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November 8

cells

what would happen if you place a blood cell in a hypertonic solution —Preceding unsigned comment added by Dmx123 (talk • contribs) 00:37, 8 November 2007 (UTC)

The reference desk it not for answering homework questions, but if you can't figure this one out on your own, you should read osmotic pressure. Someguy1221 00:49, 8 November 2007 (UTC)

... or Osmosis or Diffusion. --slakr^\ talk / 04:35, 8 November 2007 (UTC)

... or better yet, hypertonic. (The article answers this very question.) -- 20:21, 8 November 2007 (UTC) —Preceding unsigned comment added by 128.104.112.105 (talk)

navel

hey how come for some people when they're fingered in their belly button, it hurts and for others it tickled them?Jwking 01:00, 8 November 2007 (UTC)

Because some people don't know how hard to poke, and just stab you with their finger. HYENASTE 01:17, 8 November 2007 (UTC)

Reception of off air frequency standards

May be a stupid question, but why, when receiving, do you need a local oscillator to phase lock to the incoming signal? Only reason I can think is that the transmitted signal is not constant in amplitude. Why cant you use the incoming frequency dorectly? Also, why do you need a quartz Xtal osc to be locked to the incoming frequency, won't a normal vco do? —Preceding unsigned comment added by 79.76.246.62 (talk) 01:42, 8 November 2007 (UTC)

It's for FM reception. The radio signal is varying slightly up or down in frequency depending on the amplitude of the sound wave it's trying to transmit to you (Frequency Modulation). You tune the radio's local oscillator to the nominal center frequency and it's easy to produce an audio signal that's proportional to the difference in frequency of the local oscillator and the radio signal. SteveBaker 01:59, 8 November 2007 (UTC)

Er, no its AM. [1] —Preceding unsigned comment added by 79.76.246.62 (talk) 02:20, 8 November 2007 (UTC)

You need a very low bandwidth. Typically the signal will be 5kHz wide, at a frequency of 10MHz. A VCO is no where near stable for this and will drift off in a few minutes. You can see this on old cheap shortwave radios, which will need retuning every so often. The crystal oscillator is much more stable. A VCO locked to a crystal is one way to get flexibility. Another important thing for a frequency standard is low phase noise. The best way would be to have a narrow crystal filter at 10 MHz, but even so the ionosphere causes fading and phase shifts. For 60kHz standard a LO would not be needed. Graeme Bartlett 05:57, 8 November 2007 (UTC)

So you can use the ultra stable 60 kHz freq directly (or multiplied up to 1 MHz or 10 MHz or whatever)? Is that what your saying? If so, why do most designs use a local oscillator locked to the incoming frequency? —Preceding unsigned comment added by 79.76.246.62 (talk) 12:06, 8 November 2007 (UTC)

See Superheterodyne receiver. Changing the frequency of the local oscillator is what tunes the radio to a station. Its frequency is beat against everything coming in from the antenna. The resulting harmonics are filtered for the intermediate frequency, 455kHz in the case of AM. A big advantage to this system is that from there on the amplifiers need only pass the one relatively low frequency. I don't know what you mean by a crystal oscillator locked to the incoming frequency, but it has been a long time, so please clarify. --Milkbreath 03:33, 8 November 2007 (UTC)

Crystal control is absolutely not needed in an AM radio. I have owned several old AM radios which would stay tuned to a station for a year or more without re-tuning. Older car radios had pushbuttons which mechanically rotated a tuning cap to the desired stations, and did not need re-tuning for years at a time. Edison 13:10, 8 November 2007 (UTC)

For a frequency standard, a Local Oscilator is not stable enough. You would need to down convert to the intermediate frequency, filter, and then up convert to the original stable frequency to get the reference. On HF frequencies around 10 MHz drift is ten times bigger than it is on the AM band around 1 MHz that you get on an old car radio. If you just want to listen to the time pips all this extra stability is not needed, you just need to keep the radio tuned to the station. Graeme Bartlett 20:38, 8 November 2007 (UTC)

I think the answer is that the off air frequency references have excellent long term stability, but are subject to the vagaries of radio reception such as : noise, timing uncertainty due to reflections from the ionosphere etc, and unwanted amplitude modulations, breaks in reception etc. OTOH, Local crystal oscillators can be made to have excellent short term stability and low phase noise etc but are subject to long term drift from component aging. Put the two together and get the best of both worlds!

Minerals

Is it possible to make a mineral (in this case a molecule containing iron) magnetic by running an electric current through it, or around it? I have reference here to the specific mineral asbestos, or one of its several 'subspecies'?76.182.3.188 01:56, 8 November 2007 (UTC)

Very few minerals can be magnetized in a way such that they remain magnetic after the inducing field is removed. See Magnetization. A good many minerals, including some without iron, such as salt, can be made to give a diamagnetic response while the inducing field is present, but I do not think asbestos is one of them (but I'm not certain about that). Cheers Geologyguy 03:03, 8 November 2007 (UTC)

Most minerals do not conduct, so you cannot easily run a current through them. In the case of asbestos it is a good insulator so it will not conduct at any reasonable voltage. If you ran a current around it you would have an electromagnet. The atoms in the mineral would respond in some way, but most have no strong response. A few Iron minerals may respond with ferromagnetism and even be magnetised as in Magnetite Graeme Bartlett 05:52, 8 November 2007 (UTC)

See Lodestone, a mineral (Fe3O4) some samples of which are found in the ground as natural magnets. The Wikipedia article does not say it, but other sources say the magnetism may result from lightning striking the mineral [2] [3] [4]. Edison 13:07, 8 November 2007 (UTC)

microwave hyperthermia

any reason not to heat someone up (their whole body to 104 F - 107 F) with a very high tech microwave (in order to generate a healing fever)? It penetrates and is the same frequency as cell phones... —Preceding unsigned comment added by 76.168.69.208 (talk) 02:02, 8 November 2007 (UTC)

Microwaves do not heat everything uniformly, and in particular they will heat organs with a high water content more rapidly. Of special concern are the eyes where high levels of microwaves can promote cataracts and other damage.

On the more general point, I'm not sure that heat is necessarily helpful in fighting disease. Fever is one of a myriad of reactions the body produces to combat disease, but an externally produced hyperthermia might be more detrimental than helpful since it wouldn't necessarily be accompanied by the same suite of immune responses as a natural fever. Dragons flight 02:25, 8 November 2007 (UTC)

Also note: Wikipedia does not give medical advice. Anyway, the Hyperthermia article states that temperatures above 104 degrees Fahrenheit are "life threatening". I wouldn't reccomend it either, unless you know what you are doing, and I would get a second opinion (from a qualified professional) either way. I also have never heard of inducing a fever to heal by this method (or any), as fevers are caused naturally by the body as part of the immune response. SmileToday☺(talk to me , My edits) 02:28, 8 November 2007 (UTC)

He might have gotten this from last week's episode of House, in which a portion of a treatment for an individual was artificially raising his body temperature. Someguy1221 02:39, 8 November 2007 (UTC)

Microwave heating of smaller regions of the body to fever-range temperatures has been tested for various therapeutic purposes. If you're envisioning putting the entire body in a microwave oven (even a high-tech one) to heat the entire patient at once, you're out of luck. Per Dragons flight, you would get dangerous local heating effects that are very difficult to control. There are other, lower-tech methods that are just as effective. Where I have seen microwave heating employed is to do rapid, local thermal ablation of smaller volumes—a microwave antenna is inserted into a solid tumour, and the temperature elevated high enough to 'cook' the tissue.

If you go to ClinicalTrials.gov and search on the term hyperthermia, you'll find a number of trials – mostly for cancer – that are testing the use of whole-body hyperthermia as a way to sensitize the body to radiation or chemotherapy or to potentiate the immune system's response to malignant tissue (e.g. [5], [6], [7]). Techniques that have been used to achieve hyperthermia include induction heating, warm wax immersion, hot water blankets, and radiant infrared heating. Patients under general anasthesia can also be treated using extracorporeal hyperthermia—blood can be drawn from the body, warmed externally, and returned to circulation. TenOfAllTrades(talk) 12:45, 8 November 2007 (UTC)

I'll note that in non-medical contexts, there have been various suggestions to replace a home's heating system with a (low powered) microwave generator. Instead of heating the air, you heat the body directly. Supposively, this would save on energy costs. A quick Google search turns up [8]. -- 20:25, 8 November 2007 (UTC) —Preceding unsigned comment added by 128.104.112.105 (talk)

Why is water transparent?

Why is water transparent? I did some searching, and the reason for this is because water is transparent to the visible spectrum of light. But why does it exhibit this property? Does it have something to do with its hydrogen bonds that are responsible for so many of its other special properties? Acceptable 02:37, 8 November 2007 (UTC)

There’s some explanation of this in absorption spectrum. Which frequencies a water molecule can absorb depends on what possible quantum states the molecule has. The frequencies of photons that the molecule can absorb correspond to the possible differences in energy between pairs of states. The ratio of a difference in energy level to the corresponding freqency of light is known as Planck's constant. MrRedact 03:20, 8 November 2007 (UTC)

I should point out, water is slightly blue. Malamockq 03:28, 8 November 2007 (UTC)

Consider checking out Color of water. --slakr^\ talk / 04:32, 8 November 2007 (UTC)

Plasmon frequency? —Preceding unsigned comment added by TreeSmiler (talk • contribs) 03:15, 9 November 2007 (UTC)

Eyeballs are mostly made of water. This would affect what light could be seen.Polypipe Wrangler 21:24, 13 November 2007 (UTC)

Tectospinal tract

Is it safe to conclude from this picture that the superior colliculus is connected with the inferior colliculus through the tectospinal tract? Lova Falk 10:29, 8 November 2007 (UTC)

There are some axons that go from the superior colliculus to various brain locations (tectobulbar axons) but I think the vast majority of the axons in the tectospinal tract first go anterior (ventral) and then cross the midline before descending past the level of the inferior colliculus. A good neuroanatomy textbook will have a figure for the tectospinal tract showing a series of brain cross-sections, one at the level of the superior colliculus, one at the level of the inferior colliculus and several more going down to the spine. This is the best I could find online (and it is not very good). --JWSchmidt 18:51, 8 November 2007 (UTC)

Point and shoot digital camera

Is it true that camera manufacturer deliberately introduce shutter lag into PS digital camera to "encourage" their clients to buy the much much more expensive DSLR camera instead. 220.237.184.66 12:06, 8 November 2007 (UTC)

Unlikely, as not all point-and-shoot manufacturers have a DSLR in their lineup. One reason that comes to mind is that contrast-detection autofocus is slower than phase-detection autofocus. It's also possible that "something" has to be done to the CCD or CMOS sensor before taking the shot, if the sensor has been used for a live preview. For example, CCDs may accumulate charge that needs to be cleared. Since DSLRs generally do not have a live preview, they can keep the sensor ready to shoot. Maybe people know of other factors that I'm not thinking of. -- Coneslayer 13:31, 8 November 2007 (UTC)

Point-and-shoot cameras are slower because they use contrast detection rather than phase detection autofocus, and because they usually have a less-powerful focus motor to increase battery life. You can test this by pre-focusing a point-and-shoot: once it's focused, it's actually faster than a DSLR at taking the picture, because the DSLR needs to get the viewfinder mirror out of the way. --Carnildo 22:46, 8 November 2007 (UTC)

