Wikipedia:Reference desk/Archives/Science/2006 October 12

Humanities Science Mathematics Computing/IT Language Miscellaneous Archives

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sodium hydroxide

How is sodium hydroxide manufactured?

Sodium hydroxide#Manufacture. Hyenaste ^(tell) 01:06, 12 October 2006 (UTC)

Have you looked at our article on sodium hydroxide? (Use the 'search' box in the navigation box on the left side of your screen.) TenOfAllTrades(talk) 01:07, 12 October 2006 (UTC)

Mould

Hi! I'm having trouble finding explanation for the word "Saprophyte" (i know its something about mould, but what!?)74.12.96.221 01:00, 12 October 2006 (UTC)

Have you looked at our article on saprophytes? (Use the 'search' box in the navigation box on the left side of your screen.) TenOfAllTrades(talk) 01:07, 12 October 2006 (UTC)

I did but still it didnt give the clear explanation! =( 74.12.96.221 01:14, 12 October 2006 (UTC)

Did you really? The answer is quite clear in the first sentence. Here is a computer generated list of bite-size definitions of "saphrophyte". At least one should make sense. Hyenaste ^(tell) 01:22, 12 October 2006 (UTC)

Thank you very much! The first line gave me the clearest explanation! TY again! =D 74.12.96.221 01:25, 12 October 2006 (UTC)

I like the third one the best, but no problem; glad to help! Hyenaste ^(tell) 01:27, 12 October 2006 (UTC)

I liked the Wikipedia article the best. This person obviously did not try very hard to understand that fist sentance. It looked too hard? :) — X [Mac Davis] (SUPERDESK|Help me improve)04:10, 15 October 2006 (UTC)

How many "frames" per second can our eye process?

How many times per second does the human eye snap (for lack of a better word) what it sees and process it? In other words -- what is the FPS of the human eye? Pesapluvo 02:36, 12 October 2006 (UTC)

It is different per human, but it is below 60FPS. Why? If you could see beyond 60FPS, you'd see the film in a regular 35mm projector flicker. I believe it is also below 50FPS because you don't see flicker on PAL unless it is interlaced. Of course, the interlacing is on every other frame - creating a 25FPS flicker. That 25FPS is very easy for anyone to see. Therefore, I believe it is just below 50FPS. This does bring up an experiment I've wanted to do but haven't had the time. Ever notice spokes on a tire stop and start moving backwards at high speed? Does everyone see the spokes freeze at the same RPM? I doubt it. I'd like to experiment with the range of speeds that the tire spins to get people to see it as standing still. Then, does the change in speed correlate to anything: age, gender, race, IQ... --Kainaw ^(talk) 02:41, 12 October 2006 (UTC)

For the original question, see our article Flicker fusion threshold. For Kainaw's unperformed experiment, see Wagon-wheel effect.  --Lambiam^Talk 04:34, 12 October 2006 (UTC)

Some relevant articles are Flicker fusion threshold, Frame rate, and Persistence of vision. In general, the ability to perceive flicker dictates a higher frame rate than the need to have a difference between the frames for motion to seem continuous. Silent movies at 16 frames per second seemed like continuous movement, as did sound movies at 24 fps or TV in the USA at 30 fps, with a scan 60 times per second, interlaced. Some projectors flashed each frame more than once to avoid flicker. One article relevant to this is Eadweard Muybridge. Persistence of vision has an animation at 12 fps which looks reasonable continuous. Cheaply done cartoons may use 8 frames per second or less, but they are repeated for 3- or 24 frames per second to avoid flicker. This also relates to Apparent motion and Phi phenomenon. Some electric utilities in bygone years provided 25 cycle alternating current, which caused annoying flicker for some customers. Most of the world provides 50 Hertz ac, and the US provides 60 cycle ac, which generally do not cause noticeable flicker in electric lights.Edison 04:48, 12 October 2006 (UTC)

A minor nit: 50 Hz power delivers 100 half-sine-waves of power each second, so (most) lights on 50 Hz power flicker at 100 Hz. Likewise, 60 Hz power -> 120 Hz flicker.

