Wikipedia:Reference desk/Archives/Science/2011 February 20

= February 20 =

gas central heating
how much is a new natural gas central heating system cost for a 1700 square foot home --Tomjohnson357 (talk) 01:36, 20 February 2011 (UTC)


 * That depends wildly on how the house is constructed and isolated and which climate the heating system will be dimensioned for, not to mention the labor costs in whichever economy you're located in. Probably easiest simply to ask some local heating contractors for estimates. –Henning Makholm (talk) 01:47, 20 February 2011 (UTC)

2 story house, cold climate, i just need a approximate cost --Tomjohnson357 (talk) 01:53, 20 February 2011 (UTC)


 * What kind of heat? Hot water? Steam? Air? If it's hot water then it's around $2000 to $4000 for the boiler. Hot air is cheaper, probably around $1000 to $2000. I'm assuming a high efficiency model (about 93% efficient), if you go with the 80% models (if legal in your area) prices are much lower. That's for parts, installation varies a lot but is probably in the $1000 to $3000 range. Ariel. (talk) 02:55, 20 February 2011 (UTC)

it is natural gas --Tomjohnson357 (talk) 03:42, 20 February 2011 (UTC)


 * I know; you said that. But how is the heat transmitted through the house? Hot water (i.e. radiators) or Air (i.e. air ducts)? Or something else? Steam? Hydronic AKA underfloor? Do you have lots of zones? A few? Just one? Anyway, regardless of the answers to my questions, I hope my estimate was useful. If you really want, then take a high quality picture of your current heater and I'll see if I can figure it out. Take one "overview" shot, and then closer shots of any parts or pipes that look interesting. Ariel. (talk) 09:39, 20 February 2011 (UTC)
 * I had to pay $4300 for mine in November for a 95% model. Googlemeister (talk) 19:30, 22 February 2011 (UTC)

how old was yours

Lab rats
Would it be reasonable to say that lab rats are bred to be prone to getting cancer? I read once that they were so sensitive to getting cancer, that changing the amount of food they eat can statistically significantly influence their rate of cancer. 75.138.198.62 (talk) 01:51, 20 February 2011 (UTC)


 * Actually, most lab rats used in modeling cancer are actually synthetically given cancer genes. If you're a doctor studying cancer and using rats to model it, you give your rats cancer, specifically the cancer you want to study. Otherwise, generic lab rats are no less likely to get cancer than you'd expect. --Shaggorama (talk) 03:00, 20 February 2011 (UTC)


 * (ec)No, that would not be reasonable to say. However there probably is a breed of rat that is very prone to cancer, but it's specific to that breed, not lab rats in general. Plus it has nothing to do with the amount of food they eat - they just get cancer randomly. Ariel. (talk) 03:01, 20 February 2011 (UTC)


 * I think you're confusing the Oncomouse with lab rats in general. --Mr.98 (talk) 03:25, 20 February 2011 (UTC)


 * No, the OP is actually correct, lab rats do develop tumors pretty frequently without any special provocation. I don't know of any evidence that they are more prone to it than wild-types, though, but I haven't ever looked into it. Looie496 (talk) 07:49, 20 February 2011 (UTC)


 * If they are, it's probably due to being highly inbred. Most lab mice & rats aren't truly "wild-type", but are inbred strains, which are chosen specifically so that experiments are affected as little as possible by genetic variation. The principle of heterosis would say that such strains are, in general, more susceptible to most diseases/conditions/ailments than wild-types. So I'd say that, aside from specific strains like oncomouse, it's not "bred to be prone to getting cancer", as the selective pressure was for genetically sameness, not cancer susceptibility. -- 174.21.250.120 (talk) 18:00, 20 February 2011 (UTC)


 * And note that changing the amount of food people get can also affect their chances of getting certain cancers: . So, it's no surprise that this would happen in mice, too.


 * Also, if you are headed toward the conclusion that "lab rats get cancer from everything, so you can't draw any conclusions if they get cancer from substance X", then that's plain wrong, since they only look at the increase in cancer cases from an exposure to a given substance, versus the number of cases in the control group. StuRat (talk) 20:59, 20 February 2011 (UTC)

Measuring orbital and rotation periods
If there was nothing in the Universe except the Earth and Sun (in particular, no fixed stars to refer to), and the Earth's orbit around the Sun was perfectly circular, then how would we measure the length of one Earth orbit? If there was nothing in the Universe except the Earth, how would we measure the length of one Earth rotation, and how would we locate the Poles (ignoring the fact that it would be too cold and dark for life to exist)? 86.179.112.215 (talk) 03:04, 20 February 2011 (UTC)


 * People have argued that it might be impossible (Mach's principle), but I'll ignore that; you can imagine you're simply inside a dust cloud that blocks your view of the stars. In that case, you can measure the (sidereal) day length and find the poles with a Foucault pendulum (though it would be difficult to get the necessary precision), and you can calculate the year length from the difference in length between the sidereal and solar day. -- BenRG (talk) 04:16, 20 February 2011 (UTC)


 * Linear motion is relative - you can only detect it by comparing to something else, but circular motion is not. It's absolute, and you can measure it without reference to anything else. (Although you may run into measurement problems if the motion is very slow, it's just a matter of the necessary accuracy, not a problem in principle.) Ariel. (talk) 04:38, 20 February 2011 (UTC)


 * If this is a technologically advanced society, then they could also launch their own spaceships, and use those as a reference. Or, if they can measure the distance to the Sun, the length of a year could be determined by that, I believe, as each orbit has a distinct orbital speed.  (However, without any other sources of info, they may not know about this relationship.) StuRat (talk) 20:51, 20 February 2011 (UTC)

Donating blood, good for health?
Can donating blood be good for your health? The article about that suggest that it can reduce the amount of iron in your blood, which can be good in the case of people with too much of it. It also suggest that it can improve the heart conditions on men. But, why is the latter plausible? And, wouldn't the reduction of iron in all men be beneficial? 212.169.189.114 (talk) 04:09, 20 February 2011 (UTC)


 * Haven't heard about the iron idea, but I reckon that a free mini-medical checkup every three months is good for me. HiLo48 (talk) 06:03, 20 February 2011 (UTC)


 * I remember reading some years ago a study that suggested that long time blood donors replenished their red blood cells faster after a blood loss. Don't ask me to find it, though.Sjö (talk) 08:46, 20 February 2011 (UTC)


 * Iron overload - the treatment is bloodletting. I don't know if the blood would be suitable for transfusion, though. I thnk I've heard something similar to Sjö regarding blood donors recovering blood cells faster, but I've been unable to Google anything up, so I'd be skeptical. Vimescarrot (talk) 11:49, 20 February 2011 (UTC)


 * Bloodletting is, from a medical perspective, equivalent to donating blood. However, it has come into disuse as a treatment along the centuries. 81.47.150.216 (talk) 13:00, 20 February 2011 (UTC)


 * ...except for in cases like iron overload, as explained in the link that Vimescarrot gave in the very comment you are replying to. I've certainly read of cases where people regularly donated blood, and only later found that they suffered from iron overload, implying that the donating of blood had suppressed their symptoms like any other bloodletting. There was no suggestion that the blood had been rejected. 86.161.110.118 (talk) 18:20, 20 February 2011 (UTC)


 * Oh, I should also note that the article on bloodletting you link helpfully includes a few other diseases it is still considered useful for treating (although sometimes only in the absence of other treatment), including 'the fluid overload of heart failure' (?) and possibly high blood pressure. 86.161.110.118 (talk) 18:34, 20 February 2011 (UTC)


 * There's been some speculation that it might help prevent heart attacks in men. If I remember right, the hypothesis is that iron overload is the reason for the difference in heart attacks between men and women since women lose iron on a monthly basis, but no one's been convinced yet and honestly the whole gender difference there may be overblown (q.v. the red dress campaign here in the US and A).  Charity in general tends to make people feel good about themselves, which is probably beneficial for health, but so far the only people that obviously benefit from blood donation are people with specific health issues, most frequently hemochromatosis and polycythemia.  SDY (talk) 18:10, 20 February 2011 (UTC)


 * Does donating blood make you hungry ? (If you lose some portion of your blood sugar, it might.)  If this results in you eating a steak, then you might replace all the lost iron, and then some.