Inelastic collision

About Inelastic collision As we know that in inelastic collision the initial and final momentum,total energy are conserved but kinetic energy is not conservsed.this is why ,why kinetic energy is not conserved .Plz explain the example of the collsion of cars.also mentioned it that after collsion the two cars come to rest then how is initial and final momentum the same , as they are moving with a speed before collision .thanx ........usman —Preceding unsigned comment added by Star33 2009 (talk • contribs) 13:15, 8 November 2007 (UTC)

It may be useful to compare inelastic collisions with elastic collisions, whereby kinetic energy is conserved (see the first two sentences of that article, then contrast with the lede in inelastic collision). As for the car collision, bear in mind that physics is not nearly so concerned with "speed" as it is with velocity, and consider that the momentum of the two-car system is what's being conserved, not the momentum of the two individual cars. — Lomn 14:00, 8 November 2007 (UTC)

I actually ran a little experiment on this. Kinetic energy is always conserved as a physical law- if it's not, you haven't included everything in the system. So we have to know where the energy of collision goes to. I modelled this with a spring system, as springs are a good approximation for any interaction and well-established in the interaction of particles. So then we have energy initial = energy final, or $KE_{1}+KE_{2}=KE_{1,2}+E_{\textrm {spring}}$. Now note that the energy of the spring includes both the spring potential energy, 1/2kx^2, and the kinetic energy of particle 1 and 2 oscillating against the spring. It is what is called a simple harmonic oscillator. So now

$KE_{1}+KE_{2}=KE_{1,2}+KE_{1,2{\textrm {spring}}}+1/2kx^{2}$,

and so energy is conserved. The spring model has some consequences: it implies both that there are oscillations between the two particles after collision and that there is a non-zero collision time with finite acceleration. In the intro physics lab which I TA, the data showed the finite acceleration which agreed to good approximation with the theoretical acceleration of the spring (a linear differential equation - if you want me to show you how it is solved, please let me know), but no oscillations after collision. This is because the oscillations are quickly damped out as heat (another differential equation), which should also be measurable with a calorimeter, but I haven't tried this experiment. Energy is still conserved, as heat is another form of energy for which we can account.

For your second question, note that momentum is a vector quantity, so it has both magnitude and direction - that should put you on the right track. SamuelRiv 14:11, 8 November 2007 (UTC)

Momentum is a vector - but kinetic energy isn't.

If you crash two very stiff objects together (two billard balls for example) then they bounce off and the sum of the kinetic energies of the two balls is almost exactly the same before as after. When you whack two cars together, the energy is absorbed in bending and tearing metal and plastic - so they don't bounce off much - there is no kinetic energy left, it all turned into heat. If you bounce two rubber balls together, the result is somewhere between the two extremes. KINETIC energy is not conserved in real-world collisions - but TOTAL energy is. SteveBaker 03:38, 9 November 2007 (UTC)

camera pixel resolution

What is the engineering standard for stating the number of pixels that make up a camera? The actual number of individual sensors, i.e., the number of rows of sensors times the number of columns of sensors on the chip, or the number of different areas of a picture that are focused onto a single sensor or single small group of individual sensors in sequence to build the whole picture? Clem 13:54, 8 November 2007 (UTC)

Megapixel ratings are just a description of the number of sensors. The fact that more goes into a good picture than that rating alone is one of the reason the rating system is seen as being somewhat deceptive. As the page points out, in cameras this is even more deceptive, since each sensor generally registers only one color, and so the final image resolution can be easily a third less than the MP rating. --24.147.86.187 15:03, 8 November 2007 (UTC)

Here is an example of a sensor with three layers one for red, one for green and one for blue.[9] Along with the powerpoint that outlines the old technology and explains this new technology.[10] It is marketed as a 4.5 Megapixel CMOS direct image sensor with a maximum picture size of 1420 x 1064 x 3 matrix as seen in the HanVision HVDUO-5M digital camera. David D. (Talk) 23:05, 8 November 2007 (UTC)

The megapixel count on a camera is the number of pixels in the image it produces. The actual number of light-sensing elements depends on the sensor technology used and the relative influences of the marketing and engineering departments. Sensors using the Bayer pattern and the related CMYK pattern will typically have as many single-color sensor elements as there are pixels in the output image. Cameras using the Foveon sensor pattern can have one-third as many full-color sensor elements as there are pixels; cameras using Fuji's Super CCD pattern have one-half as many songle-color sensor elements as there are pixels. Cheap high-megapixel cameras will use a small sensor and scale the image up: a 20-megapixel camera might use a 5-million-element sensor and use interpolation to produce an image with more pixels. --Carnildo 23:35, 8 November 2007 (UTC)

Old cars carbon emissions

How does the average carbon output of a 1980 1.4l petrol engine compare to a recent one? I heard statistics saying that 20% of the oldest cars represent 60% of total emissions. Is this correct? In view of these statistics some countries want to tax (or are already doing it) old cars. If we add the carbon output of producing a new car and disposing of the old one to the equation, is it still so favorable to buying a new car vs. keeping an old one running longer? Should we add to this the carbon output necesssary to produce enough wealth to buy the new car or is this irrelevant? Thank you. Keria 14:06, 8 November 2007 (UTC)

Please be sure to distinguish pollution from carbon emissions. Older engines emitted far, far more pollution (unburned hydrocarbons, carbon monoxide, and the like) but the amount of carbon ultimately emitted is a strict function of the fuel economy of the car. Your 1980 1.4L car is probably emitting much less carbon per mile than my 2003 4.2L Audi (which averages about 25 miles/US gallon).

Atlant 17:20, 8 November 2007 (UTC)

You only have to compare fuel consumption numbers - that's a pretty fair comparison because the amount of CO2 you get out of a gallon of gas is about the same no matter how you burn it. My 2007 MINI Cooper'S with a 1.6l engine manages 7.5l/100km city and 5.3l/100km highway. A 1986 Honda Integra (pretty similar in size) also had a 1.6l engine and manages 7.8l/100km city, 6.8l/100km highway. So what gives? We don't do much better now than we did then...well, the MINI manages 170hp - the Integra managed about 108hp. What's happening is that we're getting more power out of engines than we used to. So whilst the consumption of a 1.6l engine isn't much better than it always was - we're able to stick a 1.6l engine into a car that would have required a 2.5l engine 20 years ago. Having said that, my old 848cc 1962 Mini manages 5.1l/100km on the highway - fractionally better than the 2007 version - but the 1962 car only has 37hp and could only manage a top speed of 72mph - the 2007 car goes MORE THAN TWICE AS FAST on the same amount of gas. SteveBaker 03:24, 9 November 2007 (UTC)

Ladder to Space

There's this infinitely tall ladder in my backyard. When I climb three or four feet and let go of the rungs I obviously fall back down to the ground. What is the minimum distance I would have to climb so that when I let go of the rungs I would never fall back down to the ground? (I think the answer is 22,240 miles where I would join all the geostationary artifical satellites of the Clarke Belt, but I'm not sure. Orbits, and rocket science in general, confuse the bejeezus out of me.) Sappysap 14:08, 8 November 2007 (UTC)

You might want to check out Space elevator. -- Coneslayer 14:17, 8 November 2007 (UTC)

It'd be geostationary orbit altitude, but only if you're also at the equator. Climbing a ladder ascending vertically from New York will not put you in a stable orbit at that altitude; you'd have to go higher. Climbing a ladder from the North Pole confers you no velocity and you'll drop from any altitude. I'm guessing the tangent of latitude is relevant to exactly how high you've got to go, but I'm not sure. — Lomn 14:24, 8 November 2007 (UTC)

Irrelevant Post Ahead: I just have to say, that Space Elevator article is extremely fascinating. Beekone 14:49, 8 November 2007 (UTC)

That is true. See Coriolis effect. We can calculate the height needed to get you in orbit - the thing about a ladder is that it's rigid, so your angular velocity at the top of the ladder is equal to the angular velocity of the earth at that latitude, so your orbit will only occur at a geosynchronous distance: see Geosynchronous orbit derivation. SamuelRiv 14:52, 8 November 2007 (UTC)

Where's a mathematician when you need one? Say your ladder is at 40 degrees north latitude. You will have to climb up to where your speed is the geostationary orbital speed. This will be higher than if your ladder were at the equator. When you let go, you will move toward the earth until you reach the geostationary altitude where you'll stay until perturbations mess things up. I'm sorry, but I don't feel like deriving the formula for all that, the relationship between angular velocity, altitude and latitude. Where's that mathematician? There are guys who can do this sort of thing in their head. --Milkbreath 17:13, 8 November 2007 (UTC)

Perhaps they're all here?

Atlant 17:17, 8 November 2007 (UTC)

Right. I had a chance to think about this just now while watching a Labrador retriever do his business (hardly a Newtonian anecdote, eh?). The only way you'll be able to let go of the ladder and just stay there is if the ladder is at the equator. So, the nerds are off the hook. (I suspect that the answer to my pointless question above is a straight line tangent to the earth at a geostationary point above the equator.) --Milkbreath 17:41, 8 November 2007 (UTC)

Not quite - there are important consequences when you get far enough out where the rotation of the Earth is too fast for gravity and you get vertical components of the coriolis effect. Integrate across the length of the ladder accounting for atmospheric effects and you'll get a mess. The ladder itself needs a counterweight to make it stable, which is why the space elevator needs such a large counterweight mass. Oh, and the math for latitude at angle phi just needs a sine term, sin(phi), to multiply through. See the formula at Coriolis effect. SamuelRiv 20:43, 8 November 2007 (UTC)

Blaise Gassend computed whether, where and how hard an object falling from any level on an equatorial elevator will hit the ground. —Tamfang 18:49, 9 November 2007 (UTC)

Answering the actual question

We were not asked how high you would have to climb on the ladder in order to enter geostationary orbit. We were asked how high you would have to climb in order to never fall back down to the ground. Which means that any stable orbit would do.

Since the atmosphere has no sharp outer boundary, there is no specific altitude above which you have to orbit in order for orbital decay due to atmospheric drag to become negligible. However, I will assume for simplicity that a distance of 8,000 km above the Earth's center (that's roundly 1,600 km or 1,000 miles above the surface) is what you need to achieve. So you want to be in an orbit with its perigee at that distance and its apogee at the point where you jump off the ladder.

The answer clearly depends on your latitude. If the ladder is at the North or South Pole, it doesn't matter how high you climb: the ladder is simply rotating around its own axis and that doesn't put you in orbit, so you'll always fall back to Earth. But if the ladder is at latitude L and you climb to a distance A above the Earth's center, then you are moving horizontally in a circle of circumference 2 pi A cos L, completing one circle per sidereal day. From here on I'll suppress units; numerical values are based on distances in km, times in seconds, speeds in km/s, etc. Then your speed is V = KA where K = (2 pi/86164) cos L.

Now, using the formulas on this page with some changes of variable names, the orbit's apogee and perigee distances A and P (from the center of the Earth), and the apogee speed V, are related by

V² = G M (2P/(A(A+P))

or in other words

V² A (A + P) = 2P G M

where G is the gravitational constant and M is the Earth's mass, and the value of GM is known to be 398,600. But in this case we also know that V = KA, so we have

K²A³ (A + P) = 2P G M

and since all the other values are known (for a specific latitude), we merely have to solve this for A. As it reduces to a quartic equation, this is not easily done by algebra, but it can be solved numerically by a simple computer program.

