Wikipedia:Reference desk archive/Science/2006 June 21

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Bismuth Tellurium Transistors used in water heating/cooling[edit]

I just recently heard about a device using bismuth telluride transistors that has the potential to boil/freeze water in seconds from the power of two flashlight batteries. I read this randomly in an old version of the CRC handbook of Chemistry and Physics. I, personally, really enjoy building things, and I would like to add this to my wanted collection, but I do not know where to begin, let alone have the schematic for such a device. I want to ask how should I build this device, and where can I gather the proper information regarding the potential power outputs and details of this machine?

Sounds like Thermoelectric cooling. DMacks 01:00, 21 June 2006 (UTC)
I don't believe there is sufficient energy in two flashlight batteries to boil or freeze any significant amount of water, even if we could release all of it instantaneously. Perhaps if you could use mains power... but you'd probably blow a fuse. Thermoelectric cooling is neat, but it isn't magic. ~MDD4696 03:39, 21 June 2006 (UTC)

Ha! Something I know a lot about! I entered this into the Junior Science and Humanities Symposium through the US DOD. This is called a Peltier junction, and I believe that page redirects into thermoelectricity (last time I was there). They are really fun! I blew mine out with more than five amperes (oh crap). What I had done was run so much current through, I either melted the bismuth-tellurium (I don't know it's melting point) or I had melted the sodder leads which flowed into the Peltier-Seebeck junction. As a note, "Peltier junctions" are used in cooling applications, and "Seebeck junctions" are used in heating. Although the heat is just "moved" using the thermoelectric effect, usually Peltier junctions are used, because there is always an opulence of heat. Peltier junctions are usually used in microprocessor cooling applications when hackers modify their processors to run at dangerously high speeds. As for the heating and freezing, in my experience with commecrially-made ones, you need at least two-three amperes before water vapor starts to directly freeze out of the air.
Construction/Parts: Ceramic plates are on each side of the sandwitch of bismuth-tellurium to create a smooth flat surface area, a freakin lot of the bismuth-tellurium alloy curved sandwitches are bent around between the ceramic (read the links already above to get this).
Building is tough! Unless you've been doing this kind of thing for decades you won't be able to. You have to have at least 20 P-N-P junctions right next to each other for it to work, and a pretty high density (I think mine was something like ~200/in2. All I remember is that you'll need a fume hood and some great smelting skills and lots of bismuth and tellurium. I tried to
Buying is your best bet. You can buy a great one off of eBay for reeeally cheap. I think mine was between forty and sixty dollars. The junction density is prime and the bismuth-tellurium structures are industrial-made. Google it correctly and there are some awesome mods people do with Peltier junctions. My favorites involve the portable mini-, single-can- beer cooler.

Flag of the United States.svg The Mac Davis] ⌇☢ ญ&#x19 B;. 04:30, 21 June 2006 (UTC)


What kind of plastic do they use in plastic surgery?-Bee(y)Ti 01:17, 21 June 2006 (UTC)

Sometimes when plastics is hurt; you send it to a hospital. If a man hits your car's plastic bumper; you must send it to a hospital to perform immediate surgery. -- Toytoy 01:33, 21 June 2006 (UTC)
Plastic surgery has nothing to do with plastic materials. The word plastic comes from the Greek πλασσειν, meaning "form", "mould", or "shape". Originally only flexible materials were called plastic, but now the word is commonly applied to all synthetic polymers, even rigid ones. Plastic surgery is surgery in which the flesh is molded into a different shape. —Keenan Pepper 01:46, 21 June 2006 (UTC)
Usually squished or pulled into a different shape. — Flag of the United States.svg The Mac Davis] ⌇☢ ญ&#x19 B;.
It would depend on the plastic individual's health insurance. I'm sure they'd do their best to match the original plastic. Peter Grey 06:09, 21 June 2006 (UTC)
Silicone. (--Shantavira 06:44, 21 June 2006 (UTC))
When a family member injured her hand slicing an avocado, we were told by the plastic surgeon that operated on her that the most common reason why people in London require plastic surgery is because of accidents to the hand while cutting fruit and veg. Avocados are particularly troublesome because of the combination of the softness of the fruit and the hardness and slipperiness of the stone. "Avocado cuts" are the number 1 cause for "middle class" families, apparently. Plastic surgery using silicone/plastic on people's chests (or wherever on their bodies) is apparently off-the-scale less common than routine repairs to skin injuries etc. --Dweller 10:13, 21 June 2006 (UTC)

Plastics are used in surgical implants, both temporary and permanent, but this is, as noted above, unrelated to the use of the term "plastic surgery". Plastics may be used in plastic surgery, as in other surgical fields. Other materials, like titanium, may also be used in any form of surgery. StuRat 13:18, 21 June 2006 (UTC)

Color and black body radiation[edit]

What is "color" molecularly? Are there any certain rules that say certian combination of particles create certain colors? How about the black body case? How does black body radiation work in this case molecularly?

  • colors depend on the frequency and color of the futons emitted by the black body--—The preceding comment was added by Beeyti (talkcontribs) . 01:58, 21 June 2006 (UTC)
I'm sorry, but what on earth do futons have to do with it? Colour is determined by the frequency of EM radiation emitted by the source. On the receiver's end, colour can be determined by the cells reacting to the radiation. See trichromatic vision for details. GeeJo (t)(c) • 02:31, 21 June 2006 (UTC)
Yes, there are rules that determine the colors of substances given their chemical compositions, but they are very complex. See crystal field theory which explains why blood is red and grass is green. Black body radiation has nothing to do with color at room temperature, because black bodies at room temperature are, well, black. —Keenan Pepper 02:41, 21 June 2006 (UTC)
However black body radiation does change the frequency and color of emissions from a body. — Flag of the United States.svg The Mac Davis] ⌇☢ ญ&#x19 B;.
At room temperature? —Keenan Pepper 05:02, 21 June 2006 (UTC)
Hehe... No... --10:17, 21 June 2006 (UTC)
Yeah, sorry. I obviously left that part out (unconciously), becuase you had just said that. Add a "at very high temperatures" to that. :) — Flag of the United States.svg The Mac Davis] ⌇☢ ญƛ. 22:17, 21 June 2006 (UTC)
Many of the relevant articles really need some additional explanation for non-scientists, but to sketch it out, you have electrons that are flying around very quickly around the atoms make up matter. These electrons are flying around so quickly that it's often better to describe their location as being within a cloud where any given electron is more or less likely to be at any one time.
The shape and size of these clouds varies according to the molecule(s) involved. Said another way, when you have two or more atoms in a molecular bond, their electron clouds combine and interact in ever more complex ways.
A side-effect of these electron clouds is that they will absorb or repel different wavelengths of light; basically different colors of light. And so the combination of the color of the incoming light, combined with the electrons in these clouds leads to what we perceive as the color of an object.
I've tried to make it understandable and avoid jargon, but certain terms really are needed for a discussion like this. Also, this is really not my field, so hopefully someone better versed in quantum chemistry will chime in.--Tachikoma 16:20, 22 June 2006 (UTC)


I've heard that after we spray a cockroach with a pesticide, the cockroache transfers its DNA that is resistant to that kind of pesticide to its offspring, am I right? If I am, how do they do it, i mean, shouldn't they be dead by then? How do they change their DNA?

The only way that resistance can be propagated in a population of cockroaches is by the following sequence of events:
  1. The presence in a population of cockroaches of various degrees of heritable vulnerability and resistance to pesticide, followed by
  2. the killing off of the susceptible ones by spraying of the pesticide, followed by
  3. successful reproduction of the surviving resistant cockroaches, followed by
  4. now a much higher proportion of resistant cockroaches in the local population.

