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January 20

What happen to helium in our sun

I know the Sun fuse Hydrogen into Helium and after billions of years eventually it ran out of hydrogen then started to fuse helium into bigger atoms and eventually collapse. What happen to helium in those billions of years? Is it just stay somewhere inside the sun? Or is it broken down to hydrogen again to start out the fusion cycle? 184.97.244.130 (talk) 00:18, 20 January 2013 (UTC)[reply]

Eventually it will become carbon through the triple alpha process.--Gilderien Chat|List of good deeds 00:25, 20 January 2013 (UTC)[reply]

That's not what I'm asking... What happen to Helium in our Sun right now as hydrogen is still abundant.184.97.244.130 (talk) 03:20, 20 January 2013 (UTC)[reply]

Nothing at all. Currently, the Sun is fusing hydrogen into helium. This produces enough energy so that the Sun cannot collapse and become dense enough to fuse helium. Therefore, the Sun is hot enough to fuse hydrogen but not hot enough to fuse helium (yet). Whoop whoop pull up ^{Bitching Betty | Averted crashes} 04:46, 20 January 2013 (UTC)[reply]

I am not sure that is correct, Whoop. I'd really like to see some sources and articles quoted. Helium three is being constantly generated and destroyed at this point. See proton-proton chain reaction. There's no a priori reason to assume that certain, perhaps low-rate, nuclear reactions with Helium four are not also going on at this point. Citations are needed. μηδείς (talk) 04:52, 20 January 2013 (UTC)[reply]

While the rate of helium burning isn't strictly zero, it's actually astonishingly close. As noted (and footnoted) in the article already linked by Gilderien, the rate of the triple-alpha process responsible for helium fusion depends on the core temperature to the fortieth power. In a low-to-middling mass star like our Sun, the rate might as well be zero right up until the helium flash. TenOfAllTrades(talk) 05:09, 20 January 2013 (UTC)[reply]

So basically helium is just there without any interaction until the Sun is hot enough to fuse them, correct?184.97.244.130 (talk) 05:31, 20 January 2013 (UTC)[reply]

Yes, it is effectively simply accumulating in the sun as an end product of the hydrogen fusion process at the moment. — Quondum 08:08, 20 January 2013 (UTC)[reply]

Is the helium mostly at the centre, or does it spread evenly throughout the mass of the Sun? I know we cannot see the centre of the Sun, but maybe some simulations have given some hints.--Lgriot (talk) 09:00, 21 January 2013 (UTC)[reply]

How well mixed together the hydrogen and helium are depends on the size of a star. For a small red dwarf, it's all mixed together evenly (which is part of the reason they burn for so long - hundreds of billions of years, compared to a mere 10 billion for the Sun). For the Sun, it's more concentrated in the core. Convection zone and Radiation zone are relevant articles, although they aren't very good... --Tango (talk) 11:29, 21 January 2013 (UTC)[reply]

Thx --Lgriot (talk) 13:35, 21 January 2013 (UTC)[reply]

why is this not used in air?

http://en.wikipedia.org/wiki/Modulated_ultrasound

why isn't this used for near-field communications (like bluetooth etc). is it because radio is so much easier? But radio requires licenses and has limited spectrum, not that this doesn't but i would imagine while no one else thinks to use it it does! :) 178.48.114.143 (talk) 00:22, 20 January 2013 (UTC)[reply]

Lol, I found this: http://www.theregister.co.uk/2012/11/08/ultrasonic_bonking/

“

A Singapore firm is punting ultrasonic sound as an alternative to NFC for short-range wireless communications, pointing out that it works with existing hardware and provides a demo app to prove it.

The software, which is on Google Play and iTunes (search for SSCconnect) in both free and ad-supported versions, allows devices to exchange contact details, URLs, and clipboard contents at the breakneck speed of 275 bytes per second.

”

But in fact this is just iphone hardware. couldn't specialized hardware be a bit better? 178.48.114.143 (talk) 00:24, 20 January 2013 (UTC)[reply]

Take a look at these:

http://alumni.media.mit.edu/~wiz/ultracom.html

http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA499556

http://www.cs.ou.edu/~antonio/pubs/conf059.pdf

In particular, look at the data rates they have been able to achieve.

--Guy Macon (talk) 00:53, 20 January 2013 (UTC)[reply]

I suspect one problem would be interference. In other words it would work fine if you were the only one using it, but what happens if you were in a crowd with many people using it. There could also be a problem with phase diffraction from multiple sources. (See also Superposition principle and Circular convolution) ~ It would probably be okay if by "short-range" you mean "a few inches". ~E:(talk) 01:17, 20 January 2013 (UTC)[Fixed header:74.60.29.141 (talk) 01:38, 20 January 2013 (UTC)][reply]

It will also drive your dog crazy. And it provides fairly limited bandwidth at plausible frequencies. --Stephan Schulz (talk) 10:14, 20 January 2013 (UTC)[reply]

oligomers and monomers

--41.203.67.133 (talk)mkm~

what are the differences between oligomers and monomers using equations and mechanisms as a basis for the differentiation — Preceding unsigned comment added by 41.203.67.133 (talk) 01:08, 20 January 2013 (UTC)[reply]