For example, suppose your backyard is at latitude L = 45°. Then we have K = (2 pi/86164) cos L = 0.00010313 and K² = 1.0635e-8. We are assuming P = 8,000, and we know GM = 398,600. So 2GMP = 2 x 8,000 x 398,600 = 6.3776e9, and we have

(1.0635e-8) A³ (A + 8,000) = 6.3776e9

so

A³ (A + 8,000) = 5.9968e17

with the numerical solution that A = 26,020 km to 4 significant digits. Taking the Earth's radius at your latitude as 6,370 km, the answer is that you would have to climb 19,650 km or say 12,210 miles in order to jump off and reach a stable orbit with the perigee mentioned above.

If your backyard is on the equator, you're in a much better position. In that case K² is larger by a factor (cos 0 / cos 45°)², which conveniently is exactly 2, which makes A³ (A + 8,000) smaller by the same factor, i.e.

A³ (A + 8,000) = 2.9985e17

This has the numerical solution A = 21,630 km. If your ladder is at the equator, you must climb a mere 15,250 km or 9,475 miles to jump into a stable orbit. And if you wanted to enter a geostationary orbit (not a bad idea if you ever intend to come back down using the ladder!), then on the equator it would be possible by climbing to the height mentioned above. Anywhere else, of course, it would not be.

--Anonymous, edited 04:49 UTC, November 9, 2007.

The effect of stars on Earth

What if a genie were to withdraw all of the stars in the universe except the sun, and the photons in transit to Earth were taken away as well. Would there be a gravitational effect? Would there be a climate change on Earth? Essentially, do the stars in the night sky play any describable role in Earth's affairs? —Preceding unsigned comment added by 150.167.179.111 (talk) 17:00, 8 November 2007 (UTC)

Well, stars provide a useful amount of light on a clear night (see night vision), but I think there's no other routinely discernable effect. The opinions of astrologers will differ, of course.

Note: It is thought that a sufficiently close supernova would emit enough gamma radiation to toast us all, but I think we all hope that won't occur any time soon, so I'm excluding that as a current effect.

Atlant 17:11, 8 November 2007 (UTC)

The stars are all far enough away that gravity is not an issue; starlight doesn't make up an appreciable amount of radiation reaching Earth so it shouldn't have any effect on the temperature, climate, etc. My bet would be "no". The stars don't play any real physical role in Earth's affairs. --24.147.86.187 18:57, 8 November 2007 (UTC)

For that matter, said genie could remove everything but Earth/Moon/Sun and we'd see no appreciable difference apart from the view in the nighttime sky. — Lomn 19:10, 8 November 2007 (UTC)

In terms of climate, the effect of the loss of every single star is essentially nil over any short or medium time scale. (Over an extremely long period of time – hundreds of millions of years – there's the risk Atlant notes of a nearby supernova explosion.) This web page talks about using single stars as light sources of known, carefully-measured intensity for evaluating the sensitivity of digital camera sensors. Interestingly, it also provides the relative illumination provided by starlight (0.001 lumens per square meter) compared to full daylight (10 .000 lumens per square meter). Making the reasonable assumption that most of the energy we receive from starlight will be at visible and near-visible wavelengths, distant stars contribute less than a millionth of the incoming radiation to Earth.

As for gravity, the effect is again negligible unless a massive star passes extremely close to the Earth-Sun system. (This would be a very rare event.) Since gravitational force follows an inverse square relationship, a star the size of the Sun only one light year away will pull on the Earth lss than one-billionth as strongly as the Sun does. TenOfAllTrades(talk) 19:52, 8 November 2007 (UTC)

On the other hand, the effect could be catastrophic. See Ice_age#Causes_of_ice_ages. While I don't believe this theory, just remember that we are in some kind of orbit around a galactic center, and therefore there is a very real gravitational effect on the Earth and Solar System. The planets and outer solar system all have enormous influence regarding the slinging of comets and asteroids into Earth's path, and there are clear measurable gravitational effects from Venus, Mars, and Jupiter. The moral of the story is that in a chaotic system like climate and ecology, you cannot just ignore the small variables. SamuelRiv 20:28, 8 November 2007 (UTC)

On the one hand, the physical effects on the planet are described above. On the other hand, there is the potential effects on society and humanity. Consider the espers' trick on Ben Reich in The Demolished Man, multiplied by six billion, or the end of "Nightfall" in reverse... the sudden disappearance of every single star in the universe would be quite traumatic. The most calm would be rightly troubled, and the least calm would revert to base fears, thoughts of religiously inspired (or even very real) armageddon. When people panic on large scales, Bad Things Happen. If everyone in the world is exposed to the same instantaneous trauma (and even something as simple as the stars disappearing can be quite effecting, I'd imagine), its a safe bet that human society as a whole would have a fairly hard time coming through and recovering from such a scenario. --Jeffrey O. Gustafson - Shazaam! - <*> 10:54, 12 November 2007 (UTC)

growing crystals of copper sulfate

how would you make a hot, concentrated solution of copper sulfate? —Preceding unsigned comment added by 86.42.210.0 (talk) 17:24, 8 November 2007 (UTC)

You would boil some water (perhaps in a kettle), pour it in a heat resistant glass or ceramic container, then add copper sulphate crystals and stir. Do not use aluminium or steel containers as copper will plate on to their surfaces. Then you decant the solution, leaving any undissolved stuff behind. Commercial copper sulphate probably has ferrous sulphate as well, so it may not be pure. As the solution cools you will get a growth of crystals. If you can hang a little crystal from a thread, you can make it grow into a big crystal. Other experiments you can do with copper sulfate solution are: add ammonia to get a dark blue solution which can dissolve cotton, add biuret to get a different dark blue solution, add a base like sodium bicarbonate to make a precipitate. Graeme Bartlett 20:14, 8 November 2007 (UTC)

Viruses

Virus is a DNA With a Protein Coat Protecting It Can a virus be destroyed if the protective protein coat is damaged so that it cant protect DNA anymore if yes then can an enzyme be used as protease to digest the protein coat thus destroying the virus ????? —Preceding unsigned comment added by 212.71.37.97 (talk) 18:04, 8 November 2007 (UTC)

Proteases generally act, well, in a general way. They either consume a protein at one of its ends, or they cleave proteins at specific amino acids. So as you can see, a protease based antiviral measure would cause quite a bit of collateral damage if used to "carpet bomb" infected tissue. It could destroy or inactivate the virus; however, some viruses are even evolved with this in mind, and cleavage of viral proteins upon entering the cell can actually activate the virus (I can't for the life of me remember what article this is in, but it came up in a previous ref desk question I can't find in the archives). Far easier to target them with antibodies. Someguy1221 19:39, 8 November 2007 (UTC)

Something like http://dx.doi.org/630030? DMacks 21:19, 8 November 2007 (UTC)

Also, your definition of "virus" is not quite right. There are many RNA viruses. And the RNA of many of those viruses can infect cells as "naked RNA"; no protein coat is required for the virus to infect cells and produce new virions. - Nunh-huh 21:18, 8 November 2007 (UTC)

ecosystem

can you show me a picture of an ecosystem (example) that a 4th grader could use to help them do a project? —Preceding unsigned comment added by 72.18.102.36 (talk) 20:16, 8 November 2007 (UTC)

I like the images and text here. But there were other suitable examples when I did a google image search. Man It's So Loud In Here 21:08, 8 November 2007 (UTC)

nuclear energy: a given

Let's face it: we are running out of oil. The federal "government" doesn't have a plan for the event known to the public (or is the coal industry now going to move in for what it's worth?). We are in a bad spot so we quickly fall back to our former nuclear technology, which could be a rescue except for the problem of "spent" nuclear fuel. So has there been any design for a facility that can "speed up the procees of nuclear decay" of a spent fuel on-site? Can a half-life be made into a quarter-life? There's energy there. LShecut2nd 23:29, 8 November 2007 (UTC)

Radioactive decay is a quantum process and as far as I know there's no way to affect it one way or another. That being said, there are plenty of other ways that one can imagine dealing with the waste problem. Unfortunately the stakes are quite high and the need for government intervention quite high as well so as a result it is a rather toxic bureacratic issue, so to speak, and progress has been pretty slow and problematic. --24.147.86.187 00:21, 9 November 2007 (UTC)

Breeder reactors are one technology that affects the quality and quantity of waste by transmuting some of the waste into other substances. The drawback is that this technology produces plutonium which is much easier to use as the core of a nuclear bomb than ordinary reactor materials. Dragons flight 01:39, 9 November 2007 (UTC)

It is a fundamental mistake to say that we are running out of oil. If (hypothetically) we were to continue consuming it at the rate we do now then we would indeed run out in 40 to 150 years (depending on the economics of pulling oil from sands and shales as the price inevitably rises). However, if we burn oil at this rate for even 20 more years, the planet will die. So given that we don't intend to kill the planet - we WILL cut our consumption. So - the problem remains - somehow we have to stop using oil. Certainly there are alternatives - nuclear isn't wonderful - but the difficulties of safely storing nuclear waste is a lot more tenable than the the problems of collecting and storing millions of tons of CO2 gas. Nuclear waste will trash the environment wherever we put it - but it's a lot better than trashing the entire planet. So let's pick a place (right in the middle of a desert someplace might be good) and dump all of the stuff there. Sure, the consequences will be nasty - but a heck of a lot less than melting ice caps, rising sea levels, increasing human misery - annihilation of some terrifying percentage of the species of life.

But you can't speed up the rate that the low level waste decays. Sure, it contains energy - but at a level that can't be economically recovered. There really isn't much you can do but let it decay. The plan ought to be to use nuclear as an emergency stop-gap. We URGENTLY need to get away from fossil fuels - and the efforts with wind, solar, wave, etc really aren't cutting it - and the one renewable resource that did anything (hydroelectric) has proven to be a problem, the dams silt up and the downstream environment suffers...argh! So we need to build a bunch of nuclear power plants - shutting down the coal, oil and gas plants as we do so. Then research - LOTS of SERIOUS research. Whatever happened to huge orbiting solar power plants with microwave downlinks? Why aren't there VAST numbers of windmills everywhere? Fusion power - perpetually "25 years away" from getting something working...we need a 'Manhatten Project' for fusion. For vehicles, we have other problems - cheap electricity could solve it - but I think ethanol may be the more likely answer.