 * Also, I would expect that donating blood could help with any excess nutrient in the blood, where that excess is harmful, and similarly could hurt where a deficiency is present. If both excesses and deficiencies exist, then determining whether it's a net benefit could be tricky.  StuRat (talk) 20:45, 20 February 2011 (UTC)


 * Perhaps this is the case only because the evolutionary design for men assumes that men would be doing hard physical work every day like running for one hour. Count Iblis (talk) 15:05, 21 February 2011 (UTC)


 * What? If you think the traditionally female roles are not hard physical work, either in a hunter-gatherer society or an agrarian society or a herder society or a modern society, I strongly suggest you try them for a week. Make sure you are actually doing all the work properly, not just your idea of the work. In many hunter-gather societies, the men's only real work is hunting, and they don't even nearly do it every day. Have you ever ground grain to flour using a saddle quern? But then, women in hunter-gather societies aren't having periods every month. 86.161.110.118 (talk) 11:32, 22 February 2011 (UTC)


 * I don't think it was true to say that men did harder work, but the nature of the work did vary. Men primarily had to do sprinting-type activities (requiring brief surges of energy), like hunting, while women had to do more marathon-type activities, like carry babies as they migrated thousands of miles.  So, as a result, men tended to have more muscle, which is good for short term power, and women had more fat, which is good for long term energy.  I suspect that additional red blood cells are more important to sprinting-type activities. StuRat (talk) 21:05, 22 February 2011 (UTC)


 * The more cell divisions there are, the more mutations there are. Does that apply here? 66.108.223.179 (talk) 20:33, 22 February 2011 (UTC)

Radiation
What dosage of radiation (only radiation, not any explosion that might come with it) would be required to kill the stereotypically-used-as-an-example healthy adult male immediately or within several seconds? What might be the source of this (I'm interested in a small, very radioactive substance such as that involved in the Demon Core accident, rather than a bomb)? I do not plan to try this at home and I suggest any readers to similarly refrain.~ 72.128.95.0 (talk) 04:15, 20 February 2011 (UTC)
 * page 2 of this NASA document for kids says that 8,000 rems (80sv) will be fatal within an hour and 10,000 rems (100sv) will cause instant death. Radiation_poisoning has some figures in Sieverts. Nanonic (talk) 05:56, 20 February 2011 (UTC)
 * See also Polonium for other figures on what is regarded as the most radioactive element. It states that a tiny tiny amount ingested or inhaled would be fatal. Nanonic (talk) 06:15, 20 February 2011 (UTC)
 * Polonium-210, which I imagine is the one you're talking about, is pretty damn radioactive, but very far from "the most radioactive". Grosso modo, the shorter the half-life, the more radioactive the isotope (although there are other variables such as the decay type and energy; for example tritium is not nearly as dangerous as its short 12-year half-life might suggest).  For Po-210, the half-life is 138 days.  But for, say, francium, the longest-lived isotope has a half-life of 22 minutes.  You could kill someone with much less francium, assuming you could somehow accumulate it and get it into the victim's system before it all decayed. --Trovatore (talk) 09:22, 20 February 2011 (UTC)
 * Indeed. Polonium has a reputation for dangerousness because it's just lasting enough to actually be useful (e.g. as a poison, or, more commonly, as a component to neutron initiators). --Mr.98 (talk) 14:01, 20 February 2011 (UTC)

Calculating Jerk in a BUS...
Hi I have observed that while travelling in a bus, when bus passes over a speed breaker passenger sitting in front side of bus feel less jerk than those sitting on rear side of bus. although I don't know the exact reason behind this but I assume that it is due to psychology of driver, as when front tyres passes over the speed breaker the driver slow down the vehicle and just after that he speedsup without caring about rear passenger........... If there is any other reason please tell me.. Now what I want to know that ... How can I calculate in such circumstenses that In bus where will be least jerk in the bus. —Preceding unsigned comment added by 220.225.96.217 (talk) 05:17, 20 February 2011 (UTC)


 * Yes, I've noticed the same effect in both buses and cars, and it happens even at constant speed. I think it is a feature of the rear suspension, perhaps combined with a coincidence of the rear wheels receiving an upwards impulse just as the rear of the bus is already rising with the rebound of the front-wheel impulse.  The effect varies with speed, and the mathematics is not simple because " jerk" is the third derivative of distance with respect to time.  I'm fairly certain that somewhere close to the center of the bus will be optimal for "least jerk", but I'll leave this for someone else to prove.    D b f i r s   08:18, 20 February 2011 (UTC)


 * Setting aside speed and suspension, which obviously affect the bounciness of everything, the rider's position relative to both tires matters. A normal (at least that I see, similar to File:NYC Transit New Flyer 840.jpg:) bus has one set of tires very near the front (often ahead of the whole passenger area) then the seats extend back to a few rows behind the rear tires. When the front tire goes up and down on a bump, it's like moving the whole bus as a lever pivoted at the back tire: the further forward or backward of the rear tire you are, the greater your vertical motion. Nobody is very far behind that pivot and the driver is the furthest forward (so maybe he's minimizing his bounce?). When the rear tire goes up and down up a bump, it's like moving he whole bus a a lever pivoted at the front tire: the further forward or backward of the rear tire you are, the greater your vertical motion. The driver is right near the pivot, so mostly just tilts forward/backward rather than actually bouncing up/down. Moving towards the rear, the vertical motion gets greater. Those behind the rear tire move even more up/down than the tire itself (being beyond it on the lever), which is a type of motion that nobody else experiences. DMacks (talk) 10:32, 20 February 2011 (UTC)


 * Jerk is the rate of change in acceleration. If F = ma is considered, then jerk is also the rate of change in force (with respect to time) per unit of mass. Use spring mechanics and lever rules to develop an approach. Plasmic Physics (talk) 11:57, 20 February 2011 (UTC)


 * Jerk as a technical term is the rate of change of acceleration. The connection to what the OP calls the jerk that a passenger "feels" is less than clear. --Trovatore (talk) 20:15, 20 February 2011 (UTC)


 * Yes, the body is most sensitive to rate of change of acceleration, but the eyes notice the maximum displacement and the velocity of the oscillation.   D b f i r s   22:04, 20 February 2011 (UTC)
 * I'm guessing (this is speculation) that the body is most sensitive to things like maximum displacement of internal organs, or the energy dissipated within the body as the organs bounce around. There is not going to be any simple relationship between those things and the rate of change of acceleration.  --Trovatore (talk) 22:10, 20 February 2011 (UTC)
 * Yes, these effects can be felt, especially if some organs resonate, but the body is most sensitive to the third derivative of displacement, and cannot feel either displacement or velocity. Acceleration is felt, but is not immediately noticed unless its rate of change is large.  "Bouncing around" produces a large rate of change of acceleration and this is the jerk that is most noticeable.    D b f i r s   23:00, 20 February 2011 (UTC)
 * My point stands, though. The human body has no "jerkmeter" per se (in the sense of an instrument that measures the third time derivative of position), and jerk in the sense of the third derivative has no special physical significance, unlike the second derivative.  The third derivative is being used as a proxy for other things; I don't know what specifically, but I made my best guess as to two of them.  The natural-language word jerk should not be confused with the third time derivative; the latter is a technical sense and does not correspond directly to any felt quantity. --Trovatore (talk) 10:00, 21 February 2011 (UTC)
 * I strongly disagree. Rate of change of acceleration corresponds to rate of change of force, and this is exactly what the body is most sensitive to.  To see this, try travelling in a high-speed lift (elevator in your country).  The high quality elevators in tall buildings have smooth acceleration that is barely noticeable, whereas the cheap lifts in smaller buildings produce a very distinct "jerk" on starting and stopping because they change very abruptly from zero acceleration to a large acceleration and back to zero.  The mathematical term  jerk (also called jolt, surge or lurch) for rate of change of acceleration was chosen precisely because it corresponds to the perceived jerk that the body is sensitive to.  An alternative experiment would be to apply a force very gradually to any part of the body.  You will hardly notice the force if it is applied very gradually (assuming that it is not large enough to cause damage), but if the same force is applied suddenly (with a large third derivative) then you will notice it much more.    D b f i r s   14:01, 21 February 2011 (UTC)
 * If you think the body is directly sensitive to rate of change of force, then I think you're just wrong. The reason you have more reaction to force that increases quickly is that you have different equilibrium positions for internal organs depending on how fast you're accelerating.  If the acceleration comes on suddenly, then they have to move to different positions, and will therefore bounce around.  It's that bouncing that you feel, not the rate of change per se. --Trovatore (talk) 17:49, 21 February 2011 (UTC)
 * I think that you are wrong, but I'll clarify. The body has sensors which directly detect pressure, but both body and brain are designed to detect changes in stimulus much more readily than a constant stimulus.  Both the chemical signals and the brain's interpretation tend to gradually reduce in the case of a constant stimulus, but are renewed whenever the stimulus changes.  Thus the body (and brain if you consider it separate from the body) is most sensitive to rate of change of force.  This is equivalent to the third derivative of displacement.  I am particularly attached to this idea (and thus biased) because, nearly fifty years ago, I asked a teacher what the third derivative represented (given that first derivative is velocity and second derivative is acceleration), and was disappointed to get just a strange look and no reply.  You might also be interested to know that fairground rides have safety limits on second derivative (g forces) and also on third derivative (rate of change of g-force) because the latter can be more dangerous.    D b f i r s   23:30, 22 February 2011 (UTC)
 * Another application similar to fare-ride design is the design of curves on railroads and highways. Instead of going straight then suddenly curving along a circular arc (higher centripetal acceleration than going straight) and then going straight again, the road gradually eases into the curve and then gradually straightens out again: even if you wind up turning the steering wheel a lot, you try to do it slowly). The gradual change in curvature is analogous to derivative of centripetal acceleration to stay in lane. And heck, this is wikipedia, so Track transition curve. DMacks (talk) 02:30, 23 February 2011 (UTC)
 * In addition to the comments above, another factor is the weight distribution on the bus, based mainly on the position of the engine. If the engine is in front, then the front will bounce slowly and the rear will move more quickly.  With a rear engine design, this should be reversed.  However, as people are often seated further behind the rear wheels than in front of the front wheels, the rear passengers might still have more of a bounce than those in front, even in this case.  StuRat (talk) 20:32, 20 February 2011 (UTC)