I'll offer you a hint: Consider what the prefixes "mono-", "oligo-", and "poly-" mean. These are all fairly fundamental concepts in polymer science. Assuming the course these homework assignments belong to is a polymer characterization course you probably want to have these kinds of definitions down cold. (+)H₃N-Protein\Chemist-CO₂(-) 02:07, 20 January 2013 (UTC)[reply]

polymerisation

what is the difference between the average degree of polymerization and the extent of a reaction using equations and reaction mechanisms to distinguish between them — Preceding unsigned comment added by 41.203.67.133 (talk) 01:12, 20 January 2013 (UTC)[reply]

Perhaps you should take a look at our article titled "Degree of polymerization". We're not going to do your homework for you, but if there are any remaining points of confusion we can probably help clarify them.(+)H₃N-Protein\Chemist-CO₂(-) 01:54, 20 January 2013 (UTC)[reply]

chemistry [polymer chemistry]

what is the difference between number average molecular weight and weight average molecular weight — Preceding unsigned comment added by 41.203.67.133 (talk) 01:17, 20 January 2013 (UTC)[reply]

You may want to have a look at molar mass distribution. These sorts of definitions are necessary to describe polymers, which are virtually always a mixture of molecular weights. Some methods such as static light scattering are intrinsically weight averaged, since scattering intensity is proportional to molecular weight amongst other things. (+)H₃N-Protein\Chemist-CO₂(-) 01:43, 20 January 2013 (UTC)[reply]

RF interference

An organization I'm in using some wireless devices that use 915 MHz. They recently replaced a wireless microphone and they say that the new one interferes with the other devices. The mike says it can be set to frequencies between 524 MHz and 865 MHz. Could the mike really be causing interference? (I used them together a few days ago and did not experience any problem.) Bubba73 ^{You talkin' to me?} 02:21, 20 January 2013 (UTC)[reply]

The frequency at which it's set isn't necessarily the only frequency at which it broadcasts. See spurious emission. Indeed, devices not made to broadcast RF at all, such as microwave ovens, frequently still do, although at a level that can only cause interference within a few feet (see Radio_transmitter_design#RF_leakage_.28defective_RF_shielding.29). StuRat (talk) 03:41, 20 January 2013 (UTC)[reply]

Microwave ovens are meant to produce RF, though. It's how they cook stuff. And lots of devices use RF at the same or very similar frequencies as microwave ovens. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 05:02, 20 January 2013 (UTC)[reply]

They aren't meant to "broadcast" RF, which is what I said. StuRat (talk) 05:06, 20 January 2013 (UTC)[reply]

WP:OR, but my wireless internet in my house goes a little jinky when the microwave is running. --Jayron32 06:03, 20 January 2013 (UTC)[reply]

To Jayron's point, Obligatory xkcd reference. Zunaid 09:02, 20 January 2013 (UTC)[reply]

It could be a harmonic, so we could try some other frequencies. I once lived in a city where you could pick up an AM radio station on two frequencies - one was half or double of their actual frequency. Bubba73 ^{You talkin' to me?} 04:23, 20 January 2013 (UTC)[reply]

For it to be a harmonic, wouldn't it have to be set to 457.5 MHz, which is outside the given range? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 04:59, 20 January 2013 (UTC)[reply]

For a 2:1 or 1:2 harmonic, yes, but there could be other harmonics, 3:2, etc, I think. Bubba73 ^{You talkin' to me?} 05:47, 20 January 2013 (UTC)[reply]

Such devices usually oscillate at the transmit freqwuncy (ie there is not internal multiplication), so only integer multiples of the rated frequency is possible - i.e., 1048 MHz, 1536 MHz, etc at the lowest channel setting. Interfering on a sub-harmonic, and relatuonships such as 3:2, 2:3 are not possible. However, radio reciveing devices can suffer from blocking - the receive circuits are overloaded and thus de-sensitised by a strong signal close to the frequency they are tuned to. How close in frequency is determined by the design of the receiver and how physically close and strong the interfering signal is. So the answer is yes - the radio mike could well be interfering, without being in a harmonic relationship. Keit124.178.178.83 (talk) 08:45, 20 January 2013 (UTC)[reply]

Thanks. Bubba73 ^{You talkin' to me?} 14:51, 20 January 2013 (UTC)[reply]

Altitude / temperature

I've hiked the Grand Canyon a few times, and one thing seems counter-intuitive: it's quite warmer at the bottom than on the rim. Consider the following: a) heat rises. b) air is thinner (less dense) at the top; sunlight should therefore be stronger. c) the bottom is in the shade much of the day, whereas the rim is not. ~ So, ~ why is it warmer at the bottom? ~:74.60.29.141 (talk) 02:43, 20 January 2013 (UTC)[reply]

My previous home was on a hill and when I was walking, it was warmer at the bottom, in the valley. I figured that at the top the air warmed by the ground got blown away by the wind, but I don't know if that is correct. Bubba73 ^{You talkin' to me?} 02:56, 20 January 2013 (UTC)[reply]