SteveBaker 02:42, 9 November 2007 (UTC)

It is quite easy to speed up nuclear decay. First, you need to chemically separate the waste into elemental components, since they must be treated differently. Then, use a neutron source to bombard the correct set of your elements. You can get the neutrons from a fusion reactor. The process can be engineered as a net power source. None of this can be done today: the science is there, but the engineering is not. This is however a way that our grandchildren can eliminate the low-level nuclear waste that we must generate to avoid killing ourselves with oil. -Arch dude 03:04, 9 November 2007 (UTC)

Well, have a look on Fusion power and ITER. Thermonuclear reactors are the next-generation-nuclear-device which would produce far less waste while consuming heavy water (D2O, more precisely, heavy isotopes of hydrogen: deuterium and tritium), i.e. they will be able to provide Humankind with cheap clean energy. —Preceding unsigned comment added by 62.63.76.14 (talk) 09:51, 9 November 2007 (UTC)

Is fusion energy really all that clean? The fusion reactor itself becomes highly radioactive. "...most of the radioactive material in a fusion reactor would be the reactor core itself, which would be dangerous for about 50 years, and low-level waste another 100. By 300 years the material would have the same radioactivity as coal ash..." If fusion energy became viable on a world scale to power everything, how bad of a nuclear waste problem would we have? Sappysap 14:24, 9 November 2007 (UTC)

Probably very little; there isn't anything important under the Sahara and Gobi deserts, or the Great Basin of Nevada. These would hold massive quantities of waste. —Preceding unsigned comment added by 98.196.46.72 (talk) 16:12, 11 November 2007 (UTC)

Also see Integral Fast Reactor, and this Q&A here. Too bad the project was killed, eh? grendel|khan 15:36, 9 November 2007 (UTC)

Reichsbrücke

The Reichsbrücke article does not state why the bridge collapsed and was anybody found guilty. Does anybody know any details concerning that collapse? Mieciu K 23:58, 8 November 2007 (UTC)

Following the third link in the References section, I find this paragraph:

Ursachen. Nachdem zunächst Gratz seinen Rücktritt angeboten hatte, übernahm der Wiener SP-Planungsstadtrat Fritz Hofmann die politische Verantwortung für den Einsturz und schied wenige Tage nach der Katastrophe aus dem Amt. Eine Expertenkommission gab kurz darauf bekannt, dass der linke Pfeiler der nach Ende des Zweiten Weltkrieges sanierten Brücke zum Teil mit Sand und "unverdichtetem Beton" gefüllt gewesen war. Durch das schlechte Material sei Wasser eingedrungen, was schließlich zu dem Einsturz führte.

Combining pieces from two machine translations of this (Google Language Tools and Babelfish) and putting the word order into something more like English myself, I figure that this says:

Causes. After Gratz [the mayor] initially offered his resignation, the Viennese FR planning town councillor Fritz Hofmann took political responsibility for the collapse and resigned from office a few days after the disaster. An expert commission shortly afterwards announced that the left column of tbe bridge, rehabilitated after the end of the Second World War, had been partially filled with sand and "uncompressed concrete". Water penetrated by the bad material, which ultimately led to the collapse.

--Anonymous, 03:29 UTC, November 9, 2007.

November 9

Robinson projection

I'm beating my head against the wall. I want to come up with a simple script that will convert latitude and longitude coordinates to x,y coordinates on a given map in a Robinson projection of a given width with a given central meridian. Ideally this would be done in Actionscript but if I had it in any sort of code or pseudocode that would be fine (which is why I am asking here and not the computing desk—it is not a computational difficulty, it is a conceptual one. Once I have an idea of what I should be doing conceptually it will be trivial to code it).

There are a few map projection projects out there but they are all extremely complicated since they are designed for exporting ALL projections; I _just_ want Robinson. I've read the article and I grok that it's about a lookup table but I still have no idea how I'm supposed to convert that table into x,y coordinates with a given map.

Can anybody help? I just want instructions on the level of "take your number from column one, multiply it by something in column two, then do something else, do something with the central meridian and the width, and presto-chango you have x and y coordinates." I'm having trouble figuring out how to use the table. --140.247.10.141 00:18, 9 November 2007 (UTC)

The general method will be to get a formula that maps the points from one projection to the other. THen you invert the formula so that it maps the points in the reverse direction. Then scan all the points on your new map, say from top left to right and then going down in a raster pattern, use the formula to get a new coordinate to look up the original map. If the coordinates are on the map copy the pixel. If its off the map stick in a "missing data " colour - perhaps white or blue. Graeme Bartlett 01:19, 9 November 2007 (UTC)

Graeme, A Robinson projection is not a formula driven mapping.

Matlab code snippet

robval = [
00 1.0000 0.0000 
05 0.9986 0.0620 
10 0.9954 0.1240 
15 0.9900 0.1860 
20 0.9822 0.2480 
25 0.9730 0.3100 
30 0.9600 0.3720 
35 0.9427 0.4340 
40 0.9216 0.4958 
45 0.8962 0.5571 
50 0.8679 0.6176 
55 0.8350 0.6769 
60 0.7986 0.7346 
65 0.7597 0.7903 
70 0.7186 0.8435 
75 0.6732 0.8936 
80 0.6213 0.9394 
85 0.5722 0.9761 
90 0.5322 1.0000 
];

robval(:,3) = robval(:,3) * 0.5072;
robval = [robval(end:-1:2,:);robval(1:end,:)];
robval(1:90/5,[1,3]) = -robval(1:90/5,[1,3]);

rvals2 = interp1(robval(:,1),robval(:,2),latitude,'cubic');
rvals3 = interp1(robval(:,1),robval(:,3),latitude,'cubic');
y = -rvals3;
x = rvals2/2.*longitude/180*2;

The above is taken from a Matlab program I wrote to generate a Robinson projection. Dragons flight 01:35, 9 November 2007 (UTC)

does it use cubic interpolation? the article does not say what kind of interpolation is used. Graeme Bartlett 01:44, 9 November 2007 (UTC)

My quick search didn't turn up specifics on the kind of interpolation (other than the phrase "simple interpolation method"), but this page shows an example Robinson map with a caption saying "...calculated with 3rd degree polynomial interpolation", for what it is worth. Pfly 06:29, 9 November 2007 (UTC)

That looks like it is in the right ballpark. Here I admit to not being able to follow MatLab's syntax with dealing with Arrays. Could someone convert it into something a little more standard, or just pseudocode? I feel like I'm on the cusp of having it but researching how MatLab deals with Arrays is something I'm not very excited about... --140.247.11.32 16:09, 9 November 2007 (UTC)

Muteness

If you're born mute due to brain damage rather than throat/vocal cord damage, as an infant, do you still cry? Assuming you can hear, what happens when you learn to understand language? Do you merely stop crying but not be able to communicate until someone teaches you sign language? In fact, is it even possible to be mute but not deaf as a result of brain damage? Or would it have to be throat/mouth damage? Kuronue | Talk 02:06, 9 November 2007 (UTC)

I'm not sure about the physiology of crying (there is no article on crying here, unfortunately), but it seems to be associated with contraction of the diaphragm, and should occur with or without vocal chords or vocalization control. Children can hear their own voices, and it's an extremely important part of their personal language development, so a mute child will probably have to learn some sort of gesture or sign language to compensate, and if one isn't taught, it will probably imitate or invent such a language from what it perceives from others (gestures, etc). Unfortunately the pages here for muteness and aphonia are also lacking. Brain damage can cause muteness if the damage is significant enough to somewhere such as Broca's area, which would affect speech production, though it would damage a lot of other linguistic abilities as well such that someone with such damage may not be able to perform sign language. A much more mild damage would be from, say, damaging one of the facial nerves that control the mouth and tongue, which would easily destroy comprehensible speech. Finally, of course, a laryngectomy of the sort that Steven Hawking went through would destroy vocalization completely. SamuelRiv 06:10, 9 November 2007 (UTC)

No other responses, huh? Why is it that there's so little information? I meant mostly the verbal part of crying - if the vocal cords function but the bit of the speech center that organizes sound into speech does not, I'd think the infant would scream but then... would the child merely become silent around 2-3 years of age when others learn to speak? Also, good point, if it's brain damage to the Broca's Area or similar, he might not be able to learn sign language... *ponders* Maybe I'm making this too hard for myself. The character I'm writing is a young boy of about 6 who cannot speak, but I want him to be able to respond to the telepathy, and he's not visibly deformed - I was going to go for some sort of umbilical-cord oxygen-deprivation at birth scenario. I oughta just have his vocal cords themselves be malformed due to mutated genes or something... Kuronue | Talk 17:06, 13 November 2007 (UTC)

asbestos -microencapsulization healing?

Once fibers get in the lung, they can't be removed., sometimes leading to cancer. When re-modeling a house, asbestos that can't be removed is encapsulated with an impenetrable material. Why can't micro-encapsulization, biological or chemical, be used likewise to neutralize fibers in situ? Wiki articles on microen. and on 'self-healing' were good. Comments? —Preceding unsigned comment added by 76.182.3.188 (talk) 02:11, 9 November 2007 (UTC)

Well, microencapsulation wouldn't remove the actual foreign particle from the lung, which I believe (not my field at all) is the primary cause of the cancer in the first place. See carcinogen. Aside from that, encapsulating the fibres should be feasible with molecules attracted the long mineral, but I'm not sure of any such engineering out there for individual strands. SamuelRiv 06:15, 9 November 2007 (UTC)

Exactly. One of the things that asbestos does (Asbestosis) happens because the particles are essentially impenetrable. The lung detects the presence of the foreign particle, triggering an immune response. Essentially it releases some localized acid to try to dissolve the particle, which does not work on the asbestos particle, but over time does start to dissolve lung tissue. Coating the asbestos in some other impenetrable shell would likely result in the same thing, unless you could use something invisible to the immune system and non-irritating to the tissues to avoid that response. Not medical advice! Arakunem^Talk 15:53, 9 November 2007 (UTC)

Time

• Will time "be faster" if an event caused the Earth to rotate faster? (and vice versa)
• Does the moon going farther out from the Earth affect the rotation of the Earth?
• Also, could this be happening now?

Time recently has been going by so freaking fast. I worship seconds now. 67.35.94.120 02:27, 9 November 2007 (UTC)

No, No, Yes. Time wouldn't change if the earth rotated faster - the length of a day would get shorter - but the definition of a second (and therefore a minute and an hour) is defined by the rate of some atomic event or other...something that will never change. The orbit of the moon doesn't significantly affect the rotation of the earth - although doubtless there is some small effect or other. The moon is most definitely gradually moving away from the earth - but not at a rate that you could possibly notice even over many human generations. Sadly, your subjective rate of time flow is something we can't do much about - but it is entirely subjective. SteveBaker 02:39, 9 November 2007 (UTC)

Actually tidal drag caused by the moon is the primary effect slowing the rotation of the Earth and is directly linked to the growth in the moon's orbit. It is still a very small effect though. Dragons flight 03:01, 9 November 2007 (UTC)

(ec) Angular momentum is transferring from the spinning earth to the moon's revolution about the earth by a mechanism involving the tides. As the moon moves away, the earth is slowing down. Yes this is happening now. This causes the day to get longer, but the effect is small. We now must add about one second per year to keep the clocks and synchronized with the earth's rotation: see leap second. -Arch dude 02:41, 9 November 2007 (UTC)

Actually, when I read the question I assumed anon was referring to special relativity effects of reduction in relative velocity to the perception of time. The centripetal acceleration that we feel makes the calculation non-trivial, but in general our time would be "slower" than that of an observer stationary relative to our rotation. SamuelRiv 03:38, 9 November 2007 (UTC)

(By a negligable amount...yes) The point is that while there are such effects, they are all quite utterly negligable compared to human timescales. Our OP is claiming that time is going by faster than it used to within his/her lifetime...that's simple not true to any measurable degree no matter what complicated science you try to throw at it. The effect of the moon, the effect of relativity - these are all UTTERLY negligable - and it is wrong to suggest otherwise just because we can. The clear answer to the OP's question is "No". SteveBaker 14:11, 9 November 2007 (UTC)

absorption refrigerator

Look at this http://www.nh3tech.org/abs.html

Since absorption refrigerator uses heat to chill things down and since CPU generates a hot of heat. Why isn't absorption refrigerator uses the CPU heat to cool down hot CPUs? 202.168.50.40 04:15, 9 November 2007 (UTC)

Expense, complexity, and environmental hazards. Basically the first two are the same reason computers don't come standard with liquid cooling. Air cooling is by far the simplest, cheapest, and safest answer for most processors. SamuelRiv 04:25, 9 November 2007 (UTC)

Another reason is that you can't use the heat from the thing you're trying to cool as the energy source for the system. If you cool the CPU, it won't be hot, so it won't be able to boil off the ammonia, so the CPU will heat up until you cool it again, at which point it won't be hot...you get the idea. And you don't want the CPU to be hot enough to power the system in the first place. I remember learning about the ammonia system when I was tutoring my landlord's kid. It's a fascinating machine. It's one of those inventions that is NOT obvious once someone's done it the way the cat door is. --Milkbreath 14:16, 9 November 2007 (UTC)

paleontology & neuroscience

If the brain of a dinosaur or other prehistoric organism were discovered, perfectly preserved so as to be in the same state as moments after its death, what facts about its behaviour, intelligence, memory capacity, etc. would neuroscientists & paleontologist be able to confidently infer from the brain & nothing else.