You have been hearing nonsense from the ignorant. Animals and plants do not consciously select the genes or change the genes they pass to their offspring. As long as you have a population with a variety of genes, and the presence of pressures that reduce the reproductive success of some of the gene types, the genetic make-up of the population will change. This is how evolution works. Why are there so many people out there who think they understand mutation and evolution and spread this kind of nonsense when they wouldn't dream of claiming knowledge of cell biology? alteripse 03:36, 21 June 2006 (UTC)

Thanks a lot!

Diluting sulfuric acid[edit]

How can you work out how hot a solution will become when diluting sulfuric acid? Eg. 98% -> 20%, how much energy will be given off?

Thanks very much 04:32, 21 June 2006 (UTC)

  • Any type of chemical process involves the transfer of energy. Any chemical has a certain amount of energy stored in its molecule. Dissolving sulfuric acid would release the energy. All you need to figure out is how much energy you need to heat the resulting solution 1 degree, then see how many degrees you can heat it with the energy you got. - Mgm|(talk) 07:46, 21 June 2006 (UTC)
  • There is no simple way to calculate this, it is one of the technical points which you learn doing practical chemistry. When I demonstrate the dilution of sulfuric acid to my students, I prepare a solution which is roughly 50%: this easily gets to excess of 80 °C. I then get the students to carefully touch the flask, which demonstrates why conc. sulfuric acid is so dangerous... Physchim62 (talk) 08:21, 21 June 2006 (UTC)
  • are you sure physchim? from a quick look around, for one mole of h2so4 mixed with n moles water: delta(Hs) (J/mol) = - 72,000(1 - exp(-0.2n)). this will give the energy released on dilution, given the specific heat capacity of water (4.2 J/g/K) (for dilute solutions, one would assume the specific heat capacity of the final solution is very similar), you can calculate the rise in temperature. diluting a mole H2SO4 to 20% w/w solution requires 27.2 mol water, which gives delta(Hs)= 71700 J/mol, which is 71.7 kJ of energy. 27.2 mol of water is around 490g, 71700/490 = 146 J/g water, 146 J/g / 4.2 J/g/K = 35 deg rise in temperature. Xcomradex 23:30, 27 June 2006 (UTC)

Train Vs Fly[edit]

Someone told me this riddle once but never left me with the answer. Perhaps someone here can help.

A fly is flying along a railway track at, say, 5 km/h. He is hit by a locomotive travelling at 100 km/h in the opposite direction. The fly quickly decelerates from 5 km/h and, considering he is now squashed to the train, travels at 100 km/h in the opposite direction to what he was initialy flying.

Re: the deceleration- he would have decelerated very (very) quickly from 5 km/h in order to travel at 100 km/h in the opp direction (5,4,3,2,1,0,1,2,3,4...100 all of which would have happened in the blink of an eye).

But for the briefest moment his speed would have been zero (the moment between travelling forward and traveliing along with the train). Now if he is not moving, and he is attached to the front of the train, surely this means the train isn't moving (if only for the briefest of moments)

Of course I know a fly can't stop a train so what is the explanation?

Sorry if I made this more complicated than it should be!

A variation on Zeno's Paradox. Speed (velocity, technically) is distance per unit time. If you're considering, not the briefest moment, but zero elapsed time, then you would be "stopped" only in the sense of considering a zero displacement, no different from if the fly wasn't there at all. Unless you wanted someone to explain motion on the scale of Planck time (that would really be cool!) Peter Grey 06:06, 21 June 2006 (UTC)
This riddle has a corollary. What was the last thing to go through the fly's head?--Shantavira 06:48, 21 June 2006 (UTC)
His ass. Johntex\talk 08:28, 21 June 2006 (UTC)
I don't think it's really a variation of Zeno's paradox; even over a finite length of time the fly will have a velocity of close to zero. The key is that implicit in the questioner's reasoning is that the the fly (and train) are point-like or rigid structures. They're not, of course. One could track the center of mass of the fly and it certainly would slow, come to a stop, then speed up in the other direction, and yes, if you neglect the effects of wind/air, this will occur after the fly has come in contact with the train. However (not to be gruesome), after the head has come in contact with the train, it will begin to be squashed towards the body. At the instant the center of mass of the fly is stationary, the back part of the fly is still moving towards the train, while the train is continuing forward squashing the head of the fly. If the fly were a point-like particle, it would bounce off the front of the change, instantaneously changing velocities. It would never have a velocity of zero; the derivative of its position at that instant would be undefined. Of course, all of this assumes classical mechanics; in reality (well, in quantum mechanics), position, velocity, and time become hazy if you probe them too deeply. — Knowledge Seeker 08:22, 21 June 2006 (UTC)
A fly can't stop a train, but a train can stop a fly - where's the paradox? Physchim62 (talk) 08:24, 21 June 2006 (UTC)
Perhaps harder to explain, but surely the same paradox; what about a cricket ball being hit by a cricket bat. Does the cricket ball (shall we say, does the centre of gravity of the cricket ball) ever have a velocity of zero? Is it only during the brief instant where both bat and ball are compressing at their mutual surface, and compressing at a rate that compensates for the combined velocity (over 100 mph?) Notinasnaid 09:48, 21 June 2006 (UTC)
We're making a mountain out of a molehill here, the answer is pretty simple: when a fly hits a train, there will be a moment when the fly is at rest. Since at that moment, the fly is in contact with the train (ie is being squashed by it), the surface it's in contact with will also be at rest. If the train were a completely solid, undeformable object, that would mean the fly would stop the train (even if only for an instant). What happens in reality is the material the fly impacts (eg the windshield) deforms on impact, and it will be momentarily at rest. Of course, it will spring back almost immediately, and oscillate a few times, which is where the sound you hear on impact comes from: the windshield vibrating.
So, the fly doesn't stop the train, but it does stop the windshield. The same goes for the cricket ball - if we treat both objects as being deformable (which they are), then the contact surface between the two will be momentarily at rest on impact. However, almost immediately, both the ball and bat will be compressed by a certain amount, and since the bat is much more massive than the ball, the bulk of its mass will continue moving forward without stopping. — QuantumEleven 12:09, 21 June 2006 (UTC)
QuantumEleven makes excellent points except for one minor (insignificant?) detail! The bat is not much more massive than the ball. The reason it continues forward is because an external force is being supplied to the bat. The bat and ball are of similar masses such that the momentum is matched (similar to Impedance matching in electric circuits - maximum power is transferred only if the masses are closely matched. Otherwise, great explanations. Nimur 18:30, 23 June 2006 (UTC)
Er, the fly never stops anything, not even necessarily the first layer of atoms on the windshield's surface. The velocity of the fly-as-point-particle corresponds to the average velocity of the fly, which is 0 when the fly's butt and head are moving in opposite directions (i.e., the fly is squishing). At no time is it necessary for the windshield (which is only touching the head, not the center of gravity) to be stationary. You could argue that the head itself must have 0 velocity at some point; then we refine it to "the windshield is only touching the front of the fly's head". Then you say that even that must stop at some point, and eventually we're arguing about individual atoms on the surface of the fly. As those are too small to be "in contact" with anything, the argument that any part of the train must have the same speed as them breaks down. (Also, remember that all velocities are relative; all of this is in a frame where the fly and train were moving in opposite directions before impact and the train's velocity wasn't negligible.) --Tardis 00:40, 24 June 2006 (UTC)
I've seen photos taken with high-speed lens/film that show different kinds of balls deforming. Tennis balls, being so squashy, are naturally a great subject for this type of shot. A cricket ball feels extremely hard, but its leather surface and (frankly rather odd) innards will nonetheless compress. --Dweller 12:14, 21 June 2006 (UTC)
  • None of the tennis ball pix I could find on Google were much good, but here's a nice one of a rubber ball deforming [1]
Also, on the impact, the windshield of the train will undergo some deformity (it will bend back a bit) and then (after some oscillations) come to its normal position. There will be the instant when the train is moving forward but the windshield (with the fly squashed against it) is moving backward. This combination (of WS+fly) can attain zero velocity. It all becomes too messy (to calculate and also physically) because the fly gets squashed -- Wikicheng 13:43, 21 June 2006 (UTC)
Please see my above post in this same section: the windshield need not move backward except relative to the rest of the train (as it must deform to apply force to the fly), and that movement can be entirely insignificant compared to the train's velocity (which can be anything, since we can fling the fly at a train instead of ramming the fly with the train). --Tardis 00:40, 24 June 2006 (UTC)