Basically the lapse rate for temperature is a function of two things: (1) solar heating occurs almost entirely at ground level; (2) the ability of the atmosphere to retain heat is weaker the higher the altitude. The way those factors trade off for something like the Grand Canyon is not so easy to work out, but you should bear in mind that the south-facing canyon walls are very strongly illuminated, pretty much all the way to the bottom. Looie496 (talk) 04:26, 20 January 2013 (UTC)[reply]

I've always understood that the main function responsible for the lapse rate is the adiabatic expansion / contraction of the circulating atmosphere. Note that adiabatic lapse rate redirects to the lapse rate article you linked. I am a bit confused by the article, though. Is the environmental lapse rate the observed lapse rate at a particular location while the adiabatic lapse rate is the predicted rate due to the modeled adiabatic processes which are the major factor? -- 41.177.85.143 (talk) 07:25, 20 January 2013 (UTC)[reply]

Following up, sources such as this give the following elevations:

South Rim (Bright Angel Trail Trailhead): 6,860ft

Colorado River (presumably at the Silver Bridge): 2,400 ft

North Rim (Kaibab Trail Trailhead): 8,241 ft

That puts the South & North Rims 4,460 ft and 5,841 ft above the river. An average lapse rate of 3.5°F/1,000 ft would estimate the temperature at South and North Rims to be 15.6°F and 20.4°F above the temperature down near the river. (Using the dry lapse rate of 5.5°F/1,000 ft would give even greater temperature differences of 24.5°F and 32.1°F.) Googling "grand canyon hotter bottom" turns up a lot of pages stating that the temperature at the bottom is 15°F to 20°F warmer than at the rim, suggesting that the adiabatic lapse rate is sufficient explanation. -- 41.177.85.143 (talk) 11:10, 20 January 2013 (UTC)[reply]

Finally, the questioner should ask themself whether they have ever wondered with equal curiosity why it is cooler on top of mountains, and if not, then why they haven't. It is the same mechanism at work. -- 41.177.85.143 (talk) 11:12, 20 January 2013 (UTC)[reply]

Yes, I realize the same principles apply to mountains, it's just that the difference seems more obvious at the canyon. Plugging numbers into a formula might provide a mathematical description, but not a satisfactory explanation of why higher altitudes are colder than lower ones (given the same approximate latitude). ~:74.60.29.141 (talk) 18:00, 20 January 2013 (UTC)[reply]

The explanation is "expansional cooling" and "compressional warming". When a gas expands it push the surrounding air out of the way (so it can expand) and that requires energy which is taken from the internal thermal energy of the gas, hence the drop in temperature. When a gas is compressed the opposite happens, the energy flows the other way, and the temperature rises. Dauto (talk) 18:19, 20 January 2013 (UTC)[reply]

That explains how an air-conditioner works; but it seems implausible that there would be an energy transference from the rim to the bottom by that means. That would require a pressure differential over time and distance, and the specific air molecules wouldn't necessarily be transferred from one place (low density) to another (high density) where the heat energy is exchanged. ~[Does that makes sense?].

I don't mean to sound unappreciative of the answers, it's just that this question has bothered me for quite some time. I'm sure the lapse rate / adiabatsis explains it, but the underlying thermodynamic principles are still unclear to me. Consider the following hypothetical: given a column of air (let's say 1km) - perfectly sealed within a thermally insulated (vertical) container. Leave it alone for a year or two; when you come back the air temperature at the bottom would be higher than that of the top, by an amount consistent with the related equations — right? ~:74.60.29.141 (talk) 21:22, 20 January 2013 (UTC)[reply]

Feeling warmer in a valley is quite common, but not all valleys are warm. All this talk of lapse rate is misleading you. As Looie496 said in his post, air is hotter nearer the ground because the air is heated by the ground, which is heated by incoming solar radiation. The air both absorbs some of the raditated heat (in both directions), and re-radiates it, in both directions. This is the greenhouse effect, resulting in a band of air at altitude that has the minimum temperature - below the air is warmer as it is close to the warm ground, and above it is warmer due to being above most of the air heat absorption.

Now, think about it. If this was the only factor, and (say) the average ground temperature at some latitude is 20 C, you'd expect the temperature at 1000 m above it to be 13.5 C, applying the satndard lapse rate of 6.5 C per 1000 m (Note that the lapse rates given in a post above are incorrect). Now, assume you are on level ground in a bloody great hole 1000 m deep. Based on the greenhouse effect, which is what causes the lapse rate, you'd still expect the temperature at the bottom of the hole to be 20 C, and the top of the hole to be 13.5 C.

It can't work that way - the temperature of air at the top of a hole cannot be 6.5 C below the surrounding air, because wind and convection will mix it to equality at the top. Any hot air at the bottom will be less dense and try to rise out of the hole, and be replaced with cooler air.

So what does cause the bottom of some but not all valleys, and the Grand Canyon, to be warmer?

The south facing walls get full sun as Looie496 said.
Some of the heat re-radiated off the south walls gets trapped by the other walls - this is a large scale example of the cavity radiator effect, well known to engineers - to find the appoximate temperature of a furnace, drill a small hole in the side - if the furnace is red hot, the hole will appear to be a hotter colour.

