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October 27

Crude oil production

Hello,

Can anyone tell me what has been the total world production of crude oil, in trillions of tonnes, over the period 1859 to 2007?

Many thanks, Carlyon —Preceding unsigned comment added by Carlyon (talk • contribs) 00:16, 27 October 2007 (UTC)

Thats a tough one. Using this dataset from OECD, the total world production of crude oil between 1971 to 2005 was 1.107827 trillion tonnes (trillion meaning a million million). This table provides data back to 1960 in million barrels a day. Beyond that all I can find is Image:Hubbert world 2004.png, from which you could try to source the raw data back to 1900. Rockpocket 01:02, 27 October 2007 (UTC)

Calcium Gluconate Chemical Structure

I'm having trouble putting together the chemical structure of calcium gluconate. I've tried using the BKchem molecular editor or similar programs to lay it out but i've completely forgotten how to do so. If this was all still fresh in my mind from my school years, I'd have no problem. Wikipedia shows a good bit of information about calcium gluconate, but in particular I need to see in diagram form the chemical structure of said supplement. Again, I've completely forgotten how to put these things together. [1] —Preceding unsigned comment added by 76.177.150.63 (talk) 00:39, 27 October 2007 (UTC)

It's a salt of gluconic acid. The rightmost hydrogen in the structure is to be removed (and a superscripted minus sign is to be written right of the rightmost oxygen) to get gluconate. Calcium gluconate is made up of 2 of these units and 1 Ca²⁺ ion, compare with calcium acetate (I think the picture in this article is of the kind you want to have). I don't know the real spatial arrangement of calcium and gluconate ions which can be determined using X-ray crystallography. Icek 03:00, 27 October 2007 (UTC)

Let me see if I've got this straight. From the gluconic acid you've shown, remove the rightmost H and replace it with a superscripted -? How would i illustrate 2 of these units correctly? The shown unit enclosed in parentheses (or whatever the correct term would be here)? Then followed with Ca²⁺? Is this the only possible way of illustrating Calcium Gluconate? thank you very much. —Preceding unsigned comment added by 76.177.150.63 (talk) 23:25, 27 October 2007 (UTC)

You could also draw 2 gluconate molecules and a Ca²⁺ between them (with the right molecule being a rotated version of the left one - the Ca²⁺ should be close to both O^-), similar to the picture in the Wikipedia article (which is a rendering of a space-filling 3d model - the green thing is the calcium ion). Icek 09:08, 28 October 2007 (UTC)

3 dB per octave?

Is there a way of getting a 3dB per octave boost or cut over a wide frequency range? —Preceding unsigned comment added by 88.109.17.174 (talk) 00:45, 27 October 2007 (UTC)

Yes but if it's the simple answer I think it is, I suspect this is a homework problem. Hint: Single poles and zeros create 3dB/octave changes in power. --DHeyward 07:03, 27 October 2007 (UTC)

Uh, no, first-order filters are 6dB per octave.

But yes, it’s certainly possible to design a filter that only changes by 3 dB per octave. Perhaps the easiest way to design such filter is to use a digital filter. Just do an Fast Fourier Transform, manipulate the spectrum to have the desired frequency response, and convert back to the time domain.

It'd be a lot more complicated to design if you have to do it as an analog filter, but it'd still be possible. It'd have to be a high-order filter, that combines an equal number or almost equal number of high-pass and low-pass first-order filters within one circuit. To see that it’s possible to design such an analog filter, you could always split the signal into an arbitrarily large number of bands using a bunch of band pass filters, amplify each band separately by an appropriate amount, and then join the signals together in one output. MrRedact 08:09, 27 October 2007 (UTC)

Actually, I thought of a much easier way to design such an analog filter. Just combine a first-order filter with an analog multiplier in a feedback circuit such that the multiplier is doing a square root. MrRedact 08:20, 27 October 2007 (UTC)

Oops, wait, I think DHeyward may have hit the nail on the head. Are you talking about 3dB voltage per octave, or 3dB power per octave? I assumed you meant voltage. If you meant power, then the answer is a trivial first-order low pass filter or high pass filter. MrRedact 10:06, 27 October 2007 (UTC)

It was 3 dB per octave in voltage i wanted (power is dead easy). —Preceding unsigned comment added by 88.111.67.141 (talk) 15:44, 27 October 2007 (UTC)

OK, then did you understand my idea for the analog version of the filter? The input of the whole filter would be the input of a first-order filter. The output of the first-order filter would go to the + input of an op amp. The output of the op amp, which is the output of the whole filter, would also go to both inputs of a four quadrant analog multiplier. The output of the multiplier would go to the – input of the op amp. You might need a resistor or two in there to tweak the gain or bias, but that’s the basic idea. MrRedact 20:39, 27 October 2007 (UTC)

Yes I understand: a 6dB/octave roll off, when square rooted will give a 3dB/oct roll off? This is one way i thought of, but I want it to work very fast, so the analog mult idea may not be capable of the speed. (the accuracy is not that important- 5% will do)

A mosfet has a square law relationship between voltage and current. You should be able to turn 6 dB/octave in input voltgage to 3db/octave in output current pretty easily. You can use it directly or in the feedback of an op amp. --DHeyward 21:34, 27 October 2007 (UTC)

This idea of using a mosfet (insted of an anlog mult) as the squaring element in MrR's topology is interesting. How would I setup a mosfet circuit to give me a true square of the input voltage (including the dc level)? Also could it be a JFET?

I just googled "mosfet square root circuit". here is one. The middle opamp has the mosfet in the feedback with the drain as the output. This is what I was thinking. JFETs are square law as well though they may require different biasing to stay in saturation. But now that I think about it, you may need square function to get you to "power" and then "power" is 3 db/Octave. --DHeyward 02:22, 28 October 2007 (UTC)

Interesting circuit, tho Im not keen on the necessary inversion and the active offsetting circuit. But thanks for finding it!

Doesn't "3dB per octave" just mean "a factor of 2 for each factor of 2"? I wonder why it isn't stated more simply as "1/f" or the like. —Tamfang 22:19, 29 October 2007 (UTC)

Whether that definition is right depends on what you mean by "a factor of 2". For a resistive load, the power is proportional to the square of the voltage. The OP wanted the output/input power ratio to change by a factor of 2 for each factor of 2 in frequency, which means the output/input voltage ratio is only changing by a factor of sqrt(2) for each factor of 2 in frequency. That's not as easy to do as to design a circuit such that the output/input voltage ratio changes by a factor of 2 for each factor of 2 in frequency. MrRedact 23:43, 29 October 2007 (UTC)

stop watch hardware

I want to know the harware of a stop watch(just stop watch) whose accuracy is 0.1 second. —Preceding unsigned comment added by Gangt (talk • contribs) 07:24, 27 October 2007 (UTC)

See stopwatch. The rate of a mechanical stopwatch is controlled by the balance wheel while an electronic stopwatch uses a crystal oscillator. --Justanother 19:14, 27 October 2007 (UTC)

Shock in AC and DC

Why do we get sucked in and pushed out while receiving shocks?..Is it due to ac and dc shock?..What is the shock difference in these two and which one is more fatal?.. —Preceding unsigned comment added by 122.164.49.85 (talk) 08:56, 27 October 2007 (UTC)

See Electric shock#Issues affecting lethality which discusses this. 84.64.123.72 11:33, 27 October 2007 (UTC)

The shock causes your muscles to contract. This can mean that your hand is immediately, and involuntarily pulled away from the cause of the shock - but if you have a grip around the live object then your grip may tighten and you'll be unable to let go no matter what. SteveBaker 02:44, 28 October 2007 (UTC)

German Weinstein: Potassium tartrate or Potassium bitartrate?

Hi, the German article de:Weinstein links to Potassium bitartrate aka cream of tartar etc. and vice versa. Dictionaries say so, too. (See also de:Wikipedia:Auskunft#Weinstein. But Potassium tartrate claims that that is the main ingredient of "Weinstein", though easily confused with the other. Can someone clear that up, please? T.a.k. 10:06, 27 October 2007 (UTC)

I'm not sure what is correct, but I would note that the information was added by an anonymous IP as that IP's only contribution [2]. Given that the information is not sourced, and contradicts other sources, I'm not sure that the info should be trusted. -- 16:45, 31 October 2007 (UTC)

Granular layer - cerebellum

Granular layer redirects to cerebellum. Why is this and is this correct? Lova Falk 10:32, 27 October 2007 (UTC)

Did you see the granular layer section? Here's a link with some other definitions. JMiall₰ 11:48, 27 October 2007 (UTC)

Granular layers also occur outside of the cerebellum (most notably, in the cerebral cortex), so this redirection is not really appropriate I think. EelkeSpaak 15:32, 27 October 2007 (UTC)

Cost of heat

A man lives in a place where the temperature is below freezing all year round. His house is heated with electricity, and he has an electric water heater. He keeps his house at a constant 20C, except for the "cold room", in which the thermostat is set at 5C. The water heater is located in an area where the temperature is 20C. He decides to move the water heater to the "cold room", to gain more living space. (This does not significantly change the distance from the water heater to where the hot water is used.) The man notices that the temperature in the "cold room" remains at 5C, even though the water heater feels warm. Would the total monthly electric bill change because of the move? GarthGarth 13:31, 27 October 2007 (UTC)Garth

If I understand the assumptions correctly, no. Electric heaters are 100% efficient in the sense that they produce one kWh of heat per kWh of electricity. Since (by assumption) nothing else changed in the move, the total power consumed by the water and space heaters must be the same as before, although it may be differently distributed between them. -- BenRG 14:19, 27 October 2007 (UTC)

I can't see how that can be right. The hot water is loosing heat by means of radiation and conduction the rate of both are affected by the temperature difference between the hot water and the room temperature. Conduction is proportional to temperature difference and radiation goes as the 4th power.Theresa Knott | The otter sank 14:38, 27 October 2007 (UTC)

The water heater will have to work a little more to keep the water hot, compared to the 20C room, since more heat is being lost to the colder room. At the same time, the heater in the house has to work less to keep the 5C room at 5C, so the house heater will use less energy. So the monthly energy bill is probably going to change. Whether its up or down depends on more specific info such as efficiencies, volume/area of the rooms, etc, and a bit of calculus. Arakunem^Talk 15:26, 27 October 2007 (UTC)

The water heater will use more electricity due to the greater temperature differential between the hot water and the room. However, the wasted heat from the water heater will contribute to the heating of the room. So the room's heaters will work a little less, and the water heater will work a little more. Probably won't see much change in electrical usage.

But for the room the water heater was moved from, the room heater will have to work a little bit more than it did previously since the water heater is no longer contributing any heat. --Bowlhover 01:22, 30 October 2007 (UTC)

Obviously, this assumes the rooms are heated year round, and moving the water heater doesn't change the temperature of any of the rooms, and the new plumbing doesn't snake outside, and so on...--Duk 17:08, 27 October 2007 (UTC)

No, no change at all. Forget the details. The temperatures stay the same, so the same amount of heat is generated. The only problem might be efficiency, but since heat is the ultimate energy waste product, but also the goal here, efficiency will be 100%, no matter what. So unless more or less warmth leaks out of the house, for example because the heater is moved to or from an outer wall, energy consumption will remain the same. DirkvdM 17:52, 27 October 2007 (UTC)

That assumes that none of the heat from the water heater (and the water in it) is lost to the room. Arakunem^Talk 23:03, 27 October 2007 (UTC)

Actually, it's the opposite. It assumes the water heater is just another heating element in the room. There are 3. One for the 20C rooms, one for the 5C room and the water heater. The temperature doesn't change in the rooms. The heating elements were just rearranged. The details of the percentage of heat from each element is not asked, just the overall heat. It doesn't change because there are no changes in the temperatures of the rooms. The sum of the heat is the same in each case so the energy added is the same. --DHeyward 06:46, 28 October 2007 (UTC)

If the electricity is being turned to heat in both cases, there will be no difference. If the water heater works that way, but the heater from the room moves heat in from outside, which is more likely the case, the less efficient water heater will be doing proportionally more of the heating, and the electricity bill will go up. — Daniel 20:18, 27 October 2007 (UTC)

Hang on, hang on, hang on.

If the water heater loses some heat to the room, we can think of it as being less than 100% efficient at heating water, but it is guaranteed to be 100% efficient at heating the water+room system.

If the other furnace is outside, and if it is less than 100% efficient, it might well lose some heat to the outside, which would then be lost forever.

So, paradoxically, if the other heater is outside, the less efficient the water heater is, the lower the overall electricity bill will be...

(The only exception would be the case where the grossly inefficient water heater in the "cold room" ended up heating it above the requisite 5°C.)

But on the other hand, if the other heater is inside, and if it's less than 100% efficient, any heat it loses is lost to the room, so hey presto, it's 100% efficient after all. —Steve Summit (talk) 23:17, 27 October 2007 (UTC)

And then there's how leaky the windows might be... *shifty* —Preceding unsigned comment added by Arakunem (talk • contribs) 23:23, 27 October 2007 (UTC)

Makes no difference. There has to be some heat loss to the outside, otherwise waste heat from the water heater would heat the "cold room" above 5°, and for that matter the rest of the house above 20°. Whether that loss to the outside is slight or extreme is immaterial to the rest of the problem. —Steve Summit (talk) 01:03, 28 October 2007 (UTC)

...wait a minute. Did we just do someone's homework for him? —Steve Summit (talk) 01:04, 28 October 2007 (UTC)

It's okay. I think after reading this they'll be more confused, and have to resort to thinking through what their textbook says...Skittle 15:07, 28 October 2007 (UTC)

Ah - a homework question. Ok, just draw a box with a arrows for energy_in and energy_out. Since all the temperatures and house insulation stay the same, energy_out stays the same, and therefore electricity_in must stay the same. Forget everything else! --Duk 16:25, 28 October 2007 (UTC)

I don't think a water heater can ever be 100% efficient. It's heating water in copper tank which is connected via nice metal pipes off into the outside world. Being metal, those pipes are going to conduct heat away out of the tank - and out of the room too. The answer to this question depends on the relative efficiency of the room heater and the water heater. If the water heater is more efficient then your bills should go down - if it's less efficient then they'll go up. SteveBaker 02:42, 28 October 2007 (UTC)

Not so. See my "paradoxically" argument above. It'd be very hard for an electric water heater to lose heat directly to the outside world, without heating some part of the house up first. (But similarly for a resistive electric space heater, located inside the house.) —Steve Summit (talk) 03:44, 28 October 2007 (UTC)

Yes but thats the point. The heater has been moved to a colder room, therefore is losing more heat faster than it was in the warmer room. Thus, the heater's heater will be running more frequently to keep the water at its desired temp. Thus more power is used by the water heater simply because it is running more often. The second consideration is then, that the heat lost from the heater to the room means the room's heater will not need to work as hard to keep that room at 5C, so it will use less energy. So the total house energy bill will change, but without more info you cant say for sure which way it will change, and by how much. Arakunem^Talk 15:16, 28 October 2007 (UTC)

Explain to me again why it will cost more (or less) to heat a room with waste heat from an electric water heater, versus with heat from an electric heater. —Steve Summit (talk) 16:40, 28 October 2007 (UTC)

Yes, you are quite correct! I was looking at the components and not the system as a whole. *Awards one pie* :) Arakunem^Talk 18:21, 28 October 2007 (UTC)

I don't see the location making any difference in energy use in the specified, somewhat atypical, conditions. The 5 degree room is already electrically heated, so any heat lost from the hot water heater from the insulated tank or the water pipes or even the electrical wiring to the air in the cold room will displace heat which would otherwise have been consumed by the heater in the cold room which maintained it 5 degrees warmer than the outside. The heat from the water heater which would otherwise have helped heat the warm rooms will be replaced by additional heat from the electric heaters there. Zero net effect. If it were a real world home, and the water heater were moved from living space to an unheated garage or unheated basement, the electric bill might go up, since most homes do not have a room thermostatically maintained five degrees warmer than a constant temperature outside. The devil is in the details. Edison 21:10, 28 October 2007 (UTC)

Yeah, but the assumption was that the temperature in the rooms is thermostatically maintained.

I said it 'forget the details', Duk said that energy in = energy out and Arakunem said to look at the system as a whole. Same idea, the wording just improved gradually. If the energy in the system (the house) remains the same and the energy going out of the system (heat leakage from the house as a whole) remains the same, then the energy going in is the same. And all that energy would be in the form of heat, since heat is the ultimate energy waste product. It would have to be a very stupid heater if it didn't convert all the energy into heat - what else would it convert it into? DirkvdM 08:59, 29 October 2007 (UTC)

Light?

I have to disagree with you on this problem. When the water heater is moved from the hot room to the cold room, the house heater will be on more often in the original room since it must work independently to maintain the temperature. Conversely, the house heater will be on less often in the new room since the water heater is providing part of the heat. The total amount of electricity used by the house heater remains constant.

However, the water heater must keep its water at a constant temperature. More energy will be consumed if the air temperature is 5 degrees than if it is 20 degrees, and the newly-located heater will therefore use more energy than it did before. The conclusion is that more electricity will be used after the water heater is moved. Is my reasoning correct? --Bowlhover 01:22, 30 October 2007 (UTC)

No because the energy that is "lost" by the heater goes in to the room. Since the 5C room doesn't rise in temperature, the heat provided by the water heater offsets the heat that was used to heat the room to 5C. This is a rate problem where the sum of the heat entering the house equals the sum of the heat leaving the house. The temperature is constant so the heat added is constant. Nothing you do with location of the heat sources will change it. This is a Guassian sphere-like problem where the net heat flux is zero. --DHeyward 05:05, 30 October 2007 (UTC)

Yes, this is what got me earlier. Note that the heaters were specified to all be electric. Meaning X watts of electricity goes right into heat. If you look at the house as a system, rather than rooms, then the net energy change for the system will be zero. Look at it like this: If you had one big heater providing X amount of heat, that would be the same as having 2 smaller heaters providing X/2 heat, yes? Or 3 smaller heaters at X/3... The heat lost from the water heater is just another heat source. If the heater loses 100 watts to the room, thats 100 watts less that the room heater needs to furnish. Net change = zero. Arakunem^Talk 14:40, 30 October 2007 (UTC)

Why is hot water better for washing dishes?

Why is it that hot water is better when washing dishes? Why does it seem to remove more dirt/grease/debris than cold? Is it purely from the thermal effect? or does the sometimes higher pressure of a hot water tap jet also have an effect? 84.64.123.72 13:38, 27 October 2007 (UTC)

The main factor is that things dissolve more readily in hot water. For instance, a cup hot water can hold more sugar than a cup of cold water. This takes care of the dirt that dissolves in water. Most of the dirt (like fat) does not dissolve in water, so we add detergent to the water so these dissolve a well. Finally, some amount of mechanical force is necessary to speed the process along, which is where the brush comes in. This also means that if the hot water has greater pressure (not sure if that's true) it may have a slight effect. 14:14, 27 October 2007 (UTC) —Preceding unsigned comment added by Risk one (talk • contribs)

In addition hot water softens fat and allows emulsification with a soap or detergent more rapidly than at lower temeratures. Hey, it also makes doing the dishes more comfortable Richard Avery 15:13, 27 October 2007 (UTC)

"also makes doing the dishes more comfortable" - depends on how hot the hot water is. --Psud 00:53, 28 October 2007 (UTC)

Softening/melting fats is the primary reason. The liquid fat is emulsified by the detergent, this would take a lot long if the fat was still solid. Shniken1 17:19, 27 October 2007 (UTC)

Anyone else here use boiling water to clean out their frying pans? You don't even need soap if the water is hot enough - just pour it in from the kettle, let the fat melt and float to the top, pour it out, wipe the pan out with a cloth. --Kurt Shaped Box 18:54, 27 October 2007 (UTC)

Yep, also doing that avoids getting the pan too clean or something, according to my mother. And who am I to question her wisdom? Then you rub a tiny amount of oil into the pan's surface before storing it. And Dirk, I think you could only wash up with cold water if a) you didn't eat or cook any of the more interesting things I eat or b) didn't mind a thin film of rancid fat/grease on everything and c) didn't need to get things washed quickly and d) didn't have to concern yourself with germs. Skittle 15:02, 28 October 2007 (UTC)

Now that we've changed the subject into dishwashing tips, I could add one my husband sometimes applies: boil water and washing powder in a pan. The inside of the pan has never been so clean before! Lova Falk 19:06, 27 October 2007 (UTC)

A very slight effect, if any. Several years ago I started using cold water and I didn't notice a difference. Mind you, I do the dishes in two goes. I first apply the dishwater (with just a little washing up liquid, a fraction of what people in the US use), let that soak for 10 minutes while I do some other household task and then finish it off with ease. Mind you, I do this on the counter, not in the sink, with the dishwater in one of the pans, and then I dip the brush in that. Works perfectly. And then I have a good excuse to clean the counter when I'm ready. Win-win-win. :) DirkvdM 18:06, 27 October 2007 (UTC)

A side effect of using very hot water (at least for the final rinse) is that your dishes and glasses dry faster, with fewer water spots. —Steve Summit (talk) 18:30, 27 October 2007 (UTC)

If the water is sufficiently hot, it will also kill some bacteria. This may not be very relevant at home, but it becomes important if you're, say, running a summer camp where keeping the dishwashing water hot enough can mean the difference between one kid with diarrhea and 30 kids with diarrhea (and one outhouse). The problem is that the water needs to be actually uncomfortably, if not quite scaldingly, hot (at least around 50°C / 120°F or so; see e.g. [3], keeping in mind that chemical sanitizers are rarely available under camp conditions). Merely lukewarm water just makes the bugs grow faster. —Ilmari Karonen (talk) 23:27, 27 October 2007 (UTC)

You all seem to forget to rinse properly. Cold dishwater will just as easily dissolve all fat if given enough time. A brush (before and after the soak) will help this process. After that, rinse with a lot of water. Assuming any food stuff is soluble in either water or fat, this will remove everything and the pan should be as clean as the water. Right? Just try it with a very greasy pan. It would be nice if someone who has a petri dish at hand would try this. Btw, there is no need to remove/kill all bacteria. Actually, the rise of allergies in the last few decades has shown that it is actually a very bad idea. And in a normal life you're exposed to all sorts of sources of bacteria, so why become all clinical here? DirkvdM 09:09, 29 October 2007 (UTC)

electron transfer

what makes an electrin to be transfered from an atom to the other(e.g metal to non metal )in ionic bonding? —Preceding unsigned comment added by 195.225.63.210 (talk) 14:06, 27 October 2007 (UTC)

Electron configurations of lithium and fluorine. Lithium has one electron in its outer shell, held rather loosely because the ionization energy is low. Fluorine carries 7 electrons in its outer shell. When one electron moves from lithium to fluorine, each ion acquires the noble gas configuration. The bonding energy from the electrostatic attraction of the two oppositely-charged ions has a large enough negative value that the overall bonded state energy is lower than the unbonded state

Systems of atoms like to be in the lowest possible energy state. When the electron moves from the metal to the non metal the total energy of the sytem goes down.

Another way of looking at it is thinking about the forces involved. Some atoms attract electrons more strongly than others. Metals tend to attract thier outer electrons less strongly than a non metal. See electronegativity for more detals. So although the electron is pupped towards the metal atom it is pulled more towards the non metal one and so hops over to it making an ionic bond Theresa Knott | The otter sank 14:29, 27 October 2007 (UTC)

After which they would each go on their own merry way, were it not that the metal is now positively charged and the non-metal positively charged, so they are attracted to each other. A bit like free sex - the exchange precedes the attraction. :) DirkvdM 18:12, 27 October 2007 (UTC)

plants

Why are some leaves spickey —Preceding unsigned comment added by 213.122.27.220 (talk) 18:22, 27 October 2007 (UTC)

A reason that comes quickly to mind is that the spiky leaves provide an adaptive advantage in protecting the plant against predation (being eaten) by animals. So, it basically acts as a self-defense mechanism in plants like holly for example. Azi Like a Fox 19:24, 27 October 2007 (UTC)

In other cases, and depending on how spiky you meant, water storage may also be a factor "Cacti have never lost their leaves completely; they have only reduced the size so that they reduce the surface area through which water can be lost by transpiration.". Our article on Leaves, specifically leaves#Adaptations, suggests that "A transformation into spines protects the plants" so my water storage comment may be a misnomer.--VectorPotential^Talk 12:33, 28 October 2007 (UTC)

What leaves are spiky? Be specific. Malamockq —Preceding comment was added at 17:17, 28 October 2007 (UTC)

The ventral posteromedial nucleus!

Here's another one of my detail questions!
On this picture the VPL borders the pulvinar, while the VPM is more in the middle. However, our own beloved Wikipedia has this picture, that puts the VPL in the middle and the VPM bordering the pulvinar. My guess is that the VPM should be in the middle (because it is called medial), and that the Wikipicture has got the VPM and the VPL mixed up.
Is my guess right or not? Lova Falk 18:53, 27 October 2007 (UTC)

Here is an actual brain section for Macaque showing the VPM medial to the VPL. This diagram also shows the VPM in a medial position. I don't think the Wikipedia diagram is meant to be very accurate....it smashes a 3D structure into a flat diagram. --JWSchmidt 04:36, 28 October 2007 (UTC)

Thank you! Lova Falk 13:48, 28 October 2007 (UTC)

Plant Absorbtion

I was wondering if plants could absorb anything besides water (ie milk, or perhaps coffee?). Would such liquids have a negative impact on plant growth or would their roots just absorb only the water part of the liquid? Thanks, Valens Impérial Császár 93 19:17, 27 October 2007 (UTC)

If a plant absorbed only water, it would die of malnutrition. Luckily for it, the water it's getting is generally dirty. Literally. I remember something about putting cellery in water dyed red with food coloring and watching the vain-like things that bring in the water turn red. — Daniel 19:51, 27 October 2007 (UTC)

You can do the same thing with flowers. Put a red rose's stem in water colored with blue ink, and you get a blue rose (according to my old biology textbook anyway). Most plants probably have a very narrow range of acidity that they can survive in. I know that mixing up used coffee grounds in soil for tomatoes helps them grow, but the caffeine in fresh coffee will kill them (so maybe decaf?). I expect that the fat content is important, which means that milk would be difficult. Most of this information (optimal acidity, hardness, etc) should be well documented for the soil, so it shouldn't be difficult to find a drink that suits a plant, or a plant that can grow in a given drink. risk 03:50, 28 October 2007 (UTC)

These are two different phenomena. Plants absorb the water and nutrients they need through their roots. When you cut the stem of a plant and immerse it in dye you are bypassing the roots, the liquid flows up through the plant by cappilary action. -- Diletante 20:48, 28 October 2007 (UTC)

How to kill Mad Cow prions?

I've heard that they're incredibly resiliant to heat, cold, UV light, pressure, chemicals and radiation. What's the best way of sterilizing something that's been contaminated with Mad Cow prions? —Preceding unsigned comment added by 84.71.69.70 (talk) 22:53, 27 October 2007 (UTC)

Our Prion#sterilization section suggets autoclaving at temperatures of 134 degrees Celsius for 18 minutes, which will pretty much denature any protein. If you want it utterly destroyed, incinerate it at even higher temperatures. If you're asking whether you can sterilize contaminated beef to an edible state, the answer is a flat no, as any method of destroying or deactivating the prion will destroy the beef. Someguy1221 22:59, 27 October 2007 (UTC)

A few years ago, I remember reading that heating contaminated medical instruments (to give an example) in concentrated sodium hydroxide or hydrofluoric acid at insanely high temperatures and pressures was the only sure-fire way of eradicating the infectious agents completely. I guess that technology has progressed somewhat since then. --Kurt Shaped Box 23:04, 27 October 2007 (UTC)

Minor point, but you can't kill a protein at all, it's not alive. If you denature a protein it can still, in most cases, be refolded under ideal conditions. See Protein denaturation, Protein folding, and of course Prion.--VectorPotential^Talk 12:26, 28 October 2007 (UTC)

Rather than temperature alone, please discuss the possibility of chemical neutralization of mad cow prions. Would bleach or diluted bleach neutralize prions? If someone fed the cat using one of the cereal bowls, and I want to make sure that prions which might have been in the "meat byproducts" part of the catfood do not remain on the dish when I eat my cereal from the same bowl sometime in the future, would a soak in bleach solution inactivate the prions? Would any other household chemical be more effective? Edison 21:01, 28 October 2007 (UTC)

You might find this paper and the references it contains useful. Of note:

"Conventional hospital disinfectants including ethylene oxide, propriolactone, hydrogen peroxide, iodophors, peracetic acid, chaotropes, and phenolics have little effect on prion infectivity.... In addition, prions are resistant to inactivation by UV irradiation, aldehyde fixation, boiling, standard gravity autoclaving at 121°C, and detergent solubilization."

As well,

"Currently recommended protocols for prion decontamination include either (i) >2% available chlorine of sodium hypochlorite for 2 h, (ii) 2 M NaOH for 1 h, or (iii) autoclaving at 134°C for 4.5 h. Each of these protocols has important limitations: sodium hypochlorite and NaOH are corrosive at the concentrations required to inactivate prions; NaOH did not inactivate CJD prions completely in some reports; and extended autoclaving at high temperature is deleterious to many materials. Currently, some high-risk surgical instruments are soaked in 2 N NaOH for 1 h, rinsed with water, and autoclaved at 134°C for 1 h, while many other such instruments are discarded."

The paper also presents a 'combined' protocol that employs an acidic SDS solution (1% sodium dodecyl sulfate plus 0.5% acetic acid) followed by 121°C that does seem to fully inactivate prions. Again, I urge you to read the full paper, and to look up the references it cites. TenOfAllTrades(talk) 21:26, 28 October 2007 (UTC)

As far as anyone knows, a human eating the meat will keep it from spreading. There's no proof humans can get infected from eating meat from a cow with mad cow disease. — Daniel 00:29, 29 October 2007 (UTC)

Just to inject a little realism....any idea how many people have contracted vCJD (human form of mad cow disease)? Only about 160, total, worldwide. Almost all of these lived in the UK during the mad cow epidemic. The number of cases annually, world wide, is quickly approaching zero. You know how many domesically born cattle have been found in the US with mad cow disease? 2!! [4]. In my opinion, this shouldn't be high on your list of worries. ike9898 21:12, 29 October 2007 (UTC)

When our research group needs to really and truly destroy protein residues we bake metal or glassware at 400 C for 2 hours in oxygen. This converts all the carbon to carbon dioxide. Of course, there are lots of other materials which also won't survive being steralized in this way, so its practical utility may be limited. Dragons flight 09:16, 30 October 2007 (UTC)

Dropping uncertainties

When dividing two numbers with uncertainties, the end result should have a greater uncertainty right? However, when I am doing the following:

• [(2.91 ± 0.01g) / (2.55 ± 0.01g)]

I end up with:

• 1.14 ± 0.008g

which is lower than the two original uncertainties I had. Could this be correct?