Mobile phones radiation[edit]

Hello. Is there any article or websaite about the radiation that is generated from different types of mobile phones? If possible, I would like to see a website that contains all brands and models of mobile phones. Thank you! --Alexignatiou 06:23, 21 June 2006

[2].... u can get som information here....--hima 09:19, 21 June 2006 (UTC)
First, you might be interested in our article about mobile phone radiation and health.
Secondly, if this worries you, have you considered buying and using a handsfree kit?
Thirdly, here's such a site, using published data from the manufacturers. I found this in about 10 seconds by using Google to search for the words mobile phone radiation comparison. Try searching first, it's quicker. --Robert Merkel 09:13, 21 June 2006 (UTC)

I did search, however I didn't find what I was looking for. And it's not something that worries me, I don't use mobile phones a lot, just because I always hear stuff... Thanks, anyway --Alexignatiou 13:04, 21 June 2006 (UTC)

What is the safety factor of a handsfree kit? Instead of putting radiation near my brain, I should but it near my balls? That makes no sense. --Kainaw (talk) 15:52, 21 June 2006 (UTC)

Answering seriously, and to say something that has not been said without jumping into the links: Cell phones do not release the kind of radiation that you may at first think they do. They release electromagnetic radiation, but it won't give you cancer. The United States government I recall as having spent some hundreds of millions or a few billion dollars on clearning up this panic in the United States about powerlines. The scare has also existed with microwave ovens. — Flag of the United States.svg The Mac Davis] ⌇☢ ญƛ. 22:23, 21 June 2006 (UTC)

Aircraft Safety[edit]

I have seen destructive videos of an blade coming loose in a turbine fan of the air intake of a jet engine.Instead,Why cant they make a composite fan and then surround all its blades with a ring,so that even if one blade brakes the plane can still land safely?Guess it would be stronger and lastlonger.

A composite fan is difficult to do at present because composites are not strong enough to widthstand the forces in a compressor fan. As for the 'ring', that's already done - all jet engines have reinforcement in the cowling around their compressor and turbine blades, so that if a blade should detach when the engine is spinning (this is called a blade-off), the engine will contain the blade and prevent it from damaging the rest of the aircraft. As a matter of fact, it's a requirement for the engine to be allowed to fly, it must contain the damage should a blade detach.
This used to be a bit of a problem with propeller aircraft - without any cowling, there would be nothing stopping a detached blade from zooming away at high speed and hitting something. Nowadays, this happens very rarely, but on some propeller aircraft (eg the ATR aircraft) you can still see the reinforcement of the fuselage wall near the propeller. — QuantumEleven 09:00, 21 June 2006 (UTC)
I read a story, either in the National Geographic or New Scientist about the design and construction of the A380. It discussed this requirement somewhat. IIRC, it was quite a scary time for them (as it would be for any aircraft design team) when they tested it... Nil Einne 14:36, 21 June 2006 (UTC)
I saw a test when they where making the big airbus thing that they had to build a complte engine and then destrucivlty test this it by blowing a propller of when the engine was at full power, which is slightly expensive as the engine was said to be worth its weight in gold --Colsmeghead 21:54, 21 June 2006 (UTC)

Just a note: Those fans have a completely insane amount of kinetic energy. You'd need QUITE a ring. I remember awhile ago I saw a program about how a turboprop pretty much fucked the whole backside of a plane with shrapnel. I think it may have been a military plane...I don't remember if they died or not... --mboverload@ 02:29, 22 June 2006 (UTC)

Yup, that's why those rings are made of some very strong stuff :) If you think the fans in turbofans are bad, the blades in turboprops are much worse: they spin at about the same speed, but each blade is considerably heavier (there are fewer of them). If a turboprop blade were to detach at full speed, it would likely go straight through the fuselage and maybe even out the other side... hence the need for, to put a name on it, armour :) (note that modern propeller designs mean that there are very few blade detachs, so no need to get worried about flying on turboprops!) — QuantumEleven 06:30, 22 June 2006 (UTC)

I did not mean a stationary ring on the body of the engine cover,but a ring attached to all the ends of the blades(Something like the rim of a cycle wheel with spokes) that rotates along with the blades

Electromagnetic Question[edit]

If an inductor were charged,but its terminals were placed far apart so that it would not short when open.Would the wave become am Electro Magnetic Wave?

Also,If light were to enter a drop of water,we know it would choose the shortest path as soon as it enters the light drop(refraction).But what if the shape of the drop had changed as the light is still in between the drop?GUESS it would again choose the path of shortest distace.BUT observing this change,is it possible to predict the future shape of the drop,or is it that shape is the cause,direction is the effect?

I don't understand your first question, but as for the second... you seem to be referring to Fermat's principle. The thing is, light rays don't have any foresight or predictive power. As a ray of light travels, it bends due to changes in the local refractive index, but it doesn't know about what lies ahead. Mathematically speaking, the extremum in Fermat's principle is a local extremum, not a global extremum. Melchoir 19:23, 21 June 2006 (UTC)
Re your first question, an inductor doesn't really get charged. What happens instead is that you store energy in the magnetic field. You may be thinking of a capacitor. But anyway you seem to be describing a steady-state situation. To have a wave you need to have some kind of oscillation. The question is rather confusing; could you describe in more detail what you mean? Arbitrary username 20:22, 21 June 2006 (UTC)

If you took an inductor and charged it, that would mean it had a net surplus of electrons (and a negative charge) or a net deficit of electrons (and a positive charge). An inductor is a good conductor, so the charge would distribute itself evenly (in the absence of an external field) so that both terminals of the inductor had exactly the same charge, and there would be no voltage difference between them, and if you shorted them, no current would flow. Thus there would be no magnetic field resulting from a current flow in the inductor. It would produce an electric field, meaning that a force would be exerted on a charged particle or object placed anywhere in its vicinity, but there would be no magnetic field. Edison 23:15, 28 June 2006 (UTC)

Better Solar Panels[edit]

Is it possible to make the sunlight induce a charge by virtue of its EMW nature rather than just cause photonic emmision and build better solar panels?

Your question is a bit confusing - perhaps you should read photovoltaic cell.--Bmk 21:38, 21 June 2006 (UTC)
The question does not make any sense. I am unfamiliar with EMW natures. If my mind is just not being sharp, that's ok, but I suggest you stay away from that kind of "scientific" material. — Flag of the United States.svg The Mac Davis] ⌇☢ ญƛ. 22:25, 21 June 2006 (UTC)

The swift answer is no (or at least unlikely). The frequency of the electromagnetic waves in sunlight (where most of the energy comes from) is way beyond the potential (unbound) electron oscillation frequency in most materials. Quite simply, the electrons can't move quickly enough, and if they can, no harnessable current is produced (unlike radio wave EM induction). --Eh-Steve 08:16, 22 June 2006 (UTC)


Is there a method to test the uv blocking effectiveness of sun glasses at home?