Supporting what I've said is the common experience of folk living in the bottom of valleys. Some valleys are warm and some are not. It depends on the alignment of the valley axis vis-a-vis the direction of the sun's radiation, and on whether or not the alignment of the valley axis allows the prevailing winds to penetrate.

Wickwack 120.145.56.251 (talk) 01:11, 21 January 2013 (UTC)[reply]

I'll have to think about this some more (later, its bedtime).. ~:74.60.29.141 (talk) 04:10, 21 January 2013 (UTC)[reply]

Hey Wickwack, can you clarify "applying the standard lapse rate of 6.5 C per 1000 m (Note that the lapse rates given in a post above are incorrect)."? I assume that you are referring to my 3.5°F/1,000 ft which is the standard value quoted in our article and is equivalent to your 6.5 C per 1000 m. -- 41.177.85.143 (talk) 07:05, 21 January 2013 (UTC)[reply]

I never looked at the Wikipedia article, as they are not intended to be trusted as a data source. They are only intended to guide novice readers to references and sources. I got the standard rate 0.0065 C/m (= 6.5 C per 1000 m) for the altitude of the Grand Canyon from a set of standard tables (Turns & Kraige 2007, page 94) I have. Note that the lapse rate is not constant but varies with altitude - at 4000 m it reduces to 1/10th this value, and at 15,000 m it is zero. At greater altitudes it increases at opposite sign, untimately reversing sign again. The value you used in your calcs, 3.5 F / 1000 Ft equates to 6.38 C / 1000 m. That is, you used a value 2% low. This is not significant in the context of the question, but I thought that someone would squawk that I used a different value. I should have explained it better - I appologise for the confusion. Wickwack 124.182.14.231 (talk) 07:55, 21 January 2013 (UTC)[reply]

Thanks. I certainly wouldn't quibble over a one to two percent difference in figures which are only given to two significant digits, but the variability of lapse rate with altitude is something which could have a significant impact on my calculations above. -- 41.177.85.143 (talk) 08:18, 21 January 2013 (UTC)[reply]

I agree that local topography, insolation, and winds can have a large effect on the actual, observed lapse rate, but the standard rate falls between the dry and saturated adiabatic lapse rates and successfully predicts the observed rate at the Grand Canyon itself, so I don't understand why additional mechanisms are needed here to explain what appears to be consistent with the thermodynamic model. -- 41.177.85.143 (talk) 07:38, 21 January 2013 (UTC)[reply]

A coincidence is not proof. I have explained why the lapse rate actually is not the full story, as relying on it alone means that the floor of the canyon, around 800 m above sea level, should have a temperature about the same as the area generally, about 2700 m above sea level, and if it did, there would be a pocket of air above the canyon at 2700 m lower in temperature than the air in the area generally, and that cannot be the case, as Louie pointed out. Wickwack 124.182.14.231 (talk) 08:04, 21 January 2013 (UTC)[reply]

The lapse rate in the Grand Canyon is specifically addressed in this (PDF 1.3 MB) 1965 Journal of Applied Meteorology article. -- 41.177.85.143 (talk) 08:46, 21 January 2013 (UTC)[reply]

Note that this paper, which is about the improved horizontal resolution of a then new satellite, talks about an apparent lapse rate, which the author's data indicates was 10 C per 1000 m at the time of measurement, well above the standard value. The article does not use the standard lapse rate to explain why the canyon floor was warmer, it merely says that there appears to be a significant lapse rate and mentions some reasons why it might be thought to occur at the magnitude that it did. The purpose of the paper was use the Grand Canyon to show how good the horizontal resolution was. Wickwack 120.145.80.46 (talk) 10:38, 21 January 2013 (UTC)[reply]

This PDF link isn't working for me (can't connect to host) but googling "Lawrence E. Stevens The Biogeographic Significance of a Large, Deep Canyon" and looking at the "Quick View" gives a 2012 work which states in section 3.4, Elevation:

Nonetheless, elevation remains an overwhelmingly important ecological state variable due to its strong negative relationship with air temperature and freeze-thaw cycle frequency, and its positive relationship to precipitation and relative humidity. The global adiabatic lapse rate is -6.49 °C/km. Analysis of paired daily minimum and maximum air temperature from 1941-2003 at Phantom Ranch (elevation 735 m) on the floor of GC with the South Rim (2100 m) produces a GC-specific lapse rate of -8.7 °C/km. The >1.3-fold steeper lapse rate in GC is likely a function not only of the dark red and black bedrock color of the inner canyon, but also to aspect. Steep, S-facing slopes in the GCE, particularly those with darker rock color, absorb and re-radiate more heat than do N-facing slopes, which often are shaded from direct sunlight, and are cooler and more humid than S-facing slopes across elevations. Overall, elevation strongly and broadly influences synoptic climate, while aspect exerts strong local control over microclimate and microsite potential evapotranspiration and therefore productivity.