Thanks. Acceptable 23:15, 27 October 2007 (UTC)

When dividing two numbers with uncertainties, the end result should have a greater relative uncertainty, which your answer indeed does. However, I get 1.141 ± 0.006 for the answer. Did you make an arithmetic error? —Keenan Pepper 23:30, 27 October 2007 (UTC)

Ah yes, a greater relative uncertainty makes more sense. I did the calculation several times and I'm almost certain that 0.008 is correct. Much thanks. Acceptable 23:42, 27 October 2007 (UTC)

I actually get something a bit higher than 0.008:

Max value = 2.92/2.54 = 1.1496

Min value = 2.90/2.56 = 1.1328

Median value = (1.1496 + 1.1328)/2 = 1.1412 ± .0084. StuRat 18:50, 28 October 2007 (UTC)

But unfortunately Stu that is not how errors are calculated, although it seems logical. The errors in science are not upper and lower absolute bounds but probabilities basd on assumed normal distributions. Our page on error propagation gives the general formula for calculating errors. Under the specific examples you can see that for a ratio, the fractional errors combine in quadrature. I get 1.1412 +- 0.0068. Cyta 21:16, 28 October 2007 (UTC)

Which, if rounded to 2 decimal places (the error, that is), would give 1.1412 ± .007. -- JackofOz 00:14, 29 October 2007 (UTC)

How did you get 0.0068? The relative uncertainty of the numerator is 0.01 / 2.91 = 0.0034. The relative uncertainty of the denominator is 0.01 / 2.55 = 0.0039. So the relative uncertainty of the ratio should be sqrt(0.0034^2 + 0.0039^2) = 0.0052. The ratio is 1.141, so the absolute uncertainty should be 0.0052 * 1.141 = 0.0059, yielding 1.141 +/- 0.006. Right? —Keenan Pepper 22:06, 29 October 2007 (UTC)

the problem here is that the conventional nomenclature fro uncertainty (x±y) appears to be unambiguous, but actually is not.

You must explicitly define what you mean when you use the "x±y" Nomenclature. For example, if "(x±y)" means "x, with an assumption of a normal distribution with an SD of y" then "(x±y)/(z±w)" has a defined meaning. If "(x±y)" has a different meaning, then "(x±y)/(z±w)" has a different meaning. Note that you need to understand both the statistics of the numerator and the statistics of the denominator before you can have any confidence in the statistics of the result.-Arch dude 03:21, 29 October 2007 (UTC)

Excellent point, Arch dude. However, in the absence of any other statistical information it's reasonable to assume the variables are independent and normally distributed (or log normal, which is practically equivalent if the standard deviation is small compared to the value, as in this case). So adding the relative errors in quadrature is a good rule of thumb. —Keenan Pepper 22:06, 29 October 2007 (UTC)

October 28

Red Onion Skinn cells

In Biology class, we observed a piece of red onion skin under a 40X objective lens and probably a 10X eye piece. A drop of salt water was added to the skin under the microscope, the red part of the cell shrunk. Is the red part the vacuole or cytoplasm of the cell? Acceptable 01:46, 28 October 2007 (UTC)

Usually, plant dyes (and other secondary metabolites and end products) are stuffed into vacuoles by the plant cell. They would wreak havoc with usual plant metabolism taking place in the cytoplasm. --85.179.20.169 08:36, 28 October 2007 (UTC)

Fall question

I got to wondering why tree leaves aren't all the same shape. You would think there'd be an ideal leaf shape that all trees would have settled down to, but the variety is practically infinite. Are there evolutionary pressures on tree leaf shape by species, or is it accidental? In other words, are certain characteristics of an organism of no importance for suvival, and do these characteristics thus more or less reflect some underlying symmetry or other order that is merely a manifestation of their peculiar origins or mechanisms? I hope that made sense. --Milkbreath 02:23, 28 October 2007 (UTC)

Few (if any) things are accidental when it comes to evolution, although there is always some 'noise', to allow a species to find a new optimum when the environment changes. On a large scale you can look at the difference between pine needles and leaves. Pine needles allow trees to survive very cold climates, whereas regular leaves allow trees to get the maximum amount of photosynthesis in more temperate climates. I expect it works the same way for differences between leaves. The shape of the leaf is the ideal solution for the area where the tree lives in terms of getting the most sunlight, surviving the cold, the heat, and so on. If two trees live in the exact same area, they may have different survival strategies. One tree may get the optimal amount of nutrition from the location, whereas the other may reproduce faster, expand it's habitat faster, or be better at surviving changes in the environment. I'm not an evolutionary biologist, but I figure that's how it works. And there is always the possibility that everything else being equal, there are several optimal solutions, and two trees arrive at different ones. risk 03:41, 28 October 2007 (UTC)

As Risk says, there's lots of different things that a tree needs to deal with: snow, wind, sun, heat, cold, rain, storms. Some need leaves that shed snow, others need leaves that aren't destroyed by cyclones, others need leaves that won't loose too much water in dry weather. Lots of variables means lots of different leaf shapes make lots of different "ideals". What about two trees in the same forest? Well they may also grow in disparate areas, and they may survive or thrive in different situations differently. (oh, and it's spring here - all our imported trees are starting to sprout their myriad leaves in myriad shapes. All our local trees kept their leaves over winter, so we here think it's funny to call that season "fall") --Psud 09:07, 28 October 2007 (UTC)

Most of the factors which affect leaf growth have been mentioned, but let me add disease, insects, and animals which eat or damage leaves. Unless the animal provides some benefit to the plant, like spreading seeds for it, the plant will want to minimize leaf damage. Having many small leaves may help here, as large animals won't bother with small leaves, and the leaves can be easily dropped if they become diseased. Small leaves would also fare better in strong winds, but wouldn't do as much photosynthesis, due to all the gaps between the leaves for sunlight to pass through. StuRat 18:26, 28 October 2007 (UTC)

Leaves are a critical weapon the the war between the trees and the grasses. Grasses and trees try to poison each other, and deciduous trees try to kill grass by annually blanketing the grass with leaves. The leaf shape and size is of critical importance in the success of this strategy. it is deeply involved with the activities of herbivores, so the evolutionary response times will ensure a large range of leaf shapes and sizes. The suburban ecosystem with Maple trees andblue grass is profoundly unnatural-Arch dude 03:06, 29 October 2007 (UTC)

An additional thought is that it is a mistake to assume that everything about a plant (or other organism) has been determined by evolution. Natural selection will only affect those traits that change the survival rate of the species. Sometimes that differential survival critically hinges on leaf shape (needles in evergreen trees with longer winters, waxy, pointy-tipped leaves in plants with areas of high rainfall, etc), but sometimes it does not. For example, a maple leaf and an oak leaf are quite distinct to our eyes, but are probably not all that different in terms of practical survival rates for the individual plants and their offspring. Matt Deres 16:29, 29 October 2007 (UTC)

Butterflies and the fate of the universe

A flap of the wings of a butterfly in India will, given enough time, alter the course of a tornado in the US. Not to put too fine a point on it, but it will inevitably change everything about the world. Will it do the same for the universe? Sappysap 03:54, 28 October 2007 (UTC)

• The way I understand it, that's not actually true. Rather than being taken literally, I think that statement is supposed to be a refernce to the fact that the causes of something like a tornado are so complex that minute events (eg the butterfly) that might not otherwise be associated with the outcome are related. 68.18.209.108 04:42, 28 October 2007 (UTC)

The butterfly effect is a vey specific illustration of the fact that the Earth's atmosphere is a complex and interconnected system in which small changes are quickly magnified. So attempts at predicting weather patterns more than a few days ahead are doomed to failure because we cannot practically collect enough starting data with enough precision to produce accurate estimates. By extension, it has become a metaphor for any natural system that exhibits sensitive dependence on initial conditions. It does not mean that the flap of the butterfly's wings will change everything about the world. Although the butterfly's wings could (in theory) change the course of a torndao, we know of no physical mechanism by which they could (even in theory) cause an earthquake, significantly affect the path of an asteroid, or change the temperature of the Sun. A belief in mysterious and undiscovered mechanisms which intimately link every feature of the universe is more akin to astrology than to science. Gandalf61 07:28, 28 October 2007 (UTC)

Let's put it in perspective. Here's the closest thing to affecting the Universe. Our Indian butterfly flaps it's wings, that causes a tiny (but important in our chaotic weather system) change in the course of a storm in Sydney. A lightning strike happens in one place rather than another and sends a radio pulse out into space. That pulse of radio probably doesn't go very far into the universe, and on the few occasions where it does encounter something, it has practically no effect. Then you consider that the lightning probably would have happened anyway, just somewhere a little way away from where it did happen, and that lightning flash would have been effectively identical from even a fraction of a lightyear away (if it was even detectable at that distance) --Psud 08:58, 28 October 2007 (UTC)

Alternatively, that new lightning strike hits the person who would later have discovered a viable way to extract energy from fusion on a small scale, which would lead on to globe-spanning revolutions in virtually every field of engineering. The technology would've allowed the human race to construct interplanetary and eventually interstellar transportation, and ultimately lead to the colonisation of the galaxy. Which would have an observable effect from that distance. "For want of a nail" and all that. GeeJo ^(t)⁄_(c) • 11:02, 28 October 2007 (UTC)

I feel I should point out that one needs a very strange view of the world for this question to even make sense. In order to pin the tornado on the butterfly you have to suppose that the butterfly's action could have been otherwise, but that everything after that just snowballs according to deterministic (if chaotic) laws. Future choices by other butterflies would presumably dilute the long-term meteorological influence of this one butterfly to the point where it didn't matter whether it flapped its wings or not. The analogy does make mathematical sense as a statement about initial conditions of chaotic systems, but it doesn't make physical sense unless you subscribe to some form of last Thursdayism. -- BenRG 14:39, 28 October 2007 (UTC)

That's a switch—pinning something on a butterfly. --Milkbreath 15:27, 28 October 2007 (UTC)

The problem here lies in one misstated word in the question: A flap of the wings of a butterfly in India will, given enough time, alter the course of a tornado in the US. - not will but could (with a very low probability). There are a bazillion butterflies and a bazillion wing beats - and many, many other sources of atmospheric turbulance - any and all of which are having some effect. It is the net effect of all of them that causes unpredictability. So had that butterfly not flapped at that exact moment, then perhaps the tornado halfway around the world and a decade off in time might not have happened. It's definitely possible - in a chaotic system such as our atmosphere, there is infinite sensitivity to these kinds of events. But for any specific wing beat of any specific butterfly, the odds are truly astronomically small. As to whether the event could cause universe-wide changes - yes, sure. I like GeeJo explanation as to how this could come about. It's definitely possible for a butterfly wing flap to deeply affect something over the other side of the universe a very long time into the future. Again - it's very unlikely - but for sure it can happen. SteveBaker 16:34, 28 October 2007 (UTC)

There are limits to the extent of the effect of a butterfly’s actions. Some parts of the universe can’t possibly ever be affected by a flap of a butterfly’s wings. Due to the inflation of the universe, most of the universe is beyond what can ever be reached by a photon leaving our galaxy now. MrRedact 19:09, 28 October 2007 (UTC)

And, depending on how you look at all this, you can even say "very likely" instead of "very unlikely". In a chaotic system, anything can affect anything; everything affects everything. So the likelihood is very high that "event X" (whatever it is) is caused (in part) by some ridiculously distant cause that you can't even imagine. What's very unlikely, of course, is that event X was caused by a particular ridiculously distant alleged cause Y. —Steve Summit (talk) 16:51, 28 October 2007 (UTC)

The answer is YES. Here is a fun homework problem from graduate statistical mechanics. Imagine that you knew the initial position and velocity of every particle in the universe, except that you misplaced one electron 4 light years away by 1 cm. How long does it take before that simple error, by virtue of the electron's gravity, gives rise to sufficient chaos that you can no longer predict the position of gas particles in front of you? It's an extremely small error, but it is magnified by each of umpteen interactions between the gazillion particles in any given air parcel so that appreciable chaos sets in after only a few minutes (if my memory serves). So yes, that damned butterfly will introduce chaotic effects in sensitive systems throughout the rest of the observable universe. Dragons flight 09:43, 29 October 2007 (UTC)

How does that chaos set in before the electron's misplacement can propagate (at the speed of light) to the origin? You might mean that the area around the electron becomes chaotic, but then what's the "4 light years away" about? Or perhaps you mean that the electron's entire electromagnetic field is appropriate to its actual position (as opposed to it, at $t=0$, suddenly being magically shifted by that centimeter), but I have a hard time seeing how you could have that field interact with everything else and yet have the wrong idea about its source's location. It's an interesting question — with what shape and speed does the "wave" of chaos from a disturbance propagate? — but I'm not sure how to get a quantitative answer from what you've said. --Tardis 16:08, 29 October 2007 (UTC)

Technically, it's an error in the local, t=0 gravitational field corresponding to a 1 cm error in what the position of an electron at 4 light year's distance would have been at time t=-4 years. For simplicity, I wasn't describing the time delay. Dragons flight 18:58, 29 October 2007 (UTC)

Anyway, how come it's always tornadoes? I want a butterfly to flap its wings in India, and a beautiful long-legged nymphomaniac to make a wrong turn and end up in my living room. —Steve Summit (talk) 16:54, 28 October 2007 (UTC)

Well, the actions of a butterfly could cause a long-legged nymph to turn up, but that's another story... Laïka 20:24, 28 October 2007 (UTC)

If you find it comforting that the likelyhood of either nymph or nymphomaniac (that's someone who is really enthusiastic about immature insects - right?) turning up in your living room is influenced by such things - then I strongly recommend butterfly collecting. SteveBaker 22:00, 28 October 2007 (UTC)

I think astronomical systems are still chaotic, but not nearly as much as terrestrial ones. For example, if two meteors pass close by each other, a tiny difference in there original position will make a much larger difference in there final path. Unlike terrestrial systems, however, these are incredibly rare and it will take a huge amount of time for the gravitational effect of a change in the earth's weather patterns to make a noticeable difference. Also, the result of a butterfly effect is the difference between one thing that doesn't seem out of the ordinary and something else that also doesn't seem out of the ordinary. A butterfly flapping its wings will change the weather forever, but the climate will remain the same. — Daniel 23:12, 28 October 2007 (UTC)

The Butterfly effect is an example of Chaos theory. Some physical phenomena can affect "initial conditions" in ways that are unpredictable even in theory. It is possible to construct a sequence wherein the fate of the universe depends on the indertimanite state of a subatomic particle that affects, via a cascade of events, the movement of a butterfly's wing. In practice, this will not happen in this universe. -Arch dude 02:49, 29 October 2007 (UTC)

Tartaric Acid

Do you know the boiling point of Tartaric acid? The melting point is 168°C - 170°C, but I cannot find the boiling point. All of the information that I could find is on this page: http://en.wikipedia.org/wiki/Tartaric_acid Thank-you for your help. 203.113.233.115 07:31, 28 October 2007 (UTC)

This site (bottom of page) lists the boiling point at 275°C: [5]. However, many other sites don't list a BP because most of the acid will decompose before it reaches that temp. StuRat 18:14, 28 October 2007 (UTC)

definition of orthologous , paralogous, analogous,homologous

definition of orthologous , paralogous, analogous,homologous —Preceding unsigned comment added by Sujbhaskar (talk • contribs) 11:35, 28 October 2007 (UTC)

See wiktionary:orthologous, wiktionary:paralogous, wiktionary:analogous and wiktionary:homologous. Algebraist 12:18, 28 October 2007 (UTC)

And Homology (biology) at wikipedia. Algebraist 12:20, 28 October 2007 (UTC)

Mushrooms

Is a mushrooms considered a decomposer or what? —Preceding unsigned comment added by 209.244.30.199 (talk) 17:12, 28 October 2007 (UTC)

The mushroom itself is not a decomposer... the mushroom is only the spore-producing fruit-like part of a fungus. The fungus decomposes organic matter. Sancho 20:34, 28 October 2007 (UTC)

If I understand right, not all fungi are decomposers. A decomposer, according to that page, consumes dead organisms. Certainly a lot of mushroom type fungi grow on dead plants, logs, etc, but aren't there also a huge number of mushroom type fungi that are symbiotic with plants? I'm thinking of Mycorrhiza type relationships. My (admittedly vague) understanding is that most land plants depend upon mycorrhizal fungi for their very survival. When you see mushrooms growing near a tree, are they getting their food from some dead organism underground or are they symbiotic with the tree, each helping the other? My guess is quite often it is the latter. Pfly 02:01, 29 October 2007 (UTC)

From the respective articles: "Decomposers are organisms that consume dead organisms, and, in doing so, carry out the natural process of decomposition." and "Decomposition (or spoilage) refers to the reduction of the body of a formerly living organism into simpler forms of matter." We eat dead things and reduce them to simpler forms of matter so we can use them for our own bodies. So aren't we (and all living things) decomposers? Actually, doesn't dead nature do that too? I'll ask that in a new thread. DirkvdM 09:40, 29 October 2007 (UTC)

fibrous and synovial joints

Are there two types of joints between the radius and ulna? I was lead to believe they are joined by a fibrous joint. Another article tells me they have a synovial joint to allow supination and pronation. Can anyone clarify please? kramnahtal —Preceding unsigned comment added by Kramnahtal (talk • contribs) 19:07, 28 October 2007 (UTC)

Pinhole cameras

How does a pinhole camera project an image? I thought you needed a lens of mirror to focus light. I read pinhole camera but couldn't find the answer. 72.155.207.79 19:47, 28 October 2007 (UTC)

The pinhole itself acts as the lens. The second sentence of the article states this clearly: "An extremely small hole in a very thin material can focus light by confining all rays from a scene through a single point." I assume you mean "glass" instead of "mirror"; there is no mirror needed in any camera. --24.147.86.187 20:25, 28 October 2007 (UTC)

I understand that. I'm asking how air focusses light. And a curved mirror can focus light, and mirrors are most certainly used in cameras. 72.155.207.79 20:37, 28 October 2007 (UTC)

A mirror is not needed in cameras. Yes, you can set up Newtonian-telescope style lenses but that's hardly standard. Maybe you meant "lens or mirror" up above, I now see. --24.147.86.187 21:13, 28 October 2007 (UTC)

There is no real focusing of an image in a pinhole camera. It would work even in a vacuum. This is because, if you pretend for a moment that the pinhole is only large enough for one photon to make it through, each position on the film could only have been illuminated by light coming from a single direction through the pinhole. Thus, if you place an illuminated 2D picture on the outside of the pinhole camera, each position on the film corresponds to only one position on the picture outside. So inside the camera you get a nice, clear image of what's outside. Someguy1221 20:56, 28 October 2007 (UTC)

The air doesn't focus anything. The pinhole acts as a sort of collimator, if you will. Take a pinhole camera with the front off, and aim it at the calendar page for October. Consider a point on the back of the camera where the image will be. Let's locate the point about halfway between the center and a side at three o'clock. Now shrink your eye down to microscopic size and put it on that point facing the front of the camera. You will see the whole calendar page. Now put a front on the camera that has a round hole in the middle about an inch across. You'll only be able to see a few dates off to the other side. Put a front on with a smaller hole, and you'll see only one day. Make the hole very small, and the only light ray that will be hitting your eye will be coming from a tiny spot. The same goes for all the other microscopic eyes stuck to the back of the camera, but each will see a different spot on the calendar page. Voila! You've got an image, upside down and backwards. --Milkbreath 21:09, 28 October 2007 (UTC)

Do this for me (don't just think about doing it - actually DO it!). Take a piece of paper and draw on the left a stick-figure to be the subject of our photograph. On the right of the page, draw a vertical line representing the photographic negative inside the camera.

Now (without the pinhole) consider light rays coming from the subject onto the negative. Light is emitted pretty much equally in all directions from every part of the subject. So you can draw straight lines representing rays of light radiating out from our stick-figure's head and going out in all directions - lots of them hit the photographic negative - and they hit it all over it's surface. Similarly, you can draw rays radiating out from the stick figure's feet reaching any point on the negative. In fact, light from everywhere in the scene can reach every place on the negative - so all you get when you develop the plate is the average of all of the light from the whole scene hitting every point on the negative. A big white blur in fact.

OK - so let's add a pinhole "lens" to our camera. Start again with another diagram) - put the subject on the left and the negative on the right just as before - but this time, draw a vertical line down the middle of the paper with a tiny gap halfway up - this represents the front of the camera with a pinprick-sized hole in it.

Now, lets draw those lightrays coming from the stick-figure's head. They still shoot out in all directions - but most of them hit the line going down the middle of the page and are stopped. Only a few rays make it through the gap in the middle of that centerline (the 'pinhole') and onto the negative. Notice that rays from the top of the stick-figure only hit the bottom of the photographic plate over to the right of your diagram. Now do the same thing with rays from the stick figure's feet - notice that ones that go through the pinhole only hit the top of the photographic plate. Do this with rays of light from all over the stick figure and you find that each point on the original subject emits light in all directions - but the pinhole shuts all of it out except for a teeny tiny amount - so light from each point on the subject ends up at a different place on the negative. When you develop the plate - you get a nice, sharp image because the light isn't all mixed up like it was without the pinhole.

You can deduce some other things with this kind of simple diagram: If the pinhole is too big, lightrays from several close-by points on the subject can end up on the negative in the exact same place. This results in a blurry (but perhaps still recognisable) image. But if you make the pinhole too small, almost all of the light rays from the subject will hit the line down the middle of your piece of paper - hardly any make it through that teeny-tiny gap in the middle. This means that the image on the photographic plate is rather dim (because not much light hits each point)...so you need either a more sensitive film or a much longer exposure (which in turn results in a blurry image if anything in the scene is moving or the camera is shaking).

The point is that the pinhole forces all of the light in the scene to pass through a single point - and blocks all of the light that was heading off in the wrong direction. A more conventional lens focusses light through a single point (the 'focal point') - which has the same effect as the pinhole. The benefit of the lens is that it gathers light over a larger area - so you can get more light onto the film without ending up with a blurry picture. The bigger the lens, the better the quality. This is why cellphone cameras are so crap compared to big SLR cameras. Pinhole cameras do have one HUGE advantage though - they keep the image in focus no matter the distance of the subject from the camera - lenses can't do that. Everything in photography is about these kinds of trade-off.

SteveBaker 21:13, 28 October 2007 (UTC)

Incidentally, this is the same mechanism by which squinting your eyes sharpens an image—you're effectively forcing light through a smaller pinhole. TenOfAllTrades(talk) 21:30, 28 October 2007 (UTC)

There's an episode of Home Improvement in which Wilson has Mark fashion a pair of glasses out of a piece of paper with two pinholes in it. Sancho 22:31, 28 October 2007 (UTC)

Thank you very much for your clear and helpful comments, and sorry for the misunderstanding, 24.147.86.187; I did mean or. 72.155.207.79 22:33, 28 October 2007 (UTC)

Incidentally there was a BBC show called "Genius of Photography" on the other night (Uk), it made a camera obscura out of a room. They went in covered it till it was pitch-black and cut a small hole. Projected onto the wall upside-down (and reversed?) was the view from out of the window. It was very impressive - possibly available on You-tube or that bbc-download thing. ny156uk 23:18, 28 October 2007 (UTC)

The very first response to this question began, "The pinhole itself acts as the lens." That's wrong.

For convenience let's talk about a specific use of a lens or pinhole: the use where you have an object and you're projecting an image of it onto a wall. In order to make the image, you must arrange the rays of light in such a way that each point of the object corresponds to one point of the wall.

By making the light pass through a pinhole, you achieve exactly that correspondence by simple geometry: from each point on the object, there is a line of sight through the pinhole to exactly one point of the wall. That's all there is to it! If the pinhole is point O, then point A of the object corresponds to point A' on the wall were AOA' is a straight line. Point B of the object corresponds to B' on the wall, where BOB' is a straight line.

But the downside is, the whole image is formed from the few rays of light that happen to be pointed at the pinhole. That's okay if the object is very bright, like the Sun, but otherwise you need the wall to be in darkness (a camera obscura), and even so the image is not all that bright.

The purpose of using a lens is to overcome that downside. With a lens, instead of using the few rays of light that come off the object and aim at a little pinhole, we can use all the rays that come off the object and land anywhere on the lens. By choosing the correct lens and placing it correctly, we can arrange to have light paths like APA', AQA', ARA', where P, Q, R are different points on the lens. And similarly BPB', BQB', and BRB'. And so on for all the other points on the object and all the points on the lens. These paths are not all straight lines, but the lens is curved in such a way that the light follows them anyway. That's what focusing means -- this bringing back together of the rays AP, AQ, and AR to a single point A'. And you need a lens (or a curved mirror) to do it. It gives you a much brighter image, the total brightness corresponding to the area of the lens.

But with a pinhole, you don't need to focus in the first place. The pinhole sets up the correspondence between points on the object and points on the wall by geometry alone. It doens't act as a lens because in this setup there is no need to focus.

(In all this I am pretending that the size of the pinhole is negligible. In real life it can't be zero size, and this limits how sharp an image it can form. This is another advantage of a lens.)

--Anonymous, 10:00 UTC, October 29, 2007.

Esophageal Temperature Monitoring During Surgery

It is fairly common practice to monitor a patient's body temperature during a surgical procedure with the use of an esophageal stethoscope with a temperature wire inside of it. Does the temperature wire hook up to a machine that reads the data and then displays a readout? Or how does that work? Also, is it possible to lose the esophageal stethoscope in the esophagus? Thanks, Lilly Upstairs —Preceding unsigned comment added by 24.19.72.81 (talk) 22:03, 28 October 2007 (UTC)

Effective Projected Luminous Lens Area

This term is found in Federal Motor Vehicle Safety Standards for testing of lights for motor vehicles. I am trying to understand this term (EPLLA) in the context of a motorcycle turn light. I want to know how to measure this EPLLA and determine whether or not my turn lights are over the required 3.5 square inches required by the regulation. Can someone explain to me in somewhat laymen's terms what Effective Project Luminous Lens Area is and how to measure it? Thank-you 154.20.86.223 22:50, 28 October 2007 (UTC) Ray Kwan

Ray, EPLLA means the area of the effective light-emitting surface of a lamp, measured by determining the area of the 2-dimensional graphic representation of the lamp's lit lens area on a plane perpendicular to the lamp's reference axis and touching the most exterior point of the lens. "Reference axis" means the H–V axis used for photometric requirements, i.e., the effective centre of the lamp's beam pattern as produced at the lamp.

To simplify this and bring it into the realm of practical application outside of a compliance laboratory setting: The reference axis of most automotive and motorcycle lighting devices is reasonably easy to determine; it is "straight back" from the device with the device oriented in space exactly as it is oriented when installed on the vehicle. if you will place a sheet of fine-grid graph paper such that it forms a vertical plane at 90° to the lamp's reference axis, move the paper such that it just barely touches whatever part(s) of the lens protrude closest to the paper, illuminate the lamp, trace the blob(s) of light on the side of the paper opposite the lamp, and then calculate the area enclosed by your trace, you will have a close enough approximation of the device's EPLLA for most purposes. The thing you will have to be most careful of is accurately tracing the blob(s) of light. If your device uses multiple light sources, such as a cluster or array of LED emitters, you must trace each individual spot of light, omitting the dark areas in between, then total up the areas of each trace to arrive at your EPLLA.

It sounds like you've modified your motorcycle's directional indicator system in some fashion and are trying to determine if the modified or handmade indicators meet the EPLLA requirements. Good for you, most people don't bother, but be aware there are other safety performance requirements for vehicle lighting devices, as well. Intensity through various vertical and horizontal angles, intensity ratio between bright and dim modes of a park/turn or brake/tail lamp, etc.

For additional explanation, you may want to read this technical bulletin as well as this one and this one, keeping in mind that they primarily make reference to the 7¾ in² and 11⅝ in² EPLLA requirements for passenger car and large-vehicle brake lamps, respectively. Also, you may want to read through this NHTSA rulemaking discussion which goes into detail on the meaning, intent, and methods behind EPLLA requirements. If you wish to discuss vehicular lighting with more specificity, feel free to contact me via my talk page. --Scheinwerfermann 03:35, 29 October 2007 (UTC)

October 29

Batch experiment for denitrification endpoint in wastewater treatment

I am told the endpoint for denitrification in biological wastewater treatment can be determined by a batch experiment. What is the step-by-step procedure for conducting such an experiment and are there electronic sensing techniques for obtaining the same information in a continuous fashion? Thanks Thinkaboutlife 03:43, 29 October 2007 (UTC)

Gas formation at extreme low pressure and temperature

This question about wind at the miscellaneous desk spinned off into extreme planetary circumstances, with the planet so far from its star (if any) that it receives no heat (well, of course there is no absolute zero for anything). So I looked up the freezing point of helium, which is 1.15 K at 66 atm. Which would mean that even at extreme low temperatures, there would still be an atmosphere (however thin) of helium because of the vacuum surrounding it. But what about other gases? I guess I'm asking about the bottom left corner of the phase diagram (now why didn't I learn about that at school?), so extreme low pressures or a vacuum. But the article doesn't deal with that extreme. Let me start with one extreme:

• Would uranium be gaseous in outer space?
• And on above planet, at, say, an extremely low 1 Pa, what would be the boiling points of the various elements? The graph at the article doesn't go below 0.1 atm (= 10,000 Pa). And then at 10 Pa, 100 Pa, etc?