I found this at Netwon/ANL's Ask a Scientist and modified it (in [square brackets]) to fit into this situation:
A way of testing the [UV] transparency is to place the [sunglasses] between the UV light source and a piece of high quality white paper. The paper will fluoresce a bright blue if there is no [filter] present. If the bright blue disappears when you put the [sunglasses] in between, the [glasses are] absorbing the UV light.
I doubt that's a very accurate method to test the filter's effectiveness, but at least it can be done at home. –Mysid(t) 09:37, 21 June 2006 (UTC)
As far as I know, glass absorbs UV rays (that is why Fluorescent lamps are safe, even though they have UV rays inside them). So the sunglasses made of glass is the simplest UV protection. Am I wrong on this? -- Wikicheng 13:53, 21 June 2006 (UTC)
Not all glass absorbs UV, otherwise how do you expect a Black light to work? The key in fluorescent bulbs is the white powder they coat the glass with, it causes the UV to (wait for it...) *Fluoresce* into white(ish) light. If you have a true UV source, what better test is there for a UV blocker than trying to stop the UV from hitting a surface and fluorescing? It's pretty obvious when it works.
The article Ultraviolet says Ordinary glass is partially transparent to UVA but is opaque to shorter wavelengths. It also says the range of UV wavelengths is often subdivided into UVA (380–315 nm), also called Long Wave or "blacklight"; UVB (315–280 nm), also called Medium Wave; and UVC (< 280 nm), also called Short Wave or "germicidal" . This means that ordinary glass blocks most of the UV (except the longer blacklight). The article also mentions Tungsten-halogen lamps have bulbs made of quartz, not of ordinary glass. Tungsten-halogen lamps that are not filtered by an additional layer of ordinary glass are a common, useful, and possibly dangerous, source of UVB light. But the surprise is Ordinary eyeglasses give some protection, and most plastic lenses give more protection than glass lenses -- Wikicheng 15:04, 21 June 2006 (UTC)

But where would i get the uv light source from?

Also, let's make clear that opaque and transparent aren't yes-no answers - there are degrees of opacity (e.g., blocks 95% of all UV rays... etc). As for a UV light source, you could use a Black Light (available in a lot of stores, esp. novelty stores, but maybe even stores such as a WalMart). You might also try a halogen lamp; they usually have a (removable) glass cover to block UV rays. Be careful as both the intense heat and the UV rays of a halogen lamp may be harmful if you are continuously exposed to either. Nimur 18:43, 23 June 2006 (UTC)

Osmosis in potatoes[edit]

Dear all I need to conduct an experiment on Osmosis in potatoes, and for that I need to write a scientific background and an extension. I know the scientific background of Osmosis, but as an extension, does the surface of potatoe affect Osmosis and if so, how could one carry out an experiment on such phenomenon? This is NOT a homework question but an extension task to my write-up

  • Have you read Osmosis? The surface is probably the semi-permeable membrane between the water and the surface cells. - Mgm|(talk) 11:40, 21 June 2006 (UTC)

Does surface area of the potato affect the Osmosis?

Absolutely. I would, in general, expect the osmosis rate to be proportional to the surface area, until the potato chunks are "soaked thru". If you have some substance you can submerge potatoes into, allow osmosis, and then detect the quantity absorbed by the potatoes, then you can design an experiment. Submerge a whole potato in a given quantity of the substance, then submerge a chopped up potato in the same quantity for the same length of time. I suggest peeling both potatoes first to avoid any effect of differing osmosis rates thru the peel than thru the rest of the potato. A simple, but less accurate, test would be to put the potatoes in iodine, then mash them and look at the color. The chopped up potatoes will presumably have much more color than the whole potato. The timing is critical, however; if you soak them too long both will be fully soaked with iodine, and show the same color. StuRat 12:56, 21 June 2006 (UTC)

Processor Architecture[edit]

If I'm installing Linux on a machine with two processor boards, should I look for a distribution designed for a 64-bit processor architecture? --Username132 (talk) 10:06, 21 June 2006 (UTC)

No, unless you know the processors are indeed 64-bit. Instead, look for the abbreviation SMP (symmetric multiprocessing), which refers to the use of multiple CPUs. –Mysid(t) 11:11, 21 June 2006 (UTC)
You need a linux kernel compiled specifically for SMP machines (otherwise only one CPU will be actually used). Some linux distributions already have one, but when it doesn't, you have to compile one yourself. You may also have to do turn on a special option if the machine has more than 1 gigabyte of RAM. – b_jonas 10:22, 23 June 2006 (UTC)

I forget where I buried my threads...[edit]

I post lots of things on lots of different forums. It would be nice to be able to use one single program to track all my threads and keep me informed of replies. I sometimes forget to check up on a thread or even where I posted it. --Username132 (talk) 11:32, 21 June 2006 (UTC)

Help:Watching pages may help. --hydnjo talk 12:06, 21 June 2006 (UTC)
I'm assuming it's for the whole net, not just WP. First, consider using either your favourites or adding links to your browser toolbar (AFAIK you can do this with most). Most BBS software has a feature where it will show the most recent 50 or so posts by a user, so rather than creating a link to every single active thread on the net, you can just create a link to your author pages on BBSs you post to.--Anchoress 12:44, 21 June 2006 (UTC)
  • I agree with Username132. How difficult would it be to modify the MediaWiki software to provide RSS feeds or email notification of replies to all wiki discussions? --JWSchmidt 01:58, 22 June 2006 (UTC)
I think it would be failry difficult, more trouble than it would worth. As this is a wiki, MediaWiki doesn't know about threads. People edit the text of pages, not add reply to "threads". If you want to be able to follow threads, you'll have to use a software more like a message-board.
Also, someone has mentioned watching pages. In addittion to that, the "my contributions" link can also help. – b_jonas 10:28, 23 June 2006 (UTC)
Well, MediaWiki does allow for sending email when a page on one's watchlist is edited, but imagine the amount of email that would ensue... it was turned off for performance and spamming reasons in Wikimedia wikis. Titoxd(?!?) 05:21, 24 June 2006 (UTC)


What are three classifications of technology and meanings of each classification.

  1. Homework questions blatantly copied directly from the book or worksheet.
  2. Homework questions poorly disguised as legitimate ref desk questions.
  3. Homework questions well disguised as legitimate ref desk questions.

StuRat 13:07, 21 June 2006 (UTC)

Broken, impractical and obsolete --Dweller 13:15, 21 June 2006 (UTC)
Stone age, space age, garbage ? Gandalf61 14:00, 21 June 2006 (UTC)
Too much, too little, and wrong kind?--inksT 21:18, 21 June 2006 (UTC)
One, two and three? --ColourBurst 04:56, 22 June 2006 (UTC)
Out of date, soon to be out of date, not yet created and thus doesn't yet have the opportunity to be out of date. --Fastfission 18:17, 21 June 2006 (UTC)
That's the winner! --Dweller 11:26, 22 June 2006 (UTC)
The question absolutely needs context, because there are no actual "three classifications of technology." Just those described in the book. Is this high school level? --The Mac Davis] ⌇☢ ญƛ. 00:10, 22 June 2006 (UTC)
Cheap, good, and fast? – b_jonas 09:46, 23 June 2006 (UTC)
Most of the technology I buy doesn't fit any of those descriptions.  freshofftheufoΓΛĿЌ  15:32, 24 June 2006 (UTC)

Lowering cholesterol levels through exercise[edit]

How is cholesterol levels lowered through exercise? What's the process? Jack Daw 13:23, 21 June 2006 (UTC)

While perhaps not completely answering your question, the cholesterol (especially the regulation section) and the hypercholesterolemia articles should help. Nil Einne 14:52, 21 June 2006 (UTC)
Yeah the latter says that exercise is one way to lower cholesterol levels, but it doesn't say how it works. Thanks though. Jack Daw 20:00, 21 June 2006 (UTC)

If I were to hazard a guess, i would say that the increased cellular turnover induced by excersize would help lower cholesterol through cell formation. Also, changes in steroid production may factor in (although, as i remember, cortisol production decreases with improved excersize habits). This segmaent, taken from the HDL article, may also shed some light on the subject:

In the stress response, serum amyloid A, which is one of the acute phase proteins and an apolipoprotein, is under the stimulation of cytokines (IL-1, IL-6) and cortisol produced in the liver and carried to the damaged tissue incorporated into HDL particles. At the inflammation site, it attracts and activates leukocytes. In chronic inflammations, its deposition in the tissues manifests itself as amyloidosis.