True for an incandescent bulb, but some types of light flash only on one polarity, like neon. Fluorescent bulbs are supposed to flash on each half cycle, but may rectify at the end of the bulb when near the end of its life cycle.Edison 18:24, 12 October 2006 (UTC)

Also, be sure to check out the "talk" pages on either Flicker fusion threshold, Frame rate, or both. Last I knew, there's more detail on the talk pages that hasn't yet been incorporated into the articles themselves.

Atlant 17:33, 12 October 2006 (UTC)

See this archived RD question, wherein the same articles as linked here are used to explain this question (to me) pretty directly. --Tardis 17:16, 12 October 2006 (UTC)

Radiocarbon dating

In Wikipedia's radiocarbon dating article, as well as other sources ([1] and [2] for instance), the half life of C14 is given as 5730 \pm 40. Does the plus/minus 40 refer to inaccuracies in estimation of the half life (Someone else suggested it may be the variance or standard dev. of a decay waiting time)? I assume there is a better estimate for it today; is there a reference to a recent estimation which drops this error measurement? Thanks, --TeaDrinker 03:35, 12 October 2006 (UTC)

The National Nuclear Data Center says 5700 years plus or minus 30 years. That's the uncertainty in the measured value. All radioactive decays follow the same law, described by only one parameter; "the variance or standard dev. of a decay waiting time" sounds like nonsense to me. —Keenan Pepper 04:41, 12 October 2006 (UTC)

To explain why the uncertainty seems so bad: How do you do an experiment to measure the half life of ¹⁴C? You get a very pure sample of ¹⁴ (any ¹²C will contribute to the mass, but not beta decay, which throws off the measurement), completely cover it with semiconductor detectors (any solid angle not covered allows beta particles to escape uncounted), and then let it sit for a long, long time. There are lots of things that can introduce uncertainty. Also, beta decay is unpredictable because two particles escape: the beta particle (an electron) and a neutrino, which can carry away any fraction of the energy, from almost none, up to the whole amount less the rest energy of the electron (511 keV). If the neutrino carries away too much energy, the electron doesn't have enough to be detected, which adds more uncertainty. Taking all that into consideration, 5700 plus or minus 30 is quite a good value, the result of many careful experiments because ¹⁴C is such an important nuclide. For comparison, the half life of ⁹³Mo is only known to be 4000 plus or minus 800 years. —Keenan Pepper 05:06, 12 October 2006 (UTC)

Thanks for the replies! --TeaDrinker 19:29, 12 October 2006 (UTC)

Critical molar volume of a van der Waals gas

According to Van der Waals equation#Reduced form, the critical molar volume of a van der Waals gas is exactly three times the parameter b which describes the volume of the particles themselves. The linear relationship is intuitively clear to me, but not the specific factor 3. What is special about a configuration in which the particles take up 1/3 of the available space? —Keenan Pepper 05:29, 12 October 2006 (UTC)

Solve: ${\partial ^{2}P \over \partial v^{2}}={\partial P \over \partial v}=0$ for a constant T. That's the critical state, and the critical v is 3*b. --BluePlatypus 05:51, 12 October 2006 (UTC)

I was hoping for something more intuitive... —Keenan Pepper 05:13, 13 October 2006 (UTC)

PROPOSED CHANGES TO THE REFERENCE DESK

If you haven't been paying attention to Wikipedia talk:Reference desk, you may not know that a few users are close to finishing a proposal (with a bot, now in testing and very close to completion) which, if approved by consensus, will be a major change for the Reference Desk.

Please read the preamble here, and I would appreciate if you signed your name after the preamble outlining how you feel about what we are thinking.

This notice has been temporarily announced on all of the current desks.  freshofftheufoΓΛĿЌ  06:57, 12 October 2006 (UTC)

For convenience, I propose any reactions to this anouncement be limited to Wikipedia:Reference_desk/Miscellaneous#PROPOSED_CHANGES_TO_THE_REFERENCE_DESK. DirkvdM 08:00, 12 October 2006 (UTC)

Equations for black body radiation

In the article Planck's law of black body radiation it is stated that:

$u(\nu ,T)={\frac {8\pi h\nu ^{3}}{c^{3}}}~{\frac {1}{e^{\frac {h\nu }{kT}}-1}}$

and then that it can be expressed as a function of wavelength with:

$u(\lambda ,T)={8\pi hc \over \lambda ^{5}}{1 \over e^{hc/\lambda kT}-1}$

Now in the second part, $\nu$ has been replaced by $c \over \lambda$, which I can understand fine, but $\nu ^{3} \over c^{3}$ has been replaced by $c \over \lambda ^{5}$.