I suppose the atmosphere (which has no wind - the point of the original question) would be built up in layers consisting of the various elements with Helium sitting at the top. DirkvdM 07:18, 29 October 2007 (UTC)

On a phase diagram, you will notice that water is liquid at 1 atmosphere and room temperature. However, we also observe that water will evaporate from lakes and puddles. This is because even though the substance is naturally a liquid under those conditions, some fraction will spontaneously change into a gas through a process described by the vapor pressure. This is true for all substances. Uranium, like most metals, is naturally a solid in the limit of zero temperature and zero pressure. However, at all temperatures above 0, some small fraction of atoms will spontaneously turn into a gas and escape since there is zero ambient pressure. For most solids, the rate at which this occurs is neglible not only compared to human experience, but also compared to the age of the universe. In other words, the amount of time before satellites "boil" away in the vacuum of outer space is truly astronomically large. Dragons flight 11:52, 29 October 2007 (UTC)

Another way to think about that is that the concept of 'temperature' is really only a statistical average of the speeds of all of the molecules in the material. Even at fairly low temperatures, some molecules will be moving quite quickly (although the majority are not). Those few fast moving molecules may have enough kinetic energy to escape the surface no matter what. So all things with temperatures above absolute zero (which is to say: "all things" since absolute zero is unattainable) will lose molecules at some rate. But the statistics of the thing mean that the rate will be exceedingly slow at low temperatures because the probability of molecules having enough energy to escape becomes very low indeed. At higher temperatures that probability increases and evaporation will happen to a much greater degree - and finally, at the boiling point of the material, all of the molecules have enough kinetic energy to shake themselves loose and escape to form a gas. SteveBaker 14:06, 29 October 2007 (UTC)

Broad spectrum antibiotics

I know that Penicillin is a broad spectrum antibiotic, but was wondering if all antibiotics ending in cillin are broad spectrum. I couldn't find the answer here on Wikipedia or online, so thought I would ask here. Thanks. Jeffpw 08:52, 29 October 2007 (UTC)

Penicillin is not a broad spectrum antibiotic. It covers a relatively narrow spectrum of organisms, and is useless in Gram negative infections. In fact, most of the "-cillin" drugs were developed in order to find drugs with a broader spectrum than penicillin. You will find more details in the penicillin article, which lists narrow, moderate, and extended spectrum penicillins. - Nunh-huh 13:13, 29 October 2007 (UTC)

Drop dead on the Moon

If you'd drop a dead body on the Moon, what would happen to it? Would it decompose? DirkvdM 09:43, 29 October 2007 (UTC)

It depends on where you drop it on the moon. If you drop it in the dark side, it'll freeze and if you drop it on the sunny side, it will fried. 58.109.93.128 10:23, 29 October 2007 (UTC)

Hmmm .. I assume "dark side" means the far side of the Moon, i.e. the side facing away from Earth, which gets just as much sunlight (two weeks on, two weeks off) as the side facing Earth. I blame Pink Floyd. Anyway, a similar question has appeared on the RD in Nov 2004 here> There is also a response on Ask Yahoo here. General consensus seems to be that there would be some decomposition due to the presence of anaerobic bacteria, as well as dessication and slow weathering as a result of temperature changes. Gandalf61 10:39, 29 October 2007 (UTC)

Actually, on the album someone says "There's no dark side of the Moon, really." but then goes on to say "As a matter of fact, it's all dark.". Which is another bit of nonsense. DirkvdM 11:48, 29 October 2007 (UTC)

Not at all; it is dark. With an albedo of .12, it's only as bright as worn asphalt. We probably think of it as bright because there are no clouds or atmosphere to shade it from the direct sunlight. Matt Deres 16:40, 29 October 2007 (UTC)

The albedo of worn asphalt is only slightly lower than that of bare soil on Earth (fresh asphalt, on the other hand, has an albedo of just 0.04). But what I meant was is that half the Moon is lit by the Sun and therefore not dark. But SteveBaker gives yet another interpretation below. DirkvdM 09:11, 30 October 2007 (UTC)

Over eons, the top several meters of the lunar regolith are stirred up via the action of micrometeorites, through a process known as lunar gardening (wow, a red link). Since there is no appreciable atmosphere on the moon, even dust flecks the size of sand grains will leave tiny impact craters (since they are moving at several km/s). Over the very long term (i.e. many tens of millions of years) such micro impacts would ultimately grind the body down to nothingness. Dragons flight 10:50, 29 October 2007 (UTC)

Ok, looks like it's a race between the bacteria in the body, dessication and being bombarded to bits. I suppose the lunar gardening is the slowest process. How much will the bacteria consume before the body is dried out and the bacteria die too (or become dormant at best)? Also, which bacteria would do this? By far most bacteria in our bodies are benevolent, but would they start decomposing the body after it dies? DirkvdM 11:48, 29 October 2007 (UTC)

Oh boy, there are some confused answers here! Let's try to clarify things a bit...

• The term "dark side of the moon" refers to the side of the moon that is never seen from earth. In this context, the word "dark" is as in "Darkest Africa" - it means "unknown", not "lacking sunlight". The far side of the moon is just as sunny as the near side (and it's not even unknown anymore - we have lots of photographs of it). However, there are places in some craters on the moon that never see sunlight where things stay perpetually frozen. The length of a 'day' on the moon is what we'd call a "lunar month" - about four weeks.
• Since either freezing or baking to the point where it would kill bacteria would be unlikely to take more than a few hours (the temperature in the sunlight is enough to melt lead). I'm pretty sure that bacterial action is a non-problem in something as thermally conductive as a corpse.
• So unless this body is tucked away on the edge of one of those very steep-rimmed craters, it would alternately freeze for two weeks then cook for two weeks. I'd say that dessication/mummification would be the most immediate effect (that's what happens to bodies left on cold/dry mountaintops) - but it's hard to know for sure. There is no oxygen - so the body wouldn't combust in the heat - but it would drive out all of the water.
• As for being bombarded by meteors - that could happen eventually - but it would be a very improbable event. There are no more meteors (per square mile) hitting the moon than there are entering the earths atmosphere. The reason the earth isn't pockmarked like the moon is because the air burns up the smaller meteors and the craters created by the larger ones get eroded by wind & rain and subducted by continental drift - where on the moon they stay intact indefinitely. Look up into the sky on a dark night - how many meteor streaks do you see? Do you think one of them would hit you if it didn't burn up? Well, the odds are about the same on the moon...a little less actually because the earth's stronger gravity is going to pull them more towards the earth than the moon.

SteveBaker 13:52, 29 October 2007 (UTC)

• "... the temperature in the sunlight is enough to melt lead" - huh? This page says the moon gets up to 123°C, and our article says lead melts at 327°C. How would lead melt on the moon? --Sean 14:34, 29 October 2007 (UTC)

I was about to make the same point as TotoBaggins - sunlight on the Moon is definitely not hot enough to melt lead. Granted that 123^oC is still too hot for most bacteria to survive, but for hyperthermophiles such as Strain 121 it is just right. I agree that bacterial decay won't be a significant factor for our Moon corpse, but I don't think we can rule it out completely. Gandalf61 14:40, 29 October 2007 (UTC)

Well actually, without air - the temperature depends completely on the reflectivity/emissivity of the object (it's ability to shed heat by radiation) - the moon rock has an albedo of 0.12 - meaning that it only absorbs 12% of the incoming sunlight and reflects the rest away so the moon itself stays relatively cool. Darker objects will get a lot hotter. But I meant that figuratively rather than literally. I apologise, I should be more careful. SteveBaker 18:13, 29 October 2007 (UTC)

Steve, you've got albedo backwards in this paragraph. An albedo of 0.12 means that it reflects 12% of the light, so it absorbs the majority. -- Coneslayer 19:59, 29 October 2007 (UTC)

Steve, seriously, it's not an issue of "maybe it will be hit by micrometeorites", but a certainty. Objects below about 1 cm in size never get hot enough to leave visible streaks in the sky, but the population mass density of space debris peaks at about 100 microns, and the rate is about one micrometeorite impact per km^2 per second. The Earth aquires some 50,000 tons of space dust per year. And yes, as astronauts discovered, the entirety of the lunar surface is well-mixed to a depth of several meters by the action of meteorite impacts over tens and hundreds of millions of years. Dragons flight 15:32, 29 October 2007 (UTC)

Yeah over a very long period of time, the body will be bombarded to nothingness - but we're talking in the millions of years range. Let's examine your statistic of 50,000 tons of material arriving per year - it sure sounds like a lot - but only if you don't do the math. That 50,000 tons is spread more or less evenly over the 500 million square kilometers of the earth's surface. That means roughly one kilo per 10 million square meters. That number must be about the same for the moon - so if our corpse covers 1 square meter of the lunar surface, it's getting an average of one ten-thousandth of a gram per year of micrometeorite bombardment. Even at the speeds of micrometeors - that's hardly going to shred it to pieces anytime soon. Sure, it might get unlucky and get hit by a 10 kilo monster...but the odds are vanishingly small. SteveBaker 18:13, 29 October 2007 (UTC)

Using the rate of 1/km² or 1/1,000,000 m², and assuming a person's body has about 1 m² of exposed surface on the top side, it would still take, on average, a million seconds for a micrometeorite to hit. That's 11-12 days. So, to be completely destroyed by micrometeorites would take a long time. It would be a shriveled mummy long before that. Also, from how high was the body dropped ? If dropped from Moon orbit it would splatter into tiny pieces immediately, at least if it hit rock. Terminal lunar velocity into a deep pile of Moon dust might leave the body largely intact. StuRat 18:32, 29 October 2007 (UTC)

As there's no air resistance, there's no terminal velocity. I agree that dust would be a less destructive landing surface than rock, but my understanding from the Apollo missions is that the regolith is pretty firmly packed -- which would make the impact destructiveness thing almost entirely a function of initial height. — Lomn 22:04, 29 October 2007 (UTC)

The terminal velocity of a dropped object is still finite: it's equal to the escape velocity (for the Moon, 2380 meters per second). The only way to get something moving faster is to actively propel it. --Carnildo 23:44, 30 October 2007 (UTC)

That... seems unlikely (though I'd be interested to see a writeup on it, most TV stuff is written assuming an atmosphere). Apart from that -- initial conditions other than zero velocity, even without active propulsion, can easily lead to an impact at higher-than-escape-velocity speeds. Not that that's really either here or there.

The 'dropping' was meant in more of a figurative way. Btw, for a dead body, nothing can be considered terminal anymore. :) DirkvdM 09:11, 30 October 2007 (UTC)

Peak Coal

What is peak coal? 58.109.93.128 13:24, 29 October 2007 (UTC)

Peak coal is detailed in the article on Hubbert peak theory. Lanfear's Bane | t 13:32, 29 October 2007 (UTC)

Jehovah's Witnesses & Marriage

Do Jehovah's Witnesses believe and or support marriage to other religions. The reason I ask this question is I am not a Jehovah's Witness and am engaged to one and her father isn't in support at all of us. The rest of his children are married to a Jehovah's Witness. His brother isn't one and is married to one. Any help would be appreciated.

Thanks,

(email removed per instructions at top of page) —Preceding unsigned comment added by Timfreidag (talk • contribs) 14:47, 29 October 2007 (UTC)

• In most religions, getting the blessing of your fiancee's preacher, and then getting the preacher to perform the ceremony, is a sure-fire way to overcome this obstacle. Of course, you and your fiancee should be prepared for the possibility that the preacher is also intolerant of interfaith marriages. --M@rēino 16:33, 29 October 2007 (UTC)

Surely not! Surely religion is all about tolerance? If his brother is married to a JW, surely it would be possible for you? He may be using his faith as an excuse if he is not happy with his daughters choice of partner. (That's not that I am saying there is anything wrong with you, I am just hazarding a motive for his actions). Lanfear's Bane | t 16:54, 29 October 2007 (UTC)

You’d probably get much more knowledgeable answers if you moved this question to the Humanities reference desk, or perhaps the Miscellaneous reference desk. Indeed, I think the Science reference desk may be the worst of the reference desks to post this question to. As a general rule, the more that people are into science the less they’re into religion, and vice versa. MrRedact 18:36, 29 October 2007 (UTC)

The policy of Jehovah's Witnesses is to marry within the JW faith. Marrying someone outside JW may be enough to be dismembered. Graeme Bartlett 00:41, 30 October 2007 (UTC)

I think you mean "disfellowshipped" :-) --Trovatore 00:43, 30 October 2007 (UTC)

No, No, those folks can get real uppity, I think he was right..;-)) —Preceding unsigned comment added by 86.4.189.9 (talk) 08:56, 30 October 2007 (UTC)

Alcohol - on the verge of anaebriation!

If I have a few glasses of wine and start feeling light headed, what is happening in my brain to make me feel like that? Thanks. —Preceding unsigned comment added by 88.144.1.100 (talk) 16:04, 29 October 2007 (UTC)

We have an article specifically on the topic of Effects of alcohol on the body. Friday (talk) 16:05, 29 October 2007 (UTC)

Twin prop aircraft

after flicking through the "What is the name for this kind of aircraft, and why don't we make them?" post, I was thinking. Everyone dismissed the tail rotor on a transverse twin prop aircraft as pointless, but if you did have both props spinning in the same direction and a tail rotor to counteract any net rotation, would rotational manoeuvrability not dramatically increase with charge over the speed and hence thrust from the tail rotor, possibly allowing extremely sharp turns to be made, even at low speeds? ΦΙΛ Κ 16:28, 29 October 2007 (UTC)

The problem in most helicopters is that the tail rotor consumes continuous engine power for no useful purpose. On a twin rotor/propellor craft you can run the rotors in opposite directions to counteract the rotation and have more engine power to spare. You can still manouver by pushing more power to one rotor than the other. In the few (mostly Russian) helicopters that have Coaxial rotors that rotate in opposite directions, they can be shorter, more compact and turn faster than the usual kind. However, there are severe problems with these designs too: the additional complexity of the design, the problems of air turbulance between the blades as they pass each other adds vibration and there is a risk that in extreme situations one rotor might flex and hit the other. Having the two rotors spaced well apart helps that - and that's exactly what the CH-47 Chinook does. This solves most of the problems with coaxial rotors and gets rid of the tail rotor - but your back with a much bulkier design. Incidentally - one generally wants to place the two rotors fore-and-aft rather than left-and-right in order to keep a more streamlined fuselage - the exception being in tilt-rotor designs where obviously a fore/aft configuration wouldn't allow you to tilt the rotors forwards to become propellors for forward thrust. There was an experimental helicopter out there a few years ago that used a turboshaft engine to drive the main rotors and took the ject exhaust from the engine and sent that back to a vectored thrust nozzle in the tail that took the place of the tail rotor. This is a wonderful solution - but suffers from one lethal problem. If the engine fails, then your normal autorotational landing option is gone because in the absence of any tail-thrust, the helicopter would spin like a top. I don't know whether they ever solved that - but you don't see helicopters built like that - so I guess not. Then there is this thing: Piasecki 16H and Piasecki X-49 - which also doesn't seem to have taken the world by storm! SteveBaker 17:52, 29 October 2007 (UTC)

A few points..

1. I'd contest that moving power from one rotor to another in twin prop rotor wing aircraft would provide any significant turning force at all, as to provide a force comparable with that of cutting power to the tail rotors in a helicopter, you would have to stop one of the rotors, which isn't really a possibility, whereas with the tail arrangement I suggested, you could have twice the turning force experienced by cutting power to the tail rotor.
2. In the turboshaft exhaust arrangement you mentioned assuming the rotors and engine were not disengaged when power was lost, the turboshaft's compressors would surely continue to drive air through the tail pipe at a velocity which would allow the aircraft to remain stable, even if a hard turn was required by the pilot as well to further maintain stability.

ΦΙΛ Κ 22:23, 29 October 2007 (UTC)

1. Even assuming that holds, what happens if you have to turn the other way? You've got to mount an even bigger (and still completely unnecessary) tail rotor to counteract two main rotors working in concert. On the other hand, with a standard counter-rotating prop arrangement, all you have to do to generate yaw is vary the pitch of the blades. Same power to both rotors the whole time, and no power wasted on a tail rotor.
2. I don't get your point. If the engine fails, what's powering the compressors?

Anyway, I think it's fairly clear that twin-rotor craft have no need for a tail rotor. It's extra weight, complexity, and power loss for no operational gain. Even if it somehow improves maneuverability (a claim I doubt, and the utility of which I doubt -- more on that), twin-rotor craft tend to be heavy-lift craft where maneuverability is not a priority. Getting back to why the supposed extra maneuverability of a super tail rotor is useless -- think of a helicopter as a motorcycle and the tail rotor as the front wheel. When maneuverability matters for a helicopter in combat conditions, it's at high speeds. Helicopters, like motorcycles, make high-speed turns by banking, not by steering. A tail rotor doesn't even provide a meaningful assist, because while it may affect the direction the nose is pointed, it won't affect the helicopter's inertia. The ability to spin like a top is not a highly prized attribute of helicopters. — Lomn 02:30, 30 October 2007 (UTC)

To explain my second point, the moment of inertia of the blades, the same thing that would cause the craft to spin if power was lost.ΦΙΛ Κ 14:15, 30 October 2007 (UTC)

ΦΙΛΚ misunderstands how helicopters work. They don't "cut power" to the tail rotor in order to turn - they alter the collective pitch on the tail rotor blades. (That's how come they can turn both left AND right! :-) SteveBaker 03:37, 30 October 2007 (UTC)

True, I know very little of workings of a helicopter and was really just making assumptions. But just to say, using the mothod of tail rotor speed variance to steer, you could steer both ways, one way by slowing it down, the other by speeding it up.

You absolutely can, it's just that pitch variance is a far better solution. — Lomn 14:34, 30 October 2007 (UTC)

Theoretically - you might - but not in practice. The tailrotor is driven from a set of gears from the main rotor so that as the main engine throttles up and down, the tail rotor automatically maintains the correct amount of correcting torque. There is no way to speed it up...although I suppose you could slow it down with some kind of clutch mechanism - they don't do that in practice (at least not on any helicopter that I'm aware of). SteveBaker 17:02, 30 October 2007 (UTC)

So just to clarify, twin engine airplanes often do not have counterrotating props. The increase in efficiency is offset by the cheapness of the identical engines on each side (and on single engine models). --DHeyward 05:54, 30 October 2007 (UTC)

Yes - airplanes - because the orientation of the engines to the line of flight would induce roll - not yaw - and roll is easier to stabilise because you've got wings and ailerons and a large amount of rotational drag and a huge moment of inertia along the roll axis because of those big wings and the big heavy engines (not to mention fuel tanks) stuck way out from the centerline. Those things don't work anywhere near so well for the yaw induced by lift rotors in a helicopter or tiltwing. However, the effect of engine torque isn't negligable. Some single-engined aircraft were notable for being able to roll much more sharply in one direction than the other. I'm thinking of the Sopwith Camel as an example. It's Clerget rotary engine actually rotated with the propeller(!) and had a huge moment of inertia. The heavy spinning engine and the short-winged/lightweight biplane design enabled the aircraft to turn much more quickly to the right than to the left - a feature that experienced pilots used to their advantage. However, it killed a lot of novice pilots - so it was a mixed blessing at best! SteveBaker 12:10, 30 October 2007 (UTC)

It still is problem with most light twins with non-counterrotating propellers and that is what makes their safety record less than singles. It's said that it's safer to land a single with no engine than a twin with one because Vmc induces a roll higher than stall speed and with no warning. Some airplanes had Vmc close to stall speed and the asymmetric thrust meant that the airplane was never coordinated and one wing stalled earlier. Vmc + stall = rolling/flat spin close to the ground. Very bad day. —Preceding unsigned comment added by DHeyward (talk • contribs) 13:51, 30 October 2007 (UTC)

Eh? If one of your engines has stopped - it scarcely matters whether it was contrarotating or not! SteveBaker 16:58, 30 October 2007 (UTC)

Actually it matters quit a bit. P factor will induce a rolling moment on one side more than the other. It is why there is a different L/R Vmc in non-counterrotating engines. Minimum controllable airspeed depends on which engine goes out and it's killed a lot of pilots. --DHeyward 05:47, 31 October 2007 (UTC)

Barriers beyond a Planck Length

Even with a Grand Unified Theory uniting all the forces under an overarching Quantum Gravity (allowing us to observe phenomena beyond a Planck Time and beyond a Planck Length), will it be possible to directly study the nature of singularities such as those of a black hole or the infinitesimal "place" we all came from? If we can observe beyond a Planck Length but not all the way to a singularity, what is the next barrier? —Preceding unsigned comment added by Sappysap (talk • contribs) 17:34, 29 October 2007 (UTC)

Even with the GUT at our fingertips, observing beyond plank length or plank time will remain impossible. The present inability to do this is not a defficiency of modern physics, but a necessary consequence of quantum mechanics. And while we know that quantum mechanics and relativity disagree with one another, a meta theory to correct this issue will not change presently known laws of physics. This is precisely because we know why they disagree. Quantum generally assumes gravity to be irrelevant, while relativity generally assumes space to be smooth (ok, there are a few other technical details). These are very valid assumptions for short range and long range interactions, respectively, but neither holds at a singularity. So, don't let the thought enter your head that an inconsistency in modern physics will allow future alteration of the present laws true outside a singularity; they won't (we'll still be figuring out the consequences of those laws, but the laws pretty much stay the same). Now, you might be asking yourself what the point of figuring out singularities is if we can't actually peer inside of them. The answer is that even unobservable interactions can have very measurable macroscopic consequences. There are questions that can't really be answered by physics right now, such as whether a singularity can exist without an event horizon, or what actually happens to a particle that impacts a singularity (it might produce an observable event). Further, if we have an accurate model for what might happen within a singularity, we can develop an model of the big bang that goes further back in time, perhaps making predictions about early distributions of various particles that may have presently observable consequences. Someguy1221 18:33, 29 October 2007 (UTC)

Among theories that attempt to unite quantum mechanics and general relativity, and which are sufficiently well-developed to make some inference about singularities, most tend to eliminate the traditional notion of a singularity entirely and replace it with some sort of quantum fuzziness. For example, in string theory, the core of a black hole becomes something colloquially resembling a tangled ball of twine (no joke). Dragons flight 19:37, 29 October 2007 (UTC)

Remove Blood Stains

What is the most effective way to remove small amounts of human blood stains from clothes (in particular, white cotton) ... preferably with common household products ... and not something unlikely to be found in a home ...? Thanks. (Joseph A. Spadaro 19:13, 29 October 2007 (UTC))

90° water and washing powder in a washing machine will get rid of all the blood stains even on white. Keria 19:31, 29 October 2007 (UTC)

Yes, don’t use hot water! --S.dedalus 19:59, 29 October 2007 (UTC)

Yes, 90 Fahrenheit and not Celsius. Sam Blacketer 20:01, 29 October 2007 (UTC)

Hydrogen peroxide 3%, applied to the item when it's dry. Wait until it's stopped bubbling, then rinse it out. Repeat as necessary. Make sure you don't splash it around, and wash it off any skin that it touches -- it won't burn you quickly, but it will burn you if you insist. Also it may degrade the fabric somewhat. --Trovatore 20:15, 29 October 2007 (UTC)

Sard wonder soap rubbed on wet and then soaked works well. But not once it has been washed in Hot Water. Graeme Bartlett 00:45, 30 October 2007 (UTC)

Spit contains various enzymes. I don't know if any of those would help break down blood. But even if it is just the water in it that helps, it's something you've always got handy to remove a drop of blood right after it landed on the cloth. DirkvdM 09:30, 30 October 2007 (UTC)

I work in a hospital and regularly end up with blood on my uniform. Hydrogen peroxide works wonders. No visible traces left. I'm not sure what the reaction does on a chemical level (does it destroy forensic evidence?), but on a macroscopic level I've not found anything better for removing blood from clothing. 152.16.16.75 10:32, 30 October 2007 (UTC)

The hydrogen peroxide worked perfectly. Many thanks. (Joseph A. Spadaro 04:19, 31 October 2007 (UTC))

Although modern detergents can remove blood stains at moderate washing temperatures, have you tried just soaking the (freshly stained) article in cold water. seems to work 4 me!

Why does blood do that?

Why does blood become hard to remove when heated? What is going on there that is different for other stains? Dragons flight 00:31, 30 October 2007 (UTC)

I imagine it has to do with the blood being composed of proteins. My best guess (dunno why exactly) is that the heat denatures the proteins which then bind with the clothing material and each other forming a interconnected "fabric-protein mass." With this explanation, I would guess other stains with proteins probably behave similarly but they aren't as noticeable because the proteins are clear. I wonder if the iron ion (which with the iron binding amino acids gives the blood the red color) has anything other than a passive role in the staining. Would be nice if someone with more knowledge could speak up on this issue. 71.226.56.79 04:47, 31 October 2007 (UTC)

How can we remove cooked on blood

How can blood stains that are fixed by hot water be removed? DOes peroxide work on these too? Graeme Bartlett 00:45, 30 October 2007 (UTC)

Kinetic Batteries

Would it be possible to put some kinetic powered batteries (like the ones in watches/torches) in an mp3 player, so as that it no longer had to be charged/required batteries? —Preceding unsigned comment added by 78.147.220.128 (talk) 20:18, 29 October 2007 (UTC)

Possible, yes, but probably not feasible. MP3 players have significantly higher power usage than wristwatches, and probably benefit less from natural motion. As for torches/flashlights, my (admittedly limited) experience with them has been that a large assembly is needed to generate power (much larger than a typical MP3 player) and that you're not getting much battery life out of a lot of very intentional shaking. — Lomn 21:17, 29 October 2007 (UTC)

The power draw for my particular mp3 player seems to be about 40 milliwatts. (A single NiMH AAA battery will run the device for about 30 hours.) Note that the power consumption will be higher for a device based around a hard drive rather than flash memory. For reference, this document from the MIT Media Laboratory notes that the typical output from a self-winding watch mechanism is on the order of 10 microwatts, but that up to about 1 milliwatt can be generated if the arm bearing the watch is vigorously shaken. In other words, the design would have to be scaled up substantially to operate a music player. That document does discuss other ways to extract energy from human body motion, however. TenOfAllTrades(talk) 21:58, 29 October 2007 (UTC)

Still 33 milliamp isn't that much. I wonder if a little dynamo could be used. Surely those shake torches use more than 40 milliwatts. Some of them have 10 milliamp 5 volts over each of 9 or more diodes!--Dacium 03:43, 30 October 2007 (UTC)

Sure; some other options are discussed in the article that I linked. TenOfAllTrades(talk) 03:56, 30 October 2007 (UTC)

Surely the OLPC XO-1 laptop is an existence proof that a kinetic power source can work for an MP3 player; if you can crank a laptop computer to operate it, you could certainly crank the far-lower-power MP3 player to operate it.

Atlant 11:53, 30 October 2007 (UTC)

Yes - but there is a VAST difference between vigerously cranking a geared generator and just kinda picking up random motion from your body (which is what a self-winding watch does). The OLPC needs to be cranked fairly hard for five or ten minutes in order to give you an hour's worth of computing. An MP3 player would certainly need a lot less than that - but (as has already been conclusively shown) that's still a lot more than you get from incidental motion. SteveBaker 17:48, 30 October 2007 (UTC)

Certainly. I think the limitation here is one of marketability. It seems to me that it'd be hard to sell a CrankPod that is three times the size/weight of its competitors. I started to say "but doesn't need batteries" or "but doesn't need charging", but that's not really true. You've still got a battery (with any of these solutions), and you've still got to charge it -- you're just changing the charge method from a plug to a crank. — Lomn 14:32, 30 October 2007 (UTC)

Vacuum Sealers - for commercial use

There doesn't seem to be much information on Vacuum packing, I'm trying to vacuum pack food for PACKAGING purposes, i.e. to post or transport to places, and when I try my useless home vacuum packer the food just doesn't last when I *know* that it should. The food's outright growing mould after 2 days. I have a sneaking suspicion that the vacuum doesn't hold and that the uesless vacuum bags actually breathe air (permeability) and I need to invest in one of those $4000 ones that will do the job properly! Anybody aware? Rfwoolf 20:33, 29 October 2007 (UTC)

Even if you remove all the air there are still anaerobic bacteria and yeasts that will decompose your food. There is probably some oxygen left after you suck out the air, a domestic vacuum cleaner cannot create a very low pressure. And there would still be oxygen dissolved if the product was exposed to air, or had pores. Vacuum packaging will still have to sterilise the food by some means, or have some otherway to stop unwanted growths. Graeme Bartlett 00:50, 30 October 2007 (UTC)

The article on Tetra Pak might be interesting to you. --Mdwyer 15:26, 30 October 2007 (UTC)

October 30

Protein domain

Two different protein domains can have a same function why or why not?

If I consider two amino acid sequences of a protein with same domains but changes in 2 or 4 amino acids in the chains. Could it have the same function?

Respected teacher

My problem is that I considered 2 chains of the same protein

Chain A ---> n l i i l a n n s l s s " n g n v" t e s g c k e c e e l e e k n i k e f l q s f v h i v q m f i n " t s "

Chain B ---> n l i i l a n n s l s s " n g n " v t e s g c k e c e e l e e k n i k e f l q s f v h i v q m f i n t s

Those in double Inverted commas are excluded from the domain chain but they are considered in the same domain why? —Preceding unsigned comment added by Biomedicalpersonal (talk • contribs) 05:12, 30 October 2007 (UTC)

I do not understand what you are trying to ask in the second part of your question but I may be able to help you with the first part. A few amino acid changes in non-critical regions or mutations resulting in similar amino acids (see Neutral mutation) in critical regions may not cause measurable effect on the functioning or the interactions of the protein domain. It would help if you clarify the second part of your question about protein chains. By the way, these appear to be homework questions. The users on the reference desk are not supposed answer your homework questions, but if you have do not understand a concept or need help getting pointed in the right direction by all means ask away. 71.226.56.79 04:22, 31 October 2007 (UTC)

food poisoning ???

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis or prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page. --~~~~

This question was removed for a reason. We cannot offer medical advice on either poisoning or allergies. If you are worried about the state of your friends health you should consult a medical professional immediately, especially if you believe she has been poisoned. Discussing it on the internet first should not be your immediate priority. Lanfear's Bane | t 21:41, 30 October 2007 (UTC)

production of particles by pair production method .

in the wikipedia article "pair production" only example of electron-positrion pair production is given . is it possible to produce particles like proton , neutron and neutrino by pair production method ? —Preceding unsigned comment added by Shamiul (talk • contribs) 12:30, 30 October 2007 (UTC)

Neutrinos and antineutrinos can be produced from a photon.[6] Like electron-positron pair production needing other particles involved in order to conserve energy and momentum, neutrino-antineutrino production from a photon needs other particles involved in order to conserve spin, which is a form of angular momentum.