Sorry I can't be more specific.Tuckerekcut 21:40, 21 June 2006 (UTC)

That's specific enough, thanks! :D Jack Daw 23:18, 21 June 2006 (UTC)


I did another article! But did I finally kick my refdesk addiction? --Zeizmic 14:17, 21 June 2006 (UTC)

I've done you a favor and rephrased your comment into an acceptable question, as you seemed to have forgotten to ask one!  freshofftheufoΓΛĿЌ  15:33, 21 June 2006 (UTC)
Zeizmic, could you rephrase your question? I have no idea what you're asking.--Yanwen 18:55, 21 June 2006 (UTC)

Shortest & longest day[edit]

According to the Solstice page, the June solstice was on the 21st at 12:26 UTC. According to my local paper in NZ, tomorrow (well today now for me), June 22th is the shortest day. The solstice was at 0:26 on June 22nd here (since we're +12) so I guess this is why the shortest day is the 22nd. Am I right that the longest day in Malaysia, UTC+8 would have been June 21st since the solstice would have been at 20:26 (on the 21st)? Or is it more complex then that? N.B. I appreciate the difference is very small, especially so in a equitorial country such as Malaysia. Nil Einne 14:44, 21 June 2006 (UTC)

Wouldn't an equatorial country have two longest days each year. Philc TECI 15:13, 21 June 2006 (UTC)
I believe you're mistaken. The summer solstice is still the summer solstice (depending on which hemisphere the country or area is in) and the winter solstice is still the winter solstice. Of course, there is no real climatical seasonal differences in an equatorial country. For example, in Malaysia, the June solstice is the summer solstice. If you happen to be on the equator, there is no longest day or shortest day I believe since there is no variation between daylength. Nil Einne 15:47, 21 June 2006 (UTC)
BTW, incase you haven't realised, I was using equatorial country in the loose sense (as it is frequently used) to mean the same thing as tropical country. Of course, even a country that is on the equator, would still have summer and winter solstices, except on the equator itself of course. Nil Einne 16:17, 21 June 2006 (UTC)
THats what I mean, on the equator itslef, the country would have two longest days, due to the leaning of the earth. Philc TECI 16:20, 21 June 2006 (UTC)
As I discussed above, unless I'm mistaken when you're on the equator there is no longest or shortest day. Each day is the same. As you move away from the equator, south or north this is no longer true (technically). It's been a long time since I took any physics but I would assume that even a metre away from the equator, there is still technically a longest day and a shortest day even if the difference is in microseconds. The longest day is still on or about the summer solstice and the shortest day is still on or about the winter solstice.
BTW, of course a country that has a part on the equator would also have parts in both the south and northern hemisphere. Therefore it would have a longest day in the southern hemisphere which is the shortest day in the northern hemispehere and vice versa. In this sense it would have two longest days, perhaps this is what you meant. But in a specific area, it's still either the longest day or shortest day (or just the solstice on the equator). Nil Einne 16:32, 21 June 2006 (UTC)
I disagree. The day is longest at the equator when the Sun appears to go directly overhead, which happens twice a year, at the vernal and autumnal equinox. However, I don't think there is nearly as much variation in the length of a day at the equator as there is in temperate zones (and, of course, the huge variation in day lengths at the poles). StuRat 16:41, 21 June 2006 (UTC)
Exactly my point, due to the tilt of the earths axis, as the year passes the sun will be directly over head first one hemisphere then the other, crossing the equator twice. Philc TECI 16:48, 21 June 2006 (UTC)
I've been trying to do this in my head, but I have a severe headache. So, on the equator, the solstice is the longest day (regardless of which solstice it is). The equinoxes are the shortest days (one shorter than the other), right? At what degree latitude do they change so that the equinoxesa are equal length days and one solstice is longest and one is shortest? What about that latitude right in the middle? --Kainaw (talk) 17:50, 21 June 2006 (UTC)
See Tropic of Cancer and Tropic of Capricorn for more information on which latitudes are the northmost and southmost of the sun's overhead travel. See Equator for that latitude right in the middle. Also, something handy for day length calculation is This navy site which will compute the sunrise/sunset times for any place on the globe. Pick a spot between the To'Cancer and To'Capricorn and see the effect on day length compared to any spot outside that.
(re-indenting for clarity) I've likewise been hurting my head with mental math, but a tropical location will still conform to the usual longest-day / shortest-day relation to the solstices. However, at the equator, the day/night cycle is precisely divided at the solstice, just as it is at the equinox. Since the length of day at the equator is a sine function with the same period as that at every other latitude, the only function that puts a 12 hour day at 0 rad and pi/2 rad is one with no magnitude; that is, the length of day at the equator doesn't vary. Sidenote: due to how the length of day is measured (any point of sun over horizon, not midpoint), the day is always longer than 12 hours. See also the graph linked from solstice at this siteLomn | Talk 18:34, 21 June 2006 (UTC)
Thanks for that, it's as I expected. I checked out the solstice article but not really the links. I also missed length of day which is equally useful. This appears to have settle the above issue, as I expected in a tropical location it's still the normal situation. Precisely on the equator, there is no variation in daylength. As you move away, there is variation but obviously e.g. a second away would have such a small difference to be meaningless. Having lived in Malaysia (KL, 3 degrees, 8 min), I know the difference is small but is not completely insignificant, about 30 minutes between longest and shortest. Of course, you do hardly notice it compared to e.g. you do here in Auckland (let alone somewhere like London).
However I'm still a bit uncertain about my original question namely how do you work out which is the longest day from the summer solstice (or the shortest day from the winter solstice). After more thinking, I'm guessing I was wrong above but it doesn't matter in most circumstances. You can't say for sure the day of the summer solstice would be the longest day since it will depend on factors such as your timezone in comparison to your actual longitude. In most case it will be but my guess here is that is actually depends which day length period is closer to the time when the summer (or winter) solstice occured.
So for example, in Malaysia since the summer solstice occured at June 21st 2026 (local time), the longest day would have clearly been June 21st since the June 21st sunset (about 1900) is far closer to the solstice then the June 22n sunrise (about 0700).
In New Zealand (or Auckland), it's a little more complex. The winter solstice occured at June 22nd 0026 as mentioned above. The sunset time on the 21st (according to a calendar I have for 2006) was at 1712 while sunrise for 22nd is at 0734. The sunrise on 22nd is closer (7h8 m) then the previous sunrise (7h14 m) so the 22nd is the shortest day but if the solstice was at 0001 for example clearly it would have been different.
Of course, it's probably easier just to work it out mathematically and in any case, given we're talking about a sine graph here, the difference between day length of the longest (or shortest) day and the day just before and just after are minute (even more so when the difference in daylength is so small as in Malaysia) but it's interestingly nevertheless.
Nil Einne 08:00, 22 June 2006 (UTC)
Here is a site with sunrise and sunset data for Quito, Ecuador, which is close to the equator: [3]. The length of each day is about 12 hours and 7 minutes, with earliest sunrise and sunsets around November 3 and May 13 and latest around July 26 and Feb 12. StuRat 03:16, 24 June 2006 (UTC)


We had recently found a small spot on our dining room ceiling. It contained Alot of little (probably 1-2 millimeters long) of black worm looking things. They had been there for a few weeks. There were almost hundreds of them. I have searched and searched, but have no luck finding any pictures or anything on them. I don't know if we need to spray our house our what? and for what? I have a baby on the way and would like to find out ASAP, please please help! Thanks, <-- email removed--> 16:58, 21 June 2006 (UTC)--

Well, it's not good, whatever it is. Get an infestation control person round asap. You could be risking the integrity of the structure of your house in the long term if this gets out of control. Trollderella 22:02, 21 June 2006 (UTC)
Ehm, irony doesn't travel well over the Internet. At least, I hope this was irony. :) DirkvdM 08:05, 23 June 2006 (UTC)