 * Let me expand on my previous comment, based on what I said earlier, a "jerkometer" could be constructed using a spring with a mass attached, and a high speed video recorder, is what is needed. Record the spring as it jerks. Using spring mechanics, calculate the maximum force experienced ar the peak of displacement in the spring. Divide this force by the known mass attached to the spring, and divide the answer by the time ellasped between equilibrium and maximum displacement. Plasmic Physics (talk) 10:53, 21 February 2011 (UTC)


 * While we're on the topic, what is it that kills a person in a fall from great height? The high deceleration or the high (dejerk?)? Plasmic Physics (talk) 10:57, 21 February 2011 (UTC)


 * It's the high deceleration that kills. Jerk is irrelevant here since it is the force that kills, not its rate of change.  I like the jerkometer, though it measures average jerk not instantaneous jerk, but with suitable spring constants it would be an excellent instrument for use at the back of a bus.   D b f i r s   14:10, 21 February 2011 (UTC)
 * If you measure displacement (either maximum or time-average), don't you still only have a forceometer, not a jerkometer? As Dbfirs says, neither rapid acceleration nor large total acceleration over a long time are easily felt as a sudden change of acceleration. Mass on a spring is a pretty good approximation of a person though (organs and stuff suspended in all the goo and stringy stuff). If we don't easily feel constant forces, we wouldn't care about a constant string stretch either. Fighter pilots care about maximum force ("g"s) for safety, but that could still be a smooth acceleration in a tight turn rather than the jerk of an aircraft carrier take-off. The jerk would be the rate of motion of the mass on the spring (per the third-derivative)...if it's bouncing up and down, so's your lunch, if it's stretched stably, you more quickly become accustomed to it. DMacks (talk) 17:37, 21 February 2011 (UTC)


 * No, change in force divided by change in time divided by mass equals jerk; it is infact a jerkometer. Plasmic Physics (talk) 08:58, 22 February 2011 (UTC)
 * "maximum force experienced ar the peak of displacement in the spring [...] divide the answer by the time ellasped between equilibrium and maximum displacement". Shouldn't the starting point for each jerk be the previous mass position, not the unaccelerated ("just gravity") equilibrium natural spring length? For example, if going to use a rocket to propel myself, I feel a jerk when the rocket ignites, which would . I do that when I'm already falling off a building (accelerating down at 9.8m/s2), is the jerk when the rocket adds to my acceleration the same as if I were in a rocket-sled on a flat track? But when falling, the mass-on-spring displacement is greater (two downward forces add). Or another way, in a rising elevator that smoothly moves faster up and then rapidly stops, isn't that a jerk at the top but seen as a decrease in spring stretch compared to it being stretched from the the vertical acceleration (and unrelated to its maximum stretch)? Rather, the jerk would be seen as the mass on the perhaps-stably extended string move up again...jerk as an absolute change or force during that particular episode, not maximum in any one direction. DMacks (talk) 17:17, 22 February 2011 (UTC)

[removed some notes to self while composing preceding comment, redundant to what I wrote] DMacks (talk) 02:22, 23 February 2011 (UTC)


 * New suggestion: use the same jerkometer, but process the data differently - plot the acceleration (calculated from F = ma) against time. Using software, estimate the tangents at each of the plot point. Take the modulus of the steepest tangent as your peak jerk. Plasmic Physics (talk) 23:51, 22 February 2011 (UTC)


 * I like it! DMacks (talk) 02:30, 23 February 2011 (UTC)

Upper bound of the size of the universe
Tom Murphy says: "All we are prepared to say for now is that over 13.7 billion light year scales, the universe looks pretty flat: it doesn’t deviate by more than 2% from being flat. But, the possibility exists that the universe is still curved on much larger scales. It’s just like the fact that the earth looks flat locally, over small scales, but is curved on the whole. The universe could be closed into a sphere, but on a much larger scale than what we can see. A 2% limit translates to a factor of 50 (it takes 50 2%’s to make 100%), so we could say that if the universe is finite, it must be at least 50 times bigger than our 13.7 billion light year horizon."

If we use the same conditions (flat universe to a certainty of 2% and assuming the universe is finite), what is the upper bound on the volume of the universe? Leptictidium (mt) 09:05, 20 February 2011 (UTC)
 * Even allowing your assumption that it's finite, why should there be an upper bound? It would be rather odd if we could say "either the universe has radius at most 10^70 gigaparsecs, or else it's infinite".  Not being an expert I can't rule out that cosmologists can make such a statement, but it seems very unlikely to me. --Trovatore (talk) 09:11, 20 February 2011 (UTC)
 * I was assuming that a maximum uncertainty of 2% also sets a maximum uncertainty for the size of the universe. Leptictidium (mt) 09:25, 20 February 2011 (UTC)
 * It doesn't deviate by more, but it could by less. So it's a minimum bound, not an upper - and the upper should include infinity. BTW I'm not convinced your math of: 2% deviation = 50 times for a circle is correct. I guess it depends on what he means by deviation. Ariel. (talk) 09:31, 20 February 2011 (UTC)
 * Lepictidium, are you aware that according to the simplest GR models, a universe that's flat or negatively curved is infinite, whereas one with positive curvature is finite? That means that "the universe looks pretty flat" means "the curvature is right on the borderline between predicting a finite universe and predicting an infinite one".  As far as I know the question is still within experimental error.
 * I believe there are more sophisticated possibilities that don't observe this strict dichotomy. Certainly it's not a geometric necessity; for example the flat torus has zero curvature but finite size.  Whether there are any GR solutions that look like a flat torus, or whether they are consistent with observations, I have no idea. --Trovatore (talk) 09:46, 20 February 2011 (UTC)
 * Murphy's argument is rather dubious. He converts a measurement of the geometry (curvature) to a limit on the topology (the size) of the Universe which is not possible. It is true that a positively curved Universe is necessarily closed (this refers to the curvature of 3D space) but the topology is, as far as I know, not unique, meaning that it does not necessarily have to be a (hyper-)sphere, as Murphy assumes. Flat and negatively curved spaces can be infinite but do not have to be (the flat torus is indeed a possibility). There are lower limits on the size of a finite Universe but they do not come from measurements of the curvature. One could also nitpick on the use of lookback time as the measure of the size of the horizon but we'll let that pass. --Wrongfilter (talk) 10:05, 20 February 2011 (UTC)
 * I don't think it is true that a positively curved universe is necessarily closed. I don't see why you couldn't take a Euclidean 3D space and assign a constant metric with positive curvature to every point.  There is in general only a very weak connection between local geometry and global topology of a space. Looie496 (talk) 17:42, 20 February 2011 (UTC)
 * This is one of the strongest connections, see e.g. Page 12 of this article. The reason is that the hypersphere S3, the universal covering space of any positively curved space, is compact. Incidentally, "Euclidean" is "flat". --Wrongfilter (talk) 17:56, 20 February 2011 (UTC)
 * There are infinite spaces that are positively curved everywhere, such as a paraboloid, but there are no infinite spaces with constant positive curvature. The data can't rule out a universe with that shape (they can't rule out much of anything, unless it's so small that a significant fraction of it fits inside the visible universe). -- BenRG (talk) 19:57, 20 February 2011 (UTC)
 * "constant" is of course important. That is implied if the cosmological principle is assumed to hold. --Wrongfilter (talk) 20:02, 20 February 2011 (UTC)
 * Our article on this is Shape of the Universe. See also Doughnut theory of the universe.  Red Act (talk) 15:39, 20 February 2011 (UTC)
 * With the doughnut theory, is space simply cut-and-pasted into a topological doughnut (like Asteroids, where when you go off one edge of the screen you arrive on the other), or is it physically doughnut-shaped, with some parts of space being longer to loop around than others when parallel lines are measured? Is spacetime is looped different ways in different directions, i.e. positive curvature around one "piece" of the doughnut, but negative around the hole?  Does this imply that there's some experiment you can do on a tabletop that will give different results depending on which way the apparatus (or the Earth and the table holding it) is oriented relative to distant stars? Wnt (talk) 19:47, 22 February 2011 (UTC)