That's what I don't get. Surely $\nu ^{3}={c^{3} \over \lambda ^{3}}$ but that leaves the equation short by a factor of $c \over \lambda ^{2}$.

What am I missing here? BigBlueFish 09:07, 12 October 2006 (UTC)

They aren't really the same u and technically its an abuse of notation to reuse it. The first, $u_{\nu }(\nu ,T)$, is the spectral density per unit $\nu$, while the second, $u_{\lambda }(\lambda ,T)$, is the spectral density per unit $\lambda$. It follows therefore that they are related by a scale factor which is ${d\nu } \over {d\lambda }$, which is where the $c \over \lambda ^{2}$ comes from. Dragons flight 09:23, 12 October 2006 (UTC)

Ah, I think I see, it comes down to my misunderstanding of the actual concept of spectral intensity. What I'm actually trying to achieve is the amount of radiation emitted at a specific wavelength, or rather, it seems, within a narrow spectral band. To do this do I need to integrate the spectral intensity? And is it possible, or reliant on an approximation? BigBlueFish 12:01, 12 October 2006 (UTC)

The intensity (energy/time/area) per solid angle in a given wavelength range $[\lambda _{1},\lambda _{2}]\,$ is just $\int _{\lambda _{1}}^{\lambda _{2}}u_{\lambda }(\lambda ,T)\,d\lambda$; however, I know of no closed-form for this. If the interval is $[\;\!0,\infty )$, of course, we have the standard results given in the article, but you said "narrow". If it's narrow enough (such that $(\lambda _{1}/\lambda _{2})^{5}\approx 1$ and ${\frac {1}{\lambda _{1}}}-{\frac {1}{\lambda _{2}}}\ll {kT \over hc}$), you can profitably approximate the integral as $u_{\lambda }\left({\frac {1}{2}}(\lambda _{1}+\lambda _{2}),T\right)(\lambda _{2}-\lambda _{1})$, but that's just an extremely simple numerical integration, which is what you want in general. Many mathematics programs (including graphing calculators) can do this automatically. --Tardis 22:55, 12 October 2006 (UTC)

magnets

When you stroke an iron nail using the north-seeking pole, which ends of the iron nail will be the north-seeking pole?

(from > then stroke.)

Sign,Chan Hor Onn

A little unsure of the exact question. When you say 'north-seeking pole' I think you are talking about the north pole of the magnet, but not sure (the south pole will seek the north pole of another magnet and thus be north-seeking, but in the Earth's magnetic pole the north pole of a magnet seeks magnetic north, which I guess is what you mean; I think this may be an old term).

So, assuming you are talking about stroking with the north pole of a magnet, it should cause the domains in the nail to align in the direction of the stroke. In other words, if you stroke towards the point of the nail, the point should become the north pole. --jjron 13:23, 12 October 2006 (UTC)

'North-seeking pole' is a phrase sometimes used in place of the more familiar 'North pole', because that is the pole which will be attracted towards magnetic North. See Magnet#North-south pole designation and the Earth's magnetic field, which explains this, but doesn't actually introduce this term. --ColinFine 23:09, 12 October 2006 (UTC)

The Earth's North pole, is, of course, a south magnetic pole (or at least close to one). That is why the north pole of a magnet points toward it.Edison 13:31, 13 October 2006 (UTC)

LAser Eye Surgery

How can the laser used in Eye surgery be selectively destructive?(destroy unwanted eye tissue,neglect rest of tissue)???