Java SDK[edit]

What is a recommended Java SDK for developing Java programs? I have the JS2E Runtime Enviorment 5.0 Update 6 installed. Is there any traditional SDK I can get similar to how the VM is the one and only? --Chris 17:39, 21 June 2006 (UTC)

There are only two Java SDK's to my knowledge. There is the official one from Sun (with multiple versions from 2.x to 5.x). Then, there is the Gnu version that is commonly distributed with Fedora Linux. Of the two, the official one would be best. I suggest using a lower version for compatability (like 4.x) and a higher one for the latest features. You appear to be asking which IDE is best used for developing Java programs. That is a completely different question. --Kainaw (talk) 17:47, 21 June 2006 (UTC)
Okay, maybe that is what I'm asking. I got my terms confused. Any answers? what is this thing called Net Beans? I programmed in Java before using jGRASP, but I never had to install the necessary components by myself. This is why I'm asking. --Chris 19:13, 21 June 2006 (UTC)
NetBeans are a collection of objects that you can use in your programs. Sun has an IDE that makes NetBeans easy to use (the NetBeans IDE). Also, the free Eclipse IDE (from IBM if I remember correctly) allows you to easily choose NetBeans (as well as many other object repositories) for code usage. If you only want to use your objects (and the default set of Java objects), than any IDE is fine. Both NetBeans IDE and Eclipse allow you to compile and run with a single button click and both have on-the-fly popups to "help" you. I put that in quotes because I hate it when boxes pop up while I'm trying to work. --Kainaw (talk) 19:23, 21 June 2006 (UTC)
There are probably hundreds of SDK's, actually - there is the OFFICIAL SUN JAVA SDK [4] as well as many other corporations or open-source groups who have created their own versions of the compiler (for various reasons ranging from business/license restrictions; running on unusual/obscure hardware; removing or adding features for performance, security, or connectivity; etc etc etc). Further, there are also multiple types of java: Standard Edition, Micro Edition, and Enterprise Edition. SE is the "standard" java you probably want. EE is for large corporate networks who have intense networking, security, and reliability needs. ME is designed for deployment on small devices such as PDAs, Cell Phones and VCRs, or embedded systems. Your best bet is the one linked above. Good luck, Nimur 18:52, 23 June 2006 (UTC)


Hi, are you ok? I hope so...

I don't know much about this, so I need your help.

I would like to know if propane lights up, does it create a expansive wave?

Could that be posible if a place's air is full of propane and it lights up?

Please... I need that info...

If you don't understand something please answer me. Thank you.

You appear to be asking if propane can explode. Yes. It can. --Kainaw (talk) 18:26, 21 June 2006 (UTC)
also, this may be obvious, but "there's no one here" -- I don't know who you were addressing, but the reference desk is empty. Volunteers edit answers on to the questions.
I'm have a tab open with this most of the day. -- The Mac Davis] ⌇☢ ญƛ. 00:13, 22 June 2006 (UTC)

What I find more interesting is: why does the propane flame in the bbq always make a loud 'wump' when you turn off the burners? There must be some air collapse, but I always thought this would be just a slow fade. --Zeizmic 19:33, 21 June 2006 (UTC)

Wow, that's a great question. There's so many things in life that we never think about. I think it was only a few years ago that they actually found out why coffee stains are light in the middle and have very dark edges =D --mboverload@ 02:24, 22 June 2006 (UTC)
Quite simple actually. Once the fuel supply is cut off, there is a higher Oxygen : Fuel ratio, so the fuel "bangs" rather than burns. This is why a room full of oxygen and propane will explode, not just burn (provided the O2 is well mixed and there is enough of it) .--Eh-Steve 08:24, 22 June 2006 (UTC)

What i mean is this.... Imagine this scenario: A place like a sears store. It's air is full of propane due to an escape. So someting lights it up. I know that the flames are going to wrap everything up and everything is gonna burn. But the "explosion" if there is in this case, may produce a expansive wave enoughly strong to blow out all the windows of the store? Or the propane light up doesn't create a expansive wave?

No, it will create a shockwave given the right fuel-oxygen mixture, otherwise it will simply burn. Thats how thermobaric explosions work. --Eh-Steve 05:14, 23 June 2006 (UTC)

how phenolphthalein was invented/discovered?[edit]

how phenolphthalein was invented/discovered?

Search for Adolf von Baeyer's experiments - especially those done in the 1870s. --Kainaw (talk) 19:40, 21 June 2006 (UTC)


Unstable radioactive isotopes subject to Beta decay which are not replenished say within a container will eventually die out in the same manner as living organisms will die out on an island which are likewise not replenished and fail to reproduce.

Is this statement true or false?

...IMHO (Talk) 20:17, 21 June 2006 (UTC)

To be more succint: probability suggests that all atoms of a radioactive isotope will eventually decay, provided they're not being replenished. However, something being true isn't a justification for "winning" an edit war, as (among other things) the relevance of the fact to the article must be considered. — Lomn | Talk 20:38, 21 June 2006 (UTC)
See latter portions of Talk:Half-life for the rationale for my "however" comment — Lomn | Talk
In that case you should have no problem answering this question rather than skirting. ...IMHO (Talk) 21:22, 21 June 2006 (UTC)
I feel that I did answer the question, I just found the analogy to living organisms cumbersome (as, apparently, did Melchior) as it seemed to be unrelated to your primary question. If I have misinterpreted, please amend my response to that of Melchior's: atoms do not decay like people.
To be blunt, though, this looks like an attempt to drag a content dispute away from the relevant talk page and/or dispute resolution mechanisms (might I suggest WP:3O?). Therefore, I tried to gently insert a suggestion along those lines. — Lomn | Talk 21:33, 21 June 2006 (UTC)
Bottom line here is revealing the truth. You need to answer "yes" or "no" to the question rather than to defer to someone else. ...IMHO (Talk) 21:53, 21 June 2006 (UTC)
Please remain civil. Melchior and I are both attempting to answer a rather vague question as best we can. As I've noted, your question, as posed, is vague enough that a "yes" or "no" isn't really sufficient. — Lomn | Talk 22:27, 21 June 2006 (UTC)
False. The fundamental principle behind the statistics of radioactive decay is that an individual atom's probability to decay is independent of its age. This is not true of the deaths of living beings! Even if the half-life of a non-reproducing human population is 50 years, within 200 years literally all of them will be dead. But for a population of radioactive isotopes with a half-life of 50 years, after 200 years 1/16 of them will have survived. Melchoir 20:59, 21 June 2006 (UTC)
Do you mind showing a table of values which with the fraction or percent of isotopes for each year over the span of the 200 years? ...IMHO (Talk) 21:17, 21 June 2006 (UTC)
Each year? I'll do it for five bucks, no less. Melchoir 21:22, 21 June 2006 (UTC)
Don't do it for me. Do it for other users. I submitted my evidence for free. Please provide the same courtesy for them. ...IMHO (Talk) 21:25, 21 June 2006 (UTC)
I get the feeling that there isn't a huge audience for a 200-row table of values whose only purpose is to verify that 2^4=16. Melchoir 21:28, 21 June 2006 (UTC)
Okay using your own criteria (above) of percentage remaining p=1/(2^(200/50)) consider this:
We want to know the percentage remaining after 200 years of any amount of anything that has a half life of 50 years.
We therefore apply the formula p=1/(2^(number of years/half-life)) or p=1/(2^(200/50)) or 1/16 or .0625 or 6.25%.
So it appears that since there is a percentage of 6.25% of the item left after 200 years that it does not matter what the item is or how much of the item there was to start off with. This is where your concept fails. For instance suppose that we start with only 10 items. How much will we have left after 200 years. That’s easy. we just multply .0625 times 10 and we get .625. Thus we have .635 items left after 200 years of the original 10 we started with. But what if the items are not divisible? What if there is not such a thing as .625 of the items we have and that any amount of the item that is less than one simply does not exist? Hummm... Since we can not divide an atom into parts by the method of Beta decay then sorry but even though we might have 6.25% of the original amount of the items left that percent is meaningless because 1.) we started with only ten items and 2.) each item is a complete and indivisible unit rather than a continuous value. If you apply half-life computation to the real world then you must take these facts into account. ...IMHO (Talk) 03:15, 22 June 2006 (UTC)
See below. Melchoir 03:50, 22 June 2006 (UTC)
I see. Then you refuse to back up your claim with data. ...IMHO (Talk) 21:56, 21 June 2006 (UTC)
Maybe take a look at the table at the beginning of the half-life article. For a half-life of 50 years, after 4 half-lives (200 years), 1/16 = 6.25% of the particles will remain. 22:39, 21 June 2006 (UTC)
Amending previous answer: After further research (i.e. the discussion on the half-life talk page), I now believe user IMHO's question is simply this: Like a non-reproducing population of humans, after a sufficient amount of time will a specific amount of a substance with a given half-life disappear completely? User IMHO's request for a table is particularly confusing given that (s)he generated exactly the requested table in an article that is up for deletion. Nevertheless, the answer seems to be that as a non-reproducing human population would not necessarily be described as exponential decay, the two processes are not the same, even though both processes may eventually end when all elements of the original collection (humans or, e.g., C-14 atoms) have decayed (barring unlikely quantum effects of a human coming back to life or an N-14 turning into a C-14). 23:16, 21 June 2006 (UTC)
The answer is false, and quite obviously so. If a non-reproducing human population decayed the same way unstable particles do, a human population of initial size would, in a period 3 times the median human life span, reduce to . Now I don't know what a good estimate for the median human life span is, but 70 years should not be too far off. The probability than a random person will still be alive after 210 years is not 12.5%, it is (for all we know) 0%!-- 02:30, 22 June 2006 (UTC)