Decay constant and half-life
We did an experiment in which there were originally $$N_0$$ fair coins. Every minute, all the coins were tossed, and those that landed heads were removed. Then the half-life of this "decay" is clearly one minute, and so $$\lambda = \frac{ln 2}{t_{1/2}}$$, roughly 0.7 per minute. However, if $$\lambda$$ is the probability that a nucleus decay in unit time, why is this not equal to 0.5? jftsang 12:07, 20 February 2011 (UTC)
 * One reaaaally hand-wavy way of explaining why there's the ln(2) term in there (i.e., the decay rate constant is greater than the half-life) is that a real exponential-decay isn't a process of "sit unchanged for the time of one half-life, then suddenly and intantly half decays", but rather half decays over the time increment. So some of the material that is observed to be gone after one half-life would actually have decayed before that full half-life time was reached. As a result, the rate of decay for all the decaying particles together is faster than just the ones in the "half that decays in the half-life" that manage to last to the end of the half-life before actually decaying. Our exponential decay article has a lot of gory math details proving the formula, if you prefer a "because the formula says so, but why is the formula correct and used at all?" type of answer. DMacks (talk) 14:15, 20 February 2011 (UTC)


 * Imagine that instead of tossing $$N_0$$ coins and than waiting a minute for the next round of tosses you were to toss a coin every (minute/$$N_0$$) for the first minute. Obviously that would lead to the same result - that is $$N_0$$ tosses after one minute. That would be a slightly more realistic representation of an actual decay process where half of the atoms don't simply decay all at once after one halflife. But then you would have to abruptly drop the rate of tossing to one toss every (minute/$$(N_0/2)$$). That abrupt change of pace shows that there was something wrong with what you were doing. You should have started with a slightly faster decay pace at the beginning (Because there were more coins to begin with and the pace of decay is proportional to the number of coins)
 * and than slowly easied that pace so there would be no discontinuity at the pace after the first minute. That means that the original decay rate must have been higher than the naive 0.5 per minute. It turns out to be about 0.7 per minute. Dauto (talk) 15:52, 20 February 2011 (UTC)

I hate to say this, but the previous answers will confuse you. From the half-life article, two equivalent formulas are


 * $$N(t) = N_0 \left(\frac {1}{2}\right)^{t/t_{1/2}}$$
 * $$N(t) = N_0 e^{-\lambda t} \,$$

Note that the first formula uses only 1/2 as a number value and is very easy to work with, but you've used the second formula, which has certain appeal to mathematicians. Now why does your lambda value come to 0.7? Because ln 2 = 0.693147181 ! The 0.7 occurs in your formula solely to cancel out the "e" (the base of natural logarithms) which has been added into it, and convert it back to 1/2. Wnt (talk) 18:06, 21 February 2011 (UTC)


 * I thought my answer was pretty good. Dauto (talk) 19:26, 21 February 2011 (UTC)

Quark Gluon Plasma Experiments - trillions of Celsius degrees ?
I was reading article about Quark Gluon Plasma and I noted that this experiments indicates that the temperatures in this phenomenons reachs around 4 trillion degrees celsius. My question is how can our devices in acelerators support this extremly high temepratures without burn out everything ? How can we keep this heat / energy isolated inside acelerators ? — Preceding unsigned comment added by Futurengineer (talk • contribs) 12:29, 20 February 2011 (UTC)


 * That temperature is achieved in a collision between two ions. It occupies a microscopic volume and last for a very small amount of time, cooling down as it expands. Dauto (talk) 15:27, 20 February 2011 (UTC)


 * I do not think that is a big problem, Quark Gluon Plasma is created when two heavy atomic nucleus collide, the volume of the plasma should be of the same order of magnitude as the volume of a nucleus 10 fm= 10*10^-15 m =0.00 000 000 001 mm. This plasma will expand and cool down very quickly, since all particles have relativistic speeds I would assume that it only exist for something like 10^-21 s = 1 zs. This happens in the centre of a cooled vacuum pipe a few centimetres across. By the time the particles in the plasma (or the produced particles) reach the pipe wall they will be very sparse. If we scale up this so the plasma is the size of a atomic bomb (r= 0.1 m approx.) then the container would have a size of the same order of magnitude as the earth's orbit around the sun. Of course Quark Gluon Plasma is much much hotter than an atomic bomb and there are several million collisions per second. I am no expert on this so other editors can maybe give a more detailed answer. Gr8xoz (talk) 15:44, 20 February 2011 (UTC)


 * An addition to what was said above: at a certain point in physics, you have to separate your "idea" of temperature from the definition of temperature, just as with every other "idea" that gets weirder as physics gets larger/smaller/hotter/colder. Temperature is defined, at the level of first principles (the basic axioms of classical physics), as the amount of heat energy needed to increase entropy (disorderly messy behavior of particles) by a given amount. So at these trillion-degree temperatures, relatively-enormous amounts of energy are needed to change the arrangement of the system just a tiny bit, but the particles in the accelerator don't necessarily "feel" hot (also because they don't really hold their own heat in like objects that we see around us do). For educational fun, if you want to reverse the definition above such that adding heat lowers entropy, you get negative temperature. SamuelRiv (talk) 19:23, 21 February 2011 (UTC)

As others said, you have to distinguish heat and temperature. Here's an experiment. A candle can easily burn you and you will perceive it as very "hot". But if you tried to warm up a swimming pool or a bath by holding a candle under it, your neighbors would laugh - it's obviously futile. Only a tiny amount of matter is "hot" in the candle's case. The heat needed to warm up a bath or swimming pool is far more than that which a candle provides even though a candle is "hotter".

Take it a step further. Imagine you could create a temperature of a billion degrees, but just for one subatomic particle. (The higher the temperature the harder it is for structures to hold together, so by the time we have a billion degrees molecules and even atoms cannot hold together which is why we get a plasma of even smaller particles and why extreme temperature is useful in studying tiny subatomic particles). Would that be enough heat to warm up your swimming pool?

It would obviously destroy any physical structure it touched, but it is so small, that the heat it contains would be easily exhausted in far less than a microsecond of contact with non-heated matter (or due to cooling by radiated heat or other mechanisms), that you'd never really notice it on a macro scale. The few million atoms it first touched would be affected but the pool as a whole would not. In research such as this, the containment system is designed to prevent heat transfer, and as others have said, the basic classical methods that must be present in the first place are removal of other particles (hard vacuum) and preventing contact with the surrounding structure (typically via laser or magnetic containment).

So the answer to the question is more like "how can we protect the few high temperature subatomic particles we produce from being almost instantly quenched" rather than "how do we protect the surrounding equipment from the high temperature particles". The answer is that if you prevent heat transfer then however hot the particles in the plasma get, the heat stays in the plasma. When it's no longer contained it is almost instantly quenched. Because the total heat is modest (though the temperature is very high) the energy can be disposed of when the beam has done its job.

That said the beam energy is not that small. You might find the article on beam dumps interesting. The problem is that the energy from repeated "bunches" of particles is disposed of in a tiny amount of time - in the case of the Large Hadron Collider that's less than 0.0000009 seconds (90 μs).