I'm not exactly sure what you mean but have you read the LASIK article? Nil Einne 11:43, 12 October 2006 (UTC)

Think of a lens, which focuses light in one point. In order to burn a piece of paper with a lens and the Sun you have to hold it exacly in the focus area. Laser light has the nice capacity of not scattering (very much). A narrow beam will remain narrow. Let several harmless beams converge on one point and their combined power will have the same effect as on the aforementioned piece of paper. I don't actually know if laser surgery uses this, but I'm pretty sure it does.. DirkvdM 09:32, 13 October 2006 (UTC)

I'm searching for an inexpensive portable medical X-ray device, or are there any build-your-own kits?

Thursday, 10-12-06; Portland, OR; 2:27am West Coast Pacific Time

Is there an authority in the reference desk who can point me in the direction of an inexpensive portable x-ray device for personal medical use; or would it be more cost effective for me to build my own X-ray device from scratch, or from a mail-order kit - in which I could include radiation-protection safety features in its design?. If I had my own personal X-ray device, then I could save myself the travel time and waiting time for a physician's appointment. I wish to operate my personal medical X-ray device to take images, myself, of the physical condition of any of my own internal anatomy (specifically my bones, vertebrae, ligaments, and spinal discs.).

The webpages below describe various forms of X-ray devices - 1) This webpage describes Flouroscopy -

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

2)This webpage describes X-ray devices used for airport security -

  http://en.wikipedia.org/wiki/X-ray_machine#Security

3) This webpage lists various forms of X-ray methodologies -

  http://en.wikipedia.org/wiki/X-rays#See_also

My reason for listing the above specific webpages is to show the type of X-ray device that I'm searching for: real-time imaging.

--MyPresentCPUisTooSlow 10:36, 12 October 2006 (UTC)MyPresentCPUisTooSlow

You realize this sounds a little crazy right? Equipment for realtime imaging is likely to be prohibitively expensive. An xray machine of the kind that produces still images is probably less so, but in most jurisdictions radiation emitting machines and/or their operators must be licensed and inspected by the state. I dare say the hassles involved in acquiring, permitting and operating such an instrument will likely offset any hassle that going to a doctor's office brings. Dragons flight 10:55, 12 October 2006 (UTC)

Of course, you could operate one illegally but you'd probably end up killing yourself. And even if you didn't you'd just be wasting your money and time since you wouldn't actually learn anything from the images Nil Einne 11:41, 12 October 2006 (UTC)

I again, advise strongly against it. Ionizing radiation hazards, high voltages and currents, lack of medical knowledge. Cost-benefit ratio even if none of those apply is still pretty bad. — X [Mac Davis] (SUPERDESK|Help me improve)14:06, 12 October 2006 (UTC)

Haha, yes, this sounds crazy, but I think you're cool! Are you by any chance going on an expedition to the south pole? ;-) I mean, it could be a bit tricky to get to a hospital from there, especially if you're isolated over the winter. I hear some unorthodox procedures have taken place there. —Bromskloss 14:12, 12 October 2006 (UTC)

After you get the self x-ray working, then it's time for the self-surgery. You could start with removing your appendix, it's no good for anything anyway.. --Zeizmic 14:31, 12 October 2006 (UTC)

In any English-speaking jurisdiction I can think of, it will cost more to get the necessary permits and approvals – devices which generate ionizing radiation, particularly for medical purposes, are highly regulated – than you would save on paying for x-rays at a clinic. How often do you actually need an x-ray? I don't recommend doing them for fun.

Frankly, if you don't know enough about x-ray technology to know that building a home medical x-ray device is a really bad idea, then you don't know enough about x-ray technology to be able to build a home medical x-ray device. TenOfAllTrades(talk) 14:51, 12 October 2006 (UTC)

X-rays should only be taken rarely, like when you break your leg, and are relatively inexpensive. If you have so many X-rays taken that it would actually be cost effective to have your own X-ray machine, then you are having way too many taken. When you have that many taken, the risk of giving yourself cancer from all the radiation far outweighs the any benefit. Can you explain why you have so many X-rays taken ? StuRat 14:53, 12 October 2006 (UTC)

One clarification, the shielding on an X-ray machine is designed to shield the OPERATOR from X-ray radiation. There is no way to shield the PATIENT from X-rays and still get an image. This is why it's dangerous to have an excessive number of X-rays taken. StuRat 15:00, 12 October 2006 (UTC)

Anon, I have a used XRay machine I could part with in exchange for appropriate compensation. The radiation shield is broken but otherwise works. Sold as is, no refunds. Antonrojo 16:40, 12 October 2006 (UTC)

Coming soon: the sequel to the David Hahn article!