I find it interesting that he does not mention the arid climate as a factor in the lapse rate as the dry adiabatic rate of 9.8 °C·km^-1 is greater than that observed in GC. A paper on the subject written by a climatologist would carry more weight than a passing mention by a biologist, but I'd be inclined to pay more attention to either than to anything said here. -- 41.177.85.143 (talk) 11:03, 21 January 2013 (UTC)[reply]

►The bottom line: the primary principle involved would be adiabatic lapse which relates to air density, which relates to barometric pressure (?). Therefore, the barometric pressure at the bottom is consistently higher than at the top. [?]

Localized topography, azimuth orientation (etc.) primarily accounts for the variety of micro-climate conditions.

~Eric _the OP:74.60.29.141 (talk) 22:04, 21 January 2013 (UTC)[reply]

why when people have fever the feel colder?

seem a bit pardoxic. i had fever today, and i felt cold, anything i wore almost gave no heat. why is that? — Preceding unsigned comment added by 79.176.113.107 (talk) 03:30, 20 January 2013 (UTC)[reply]

Because your bodies ability to sense temperature is out of whack. Your body is not an accurate thermometer, and your general sense of warmness and coolness is not really directly connected to the internal or external temperature. Wikipedia has an article on thermal comfort which is a bit bloated, but has some information. Physiologically, your sense of temperature is wrapped up in your Somatosensory system. --Jayron32 03:35, 20 January 2013 (UTC)[reply]

It's common to swing back and forth between feeling hot and cold, when you have a fever. StuRat (talk) 03:37, 20 January 2013 (UTC)[reply]

My understanding is that a fever is a response to infection in which the body tries to raise its temperature in order to increase the activity of the immune system. It does that by raising the homeostatic "set point", and anything below the set point is going to feel cold to you. Looie496 (talk) 04:16, 20 January 2013 (UTC)[reply]

It's not really that the body's ability to sense temperature is out of wack, but that the thermostat has been reset either by the immune response or the pathogen itself. See Pyrogen (fever). One theory of fever is that whichever agent induces the fever, the pathogen or the body itself, will function better against its opponent with a raised temperature. Many proteins function best within a set temperature range, and leaving that range can have a huge effect. μηδείς (talk) 04:47, 20 January 2013 (UTC)[reply]

Looie496 is right. Your body has a "normal" temperature. When you get a fever, the "normal" temperature rises. Therefore, what originally was at the "normal" temperature is now colder than the "normal" temperature, and consequently feels cold. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 04:52, 20 January 2013 (UTC)[reply]

For completeness sake, the homoeostatic set-point for temperature is set by the hypothalamus. Fgf10 (talk) 12:03, 20 January 2013 (UTC)[reply]

Fevers occur in reptiles, fish,[1], and even insects.[2] Fever is stimulated by eicosanoids and therefore would seem to be conserved with mammals. [3][4] Amazingly, this means that the fever response dates back almost to the Urbilaterian, maybe further. All this time fever has been a matter of behavior, and only in the most recent times does it control internal temperature directly. What's odd is that Hollywood appears unalterably convinced that people with fever are hot, and must have shown this idiocy in a thousand movies. Don't they get sick in celebrityland? Wnt (talk) 22:46, 20 January 2013 (UTC)[reply]

I think there is a big misunderstanding here. Your body doesn't directly detect the temperature of the air around you.

Let's do an experiment: Touch a piece of metal with one hand and a piece of plastic with the other. Which feels warmer? The plastic - right? But in truth, they are both at the exact same temperature (you can check...use a thermometer!). That's because, what we actually detect is the amount of heat energy that is drawn out of our skin. Metal feels colder than plastic because it conducts heat away from your body very efficiently, where plastic doesn't.

Now, consider Newton's law of cooling: "The rate of heat loss of a body is proportional to the temperature difference between the body and its surroundings.".

That means that when your feverish body is hotter (compared to the surrounding air) than it usually is, you lose heat to the environment more rapidly than you normally would...and because "rate of heat loss" is what our skin actually measures, you feel colder because you're losing more energy to the outside world than usual.

When I have a fever, I feel cold because my body is hotter compared to the surrounding air. Seems backwards - but it's not. SteveBaker (talk) 17:25, 21 January 2013 (UTC)[reply]

Check your understanding against Transient receptor potential channel, especially TRPM and TRPV. The proteins act as thermometers, and being in exposed regions of skin which may vary from core body temperature, they can sense a temperature that is influenced by the outside air. Likely someone can remedy my ignorance, but I am having trouble thinking of an instrument that measures heat flow directly. Wnt (talk) 16:55, 22 January 2013 (UTC)[reply]

I think both of you are basically correct. The receptor activity is determined by the temperature of the skin, which is largely determined by the rate of heat loss from the skin (also by the amount of blood flow reaching the skin). Looie496 (talk) 17:08, 22 January 2013 (UTC)[reply]

There is no need to argue - do the experiment. Metal feels cold, plastic feels warm. How can you explain that other than by the fact that we measure rate of heat loss as opposed to temperature? SteveBaker (talk) 21:03, 22 January 2013 (UTC)[reply]

