (talk) 23:46, 28 September 2011 (UTC)[reply]
I outright reject the usurpation of the term "Cantenna" for some upstart waveguide doo-hickey used with Wi-Fi. The word has a long and honorable history as a descriptor of a RF dummy load embodied by a 50 Ohm resistor in a gallon can of mineral oil, used to adjust an amateur radio transmitter, and able to dissipate a kilowatt. Can the upstart handle a kilowatt? Hunh? Edison (talk) 04:23, 29 September 2011 (UTC)[reply]
September 29
Criteria for mountain pass
Is there some sort of defined criterion or criteria for what does and doesn't constitute a mountain pass? Our article on the topic says that there are precisely 42 passes in the 5300-km-long border between Argentina and Chile; it would seem that some sort is necessary in order to say that most low spots between mountains aren't truly passes. Is there perhaps a minimum amount of topographic prominence for the mountains on each side of the pass? Nyttend (talk) 01:11, 29 September 2011 (UTC)[reply]
There could be within some limited context, but not in general. For a very non-pass-looking-pass, take a look at Deadhorse Pass, File:CupLakeDeadhorsePass.jpg—it's the dip on the right side of the photo. I wonder about the claim of "precisely 42 passes" on the Argentina-Chile border. Seems fishy. Perhaps there is some context, like "named passes" (although I would think there are many more than 42), or some arbitrary cutoff prominence. Or perhaps Chile and Argentina have some official standard they abide by for whatever the term would be in Spanish... Pfly (talk) 01:33, 29 September 2011 (UTC)[reply]
I would think that one criterion -- maybe the only one -- is that at least a few people have actually used it to get from one side to the other. Looie496 (talk) 01:41, 29 September 2011 (UTC)[reply]
"For passes with roads, it is also customary to have a small roadside sign giving the name of the pass and its elevation above mean sea level. An example of this is Argentina and Chile that share the world's third longest international border, 5,300 kilometres (3,300 mi) long, running from north to the south through the Andes mountains, having a total of 42 mountain passes between them"
--emphasis mine. I think the article is saying that there are 42 passes with roads, each of which has a sign with a name and elevation. I have no idea if that's true, but it does seem to imply how the count was arrived at. SemanticMantis (talk) 01:55, 29 September 2011 (UTC)[reply]
mental excercise
What mental exercises ramp up the burning of calories to the max such that a person who is wheel chair bound can burn off as many calories as possible? --DeeperQA (talk) 02:12, 29 September 2011 (UTC)[reply]
CT scans are said to be used to find places in the brain where activity has increased as the result of seeing an appealing human form which have been said to cause excitement. The method I am told relies on the places in the brain where the greater activity is related to increased cerebral metabolism, i.e., the burning of calories. --DeeperQA (talk) 03:41, 29 September 2011 (UTC)[reply]
Perhaps playing a musical instrument ? Dealing with music seems to be a global process in the brain and presumably burns slightly more glucose. The movements involved in playing it probably burns more calories though I suppose. Sean.hoyland - talk04:05, 29 September 2011 (UTC)[reply]
I'm pretty sure that the brain is like a TV, at least in that what's on the screen doesn't really influence how much power it uses.. Overall, your brain consumes about 20% of the oxygen you inhale, I've never heard of anyone "puffing them selves out" on a particularly strenuous mental activity. Vespine (talk) 04:13, 29 September 2011 (UTC)[reply]
I'm not so sure. Try doing something which is likely to use as many different parts of your brain as possible: translate this article Polynesian navigation into medieval French (or whatever obscure language you prefer) while balancing a spinning plate on a stick held in your mouth - none of this will involve a great deal of physical effort, but your brain will probably tire rather quickly. If Vespine has never 'puffed himself out' from mental activity, he has either been exceptionally lucky, or has more mental grunt than the rest of us. AndyTheGrump (talk) 04:27, 29 September 2011 (UTC)[reply]
The following lists the amount of energy used during various activities in kJ/kg/h. (Sorry it's not in calories. We're metric here.)
Sitting quietly 1.7
Writing 1.7
Standing relaxed 2.1
Driving a car 3.8
Vacuuming 11.3
Walking rapidly 14.2
Running 29.3
Swimming (4km/hour) 33
Rowing in a race 67
It's the fourth line that's important - Driving a car. We do that while sitting down. Many wheelchair users drive. We all know we get tired after driving for a long time. Obviously it uses more energy than just sitting there. HiLo48 (talk) 04:35, 29 September 2011 (UTC)[reply]
Let's say there were two planets that were both identical to our earth. The only difference is that they were close enough that their exospheres touched (The shortest distance between the two planets' surfaces would be 1 megameter). Let's assume everything else is the same as far as our earth is, except that this double planet shares a single moon which orbits the two earths' center of mass. Now, obviously they would be tidally locked to each other. But how stable would this configuration be? How fast would they be rotating around their center of gravity? Would humans or any life forms be able to survive on this double planet? Assuming that they could, what would be the physics behind traveling from one planet to the other? In other words, given our current technology, how feasible would it be to design aircraft (or would it be spacecraft?) that would be able to fly from one planet to another?
Can intermediate or stellar mass black holes be detectable gravitational lenses?
Could an ordinary black hole or an intermediate mass black hole (say less than 50,000 stellar masses) produce detectable gravitational lensing? I ask because [6] says "Micro-lensing shows no change in the lensed object's shape, just in brightness." 69.171.160.131 (talk) 05:54, 29 September 2011 (UTC)[reply]
Please do your own homework.