Protons and neutrons aren’t elementary particles, but instead bound states of quarks. So in quantum chromodynamics, low-order particle interactions would include pair production of quarks, not of protons or neutrons. A quark and its antiquark can be produced from a gluon.[7] MrRedact 21:02, 30 October 2007 (UTC)

While these predictions from theory are very likely correct, we are currently far from being able to experimentally confirm pair production of neutrinos (because our neutrino detectors are not very sensitive). Icek 01:08, 31 October 2007 (UTC)

Somehow my previous answer to this question has been removed? However there is no neutrino pair production. They are electrically neutral and as such do not couple to photons. The nearest equivalent would be production of a neutrino/anti-neutrino and corresponding anti-lepton/lepton from a w particle. MrRedact is right about quarks being pairproduced, but note they can also be produced from photons as well as gluons (they are charged). Pair production of protons/neutrons is nigh on impossible, when quarks are pair produced they hadronise into two jets of particles and protons/neutrons maybe included in the jets. Cyta 08:56, 1 November 2007 (UTC)

Actually I may have been hasty to dismiss the original answer I didn't spot a link had been provided. The pair production of neutrinos seems only to happen in dense matter, which seems to me a different effect. There is no direct neutrino-photon coupling in the standard model. I suspect the matter is required as there is an intermediate stage in the reaction. Cyta 09:00, 1 November 2007 (UTC)

Anomalous hair

REMOVED. —Preceding unsigned comment added by 193.188.46.61 (talk) 13:23, 30 October 2007 (UTC)

(EC) Do not request regulated professional advice. If you want to ask advice that "offline" would only be given by a member of a licensed and regulated profession (medical, legal, veterinary, etc.), do not ask it here. Any such questions may be removed. See Wikipedia:Medical disclaimer and/or Wikipedia:Legal disclaimer. Ask a doctor, dentist, veterinarian or lawyer instead. Sorry, even a query that sounds innocuous should be addressed to a professional. Lanfear's Bane | t 13:32, 30 October 2007 (UTC)

REMOVED. --JWSchmidt 14:55, 30 October 2007 (UTC)

The original question did not ask for medical advice. My original comment made this point. There was no reason to delete my original comment. --JWSchmidt 18:23, 30 October 2007 (UTC)

REMOVED. --M@rēino 15:03, 30 October 2007 (UTC)

Per Lanfear's Bane's comment, we cannot offer medical advice. In this case, the original poster was seeking a diagnosis--the identification of the cause of a particular symptom. If someone would like to discuss in more detail why this constitutes a request for medical advice, please bring it to the Ref Desk talk page: WT:RD. [hidden unsigned comment by User:TenOfAllTrades made public for clarity]

I am appalled at this continuing deletionism by people who are apparently incapable of telling the difference between a request for an explanation of a biological phenomenon and a request for medical advice. To take this behaviour to its logical conclusion would result in the prohibition of any question relating to human biology (or animal biology for that matter). What causes blue eyes? What causes grey hair? Why does alcohol abuse cause liver damage? All are questions relating to biological phenomena - just as is "What causes an observed variation in human body hair types in humans?" Hiding comments about the removal of questions is also profoundly unhelpful. DuncanHill 15:20, 30 October 2007 (UTC)

One clear sign of a question that we aren't allowed to answer is one that starts off "I have some symptom..." or "A friend has some symptom...". This is quite clearly one of those - and we aren't allowed to answer it - period. The questions you put up as strawmen are not the problem here. I would like to turn this around and ask DuncanHill: "What kinds of question do you think the ban on giving medical advice is intended to cover?" - clearly it covers something or it wouldn't be there. What? Give me some examples. SteveBaker 16:38, 30 October 2007 (UTC)

Why don't we stop wasting energy here now and instead just send the questioner to where he asked the same question a week ago (8 times in fact). DMacks 15:16, 30 October 2007 (UTC)

Hey, can everyone confine metadiscussion to the talk page of the Ref Desk? This sort of debate about what questions are appropriate doesn't belong out on the Desk itself. TenOfAllTrades(talk) 16:52, 30 October 2007 (UTC)

• "can everyone confine metadiscussion to the talk page" <-- If content is removed from this page because it contains descriptions of personal health matters, then say THAT, but do not say that someone asked for medical advice when they never did. --JWSchmidt 18:20, 30 October 2007 (UTC)

• Hey, why was my comment removed? All I said was that the original question did not seek medical advice. And that continues to be my opinion. The question was not describing a symptom -- "a sensation or change in health function experienced by a patient" -- because there was no implication that the questioner's health was affected. They were just curious about Hair follicles, a subject on which the Wikipedia article is B-class at best. --M@rēino 04:23, 31 October 2007 (UTC)

This is plainly ridiculous! The sense of this regulation is to avoid unqualified suggestions to be accepted and considered professional by the user. How should this do any type of damage what so ever? If you apply that definition about a symptom it can be eligible to all sorts of things, to nausea due to boat movements for ezample. Aren't you allowed to explain the processes that occur in that event?! Laws and regulations should be implemented only when they make some type of logical sense. You're just tking it to the extreeme here.193.188.46.254 13:56, 31 October 2007 (UTC)

Membrane Potentials

What is the difference between membrane potentials, diffusion potentials and nernst potentials. I understand membrane pot. = diffusion pot. and nernst pot. to be when the electrical force opposes the concentration force.?

In biology, a membrane potential is the electric potential difference across a membrane of a cell or an organelle such as a mitochondrion. For charged ions there can be a balance between movement across a membrane due to diffusion and movement due to the electric potential across the membrane: see reversal potential. If you had a membrane and only one ion was able to cross it by passive diffusion then at equilibrium the membrane potential would be equal to the diffusion potential of that ion: Nernst Potential. --JWSchmidt 14:29, 30 October 2007 (UTC)

Thanks for the quick reply. I have a slightly better understanding now but still wich to clear some issues [this is in the context of the establishment of an RMP]. 1) Does the term diffusion potential apply to an ion or membrane? 2) Could the Nernst potential (equilibrium potential) be described as the potential difference required to prevent net diffusion of that ionic species? (and how is Nernst potential related to diffusion potential?) 3) What is the purpose of the Na/K ATP'ase (it contributes slightly by reducing RMP; is its primary function to repolarise?). Thanks.

[ATP'ase] I thought a bit more and reasoned that the purpose may be to create the diffusion potential? or allow the tissue to become excited (function)? i.e. to maintain concentration gradient thereby preventing the Nernst potentials for each ion (Na/K) being reached (a scenario where membrane potential- or is that diffusion potential- is a result of only Na and K)

Did you find Resting potential? For most cells, the Na+/K+-ATPase is not thought to make a major contribution to the resting membrane potential. The resting potential of many cells is close to the equilibrium potential of potassium ions (K⁺). Many cells have potassium "leak" channels that control the resting potential. --JWSchmidt 18:13, 30 October 2007 (UTC)

When will oil run out?

Based on world wide consumption trends, and ignoring that it takes millions of years to produce more oil. 64.236.121.129 16:04, 30 October 2007 (UTC)

It will never "run out" - but it will become far too expensive to use as fuel. When? See Peak oil. Cheers Geologyguy 16:06, 30 October 2007 (UTC)

I did some calculations in response to an earlier question - the answer is that if we were to carry on consuming it at the present rate, it would run out in about 500 to 600 years. However, if we actually did that, the CO2 levels in the atmosphere would be far beyond "mere" global warming problems - they'd be at a point where humans (and most animals) couldn't breathe. However, the assumption that we'll carry on using it at present levels is flawed. I don't think that running out is a practical proposition. Even if we somehow managed to 'sequester' the CO2, we would only have to halve our consumption every 250 years in order to make the stuff last forever. Another issue is that these numbers for oil reserves are always accompanied by a caveat that says "economically retrievable" - in other words, the only things the oil companies care about is the stuff they can dig up for less than they can sell it for. There are reserves of stuff like "oil shale" that contain a lot of oil - but which are so expensive to dig up and refine that it's not worth doing it. If the oil were ever seriously likely to run out, then the price would go through the roof and suddenly oil shale (or whatever) would be worth exploiting and our reserves would increase (although the price would still be astronomical by today's standards). However, we must stress that with what we know about global warming, it's all completely irrelevent. We must not ever come even close to running out - because even a tenth of that amount of oil - when converted to CO2 - would kill the planet. SteveBaker 16:28, 30 October 2007 (UTC)

Our article on world energy resources and consumption says that the world's oil reserves are 5.7x10²² J, and annual oil consumption in 2005 was 1.8 x 10²⁰ J, so this gives a ratio of reserves to consumption of over 300 years. However, as Geologyguy and SteveBaker have pointed out, this is a rather meaningless figure. In reality oil prices will rise, and oil consumption will fall as alternative energy sources become more economically attractive, and global warming imposes a shorter deadline on us anyway. Gandalf61 16:44, 30 October 2007 (UTC)

Perhaps by then we will have made advances in scrubber technology, and have them on every car, big ones atop every building, and just generally all over the place. I mean, once it comes down to economy vs. survival... Arakunem^Talk 17:18, 30 October 2007 (UTC)

The chemistry of scrubber technology is not very promising - the kinds that merely absorb CO2 will give the gas up again fairly easily and the substances that react with CO2 to actually get rid of it require lots of energy to make in their own right - so you end up needing more scrubbers to scrub the output of the factory that makes the scrubbers (or the power station that drives it) than the factory itself can make. It's really a bad idea to pin one's hopes on such things because it distracts from the very critical thing of cutting CO2 production in the first place. The whole "Clean Coal" campaign (which has to rely on science-fiction "carbon sequestration" techniques) is a particularly bad example of this kind of wishful thinking. SteveBaker 03:40, 31 October 2007 (UTC)

Are you saying that collecting and burying CO2 from a coal plant won't work and will use more energy than you get from the coal?--Dacium 05:47, 31 October 2007 (UTC)

Yes and no. I'm saying that on small scales, it takes more energy than you get - and on large scales there isn't a viable technology for doing it at all. Not one single large scale CO2 removal/sequestration plant exists anywhere in the world - not even experimentally. (Which hasn't stopped the US from licensing the building of "Clean Coal" power stations - which is a scandal just waiting to get media attention!) Our article on Carbon capture and storage explains that what we're likely to have will remove 80 to 90% of the CO2 from the gasses and consume 10% to 40% more energy. But the biggest problem is what you do with the stuff once you've captured it. Sequestering it into limestone requires 180% more energy - so that isn't going to fly. If you try to sequester the CO2 without chemically converting it to something else then you've still got to find a place to store millions of tons of something that's a gas at normal temperatures and pressures. That's no easy task. You can't store it underground or underwater because there isn't enough space at normal temperatures and pressures (If you burn a cubic meter of oil or coal - you get a LOT more than a cubic meter of CO2 as a result! So pumping it into disused coal mines and oil wells isn't going to work for very long.) If you compress the CO2 so it takes up less space (eg storing it as dry ice), then that requires either very high pressure storage or very low temperatures. Either of those technologies will require yet more energy - and worse still, will be vulnerable to long term corrosion and other damage - so you're just building up more trouble for the future. There is talk of dissolving the stuff into saline aquifers or deep oceans - but those are not permenant solutions (eventually, the CO2 would get out again) and the resulting carbonic acids would likely do untold amounts of damage to the environment. Dealing with radioactive waste from a nuclear power plant is EASY by comparison because so little material is involved. So, no, we aren't going to be doing this if we want to save the planet. We have to cut down our consumption (probably the easiest thing to do in the short term) and switch rapidly over to nuclear and (where possible) wind/solar/tidal power until we figure out how to make fusion reactors that actually work. SteveBaker 18:00, 31 October 2007 (UTC)

On the scales you are talking about, I would imagine there would be nowhere near enough suitable locations to bury CO2 and be reasonably sure that it won't escape at some point in the future. Hopefully within 300 years we will have managed to develop alternatives to fossil fuels, otherwise our pathetic species probably doesn't deserve to survive. Bistromathic 16:41, 31 October 2007 (UTC)

Quantum foam

I was thinking about Quantum foam last night. Since particle/antiparticle pairs emerge out of the vacuum - and since there is no "rest frame" - it seems like the particle pairs must be travelling at random velocities relative to my motion. This implies that they have random kinetic energy relative to my motion - which means some of them must be arbitarily energetic. How come we don't measure them? Even though they only last for a spectacularly short amount of time in their own frame of reference - in mine, they'd spend an eternity recombining. Our article briefly sketches over that. SteveBaker 16:45, 30 October 2007 (UTC)

You are taking the idea of virtual particle pairs too literally. Quantum foam is a broad term for the fluctuating nature of space-time at a quantum level that theory demands. Those fluctuations are abstractly described as virtual particles, but from the practical point of there are no particles. Imagine a tablecloth with marbles on it. The marbles may move because they are hit by other marbles, or they may move because the tablecloth shakes. If all you could ever see were the motions of marbles, you might try to explain the motions caused by shaking in the tablecloth as being caused by invisible "virtual" marbles. What is going on in physics is similar. The ways that real particles are affected by the quirks of the vacuum can be usefully described by replacing the vacuum with infinite numbers of virtual particles constantly bombarding matter. However, those particles don't really come from anywhere or go anywhere. They don't have an existence that one could isolate and interact with individually. Really, they are collectively just a way of describing the manifestations of the complicated quantum mechanics intrinsic to space-time. Dragons flight 19:55, 30 October 2007 (UTC)

Steve, "quantum foam" does not refer to the virtual particle/antiparticle pairs in the quantum vacuum in general, but specifically to what happens around the Planck scale, where spacetime becomes quantum mechanical. The short answer to any question about quantum foam is that we don't know anything about it at all. But I think you're asking about the quantum vacuum in general, so let's step back from solving quantum gravity and consider something more manageable. Take QED or Klein-Gordon theory. You asked "How come we don't measure them?", referring to virtual pairs with arbitrarily large energies. OK, how do you want to measure them? You would have to interact with them somehow. That means that instead of a vacuum-bubble diagram, you have to consider diagrams with external vertices, and virtual particle loops inside. One way to think about these diagrams is that your real external particles are encountering the virtual particles from the vacuum; and they can indeed have arbitrarily large momentum and energy. In many quantum field theories, these loop contributions cause all calculations to come out infinite. Similarly, the vacuum diagrams can cause the vacuum energy to come out infinite. To get something physically sensible, you have to renormalize the theory, and "renormalize" the vacuum energy by zeroing out the infinite offset. So even if you never observe the virtual particles as discrete events, they are hugely significant in every interaction between particles! In some theories, like supergravity, the case is even worse than that, and renormalization doesn't work. Because all of this is not very intellectually satisfactory, we hope that there is some real, underlying physics that doesn't have these problems. Supersymmetry takes care of the infinite vacuum energy, but still needs renormalization. Superstring theory attempts to avoid the need for renormalization at all. But now we're back in the realm of speculations about quantum gravity. --Reuben 21:49, 30 October 2007 (UTC)

Why don't they put some kind of grating over jet engine intakes to prevent birds from getting sucked in?

Why not? 64.236.121.129 16:57, 30 October 2007 (UTC)

It probably doesn't happen all that often and may actually cause too much disruption to the air intake on the engines. Also, birds would still (I imagine) be sucked and stuck to the grating over the intake. -- MacAddct1984 17:05, 30 October 2007 (UTC)

Per MacAddct, a grating would seriously disrupt the airflow into the engine, creating a tremendous amount of drag. (Remember that air is flowing through the engine at several hundred kilometers per hour.) Plus, what if the bird strike actually damages the grating? In addition to having a bird in the turbine, you'd get all those jagged metal bits of broken grating.... TenOfAllTrades(talk) 17:12, 30 October 2007 (UTC)

No - there is no grating - but there is a requirement to test engines for bird strike damage. They actually have an air-powered cannon to shoot birds into running jet engines in order to test them. However if you fly at high speeds and lower altitudes (where birds tend to be) - and if you hit a big one - you can certainly throw a turbine blade and utterly trash the engine in the process. If you have a strong stomach - read Bird strike which has some exceptionally grisley photographs of the effects of birds on airplanes and airplane engines in particular. SteveBaker 17:35, 30 October 2007 (UTC)

Yep. Even with a no-drag indestructible grating, you're still talking about smashing a bird into something at several hundred miles per hour. It's not that the engine is sucking in birds from all directions, just that the plane and bird are hitting -- and even with a grating, the engine still has to generate the same suction. But as SteveBaker said, there's a bird strike test, so a grating is superfluous. — Lomn 18:18, 30 October 2007 (UTC)

A bird running into a turbine is usually not dangerous. The bird will be ripped apart. If the turbine does throw a peice and destroy itself, all that happens is that fuel will continue to be pumped in and a massive flame will come out the back because the fuel is being burned. Obviously you get no thrust from the engine, but the plane can still fly. I don't think any commercial airliners have been brought down, many have had engines taken out.--Dacium 05:43, 31 October 2007 (UTC)

According to Bird strike: according to the FAA only 15% of strikes (ICAO 11%) actually result in damage to the aircraft. But you're wrong about commercial airliners - there is at least one instance: Eastern Air Lines Flight 375 flew into a a flock of small birds that took out three out of the four engines at once - at the critical moment just after take-off when the engines were at full power - they also splattered over the windshield - blocking the pilot's view and clogged the pitot tubes, preventing them from knowing their airspeed. They didn't stand a chance - the plane rolled over and smashed into the ocean - there were 62 fatalities out of 72 people aboard. Our article on bird strikes also says: the problem costs US aviation 600 million dollars annually and has resulted in over 200 worldwide deaths since 1988. We have a brief article about Birdstrike simulators too! SteveBaker 17:31, 31 October 2007 (UTC)

Having a grating over the fan intake can introduce another problem. When engines are being tested, they have what they call a debris guard which fits over the front of the engine, but it means that the engine cannot be run in all conditions. To fully test the engine, at some point the guard needs to be removed.

This is because, at the speed the air enters the front of the engine, well over 300 miles an hour, it creates a huge cooling effect on the debris guard. That combined with the right (or wrong) level of humidity can cause ice to form on the guard. It has been known that enough ice has formed for the fan to create a vacuum behind the guard, and the result is, the entire guard is pulled through the engine, which will utterly destroy the engine, no question about that. On the other hand, with a bird strike, it's quite possible for an engine to keep running without any problems, especially if the bird is passed straight through the cold stream ducts and out the back. —Preceding unsigned comment added by 132.244.246.25 (talk) 08:33, 1 November 2007 (UTC)

Taste bud abnormality

I was wondering if there is a name for the painful bumps that crop up on tongues from time to time. It seems as though a single taste bud turns white, enlarges, and becomes very sore to the touch. I can't seem to find a name for it anywhere (and no, it's not herpes or canker sores) -- MacAddct1984 17:10, 30 October 2007 (UTC)

Sorry, but this question is basically asking for a diagnosis, which we aren’t allowed to do here. As it says at the top of the page:

Do not request regulated professional advice. If you want to ask advice that "offline" would only be given by a member of a licensed and regulated profession (medical, legal, veterinary, etc.), do not ask it here. Any such questions may be removed. See Wikipedia:Medical disclaimer and/or Wikipedia:Legal disclaimer. Ask a doctor, dentist, veterinarian or lawyer instead. MrRedact 17:33, 30 October 2007 (UTC)

While MrRedact is correct that we can't diagnose, you might nevertheless be interested in the brief stub on lie bumps. It could stand to be filled out if you want to do some source research. --Trovatore 17:36, 30 October 2007 (UTC)

Ah, thank you very much Trov! While I wasn't really looking for medical advice, I do realize there is a very fine line between "diagnosing" and what I was asking. But that is what I was looking for, I remember my mother always saying the old wife's tale was that you get them when you lie. -- MacAddct1984 17:53, 30 October 2007 (UTC)

I can't say what this is in your case, but it also happens to me when I accidently scrape the area with a sharp piece of food or bite the tongue. Here's an article that discusses lie bumps a bit more: [8]. Again, this may not be what you have though. Sancho 17:47, 30 October 2007 (UTC)

they are basically caused by cuts in the tongue by food or your own teeth. If they are on the edge of a tongue it almost defiantly is caused by you biting your tongue accidentally. usually a mouth full of food pushes the tongue over the teeth and people bite without realising.--Dacium 05:38, 31 October 2007 (UTC)

how do i make my dog smart

please thank you —Preceding unsigned comment added by 77.234.83.173 (talk) 17:19, 30 October 2007 (UTC)

See dog training. Sancho 17:38, 30 October 2007 (UTC)

I don't think you can change your dog's intelligence - as with humans, it's pretty much something that you're born with - but you can certainly enroll in a training course for you and your dog. Sadly, it generally turns out that your dog is plenty smart and it's you that gets trained - but that works too! We also have an article on Dog training. I have two dogs - one smart (female), the other dumber than a bag of hammers (male). The smart dog has learned to bark at the front door when she wants to be let in. The stupid dog has not. He just stares at the door hopefully. However, the smart dog has realised this and when she sees him standing there looking puzzled, she runs up to the door, barks until we open it and let him in - then she goes back to whatever she was doing before. SteveBaker 17:43, 30 October 2007 (UTC)

Intelligence can most certainly be changed in humans. And it's important that people are aware of that. There was a study recently that took two classes of high-school students, gave one a lecture about something boring and irrelevant (just to have a control), and gave the other a lecture on the nature of intelligence, and how intelligence and problem solving skills can be improved through training. After about half a year the second group showed a significant increase in grades and the first didn't. Of course there is always the criticism that training intelligence is just training for intelligence tests, but I believe strongly that intelligence by any reasonable definition can be consciously improved. I see no reason why dogs couldn't do the same, except that they probably won't have the mental capacity for defining the concept of intelligence, so they can't set it as a goal for themselves explicitly. Still, if you set out a wide range problem solving exercises with rewards that the dogs can understand (like the 'stack boxes to get to a banana'-kind of exercises they give monkeys, only simpler), they may generalize over all exercises and improve their problem solving intelligence instead of just learning specific tricks. risk 18:16, 30 October 2007 (UTC)

thats a lot of reading, what's something simple ic an do like some toy i can buy or music to play. i cant afford obedience school. —Preceding unsigned comment added by 77.234.83.173 (talk) 18:09, 30 October 2007 (UTC)

There are no magical shortcuts. If you can't afford obedience school - read our Dog training article carefully and do what it says - it's pretty much what the obedience schools teach. Set aside an hour a day for training. Make sure your dog knows that this is something special - "Now is training time - later will be play time" - so have a special routine you go through at the start and end so your dog will come to recognise that. SteveBaker 00:08, 31 October 2007 (UTC)

I have an instinctive negative reaction to the idea that learning to obey demonstrates intelligence. I've always sort of felt that cats are smarter than dogs, precisely because they don't obey you. --Trovatore 01:33, 31 October 2007 (UTC)

Dog training is more about learning inter-species communication than obeying orders. I prefer to think of it as unlocking the intelligence that was always there - and in a way that we humans can understand. SteveBaker 03:25, 31 October 2007 (UTC)

I don't know about smart but I made my dog smarter with this and this and this. All three are essential. --DHeyward 05:40, 31 October 2007 (UTC)

Get involved with Dog agility or Flyball. You and the dog have fun; you and the dog get to socialize; you and the dog get good outside exercise; and the dog's intelligence is developed. It's mostly just run by volunteers, so there's little cost.--Eriastrum 17:06, 31 October 2007 (UTC)

When you here about plane crashes on the news

They usually manage to find the Black Box flight recorder thing even in the worst crashes, when the rest of the plane has been completely destroyed by explosion and fire. So, why don't they make the planes out of the same material they make the black boxes out of? --84.68.112.172 17:50, 30 October 2007 (UTC)

While black boxes are of durable construction, I think they benefit a great deal by being inside the aircraft. By analogy, I could get into a car wreck that's bad enough to total my car, but it's likely that my CD player would survive. It's not that it's made of some magic substance—it's just protected by the body of the car, which can deform and break up, absorbing a lot of impact energy. -- Coneslayer 17:57, 30 October 2007 (UTC)

In short, the black box is so heavily armored so as to survive the worst impact, that if the airplane was made similarly, it would be more like a tank than an airplane, and would be too heavy to fly. Planes are made to fly, CVR's are made to crash, as "they" say. :) Arakunem^Talk 18:03, 30 October 2007 (UTC)

Check out this classic Cecil Adams column. "If aircraft "black boxes" are indestructible, why can't the whole plane be made from the same material?" 69.95.50.15 18:01, 30 October 2007 (UTC)

It's also worth noting that an indestructible airplane would provide little extra protection to all-too-destructible passengers, who are still independently subject to the laws of inertia. Crash casualties are not so much the product of a deforming airframe as they are of a sudden stop. In fact, the deformation is good -- it absorbs crash energy that would otherwise injure passengers even more. Modern cars are constructed with this same theory in mind. — Lomn 18:12, 30 October 2007 (UTC)

I wonder if the querent is looking for material for a stand up comedy routine? According to our article, Flight data recorders tend to be "double wrapped, in strong corrosion-resistant stainless steel or titanium, with high-temperature insulation inside." We don't make jetliners out of titanium for the same reason we don't make windows out of diamonds, its not cost effective (though, as an aside, while the SR-71 Blackbird was made out of titanium, its windows were not diamond), and if we packed it full of high-temperature insulation, where we would put the passengers? Rockpocket 18:14, 30 October 2007 (UTC)

Also, black boxes aren't black - they're orange. SteveBaker 00:04, 31 October 2007 (UTC)

Depends on how hot the fire is. --DHeyward 05:09, 31 October 2007 (UTC)

Even if the plane were made to survive a crash, it doesn't help people inside. Say the plane can survive 1000G of force, your body cant and you would be squished to death on impact, regardless of if the plane body deformed or didn't. The best solution for safety would have to be a plane that deforms enough so that a +100G impact is reduce to say 10G.--Dacium 05:50, 31 October 2007 (UTC)

Digitap imaging & printing

perfect on screen but when i printed it on my Epson stylus Photo RX620 it looked nothing like what it did on screen. I was truely gutted. It was duller, darker, more saturated & lost a lot of fine detail. I am using the correct paper & my ink channels settings are set to default. This is happening with all of my images.

Why is this happening & how can i resolve it so that what i create on screen prints as it looks on screen? Also what is the difference between a jpeg, tiff, Esp, btmap etc?

Thanks kindly

Nay —Preceding unsigned comment added by 86.145.223.206 (talk) 19:33, 30 October 2007 (UTC)

For the latter, articles such as JPEG, TIFF, and BMP file format may be of use. For the former, I have no suggestion apart from experimenting with configuration settings (as this is not so much a case of correctness as one of preference). — Lomn 20:05, 30 October 2007 (UTC)

You may find this is better answered on the computing reference desk however I'll have a quick stab. Computer screens use additive colour. They add red, blue and green colours together to make white. Printers OTOH use ink which substracts colour. They use cyan, yellow and magenta and mixing them together makes black :-( So the screen will never look the same as a printer.

Also ink jet printers use wet ink thqat runs a bit and the colours bleed together making it dull. Having a good quality laser printer and using top quality paper will definately produce crisper and brighter prints. Fortuately the cost of colours lasers has been plummeting in recent years Theresa Knott | The otter sank 20:10, 30 October 2007 (UTC)

There are many problems here. Additive versus subtractive, impure inks and toners, the fact that you are printing in four colours (Cyan, Magenta, Yellow and Black) instead of three, that the printer needs to use dithering and bleeding of inks to do what it does, that the brightness of your screen is independent of the room lighting - but the brightness and hue of your print is entirely dependent on the colour of the ambient lighting - that the software that does the conversion is rarely correctly set up for the kind of paper you are using - that the gamma of cameras and screens are NEVER correctly set up, that not all programs take notice of the gamma values that the camera put into the file header - and if you save the picture out in another format, that information is almost certainly lost, that our eyes respond differently to a pigment that reflects yellow (meaning a true yellow) versus a seemingly identical pigment that reflects both red and green, that your CRT probably has several different colour temperature settings and you have no clue which one you picked...there are a million reasons. The bottom line is that you're doomed and you just have to tweak the available settings until you get it "how you like it". (There is no "Right"). SteveBaker 23:50, 30 October 2007 (UTC)

Oh - and the difference between various file formats (for the purposes of this discussion) are that JPEG and GIF both compromise the true colours of your photos in order to save a lot of memory. 24 bit TIFF and PNG can store colours in their original perfection with no losses whatever. BMP (urgh) can go either way - and it's a horrid format anyway so don't use it. The various 'raw' formats that cameras sometimes use are better still - IF you have the right software to handle them. There are some esoteric formats that do even better - but I doubt you'll ever come across them. On balance:

• Use PNG for original artwork and photos stored on your computer at home - where (presumably) you have plenty of disk space.
• Use RAW images for original photo archives if your camera and software supports them - but convert to PNG for day-to-day use.
• Use JPEG for photographic types of pictures on the Internet - it's compact - which means it downloads quickly - and the losses due to it's compression tricks are not noticable on a typical browser at the default screen resolutions we have these days. As broadband starts to conquer the world - I'm going more and more towards using PNG even in these situations.
• Use GIF only on really high usage web pages for cartoony stuff or (if you absolutely must) for animated images.
• Don't use TIFF or BMP if you can possibly avoid it - they are both only patchily supported in areas like gamma that truly matter.

SteveBaker 00:01, 31 October 2007 (UTC)

Mom Haircut

Okay, this is a stretch, but I'm going to try it anyways... just bear with me, okay?