(Disclaimer, not a physicist, those who are please feel free to elaborate or correct any of this) FT2 (Talk 15:58, 22 February 2011 (UTC)

cell wall in plant cells
I would like to know what is the role of cell walls in lifespan of a plant cell? Does thicker cell wall mean longer life or it is not so? And can plant cells live for 100s of years? If yes then how do they stay alive for such long time and is there any role of cell walls there?? — Preceding unsigned comment added by Ptamhane (talk • contribs) 16:04, 20 February 2011 (UTC)


 * We know that plants can live thousands of years so it stands to logic that plant cells can live thousands of years (Unless you meant live that long without going through cell division in which case I don't know the answer). I don't know whether cell walls play a role at all. Dauto (talk) 17:21, 20 February 2011 (UTC)


 * That is not necessarily true. Many trees shed their leaves every year. The living part of the bark may expand outwards every year, with the inner part dying. So a tree could live a thousand years, but that need not imply that any cell within it lives for that time. In the same way, a Chinese dynasty may exist over a thousand years, but none of its individuals lives a thousand years. 92.28.245.90 (talk) 17:51, 20 February 2011 (UTC)


 * The disagreement here are over the definition of the age of a cell. When a cell divides in to two do you get two new cells or two cells that each are as old as the original cell. If I split a stone in halves the halves are as old as the original stone but if we calculate the age of cells in the same way then we can argue that all cells are 3.7 billion years old and that does not seems to be a useful definition. (It is unclear how such a definition applies to sexual reproduction.) The problem are that after the cell division there are no distinction between child and parent. Gr8xoz (talk) 19:10, 20 February 2011 (UTC)


 * If we say that all cells are the age of their parent, plus that cell's parent, etc., then aren't all cells the same age (of approximately 3-4 billion years), dating from the first cell ? Perhaps you had in mind an exception for the reproductive cells in sexual reproduction, but some plants reproduce asexually, so this doesn't seem like a sensible way to measure the age of cells, in such a case. StuRat (talk) 20:20, 20 February 2011 (UTC)


 * As I said "if we calculate the age of cells in the same way then we can argue that all cells are 3.7 billion years old and that does not seems to be a useful definition.". I just wanted to explain why the previous answers were contradicting. I do think it is more useful to measure from the latest cell division just as you seem to do. I do not how ever feel that this must be obvious for everybody so I think the age (lifespan) of a cell need to be defined before the question can be answered. Gr8xoz (talk) 21:10, 20 February 2011 (UTC)


 * The length of time a cell is viable is programmed into it. Programmed cell death in plant tissue. Look at very old trees and you'll see that they are hollow with just the outer layer still living. The wood itself is dead. --Aspro (talk) 17:36, 20 February 2011 (UTC)


 * Exactly. Only the layer between the bark and the wood is alive. And some of the cells in that layer may have been there for a long time. The question is how long? Also, How long do they live between cell division cycles? I'm not sure what exactly the OP has in mind. Dauto (talk) 18:04, 20 February 2011 (UTC)
 * The OP is simply asking if its the morphological  characteristic (e.g. thickness of cell walls) or something else.  --Aspro (talk) 18:22, 20 February 2011 (UTC)


 * I'm not sure about any direct relationships between cell wall thickness and longevity, but the trees with the longest living leaves tend to have thicker cell walls. This is because they live in dry (including cold) or nutrient-poor habitats where they are only able to photosynthesise very slowly, so to end up with a positive carbon gain over the lifetime of the leaf, they need to be kept for a long time. Thick cell walls help with this because they are harder for microbes to penetrate, deter herbivores as they are relatively inedible and in cold areas withstand ice blasting and allow water to be stored in the apoplast where it can freeze without damaging the inside of the cell. These papers discuss the inherent trade-offs that plants make when 'deciding' how to construct a leaf and sort of shows what I mean. I can't find any references, but I'm fairly sure such cells will be the plant cells with the longest life spans - most of a tree trunk is dead - as has been said, and the same applies to the large parts of the roots. I do have a paper reference showing that some leaves live for up to 12 years so this is probably the upper limit, unless you want to include embryos in seeds. SmartSE (talk) 23:46, 20 February 2011 (UTC)

Gholson, Mississippi
What is the population of Gholson, Mississippi, and what is the population? -- Perseus  8235 17:02, 20 February 2011 (UTC)


 * The United States Census Bureau does not recognize "Gholson" as a census designated place (nor as an incorporated town, city, or other municipality) in Mississippi. You can verify this at the official Census Factfinder website.  There is a Gholson, Texas, nowhere near Mississippi.  If an unofficial place informally called "Gholson" is located in Mississippi, no authoritative statistics exist for its population.  Nimur (talk) 17:33, 20 February 2011 (UTC)
 * Before posting a request for help here, Perseus began an article about this community. I've found some sources and added them to the article, but a population source is not among them.  As an unincorporated community, it has no official boundaries: as a result, it can't possibly have a specific population.  Nyttend (talk) 21:19, 20 February 2011 (UTC)

Why Doesn't the recoil from Handguns hurt?
Although I have never been shot by a bullet, I've read anecdotes on the web that being shot with a handgun bullet while wearing a bulletproof best is similar to being struck by a sledgehammer or a baseball bat. From Newton's Third Law would this not mean that the recoil of the gun would feel equal or more than being hit by a sledgehammer/bat? Having fired handguns, I do not feel like I'm being hit by a sledgehammer each time I pull the trigger. Acceptable (talk) 02:50, 20 February 2011 (UTC)


 * This was asked on the misc desk, some of the regulars here might have some insight. CS Miller (talk) 17:40, 20 February 2011 (UTC)


 * Because the bullet is accelerated over the entire gun length, but decelerated over a much shorter distance (and hence in a much shorter time). 213.49.110.216 (talk) 18:04, 20 February 2011 (UTC)


 * And that's why powerful guns need longer barrels. Dauto (talk) 18:07, 20 February 2011 (UTC)


 * I do not think that is correct, I think the bullet leaves the barrel before any significant momentum is transferred to the body so the velocity that the gun hits you with are dependent on the generated momentum and the mass only. See Gr8xoz (talk) 18:56, 20 February 2011 (UTC)


 * I believe longer barrels are for increased accuracy, since that ensures that it will leave the barrel at a more precise heading towards the target. Some rather high-powered guns, such as mortars, have rather short barrels, but aren't as accurate.  StuRat (talk) 20:35, 20 February 2011 (UTC)


 * Newton's Third Law states that if two objects interacts directly then they will be affected by forces of opposite direction but the same magnitude. This apply to direct interactions with no time delay such as the force between the bulletproof vest and the bullet, it does not apply to indirect interactions such as between the gun and the bulletproof vest.


 * Newton's Second law is more interesting, it states that the same momentum but of opposite direction is needed to accelerate and to stop the bullet. (We will ignore air resistance and recoil from combustion gases.) Momentum can be calculated as mass multiplied by velocity or force multiplied by time. Since the bullet plus the moving part of the bulletproof vest has less mass than the gun it follows that the bulletproof vest will hit the body with a higher velocity than the gun. If we assume that the forces on the bodies in both cases is the same (not necessary to be true but hard to estimate and it illustrates the point.) then it will take the same time to stop the gun and to stop the bullet and the moving part of the vest. Since the vest hit the body at higher velocity it will move longer in that time so it will do more damage to the body. Gr8xoz (talk) 18:44, 20 February 2011 (UTC)


 * In addition to the force of the bullet strike being more compressed in area and time, I can think of two other factors:


 * 1) The bullet may strike in an area less able to withstand high forces than the hand.


 * 2) The victim is more likely to be surprised by the shot, and the lack of psychological preparation may make it seem worse than it would otherwise. StuRat (talk) 20:08, 20 February 2011 (UTC)


 * So you think that the human hand is more able to withstand a bullet strike than other parts of the body, and a gunshot will do less damage if one is psychologically prepared for it? I would not recommend putting either of those postulates to the test.  Sp in ni  ng  Spark  21:01, 20 February 2011 (UTC)


 * I said it would "seem" worse, not "be" worse. Do you not know the difference ?  As for the hand being more able to withstand forces than some other parts of the body, then that certainly is the case for some body parts, like the eyes.  And I'm not talking about a direct bullet strike, but rather one felt through a Kevlar vest. StuRat (talk) 22:28, 20 February 2011 (UTC)


 * The forwards momentum of the bullet hitting the body is the same as the backward momentum of the gun (assuming no loss in speed from air resistance), so the impulse on the body is the same. What the questioner has not taken into account is that this impulse is really quite small, initially moving the body at less than an inch per second.  The effect is noticeable only if it occurs as an impact over a small area, as with a bullet hitting a kevlar vest or a gun held lightly an inch or two from your face.  Normally, a gun is held firmly so that there is no impact with a part of the body, and the recoil momentum is hardly noticeable as it is transferred to the body, though a shotgun recoil can be painful on the shoulder if not held firmly.  If the bullet is caught in a thick pad that absorbs the energy over a sufficient distance and area, then the effect on the body will be similar, though perhaps more noticeable if unexpected.  The film cliche of victims being thrown backwards by a bullet is entirely faked.  It would take a hit from a missile launcher to produce the momentum effect depicted.    D b f i r s   21:41, 20 February 2011 (UTC)


 * I thought of another reason:


 * 3) The hand, being on the end of the arm, is more free to "recoil" than most of the body. Thus, the force is more gradually absorbed by the hand, wrist, elbow, and shoulder, instead of all being absorbed by the hand immediately.  I suspect that if the hand was held firmly in a fixture while the gun was fired, it would hurt more and do more damage.  As for shotguns held against the shoulder, the body likely still recoils somewhat, although not as much as the entire arm. StuRat (talk) 22:34, 20 February 2011 (UTC)