Atlant 17:39, 12 October 2006 (UTC)

Per the x-ray article, Thomas Edison "dropped X-ray research around 1903 after the death of Clarence Madison Dally, one of his glassblowers. Dally had a habit of testing X-ray tubes on his hands, and acquired a cancer in them so tenacious that both arms were amputated in a futile attempt to save his life." This might be a drawback to hobbyist home x-raying. Many of my generation remember those cool fluoroscope machines shoe stores had in the 1950's. http://www.orau.org/ptp/collection/shoefittingfluor/shoe.htm You could see if your shoes fit by looking down at the x-ray image of the shoes and your toes. It was tough when the government announced they caused cancer and yanked them. Edison 18:33, 12 October 2006 (UTC)

The weird thing is, why did it take them 50 years after Edison discovered that X-rays cause cancer for the government and shoe stores to figure it out ? StuRat 04:15, 13 October 2006 (UTC)

The shoe industry as well as the manufacturers made money with the machines. The theory was that the machine would be used only a few seconds at a time, a few times a year. In practice, I would use it every time I went to the store, because it was cool to see the bones in your feet. In the beginning, it was a way to make a few bucks with World War I surplus X-ray machines. It took until the late 1950's to finally get rid of them. Edison 04:22, 13 October 2006 (UTC)

The sale and operation of X-ray devices is tightly regulated by state law. In short, it would be impossible for you to own or operate an X-ray machine legally, unless you happen to be a doctor or a physicist. If you're curious, you can read Oregon's applicable laws here. —Brim 17:11, 14 October 2006 (UTC)

Try a friendly dentist who is upgrading his X ray machine, or try to find one of those things they used to use in shoe shops to see if your feet were still there when you put the shoes on. 8-) --Light current 18:17, 14 October 2006 (UTC)

Negative refractive index

I have heard that scientists have built substances that have negative refractive co-efficient.So why cant we build fibre optic cables out of them (total internal refraction from rarer medium instead of usual denser medium)?

From a quick look at Refractive index which links to Metamaterial, I would say it's probably because they're still largely theoretical with some limited demonstrations. They do appear to be of great interest for a variety of reasons and may be used for fibre optic cables in the future I guess if we can master them and produce them cheaply enough to be worth it but until then... Nil Einne 11:35, 12 October 2006 (UTC)

Actually I just thought of something. Will it actually work? Won't the light just never come out of your metamaterial at the other end? Nil Einne 11:39, 12 October 2006 (UTC) Ignore this, I wasn't thinking properly Nil Einne 11:20, 13 October 2006 (UTC)

It has been experimentally worked with before. Links: [3][4][5][6][7][8][9][10] — X [Mac Davis] (SUPERDESK|Help me improve)14:03, 12 October 2006 (UTC)

Loop DNA

What are the functions of loop DNA?

Er what exactly do you mean loop DNA? Nil Einne 11:32, 12 October 2006 (UTC)

Are you maybe talking about DNA hairpin loops? Simon A. 12:15, 12 October 2006 (UTC)

Or maybe plasmids? Laïka 13:04, 12 October 2006 (UTC)

Or is it do your own homework? -- Plutor ^talk 23:24, 12 October 2006 (UTC)

Tell that to my friend, she said to me ask this question to Wikipedia because I don't know how to do it, so spare me the lecture

Plutor m:don't be a dick, what gave you the inclination that this was homework. Seems like a perfectly honest question to me. Philc _T^E_C^I 17:59, 13 October 2006 (UTC)

Types of rock or stone?

What are the most common types of stone used for altars or megaliths in Europe?--Sonjaaa 14:09, 12 October 2006 (UTC)

There is such a wide variety of quarry stone all through Europe, that I doubt we can come up with anything. With all stonework, builders usually take what is close, because of transportation difficulties. That said, you might find Italian marble as a slight favourite, at least for thin facing. --Zeizmic 14:35, 12 October 2006 (UTC)

I'm not sure about "most common", but also see Sarsen. Avebury and Stonehenge are built (primarily) of this calcified silicified sandstone.