Hypothetically-speaking; please disregard whether such a brain could possibly be found in reality. 3170s228 04:42, 9 November 2007 (UTC)

From the perspective of neuroanatomy, an awful lot. From a cognitive perspective, next to nothing. Animal brains (which I know nothing about) are a lot different from human brains, but in general we could find out most innervation and structure of the brain, which would tell us quite a bit about a variety of processing abilities. Intelligence is usually measured in terms of cerebrum size, so we'd have an estimate of that. We may be able to infer some things about memory if limbic structures are present, including the apparent structure of long-term memory and perhaps even its synaptic plasticity. With present understanding, however, we could infer next to nothing about behavior - I don't see any real viable correlation of brain structure to behavior in a class of animals like reptiles. SamuelRiv 05:33, 9 November 2007 (UTC)

Animals tend to have similar brain parts. Knowing the existence and relative sizes of different parts of the brain can tell you a lot. They hardly need a perfectly preserved brain. The cavity in a fossilised skill would do. I remember seeing something where they showed that the brain of some kind of dinosaur was similar to that of an alligator and that it probably behaved like one. — Daniel 01:29, 10 November 2007 (UTC)

How would one find out about the kind of society of prehistoric humans?

How would someone find out about the kind of society that existed among prehistoric humans 50,000 years ago? I am not asking what the answer actually is but rather how you would find the information. I do not mean by like research but how would like scientists find out this information. Any help would be appreciated. —Preceding unsigned comment added by 69.181.131.67 (talk) 05:16, 9 November 2007 (UTC)

Great question! As you might guess, information about the social structure 50,000 years ago will be fairly limited. What kind of inference you make will be based on whatever physical evidence is preserved. By examining what kind of tools they had, and what wear was associated with them, along with bones or other trace detritus, it may be possible to guess at what they ate (some other techniques, such as Isotope analysis also can be telling in this regard). Sometimes human skeletal remains can give clues about how the person lived, although this is tricky at long ages. Occasionally a glimpse of the "soft culture" is given; One (Neanderthal) skeleton recovered from the Shanidar site indicates injury and healing. A reasonable guess, since he would have been unable to fend for himself, is that he was cared for by his companions. Was there a specific question which prompted the interest? --TeaDrinker 05:37, 9 November 2007 (UTC)

Note bones also tell about religious practices and superstitions. Garbage (discarded bones of killed animals tell what they hunted and how) and housing are other important finds. Then one can look at probable migration patterns or recorded history to see where that society went: if a descendant society still exists, one could learn a lot from examining them and their own perception of history. Linguistic history of any kind could almost certainly not be traced that far back, unfortunately. SamuelRiv 05:41, 9 November 2007 (UTC)

An overly broad and probably not very helpful answer: Look at the archaeological record and, using everything we know in modern science (particularly medicine) and current cultures with deep roots, make some "educated" guesses. --Bennybp 07:38, 9 November 2007 (UTC)

• A while back they did a reconstruction in Germany. Build a prehistoric village and have people live there in the way the prehistoric people did. They're interactions changed significantly from what they were used to in modern life. You can Google on SWR and "Das Steinzeit Experiment". - 131.211.175.100 12:26, 9 November 2007 (UTC)

First Solar

According to this

According to Deutsche Bank analyst Steve O’Rourke, the company’s results came in well ahead of expectations as its German production facility ramped ahead of schedule, and incremental improvements in efficiency, production throughput, and currency exchange rates drove panel cost per watt to a new low of $1.19

Does that mean I can buy a 65 watt solar panel from First Solar for only $78 dollars???

5kW worth of Solar Panels for only $6000 + installation cost?

220.237.184.66 06:14, 9 November 2007 (UTC)

Well, perhaps - but maybe they don't make them in 65W sizes. You are assuming that this is the cost of a complete unit that you can bolt onto the roof of your house (or whatever) and use - that may not be the case. I suspect they are talking about the raw silicon wafers - it may cost considerably more to assemble them into weatherproof units, add controller electronics, etc. Is that the retail cost or the wholesale cost - if you're going through a middle-man, it could be more. In short, no, it doesn't necessarily mean that...although it might. SteveBaker 14:03, 9 November 2007 (UTC)

gryroscopes

Can two gyroscopes spinning in opposite directions counteract each other? —Preceding unsigned comment added by 204.251.179.61 (talk) 07:05, 9 November 2007 (UTC)

Sure. Why not? It all depends on what you mean by "counteract". -- kainaw™ 13:17, 9 November 2007 (UTC)

From my interpretation of the question, it seems you're asking if two gyroscopes in the same system but spinning on different axes and otherwise not interacting counteract, and the answer is yes and no. No in the sense that they both will resist changes in their absolute spin axis to conserve angular momentum, so they will still keep an object on a stable trajectory. However, a single gyroscope can have its system rotated about the gyroscope's axis without problem, so you need a second gyroscope in a different direction to restrict all rotation. In this way, they do have an interaction effect. SamuelRiv 13:56, 9 November 2007 (UTC)

If they are bolted together then as you try to turn them, each gyro will exert a force onto their common framework - the resultant force will apply. It's no different than if you bolted two cars together and tried to steer them in opposite directions. Gyroscopes (like magnets) have taken on a mystery that they don't truly deserve. SteveBaker 13:59, 9 November 2007 (UTC)

Man, magnets are really, really awesome, and do deserve every bit of mystery that they're given! Invisible forces! Immense power in tiny pieces of metal! Wow! I'm still amazed by magnets, they're the closest thing to magic that I interact with on a daily basis! --24.147.86.187 21:25, 9 November 2007 (UTC)

I dunno - gyroscopes are pretty weird too. Take the front wheel off your bicycle, hold it by the axles and sit in a swivel chair - now have someone spin the wheel up as fast as they can get it to go - then tilt the wheel back and forth...tell me that ain't freaky on the scale that magnets are. But you're wrong about the magnets having "immense power" - they don't store any energy in the form of some kind of "magnetic power" in the way that (say) a battery stores electrical energy. SteveBaker 21:52, 9 November 2007 (UTC)

All I really mean by "immense power" is "electromagnetic forces are so much more powerful (on small scales) than gravitational ones, which is awesome to think about. A magnet can totally kick earth's ass when it comes to attracting a piece of metal." --24.147.86.187 06:04, 14 November 2007 (UTC)

As a related question, if you've got a sealed box containing two counter-rotating coaxial flywheels, is there any measurement you can do to tell if the flywheels are rotating or not? How about if the axises of rotation are parallel but not co-linear? --Carnildo 21:43, 9 November 2007 (UTC)

So it won't matter if the axes of rotation are not collinear at all. You can translate your whole spinning system arbitrarily and nothing will change. The angular momentum vectors of the two flywheels won't cancel - both will resist angular momentum changes of their respective systems. So if you turn the box, you will feel twice as much restorative force from the conservation of ang mom as if you only had one flywheel, whether or not they're spinning in the same direction. SamuelRiv 23:38, 9 November 2007 (UTC)

This problem is a little more subtle than first meets the eye. The behavior of a gyroscope is normally just governed by the conservation of angular momentum. However, if you have two identical rotors with collinear axes in one rigid frame, rotating with equal speeds in opposite directions, the angular momentum of the two rotors cancel out, and the net angular momentum of the system is zero. So it would appear at first glance that you should be able to rotate the system as easily as any other object with the same moment of inertia, since there is no net angular momentum to be conserved. If you apply a torque to the system, the precessions due to the two rotors are in opposite directions, so there is no net precession.

However, there is a complication. If you apply a torque to the system, the tiny bit of precession that each rotor is able to perform will cause stress on the frame – compressive stress on one side of the frame, and tensile stress on the other side of the frame. The resulting strain in the frame will cause the frame to act like a spring, and resist the rotation you’re trying to apply to the system. If you remove the torque from the system, the frame will spring back to its original position.

In short, the answer to the original question is "no", and the answer to Carnildo’s question is "yes."

In the real world, there would of course also be other complications such as friction between the rotor and the frame, a finite yield strength of the frame, etc., that I’m ignoring in the treatment above. MrRedact 11:10, 12 November 2007 (UTC)

LEL and UEL of gases.

I looked at a table on your website which states the LEL and the UEL of various gases. How do I know that the information seen on your website is correct? Is this ever checked or is this just something that someone put together? I'm keen to use your website but only if the information has been verified and is correct. Thank you. Mr Sangster. —Preceding unsigned comment added by Andy Sangster (talk • contribs) 10:38, 9 November 2007 (UTC)

If you're referring to Explosive limit, then you should check the references at the bottom of the page. All information on Wikipedia should be properly sourced, so you can go to the sources if you want to be sure. -- JSBillings 13:13, 9 November 2007 (UTC)

To be honest - there are no guarantees. Whilst Wikipedia editors are supposed to provide references for all of the facts they list, this is far from typically the case - and even if there are references, you can't know for sure that the author of the article typed them in correctly from those references. Worse still, some annoying little kid could have come along and messed up all the numbers for a joke. That is actually very rarely the case - but if you want absolute guarantees - there aren't any. In the case of this article, they have not indicated specifically where that particular information came from - but merely listed the two references were used in writing the article. In order to check that this particular set of information is true, one would have to look into the references and check to see that the information in the article actually agrees with them. In a better referenced article (such as the ones I wrote for the Mini and Mini Moke cars), you'll see little blue tags that look like this: ^[25] which you can click on to take you to the exact reference where that particular fact came from. However, as I said, sadly not all articles are referenced to that standard. As a practical matter, most scientific Wikipedia articles such as Explosive limit are reliable. I wouldn't use the information (without checking the references first) if someones life depended on it - but for more casual uses it's convenient. Wikipedia has been shown in several surveys to be more accurate than printed encyclopedias such as the Britannica. So I guess you can trust Wikipedia to the same degree that you'd trust any general encyclopedia. SteveBaker 13:53, 9 November 2007 (UTC)

Wikipedia is inherently no more or less reliable than any other encyclopedia. Per Wikipedia's Risk disclaimer and General disclaimer, you should place the same trust in this material as you should in material from any general encyclopedia: none at all. If you are going to be handling flammable gases, contact the manufacturer for up-to-date information (UEL and LEL are often listed in the MSDS that ships with products), or perform your own tests. Depending on your location, there are any number of independent testing labs that can measure vapour pressures and explosive limits, as well. TenOfAllTrades(talk) 14:23, 9 November 2007 (UTC)

Of course Wikipedia is less reliable than a real encyclopedia. It has advantages in other areas, of course, but professional editing counts for something. --Anon, 02:00 UTC, November 9, 2007.