 * I've never shot a handgun, but I always thought that best advice was to hold it firmly, thus transferring momentum to the body rather than allowing the gun to recoil freely and hitting a part of the body. You would notice the recoil if you shot with an outstretched arm with the direction of firing at perpendicular to your arm, but you will normally only feel pain if you allow the recoiling gun to hit you or you allow your recoiling hand to hit a hard fixed object.    D b f i r s   22:53, 20 February 2011 (UTC)


 * I have shot several handguns, including 9mm, 0.45 and a 50 Cal revolver which would better be described as a hand cannon then a hand gun:). My take is, what sturat mentions just above, that your arm actually makes for a very good shock absorber: when you fire, your arm, not having a lot of mass, recoils with the gun over a certain distance. With the 50 cal in particular, you are strongly advised to take your head out of the way before you fire because your hands are almost guaranteed to travel backwards past the point of your head/face, regardless of how firmly you are gripping. When you get hit by a bullet however, your body, having a lot more mass, doesn't travel back very far at all, so the distance over which the energy is transferred is much shorter. Vespine (talk) 23:19, 20 February 2011 (UTC)


 * I would look at it in terms of "kinetic energy transfer" (though the "Force/area" issue and other points mentioned above are also relevant, I agree). So wouldn't the following reasoning be valid?
 * Because the momentum of the handgun must equal the momentum of the bullet, the energy transferred by the handgun to the hand is much less than the energy transferred by the bullet to the target.
 * Consider the M1911 pistol which weighs about 1100 grams and fires .45 caliber slugs which weigh say about 11 grams:
 * $$m_{slug}=11grams$$
 * $$m_{gun}=1100grams=100m_{slug}$$
 * and since
 * $$m_{gun}\mathbf{v}_{gun}=m_{slug}\mathbf{v}_{slug}$$
 * we also have
 * $$\mathbf{v}_{gun}=\frac{m_{slug}}{m_{gun}}\mathbf{v}_{slug}=\frac{1}{100}\mathbf{v}_{slug}$$
 * The kinetic energy of the slug is
 * $$E_{slug}=\tfrac{1}{2} m_{slug}v_{slug}^2 $$
 * and that of the gun is then
 * $$E_{gun}=\tfrac{1}{2} m_{gun}v_{gun}^2=\tfrac{1}{2}100m_{slug}(\frac{1}{100}\mathbf{v}_{slug})^2=\frac{1}{100}(\tfrac{1}{2} m_{slug}v_{slug}^2)=\frac{1}{100}E_{slug}$$


 * So, in this example, the kinetic energy of the handgun is only one-hundredth that of the slug. The kinetic energy absorbed by your hand is therefore far less than the kinetic energy absorbed by, say, a human body-part struck by the slug, which is why there is less tissue damage to your hand than there is to your target. WikiDao    &#9775;  00:35, 21 February 2011 (UTC)


 * I concur with WikiDao completely. The transfer of energy to the shooter yields the perceived recoil and the shooter must present adequate force (F=ma) to counter the effects...a product of body mass and resistive force (muscular). But who says they don't hurt?


 * Nice slow-motion video of the transfer of momentum energy to the human body.


 * Greater ballistic energy transferred to smaller masses yields increased recoil...or another way of saying it, the bigger the load, and the smaller the body, the bigger the kick. ⋙–Berean–Hunter—►  ((⊕)) 01:49, 21 February 2011 (UTC)


 * Since the question was about the impact of the bulletproof vest on the body and not about the impact of the bullet the mass of the moving part of the vest should be added to the mass of the bullet in the calculation. So if $$m_{slug} + m_{vest}=55grams$$ then the vest would impact the body with 20 times the energy of the guns recoil. Gr8xoz (talk) 04:04, 21 February 2011 (UTC)


 * (... but much less than the energy of the bullet.) The vest is designed to absorb and spread energy, so probably much less than you calculate because energy is lost in the collision of the bullet with the vest.    D b f i r s   07:53, 21 February 2011 (UTC)


 * My calculation is correct, I assume a fully inelastic collision between the vest and the bullet, this means that the vest absorbs as much energy as possible. After a inelastic collision with a stationary object the bullet and the object will have the same momentum as the bullet had before and this can be used to calculate the velocity and kinetic energy. The problem is that the west does not move as a solid object hit in the centre of gravity so "equivalent" mass of the part that move has to be calculated. To do this in detail you need a detailed simulation of the deformation of the west. I assumed 44 grams for a soft vest, a ceramic plate has of curse higher mass. I assume that the collision between the vest and the bullet is much faster than the collision between the vest and the body. Even if the vest is "designed to absorb and spread energy" it must follow the laws of physics.--Gr8xoz (talk) 12:22, 21 February 2011 (UTC)


 * Yes, I initially mis-read your reply, and I agree with your calculation if the bullet hits the vest and the vest then hits the body, but the interaction is, as you say, more complex, with the vest being, to some extent, molded to become part of the body. Energy transfer is diffused, but transfer of momentum is simple and small.    D b f i r s   13:39, 21 February 2011 (UTC)


 * It seems like there is a hidden punch in every bullet. Because when one person punches another, both hand and ribs suffer the same force of impact; yet the hand usually suffers less.  And of course, "punching" with the barrel of a pistol makes the inequality greater.  Wnt (talk) 17:54, 21 February 2011 (UTC)


 * Something else that helps to visualize this is an ice pick. If you hit someone with the business end, it will do far more damage to them than to you, both because of the decreased area (and thus increased pressure) at that end, and because the force is all applied to them at once, versus over maybe a half second to you, if you gave it a full swing. StuRat (talk) 19:44, 21 February 2011 (UTC)

Basic Phsyics
If I am going to shoot a bullet from 5m directly at the ground. At the same time as I shoot the bullet, I will drop a bullet from 5m towards the ground. Assuming same weight bullets and the rifle won't slow down the bullet, will both bullets hit the ground at the same time? —Preceding unsigned comment added by 12.180.137.195 (talk) 19:13, 20 February 2011 (UTC)


 * Yes they should. Both the forward moving and horizontal moving bullet will arrive at the ground at the same time. *Ignoring air resistance* See http://media.pearsoncmg.com/aw/aw_0media_physics/hewittvideos/projectile_demo.html --Tyw7 (☎ Contact me! • Contributions)   Changing the world one edit at a time! 20:08, 20 February 2011 (UTC)


 * You've misread the question. It said the bullet is fired "directly at the ground", not parallel to it. StuRat (talk) 20:12, 20 February 2011 (UTC)


 * No, the bullet fired from the rifle will hit first, as it will leave the rifle going faster than the dropped bullet. The final velocity depends on both the acceleration due to gravity and the initial velocity. Note that I am assuming that "5m" means 5 meters.  If you meant 5 miles, then both bullets would be going at terminal velocity by the time they hit (the dropped one having accelerated to that point and the fired one having decelerated).  However, even in this case, the fired bullet would have moved faster initially, so would still hit first. StuRat (talk) 19:58, 20 February 2011 (UTC)


 * I think the OP has misunderstood a common thought experiment where the bullet is fired horizontally at the same moment as a other bullet is dropped. The air resistance can complicate things a little bit but essentially it is expected that the bullets hits the ground simultaneous as long as the ground can be approximated as flat. See Gr8xoz (talk) 20:13, 20 February 2011 (UTC)


 * See the article External ballistics for more background. ⋙–Berean–Hunter—►  (<b style="color:#00C">(⊕)</b>) 01:53, 21 February 2011 (UTC)


 * If the bullet was dropped compared to when it was shot from a gun, they would arrive at the same time. However, if it was shot downwards, then the bullet that was shot downards would arrive first. --<i style="font-family:Kristen ITC; color:green;">Tyw7</i>  (☎ Contact me! • Contributions)   Changing the world one edit at a time! 05:46, 22 February 2011 (UTC)


 * Assuming you meant shot horizontally, then that is correct. However, if shot even slightly upward it would arrive later, and if even slightly downward it would arrive earlier. StuRat (talk) 08:02, 22 February 2011 (UTC)

Technically if the ground was absolutely smooth and the air absolutely still, and one bullet dropped and the other fired horizontally parallel to the ground, you'd still have to allow for the curvature of the earth. Microscopic effect though. FT2 (Talk 14:39, 22 February 2011 (UTC)

See. Air resistance has an effect, but the effect is surprisingly small. Wnt (talk) 19:37, 22 February 2011 (UTC)

Light waves/photons
Are light waves composed of photons moving in a wave light pattern? --<i style="font-family:Kristen ITC; color:green;">Tyw7</i> (☎ Contact me! • Contributions)   Changing the world one edit at a time! 19:55, 20 February 2011 (UTC)
 * Unfortunately it is not that easy se Wave–particle duality Gr8xoz (talk) 20:03, 20 February 2011 (UTC)