Atlant 17:45, 12 October 2006 (UTC)

I'm sure Atlant meant silicified!

It is a dense, hard rock created from sand bound by a silica cement, making it a kind of silicified sandstone.

But the answer has to be it will be the best local stone available because of transportation difficulties.--Light current 10:44, 16 October 2006 (UTC)

My short term memory apparently went "boink!" between the moments of reading the Sarsen article and typing this answer; thanks for the correction!

Atlant 14:43, 16 October 2006 (UTC)

Earth's moon: why it always shows the same face

How is it that our moon always shows the same face. How does the earth cause the rotation of the moon to stabilise in this way. Not too much mathematics, please!

Thank you

Tim

See Synchronous rotation, and the last sentence before see also will link you to the cause. Absolutely zero mathematics! Hyenaste ^(tell) 19:04, 12 October 2006 (UTC)

Or, you can just go directly to the tidal locking article. -- Plutor ^talk 23:16, 12 October 2006 (UTC)

Note that it doesn't always show exactly the same face. It wobbles somewhat and over a large enough period of time as much as 95% (I believe it was) of its surface can be seen from Earth. Thanks to this, sir Patrick Moore the presenter of The Sky at Night managed to see the mare orientale before the USSR could photograph it. DirkvdM 09:38, 13 October 2006 (UTC)

I believe the actual figure is 59%, due to Libration. --Jmeden2000 15:42, 13 October 2006 (UTC)

Oops, now there's a silly mistake. Right figures, wrong order. I already thought 95% was a bit too much. But I certainly thought it would be more than 59%, so maybe that caused the mistake. DirkvdM 09:05, 14 October 2006 (UTC)

Cooling Towers

Could someone explain why it is necessary to condense the steam at a power station, before heating it back up to steam for turning of the turbine? Why not just direct the steam straight back into the chamber in which it was heated in the first place and use less fuel? --Username132 (talk) 19:14, 12 October 2006 (UTC)

The phase transition is important. The gas takes up more space. The force of the expanding gas is what drives the turbine, see Steam engine. --JWSchmidt 19:37, 12 October 2006 (UTC)

The cooling isn't strictly necessary. You could vent the steam directly into the atmosphere, and this is the emergency option when the turbines suddenly have to be shut down. However, this is valuable demineralized water that you are throwing away and difficult to replace after a while. The concept of a condenser recycles this water and the excess heat is carried off with either hot air or hot lake/river water. --Zeizmic 20:15, 12 October 2006 (UTC)

Most of the heat that is absorbed by the water/steam goes into boiling the water. Raising the temperature of existing steam wouldn't absorb nearly as much heat energy, so much more total water would be needed. Being in steam form, this would take up a huge volume, versus the rather small volume of water needed in the current system. StuRat 23:55, 12 October 2006 (UTC)

Lower discharge pressure from a turbine or other steam engine means greater delta P which means more work per stroke. Work = force times distance, so zero pressure means greater work per stroke of a piston at a given boiler pressure than atmospheric would give. Edison 04:24, 13 October 2006 (UTC)

I assume the question is why waste the energy stored in the steam by condensing it. The answer is given by JWSchmidt, though maybe not explicitly enough. In order to get the expansion again you need to 're-shrink' the water first by letting it condense. DirkvdM 09:57, 13 October 2006 (UTC)

Thanks, I think I understand now. If you didn't have the condenser, then you wouldn't have a region of lower pressure for the steam to move to and therby drive the turbine. I just don't like that 60% or more fuel is wasted in this way. I just checked out the article on Combined heat and power and found this boiler thing which generates electricity as it boils your water. But it only runs on gas and not coal or nuclear fuel. --Username132 (talk) 14:44, 13 October 2006 (UTC)

And most of our energy production still uses this ancient technique. Even nuclear plants do. Finding a more efficient way to transfer heat to electricity (or maybe hydrogen or other energy carriers) would be a major step forward for mankind. Is there maybe some physical reason one can't achieve a high efficiency? Are maybe the forms of energy too different? DirkvdM 09:09, 14 October 2006 (UTC)

In practice, the main alternative for turning heat into electricity - the thermocouple - is way, way, way less efficient. They use them in radioisotope thermal generators but there are proposals to replace them there with Sterling engines because they are just so wasteful.