Well, I did do the experiment - and found that you are wrong. I got a thick piece of aluminium (about 150 x 30 x 25 mm), an excellent conductor of heat. I mounted on it two PT100 platinum film temperature transducers, on opposite sides. These are available in very thin forms - the ones I have are about 100 x 10 x 1.5 mm mounted. I got a similar piece of Tufnol, a type of composite filled plastic, moderate conductor of heat, and attached two PT100 transducers to that. I also got a similar size piece of polystyrene, a poor conductor of heat, and mounted two PT100's to that. I connected all the PT100's up to electronics to read the temperatures in deg C and after allowing a settling time they all read the same temperature - 23 C. Then I put two fingers on the top PT100 on the aluminium block, two fingers on the top PT100 of the Tufnol, and two fingers of the other hand on the top PT100. Sure enough, very quickly I felt the aluminium as cold, the tufnol a bit warmer, and the polystyrene as warmest. None of the bottom PT100's changed. But while the top PT100 on the aluminium did not change its temperature while my fingers were resting on it, the top sensor on the tufnol went up 4 C or so, and the polystyrene top sensor quickly went up 9 C. In other words, plastic feels warm not because of a low flow of heat, but because its surface in contact with your skin gets increased in temperature. Wickwack 121.221.216.91 (talk) 14:41, 23 January 2013 (UTC)[reply]

Out of battery detonation

What are the most common causes of out-of-battery detonations in firearms? 24.23.196.85 (talk) 05:45, 20 January 2013 (UTC)[reply]

Human error? ~:74.60.29.141 (talk) 05:48, 20 January 2013 (UTC)[reply]

See slamfire. Zoonoses (talk) 06:04, 22 January 2013 (UTC)[reply]

Do you mean that accidental slamfires are the most common cause of out-of-battery detonations? 24.23.196.85 (talk) 06:13, 22 January 2013 (UTC)[reply]

Isn't slamfire intentional and beneficial, due to its allowing the user to spray shells onto the enemy by simply holding the trigger and operating the pump? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 21:44, 22 January 2013 (UTC)[reply]

I meant "see slamfire," nothing more. After reading slamfire, you should have the answer to your original question. Zoonoses (talk) 06:49, 23 January 2013 (UTC)[reply]

Electrical conduction

Could somebody please tell me the answer to this, I can't seem to find anything about it in Wikipaedia articles. If I connect an ordinary multi-strand copper wire to a 12volt DC connection, but only get half the strands in the connector, then connect the other end of the cable to a device (like a light for example) but only get the other half of the strands in that connector, how efficient will the power transmission be? Thanks in advance. 124.191.177.1 (talk) 07:21, 20 January 2013 (UTC)[reply]

Assuming that the individual strands of the conductor are not corroded, oxidized or coated in any way, the inter-strand resistance (per metre) between copper strands will be low. Thus, after as short distance (a few diameters of the copper of the wire), the current should be effectively uniformly distributed over the cross-section of the copper. Thus, one could say "about as efficient as using all strands, but with a small added length to the cable". For most practical purposes, this means the effect will be insignificant. — Quondum 07:59, 20 January 2013 (UTC)[reply]

(agreed) ... and it's very unusual to get corrosion or significant oxidation in the middle of a wire if the insulation is undamaged. For wires carrying a high current (for their cross-section), then it is obviously better to connect as many strands as possible, but even one connected strand at each end (and not the same strand) will usually give efficient power transmission, though I wouldn't recommend the practice because the single strand at each end might get hot, and it will act like a fuse wire, burning out at a certain high current. Dbfirs 16:51, 20 January 2013 (UTC)[reply]

Thankyou, the multistrand cable I am using is tinned copper, so I assume this makes no difference to the interstrand resistance.124.191.177.1 (talk) 07:23, 21 January 2013 (UTC)[reply]

~~Tin is actually an even better conductor than copper, so using tinned copper helps. In addition,~~ the tinning prevents oxidation of the copper, so largely prevents the problem of corrosion mentioned by Quondum above. Dbfirs 10:22, 21 January 2013 (UTC)[reply]

Dbfirs: are you sure that tin is a better conductor than copper? I seem to remember copper is a better conductor, and the table at Electrical_conductivity#Resistivity_of_various_materials seems to support this. – b_jonas 16:03, 21 January 2013 (UTC)[reply]

Sorry, my error. I picked up the conductivities from a table elsewhere on the internet and either it was wrong or I mis-read it (the latter being more probable!) In fact pure tin has only one seventh the conductivity of copper. I've striken my erroneous comment above, leaving only the second part that remains valid and is more significant in the situation being considered. Tin does have a higher conductivity then oxidised copper, and bare copper corrodes easily. Dbfirs 22:45, 21 January 2013 (UTC)[reply]

You're right that tin is a worse conductor - only silver is better than copper. But "tinned copper" doesn't necessarily mean literally "copper coated with tin" - it's probably a lead/tin alloy with other metals involved in the mixture. The practical function is twofold - excluding air and thereby reducing the corrosion of the copper - and (because lead/tin alloy is so soft and because it flows into the gaps between the conductors) improving the contact area between the strands and between strands and terminal block. Those two things taken together greatly improve the quality of the contact. SteveBaker (talk) 17:12, 21 January 2013 (UTC)[reply]