Surprisingly, (actually, AMAZINGLY SURPRISINGLY) that's not true. Wikipedia is actually more reliable than almost any other encyclopedia - the venerable Encyclopedia Britannica is normally considered the 'gold standard' of encyclopedias - and Wikipedia is about on a par with it in terms of accuracy and miles ahead on breadth of coverage. Wikipedia is by any measure most certainly a 'real' encyclopedia - whatever that means. As to whether "professional editing" matters - I'd say that results say not. Here are some often-quoted facts:

• A 12-year-old kid [11] found five errors in Britannica after just a couple of days of checking. His only recourse was to write to the editor, and the errors may be corrected in print in a few years.
• Nature magazine did an extensive study of science articles by experts in those fields and found: "Our staff compiled lists of factual errors, omissions and misleading statements that the reviewers pointed to (we had 42 usable responses) and tallied up the total number for each encyclopaedia: 123 for Britannica, 162 for Wikipedia." - of course all 162 of the Wikipedia errors were fixed within a few days of the list being made public. You still can't buy a copy of Britannica with the 123 errors fixed up.
• Wikipedia:Errors in the Encyclopædia Britannica that have been corrected in Wikipedia
• Wikipedia:External peer review

So don't write off Wikipedia as "obviously" less reliable - there have now been half a dozen proper academic studies - and they all sow that Wikipedia and Britannica are about equal - and both are miles ahead of any other general-purpose encyclopedias out there.

SteveBaker 18:56, 10 November 2007 (UTC)

It's safe to say Britannica and the other "professionally edited" encyclopedias don't have quite the same vandalism problem, though... —Steve Summit (talk) 20:39, 10 November 2007 (UTC)

Also, if only to play devil's advocate, those "often-quoted facts" might seem to an impartial outside observer to be a wee bit biased and one-sided. The first doesn't say anything about Wikipedia's error rate; the second says that Wikipedia has (or had) more errors than Britannica. The third and fourth come from Wikipedia itself, so despite Wikipedia's vaunted principle of NPOV, they can't be assumed to be entirely balanced.

(I bet I could find five errors in Wikipedia after only a couple of minutes of checking -- what does that say?) —Steve Summit (talk) 20:46, 10 November 2007 (UTC)

You aren't a 12 year-old. SteveBaker 23:29, 10 November 2007 (UTC)

To answer the middle part of the question: yes, the website was "just put together by someone" -- quite a few different someones, in fact -- and it has also been checked, if not 100% systematically, by other someones.

Now, for any given fact, was it entered correctly by the someone who entered it, and has it ever been checked by someone else? Maybe. Maybe even probably. But definitely not definitely. —Steve Summit (talk) 23:18, 9 November 2007 (UTC)

methylphenidate Q.

http://en.wikipedia.org/wiki/Ritalin#Known_or_suspected_risks_to_health

The R.A. El-Zein study says it will cause chromosomal abberations. I'm no biologist and dont really understand these things, but does this just refer to normal body cells and the only problem is that it could possibly cause cancer, or will it mutate gamete cells as well and these mutations passed on to offspring... thank you Rocktruly21 14:43, 9 November 2007 (UTC)

That study only tested lymphocytes for chromosomal abberations, so this would be called "need for further testing." It should be noted, however, that this study's results were not reproduced to any extent by a much larger but otherwise identical study, as mentioned in the same section you linked to. Someguy1221 18:23, 9 November 2007 (UTC)

SILVER RINGS

hi again.. please choose one of those 4 choices as an answer. question= SILVER RINGS ARE... choices= a) homogenous mixture. b) compound c) element please justify to me your answer because i have a test coming up in this and i REALLLY need to know since my book doesnt provide an answer. thank you alot, jimmy —Preceding unsigned comment added by 212.71.37.73 (talk) 20:00, 9 November 2007 (UTC)

Do you mean rings made of silver? If so, then, if they were not rings what would they be? —Tamfang 20:09, 9 November 2007 (UTC)

Jewelry is not generally made from pure elements, even though they are named to make you think they are; pure gold or silver just don't have the necessary strength. RIngs are made of alloys, which are homogenous mixtures. When you're talking about jewelry, "gold" = gold + copper or nickel or palladium; the actual gold content will be specified in terms of karats, with 24 karat being pure gold. "Silver" = silver + copper or germanium, zinc, platinum, silicon or boron.; silver content is specified in terms of percentage. (a) is the answer you're looking for. - Nunh-huh 20:26, 9 November 2007 (UTC)

Silver jewelery is most often an alloy called sterling silver. Graeme Bartlett 20:43, 9 November 2007 (UTC)

Since this is a test - I'd bet the answer is 'Element' because silver is an element. There are practical matters to do with how jewellery is often made of an alloy of silver and something else - but unless you are doing a test in jewellery-science, they won't expect you do know that. The point is that 'silver' is not like 'bronze' (which is a homogeneous mixture) or 'sugar' (which is a compound). SteveBaker 21:48, 9 November 2007 (UTC)

On the contrary, "silver" in the mouth of a jewelry dealer means "mostly silver" = silver alloy. If the answer isn't (a), demand a recount. - Nunh-huh 22:55, 9 November 2007 (UTC)

Umm, what is (d), the fourth choice? hydnjo talk 23:38, 9 November 2007 (UTC)

the scifi kitchen

If you were to cook beef in an oxygen-free atmosphere, would it turn brown? —Tamfang 20:08, 9 November 2007 (UTC)

I'm not a chemist, but the browning reaction is the Maillard reaction, in case that helps anyone. -- Coneslayer 20:17, 9 November 2007 (UTC)

Yes it would still go brown, and black if you over heated it! Think what happens when you deep fry food. There is no oxygen in the liquid hot fat, but things still go brown or black. Graeme Bartlett 20:40, 9 November 2007 (UTC)

Ever heard of "ear music" heard by hearing impaired people?

A person who worked with hearing impaired people remarked at a seminar recently that there was such a thing as what she termed "ear music," if I recall the term correctly. I haven't found it under similar terms or this, but anyway, she said that it was sounds, almost like phantom limb of the ear, that the nerve endings pick up when it's really quiet that aren't there, as if a low radio were playing in another room or something. I asked specifically if tis might be hypnagogia, and she said this was something different, it wasn't just when one was nearly asleep. Have you ever heard of it? I found it odd there was an article under phantom eye syndrome here (which i'd never heard of myself]] but not phantom ear, but she didn't call it phantom ear. (In fact, if she had, I'd have guessed from context what she meant right away.)Somebody or his brother 20:15, 9 November 2007 (UTC)

Sounds like something similar to hallucinations experienced during sensory deprivation. Information on the web on this seems scant- most seems related to various mental disorders and brain lesions, none of which relates to hearing impairment as a whole. But I'm not sure. Maybe check the references on that page, or see hallucinations in the sane. SamuelRiv 21:36, 9 November 2007 (UTC)

Like Tinnitus? SteveBaker 21:42, 9 November 2007 (UTC)

You might be interested in Oliver Sacks newest book, Musicophilia: Tales of Music and the Brain, which devotes a whole chapter to this topic. It is apparently a lot more common than once thought, especially among the hearing impaired. Sacks calls it musical or aural "hallucination", while noting that some people object to the word "hallucination". Pfly 09:02, 10 November 2007 (UTC)

(later addition) I skimmed the chapter this morning for a bit more specificity. It's chapter 6, titled "Musical Hallucinations". Some of the things Sacks says about it -- musical hallucinations were once thought rare and perhaps associated with temporal lobe epilepsy (TLE), but in recent years it has been recognized as more common and only rarely associated with TLE. Musical hallucination is not a psychosis, not "mental illness"; rather it is "real", "physiological", and benign. Sacks stresses this point -- some people have suggested a similarity between musical hallucinations, "hearing music", and schizophrenic "hearing voices", but Sacks shows how very different the two are, both physically in the brain and in the way they are experienced. Musical hallucinations can take many forms, but common aspects include: hearing music "for real", not just "in your head"; often associated with hearing loss and "emerging" from humming and buzzing type noises (eg, humming refrigerator, tinnitus, etc); more common in the elderly but can occur at any age; when caused by something like a stroke, tends to die away with recovery, otherwise musical hallucinations tend to be "very persistent" and "chronic". Some of the striking differences mentioned: for some people the music is hear very loudly, while for others it is soft and vague. For some it can be very annoying and even intrusively disruptive, while for others it can be pleasant and easily ignored. For some the music tends to be "whole pieces" or at least whole melodies, while for others the music "fragments" into tiny bits that skippingly repeat endlessly. Most people cannot control it, but some are able to "direct" it to some degree. It is not very well understood, neurologically. There is no cure. People for whom it is life-disrupting, a doctor might be able to find ways to reduce its strength. Sacks writes that of the people he knows of who have musical hallucinations, about 80% also have some kind of hearing impairment. Also, of all people with hearing impairment, about 2% develop musical hallucinations. There's lots more in the book, and as always Sacks writes very engagingly. Pfly 19:59, 10 November 2007 (UTC)

Why is chicken, rabbit and pork meat a different color from cows, deer,elk etc?

This isn't a joke, I am truly wondering why creatures with the same basic physiology have different colored meat. Respectfully, curious —Preceding unsigned comment added by 209.237.84.150 (talk) 21:00, 9 November 2007 (UTC)

It's determined by the concentration of myoglobin in the muscle cells. See white meat and red meat. --Carnildo 21:40, 9 November 2007 (UTC)

It's not just the myoglobin but also the concentrations of mitochondria. Dark meat has more mitochondira per cell, while white meat has less. David D. (Talk) 22:23, 10 November 2007 (UTC)

As a sidenote, nitrite is used to prevent the myoglobin from decaying in cured meat. Icek 04:09, 11 November 2007 (UTC)

November 10

What is the oxidation number of carbon in methanol?