 * (ec) No, "one photon" is just a certain amount of the energy in a light wave. It doesn't have a location in the wave. It's similar to asking whether your checking account is composed of one-cent/penny/yen coins. -- BenRG (talk) 20:10, 20 February 2011 (UTC)
 * That's not entirely true. For example, if you shoot one photon at a target, there will be exactly one collision. If you shoot half of two separate photons at it, there could be anywhere from zero to two collisions. The difference is more pronounced with fermions, which have a probability of zero of being in the same place, affecting the wave-form. — DanielLC 21:04, 20 February 2011 (UTC)

The easiest way to imagine wave-particle duality is that the wave is chopped into small sections, each with a fixed energy.<BR/> Actually, the wave is a probability wave, and the square of its amplitude is the probability. So, the concept of "photon" only comes in place only when you know where it is for sure, i.e., the probability at that point is 1. Otherwise, the photon is nowhere and everywhere at once, so its better to think of it as a wave. The same applied for matter. Manish Earth Talk • Stalk 13:34, 21 February 2011 (UTC)

Faking cold fusion
On the JANUARY 15th FOCARDI AND ROSSI PRESS CONFERENCE an demonstration was done that supposedly shows cold fusion. The researchers them self present it on. They have got some media attention such as, and. Many images can be found at. They demonstrated 12 kW heat generation by boiling water in about one hour. This is so much energy that it seems to rule out a mistake. This leaves two possibilities:
 * 1) A fraud.
 * 2) An interesting nuclear process such as the fusion of nickel and hydrogen that the researchers suggests.

If this is true it will be truly revolutionary but I think experience show that this type of demonstration is probably some kind of fraud. It is also hard to explain theoretically how it can be possible to get cold fusion and why there are so low radiation levels.

Some "experts" have calculated that it would be hard or impossible to hide enough chemical energy in the device to fake it that way. (They calculated the amount of hydrogen and oxygen needed.) One suggestion on how to fake this that I have read in an article comment is that they did not feed it with water, instead they could have used hydrogen peroxide. Hydrogen peroxide decomposes in to oxygen and water vapour when it come in contact with a catalyst. Would the audience detect the difference between water and hydrogen peroxide by smell, viscosity, colour and so on??? Does anybody has any other suggestions in how they can have faked this? Gr8xoz (talk) 20:59, 20 February 2011 (UTC)


 * I saw mention of nickel powder - if they used aluminium or magnesium powder I doubt anyone could tell the difference and those would have provided plenty of energy to boil water. As a general point of view scientists are rather bad at detecting deliberate fraud. A professional magician would be much better at that. Ariel. (talk) 21:36, 20 February 2011 (UTC)
 * As I understand it it was claimed to be a rather small amount of nickel powder but since the inside of the device was not inspected nobody except the people that build the device knows if there are aluminium or magnesium powder in it and in what amount. Do you mean the heat generated when magnesium reacts with water? Would that really be enough energy? Would not the magnesium compounds be easily detected in the steam, a white powder? Gr8xoz (talk) 22:17, 20 February 2011 (UTC)
 * I was just saying that measuring the volume of hydrogen needed is pointless - there are plenty of other fuels you can hide that are much smaller. Also, what was in the compressed gas cylinder? It was labeled as nitrogen - but that doesn't mean that it's the truth. Plus did anyone check that the amount of steam really included ALL the water? Perhaps they drained some water and only made a small amount of steam? Ariel. (talk) 00:19, 21 February 2011 (UTC)
 * The device was connected to a hydrogen bottle during the experiment(suposedly fuel for the fusion proces H+Ni->Cu), it was weighted by supposedly independent persons before and after the test and lost about 1 g H2. The nitrogen was connected afterwards to shut-down or clean the device. I think the calculation assume that since the device have no air inlet or exhaust pipe it would need to contain any oxidizer used and that the exhaust would need to be undetectable in the steam. As I understand it the device is rather small and mounted so that a drainage pipe could not have been hidden. I have not checked all the details. Gr8xoz (talk) 03:37, 21 February 2011 (UTC)
 * I skimmed the 3 videos and I didn't see any of that. You should utterly ignore what they said, and only go by what you can actually verify. Ariel. (talk) 04:07, 21 February 2011 (UTC)
 * Thank for your answers. I can not verify anything, the videos could easily been faked. It could of curse be so that all journalists and "independent" observers there was in on it or that some was easily fouled and the rest was in on it. If we assume the observers were not in on it how cold it be faked? Given that respectable magazines like this (Swedish) write about it I assume they could not just make up the names. So I think the report from the observers should be given some credibility. (Not that I blindly trust it but I am much less suspicious on that than something Rossi him self has written.) --Gr8xoz (talk) 11:55, 21 February 2011 (UTC)
 * Maybe I missed it (I was skimming since I don't speak the language in the video), but weren't they not even in the room? It looks like they did the demo by video? Also, journalists do not attempt to detect fraud, they simply want to report what happened, and the more interesting the better. Please remember: I have NO idea if this is real or not, all I can tell you is that there were opportunities for fraud. But I can't tell if there actually was fraud. Ariel. (talk) 21:09, 21 February 2011 (UTC)
 * Since the observers report that they used there own power meters, thermometers, scale and radiation detectors I would assume that the demonstration was not just a video recording. Obviously some videos show a presentation and some show the actual demonstration. --Gr8xoz (talk) 22:42, 21 February 2011 (UTC)
 * I know that James Randy says that magicians are better than scientists in detecting fraud but I think that in cases like this you need both understandings of illusions and science to be qualified to detect any fraud. Gr8xoz (talk) 22:24, 20 February 2011 (UTC)
 * It's not so much that scientists can't detect fraud as that they don't evaluate it, preferring a different standard. Randi and the scientists would both agree that in order to start looking for fraud you'd have to be able to get there, look over the equipment, see where the hydrogen comes in from, test the amperage in the electric cables and so on.  But scientists have the more exacting standard that someone else has to be able to simply read the report about the experiments, construct his own apparatus, and create cold fusion on his own.  This clearly makes fraud difficult to perpetuate - though it is true, often "irreproducible results" are discounted by scientists without anyone ever knowing whether they were the result of fraud or simple contamination or error. Wnt (talk) 17:40, 21 February 2011 (UTC)
 * Since they don't want to release some details due to issues regarding the patent application it can not be independently reproduced. Supposedly independent researchers where there and did "see where the hydrogen comes in from, test the amperage in the electric cables and so on." They did so using their own instruments but they were not allowed to look inside the devise. There are three explanations: 1. The observers was in on it in this case the fraud is very simple and uninteresting. 2. They were fouled, I think it is interesting to think about how this could be done. 3. The device do actually work as advertised. --Gr8xoz (talk) 22:53, 21 February 2011 (UTC)


 * If they don't give independent labs the ability to reproduce the experiment, then it's very likely fraud. There are lots of ways they could enter into contracts with said labs that would render no threat to their patent applications, the same types of non-disclosure agreements that the FDA enters into with drug companies who have patent-pending medicines. They are either entirely non-confident in their method, or they are trying to pull a fast one (get a patent, use it to bark up funding, then get out of dodge before the number of non-confirmations becomes impossible to ignore). --Mr.98 (talk) 14:05, 22 February 2011 (UTC)


 * I can understand that someone with the Answer To The Energy Crisis could be paranoid about someone else taking their invention, but unfortunately, it's not science until it's reproduced. (I say someone could take their invention because, for example, there might be some component to their device that isn't optimized or has an error in it, and some company could jump in and patent the right way to do it, then hold the whole invention hostage, refusing to license it unless they get a large chunk of the profits - indeed, effectively mandating a high profit margin even if the inventors had been philanthropically minded toward a larger market at a lower price.  Even if a NDA has been signed, no one can prove that the sudden rush of activity in improving the component was actually related to an illicit divulgence of the details) — Preceding unsigned comment added by Wnt (talk • contribs) 14:32, February 22, 2011

Collapsing pyramids?
In referring to ground stone used somewhat as mortar, Egyptian pyramid construction techniques says the following:"The filling has almost no binding properties, but it was necessary to stabilize the construction."Why would filling be needed for the construction to be stable? How possibly could a pyramid collapse, since it's not top-heavy and it's made of solid limestone blocks with no holes larger than a burial chamber? Nyttend (talk) 21:23, 20 February 2011 (UTC)


 * If no filler was used between roughly cut stones, it would be impossible to get each to be level, and this would become more exaggerated with each new level. Thus, the size of gaps would increase near the top, and this would allow water infiltration, which, if it froze (does it get below freezing there ?) would tend to drive the stones loose, over time.  There's also probably the "comfort factor", that the didn't want dripping water to ruin the pharaohs' afterlives. StuRat (talk) 22:19, 20 February 2011 (UTC)