(rewrap reply to 03:15, 22 June 2006) If you start with 10 items with a half-life of 50 years, then after 200 years, there is approximately a 52.4% chance that all 10 decay, 35.0% that 1 is left, 10.5% that 2 are left, 1.9% that 3 are left, 0.2% that 4 are left, and the rest are insignificant. Now compute .350 + 2*10.5 + 3*.019 + 4*.002 and tell me what you get. Melchoir 03:40, 22 June 2006 (UTC)

You must first provide the basis for your statements. I can say that 1 = 0 but if I do I think whoever I say it to might be entitled to the courtesy of my telling them why and showing them step by step my reasoning. Otherwise they have every right to reject what I am saying and interact with me no more. ...IMHO (Talk) 03:57, 22 June 2006 (UTC)

I'll tell you how I got 35.0% if you tell me the result of the computation I asked for. Melchoir 04:01, 22 June 2006 (UTC)
I don't care how you got it and I am not interested in solving puzzles or playing games. Please find someone else to play games with. ...IMHO (Talk) 04:07, 22 June 2006 (UTC)
Rather ironic, from someone who seems to have quite a grasp of mathematics. Here you go, though:
First, Melchoir has an error ("2*10.5" should be "2*.105"). After that, though, it's just summed probability. The average number of atoms remaining is equal to P(0)*0 + P(1)*1 + ... P(n)*n, where in this case n=10.
The probability P(x) is C(n,x). Sum all those terms and you get 0.625 atoms remaining, on average, in your example. As I noted way back in the first response, though, this is probability, not reality. The number of atoms must, of course, be integral, and is most likely 0 or 1 (an 82.5% chance, if my math is correct).
Does that clarify things? — Lomn | Talk 04:13, 22 June 2006 (UTC)
Then stop acting as if you knew what you're talking about. Melchoir 04:10, 22 June 2006 (UTC)
He's not playing games; he's trying to explain to you why not all of the particles will necessarily have decayed even after the expectation value generated by the half-life calculation falls below 1.
Given a half-life of 50 years and an elapsed time of 200 years, we have a total of 4 half-lives elapsed. The probability of a given particle undergoing radioactive decay during any fifty-year period is 0.5, by definition. The probability of a particle surviving four such consecutive periods is therefore (0.5)4, or 0.0625 (6.25%); the probability of one particle decaying is therefore 1-0.0625, or 0.9375. The probability that all ten particles will decay is therefore 0.937510, or 0.5244 (52.44%): the value Melchior gives above. Similar calculations give the remaining figures (it's an example of a binomial distribution). There's even a small probability (about 10-12) that none of the atoms will have decayed. TenOfAllTrades(talk) 04:22, 22 June 2006 (UTC)

At this point in the discussion the issue seems to be one of certainty versus probability. In fact I think we agree that they are opposites. When one is uncertain one turns to probability for a reasonable answer but when one is certain one would be foolish to turn to probability for the answer. I am certain that isotopes which undergo radioactive or Beta decay as individuals and therefore as a population are not immortal unless you have an infinite supply. Thus if you have a closed container of Carbon-14 which can not be penetrated by cosmic radiation or influenced in any way such that Nitrogen-14 within the container will be converted to Carbon-14 that the Carbon-14 contained within the container will eventually perish, die, drop dead, cease to exist, go kapoof, say bye bye, bite da bullet, kick the bucket, fall off the cliff, go kaplunk, join King Neptune, ride the ferry, antagonize da horn, get whacked, meet it's maker, build a pyramid, get blue screened, sweet repose, see the henchmen, walk the plank, push up the lillies, etc. ...IMHO (Talk) 12:16, 22 June 2006 (UTC)

You still appear to misunderstand the vital role probability plays in radioactive decay (in fact, it's solely and entirely driven by probability). While it is exceedingly likely that radioactive decay eventually consumes all non-replenlished atoms, that cannot be said with mathematical or scientific certainty. Aside from that minor quibble of vocabulary, I think we've now resolved your original question. — Lomn | Talk 13:02, 22 June 2006 (UTC)
How about this thought experiment: Assume that radioactive decay is random (as it appears to be) and that each atom's decay is independent of each other atom (as it appears to be). The universe is very large and there will be a *lot* of Carbon-14 (for example)—enough Carbon-14 to last a *very* long time. If we take 15 atoms of this large amount of Carbon-14 in the universe and put the in an inpenetrable container (no gamma rays, etc.) what will happen? The answer is that we cannot say for certain. Even though we know that Carbon-14 is unstable we cannot predict (with certainty) when or if these 15 atoms will ever decay. We might have, by chance, picked the few atoms that basically out-last the universe. We will find (at the least) that our small number of atoms on the whole of human experience appear to never decay--that they last forever. This is the problem with your desired example. You (IMHO) are applying your intuition to a random process and humans, generally, have poor intuition when it comes to randomness—it appears to not be in our genes. —Bradley 17:45, 22 June 2006 (UTC)
The question asked if isotope atoms die out in the same manner as living things and that's clearly false. Even as a metaphor the comparison is weak. Peter Grey 22:29, 22 June 2006 (UTC)
I think they won't die out. The lifetime of atoms of a radioactive material have an exponential distribution which will cause that you need lots of time for many atoms together to all die out. This is quite unlike living organisms, which age with time and die more easily as they're older. Radioactive atoms don't get older. – b_jonas 09:24, 23 June 2006 (UTC)

DVD decoder?[edit]

Having installed a DVD optical drive and Windows XP Media Center 2005 Edition on a new computer, I am unable to play content-protected DVD's (although unprotected DVD's can be painlessly played). Windows Media Center, Windows Media Player as well as other DVD software show various error dialogs, complaining about problems with the decoder. My guess is that the problem may be resolved by purchasing either of the following products: CinePlayer DVD Decoder Pack or NVIDIA DVD Decoder. I am right? Is it worth a try? --Andreas Rejbrand 20:45, 21 June 2006 (UTC)