If you think about it, even if you get all of the conductors stuffed into a "screw terminal", only the outer strands are actually in contact with the terminal block - so you're relying on inter-strand conductivity to make the other connections anyway. The big problem with only connecting a few of the conductors is the very short distance between the terminal block and the point where all of the conductors meet. It doesn't much matter how short that distance is - the wires can still overheat at that spot and cause problems. SteveBaker (talk) 16:26, 21 January 2013 (UTC)[reply]

The reason why wires are made multistranded is to improve tolerance of vibration and to make the wire easier to install, which usually requires sharp bends. If not all the strands are soldered or clamps at the termination, you have an increased risk of the wire breaking. The reason for the tinning is to improve solderability. Wire intended for screw clamp connection is generally not tinned. Oxidation is not a problem as it won't occur where the strands are under clamping pressure at teh termination, and should not occur within the plastic insulation. Ratbone 124.178.141.66 (talk) 01:12, 22 January 2013 (UTC)[reply]

I agree that oxidation does not normally occur within the plastic insulation, even after 40 years in damp conditions, but it can certainly occur around the clamped wire, and if the clamping is not as firm as it should be, it can spread under the clamp and increase resistance at the contact. I've seen it happen, but I agree that it shouldn't under a really secure clamp. The problem is more serious in telecommunications and data wiring where insulation-displacement connectors are used. Tinned wire is standard in these applications. Dbfirs 08:17, 22 January 2013 (UTC)[reply]

tDCS for insomnia?

According to this article: "I’ve been told that [transcranial direct current stimulation] is handy if you have racing thoughts at bedtime." Have any controlled trials of tDCS confirmed this? If not, with how much confidence can it be inferred given results with TMS? Neon Merlin 08:22, 20 January 2013 (UTC)[reply]

I can't find any details of any such trials. It may be of interest to you to note an article in the Journal Of Psychiatric Research (2013 Jan; Vol. 47 (1), pp. 1-7), which is 'Clinical utility of transcranial direct current stimulation (tDCS) for treating major depression; a systematic review and meta-analysis of randomized, double-blind and sham-controlled trials.' This concludes that there is at this stage no clear evidence for the clinical utility of this method for the treatment of MD, and more extensive trials will be required. ---- _nonsense ferret 14:27, 20 January 2013 (UTC)[reply]

Nuclear fusion to produce phosphorus

What temperature would be required for the fusion reaction ²⁸Si + ⁴He → ³¹P + ¹H (which [jtgnew.sjrdesign.net/stars_fusion.html this site] says is part of the stellar oxygen-fusion chain) by known methods? Could any known fusion reactions produce ³¹P at a lower temperature from isotopes abundant on Earth? Neon Merlin 13:56, 20 January 2013 (UTC)[reply]

You misunderstood what they said. They mean that the Oxygen-burning process has several possible outcomes:

¹⁶O+¹⁶O= ³¹P + ¹H or

¹⁶O+¹⁶O= ²⁸Si + ⁴He

among them. The required temperature is more than 1 billion K. Ruslik_Zero 17:48, 20 January 2013 (UTC)[reply]

Electric current through gases

At normal pressure air or any other gas is a nonconductor of electricity, but at low pressure the gas become conductor of electricity. How does this happen?
In a discharge tube, sparking is accompanied by crackling noise. How this noise is produced? — Preceding unsigned comment added by Want to be Einstein (talk • contribs) 14:02, 20 January 2013 (UTC)[reply]

Non-ionised gasses are not conductive at any pressure, low or high. What makes you think they are conductive at low pressure? Conversely, an ionised gas is conductive at all pressures.
When an electrical discharge occurs, there is local heating, which causes an increase in pressure. The over-pressure travels outwards and thus constitutes a sound wave.

Wickwack 120.145.81.211 (talk) 14:48, 20 January 2013 (UTC)[reply]

Some links you might find useful: Plasma (physics)#Generation of artificial plasma, dielectric strength, and for a more detailed explanation of the mechanism, see Paschen's law. — Quondum 15:05, 20 January 2013 (UTC)[reply]

I have seen that gases become more ionized at low pressure than at high pressure. How a gas becomes ionized at low pressure ? Want to be Einstein (talk) 17:07, 23 January 2013 (UTC)[reply]

Same way as at high pressure. Energy must be supplied to knock electrons off the atoms or molecules. See Paschen's Law, cited by Quondum. Wickwack 121.221.91.193 (talk) 01:47, 24 January 2013 (UTC)[reply]

Hydrolysis of Phosphodiester bond in DNA

When the phosphodiester bond in the Phosphate-backbone of the DNA is hydrolysed by DNAses enzymes, which one of the P-O bond is broken? Dnakid (talk) 17:59, 20 January 2013 (UTC)[reply]

Depends on the DNAse. Deoxyribonuclease I leaves 5' phosphates, whereas Deoxyribonuclease II leave 3' phosphates. -- 71.35.98.191 (talk) 20:24, 20 January 2013 (UTC)[reply]

compressing gas into a cylinder

would a very light-weight "pod" floating on a helium baloon with a line into it, be able to consist of also a battery, light empty canister, and pump, and descend when it wishes by pumping (compressing) the helium (or enough of it) back into the cannister to descend?