 * The article states ...stones forming the core of the pyramids were roughly cut..., which means that any upper course of stone blocks would only have had a few points of loadbearing contact to the lower course. The resulting uneven stress (receive load from top / transfer load to bottom) in any given block would have caused it to split and crumble, resulting in the gradual collapse of the entire face of a pyramid (which you can clearly see on some of them).  A filler (binding or non-binding) has the simple purpose of distributing this load equally.
 * Compare this with you sleeping on a hard surface. Your entire weight will be resting on a few "protruding" bits, the head, shoulder blades, pelvis, etc.  If you were to sleep on a beach - on a mattress of sand - a bit of wriggling about would evenly distribute your weight to the  delight of Morpheus.  --Cookatoo.ergo.ZooM (talk) 23:06, 20 February 2011 (UTC)


 * "Collapsing" here doesn't mean that they unfold and pour out the edges; it is more like mine subsidence (wow, that's a red link?). Even a multi-ton block of stone can break if you lay it on an uneven surface with half a pyramid on top of it, and the same applies to the one on top of it and so forth. Wnt (talk) 17:22, 21 February 2011 (UTC)
 * We have Subsidence, I'm not sure how to create a redirect yet, if i learn in the next 30 minutes i might fix it. Vespine (talk) 23:01, 21 February 2011 (UTC)
 * Well, THAT was easy!:)Vespine (talk) 23:04, 21 February 2011 (UTC)
 * Actually, is that actually the OPPOSITE of what mine subsidence means? Vespine (talk) 23:04, 21 February 2011 (UTC)

General Relativity and Conservation of Energy
I've heard energy isn't conserved under General Relativity. For example, cosmic background radiation is being redshifted due to the expansion of space with the energy going nowhere. Does the stress–energy tensor act as pretty much the same thing, or would it be possible to take advantage of this? If you can take advantage of it, how hard would it be to build a perpetual motion machine? Would it require a solar system? A black hole? — DanielLC 21:35, 20 February 2011 (UTC)
 * Redshift is not due to a loss of energy, it is due to a change of reference frame. Imagine a car of mass m driving past you at speed v; you say, the car's kinetic energy is $$1/2 m v^2$$. Now imagine you're sitting in the car. Now the kinetic energy is 0, because the speed v=0. Where did the energy go? Nowhere, you're just measuring it in a different reference frame. The fact that energy is not invariant to changes of reference frames is on the same level as time dilation and length contraction in special relativity. Conservation of energy always holds for local interactions like collisions or particle decay, if consistently described in the same reference frame. --Wrongfilter (talk) 22:01, 20 February 2011 (UTC)
 * I thought it's also partially due to the expansion of space. If a photon has a wavelength of 500nm, and space doubles in size, it will have a wavelength of 1000nm. — DanielLC 23:07, 20 February 2011 (UTC)
 * Change of reference frame in an expanding universe. Better? I should stress that I did not mean to imply that there is a one-to-one correspondence between the car example and cosmological redshift. I was just targetting the notion of energy not being conserved rather (that's what the example can do) than trying to explain what redshift is (it can't do that). --Wrongfilter (talk) 23:10, 20 February 2011 (UTC)
 * The problem with the expansion of space, is that it's the same everywhere. If you want to harness the energy, you have to have a difference in the expansion thus a harnessable "potential difference". Anyways, a perpetual motion machine built on this would be driven by an external force (cosmic expansion), and thus it's no longer a perpetual motion machine, rather a power cell rather like a solar cell. Manish Earth Talk •  Stalk 13:26, 21 February 2011 (UTC)
 *   I don't think redshifted light is losing energy. Remember, the light is redshifted because the universe expands.  You may have light with 1/5 the frequency and 1/5 the energy as it had before, but meanwhile the universe has grown 5 times bigger and the same amount of light now covers a trail 5 times longer than before.  I think...   there are some aspects that confuse me.  (For example, if a single photon is emitted from a source moving away from you at great velocity, do you still receive a single photon? does the redshifted photon split somehow? is the existence of the single photon at the far end sort of blurred into a quantum haze for you, like a cat in the box?) Wnt (talk) 17:49, 21 February 2011 (UTC)
 * 1st: Yes, the volume increases and the energy density decreases but those two effects don't cancel each other completely. Defining $$a\,$$ as the expansion factor, the volume scales as $$V \sim a^3$$ and the photonic energy density scales as $$ \rho_E \sim a^{-4}$$, so the energy scales as $$ E = \rho_E V \sim a^{-1}$$. 2nd: No, The photons don't split. What the OP is neglecting here is that since there is a change in volume and there is pressure, work is being performed and it is not surprising that there is a drop in energy. Dauto (talk) 18:12, 21 February 2011 (UTC)
 * Hmmm, so you're saying that the photons, indeed, lose energy according to the scaling factor, i.e. as their wavelength increases their energy decreases. And this energy somehow goes into expanding spacetime?  Does this mean light's energy applies pressure and works like dark energy?  I haven't seen this point often made, and it does hint at some more intimate relation between electromagnetic energy and spacetime than I realized... Wnt (talk) 02:40, 25 February 2011 (UTC)

Electrolysis of urine/Urine powered cars
I've read that performing electrolysis of urine produces far more hydrogen than water does.

http://www.nydailynews.com/lifestyle/health/2009/07/15/2009-07-15_urine_power_hydrogen_produced_from_urea_could_be_used_to_power_cars_houses.html http://www.wired.com/autopia/2009/07/pee-powered-cars/

These articles (among others) claim that one day we may be able to run our cars on hydrogen produced from the electrolysis of urine alone. Is this possible? I know that with water, you are putting more energy into electrolysis of water, than you are getting out of the hydrogen produced. But what about urine? ScienceApe (talk) 22:58, 20 February 2011 (UTC)
 * The electrolysis decompose urea (NH2)2CO to get hydrogen H2. Urine contains about 1% urea, a human would leave about 25 grams per day. I do not know the mileage but it would probably be worse than for gasoline so each person would be able to travel less than 250 meters per day. It can be a small contribution to the fuel supply but it will be very limited. Biogas from human feces would give a larger contribution. Gr8xoz (talk) 23:42, 20 February 2011 (UTC)


 * Let's make things clear. Hydrogen is a means to deliver energy, not a source of energy. Any method you chose to produce hydrogen is going to consume more energy than you get from the hydrogen at the end of the day. Dauto (talk) 02:11, 21 February 2011 (UTC)


 * Too bad. It would have been nice to start a long trip by buying a couple cases at BevMo rather than tanking up at Chevron. PhGustaf (talk) 02:41, 21 February 2011 (UTC)


 * It's not terribly uncommon for human waste to be turned into energy for use. The typical method is to let bacteria break poop down into biogas, and then using that for heating or cooking:,, .  Running a car on biogas is probably feasible, but I don't get the point of trying to break down urea for energy, a relatively small part of human waste, when we have all sorts of great fuel coming out our other hole. Buddy431 (talk) 03:03, 21 February 2011 (UTC)


 * In either case, I think the problem is that the total quantity of energy is low, relative to what cars need, so it wouldn't pay to set up an infrastructure to extract, store, and deliver such a fuel. It would make more sense to extract the energy at the waste treatment plant and use it directly there. StuRat (talk) 05:25, 21 February 2011 (UTC)


 * Of curse the processing will not be done in the car but biogas is already used in cars and buses on a rather large scale and hydrogen has often been proposed as a car fuel due to its high energy-content, it can also be used to produce liquid fuels. --Gr8xoz (talk) 12:01, 21 February 2011 (UTC)


 * LOL @ "Of curse".... yes it would be a curse to have your car storing manure while it decomposes into gas. I don't agree that bio-gas is used on a large scale now.  There are a few experimental programs here and there, but you can't just drive up to your average gas station and expect to get bio-gas there.  And if the source is as small as human (and even livestock) waste, it would never pay to set up such an infrastructure, at least in most places.  A location that has lots of manure available, like a cattle feed lot, might want to use bio-gas for their tractors and trucks. StuRat (talk) 19:36, 21 February 2011 (UTC)


 * I do not know where you set the limit for "rather large scale", obviously it is not near the scale of gasoline but here in Sweden some cities run their city buses and garbage trucks on biogas. In each of the bigger cities there are a few gas stations with biogas that are open for private cars. I think the upgraded biogas and natural gas are feely mixed since they have nearly identical composition and the difference in buying the one or the other is only a difference in accounting. If you buy biogas someone have to put in the same amount of biogas in to the network, similarly to "green electricity". So the infrastructure are largely there already. There are more than 30000 gasdriven cars. I do think we use only a small fraction of all the usable biological waste to gas production so the potential are much bigger but it would not be nearly enough to provide all the fuel. See Tables of European biogas utilisation and Biogas --Gr8xoz (talk) 22:24, 21 February 2011 (UTC)