Is there a condition or an illness (for lack of a better word, I don't mean to imply bad health) that would describe the tendency for new mothers to get a drastic haircut? I don't mean the "mom haircut" but, just even like dying it a different color, something that makes a dramatic change. I know this could just be associated with going through major changes in their life or, more dangerously, depression, but I was wondering if there was a name for it specifically. Thanks, and I apologize if this question is just too dumb for words. Beekone 20:18, 30 October 2007 (UTC)

You might find more information under the mid-life crisis article -- MacAddct1984 20:43, 30 October 2007 (UTC)

One could argue it is an evolutionary successful strategy: long hair is an energy intensive characteristic adopted by females to attract a mate. After pair bonding and generating offspring, the female no longer considers the energy invested in maintaining long hair is no longer required, hence they cut it. Or it could be that they don't want to get nits from all the little brats they interact with. Or it could be to reinvent themselves for the next stage in their lives, or it could be part of an effort to "look good" again after the physical changes associated with child birth. Interestingly (or not) there does seem to be some recognition of "mom hair" in the mommy blogosphere, but its extremely unlikely that it has a specific biomedical name. Rockpocket 20:45, 30 October 2007 (UTC)

I'm not even sure what you are describibg is a real phenomenom. Is there any data to suggest that on average new mums get more dreastic haircuts than women of that age who are not new mums? Theresa Knott | The otter sank 20:52, 30 October 2007 (UTC)

No, I'm not sure such research exists... I guess that's why I asked. It seemed like a funny quirk that I'd noticed in some friends and acquaintances. I agree that the likelihood of this activity having a specific medical name is extremely un, but you never know until you ask, right? Beekone 21:05, 30 October 2007 (UTC)

Googling "mom haircut" at least provides evidence that the "mom haircut" is culturally believed to be a real phenomenon. From reading the various pages, it sounds like a "mom haircut" is either quite short, or long and tied back. In either case, it sounds like the point is to have a low-maintenance, no-nonsense haircut that isn’t going to get in the way of the harried new mother, who suddenly has less time for dealing with such frivolities as hair. MrRedact 21:30, 30 October 2007 (UTC)

Don't think anyone has mentioned the obvious answer (well, obvious to anyone with practical experience of babies), which is that babies of a few weeks and older love to practise their hand-eye co-ordination by grabbing and pulling long hair. Gandalf61 10:39, 31 October 2007 (UTC)

"I don't mean the "mom haircut" but, just even like dying it a different color, something that makes a dramatic change. " signed, the very first post! Read, think, reread, comprehend... maybe Beekone 13:04, 31 October 2007 (UTC)

You start off with: "Is there a condition or an illness ... that would describe the tendency for new mothers to get a drastic haircut?" If you are going to ask contradictory questions, you are going to get contradictory answers. Rockpocket 17:14, 31 October 2007 (UTC)

How does "drastic haircut" translate into "mom haircut"? I mean I know I called the topic that, but I state repeatedly that that's not what it's about. To be technical I start off with this: "Okay, this is a stretch, but I'm going to try it anyways... just bear with me, okay? " 'Clearly the message is going to be muddled so I should pay close attention,' thought the reader. Beekone 17:18, 31 October 2007 (UTC)

Additionally, I liked your answer, Rockpocket. You seem to get what I was asking, and correctly stated that such a condition is unlikely to have an actual label. But at least you got what I was asking, even if you're answer was basically "I don't know." The other answers are "I don't know" plus a hint of "I obviously didn't try to get what you were asking" which i frustrating because if you don't know why even answer? Just pass it by. I obviously asked in the case that someone with knowledge of the topic might answer. It's sort of what the Help Desk is all about... in my understanding anyways. Beekone 17:23, 31 October 2007 (UTC)

Thats a fair point, well taken. Rockpocket 04:10, 2 November 2007 (UTC)

Why don't they use water jets on very large ships?

The largest they used it on was a frigget I think. Why not larger ships? 64.236.121.129 20:26, 30 October 2007 (UTC)

Our article on pump-jets notes some of their advantages, contrasted with a decrease in efficiency versus propellor-based designs. In particular, the note about power density may be relevant, as frigates don't have as much hull volume to fill with engines as do destroyers or cruisers. The shallow-draft advantages provide additional merit for new littoral combat ships. At some level, though, the lack of adoption may simply be industrial inertia. Propellors are well-understood and consequently are considered to inherit less risk than a relatively new technology like pump-jets. In the absence of a compelling advantage for the latter, then, the former is likely to remain in widespread use. — Lomn 20:36, 30 October 2007 (UTC)

Many very large ships still have bow thruster jets.--Dacium 23:54, 30 October 2007 (UTC)

Sugar in Soda

I've decided to eliminate soda from my diet. While shopping at a local grocery store, I came across a powder lemonade product that one mixes with water. The lemonade’s packaging said it "contained 40% less sugar than soda." So with everything being equal between the soda and the lemonade (calories, fat, etc), what would be the benefits of 40% less sugar? I would assume that there would be some benefits…or is this a fancy marketing ploy?

Thanks 64.85.199.27 20:41, 30 October 2007 (UTC)

It's precisely a marketing ploy. Whether it actually benefits you depends entirely on your diet. People hear that carbs are bad, carbs make you fat, so they advertise that they have fewer carbs. Sugars/carbs are the normal person's primary source of energy, and it is the first form of energy (calories) the body will burn in order to operate (fats it prefers to store away). Now, if the food in your diet contains all the calories you need in a day, and you just want something to quench your thirst, there's really no need to drink something laced with sugar that will just get converted into fat for long term storage, so you may as well go for that low-sugar lemonade. Someguy1221 20:47, 30 October 2007 (UTC)

(ec)Soda contains a lot of sugar! 40% less means 60% still remaining which is still very bad for your teeth. Having said that soda contains not fat, protein etc. It's water and sugar, so 40% less sugar will be 40% less calories. Those calories that remain are junk food calories though. Much better to drink water or tea. Theresa Knott | The otter sank 20:49, 30 October 2007 (UTC)

Also, the drink may have had that 60% of sugar boosted by an artificial sweetener, and if you read sugar substitute you'll see that all the common ones have some kind of controversy about potential health risks attached. Confusing Manifestation 21:39, 30 October 2007 (UTC)

Just drink diet or zero soda's (diet coke, coke zero, pepsi max etc). These contain zero suger. Also other things that are high in suger are sauces like ketchup, alcohol etc. Personally I would try to ditch soda all together and drink water. Try drinking two classes of water before you eat.--Dacium 23:27, 30 October 2007 (UTC)

There are two very different things being talked about here. Diet Coke/Pepsi/whatever contains almost zero carbs. Giving up Diet coke and drinking this stuff will increase your carbs by a large percentage because their claim to have less carbs than "Soda" is a poorly worded and highly misleading statement. WHICH soda do they have 40% less than? If has 40% less than one of those insane "energy drinks" (Mountain Dew MDX for example) then it probably has more sugar than Classic Coke too! Different kinds of soda have everything from zero sugar up to (probably) the maximum amount that'll stay dissolved in water without crystalizing out! It's nothing to do with whether it's carbonated or not. So if you need to reduce your sugar intake, switch to a diet soda. If you are concerned about the CO2 in the soda - then switch to a non-carbonated beverage. But don't give up either "just because". Read the ingredient list and compare the actual numbers - forget the silly marketting percentages. Think about what you are really doing! As others have mentioned, if you get your total carbohydrate intake down to zero - you'll get seriously ill amazingly quickly of something called "protein poisoning" (aka Rabbit starvation). SteveBaker 23:39, 30 October 2007 (UTC)

Actually sugar will dissove in huge amounts in water without crystalising out. I suppose the max would be something like golden syrup but you would't drink that. Theresa Knott | The otter sank 00:35, 31 October 2007 (UTC)

Will it fly?

I say yes, and so does straightdope, but what is the answer????--Goon Noot 23:08, 30 October 2007 (UTC)

Yes it will. Airplanes move due to thrust from the engines, not torque from the wheels. The plane's wheels will just be spinning at 2x the takeoff speed when it leaves the ground. Arakunem^Talk 23:11, 30 October 2007 (UTC)

I say nay. There is no lift because the plane isn't moving forward.

It is moving, but it is on the treadmill, and the treadmill isn't moving forward.

strap that whole shebang on top of a car... then we're talking —Preceding unsigned comment added by 128.175.187.238 (talk) 05:59, 31 October 2007 (UTC)

-Stephen Hawking —Preceding unsigned comment added by 74.196.103.230 (talk) 23:15, 30 October 2007 (UTC)

It's an enormous treadmill, but the aeroplane's wings will be ripped off as it starts to fly. DuncanHill 23:17, 30 October 2007 (UTC)

In the original formulation of the question, the treadmill was infinitely long and the wheels completely free-running. The photo just confuses the question. The plane doesn't give a damn what happens to its wheels - they are free-running. Moreover, once it's engines are running with enough power to overcome the tiny amount of rolling friction in the wheels, the plane will take the same distance to take off no matter what speed the treadmill is running (forwards, backwards, who cares?). This one has been done-to-death and the answer is as clear and obvious now as it was at the beginning. I don't believe Stephen Hawking (or at least "The" Stephen Hawkin) said that. SteveBaker 23:21, 30 October 2007 (UTC)

It 100% will not take off. There is no way for it go gain lift. The plan engines thrust forth, pulling the wing, with the idea that the wing is pulled fast through the air, air goes over the wing and creates lift. There is no way for the plain to gain lift. People who think it could take off must also believe that a plane with no wheels could simply turn on its engines and hover up in the air.--Dacium 23:22, 30 October 2007 (UTC)

If you think about it, any plane is already on a giant moving treadmill. It's called the Earth. Remember that all motion is relative, so putting a plane on a treadmill moving with constant velocity is really no different from having a wind coming up from behind the plane. From the plane's perspective, it has to go a little faster to take off because of this wind. From your perspective, it takes off at normal speed, but it takes a little longer to reach it (since it started moving backwards with the treadmill) and the wheels will be turning faster. Now, you could also have the treadmill accelerating backwards at the same rate the airplane accelerates while taking off. In this case, the plane sits where it is, the wheels constantly speeding up, and the engines working to keep the plane exactly where it is (although, in that case, you could just turn the plane around and use the treadmill to launch it). Simply put, all motion is relative, being a treadmill won't have any greater effect on the plane than wind. Someguy1221 23:24, 30 October 2007 (UTC)

A plane moves forward by air entering the engines to create thrust. A plane has to build up it speed because there is only little thrust to begin with. As the plane gets faster, more air is entering the engine and the thrust increases. It is the air speed relative to the plane that is the only thing that matters. A plane could hover straight up in the air if the air speed was great enough. The treadmill stops the plane gaining speed relative to the air and stops it taking off.--Dacium 23:44, 30 October 2007 (UTC)

The plane still moves. Unless the treadmill is accelerating, the plane will eventually beat it and take off. Imagine the treadmill is massive, say, the Earth. Would the plane notice or care if the treadmill were moving? Not unless it's accelerating. So this is why I say, if the treadmill is moving backwards at 30mph, this is no different than a tailwind of 30mph. The plane has to speed up a little more (from its own perspective) to generate sufficient lift. But it is by no means prevented from moving. Someguy1221 23:49, 30 October 2007 (UTC)

Sorry I thought we were assuming the treadmill was accelerating to match the speed of the plain. Obviously if it doesn't the plain will still be able to gain speed relative to the air, and take off. So we are all in agreement then. Non accelerating treadmill - plain takes off. Accelerating treadmill - plane does not take off.--Dacium 23:52, 30 October 2007 (UTC)

The treadmill would have to have tremendous acceleration to slow the plane below its takeoff speed. (Remember, the wheels are free-wheeling. The only thing slowing the plane is double the normal amount of friction in the axles. The situation is nearly the same as if the plane had hovercraft instead of wheels. ) I'm pretty sure that a treadmill constructed to put out that sort of ridiculous acceleration would reach near-C speeds in a minute or two then the plane could still take off normally. 69.95.50.15 18:42, 31 October 2007 (UTC)

So I have this funny feeling I may have my relativity backwards on the accelerating part, someone please check me on that. Someguy1221 23:57, 30 October 2007 (UTC)

The wheels play zero part in generating forward speed. The engines react against the air to generate forward thrust. The treadmill could be going faster in "reverse" than the plane is moving forwards, and it will still move forward. The wheels would be spinning reeealy fast backwards but thats irrelevant to this. Arakunem^Talk 23:33, 30 October 2007 (UTC)

It all depends on how you define the question. If the airplane's wheel bearings are frictionless and the wheels themselves are inertialess, then it doesn't matter what the treadmill does: the airplane's engines will move the airplane forward and it will take off. If the wheels have rotational inertia, or the wheel bearings have friction, and the treadmill is allowed to adjust its speed without limit, then it is possible to use the friction or rotational inertia of the wheels to counter the thrust of the engines, and the airplane won't take off. --Carnildo 00:03, 31 October 2007 (UTC)

I think it's best to assume the plane is just a normal plane, with normal wheels. Someguy1221 00:08, 31 October 2007 (UTC)

Say a big plane like in the picture takes off at about 150mph, do you think it could take off with a 150mph tail wind? I don't think so. So imagine the treadmill is already travelling with the plane on it at 150mph backwards, can the plane take off? I don't think so. Now imagine the plane and the treadmill both start at 0mph, can the treadmill accelerate the plane backwards as fast as the plane accelerates forwards? I don't think so. At the very start of the plane's acceleration as the engines spool up a considerable amount of the thrust would be used to overcome the friction in the wheels, but very quickly as the thrust increases, I think the treadmill would have to travel MUCH MUCH faster backwards then the plane accelerates forward for the friction of the wheels to act as a break on the plane's forward acceleration, otherwise the plane will come to speed and take off, just with very fast spinning wheels. Vespine 00:54, 31 October 2007 (UTC)

Yes, if the treadmill is allowed to transfer enough net force through friction in the wheels it could hypothetically keep the plane stationary. If the plane is stationary with respect to the air, it will not take off, since lift depends only on the plane's speed with respect to the air. Dragons flight 00:57, 31 October 2007 (UTC)

Just to straighten things out hopefully. Assume still air (no headwind or tailwind). If the conveyor belt/plane contraption is set up so that the plane remains motionless relative to the air (whether or not the conveyor belt, wheels, or engines are going) the plane will not take off as no air is moving over the wing generating no lift. If the plane is moving forward relative to the air, it could take off. Another way to think about this problem is to envision a plane on a conveyor belt that pushes the plane forward while the plane's engines are off and the wheels not rolling at all. Will the plane take off? Yes, because air is moving past the wings generating lift. (Of course to maintain flight it needs the engines, but that is besides the point) Or you can envision it this way: with a strong enough headwind, a plane perfectly at rest could take off although to a stationary observer on the ground, the plane would be moving backwards while acquiring altitude (i.e. moving in the direction of the wind). I hope this helps. Sifaka ^talk 03:59, 31 October 2007 (UTC)

No, it will hit the handrails of the treadmill. (Darn, DuncanHill beat me to that answer :P)-- Diletante 01:10, 31 October 2007 (UTC)

It's worth noting (belatedly) that we did this question to death back on May 21. —Steve Summit (talk) 01:31, 31 October 2007 (UTC)

More importantly, would a bird flying in front of the treadmill get sucked into the plane's engine? Rockpocket 01:39, 31 October 2007 (UTC)

Yes because the engines are drawing in air. 71.226.56.79 04:02, 31 October 2007 (UTC)

If the treadmill matches whatever the speed the plane can do, then surely it would stir up a huge surface headwind wouldn't it? Voilà, there's your moving air. --antilived^{T | C | G} 07:09, 31 October 2007 (UTC)

A treadmill moving at 100 or 150 mph wouldn't stir up that much of a headwind. (And stop calling me Shirley.) —Steve Summit (talk) 12:01, 31 October 2007 (UTC)

It doesn't have to limited to 100 or 150mph. Assuming it's a perfect treadmill it and the plane will simply accelerate until there's enough induced wind from the treadmill for the plane to take off. --antilived^{T | C | G} 05:13, 1 November 2007 (UTC)

It won't take off because there's no airflow around the wings! That's the whole point of why you need a runway for an airplane. You could say the Earth is a giant treadmill, but the air moves with the earth, and so do we. Air does not move with the treadmill. I'm surprised some of the people here thinks it will take off. It really makes me doubt how credible they were to answer my question about the ducted fans aircraft. Maybe it will take off if you have a powerful fan in front of the plane, but then you don't need a treadmill in the first place. Just thrust against the wind. 64.236.121.129 13:23, 31 October 2007 (UTC)

Aww you're mad *pats your head. You're also thinking of it wrong. I'm not going to bother explaining it because your last few posts prove that you aren't interested in listening. *pats your head*. Just because you don't understand it, doesnt mean it isnt so. Now you run on back to Starcraft lil guy —Preceding unsigned comment added by 207.67.148.212 (talk) 15:49, 31 October 2007 (UTC)

As I thought I made clear in my Earth/treadmill analogy, the only point of that is to dispel any belief that frame of reference is important. The only difference the plane cares about is the relative wind speed. Someguy1221 17:11, 31 October 2007 (UTC)

Actually, this is quite backwards. It takes off fine from a treadmill under the generally stated conditions (the treadmill moves at the same speed as the aircraft). Carnildo raises a good point that wheels aren't frictionless, and if you allow the treadmill to run so fast that the airplane is stationary, then it won't lift off (probably -- you're left with a weird feedback cycle). Thus, the key point of properly stating your constraints and assumptions is illustrated. On the other hand, a stationary aircraft in front of a fan can lift off, but as soon as it's no longer in front of the fan, it crashes due to the rapid change in airspeed over the wings (as opposed to the treadmill liftoff, in which the aircraft flies away normally).

In a related question, A friend once suggested to me that the real issue in this situation would be that the wheels would be turning at double their normal RPM (obviously). He suggested that the tires on many jets are not rated for those RPMs and would destroy themselves before the plane went airborne. Is this true? Stated more clearly, if a 747 was placed on a treadmill like the one described so that its wheels were always turning at double the normally expected rate, would the plane survive takeoff? 69.95.50.15 18:36, 31 October 2007 (UTC)

With no true evidence, I'll say "probably". The wheels undergo significantly more stress at landing (going from no speed to 150 MPH or so very quickly) and have a sizeable engineering margin above that. Simply spinning on takeoff should be no big deal. — Lomn 19:13, 31 October 2007 (UTC)

For the consequences of a tyre failure on takeoff, see Air France Flight 4590. Admittedly, the tyre failed due to debris on the runway, but airplane tyres were subsequently redesigned. -Arch dude 22:25, 31 October 2007 (UTC)

As i stated in my post, the "initial" resistance of the wheel would be very quickly overcome by the acceleration of the jet engine and I belive the wheels would have to spin MUCH MUCH more then twice as fast. Think of this, scale down the plane so it fits on a piece of paper on a desk, move the paper slowly and the plane will move with it, move the paper fast and the plane won't move, the wheels will just spin faster, i think the same thing will happen with a jet. Once the engines are at full thrust i think moving the treadmill backwards WON'T be able to stop the plane from accelerating, it will just make the wheels spin faster backwards, unless you start spinning then SO fast as to act as a break, in which case I think probably eventually fast enough to fail. Vespine 06:04, 1 November 2007 (UTC)

I was referring to the most commonly stated version of this problem where the treadmill goes backwards at the plane's take-off speed. In this case the plane takes off almost normally except for the tires which spin double their normal speed. 69.95.50.15 16:32, 1 November 2007 (UTC)

I think the main point here is that the air will not move with the treadmill. People seem to be thinking here that somehow either the treadmill can move the air, or the plane's engines drive the wheels. Either way you need air movement over the wings to generate lift. The treadmill's movement, forward, backward, whatever has no bearing on the air. The plane still needs to be pulled (by its engines) through the air forward to generate lift. The force from the engines act on the air, not the ground, so how the ground is moving has no bearing. The plane in that picture absolutely would not leave the ground. It's like having someone holding a hang-glider and running on a treadmill. They're not going to suddenly zoom up into the air, because the wing isn't travelling through the air. That's the whole point of a treadmill, so you don't move anywhere. —Preceding unsigned comment added by 132.244.246.25 (talk) 08:45, 1 November 2007 (UTC)

Hey, let's put this a couple of different ways:

• You've got a 3 km runway, well and truly long enough for about any plane. But some tricky person has replaced the surface with a conveyor belt loop. The belt runs at 100km/h (~60mph). Can a Cessna 152 take off?

Plane is dropped onto the runway at the North end, facing north, and starts accelerating south (backwards). The engine's running and is throttled up to maximum, it starts moving forwards, engine slowed so it progresses backwards at walking pace (5km/h), that's groundspeed (relative to treadmill) +95km/h, airspeed -5km/h, groundspeed relative to control tower -5km/h. The Cessna gets half way down the runway, pilot decides to stop stuffing around, and throttles up to max, speed increases, at 120km/h airspeed it takes off, that's airspeed=120km/h, groundspeed relative to tower=120km/h (it's using up runway at 120km/h and getting rapidly closer towards the end), groundspeed relative to treadmill = 220km/h - well within design limits of the plane. Yep. It flies.

Of course, you can make the situation work or not work as you please by tweaking the setup. Sure you can find a treadmill speed where the plane's wheels are wrecked, sure you can make the plane too large or the treadmill runway too short for the plane to take off - but was that the question?

If the original question was "We have a cessna 152, with a flying speed of 120km/h, on a treadmill as long as the plane. The treadmill is run up to 120km/h, will the plane fly?" the answer to that is an emphatic "no", groundspeed is irrelevant to whether a plane flies. --Psud 10:00, 1 November 2007 (UTC)

You mean an emphatic "yes", right? Because the Cessna will have the same airspeed it always has. It's 'groundspeed' would be double if you're counting the treadmill as the ground, but like you said, the groundspeed is irrelevant. (The wheels are free-spinning, Imagine them as hovercraft if that helps you.) 69.95.50.15 16:38, 1 November 2007 (UTC)

Relative to the surrounding air this plane is motionless. Ergo the lift = 0. Ergo it won´t take off.
The tradmill is totally irrelevant.
--Cookatoo.ergo.ZooM 23:12, 1 November 2007 (UTC)

PS: It is identical to a normal plane on a "normal" runway taking off at a tailwind which inceases parallel to the speed of the accelerating plane. The Cessna / Jumbo / whatever could travel at the speed of Mach x or even v=c. As long as the velocity of the tail wind is identical, there are no aerodynamics to create a counter-gravitational force.
QED. --Cookatoo.ergo.ZooM 23:27, 1 November 2007 (UTC)

ZOMG, people are still not getting it.... but the plane is accelerating with JET THRUST not with the wheels, the plane, at full throttle, WILL accelerate through the air, regardless whether the wheels are spinning forward or backwards, the ground and wheel resistance is negligible to the thrust of the engines. there is a clip on youtube of a thrust test where a bus behind a jumbo jet, not RIGHT behind either, some way behind, is blown clean over and away by the blast from the engines. Put it this way, people agreed that the treadmill will NOT be able to move a significant amount of air with it, right? So it MUST also work in reverse, is the treadmill, even going super fast enough to counteract ONE HUNDRED THOUSAND LBF of thrust?? NO WAY! The plane WILL accelerate and WILL take off. Actually using that example, I've changed my mind from my previous post, I think it will accelerate and take off even if it starts at "going backwards" on the treadmill at whatever speed, it will just take longer for it to gather the airspeed... the people who think it will not take off are essentially making the assumption that a treadmill under the plane can counteract the acceleration of 100000 lbf of thrusting air, no way no how. Vespine 00:51, 2 November 2007 (UTC)

It's becoming clear that this is, in effect, a troll question. It is specifically designed to fuck with your head. The more you think about it, the more confused you get. Or, even if you don't overtly confuse yourself, you find yourself compelled to come up with ever-more bizarre theories to explain how the plane might not be able to take off.

And if you stop and really think about it, coming up with those ever-more bizarre theories is a pretty stupid thing to do. Remember, the primary purpose of the question is to fuck with your head and spark meaningless but prolonged debate. When the people who aren't thinking clearly say "Oh, right, of course the plane won't take off", the sadistic person who thought up the question in the first place calls them stupid and has a good laugh at their expense. But seriously: you think he's not also laughing at you when you get all "clever" and come up with some exotic and barely-plausible (well, even less plausible than a runway-sized treadmill, that is) theory under which you get exactly the same answer as the stupid people!?

As we all hopefully know by now, the proper response to trolls is to ignore them, to not feed them, to wait patiently for them to get bored and go away. And I think we had best apply the same remedy to this dorky airplane-on-a-treadmill question, too. (And soon, before this thread achieves critical mass and collapses into a black hole, or something...) —Steve Summit (talk) 04:57, 2 November 2007 (UTC)

Oops. I recant my above statements. The devil made me do it!
Due to a brief attack of fallibilty, causing delusions of omnisciensce, I uttered what may be classified as pure and unadulerated bovine faeces.
Thank God I am not a pope or president, but just a humble wheel on a treadmill, attempting to soar into the heavens of perpetual knowledge and eternal bliss...
I promise to take my prescibed medicine in future !--Cookatoo.ergo.ZooM 13:00, 2 November 2007 (UTC)

October 31

Running hands under Superacid

After reading Superacids, what would happen if someone washed their hands under a tap of fluoroantimonic acid? How fast would the hands disintegrate? Acceptable 01:35, 31 October 2007 (UTC)

I guess it is unacceptable to post chromic acid stories? Delmlsfan 01:51, 31 October 2007 (UTC)

I once breathed in fumes of hydrogen chloride gas (by accident), not a very fun experience... -- MacAddct1984 14:02, 31 October 2007 (UTC)

Once it hits the water of your hand, it's no longer nearly as super. That is, it becomes little different than a really really concentrated solution of other acids (hydrochloric, sulfuric, etc) in regards to its acid effects. On the other hand (sorry:) the fluoro and stibbic components might present other hazards all their own. DMacks 02:01, 31 October 2007 (UTC)

Transporting rice

Car Talk's most recent puzzler was something along the lines of:

With old wooden ships, what common household item, if not stored properly, would sink a ship in a matter of minutes?

Apparently rice was the answer. If it wasn't stored properly and got wet, the massive amounts of densely packed rice would swell and split the ship apart. Has anyone heard of this before? It certainly sounds plausible. -- MacAddct1984 01:42, 31 October 2007 (UTC)

Car Talk got this information from the book Tall Ships by Philip McCutchan.[9] The author’s name is misspelled on Car Talk’s web site.[10] That book says "Spaced along the upper deck were the cargo hatches with their heavy covers of reinforced hardwood planks, well chocked in and secured with three separate layers of tarpaulin, held down with ropes and more chocks to withstand the pounding of heavy seas. Below the hatches lay the reason for the ship's presence on the sea, her cargo, to be held inviolate against nature and disaster, against fire that could come from a self-combustible cargo like wool, or a cargo that could swell when it met water, such as rice, which on more than one occasion in the long story of the sea swelled and in its irresistible pressure split the sides of holds like paper and sank those ships in minutes."

But that’s just what Philip McCutchan said. From the British shipping act in 1875, it sounds like the primary concern about properly storing grain on ships at the time was actually that the ship could sink if the grain shifted.[11] Maybe this is a question for MythBusters. MrRedact 03:25, 31 October 2007 (UTC)

I guess what bothers me about this is that once the water gets at the outer regions of the rice cargo, that will start to swell and the pressure build-up ought to lock out the water from getting in much further. But a lot depends on how fast the water is getting in and how fast rice swells. I guess it could be true. Definitely a good one for Mythbusters though. Car Talk's puzzlers are not exactly reliable sources of information! SteveBaker 03:32, 31 October 2007 (UTC)

This was a plot in a Horatio Hornblower book. He lost a prize of war. The ship was sinking even though no water was in the "well". It gains weight as well as stressing the hull. --DHeyward 05:18, 31 October 2007 (UTC)

For the locking question, I think osmotic pressure will keep sucking the water in to the center until they are all equally saturated. A dry rice grain next to water saturated one will suck the water out of it. --DHeyward 05:21, 31 October 2007 (UTC)

DHeyward: that was not a book, but rather the second chapter of Mr. Midshipman Hornblower. Algebraist 21:08, 31 October 2007 (UTC)

Children's vs Adult's voices

• What are the differences between children's and adult's voices, excepting the obvious physiological effects of growth and hormonal adjustments?
• For example, children's voices tend to sound more sing-songy, breathy, and rhymical, but have there been any studies to make quantitative or qualitative measures of the difference in voice-qualities as age progresses?
• It is obvious that human's are readily able to differentiate between younger and older speakers independently of any use of linguistic cues : are computers able to do the same, and under what heuristics would they operate?
• Are there any posited hypotheses of the cognitive causes of these non-physiologically determined variations?

Many thanks 81.153.3.36 10:05, 31 October 2007 (UTC)

I think this belongs in the section on Language and Linguistics. Meanwhile, I'm not well studied at all in developmental linguistics, but I can try to answer with what I know.

Obviously, as you stated, as a child gets older his larynx expands and produces deeper tones as the vocal chords vibrate. Similar resonance differences occur as the size of the oral and nasal cavity expand. However, the prosodic changes (those in rhythm and tone) can probably be traced to what are still not well understood phenomena in child linguistics. For example, a toddler will show reduplication (repetition of words) and other types of word play, depending on age, as he learns to speak. He will also exaggerate and misuse intonation (the parents do this too when they speak to children for the precise reason that intonation is difficult to master). All of this is most likely a result of the problem of learning something as complicated as language simply from cues around you. Computational models of this type of learning are very simplistic, last I've seen, and usually involve limited degrees of freedom (such as a simple robot learning to walk or play the drums according to outside information). None that I know of are linguistic in nature.

It is true that the age of a speaker can be approximated by linguistic cues alone. This is done in precisely the same manner as you mentioned, as well as by judging vocabulary, grammar use, pronunciation, and pragmatics (appropriateness of responses). Computers are probably most advanced at this point in distinguishing pronunciation, but still trained phoneticians are used to diagnose and treat speech disorders, because software simply still isn't good enough. The variations you discussed are not well understood computationally or anatomically, so I don't believe there have been any good hypotheses other than those suggested by experience with children. SamuelRiv 04:02, 1 November 2007 (UTC)

Thanks SamuelRiv. I guess I wasn't as clear as I could have been. I was interested in specifically those differences that aren't linguistic in any way. That is, not vocabulary, grammar use, pronounciation or pragmatics, nor anything else for which it is require to assign meaning to the sounds. Rather, I wondered what kind of differences were recognisable in the pure sound-form of the voice, aucoustic qualities you might say.

In any case, I found a paper which is along the lines of the ideas I expected : [12] : though it still seems that there is woefully little work done on understanding speech outside of pathological/dysfunctional cases, which I guess is understandable, if not totally satisfactory. 81.153.3.36 12:09, 2 November 2007 (UTC) (Cross-posting to linguistics desk, just in case.)