What is the oxidation number of carbon in methanol? According to the article oxidation number, I get 4 (four shared pairs of electrons, in which one of each pair belongs to carbon, results in a charge of +4 upon removal). --212.204.150.105 21:14, 7 November 2007 (UTC)

Well, the oxidation state is the charge that carbon would have if the bonds were all ionic, so you need to know, for each bond, which atom is more electronegative. Given that this could be a homework question, I'm not going to tell you the answer. Someguy1221 21:29, 7 November 2007 (UTC)

I know that the oxidation state is -2 (oxygen more electronegative and three hydrogens each less electronegative), however my question pertains to the oxidation number which according to the WP article can sometimes be different. I expect, that the oxidation number should also be two, but when I follow the instruction given in the first sentence of the WP article, I get 4, as described above. Is the article wrong or is my interpretation wrong (and if so, how)? --212.204.150.105 21:44, 7 November 2007 (UTC)

I belive that's only in reference to coordinate bonds. Oxidation numbers are usually only used for metals. Someguy1221 21:55, 7 November 2007 (UTC)

The article on oxidation state says Redox (shorthand for reduction/oxidation reaction) describes all chemical reactions in which atoms have their oxidation number (oxidation state) changed - firstly, I don't think it should say oxidation number (especially not preferentially to oxidation state) and thirdly, I read elsewhere that it should take into account the electronegativity - I think it's possible for an atom to be reduced even without a change in oxidation number, so long as the ligands of the product are less electronegative than the ligands before. Thus I think the article is at best misleading, if not wrong. --212.204.150.105 22:31, 7 November 2007 (UTC)

Slowing time

Can you make time for a particular area seem to go slower by moving away from it? That is, lets say, the TV is on, and I start moving away from it at half the speed of light, would the video on the TV appear to be going half its normal rate because it is taking longer and longer for the light to reach your eyes? What if it moved away from a given spot and I move away in the opposite direction at half the speed of light, would it appear to freeze. Of course, a TV that big that my eyes would be able to resolve it is impossible, but just asking... 208.63.180.160 00:01, 10 November 2007 (UTC)

If you moved away from the TV, in addition to something similar to the doppler effect causing it to appear in slow motion, and the actual doppler effect making everything have a lower wavelength (redshift), the relatively slower speed of light through you would make time pass slower for you. It wouldn't counter it out perfectly. Two objects moving at half the speed of light away from a certain point aren't moving the speed of light away from each other, due to the time shift. This is all special relativity stuff. If it's too advanced for you, you may prefer Introduction to special relativity or the simple english article for it. — Daniel 01:42, 10 November 2007 (UTC)

Moving either towards or away from the TV (or the TV moving towards or away from you) has the same effect, the speed of the show on the TV would change by a rate determined by the square root of 1-v²/c². If your speed is half the speed of light then you have v=0.5c - so the time distortion would be sqrt(1-0.25) or about 87% of normal speed. —Preceding unsigned comment added by SteveBaker (talk • contribs) 18:36, 10 November 2007 (UTC)

Hold on... I obviously have some sort of misunderstanding here. If the TV is moving towards me, will the photons not hit me a a higher rate than if it is moving away? If a tennis ball launcher throws a ball every second, and it is moving towards me, I will be hit by balls at a rate of greater than one per second; isn't the concept the same, so I will see the events on the TV speed up by moving towards it because the light has to travel a progressively shorter distance and thus I see "newer" light quicker than I would standing still? (I have no understanding of special relativity, and I'm not understanding the articles). 208.63.180.160 01:39, 11 November 2007 (UTC)

Your view of the way light (photons) work is incorrect - and that one single fact is the entire reason we have relativity and everything that goes with it. Unlike tennis balls, the speed of light is a constant - irrespective of the motion of the source or the viewer. This was shown most clearly by the Michelson–Morley experiment. Hence, the photons coming from the TV will hit your eyes at exactly the same speed regardless of whether you are moving closer to the TV or further away from it. You'll see 'red-shift' (if you are moving away from the TV) or 'blue-shift' if you are moving towards it - but those are changes don't alter the speed. What's happening here is MUCH weirder than the 'classical' physics view that you are taking here. When you move at half the speed of light relative to the TV, the passage of time, distances and masses all change by that 87% factor - even after you take into account the fact that you are getting closer to it so that the light waves have a shorter distance to travel. SteveBaker 02:18, 11 November 2007 (UTC)

Actually there is a speed-up effect when the TV is coming towards you at a good portion the speed of light, Steve. Imagine this isn't a TV you're looking at, but just a strobe light that pulses once a second in its own reference frame. Yes, if it's approaching you at 50% the speed of light, it only strobes once every 1.155 seconds from your perspective. BUT, since the strobe is coming towards you, the pulses will actually arrive faster than you'd expect. This isn't exactly the doppler effect, but it's very similar. I don't feel like precisely calculating which is more important, but its quite obvious if you look at extreme situations. Imagine the strobe begins transmitting from 1 light year away, but is travelling towards you at a Lorentz factor of one million. The strobe in this case is trailing so close to its own transmission that the entire 1 year lifetime of its transmission suddenly arrives at you in a 32 second spurt just a year after its light first reaches you. Now, it so happens that due to relativistic slowdown, the strobe only actually managed 32 strobes in this time, so the visual speed-up effect cancels out the relativistic slow-down effect (the former is entirely a deficiency of observation, the latter is physically happening). And so at best this can simply cancel out the appearance of relativistic slowdown, though certainly it would exacerbate such if it were traveling away from you. An effect that is not cancelled out by relativity, however, is the appearance that the strobe was traveling at a velocity equal to its Lorentz factor, in units of the speed of light. This actually has to be taken into account in observations of the plasma ejections of quasars, some of which have an "apparent velocity" greater than that of light, before this effect is accounted for. Someguy1221 02:50, 11 November 2007 (UTC)

What you are describing is exactly the Doppler effect. Its magnitude is given by the Doppler formula, the longitudinal case of which is quoted in MrRedact's reply below. -- BenRG 00:17, 12 November 2007 (UTC)

Steve made a mistake. Your (the original poster’s) intuition about the case where the TV is moving away from you is almost correct. If you’re watching a TV moving away from you at half the speed of light, it will look like the show is being shown at 0.577 times as fast as its normal rate. In the limit as your speed away from the TV approaches the speed of light, the rate at which the show looks like it’s being shown approaches 0, i.e., it approaches looking like the TV show is stuck on one frame.

If you’re watching a TV coming at you at half the speed of light, it will look like the show is being shown at 1.732 times as fast as its normal rate. In the limit as your speed toward the TV approaches the speed of light, the rate at which the show looks like it’s being shown approaches infinity.

In general, if you’re approaching the TV at speed v, the apparent frame rate of the TV show is proportional to (1+v/c)/sqrt(1-v²/c²), where c is the speed of light. The same expression also works for the case that you’re moving away from the TV, in which case v is negative.

In reality, the colors of what's on the TV are affected if you're moving relative to the TV, so if you're moving too quickly, the colors would change so much that your eyes wouldn't be able to see them. MrRedact 08:18, 11 November 2007 (UTC)

Has any tried watching a TV while moving at a fraction of the speed of light? Most of the time the screen will be too tiny to see! Even at 0.000001% of the speed of light you will only get a few seconds viewing out of it! (joking here). Graeme Bartlett 20:23, 11 November 2007 (UTC)

Those who claim I made a mistake didn't read my reply - I specifically pointed out that you get the time dilation effect even after you take into account the fact that you are getting closer to it so that the light waves have a shorter distance to travel. SteveBaker 02:20, 12 November 2007 (UTC)

Not sure what you mean by "even after" -- the answer is that the frames will appear to move faster, not slower. By the way MrRedact's expression (1+v/c)/sqrt(1-v²/c² simplifies to the nicely symmetrical-looking ${\sqrt {\dfrac {1+v}{1-v}}}$ (where of course I'm using c=1, as is generally a good idea when talking about relativity).

Now in practice, of course, if you're going fast enough to notice any effect at all, you'll be going far too fast to process even a single frame from the TV. --Trovatore 02:28, 12 November 2007 (UTC)

Your confusion is consistent with misremembering what an "observer" is in relativity. In relativity, an "observer" isn’t one person at one location, but effectively actually a whole system of people spread throughout space, at rest relative to each other, who have a set of synchronized clocks. The time at which at event occurs is measured at the location at which the event occurs. So the difference in time between two events, which is what time dilation measures, doesn’t account for any time it takes for light to travel anywhere after either of the two events.

Suppose the TV is traveling at half the speed of light in the x direction relative to the lab frame, and suppose the TV has a mirror or two attached to it such that the screen can be seen by someone either in front of it or behind it. We won’t even need a Lorentz transformation for this; in the following, all coordinates are lab coordinates. Pick units such that c=1, and the period between frames of the show (as measured in lab coordinates) is 1. Pick the origin such that the TV starts to show frame number 0 when the TV is at x=0, t=0. At time t=1, when the TV starts to show frame number 1, the TV will be at x=0.5.

To someone at rest in the lab frame sitting at the origin, the light from the start of frame 0 will reach them at t=0, and the light from the start of frame 1 will reach them at t=1.5. To someone at rest in the lab frame sitting at x=0.5, the light from the start of frame 0 will reach them at t=0.5, and the light from the start of frame 1 will reach them at t=1. So the apparent frame rate as seen by the person in front of the TV is 3 times faster than the apparent frame rate as seen by the person behind the TV. This is consistent with the ratio 1.732/0.577 of the numbers given in my post above. MrRedact 07:29, 12 November 2007 (UTC)

Greenhouse Gases

Are Greenhouse Gases good for the environment or bad? —Preceding unsigned comment added by 70.171.192.2 (talk) 02:24, 10 November 2007 (UTC)

Have you consulted the article greenhouse gas? By themselves in moderate amounts they are not "bad" but in excess, at the rates that they are currently produced by human activities, they increase the greenhouse effect which leads to global warming. Which is bad. --24.147.86.187 02:31, 10 November 2007 (UTC)

We need to have some, otherwise the earth would be freezing cold all over. Carbon dioxide is essential for the life of photosynthetic plants. Graeme Bartlett 12:09, 10 November 2007 (UTC)

The 'normal' amount (about 0.038%) of CO2 is essential - less than that would be bad because the earth would freeze, more than that would be bad because of global warming. Right now, we have all the CO2 we need - and it's going up - so adding more greenhouses gas is definitely bad. Even if humans ceased to produce greenhouse gasses at all, the natural amount from animal respiration and volcanos would be plenty to keep the earth running OK. SteveBaker 18:31, 10 November 2007 (UTC)

I have no idea where the idea that global warming is bad came from. An increase in global temperature has a positive net effect on humanity. Look at climatic changes of the middle ages in europe and you will see that the warm periods were better recieved by the population than the cold ones. The main cause are reduced heating requirements and increased agricultural yields. Putting this to a global perspective: there is not a single area on earth to hot to live in. Too dry, yes, but not too hot. However there is a whole uninhabitable continent at the south pole, covered in mountains of ice.

So its good. The real problem is, at some point it will be too much. We don't know how the ecosystem reacts to really dramatic increases in temperature. It might be fine with 8° and collapse at 9°. Who knows? So the reason for reducing CO2 emissions is: we do not want to find this out the hard way. —Preceding unsigned comment added by 84.187.90.130 (talk) 00:49, 11 November 2007 (UTC)

WHAT! WHAAAT! Where have you been the last decade? This is quite the most ill-informed, unthinking response I've ever read on this topic! Firstly, the change in the middle ages was small compared to what we're talking about - also it was over fairly quickly and this time around it's going to be permenant. Secondly, while crop yields increase in the extreme latitudes, they decline sharply in the equatorial regions - the net effect will be disasterous. Reduced heating requirements are replaced by increased air-conditioning requirements (you've never lived in Texas have you?!). The antarctic region might become free of ice - but whats uncovered will be bare rock - no soil. The sharply increasing sea levels will drown all of the nice flat primo agricultural plains around the coastlines. You suggest that an 8 degree rise "might be OK" - but I can tell you that a 4 degree increase will be plenty disasterous. Please - do some reading of the proper scientific literature on this subject...or if (as it seems) you are unable to learn from these sources - rent a copy of "An Inconvenient Truth" - it's very comprehensible. SteveBaker 02:08, 11 November 2007 (UTC)

I don't think anyone is talking about an 8C rise. I think a 4C rise was on the tall end of the IPCC report for a 100 years. Sea level rise will probably affect the salinity of river deltas and a lot of food production areas but to say it will drown all the agricultural plains around the coastline is a little too pessimistic. As for AIC, any film that links smokestacks to hurricanes as AIC does on it's cover and in the material is not science. Stick to IPCC. Summary reports if you don't have time. Read about the discrepancies and conflicts if you do. Cloud cover and carbon sinks in the ocean are too big ones that have a lot of research and understanding yet to be done. --DHeyward 06:36, 13 November 2007 (UTC)

0.038% is the current CO₂ concentration, the pre-industrial level was about 0.027%. A question to the experts: How much higher would the equilibrium (glaciers need time to melt...) temperature be at 0.038%, compared to 0.027%? Icek 04:04, 11 November 2007 (UTC)

What do you mean? I thought I understood your question except for the glacier part. Glaciers have their own cycles for formation and melting. They've been melting on average for 10,000 years. Melting has been accelerated due to global warming but as I understand it, there was no equilibrium before. Glaciers can grow and shrink in both warming and cooling periods. --DHeyward 06:41, 13 November 2007 (UTC)

Well, I thought that the melting of the glaciers would be a heat sink for some time, keeping temperature lower as long as there are large glaciers. As the glaciers get smaller, the heat sink would get smaller, and the temperature would rise. Therefore I thought we have to think about the glaciers if we want to compute an equilibrium temperature. I see that the time for reaching such an equilibrium is quite long, maybe longer than climate cycles due to changes in Earth's orbit, obliquity, and equinoxes.