Of course, the battery would have to be recharged after a while, but in this way helium could be conserved. The idea is that when it wishes to ascend again, it can refill the baloon from the helium it has just pumped into the cannister.

Would this work? Thanks. 178.48.114.143 (talk) 18:28, 20 January 2013 (UTC)[reply]

Yes, in fact there is a (proposed) aircraft which uses a similar principle -very interesting concept- See "Gravity Plane": http://www.youtube.com/watch?v=0QZ1KzveIic (WP won't allow direct link to YouTube) ~E:74.60.29.141 (talk) 18:56, 20 January 2013 (UTC)[reply]

More directly related to what you're thinking about: Aeros Flight Buoyancy Management - "By compressing and decompressing helium, the density in the ship can be varied as a means to control the ship’s static heaviness."

On a much smaller scale (your balloon) the limiting factor would be the weight of your "pod". ~E:74.60.29.141 (talk) 21:52, 20 January 2013 (UTC)[reply]

A very rough rule of thumb is that you need one cubic meter of helium to lift one kg of craft. That has to include the weight of the balloon itself. SteveBaker (talk) 02:36, 21 January 2013 (UTC)[reply]

What you are describing is "sort of" an aerial version of a Stab jacket, similar principle anyway. Vespine (talk) 05:25, 21 January 2013 (UTC)[reply]

Except that while gas is added from a cylinder to a stab jacket to increase buoyancy, it is vented to decrease buoyancy. It would be as if there was a pump to move gas from the jacket back to the cylinder when less buoyancy is desired. -- 41.177.85.143 (talk) 07:50, 21 January 2013 (UTC)[reply]

In order to link to youtube use this markup in edit: [http://www.youtube.com/watch?v=0QZ1KzveIic gravity plane vidoe] to produce this link: gravity plane video. The video itself is absurd, since the plane is presented as a perpetual motion machine, and its capacity for failure seems unlimited. μηδείς (talk) 07:30, 21 January 2013 (UTC)[reply]

No, that doesn't work. The canister strong enough to contain the helium always weighs more than what that much helium can lift. – b_jonas 15:58, 21 January 2013 (UTC)[reply]

What makes you say that?!? I certainly don't believe it's true. The volume of a sphere is a function of the cube of the radius - the surface area increases as the square of the radius. Since the amount of lift is proportional to the volume - but the weight of the envelope is a function of the surface area. Double the size of the balloon and you can double the thickness of the envelope without changing the pressure it has to withstand. Hence it's always possible (in principle) to build a bigger balloon to make it light enough to fly. (Recall the Mythbusters episode where they made a balloon out of lead foil - and it actually flew!) But the amount of additional pressure required to make the balloon lose altitude doesn't have to be that much - providing you don't need to lose altitude rapidly. From that point of view, it's a plausible machine. SteveBaker (talk) 16:17, 21 January 2013 (UTC)[reply]

The plane in the video is (effectively) powered by compressed air. They say that compressed air is used to create thrust - it's dumped as ballast - and it's used to re-compress the helium so the plane can return to earth. What replenishes that compressed air? All of the energy for the flight is contained in that compressed air...but it's not a perpetual motion machine. That said, the video is confusing...why would this plane require wings? Why folding wings? Lighter-than-air craft aren't magical - they are like up-side-down aircraft. With a regular plane, ascent costs lots of energy, but coming back down again is free. With a conserved-gas lighter-than-air craft, it's the reverse...going up is free, but it takes energy to get back down again. The only "free lunch" comes from venting gas at altitude - effectively switching from being lighter-than-air to being heavier-than-air and thereby getting a free ride both up and down. But venting gas is a problem: Helium is expensive, and it's a non-renewable resource that humanity is rapidly running out of - so simply venting the stuff to get you back down again is not a great way to go. That leaves you with other potential lifting gasses - but they are either rare and expensive (helium, neon), potentially dangerous (hydrogen, methane, ammonia), energy-intensive to produce and maintain (hot-air, steam), only marginally light enough (nitrogen, neon) or nasty green-house gasses (methane).

Recompressing helium at altitude (as suggested by our OP) is a good approach - but it still consumes energy.

I like the idea of using a Rozière balloon - which is a hot air balloon with a helium balloon inside it. If you paint the upper envelope such that one side is black and the other silver. By turning the balloon with the black side facing the sun, the hot air will expand - causing the balloon to rise. Rotate the balloon with the shiney side towards the sun and the hot air can cool. With the majority of the lift coming from the helium, you only need the hot air to provide altitude control. SteveBaker (talk) 16:17, 21 January 2013 (UTC)[reply]

Put thin film solar panels on the blimp and and make the battery life infinite. Have they ever tried acid and metal filings-based buoyancy storage? Vent the hydrogen to go down. Sagittarian Milky Way (talk) 18:16, 21 January 2013 (UTC)[reply]