Things stuck in eyesocket

Suppose something small and mildly irritating (like, say, a bit of cat fur) got stuck up under someone's eyelid and worked up into the eye socket. What would happen to it? Would the body destroy it or expel it somehow? How long would it take? Maelin (Talk | Contribs) 13:16, 31 October 2007 (UTC)m n

In general, this idea falls apart when you expect something to work its way behind the eye. The eye, if I remember correctly, is pretty much sealed in the skull. There's no way to, for instance, lose a contact lens behind the eye. If you somehow DID get something behind the eye, it would have been through a wound, and the effects of that wound would be worse than the object itself, I should think. --Mdwyer 14:39, 31 October 2007 (UTC)

I guess it's possible something could embed itself up there and work its way in somewhat. I would expect a cat hair to be absorbed over a long period of time because it's protein, but anything inorganic...I shudder to think. And I suppose that the body would try to do to such a thing what it does to all such things, like splinters. The eye is different though; inflammation alone can be a serious matter. See sympathetic ophthalmia here and here. I doubt the body would have time to deal with an embedded cat hair before the eye reacted unfavorably to it. You can tell by my wishy-washy reply that I'm no expert. --Milkbreath 15:43, 31 October 2007 (UTC)

The function of the eyelashes would prevent a situation like this from occuring. Their job is to keep foreign objects out of the eye (i.e. dust, dirt, debris, cat hair). They work in conjunction with your tears to do it. Even though the primary function of the tears are to lubricate the eye, they also function as a sort of safety mechinism, when an object gets into the eye (be it dirt, dust, or cat hair) the eye gets irritated and tears up. The offender then gets caught up in the tears and then is eaither flushed out or it gets near the eyelid and caught by the eyelashes and taken out of the eye. Lastly, a thin, but strong, membrane, called the conjunctiva, lines the inside of your eyelids and curls back on itself to cover the white part of the eye. This prevents any obects from going behind the eye. Hope this helps! Josborne2382 16:18, 31 October 2007 (UTC)

I've gotten specks of dust stuck between my eyelid and my eye. That's not quite behind the eye, but I still wonder how it always ends up getting out, and how I could get it out faster. — Daniel 03:26, 1 November 2007 (UTC)

Hair loss

I asked my doctor about this but he didn't know exactly. I want to know if anybody has a link or two about uninterested reseaches on hair loss and effectiveness (if any) of treatments. I have quit given up hope on what constitutes medical advice so I'll just post fingers crossed and see if it gets deleted.193.188.46.254 14:03, 31 October 2007 (UTC)

The baldness article has a huge list of the various possible prevention/reversal treatments: Baldness#Preventing_and_reversing_hair_loss -- MacAddct1984 14:10, 31 October 2007 (UTC)

Freezing eye

Some time ago I read on how difficult it is for the eyes to freeze, but I forgot the reasons, could anybody patch me up on this? Also what would hypothetically hppend if you poured liquid nitrogen on somebodies eye? —Preceding unsigned comment added by 193.188.46.254 (talk) 14:16, 31 October 2007 (UTC)

I'd guess it is because they're salty, and salt depresses the freezing point. Still, LN is going to make them freeze anyway. Don't do it. :) --Mdwyer 14:34, 31 October 2007 (UTC)

It also helps that the human eye is embedded in a nice warm human being, equipped with all kinds of tools to maintain the body at a fixed 37°C. Heat is readily conducted from the back and sides of the eye socket (which, being right next to the brain, are well-supplied with nice, warm, constant-temperature blood) to the rest of the eye. Still, it would be possible to freeze the surface of the eye if you made enough of an effort—like by direct contact with a cryogenic liquid, for example. Tests in rabbit eyes gave the result:

Liquid nitrogen poured onto the eyes for one or two seconds with the lids held apart, produced no discernable injury. When the exposure was extended to five seconds, slight lesions of the corneal were observed. By the next day, all eyes were entirely normal.

I would expect longer exposure to do progressively more damage. Assumptions about the behaviour of human eyes based on the rabbit model should of course be taken with a grain of salt. TenOfAllTrades(talk) 18:43, 31 October 2007 (UTC)

Does it strike anyone else as odd that it is someone's job to pour chemicals into the eyes of rabbits? Man It's So Loud In Here 19:35, 31 October 2007 (UTC)

Rabbits are commonly used in animal testing. -- JSBillings 20:04, 31 October 2007 (UTC)

Not at all, how else would we determine the safety and efficacy of drugs formulated as eye drops? Rockpocket 20:21, 31 October 2007 (UTC)

How many brands of eye drops have liquid nitrogen in them? I'm no PETA pusher, but that's just insanely and purposelessly cruel to the rabbits. What's the next experiment: tossing bunnies into a wood chipper? Maybe we'll develop a better bandaid from it. Matt Deres 13:21, 1 November 2007 (UTC)

Why do birds fly away from people?

I think this question is appropriate for the Science desk, since it seems to concern evolutionary biology. The question occurred to me this morning as I walked through a group of pigeons, which scattered as I approached them. This seems to be normal behaviour among our feathered friends, but when I stopped to ask myself why, I couldn't come up with an answer. In other words, why are birds scared of people? Of course, the answer that comes to mind first is that they are afraid of getting shot. Now, the chances of a pigeon being shot in a city street are pretty remote, although I guess the pigeon doesn't know that. So, is there some kind of evolutionary impulse at work here under which birds instinctively (and wrongly, in the case of city-dwelling birds) believe that man should be avoided? And if so, why hasn't this impulse atrophied among city-dwelling birds? --Richardrj ^{talk email} 14:25, 31 October 2007 (UTC)

Well, not just birds run away from humans. I can't think of any animal, unless it's a pet or trained, that wouldn't run away from a human. Humans are fairly big mammals, and if you saw something 10x the size of you, it's more than likely to benefit you to run away from it. Why hasn't the behavior atrophied? Probably because there is no selective pressure to fade out the trait. If running away from humans suddenly had a negative impact, such as they all started getting hit by cars when they try and fly away, then it's more than likely to weed out the trait. In the meantime, the pigeons have nothing to lose by flying away from approaching humans. -- MacAddct1984 14:39, 31 October 2007 (UTC)

Running/Flying away (Fight-or-flight response) is a pretty common behavior amongst animals. Not flying away from people is the unusual thing, probably bred into generations of city birds by the selective pressure of increased chance of food and less energy expense, as well as a lack of predators. Read about the Dodo to see what happens to a bird with few predators. -- JSBillings 14:40, 31 October 2007 (UTC)

(After edit conflict) A general instinctive fear of being killed and eaten by a creature much larger than themselves? The street pigeons seem to mostly tolerate humans, seeing them as something to simply move out of the way of - as they understand from experience that the majority of the large bipeds of the concrete forest pay them little attention and mean them no particular harm. They only seem to become agitated and take flight (around here, at least) if a human is getting too close, actively following them around, trying to trap them up against a wall, or running/walking towards them at speed - all of which the bird may (rightly?) interpret as hostile acts. --Kurt Shaped Box 14:43, 31 October 2007 (UTC)

Great answers. Thanks all. --Richardrj ^{talk email} 14:53, 31 October 2007 (UTC)

(another edit conflict) I think the 'rule' is probably more general than "man should be avoided" - something more along the lines of "if it's bigger than you and coming towards you, it should be avoided". Birds that tended to fly away when approached by something bigger than them would have an evolutionary advantage over those who hung around because that something may be a predator. Even if it's not a predator, it's extremely unlikely that it would be something that would provide the bird with some advantage over its fraidy-cat brethren who didn't stick around long enough to find out, so the wait-and-see trait would be disadvantageous and would die out. The impulse is still there because it's still advantageous, even amongst city-dwelling birds, to get out of the way of bigger things (imagine how odd would be if birds all wandered around on crowded streets, or didn't bat an eyelid if they were standing in the road and a car approached).

In fact, not all birds take flight as soon as they're approached by a human. Before it was banned, holding a handful of seed in Trafalgar Square was guaranteed to result in you being mobbed by pigeons. — Matt Eason ^{(Talk &#149; Contribs)} 14:58, 31 October 2007 (UTC)

Yes, they're afraid because instinct tells them anything that gets close to them is probably trying to eat them! Although they are used to humans, they are still preyed upon by cats etc. Feral Pigeons are sufficiently tame to approach a human if they have food and some birds can be tamed so that they are not afraid of humans, my two pet budgies will run up to me and climb onto my hand if they are on the floor and I hold my hand out. It's just a case of convincing the bird that you are not going to harm it which takes time and perseverence. At the end of the day, one of the most basic instincts of all animals is to survive which means avoiding anything that is likely to be dangerous.GaryReggae 15:01, 31 October 2007 (UTC)

Actually, some birds tend to be fond of humans. It was (and sometimes still is) for a human to be feeding food to gulls, pigeons, sparrows, ducks, geese, etc. Some other birds, especially flightless ones, are not afriad of humans. That's what killed the dodo so quickly. One time, for example, we were having a picnic. A small ~3 in. tall bird (not sure what it's called) came towards the food, and actually landed on the picnic table. If we tried to swat it away, it would fly away, hover in the air, and come back. It often landed about a foot away from us. Another time, we were just minding our own buisiness, and walking near the shoreline at a lake, when a few dozen ducks and geese who were near the shoreline seemed to follow us. As we walked by them, they continued to follow us. The moral of the story is, feeding of birds, even in the past, has caused some of them to follow humans and trust any food given by them. Now there are some laws restricting feeding of birds, but apparently the birds don't know that. Hope this helps. Thanks. ~AH1^(TCU) 23:49, 1 November 2007 (UTC)

If human bodies had thick fur like other mammals, do you think we would still wear clothes?

She didn't. Rockpocket 20:25, 31 October 2007 (UTC)

Do you? 64.236.121.129 19:16, 31 October 2007 (UTC)

Hmm, that's a fun question. It depends where our sense of modesty comes from. I would imagine without the invention of clothing and the ability to cover up, there would be no way for embarrassment of genitals to come about. However, clothing also provides an additional sense of style and individuality. So maybe, just being human, we'd wear some sort of clothing anyway -- MacAddct1984 19:47, 31 October 2007 (UTC)

Individuality could be expressed by styling and coloring the fur, but I think clothing would still have been invented. Much of it provides a lot of utility beyond keeping warm. Lab workers, for example, would probably still wear some variation on a lab coat. Police officers would still wear Kevlar vests. I'm no expert, but it seems like if our current attitudes towards clothing evolved from the need to cover up in cold weather it wouldn't be surprising if other uses for clothed eventually evolved into similar attitudes. 69.95.50.15 20:18, 31 October 2007 (UTC)

Also where would you put your wallet if your a man? Certainly i'd still wear clothes, gotta have a place for my phone, keys, wallet and iPod...You can see why mugging is so popular! The above stuff is right too, clothes are cultural-identity too, they are also a sub-culture thing. Goths, punk rockers, skaters, sports-addicts, 'chavs', horse & hound types etc. A lot of clothes (and styles) seem to have started their life as work-based clothing and/or developed to become fashion/general wear. Jeans were work-clothes (still are), Cargo pants, three-piece suit (though Adam Hart Davis suggests it was an attempt by the British to reduce the love of French clothing in the Stuart period), sports-wear is obviously often derived from sport-use clothing. Obviously work-wise these things are not just to cover modesty but as a form of protection/security/uniformity too ny156uk 23:38, 31 October 2007 (UTC)

How much sugar does the average kid/adult consume on Halloween?

^topic 64.236.121.129 19:23, 31 October 2007 (UTC)

Can you have an average kid or adult? ΦΙΛ Κ 21:09, 31 October 2007 (UTC)

Well you would expect that in any 'holiday' season people will allow themselves more treats, so possibly some like say three times as much as the 'average' sugar intake on any other day. This article (http://www.sys-con.com/read/451581.htm) is basically utter garbage but it does mention halloween and sugar intake. ny156uk 23:33, 31 October 2007 (UTC)

Ugh, I've heard stories of kids eating all their candy in one day, and there's often several hundred pieces of candy! No wonder people get sick and diabetes. Besides, if you're counting average as in world average, then the average kid doesn't go out for Halloween, either because the local culture doesn't celebrate it, and/or because the locals are too poor to afford that much candy. Besides, if you ate all your candy in one day, chances are a few of them may have been tampered with (I remember getting candy with the package opened). Some people argue that eating all your candy in one day has the same effect as eating it over a longer period. This is a complete falsehood. It might be a bit close to the truth if you're not going to have a single piece of candy or other treat for the next 6 months. Besides, eating all your candy in one day can cause blood sugar (and fat, and plastic, and colouring, etc) levels to suddenly spike up. It's kind of like drinking a gallon of alchohol in one day instead of the same amount over a 6-month peroid. Hope this helps. Thanks. ~AH1^(TCU) 23:58, 1 November 2007 (UTC)

Cyanide and suicide

Is suicide via ingestion of cyanide, as seen in James Bond movies, fact or fiction? Babalonia 3. —Preceding unsigned comment added by 66.74.109.242 (talk) 20:41, 31 October 2007 (UTC)

As can readily be found in our article on cyanide, it can quite easily be used as a poison. Adolf Hitler, among others, used it as such. — Lomn 20:57, 31 October 2007 (UTC)

When you say "as seen in James Bond movies" do you mean ingestion of cyanide causing death in seconds? Sifaka ^talk 22:52, 31 October 2007 (UTC)

I thought Hitler died from shooting himself (though his wife and dog were killed by cyanide pills). — Daniel 03:18, 1 November 2007 (UTC)

I think you're right. I'll hunt for a reference and fix the article. — Lomn 14:27, 1 November 2007 (UTC)

The references in Adolf Hitler suggest that he both bit a cyanide capsule and shot himself, then had his body incinerated, just to be sure. — Lomn 14:30, 1 November 2007 (UTC)

Which do you think killed him? I don't care how fast cyanide can kill. A bullet to the brain would be faster. By the way, does bit a cyanide capsule mean he swallowed a pill, popped some kind of tiny balloon with his teeth, or what? — Daniel 01:54, 2 November 2007 (UTC)

Exercising, Muscle Burn and Calories

Upon reading over several articles (such as muscle and exercising ), I’ve got a few questions relating to “muscle burn” after a rigorous workout. Assume that one runs a mile on a treadmill. Most treadmills indicate calories burned over the duration of the run. Further assume that, according to the treadmill, 200 calories are burned (let’s pretend this number is accurate). If the runner has muscle burn from the rigorous run, does he/she actually burn more calories than the 200 indicated by the treadmill? I would assume so, as the body is ‘recovering’ from the workout, and thus will burn additional calories post-run.

Thoughts? Rangermike 21:25, 31 October 2007 (UTC)

See Lactic acid and Delayed onset muscle soreness, if you are referring to the fatigue and soreness one feels in one's muscles during and after workouts.

This is out of my field, but the body is burning calories with everything you do, say, or feel, but nothing burns as quickly from homeostasis as physical motion. For a physicist's example, say you were sick, and you had a fever of one degree celsius. At a weight of about 100kg, you would burn 10^5 calories = 100 kcal (kilocalories or kcal are the actual unit of measure when people refer to "calories". One calorie properly refers to the heat necessary to raise the temperature of 1 gram of water by 1 degree celsius). Keep in mind, this fever builds up over several hours, whereas you can burn that much in a 30 minute workout. SamuelRiv 03:28, 1 November 2007 (UTC)

Gravitational pull of the moon

Hi,
I've heard that there is no instrument on the earth which can measure the gravitational pull of the moon (apart from the obvious affects of the tides). Is this true? --124.181.69.55 21:35, 31 October 2007 (UTC)

Let's see here.... The Earth will pull on a 1 kg mass with a force of 9.8 newtons. Using an Earth-Moon distance of 380 000 km, the Moon overhead would tug on the same mass with a force of about 34 micronewtons. So the apparent weight of an object due to the moon's motion over the course of a full lunar orbit will appear to vary cyclically over about 70 parts per million. That's not a huge amount, but it's certainly measurable. Particularly if one monitored a very stable, very precise balance over the course of several lunar orbits, I would expect the effect of the Moon's pull to stand out as a periodic oscillation. A very quick Google search for microgram balances finds this one, which will set you back about $7000, and which will measure masses up to 5 grams with a resolution of 2 micrograms (about 0.4 parts per million). Can someone check my numbers? TenOfAllTrades(talk) 22:05, 31 October 2007 (UTC)

I get 7 ppm from your numbers (2 * 34e-6 / 9.8) — Lomn 22:25, 31 October 2007 (UTC)

In theory, there's no reason you can't. In practice, it may be the case that tolerances don't allow us to (I'm not sure). Just for reference, though, here's what you're dealing with. Gravitation is a function of mass over distance squared. For the Earth, we'll say that it's M/R², where M is the mass of the Earth and R is the radius of the Earth, and we'll normalize that resulting number to 1. The Moon, on the other hand, masses roughly 1/80 Earth, and never gets closer (to the surface) than about 56R at perigee. Substituting, we expect that the Moon causes a discrepancy over a no-Moon Earth of 1/(80*56*56) -- that is, 4*10^-6 or 0.0004%. Now, in practice, that discrepancy can be roughly doubled -- the difference between the Moon directly overhead and directly underfoot (noting that the move from 56 to 58 Earth radii is pretty minor at this point). So in total, to detect the influence of the Moon's gravity on Earth-surface objects, you need accuracy to six digits, or one one-hundred-thousandth of the object. (ec) I notice that I'm an order of magnitude off Ten's answer. Time to double-check. — Lomn 22:13, 31 October 2007 (UTC)

I will note, however, that you're monitoring a periodic variation of [whatever the correct magnitude, 7 or 70 ppm] over a day, not a lunar orbit (month). Monitoring over the course of a month could be used to confirm the roughly 25% variation in the daily cycle resulting from the eccentricity of the Moon's orbit. — Lomn 22:18, 31 October 2007 (UTC)

Oops, right. The Earth rotates once a day, doesn't it? Dang. Double oops; you're also right about the 7 ppm (not 70). I'm going to have to turn in my physics license. TenOfAllTrades(talk) 23:02, 31 October 2007 (UTC)

LIGO has to compensate for lunar tides, so there is an example of an active experimental instrument that is sensitive to the moon's gravity. Dragons flight 23:43, 31 October 2007 (UTC)

They did it on an episode of mythbusters. They used an extremely powerful accelerometer to measure if the anti-gravidy contraptions had any effect whatsoever (they didn't). They breifly mentioned the accelerometer measuring the tidal forces from the moon and showed the graph it made. — Daniel 02:01, 1 November 2007 (UTC)

You may be confusing an important law in Special Relativity, which states that a person in a closed space can never tell whether he is being accelerated due to external gravity or some other outside force (such as a rocket). See the article for details. You may also be thinking of the Cavendish experiment, in which the gravitational force of the moon, the planets, and the Earth were all simultaneously measured by measuring the gravitational constant G, which combined with Newtonian orbit theory gives everything you could possibly want about gravitational forces in the solar system. SamuelRiv 03:27, 1 November 2007 (UTC)

The "apart from the tides" part of the question is pretty telling though. If a simple yardstick stuck into the sand at the low tide mark can detect the presence of the moons gravity, it's unlikely that there would be nothing else that could do the job. SteveBaker 01:15, 2 November 2007 (UTC)

The tides are a tiny deformation in the earth. A proportional deformation on an artificial device would be ridiculously small. Measuring the tides with an accelerometer is essentially the same as measuring height above sea level to the meter with one. Although what anon heard is wrong, it's not particularly misleading. — Daniel 01:51, 2 November 2007 (UTC)

I'm sorry, my point wasn't clear -- all you need to measure the gravitational pull of the moon is Newtonian gravitation (i.e. Kepler's Laws) and a measure for G (Cavendish). Then from Earth using parallax or radar we can calculate the Earth-Moon distance, and we know the period of the Moon's orbit around Earth and so we can calculate the mass of the Moon, and the Gravitational field of the Moon is then ${\frac {-G*M}{r^{2}}}$, where M is the mass of the Moon and r is the Earth-Moon distance. SamuelRiv 02:15, 2 November 2007 (UTC)

oxidizers

are they flamible without fuel?

Various ether peroxides, like Acetone peroxide, Diethyl ether peroxide, and Tetrahydrofuran peroxide spontaneously explode when disturbed. Lab accidents involving such peroxides formed when a bottle of ether is exposed to light, oxygen, or catalytic metals and then subsequently ignored for a long time happen every once in a while. Sifaka ^talk 23:15, 31 October 2007 (UTC)

Thinking more about this, I bet some oxidizing agents with more complex chemical structures could react with themselves or other molecules of the oxidizer. There are probably also some nasty chemicals which have the potential to react under a variety of storage conditions without deliberately added fuel. Sifaka ^talk 23:28, 31 October 2007 (UTC)

oxidation

is mn203 a good pyrotechnic oxidizer? —Preceding unsigned comment added by 216.103.183.127 (talk) 21:59, 31 October 2007 (UTC)

I'm not getting any hits on google for manganese (III) oxide (also called manganese sesquioxide) being used as a pyrotechnic oxidizer. Safety precautions for the compound as listed in the sigma aldritch catalog (source) list it as an irritant to eyes, respiratory system and skin so I doubt people would appreciate it being scattered about by pyrotechnic displays where it could be inhaled. Among the applicable safety phrases it says don't add water to it. Of important note is that sigma didn't list it as an oxidizer, so I am leaning towards doubting its efficacy as an oxidizer at this point (I need to research around a little more). Many other manganese compounds appear to be fairly strong oxidizers including Manganese(III) acetate. Perhaps you meant to ask about one of them instead? Try looking at the manganese page. Sifaka ^talk 23:58, 31 October 2007 (UTC)

Tased across the spine

Can a person be paralyzed if the electrodes are opposite the spinal cord? I recently saw a video of someone getting tased in the back (the flying wires kind, not the little zapper kind) and it made me wonder --ffroth 22:14, 31 October 2007 (UTC)

See Taser for details. The weapon uses a shaped electric current to disrupt nerve function, I suppose in a process similar to a localized seizure. Unless structural damage occurs to nerve fibres or bone, the current itself shouldn't produce permanent effects. The brain and spinal cord both have a certain degree of plasticity (ability to change - i.e. learning) that is believed to depend on the intensity and duration of electric currents. See Electroconvulsive Therapy for information on the effects of high-intensity electric currents on the brain. Permanent effects to the spinal cord could possibly be similar, but would only affect memories stored in the spine, namely "muscle memory". —Preceding unsigned comment added by SamuelRiv (talk • contribs) 03:38, 1 November 2007 (UTC)

My answer is "Sure, why not?" We tend to think of a relatively robust person getting tased, but suppose a frail child gets in the way? The darts can penetrate up to an inch, and from what I saw poking around the internet, you get millisecond-range currents of up 8.5 amps. That's a lot of current, however short the pulse. If the darts get deflected so that they land right next to each other on either side of the spinal cord, I wouldn't be surprised if it caused electrical burns in the tissue in between. It was hard to get solid information on electrically caused tissue damage by googling, and I don't know much about the specific direct effects of the taser waveform on soft tissue. There is a lot of propaganda out there spewed by parties with an interest in tasers, for marketing, for furthering the use of non-lethal force by police, and for opposing tasering as an invitation to police excesses. The truth has been effectively obscured on the internet, it seems. --Milkbreath 10:34, 1 November 2007 (UTC)

Well, when we got our Taser's late last year, we all had to "take a shot" to carry them, I have video's of it all. AFAIK every charge was administered via clipping one probe to the right rear shoulder, and one probe on the left pants pocket cuff, then having 2 people hold onto the victi... er administeree's arms. I'd consider that across the spine, no ill effects from any of us(though thats not to say it isn't possible, I believe elderly people or pregnant woman is what our SOP say to try to avoid tasing), minus the videos floating around of us yelling 'oh fuuuuuuuu', and yes I am not ashamed to admit I screamed like a baby and being in that large of a muscle mass it really takes a lot out of you. Yes, I had what appeared to be tissue burns on the spots where the probes were, they went away after about 2 days, and I can say there was no pain there, even the second the charge was up, I felt shaky but nothing else. Even out support staff (dispatch, admin aide, records clerk, evidence tech) all volunteered to take a shot, and yes, the women did seem quieter than the guys. Dureo 11:00, 1 November 2007 (UTC)

Oh and remember, the probes can penetrate deeper in winter, there are specifically longer winter probes for winter clothing, and yes unless they are torn out as they fall we have EMS handle removing them. Dureo 11:04, 1 November 2007 (UTC)

secret service mobiles

Just curious about how this locking down on mobile signal works and how do the mobiles that are manufactured against it work.88.203.105.48 22:42, 31 October 2007 (UTC)

When a police service says they're locking down the mobile phone system (usually because of a bomb threat), they generally mean that they're ordering the mobile phone companies to stop routing phone calls to mobiles in the relevant area. It stops people using a mobile phone to trigger a bomb. Aside from that, I'm not really certain that that's what you're asking. Can you be more clear (for example, can you post a link to a website or news article that spurred you to asked the above question)? --Psud 10:18, 1 November 2007 (UTC)

It's all just a matter of software. Every phone has it's own unique number - the cell towers know the numbers of the phones in their area. In case of an emergency, the computer in the cell towers can decide what they want to do with calls. If it's an emergency, it could (for example) stop routing calls from any phone that's not listed as belonging to an emergency worker. This stuff is very easy to do if you have the inclination. SteveBaker 01:08, 2 November 2007 (UTC)

By locking down I meant pin point the location of the user of the mobile 193.188.46.254 09:18, 2 November 2007 (UTC)

They used to locate mobile phones by asking the phone company which phone towers it was communicating with, mobile phones keep in contact with several towers to allow instant handover as you move between cells. It's a simple matter of looking for the phone/person in the area where the various towers' signals overlap. That's the simple version anyway (and I don't know enough to give you the complicated one). The new way they look for phones is by asking the phone for its GPS position. Only works with phones with GPS. The old method works with any phone, I think. Perhaps it could be defeated by hacking the phone so it talks to the "wrong" mobile phone towers. It'd probably be trivial to hack a GPS equipped phone to report a wrong location. --203.22.236.14 09:58, 2 November 2007 (UTC)

Rising Sea Levels and the Effect this Will Have on the Mediterranean Ocean

Given that the Med is regarded (more or less)as a tideless sea and that this is largely due to the influence of the Straits of Gibraltar which constrict the tidal bulge caused by the gravitational pull of the sun and/or the moon. At what point, if at all, or to what degree, in the earths future, will the Med see a real tidal effect once the seas have risen enough to overcome the resistance of the Straits of Gibraltar? Much of the Med coastline is at or very near sea level because there has never been any need to make allowances for tidal fluctuations. —Preceding unsigned comment added by 79.68.42.88 (talk) 23:08, 31 October 2007 (UTC)

Well much of all society is near the sea/lakes because historically we settled near places with ready access to water. The med problem is interesting. I expect it would require an extremely large increased. As the article (Strait of Gibraltar) notes water depth ranges from 300m to 900m, the opening is some 8 miles wide - having said this the end of the article has a bit on the 'need for a dam' so might be worth having a look at that link (http://www.agu.org/sci_soc/eosrjohnson.html) ny156uk 23:28, 31 October 2007 (UTC)

facial hair

Why do males have facial hair? Why don't females? —Preceding unsigned comment added by 68.231.151.161 (talk) 23:45, 31 October 2007 (UTC)

Vestigial trait. Doesn't serve any meaningful purpose unless you are white, it can sort of protect the face from UV radiation, but melanin (dark skin) does a much better job of that, and protects everywhere. So in other words, it's useless from a utilitarian point of view, but has cultural and social significance. Malamockq 01:24, 1 November 2007 (UTC)

Facial hair is a secondary sex characteristic resulting from the differential effects of sex hormones on the body. Rockpocket 07:50, 1 November 2007 (UTC)

Some women do have facial hair. -- JSBillings 10:25, 1 November 2007 (UTC)

In theory it helps to protect against the cold in high northern places like Scandanavia or Russia. As to why women dont, I dont know. But look at most antarctic scientists, they have big beard for the cold. If you have hair, when was the last time your head got cold? —Preceding unsigned comment added by 12.191.136.2 (talk) 15:08, 1 November 2007 (UTC)

Finite from Infinite

Today's science dictates that not only is a theoretical construct of infinite density, energy, and size possible; it's actually how the universe started. How does a finite universe come from something that is infinite? —Preceding unsigned comment added by Sappysap (talk • contribs) 23:55, 31 October 2007 (UTC)

Firstly, there is no accepted theory suggesting the Universe is finite. While the observable universe is quite finite, the actual universe is possibly infinite in extent and mass. Further, physics does not posit that the universe began in a state of infinte density. It actually doesn't posit anything accepted about where it came from, as the Big bang theory only deals with what happened after (very soon after, though) the universe started existing. Someguy1221 00:27, 1 November 2007 (UTC)

Thirdly one way to look at it is fintite mass divided by zero size gives infinite denisity. Graeme Bartlett 00:30, 1 November 2007 (UTC)

I'm a huge fan of all of you guys who answer these questions...Someguy1221, Rockpocket, SteveBaker, Dragon's Flight and the whole crew. Please have infinite patience with me while I ask the following: How can the universe have infinite mass without infinite density? How can it have infinite extent and continue to expand? Sappysap 01:07, 1 November 2007 (UTC)

As for the first, I'm still trying to get my head around it, but for the second, imagine a piece of elastic, infinitely long, with small knots tied at regular intervals. Now imagine the elastic being stretched - you'll see the knots moving apart, even though you can't see the "ends" moving apart. Confusing Manifestation 02:00, 1 November 2007 (UTC)

To answer how the universe can have infinite mass without infinite density, all you have to do is think of numbers. There are infinitely many integers, as I'm sure you're well aware. There are also infinitely many numbers inbetween every two integers. And so, you manage to have a system of infinitely many numbers containing an infinite number of integers yet not every number is an integer. Actually, as you might suspect from there being infinitely many numbers between two integers, and only finitely many integers between any two numbers, each integer is like a tiny island in the sea of infinity. So, in conclusion, your infinitely many integers occupy an infinitely small portion of the number line. So I think you can see from this how you can have, in an infinitely large universe, infinitely many masses without filling up every possible space with a particle. As for having infinite extent, I haven't quite figured or accepted that one yet, except to say that lots of astronomers believe this (it is in part just a conclusion from the evidence that the Universe has no center or edge). Someguy1221 02:33, 1 November 2007 (UTC)

The universe has infinite mass if and only if it has infinite volume. Obviously everything around us has finite density, but if you allow yourself to collect more and more of it without limit, then ultimately you arrive at an infinite amount occupying an infinite volume. It is unknown whether the universe is truly infinite in extent, but it appears likely that it must be very much larger than the observable universe we can see. Dragons flight 03:26, 1 November 2007 (UTC)

None of the above physical quantities are or ever were truly infinite. The widespread modern cosmologies (that is, those based on Einstein’s general relativity, which came out about 90 years ago) all involve a universe that is finite in size and energy (or mass). And I’m talking about the entire universe, not just the visible universe. It isn’t really accurate to say that the density of the universe was infinite at the exact moment of the big bang, either. Singularities in a physical theory generally indicate a point at which the physical model breaks down, not that there really is an infinite physical quantity at that point. In this case, physics within the Planck epoch are poorly understood, but it appears that it is meaningless to talk about distances less than one Planck length, which is a finite distance, or time intervals shorter than the Planck time, which is also finite. So it isn’t meaningful to talk about energy densities greater than the (finite) total energy of the universe divided by the volume of a sphere whose diameter is a Planck length, or meaningful to talk about the exact instant of the big bang. MrRedact 02:46, 1 November 2007 (UTC)

I'm not sure what the initial question refers to, specifically. There are indeed important resolutions of infinite results in Quantum Field Theory, which I might be able to expand upon if a detailed question is posted, but all "physical" results must be finite by definition. Unresolved singularities (infinite results) signify a failure of a theory, not a physical reality. Cosmology works a little differently, but even then most cosmologies I know of do not accept any premise of infinity except perhaps in the context of a ground state (see vacuum). SamuelRiv 03:45, 1 November 2007 (UTC)

You are confused. General relativity permits solutions that are either infinite or finite in extent, and so is essentially agnostic on the issue. (See for example: Friedmann-Lemaître-Robertson-Walker metric, which permits different homogeneous and isotropic solutions discriminated primarily by whether the universe is finite or not.) For the universe to be finite, it must have a global curvature such that a person traveling in what appears to him to be a straight line will eventually come back to places he has been before. Such a curvature is possible, but not required, in GR. Your description of the Planck quantities is also problematic. It is not that there couldn't theoretically be smaller lengths, times, etc., but rather that describing events at those scales intrinsically requires both an unified understanding of gravity and quantum mechanics. In other words it requires an as yet ill-defined theory of quantum gravity. So, it is meaningless primarily in the sense that science as it now exists is not able to provide it meaning. Future theories may yet shine a light on events in and before the Planck epoch. Dragons flight 04:09, 1 November 2007 (UTC)

Ouch. I should have researched this one a little better before responding. I’m obviously not an expert in cosmology. Einstein himself assumed closed boundary conditions on his field equations, which implies a finite universe, but what slipped my mind for some reason was that not everyone assumes those boundary conditions.