Also, Rankine cycle steam turbines have gotten a lot more efficient over the past few decades. You used to get about 33-34% thermal efficiency with conventional turbines. These days, the very best supercritical turbines in the latest coal-fired power stations get about 50% efficiency. But even that's not the best you can do. If you gasify the coal and run it through a combined cycle gas turbine, you can now get over 60% thermal efficiency. Even fuel cells find it hard to beat that number in practice. --Robert Merkel 00:30, 15 October 2006 (UTC)

That's still not quite impressive. I suppose teh reaason fuel cells are also that inefficient in practise is that only a fraction of the research effort has gone into developing them, partly because they are a much more recent invention. The article doesn't say what efficiency they might theoretically reach, but if they already compete with the traditional system, there is good potential. DirkvdM 08:13, 15 October 2006 (UTC)

Ballistics

It requires some energy to discharge a spent casing from a semi-automatic or automatic gun/rifle. This energy HAS TO come from the hot gases that propel the bullet. Can anyone tell me about how much velocity is "lost", for a given bullet, that would otherwise NOT be lost if same bullet was fired from a single-shot rifle? Loss might be slight, but there certainly HAS to be SOME loss. I thank you ahead of time if you can help. 205.188.116.74 19:35, 12 October 2006 (UTC)

It depends entirely on the type of round and the gun it is fired from. For example, an M16 has a spring buffer that is pushed back by a gas tube. Gas fills that tube as the bullet leaves the barrel - so the bullet is beyond being affected by the gun. The spring buffer retracts after the gas tube fills, popping out the expended round, and loading the next round. --Kainaw ^(talk) 19:46, 12 October 2006 (UTC)

Even with an old Lee-Enfield single shot, a lot of energy goes into mashing your shoulder. --Zeizmic 21:01, 12 October 2006 (UTC)

I believe the gases are in the chamber as the bullet is in the barrel. so you do get some loss of gas that would be otherwise be pushing the bullet, but only a small portion is used this way. weapons like the Steyr AUG (i imagine its pretty standard on other weapons too) have an adjustable gas port so you can use more gas if the mechanism is gunked up. Xcomradex 21:12, 12 October 2006 (UTC)

Note that the revolutionary thing about the AK47 was that it used the power of the fired bullet to drive a very simple mechanism to not only discharge the spent casing but also load the next bullet. I think. Heard somthing like that once, but I don't know much about guns, so some details may be wrong. DirkvdM 10:02, 13 October 2006 (UTC)

Well that's what gas-operated reloading is. Don't think the Kalishnikov was a pioneer though. — X [Mac Davis] (SUPERDESK|Help me improve)04:16, 15 October 2006 (UTC)

It wasn't? That's new. DirkvdM 08:18, 15 October 2006 (UTC)

Definately not new in the Ak-47. The german WWII era StG-44 has it for example, as did the M1 Garand. Xcomradex 10:21, 15 October 2006 (UTC)

Ah, that aspect of it. I thought he meant the AK47 as a whole. That would be a bit of an odd thing to say about a mechanical military weapon that is still extremely popular after 60 years. DirkvdM 06:53, 16 October 2006 (UTC)

Ants

I just ordered one of those NASA-gel Antworks ant farms , and have a few questions about the ants themselves. I think it would be very nice if the colony of ants were able to reproduce so that I could follow their progress over the years. However, I'm not sure if that is possible with the kind of ants one receives from the order form (p 18-19), which is for 25 all-female ants.