It might be that your DVD drive is set to the wrong region. Do the messages give any hint of this? See if you can check the region. Many DVD drives allow you to change the region up to five times; after that (a hardware limit) you are stuck in the last region you chose. Notinasnaid 21:11, 21 June 2006 (UTC)
Thanks for your respons, but, actually, I've already checked that, and it is valid. --Andreas Rejbrand 21:20, 21 June 2006 (UTC)
The software you've listed may fix the problem. I notice that the NVIDIA one has a 30-day trial, might I suggest you see if that fixes the problem before you spend money? — Lomn | Talk 21:24, 21 June 2006 (UTC)
Seems very wise indeed. Thanks! --Andreas Rejbrand 21:31, 21 June 2006 (UTC)
You might try using VLC, which is usually not picky about playing things. If that works, you can rule out issues with your drive, and probably won't want to continue to mess with Windows Media Player. Trollderella 21:59, 21 June 2006 (UTC)
Thanks for your answer. (Actually, however, I think Windows Media Player and particularly Windows Media Center are great software.) --Andreas Rejbrand 22:05, 21 June 2006 (UTC)
Regardless, the advantage with VLC is that you do not have to worry about codecs issues and the like. This will help to establish a base line to work out what the problem is. If your drive or the disc is broken, then it's no use fooling around with other issues Nil Einne 08:13, 22 June 2006 (UTC)
De gustibus non est disputandem. Did you get it working? Trollderella 22:18, 21 June 2006 (UTC)
It works in VLC, but that does not completely solve my problem. I want it to work in WMC. --Andreas Rejbrand 10:34, 22 June 2006 (UTC)
WMC does not have its own DVD decoders. You are required to get third party codecs for it. I know Dell computers come with Cyberlink PowerDVD preinstalled, but I decided to uninstall it and as much preinstalled software as I could, and then eventually reformatted. This is the official WMC DVD decoder page, but they cost money. This is why I use VLC religiously. --Chris 15:23, 22 June 2006 (UTC)
I see. Well, at least they aren't excessively expensive. --Andreas Rejbrand 15:47, 22 June 2006 (UTC)
Your official latest video card driver's from your manufacturer (ATI, NVIDIA, etc.) may also provide this functionality to WMP. -- 20:41, 27 June 2006 (UTC)
Interesting idea, thank you --Andreas Rejbrand 19:50, 28 June 2006 (UTC)

Metal Plate in Head Questions[edit]

1. First of all, if someone is shot in the head and miraculously survives, is a metal plate inserted into the skull the normal procedure to repair the damage?

more importantly,

2. Let us assume hypothetically that a person has been shot in the head by a 10 mm bullet, and survives. Let say it 'grazed' his head and pulled some of his skull off, but left his brain competely intact. Would a doctor use a metal plate in this instance to repair the missing skull fragment(s)? How would he go about inserting it? How long would it take for the wound(s) to completely heal and for the plate to be completely integrated into his skull?

and finally,

3. If a man with a metal plate in his skull was shot directly, at a 90 degree angle of impact from the tip of a 9 mm full metal jacketed bullet, in the same spot as the plate is inserted in his skull, would the metal plate protect him from the bullet? What would the effects be on the man? Would he be knocked unconscious, or not? Would the bullet ricoshet? If the plate does protect him from a 9 mm bullet, what is the maximum full metal jacketed caliber that the plate would protect him from (.357, .44, etc.)?

Thanks, 22:10, 21 June 2006 (UTC)

None of these questions are easy to answer, since traumatic head wounds are very unpredictable. The effect, recovery and treatment depend greatly on the specifics of the injury. The balistic properties of metal plates in the context of head injury are likewise not easy to predict. I assume that you are researching a novel, since I don't want to contemplate the other possibilities. Go for it - it's certainly not impossible that a man with a plate in his head could be saved by a freak balistic trajectory. Trollderella 22:17, 21 June 2006 (UTC)

To answer:

1) Yes, if the original skull fragments are unworkable.

2) It depends entirely upon the injury. Titanium plates are custom made to fit the skull (also full oseointegration occurs), and some insight can be given here at this link. i.e. the plates are screwed into existing bone (pictures of the screws here)

3) Bullet deflection will only occur at grazing incidences with such thin metal plates. At 90 degrees, the titanium plate probably won't even stop a .22 round. However, it is possible for a 9 mm para at a high incidence (i.e grazing) to be deflected. This has been demonstrated with the guy who shot his friend's XBOX with a 9 mm parabellum and it wasn't even scratched. As for higher calibres (including FMJ), it's even less likely, and for a .357 magnum to be stopped by this is just silliness.

Reply if you want further details. --Eh-Steve 08:43, 22 June 2006 (UTC)

As for being shot directly by a 9 mm full metal jacket, a thick metal plate (you are talking inches, not millimeters) will absorb the strike, denting inward. However, to push against the bullet coming in, it also has to push against the skull where it is attached - likely crushing it on one side. So, assuming the plate is thick enough to keep the bullet from punching a hole right through it, the likely end-scenario would be that the weak side of the skull would cave in, causing the plate to fold inward (into the brain) and the smashed bullet fragments would go into the brain as well. I would expect more damage to the brain than having a 9 mm hole punched through it. --Kainaw (talk) 14:18, 22 June 2006 (UTC)
Thank you to all for responding so throughly. In response to Trollderella's comments, I am indeed researching a novel, not planning an assassination or some such violent foolishness. From what I have read in the responses and through other research, it seems unlikely that a metal plate inserted in the skull via the normal medical procedures would protect a person from bullets flying at his or her head. However, seeing that I am writing a fiction novel, I may be able to exaggerate the strength of these plates, and a person's ability to stay conscious from trauma to the brain. Hopefully I will not have to exaggerate these factors to the point of rediculousness or, even worse, to the point that it breaks the readers suspension of disbelief. Thanks again to all. 21:45, 22 June 2006 (UTC)
Well, speaking of novels, some novel of Rejtő Jenő features a men who was protected from a gunshot to his heart by a metal plate he's placed there deliberately because he knew the antagonist would always aim there. – b_jonas 09:17, 23 June 2006 (UTC)

'Good' cholestrol[edit]

I want more of it. I read the article, which told me what it is, but not how to get more of it. What should I be eating more of? Thanks! Trollderella 22:14, 21 June 2006 (UTC)

Eat more: vegetables, fish, nuts, high glycemic index carbs, and monosaturated fats. Eat less: polyunsturated fats, hydrogenated fats, junk food. Consider taking omega fatty acid supplements (fish oil, evening primrose oil, seabuckthorn oil) and lecithin.--Anchoress 22:24, 21 June 2006 (UTC)
Have polyunsturated fats had multiple comments by me removed from them ? :-) StuRat 03:23, 24 June 2006 (UTC)

Thank you very much - it might be good to add these to the articles on LDL and HDL. Trollderella 22:26, 21 June 2006 (UTC)

Garden of Eatin' chips and Lance crackers have good fats in them. —Keenan Pepper 00:56, 22 June 2006 (UTC)
What about Steakhouse of Eatin'? — The Mac Davis] ⌇☢ ญƛ. 01:38, 22 June 2006 (UTC)
I had a very low HDL (less than 10). I increased it through excercise and drastically increasing fish in my diet. By fish, I don't mean shrimp, crab, oysters... I mean things like salmon, whitefish, and bass. After a year of using the stairs (I work on the 12th floor) and having fish a minimum of 3 days a week, my HDL is now 22. --Kainaw (talk) 18:29, 22 June 2006 (UTC)
Becareful with too much fish though. They retain lots more toxins than the water they live in retains. Mercury is one of the things you'll hear about today, though I don't know how much truth there are behind this. --Russoc4 21:02, 22 June 2006 (UTC)

Certain medications can increase the level of HDL: fibrates, statins, and niacin. - Cybergoth 21:22, 26 June 2006 (UTC)