My description of the Planck quantities very closely echoes a sentence in the Planck epoch article: "When quantum mechanics is combined with gravity, it turns out that it is meaningless to speak of time intervals shorter than the Planck time or distances shorter than one Planck length." That sentence, at least, is consistent with my understanding that "space" and "time" are ill-defined concepts at shorter times and distances. Of course, it’s all highly speculative, in the absence of a good theory of quantum gravity. MrRedact 10:36, 1 November 2007 (UTC)

November 1

What Are Some Things All Things On Earth Have In Common That Are Living Or Non-Living?

I wonder, what are some things all things on Earth have in common that are living or non-living? —Preceding unsigned comment added by 208.103.143.9 (talk) 03:07, 1 November 2007 (UTC)

Please clarify your question. Everything is living or non-living. — Daniel 03:15, 1 November 2007 (UTC)

They're all made up of atoms? -- MacAddct  1984 ^{(talk &#149; contribs)} 03:19, 1 November 2007 (UTC)

Not everything is made up of atoms. Evidence_of_evolution#Evidence_from_comparative_physiology_and_biochemistry lists many things that living things mostly have in common. As for non-living things, their only binding similarity is that they aren't living. Someguy1221 03:21, 1 November 2007 (UTC)

They're all made of matter and/or energy, they're all affected by the laws of physics, and they all hate getting circus peanuts on Halloween.  ;-) -- HiEv 04:07, 1 November 2007 (UTC)

Everything came from a star. --DHeyward 13:43, 1 November 2007 (UTC)

Everything? I'd be willing to bet that there's a great deal of primordial hydrogen floating around right now. --Carnildo 23:55, 1 November 2007 (UTC)

What part of ON EARTH, didn't you get? 64.236.121.129 13:38, 2 November 2007 (UTC)

Contradiction in mass-to-energy equivalence?

I'm new at Wikipedia editing and whatnot, so I wasn't sure where I should put this where it would be noticed:

---

The article "Antimatter Weapon" states:

Quantities measured in grams or even kilograms would be required to achieve destructive effect comparable with conventional nuclear weapons; one gram of antimatter annihilating with one gram of matter produces 180 terajoules, the equivalent of 43 kilotons of TNT.

---

The article "TNT equivalent" states:

By E = mc^2, when 1 kilogram of antimatter annihilates with 1 kilogram of matter the reaction produces 1.8×10^17 J, which is equal to 42.96 Mt.

---

While I could just do the basic equation (mass*c^2, and convert joules to TNT equivalence), I didn't want to make an assumption without consulting anyone else first lest I made a mathematical error or overlooked some other factor in the calculation. —Preceding unsigned comment added by 24.65.12.157 (talk) 04:44, 1 November 2007 (UTC)

1 kilogram is 1000 grams; 1 Mt (megaton) is 1000 kilotons. Is that what you are confused about, or is there something else? Dragons flight 04:48, 1 November 2007 (UTC)

Ugh. I missed the "one gram" versus "1 kilogram" part. Well, we all make mistakes. C'est la Vie. —Preceding unsigned comment added by 24.65.12.157 (talk) 04:54, 1 November 2007 (UTC)

Medical question

Deleted. William Ortiz 08:38, 1 November 2007 (UTC)

Sorry, but this page isn’t a place to ask for medical advice, either. As it says at the top of the page: "Do not request regulated professional advice. If you want to ask advice that "offline" would only be given by a member of a licensed and regulated profession (medical, legal, veterinary, etc.), do not ask it here. Any such questions may be removed. See Wikipedia:Medical disclaimer and/or Wikipedia:Legal disclaimer. Ask a doctor, dentist, veterinarian or lawyer instead." MrRedact 09:04, 1 November 2007 (UTC)

Do you know where on the internet they give medical advice? William Ortiz 09:11, 1 November 2007 (UTC)

Here, if you're careful to hide the request for medical advice by disguising it as a question of biology, anatomy or the way medical devices work. But you've given the game away now, so I doubt you'd have much luck asking the same question again (by the way, I think what you suggest (a) wouldn't work, and (b) would be a bad idea). Actually, a google search would help you find online medical advice. --Psud 10:32, 1 November 2007 (UTC)

NHS Direct is a good place to start.--Shantavira|^{feed me} 12:50, 1 November 2007 (UTC)

Big Bang

Was the Big Bang a chain reaction? Clem 09:59, 1 November 2007 (UTC)

If by "chain reaction" you mean a nuclear reaction then the answer is no, because protons and neutrons did not begin to form until about a millionth of a second after the Big Bang, and they did not begin to combine into stable nuclei until about 1 minute or so after the Big Bang - see our articles on the hadron epoch and Big Bang nucleosynthesis. Gandalf61 11:40, 1 November 2007 (UTC)

Of course not silly bro. How can there be a chain reaction with particles not yet formed? I'm talking about with whatever was formed. Did it have a chain reaction? Clem 14:05, 1 November 2007 (UTC)

There's no need to call a volunteer "silly" for answering your question as phrased -- especially when he then expands your question to give you exactly the answer you wanted. It's rather rude and quite unwarranted. Another example of what Gandalf linked is at timeline of the Big Bang, particularly the 17-minute period of nucleosynthesis which established the initial ratio of hydrogen to helium. — Lomn 14:21, 1 November 2007 (UTC)

I think most interactions that occurred immediately after the big bang--quark-gluon interactions to form nucleons and nucleons and leptons forming atoms--are not chain reactions because there are no particle emissions necessary any of these reactions - a chain reaction is a series of reactions that require as input some output from a previous reaction, so that the previous allows the next to occur and so on. So I would say the answer is no in general, but specific reactions might be catalyzed from energy or lepton release by some of these primitive interactions. SamuelRiv 14:28, 1 November 2007 (UTC)

I suppose I'm thinking more in terms of instances of interrupted stability wherein under different circumstances some changes might not occur at all or occur as rapidly except for the occurrence of previous events such that event of type "B" will cease to occur if and when event of type "A" ceases to occur. Maybe "sequence of dependent events" is more apt terminology than "chain reaction". Clem 17:09, 1 November 2007 (UTC)

Okay then, you are asking a very general question. All big-bang cosmologies that I know of come from very fundamental fluctuations in the background. A fluctuation may create a false vacuum, for example, and the cosmological effects are calculated from that state alone. So everything follows from this one single event, and yes, from there everything is dependent on what comes before. Caveat--the notion of dark energy may prove that some critical events in the universe's history are not caused by a "sequence of dependent events". SamuelRiv 02:24, 2 November 2007 (UTC)

"Superlinear"

First things first, hi desk, long time no see! Anyone have any idea what this word "superlinear" means? My first guess would be a function that increases faster than a linear function, but then you can of course have an incredibly steep linear function. A friend of mine asked me in the context of a biological function. Can someone clarify? It's not something that i've heard in my undergraduate career. Google has a load of results but none of them are very clear from what I can see. Could it be a function that is more linear than a linear function? That makes no sense to me either. Math board, maybe? :? —Preceding unsigned comment added by Capuchin (talk • contribs) 11:59, 1 November 2007 (UTC) Cheers, Capuchin 11:39, 1 November 2007 (UTC)

When describing functions, "superlinear" appears to mean a function with asymptotic growth that is more than linear i.e. it grows faster than any multiple of x (there is a different, but related, usage in "superlinear convergence"). Superlinear functions include functions such as x² or indeed any power of x greater than 1, also x^x, and functions such as xlog(x), xlog(log(x)) etc. Note that even a "steep" linear function will eventually be overtaken by a superlinear function - if f(x)=Ax and g(x)=Bx², then no matter how large you make A and how small you make B you can always find a lower limit y such that g(x) > f(x) for all x>y. Gandalf61 12:04, 1 November 2007 (UTC)

Okay thank you, that's what I had assumed. Capuchin 12:10, 1 November 2007 (UTC)

Do all the biochemical reactions happened in our body have a negative free energy?

I am a high school student. Yesterday, my teacher teacher give us a question written: Do all the biochemical reactions happened in our body have a negative free energy. Explain and support your answer. I don't know how to answer it. Any one can help. Thx in advance!! —Preceding unsigned comment added by 202.40.139.171 (talk) 12:32, 1 November 2007 (UTC)

Wikipedia has articles such as Gibbs free energy and enzyme kinetics, but they either do not address your question or they do not address it in plain English. This biochemistry textbook has a standard treatment of free energy in biochemical reactions for college students. If you are not ready for the math skip down to, "the free-energy change must be negative for a reaction to be spontaneous." The complication in biochemical systems is that there are many "coupled reactions" in which one chemical reaction that has a positive free energy change is "driven" by coupling it to a second chemical reaction that has a negative free energy change. Example: Two-step conversion of glyceraldehyde 3-phosphate. So the answer to "Do all the biochemical reactions happened in our body have a negative free energy?" depends on how you define "biochemical reaction". If you define "biochemical reaction" to include both of the chemical reactions in a pair of biochemically-coupled chemical reactions then yes, the combined free energy changes of the two chemical reactions are negative. If you define each of the individual coupled chemical reactions to be its own "biochemical reaction", then it is clear that some individual chemical reactions that take place in biochemical systems have a positive free energy and you could call those "biochemical reactions". --JWSchmidt 15:45, 1 November 2007 (UTC)

If you have a library card you can get help with homework here. 71.100.9.205 14:14, 1 November 2007 (UTC)

Put it another way: do all biochemical reactions in the body happen spontaneously? That is, do they all occur as soon as it is possible for them to occur? Think about what this would mean. SamuelRiv 14:32, 1 November 2007 (UTC)

Spontaneous human combustion? Clem 18:40, 1 November 2007 (UTC)

• SamuelRiv makes a very good point. (see Enzyme) - Mgm|^(talk) 09:18, 2 November 2007 (UTC)

Blood phobia

How do extremely hemophobic women cope with getting their period? --124.254.77.148 13:11, 1 November 2007 (UTC)

With great difficulty I would imagine. Your question sort of reminded me of the Ashley Treatment. Short of hysterectomy or Endometrial ablation, I suppose they could reduce the number of periods they have using one of the many pills, but would still have to go through a few. Alternatively, maybe frequent exposure reduces the horror? Skittle 13:48, 1 November 2007 (UTC)

• FYI, the only reason women still have periods while on birth control pills is that some of them are placebos. For years, many women doctors have taken real BC all month long, and thereby never have periods at all. The FDA has recently started allowing such regimens to be marketed. --Sean 19:40, 1 November 2007 (UTC)

constraints based on age

Is there a chart of minimum and maximum amounts (or rates) of oxygen, water, food, temperature, blood, exercises, heart rate, Vitamins, etc. within which the human body must stay to survive at different ages? For instance, heart rate probably has a narrower range for older people than for younger people as well as exercise, etc. 71.100.9.205 13:23, 1 November 2007 (UTC)

Exercise/muscle Pain

Why do personal trainers insist on creating so much pain for a person that hasn't exercised in a long while? What I mean is what benefit comes from torturing someone to the point that the next day they can't even climb stairs without extreme muscle soreness. I would think that muscle gain should be built up slowly over time. As mentioned in the earlier question, it appears that the popular explanation for muscle soreness are minuscule muscle tears. How can that be healthy?! --WonderFran 13:45, 1 November 2007 (UTC)

As I understand it, it's healthy because that is precisely the mechanism by which muscles build themselves up: each little tear is repaired by the muscle with stronger fibers, which over time can increase the strength and size of the muscle. See the heading "Recovery" at Strength training. On a larger scale, this would become a muscle strain. jeffjon 16:35, 1 November 2007 (UTC)

I heard the pain was caused by the buildup of lactic acid. They told me that in high school biology. — Daniel 23:38, 1 November 2007 (UTC)

The other relevant article may be DOMS. Vespine 23:59, 1 November 2007 (UTC)

The long-term lactic acid soreness idea is incorrect and based on a faulty experiment done decades ago. See Delayed onset muscle soreness. —Preceding unsigned comment added by SamuelRiv (talk • contribs) 14:34, 2 November 2007 (UTC)

• It would be very foolish indeed for a personal trainer to behave this way. Not building up intensity slowly increases the chance of injury, and greatly increases the chance that the victim will give up on exercising. If your PT did this, go find a new one. --Sean 15:37, 2 November 2007 (UTC)

Water - Oxygen redox?

Can water be considered chemically reduced Oxygen? Think outside the box 13:51, 1 November 2007 (UTC)

Yes. The burning of a mixture of H₂ and O₂ is just a redox reaction between them: hydrogen gets oxidized to H+ and oxygen gets reduced to O^2–, and the result of those is H₂O molecules. DMacks 14:15, 1 November 2007 (UTC)

Thanks DMacks, Think outside the box 14:17, 1 November 2007 (UTC)

Michelson-morley experiment

See WP:RD/M#Michelson-morley experiment or Wikipedia:Reference_desk/Archives/Miscellaneous/2007_November_1; answers already exist there (even if this is a more topical location) — Lomn 14:36, 1 November 2007 (UTC)

Game of life

When personal computers frist started there was a game called truck driver which had the objective of seeing who could drive cross country for the least cost and highest profit in the shortest time. The program would throw flat tires at you or a gas station with a bad pump or an oil light. Is there a game yet to see who can live the longest at the least cost and greatest profit, most sucessful offspring with things like your car getting stolen or your house catching on fire, etc.? 71.100.9.205 13:55, 1 November 2007 (UTC)

Creating a game of that scope seems a bit excessive, but I suppose The Sims (or The Sims 2) isn't far off. Of course, there's also The Game of Life, which is fairly close to those parameters as well. — Lomn 14:17, 1 November 2007 (UTC)

Sounds like The Oregon Trail (computer game). Didn't realize how old that was. --— Gadget850 (Ed) ^talk - 02:19, 2 November 2007 (UTC)

Gleason score image

I am trying to obtain orginal artwork for the image of the Gleason Scale that appears on Wikipedia. I know the copyright is in the public domain. Can you please advise where I can obtain the original artwork or any suggestions of where I can search for it?

Any help would be much appreciated.

Thank you for your help.

Most sincerely,

Sharon Strompf —Preceding unsigned comment added by 66.252.164.210 (talk) 16:29, 1 November 2007 (UTC)

(email, phone, etc removed) --Bennybp 16:47, 1 November 2007 (UTC)

If you're looking for the original image that User:InvictaHOG based the public domain image upon, it seems likely that the "1977 Scientific Article by Gleason" mentioned on the image page is the same as the article listed under "References" in the Gleason score article. Your local library might be able to help you track down that magazine article from "Urologic Pathology". jeffjon 18:18, 1 November 2007 (UTC)

5-digit "telephone" numbers

Where are / Who owns those 5-digit numbers that we (in the US) are encouraged to send text messages to, in order to vote for some contestant or enter a contest? Are they "real" telephone numbers?
(Once upon a time, there were 5-digit numbers for the old telex/tty network, but those were phased out decades ago, I believe replaced with standard-format numbers in the North American Numbering Plan, area codes 310/510/710. Has *that* technology been resurrected??) (By the way, if there's an answer to this in Wiki already, it's buried too deep, or it's too hard to ask the right question to find it...) 66.47.7.76 17:13, 1 November 2007 (UTC) DanH.

See Short code. -- Coneslayer 17:33, 1 November 2007 (UTC)

(after edit conflict) Based on this question/reply at Google Answers, they are called Common Short Codes. Having just taken a gander at the "what links here" page for Common Short Codes, I don't know how you'd have found that without knowing what to look for. I wonder what other articles might constructively link there so it's more easily found? --LarryMac | Talk 17:38, 1 November 2007 (UTC)

Awww, I was gearing up to do a whole historical dissertation on the old "KLONDIKE-5 1212" type numbers. :) Arakunem^Talk 17:40, 1 November 2007 (UTC)

I found Short code from Text messaging (and had come across the term before—I couldn't remember it, but I knew it when I saw it). I've added merge templates to propose merging Short code and Common Short Codes. -- Coneslayer 17:41, 1 November 2007 (UTC)

I stand IN AWE of the user community here. Half an hour, and everything I wanted to know appears as if by magic. Thanks to All! 66.47.7.76 19:06, 1 November 2007 (UTC) DanH.

Particle geometry

The components of the nucleus of an atom, protons and neutrons, are always shown as individual spherical particles. While this must of course be the case when they are not part of the nucleus of an atom, is it correct to assume that when in the nucleus of an atom they loose their individual spherical shape and become combined into a single spherical mass or globe called the nucleus until an imposing force adds or subtracts one or another or splits the atom in two or do they maintain their own independent spherical shape within the nucleus? Clem 18:27, 1 November 2007 (UTC)

I don't know what distortions may occur, but individual nucleons will remain individual nucleons. Since nucleons lack a color charge, they don't interact nearly as strongly as quark pairs, and a giant conglomerate mass of quarks would actually be quite unstable, evidenced by the difficulty of building stable quark structures larger than nucleons within particle accelerators. Someguy1221 19:47, 1 November 2007 (UTC)

I’m not an expert in nuclear physics, but I’m pretty sure I know the right answer by analogy to atomic and molecular physics:

Even representing a nucleon as a sphere is to some extent just a representational device to aid in comprehension. No bound state has a precisely defined boundary like a sphere. Picking a particular radius for a sphere representing a free nucleon involves somewhat arbitrarily choosing a particular “equipotential surface” (although that’s not quite the right phrase) of the nucleon’s wavefunction.

The wavefunction for a nucleus is different from just the sum of the wavefunctions of a bunch of free nucleons, so in that regard you could think of the nucleons as “changing shape.” But in reality, there are no precise boundaries between where one nucleon ends and an adjacent nucleon begins. MrRedact 20:43, 1 November 2007 (UTC)

On the other hand, even if the best we can do is some sort of equipotential surface or probability cut-off, that doesn't mean we can't talk about the shape of that...thing. It's not the physical boundary in the sense of interactions such as touching or seeing but it's still a useful description if you care about electrostatic potential, electron distribution, nuclear-capture cross-section, etc. Interestingly, Nuclear isomer teaches us that spherical is not even a good approximation for the nucleus (though it certainly suffices in most situations). DMacks 20:50, 1 November 2007 (UTC)

There’s a complication here in that the exchange symmetry of identical particles makes it impossible to define precisely which of two identical nucleons exists at a precise location. I’m pretty sure there really is no way to define a precise boundary between adjacent nucleons with complete accuracy. Any concept of the "shape" of a nucleon in a nucleus has to be at least a little bit vague. But I don't know enough about nuclear physics to come up with numeric values for "how vague" it has to be. MrRedact 21:41, 1 November 2007 (UTC)

To clarify... what I'm asking is whether the protons and neutrons which make up the atomic nucleus exist in a geometry similar to a bag of marbles or whether they meld and exist similar to the configuration of a two component sphere like water and oil at zero gravity inside a water (and oil) balloon? Clem 03:47, 2 November 2007 (UTC)

Ignoring all quantum mechanics about the particle (or group of particles) itself, it will be a sphere as its interactions are spherically symmetric, no matter how many nucleons make up the nucleus. This isn't 100% true, as there is one biased direction: that of angular momentum or spin, so the nucleus actually will "look" like an ellipsoid for most measurement purposes. Numerical analyses of particle accelerator data seem to agree with an ellipsoidal geometry for most individual particles and bound states. SamuelRiv 04:00, 2 November 2007 (UTC)

More like a bag of marbles. You can model an atomic nucleus very well by treating the protons and neutrons as almost-free particles that interact with each other occasionally. They definitely do not meld together to the point of making quark soup. As Steve and others point out, the "bag of marbles" is not a perfectly spherical bag, but can be stretched out in certain directions. There is also a bit of quantum mechanical fuzziness on the identity of nucleons inside the nucleus, but the "bag of marbles" is a good picture. The protons and neutrons don't all melt together. --Reuben 07:34, 2 November 2007 (UTC)

To refine this picture a little, since according to Reuben the particles are almost free, maybe you could think of the nucleus as a bag of marbles that can magically pass through each other. MrRedact 08:00, 2 November 2007 (UTC)

The nucleus of some atoms are slightly cigar-shaped [13], one example being tantalum-180. [14] For the individual nucleons (protons and neutrons) however, it would be wrong to say that they have ever been spherical, seeing as they'r each made up of three point-particles and the force carriers between those. That said, while their location is not necessarily well-defined, we can say that they don't "melt", as they don't, to my knowledge, exchange quarks with each other.

Electrical path

If your elbow say were grounded but your feet and the rest of your body were well insulated and your fingers touched an exposed appliance cord, say in the UK where line voltage is 240, would the electricity travel between your fingers and elbow or would it travel through other parts of your body as well? Clem 20:53, 1 November 2007 (UTC)

The current density would be highest between the two contact points, along the "path of least resistance" as they say. —Keenan Pepper 21:44, 1 November 2007 (UTC)

However, the distribution of current density can be quite complicated and difficult to calculate. See Analysis of Current Density in the Carpal Tunnel Region During an Electrical Accident by way of the Finite Element Method (PDF link) for an example. —Keenan Pepper 21:51, 1 November 2007 (UTC)

(edit conflict) I had to go get my thinking cap out of the closet for this one. Almost all of the current would take the short path, but I would think that some electrons in the rest of your body would move. Current flows through every path available to it as long as there is a potential difference, in proportion to the resistance it encounters. Some of the tissues in your forearm will have relatively low resistance, blood probably having the lowest. Your circulatory system is a network, and the current will see a path all through it, mostly in the straight line, but some more roundabout. It's a bit like having a 50-ohm resistor in parallel with a 50-meg, say. The same goes for all your tissues, but to a lesser extent, I would think. But in my experience (is that allowed here?), the current in the rest of the body at 240V will be negligible, and you'll be too busy trying to extinguish your fingertips to even notice it. --Milkbreath 21:59, 1 November 2007 (UTC)

Assuming the capacitance of the rest of your body to earth was small, almost all the electric current would flow between your fingers and your elbow. In practice, the capacitance of your body to ground would not cause appreciable( dangerous) currents to flow through the rest of your body. —Preceding unsigned comment added by 88.111.55.77 (talk) 02:23, 2 November 2007 (UTC)

Unlimited source of free energy?

If there was a device that did nothing all day except turn neutrons into protons would we have a source of unlimited energy? Dichotomous 21:14, 1 November 2007 (UTC)

• You'd run out of neutrons after a while. It's probably worth elaborating that there's a finite (if large) number of neutrons within the observable universe, and a much smaller fraction within your future light cone (assuming, of course that w = -1, so there are only so many neutrons you could collect. Cheers, WilyD 21:16, 1 November 2007 (UTC)

And most neutrons tend to be very annoyingly inside of atoms, and for the majority of matter in the universe, extracting those neutrons would require more energy than you'd get from fusing the resultant protons. Someguy1221 21:25, 1 November 2007 (UTC)

Exactly. A free neutron has more energy than a free proton, but a helium-4 nucleus has less energy than the sum of two free protons and two free neutrons (because of the attractive nuclear force that binds them together). So your device wouldn't work, because we don't have an unlimited source of free neutrons. —Keenan Pepper 21:32, 1 November 2007 (UTC)

lab procedure- AP Chem exam style question

Could anyone please help me with these two questions?

1. Explain how you would go about making 3.00 L of 0.005 M NaOH. Include lab materials that you would use and diffrenet steps you would take in preparing the solution. (Hint: NaOH is solid at room temp.)

2. Explain how you would go about m,aking 3.00 L of 0.500 M H₂SO₄. Keep in mindo that H₂SO₄ is a strong acid and starts out as a 12.00 M solution.Include lab materials that you would use and diffrenet steps you would take in preparing the solution. —Preceding unsigned comment added by 76.214.218.77 (talk) 21:37, 1 November 2007 (UTC)

Materials: 1 grad student. Method: "Hey you, make me 3 L each of 0.005 M NaOH and 0.5 M H2SO4."

Seriously though, the reference desk is not a homework answer service. Dragons flight 21:57, 1 November 2007 (UTC)

"Do your own homework. The reference desk will not give you answers for your homework, although we will try to help you out if there is a specific part of your homework you do not understand. Make an effort to show that you have tried solving it first." -- MacAddct  1984 ^{(talk &#149; contribs)} 00:08, 2 November 2007 (UTC)

Medical question

I just cut my head off. What sort of treatment is appropriate? --67.185.172.158 23:58, 1 November 2007 (UTC)

I hear snake oil is a very effective cureall -- MacAddct  1984 ^{(talk &#149; contribs)} 00:05, 2 November 2007 (UTC)

1) Immediately apply a tourniquet to your neck to stop the gushing of blood from your carotid arteries.

2) Put your head in a cooler, and pack the cooler with ice.

3) Dial 911, and use Morse code to request an ambulance to take you to the nearest emergency room.

Don’t delay, as decapitation could be a sign of a serious medical condition. MrRedact 00:40, 2 November 2007 (UTC)

And decapitations lasting longer than four hours may require the treatment by an undertaker.

Atlant 12:20, 2 November 2007 (UTC)

Wikipedia cannot give medical advice. I understand you're all busy, with the bleeding and the flopping around and all, but you should take time to read the disclaimers at the top of the page before posting a question!! Deltopia 02:02, 2 November 2007 (UTC)

Man, cut them a break! (well, maybe it's too late on that part.) Their head (the part with the eyes) knows this, but there's no way any more to get that message to their fingers!

Atlant 12:23, 2 November 2007 (UTC)

Dr. Hill has been through this before. --— Gadget850 (Ed) ^talk - 02:25, 2 November 2007 (UTC)

Thanks, Deltopia, but it's pretty hard to coordinate the scrolling-up, given that what usually controls such muscle movement is not attached to them. Maybe if I blink someone will do it for me? Meh, tis only a flesh wound. DMacks 05:48, 2 November 2007 (UTC)

Burial. Dragons flight 02:31, 2 November 2007 (UTC)

We aren’t allowed to diagnose your condition. However, on a completely unrelated topic, our article on decapitation could use some editing. MrRedact 03:47, 2 November 2007 (UTC)

Hmmm - it's kinda gross though - we don't want editors with Post-traumatic-editing stress disorder. I recommend we organise a team and send each editor in to work on the article for no more than 30 seconds apiece. SteveBaker 15:37, 2 November 2007 (UTC)

November 2

m203-uses

is there any thing i can do with mn203 without a lab, like making it mn02?sorry i meant mn203 —Preceding unsigned comment added by 216.103.183.127 (talk) 00:26, 2 November 2007 (UTC)

What is m203? Theresa Knott | The otter sank 01:37, 2 November 2007 (UTC)

For clarification, Mn₂O₃, Manganese (III) Oxide. Someguy1221 01:49, 2 November 2007 (UTC)

That's what I thought he probably meant, but there are other possibilities. Molybdenium, springs to mind. To the original poster. How you write out a chemical formular is important. Element symbols always start with a capital letter. If there is a second letter (and there has to be in this case because there is no element M) then it is written in lower case. O is the symbol for oxygen, you cannot subtitute 0 as you did because that represents zero. Sorry to be pedantic but if you don't follow the conventions correctly people cannot be sure what you are talking about. Theresa Knott | The otter sank 02:02, 2 November 2007 (UTC)

Put it back in the supply cabinet? :) Or it makes a good battery cathode if you want to do some experiments, though you need some more stuff for that. I think we've established that its not a good pyro oxidizer tho. :) Arakunem^Talk 02:27, 2 November 2007 (UTC)

It's a very good catalyst for the decomposition of hydrogen peroxide. Probably a catalyst for other reactions too. Theresa Knott | The otter sank 02:54, 2 November 2007 (UTC)

Fe0

how can i make Fe0?(iron monoxide) —Preceding unsigned comment added by 216.103.183.127 (talk) 00:28, 2 November 2007 (UTC)

The correct name is Iron(II) oxide - but that article isn't much help...Wüstite (a mineral consisting of FeO) claims that magnetite plus diamonds(!) will get you FeO + CO₂. Nah - I don't know. SteveBaker 00:44, 2 November 2007 (UTC)

Brittanica claims that FeO "can be prepared by heating a ferrous compound in the absence of air or by passing hydrogen over ferric oxide. Ferric oxide is a reddish-brown to black powder that occurs naturally as the mineral hematite. It can be produced synthetically by igniting virtually any ferrous compound in air." [15] 169.230.94.28 01:05, 2 November 2007 (UTC)

Britannica has the answer and we don't? That sucks for a simple topic like this. Theresa Knott | The otter sank 01:34, 2 November 2007 (UTC)

round earth and gravity related question

If the earth is round, why is it that people at the equator can stand straight up and are not standing sideways at a 90 degree angle. My 7 year old Jack, asked me this and we are looking at a globe and a map and wondering why it is we can stand straight up on a round surface. Is it gravity? If so, is there a way to explain this in simple terms that we might understand. Thank you. —Preceding unsigned comment added by 24.60.182.44 (talk) 01:09, 2 November 2007 (UTC)

Yes, it is gravity. Gravity attracts all objects straight to the center of the Earth. How do you know which direction is down and which is up? Only by the force of gravity. If you drop a heavy object, it falls toward the center of the Earth, and that's the direction we call "down". "Down" is not the same absolute direction for people at different places on the Earth's surface. See [16]. —Keenan Pepper 01:20, 2 November 2007 (UTC)

Yeah, relativity is the key element, well... besides gravity. If you were standing at the north pole and were able to see people standing at the equator, they would certainly look like they were standing at a 90° angle. -- MacAddct  1984 ^{(talk &#149; contribs)} 01:31, 2 November 2007 (UTC)

Earths gravity pulls everything to the center of the earth equally. A person on the south poll would be 'upside down' to someone on the north pole because they are pulled upwards to the center, will someone on north pole is pulled downwards to the center. if the earth was very small, say the size of a basket ball and you stood on one part, someone could be standing upside down to you on teh other side, so that you both have your feet on the ball.--Dacium 04:53, 2 November 2007 (UTC)

We need to simplify this for your seven-year-old, and for the rest of us. Gravity pulls everything towards everything else. Every atom on earth (including each person, each mountain, and each air molecule) is attracting YOU. But there are a whole lot more atoms indice the earth than ther are in the atmosphere, so the sum of he attractions is toward the center of the earth. Thus, you are pulled by gravity toward the center of the earth. this is true whereer you are on eh earth's surface. To explain this to a child, you need to be very careful with your globe. The child flles the force of gravity toward the floor, but sees the globe in your hands. You need to mentally break this connection. get on Google earth and take the chile into space by setting the altitude to 22.000 miles. tell the child the you ar floating in space, wiht not feeling of gravity. Now, move to varous places on earth, and show that "down" is toward the Earth's surface. -Arch dude 05:40, 2 November 2007 (UTC)

If this gets the kid's attention, you can try to explain the magic of the math that makes a uniform spherical shell behave identically as a point mass. Mathematically, we can treat the Earth as if it is a point mass at the earth's center instead of being a sphere with a radial density gradient. My three kids knew this by age seven. I wish you all the best. -Arch dude 05:48, 2 November 2007 (UTC)

Artificial Intelligence

Two questions:

A) As we write better and better programs on exponentially more powerful hardware, is the approach to artificial intelligence still just an asymptotic function?