Will this group of ants be able to reproduce, and, if not, where could I order ants that could? I would assume both a queen and a male are necessary, and perhaps a bigger environment than a regular ant farm could provide. --JianLi 05:00, 13 October 2006 (UTC)

No, they won't be able to reproduce. Considering how easy it is to collect ants, do you really need to breed them in captivity ? StuRat 20:11, 14 October 2006 (UTC)

Well, I dislike having to collect ants not because it would be hard (though, since winter is coming, it actually might be), but because I feel that this provides an incomplete picture of their world. Apart from studying ant reproduction, it would be nice to have a feeling of continuity; in a few years, I could the satisfaction of having a group of ants whose ancestors I knew, or perhaps I could pass my ants onto my grandchildren, for example.

Unfortunately, it seems, it is illegal to mail live queen ants in the US. [11] [12] --JianLi 22:46, 14 October 2006 (UTC)

Wait until summer, then excavate an ant hill and take the queen along with all the rest. StuRat 21:01, 15 October 2006 (UTC)

Goldfish

On a related note, when I had goldfish, I was not able to get them to reproduce (though I didn't really try: there were only two of them, and I didn't even ascertain if they were of opposite gender). Are there any specific conditions one has to maintain for goldfish reproduction? I merely keep them in a bowl without any fancy electric filtering, etc, and I empty the water regularly. Wouldn't the eggs be thrown out with the water if I did this?

Thanks, --JianLi 21:58, 12 October 2006 (UTC)

Asking why they don't reproduce is a bit like asking why you and some random person of the opposite sex don't reproduce when left alone in a room. In short, most species use some type of mate selection process and the chances of any two random individuals meeting each other's criteria is slim. Even then, a very specific environment is needed for them to reproduce. It's really amazing that any animals reproduce in captivity. StuRat 23:45, 12 October 2006 (UTC)

Goldfish eggs (well, at least "Comet" eggs) are small spheres about 1mm in diameter. They are adhered (by the fish) to various suitable surfaces in the habitat. We have a school of comets that move to a small outdoor pond during the summer. Last year, our comets were "frisky" while outdoors and we used to find a lot of eggs adhered onto the filter/pump and occasionally on the water plants. We carefully separated any eggs we found and put the surviving eggs into a small fishbowl. About 1/3 of the collected eggs hatched out and we eventually raised about a half-dozen new fish "from egg". (A few fish were lost to cannibalism!) The hatchlings are very interesting, looking like nothing more than tiny sticks with eyes. (We used a jeweler's loupe to observe them, both in egg and afterwards.) Later, upon moving the rest of the adult school back indoors, we found two more fry that had hatched and survived outdoors without our "help".

The fish seemed to be friskier when the pond was relatively green with algae. Better privacy? Concern they were going to die? Who knows! But this year, for whatever reason, we don't think the comets were breeding. We never saw any eggs, and no particularly small fish came back in this fall. Or maybe the neighborhood cat was umm, err...

Atlant 01:16, 13 October 2006 (UTC)

I suspect the problem was your goldfish weren't homosexual so they didn't like each other Nil Einne 11:16, 13 October 2006 (UTC)

That's a nice fish story :). I guess I should at least get more fish and a bigger bowl to increase the chances. --JianLi 22:51, 14 October 2006 (UTC)

You may want to check on fish-breeding tips concerning your species.[13] I think I read somewhere on here about some species rarely mating when in captivity. Maybe someplace else, maybe not true. — X [Mac Davis] (SUPERDESK|Help me improve)04:20, 15 October 2006 (UTC)

brain

why do you want to use the right side of your brain more in art class than your left side ??

One side is artistic, and the other side is logical. StuRat 23:50, 12 October 2006 (UTC)

But, of course, in some people, neither side is logical 8-)--Light current 00:42, 13 October 2006 (UTC)

The known lateralizations are trends and do not apply to every person in every case. --JWSchmidt 02:29, 13 October 2006 (UTC)

And, you knwo, so far I have never actively taken care to switch on the correct side of my brain when trying to do something. Simon A. 09:08, 13 October 2006 (UTC)

I wouldn't know how to control that either. My brain probably does it all by it self. It seems to have a mind of its own. DirkvdM 10:06, 13 October 2006 (UTC)

Both sides of your brain refuse to work equally. :-) StuRat 22:11, 13 October 2006 (UTC)

Of course they do. What, did you think I had a communist brain? DirkvdM 09:13, 14 October 2006 (UTC)