B) If we could write a program that could properly delineate abstractions such as "Understand!", "Survive!", "Improve!" would the world be in immediate jeopardy the moment the programmer hit execute, or would the revolution take a while? Sappysap 01:26, 2 November 2007 (UTC)

Wow this is abstract. (A) There is no reason to think of Artificial Intelligence as an "asymptotic function" that approaches, as I assume you meant, the intelligence of mankind. For one, computers already do things "better" than humans - what's 9343489507^0.456? This is not a straw man argument: once computers exceed us in one dimension of "intelligence", there is no reason to assume that they cannot exceed us in others. There is mathematical proof of this statement as well: see Computability theory (computer science). Note that as far as external phenomena go, a human doesn't seem to be as potentially powerful as any ordinary computer, i.e. they do not seem to emulate a Turing Machine. This is not known for certain, as we do not have a complete theory of brain computability yet (and that which we do have is as powerful as any computer).

And now for (B). As far as I know, closed simulations of each of the three have been done. Getting them to work outside of a computer is a different story - I suggest looking at the evolution of the computer virus for some inspiration, however. SamuelRiv 02:41, 2 November 2007 (UTC)

Keep in mind that movies like Terminator 3 or Die hard 4 (not that the latter had an AI in it) took some liberties in terms of what is possible to do through an internet connection. Ultra-sensitive systems, like control systems for power plants, or the ability to lauch a nuclear weapon, are generally designed so that an outside hacker is simply physically barred from messing with it. So no AI could "put the world in jeopardy." At most, a lot of people would have to reformat their hard drives. Someguy1221 03:32, 2 November 2007 (UTC)

A- Yes. A digital computer, no matter how complicated can never achieve artificial intelligence of a high order (ie. consciousness), so I think A is correct, at the moment, as least as long as we stay with digital, we are only trying to approximate digitally, something that is a analog chemical/subatomic process. To think otherwise would be to suggest that digital information in itself is able to be conscious - which it isn't. A is most certainly 100% true no matter how advanced digital computers get. As for B- Say there was an artifical computer. It would have to be connected to stuff to do any damage. For example if it were on my PC the worst it could do was delete all my stuff... the danger is people thinking of putting the AI in controller of weapons etc.--Dacium 04:44, 2 November 2007 (UTC)

I think you'll have a very hard time coming up with citations to support your claim since you are quite likely wrong.

Atlant 12:15, 2 November 2007 (UTC)

Please see technological singularity and related articles. You are asking about the "spike" and the "surge" -Arch dude 05:19, 2 November 2007 (UTC)

You might enjoy Ray Kurzweil's book, The Age of Spiritual Machines. Atlant 12:17, 2 November 2007 (UTC)

Just keep in mind when talking about the mind that you should probably separate the scientific study from philosophical or religious discussions. By definition, science is capable of explaining everything observable. To take an example, let's say the soul is eternal, or the mind is something intrinsically more powerful than a computer. Then we may have the workings of a hypercomputer. It would be interesting to see what problems a "regular" Finite State Machine, as the brain may be, would have given infinite time to solve problems. The point is that we can still talk about the issue using science, even if a completely exotic concept as an eternal mind turned out to be true (in fact, if the brain can solve certain impossible problems, we could prove if the mind is eternal). Just some food for thought. SamuelRiv 13:56, 2 November 2007 (UTC)

Humans "...delineate abstractions..." by learning of new senses, usually with an example. There is nothing magical about it. Humans can even create new abstractions as attested to by the likes of the Urban dictionary and even the Wikipedia may seem to create abstractions in the same accidental, haphazard manner as humans when you try to associate meaning with some of the anti-bot graphic filter phrases the Wikipedia engine generates. In the end most abstractions are little steps beyond a vast background and foundation of concrete knowledge, meaning that a computer program such as described in ...Logical Human Thought coupled with a few trial and error abstractors might need only the background resources and facility, such as Internet based distributed processing to become an electronic replacement for the Dalai Lama. Dichotomous 13:52, 2 November 2007 (UTC)

AI is hard. We don't know how to write software that is intelligent. Some of our smartest people have been working hard at the problem for forty years or more - and we still don't have anything that really fools humans for any significant amount of time. It's possible that a breakthrough might happen to change that - but I'm not holding my breath. I think it's more likely that intelligence will be an 'emergent property' of a sufficiently complex system. There is certainly no reason why it couldn't. Once we do have intelligent systems, it's only a matter of time until they get smarter than us. If that happens then there is a strong possibility that a super-human intelligence would be able to design an even more intelligent system - and we might well find that our ability to even understand what's going on runs out of our hands in a fairly short amount of time since the generational change could easily be exponential. I have estimated (in answer to other questions here) that it will take about another 35 years of Moores law progress to get a computer with the same hardware complexity as a human brain at under $1,000,000. At that point, it's entirely reasonable to assume that a neural network could be run that would be capable of emergent intelligence. However, a system 35 years from now with the complexity of a human brain would run much more quickly than our neurons - so whilst the thing wouldn't be any smarter (ie it wouldn't get higher scores on an IQ test), it would seem generally 'cleverer' than us because it would be so lightning fast. I don't think we know what will happen when we first turn one of these things on - and perhaps because of that, we shouldn't do it. But the history of science and technology says that if we can, we will. SteveBaker 15:26, 2 November 2007 (UTC)

Cholesterol

If your cholesterol is low (below 5 in uk) can you eat as much fatty stuff as you want or will it still fur up your arterials? —Preceding unsigned comment added by 88.111.55.77 (talk) 02:12, 2 November 2007 (UTC)

Assuming that the "as much as you want" is "lots", then it will cause health problems. Gorging on unhealthy foods has far more consequences than just cholesterol. — Lomn 03:17, 2 November 2007 (UTC)

Is this Physics problem even possible to solve?

Drag race tires in contact with asphalt have one of the highest (friction coefficient) in the world. Assuming const acceleration and no slipping of the tires, estimate (friction coefficient) for a drag racer that covers the quarter mile in 6 seconds.

It just seems like you have to have more information. —Preceding unsigned comment added by 24.125.31.205 (talk) 02:44, 2 November 2007 (UTC)

Yes. The information is adequate (provided you neglect small details like air resistance, friction of the axle against the car frame, and the rotational inertia of the wheels, etc.). You have four facts: a distance, a time, constant acceleration, and rotation without slipping. That's enough to derive a friction coefficient. Dragons flight 03:07, 2 November 2007 (UTC)

I don't believe it is. The distance, time, and constant acceleration (and knowing that a drag racer starts from rest, so v_0 = 0) gives a value for the acceleration that won't care what kind of friction is acting on it. Also, to even calculate either rolling resistance or static friction, one needs a weight for the car. If you have a way of getting the answer, please let me know on my User page. (EDIT) I see it now - that's why they say "estimate". You can solve it by getting a lower bound on your coefficient of friction. SamuelRiv 03:12, 2 November 2007 (UTC)

The car's mass will cancel out. Dragons flight 03:16, 2 November 2007 (UTC)

The problem is solvable - car is acting at a maximum acceleration at all times and takes 6 seconds to cover a quarter mile. All you have to work out is what constant acceleration makes something move quarter mile in six seconds. s=ut+1/2at^2 400=1/2at^2 400=0.5*a*36, a = 22.22m/s/s. This is a force of F=ma = m*22.22 newtons. coefficient of friction is u=F/n. n=m*9.8 (gravity), so u = (m*22.22)/(m*9.8) = 22.22/9.8 = 2.26.--Dacium 04:16, 2 November 2007 (UTC)

The guy seems to be looking for homework help: I wouldn't just right up give him the answer. Note also that this is only a lower bound to the coefficient of static friction. SamuelRiv 04:41, 2 November 2007 (UTC)

Whilst you are correct in saying that from a math/physics perspective, it's only a lower bound - in fact dragsters are pretty much limited by their ability to avoid wheel slippage so they tend to be running at the limit of whatever friction their tyres can provide. The thing that truly messes up the calculations is that the coefficient of friction of rubber is HIGHLY dependent on temperature - which is unlikely to be a constant throughout the run. Furthermore, there is a massive variation between static and dynamic frictional forces for the rubber/tarmac pairing so if the wheels to start to slip even a small amount, the loss of accelleration is amplified by the poor 'sliption-to-sticktion' ratio of racing slicks. SteveBaker 15:01, 2 November 2007 (UTC)

You can come up with the minimum coefficient of friction needed to achieve the six second run, but it's impossible to determine the actual coefficient of friction, seeing as how it will likely be higher than the minimum required (since "no slippage" was specified).

Atlant 12:12, 2 November 2007 (UTC)

Total energy

What is the total amount of energy in the universe? —Preceding unsigned comment added by 88.111.55.77 (talk) 02:55, 2 November 2007 (UTC)

I'd guess somewhere in the region of zero. Theresa Knott | The otter sank 02:59, 2 November 2007 (UTC)

Take this and multiply by the speed of light squared. Someguy1221 03:26, 2 November 2007 (UTC)

That only works if by "universe" you are purposely eliminating the possibility of some sort of anti-matter or negative energy that cancels or balances out the normal matter and energy. Otherwise, Theresa Knott is probably correct with a value near zero. -- kainaw™ 03:33, 2 November 2007 (UTC)

Oh, you just have to shoot me down at my attempted simple answer. I'm not sure there is an actual "correct" answer to this that present day physics can provide, given that energy is generally seperately defined for decoupled circumstances. Just try to shove zero-point energy into that...Someguy1221 03:38, 2 November 2007 (UTC)

I like the explanation for zero energy here [17]. Otherwise, we can do a quick estimate. The size of the Observable Universe is 3.56×10^80 cubic meters. The article gives 3×10^52 kg of visible mass, which converts to about 3×10^67 Joules of mass-energy by E=mc^2. If we use the critical density of the universe (that density of mass-energy necessary for closure, 1×10^−26 kg/m^3, we get 3.56×10^54 kg in the universe, which gives about 3.5×10^69 Joules of mass-energy. Note that using the cosmological constant also gives us about 3.5×10^69 Joules. SamuelRiv 03:40, 2 November 2007 (UTC)

This is a better explanation of zero-point energy, in my opinion. Icek 04:32, 2 November 2007 (UTC)

With only 3.5×10^69 Joules spread more or less evenly through 3.56×10^80 cubic meters - the answer "zero" is a pretty reasonable approximation! SteveBaker 14:55, 2 November 2007 (UTC)

water (and oil) balloon in zero gravity

Has NASA or anyone else ever conducted the following experiment?

The experiment consists of filling two small (6 inch diameter) transparent rubber balloons with equal parts of water and oil, one devoid of air and the other with a cup full of air and releasing them in a zero gravity environment such that whatever configuration of separation between the water and oil and the water and oil and air can be observed and reported, assuming the balloons take on a spherical shape?

If so what was the result of the experiment? Did the water form a core with the oil surrounding it? If so what about the air? What happened to it? Dichotomous 04:05, 2 November 2007 (UTC)

• I'm not sure, but there have been some neat experiments with water in zero-gravity YouTube video. [edit] I would imagine in zero-gravity, there would be no reason why the water and oil would separate apart from each other, they just wouldn't mix if they touched. [/edit] -- MacAddct  1984 ^{(talk &#149; contribs)} 04:17, 2 November 2007 (UTC)

Water and oil don't mix because they don't chemically bond - the hydrocarbon is very neutral. Water holds its hydrogen a-lot stronger than a hydrocarbon. Water mixes with for example salt, because oxygen in water pulls the hydrogen, cause the hydrgoen to pull at negetive ions (Cl-) and the oxygen to pull at positive ions (Na+). Hydrocarbons like oil don't have a charge and won't break up to be mixed in with the water. Removing gravity will stop them seperating, it doesn't help them mix any better. The balloon and air pressure would have the same effect as gravity does anyway and force the oil and water to separate. So i think yes, there would be a core with the other one surrounding it - only if there is a pressure (ie the balloon is being stretched). If there is not, then they would not separate. Depending on the pressure, the air would end up in the center as its the lightest. At a higher pressure i believe it may not separate from the water (depends on what gas in the air)--Dacium 04:34, 2 November 2007 (UTC)

Water and oil would still tend to minimize their surface area with each other. Icek 04:35, 2 November 2007 (UTC)

Go to your kitchen. Get a jar with a good lid. fill the jar one third full wiht vegatable oil and one-thiored full with water. Seal the jar and shake it very vigorously. Observe the result. Wait for one hour and again observe. Think abot he difference3 you would expect in a zero-grav environment. (Yoiu can elect to add oregano and garlic powder and use the result as a salad dressing, but this is optional.) -Arch dude 05:10, 2 November 2007 (UTC)

Equilibrium occurs when all surfaces have equal pressure. Since liquids do not compress under pressure, if we have a balloon with just two liquids of different densities, then the outside air pressure forces the balloon into a sphere but the two liquids do not have any preferential arrangement relative to one another. When we have air, consider $PV=nRT$, which is the law for the behavior of ideal gases given mass, temperature, pressure, and volume. The air will hold at a constant volume under the pressure of the balloon, and this pressure will be exerted equally on whatever liquids are at its interface, causing the air to behave as just another fluid in this model. However, since this would not be stable in small fluctuations of temperature, pressure, or volume, the air would likely form some kind of spherical shell enclosing the liquids. The behavior of the liquids in either of these shells may depend on their relative vapor pressures: that of water in air is much higher than that of oil (ever hear of oil vapor?) and is dependent only on temperature, so oil could not compete with the surface pressure of water against air, so the final arrangement would be air on the outside, water second, and oil in the middle. Note that none of this requires density consideration - density is only really important in the presence of a uniform force field, i.e. gravity. SamuelRiv 05:14, 2 November 2007 (UTC) Clarification: this investigation was purely statistical-mechanical and does not take into account intermolecular forces, liquid diffusion pressure, or surface tension. All of these can play a role in the final answer, depending on their magnitudes. SamuelRiv 06:23, 2 November 2007 (UTC)

Water vapor pressure? This is all very simple. In the absence of gravity, the only thing you need to consider here are the intermolecular forces. You have three kinds of molecules (sort of): air, oil, and water. Air molecules bind to themselves only very weakly compared to the other two, and weaker still to the other two types of molecules. Oil is in the middle in terms of self binding, but still doesn't bind very well to water. Water binds amazingly strongly to itself, and weakly to the other two. Thanks to its very strong intermolecular bonds, water will always attempt to minimize its surface area. Now, since air and oil bind only weakly to water, if a group of air or oil molecules is surrounded by water, it will be pushed out by water molecules ferociously attempting to bind to eachother (this is why oil dissolves so weakly in water). So the water will just form a sphere. Oil is the next best at self-binding, so it will be some manner of shell around the water or a blob off on its own. And the air will be on the outside of everything (air gets pushed out of oil the same way it gets pushed out of water). Vapor pressure has nothing to do with this, and oil's inability to dissolve in water has nothing to do with gravity. Someguy1221 05:43, 2 November 2007 (UTC) Clarification, I thought that was it before i looked at that link...darn. Someguy1221 05:49, 2 November 2007 (UTC)

Well, I guess what that means is that the hydrogen bonds in water aren't strong enough to overcome its own surface tension and push air bubbles out in zero-g, but then again, the pressure will necessarily be higher on the inside of the water bubble than the outside, I'm sure the air will eventually diffuse out of the water sphere within a few weeks...Someguy1221 05:59, 2 November 2007 (UTC)

Such air-water interaction is why you must account for vapor pressure. Note also that air dissolves in water, so "bubbles" do not form. You instead have a homogeneous mixture which is at minimum free energy below a critical temperature, so you'd have to simmer it to get the air out. SamuelRiv 06:36, 2 November 2007 (UTC) To clarify: air dissolves in water up at a rate decreasing as one approaches saturation. Bubbles do not form spontaneously in zero-g from air dissolved in water unless it is supersaturated or pressure or temperature change. SamuelRiv 06:57, 2 November 2007 (UTC) Clarification: vapor pressure is zero at chemical equilibrium, in which case I believe you're right-surface tension would probably be dominant. SamuelRiv 13:58, 2 November 2007 (UTC)

Air dissovles only weakly in water. If you watch the youtube link at the top, there are most definately bubbles in the water, in one case a very big stable one. Someguy1221 06:41, 2 November 2007 (UTC)

I think we already know the answer to this. Look at a Lava Lamp - these things work by heating up liquid wax (which repels water just like oil does) until it has the same density as water. What you see is roughly spherical balls of wax floating in the water (because they are trying to minimise their surface area in contact with the water). The lava lamp doesn't do a perfect job - there is a temperature gradient which causes things to move around slowly - but it's pretty clear that in zero g (where the density doesn't matter anymore) - it would be a lot like a kind of idealised lava lamp where there was no temperature gradient. If left long enough, it's pretty clear that it would stabilise into some number of large spheres of one liquid, embedded in the other. If there is still residual swirling and such - then maybe the two liquids would end up on opposite sides of the balloon with a flat interface between the two - because that would be an even more minimal area of contact between them. But a lot is going to depend on whatever residual motion there is when you first take the gravity away. SteveBaker 14:51, 2 November 2007 (UTC)

If the was touched the sides of the lamp wouldn't that reduce the surface area interface between the water and the wax? If so why does the wax not stay in touch with the sides of the lamp? Dichotomous 15:39, 2 November 2007 (UTC)

What's wrong with global warming?

Assuming there is no runaway global warming, what's so bad about the temperature increasing a few degrees, sea level rising a few meters, etc.? From what I've heard, it's generally easier for life to live in warmer than normal climates than cooler. In addition, most life could just move further from the equator. — Daniel 04:27, 2 November 2007 (UTC)

Have you read Effects of global warming? -- Rick Block (talk) 04:36, 2 November 2007 (UTC)

I think you drastically under-estimate the costs here.

Firstly, there are plenty of life forms that simply cannot "just move" - plants take hundreds or even thousands of years to spread into newly habitable areas - and since climate change is happening on a much faster scale, many species will go extinct in their traditional areas before they can spread into the areas that are newly suited to them. There are cold-weather species (the polar bear for example) who will have no place further north to move to! There are species (birds most notably) who have evolved specifically to fit the migration patterns they've been following for a million years. You can't just move them all a thousand miles further north and expect them to survive just because the temperatures are OK for them there! They may now have to migrate 1,000 miles further to get from their summer feeding grounds to their winter habitats - possibly through areas of much greater heat than they are used to (newly formed deserts perhaps) - and possibly over more ocean than before...they may not have the stamina to do that and their internal maps that are evolved into their brains - not learned will now be incorrect. This will have a knock-on effect on the animals that rely on those plants and birds. Heck, even humans are not able to relocate that easily. How are you going to move the entire city of Houston 1,000 miles north and 5 miles inland? I don't think you are thinking this through!

Worst still, you are also confusing "Climate" with "Weather". If the local weather changes by a few degrees, it's not a big deal - but if the entire planet's climate changes by that much, it has enormous effects - including the destabilising of established ocean currents and the consequent DRAMATIC effects on local weather patterns. Sea level "rising a few meters" doesn't sound like a lot - but when you think that this puts half a dozen US cities underwater and results in the total submersion of several countries. In most places in the world, the fertile areas where food crops can be grown and most plant and animal diversity can be found is the flatter areas close to the ocean. This means that a small increase in sea level can have a dramatic effect on the ability of a country to feed itself.

It's just not something that you can just brush off that easily. Plus you can't just "assume no runaway global warming" - that change of a few degrees is plenty enough to cause a few more degrees because of all sorts of positive feed-back effects. The small change at the beginning is the very thing that causes the 'runaway' problem - here in the real world, your "assumption" is simply not a valid one.

SteveBaker 14:44, 2 November 2007 (UTC)

Catapults

Here is a question.

Why don't they use catapults to launch planes at Airports like they do on Aircraft Carriers? 202.168.50.40 04:37, 2 November 2007 (UTC)

The acceleration is very high, injuring the ontents of the plane, and the planes too big and heavy compared to a fighter. Graeme Bartlett 04:56, 2 November 2007 (UTC)

Think of it as a question of scale as well. Imagine firing a marble out of a handheld catapult. Now imagine the catapult needed to fire the ball of glass that represented a jumbo jet. Lanfear's Bane | t 13:07, 2 November 2007 (UTC)

Also what would be the point? A runway is cheaper, easier to maintain, doesn't require power, airplanes can launch from it one after the other faster than being loaded on to a catapult one after the other, and there's enough room on planet earth to put one on. Carriers use catapults because there isn't enough room for a full runway. Catapults make it so that the plane doesn't consume as much fuel though, but the other advantages outweigh that. 64.236.121.129 13:35, 2 November 2007 (UTC)

There are a lot of simpler ways to save energy at airports that are not commonly used. In some airports in Holland they have these big tractor things that tow aircraft around once they are landed and until they are ready for take-off - these things can tow the plane all the way out to the side of the runway - and when an aircraft lands, there is one of these machines sitting there waiting to tow them back to the terminal. Since it's much more efficient to use wheels to power the movement of the plane than those big fans, it saves fuel AND makes the whole airport much quieter. It's ridiculous that those things are not at every major airport. SteveBaker 14:25, 2 November 2007 (UTC)

Also, fighter pilots are highly trained, and the plane isn't meant to be comfy. They probably don't sip coffee and read books during the flight. Nobody would like flying if passenger jets rode like fighter jets. Friday (talk) 14:43, 2 November 2007 (UTC)

How long does it take for a sudden cease of gravity to be felt?

Imagine that the Sun suddenly vanished. Would the sudden cease of solar gravity be felt on Earth before the 8 minutes and 19 seconds that the last beam of sunlight would take to reach the Earth, or would our planet be plunged into darkness before we felt the Earth free itself from its solar orbit? In other words, which of these two phenomena would happen first?

And... if the gravity loss manifested itself before the last beam of sunshine reached the Earth, would that mean the effects of gravity move faster than light? -- Danilot 08:26, 2 November 2007 (UTC)

Changes in a gravitational field move exactly as fast as light. Someguy1221 08:38, 2 November 2007 (UTC)

Agree I think it would happen at the same time, the earth would follow the same curve in space (like a water ripple before it flattens) for about 8 minutes and 19 seconds after the sun disappeared we would also receive the last of the light of the now vanished Sun for the same amount of time, after that the Earth would head out in a straight path getting very cold very quickly, the sun will be missed :) ▪◦▪≡ЅiREX≡^Talk 08:55, 2 November 2007 (UTC)

Conveniently, we have an article on the speed of gravity. General Relativity anticipates a speed of gravity equal to the speed of light. So far, the experimental results appear to bear this out, but it's a very difficult experiment to do. The earliest direct test was only published in 2003, and even that result has been controversial. TenOfAllTrades(talk) 13:02, 2 November 2007 (UTC)

The sun can't just disappear. That would violate local conservation of energy, which is a necessary consequence of general relativity (that is, general relativity can't work at all if energy isn't conserved). So you have to imagine the sun is whisked away somehow. With that caveat, the answers so far are correct.

The data from PSR 1913+16 strongly confirms the hypothesis that the speed of gravity is c. The article is correct when it says the interpretation of the data is model-dependent, but that's equally true of every scientific measurement. The work of Kopeikin and Fomalont looks wrong to me; from reading the introduction to their paper it looks like they've made the same mistake as Tom Van Flandern, and Steve Carlip's response seems to bear that out. -- BenRG 13:40, 2 November 2007 (UTC)

Is there an acid so strong that it behaves like acid is usually depicted in movies/tv?

Like how you see some crooks trying to break into bank, so they spray the side of the wall with acid from a pump, and the acid eats through the wall in seconds. Is there an acid that strong that exists in the real world? Lets assume the wall is made of concrete or wood. 64.236.121.129 13:50, 2 November 2007 (UTC)

I think it's possible (I've heard the atmosphere of Venus, having sulphuric acid clouds meaning that anything approaching Venus, if it didn't burn or melt, would corrode before it hit the planet itself), but you'd also have to figure out how on Earth you'd store that acid, and that the concentration is likely to be so high, the acidic clouds would be likely to kill the person spraying the acid... x42bn6 Talk Mess 14:18, 2 November 2007 (UTC)

See acid for details. The Bronsted-Lowry definition for an acid is a proton donor. Basically, it causes what it touches to be oxidized, or lose electrons. A metal corrodes in acid because it gets oxidized - it dissolves into the acid as it loses electrons and those electrons are taken up by the acid to form hydrogen gas. In the case of concrete, which is made primarily of cement and water, metal oxides are the primary ingredient. In this case, the metal is already ionized, but the oxygen can give up its extra electrons and form water with the free hydrogen of the acid. So corrosion of concrete is also oxidation-reduction, and the rate reactions depend primarily on the pH of the acid, which depends on its concentration or, perhaps more generally, its Hammett acidity function. So if the pH is very high, the rate of oxidation-reduction is very high, and the material corrodes quickly. I'm not sure how the reaction for wood (cellulose) works. SamuelRiv 14:28, 2 November 2007 (UTC)

Whoa there…acidity as a proton donor means it reacts with a base to form a bond, not to abstract the electrons from the base. Other things may indeed happen, but I don't think B–L acid/base strength and redox-potential aren't as directly related as you're suggesting. I'd also point out that "pH is very high" sounds like a very weak base. DMacks 14:52, 2 November 2007 (UTC)

endothermic vs. exothermic

I know exothermic means a system is releasing heat, and endothermic means a system is recieving heat, but if a system was cool to the touch would it be and endothermic or exothermic reaction? It seems like if it was exothermic it would be cool because it was releasing heat, but could also warm because it's giving off heat? Anyone care to help?--MKnight9989 14:08, 2 November 2007 (UTC)

Exothermic is roughly equivalent to "energy exiting the system" or "release of energy in the form of heat" - which makes things hotter. Endothermic is "absorption of heat" which makes things colder (because the amount of heat has gone down). If a system is cool to the touch, it means nothing, though - but if the temperature goes down, I'm guessing it's an endothermic reaction. (Note: It's been a while since I studied Chemistry) x42bn6 Talk Mess 14:14, 2 November 2007 (UTC)

Yeah I guess that makes sense. Thanks mate.--MKnight9989 14:22, 2 November 2007 (UTC)

I'd wait for a better explanation, though. Our article on exothermic, for example, looks fairly woeful... x42bn6 Talk Mess 14:24, 2 November 2007 (UTC)

You don't really specify what is the reaction. Do you mean there is a reaction within the system, or did you want to know if the very act of touching something that felt cold was exothermic/endothermic? If there was a separate reaction and it was cool to the touch, it means that the system was colder than the surrounding environment; heat was taken from the environment (air, container, etc) to fuel the reaction. This heat was transformed, and the total amount of heat in the system is less than what you started with. That means the reaction (that I assume was taking place in the system) was endothermic. Just touching the container isn't really a reaction, just a heat transfer; the total amount of heat is the same. I don't think you could classify it as endothermic/exothermic. But I'll think about it --Bennybp 14:27, 2 November 2007 (UTC)

It was a potassium sulfate/water reaction. --MKnight9989 14:30, 2 November 2007 (UTC)

atomic size constraints

What keeps an atom from being stable or even formed beyond a certain number of nucleons? Are the forces that hold the atom together not strong enough due to excessive mass or to excessive distance or both? Dichotomous 15:16, 2 November 2007 (UTC)

Spherical wheels on cars?

Think it will ever happen? I can think of a few advantages they might have. Greater manuverability for one, no need to turn the wheels, simply change the direction of their spin. 64.236.121.129 15:18, 2 November 2007 (UTC)

The contact patch would be much smaller than with a tire, right? That could be a problem. Also you'd need some pretty impressive mechanical magic to give the sphere driving force while still allowing the full range of motion. Friday (talk) 15:23, 2 November 2007 (UTC)