Restating my question: Averaged over the long-term climate cycles of the next few 100,000 years, how much higher would the temperature be if we stopped burning fossil fuel now? And how would the CO₂ concentration develop? Icek 07:45, 13 November 2007 (UTC)

I'm not an eco scientist but I do Know that the UK's climate relies heavily on the Gulf slipstream. If the polar caps melt, the sea temperature drops by a degree or so and the slipstream stops. That would be the UK buggered! 88.144.64.61 08:40, 16 November 2007 (UTC)

Battery charge

1. How do rechargeable devices like mobile phones and mp3 players measure the amount of charge/energy left in the battery? Is it by measuring the terminal voltage? (which would be inaccurate due to polarization effects)? Or do they have ampere hour meters inside them?
2. When we operate them continuously the meter suddenly drops low and after switching off for some time the bar goes back up a little; which would explain the drop in voltage due to polarisation. Am I correct in assuming that?

59.93.9.23 05:35, 10 November 2007 (UTC)

Last I heard, the battery life indicator uses the voltage at the terminals. It is not accurate when it comes to predicting how long until the device quits. Each battery will be in a different stage of its life, and an older one will usually die sooner at, say, two bars. The graph is not linear even for any one battery, the change in voltage is very small, and current demand is unpredictable. You would need laboratory-standard equipment to make the thing at all accurate even under controlled conditions. I'm sure the phenomenon you mention in 2 is due to what is called polarisation, though the battery will drop some voltage internally under load anyhow. If it's right on the line between bars, you'll see a change upon power off just from that. The indicator will have some kind of anti-hunt designed in to keep the display from jumping around, so it will itself be slow to react. --Milkbreath 16:43, 10 November 2007 (UTC)

This is a good question, which I wish I had some more definitive answers to. (But since when has lack of definitive information stopped an armchair RD reader from speculating?)

There are at least four things a battery-charge indicator could look at in trying to make a determination of how much life the battery might have left:

1. Terminal voltage. A battery is, of course, a two-terminal device, so fundamentally, this is all you've got access to. Unfortunately, by definition, a battery is supposed to be a constant-voltage device, so its voltage shouldn't (and doesn't) change much over its lifetime. A theoretically ideal 1.5-volt battery would give 1.500000000 volts for its entire lifetime, then crash precipitously to 0 -- so a charge indicator looking only at voltage would have nothing to go on!
2. Of course, we don't have to limit ourselves to looking at instantaneous voltage; we can also look at rate of change (dV/dt). I'm pretty sure the voltage drops at different rates at various points during a typical (non-ideal) battery's discharge curve.
3. History. Top-end, full-featured batteries (such as the ones used in modern laptops) contain their own microprocessors. These can learn what that particular battery's discharge curve looks like, and use that knowledge to make a much better estimation of how much life is left based on where in the (now known) discharge curve it looks like we are.
4. Current draw. If the device has a built-in ammeter so that it can measure how much current is being drawn, and if it knows (perhaps based on historical information discovered by #3) what the battery's capacity in amp-hours is, it can make a very accurate estimation of how much life there is left. Of course, that estimate can and will vary if the current draw changes. I believe that's one reason why the bar can go back up. For example, I regularly notice my laptop's expected lifetime jump back up just after I stop doing something CPU-intensive.

Of course, this is all complicated by real-world considerations. Rechargeable batteries have a limited number of charge-discharge cycles, and can't hold a charge for as long the more cycles they've gone through, and are also prone to notorious "memory" effects. (I don't know how hard microprocessor-based batteries work to assess these effects; though they certainly could.) Also, most batteries show a sort of "rejuvenation" effect when they've been given a chance to rest after working hard (i.e., just like you or me). That's the other possible explanation for the bar-going-back-up phenomenon you noted. —Steve Summit (talk) 23:04, 10 November 2007 (UTC)

Thanks for the replies. Googling gives (what seems to be the terminology for this sort of thing) State_of_charge. Some websites and the wiki article say that in laboratory conditions they measure the concentration of the electrodes/electrolytes The results also include the microprocessor thingi. It seems to be called Charge_controller device. 59.93.9.69 04:35, 11 November 2007 (UTC)

Freezing Water Question

What would happen if you tried to freeze water it it were confined so it could not expand? For instance if a quantity of water were enclosed in a solid block of steel and the whole thing were subjected to low temperature and the water could not expand as it would when freezing, what would happen to the water? Would it stay liquid or freeze solid without expanding???? —Preceding unsigned comment added by 207.69.137.23 (talk) 05:57, 10 November 2007 (UTC)

Depends on where it is on the phase diagram. Freezing water in a confined space creates pressure on the order of several atmospheres (you can drive a go-cart with it) which may push it back into liquid phase, but again, it depends on both pressure and temperature. The way the ice crystals form is also dependent on this diagram, so you should really just check out the phase diagram article. SamuelRiv 06:11, 10 November 2007 (UTC)

Better yet, check out the responses from the last time we had this question. In brief, as the temperature is lowered, at first the water will remain liquid and generate an increasing pressure. If the temperature continues to be lowered, eventually it will freeze into a form (phase) of ice denser than the everyday kind. --Anonymous, 06:16 UTC, November 10, 2007.

anti inflammatory

do anti inflammatory medicine (gen-naproxen to be specific) affect your mood? —Preceding unsigned comment added by Morvarid rohani (talk • contribs) 08:08, 10 November 2007 (UTC)

In the list of side effects of naproxen (see here), no mood disorders are listed as frequently reported. However, depression has been reported in 1% to 10% of patients, and anxiety has been reported in <1%. Reports do not necessarily indicate causation. The fact that a drug causes a side effect in some patients does not mean that it is the cause in a specific instance; any question of whether a drug is responsible for a specific clinical condition should be discussed with a physician. - Nunh-huh 08:45, 10 November 2007 (UTC)

Bullets

Hi. How exactly does a bullet to the brain kill? Why do some people shoot the mouth, yet others shoot the temple? At what point does life cease? 203.124.2.43 11:44, 10 November 2007 (UTC) Adam

I used to have a great animation of a bullet going through the brain, bouncing off the other side of the skull and basically making a soup of the brain matter after bouncing around several more times, but I can't find it. When a bullet enters one side and exits the other, the entry and exit take a lot of compressed gases and brain matter with them making a small explosion in those areas, which can greatly increase trauma (that mostly depends on the bullet head shape). Sometimes bullet injuries leave people alive, like Manfred von Richthofen (the Red Baron) and Phineas Gage (an iron rod, not a bullet). In Phineas's case, the spike damaged mostly one area of the frontal lobes, which govern a lot of higher-order reasoning and personality, but not so much in terms of low-order processing.

So let's get to the meat of your question: assuming a pointed bullet with enough speed to not bounce around the brain so that extra trauma is minimized, how do you kill someone? Shooting through the temple kills mainly by hitting the limbic lobe, which contains a lot of mid-order processing (thalamus), memory, and important regulatory glands. It is a guaranteed kill in some sense because you can destroy everything a person perceives about the outside world, whether or not their heart actually stops immediately. Shooting through the mouth is more appropriate because on the other side lies the brainstem and cerebellum, both of which control low-order function like breathing and heartbeat, with the brainstem being the connection of the brain to the body. So the kill in this case would be roughly instantaneous, though actual brain death would occur a bit later, because there would be some latent blood flow.

One more thing - severe brain trauma results in a couple of defense mechanisms by the brain. One is the coma, which for most bullet injuries would set in quickly, so the person would be incapacitated in any case. The other is a release of something (I forget what - I believe glucose) into the cell ether which results in a massive killing of brain cells. My neuroanatomy professor referred to it as a "self-destruct mechanism", but we don't know yet why it exists. Regardless, that can easily make brain trauma much more damaging than that of the actual impact. External links: [12] and [13] SamuelRiv 14:07, 10 November 2007 (UTC)

• 1. A brain can't absorp the shock of the bullet's entry without sustaining damage. 2. If that doesn't kill, bleeding will. - 87.211.75.45 19:32, 11 November 2007 (UTC)

Gold-labelled antibody = "fusion protein"

Is it appropriate to cover protein-non-protein ligations in the fusion protein article? I was just starting an article called conjugation (biochemical) and want to determine whether its warranted or not. Conceivably, one article should cover conjugation of proteins with other proteins, non-protein molecules, and possibly even non-protein-non-protein conjugations (none spring to mind). Perhaps the article protein engineering is more suitable for this? In which case, I can add a link at the conjugation disambiguation page? --Seans Potato Business 13:17, 10 November 2007 (UTC)

Wikipedia probably has many stubs such as Radioiodinated serum albumin and Biotinylation that could be collected into Protein labeling. I suggest not placing most "protein labeling" methods in the fusion protein article, but some fusion proteins are used to attach a label to target proteins. Sadly, Green fluorescent protein only seems to have an external link for the important topic of GFP fusion proteins. Maybe Conjugated protein could be expanded to include both natural and artificially conjugated proteins. I'd leave protein engineering to itself. --JWSchmidt 14:52, 10 November 2007 (UTC)

(EC) As a biochemist, I've never seen the term 'fusion protein' used to describe a non-protein combination. 'Fusion protein' is a bit more specific; it implies that a change has been made (insertion, replacement, and/or concatenation) to the primary amino acid sequence of the protein. This runs the gamut from adding a little tiny His tag through humanizing an antibody to attaching a whole additional protein (like GFP or a second part of an enzymatic complex). I would include under the 'fusion protein' definition the addition of domains that will dock particular prosthetic groups (a domain that binds a single metal atom, for instance).

For something like a gold-labelled antibody, you could use exactly that term. 'Immunogold', 'gold-tagged antibody', and even 'gold-conjugated antibody' come up a fair bit, too. For a general term to describe 'sticking something interesting to a biomolecule', 'conjugation' is probably as good a word as any.

As an aside on the topic of 'non-protein-non-protein conjugations' you need to remember the other important classes of biomacromolecules: DNA, RNA, and polysaccharides. All of them can be (and often are) modified with various sorts of labels (fluorescent, radioactive, immunogenic) to allow them to be studied. Conjugates that modify their function are also used sometimes, though this is perhaps less common. On a terminology point, the DNA equivalent to a 'fusion protein' would probably be 'recombinant DNA'. TenOfAllTrades(talk) 15:11, 10 November 2007 (UTC)

So perhaps conjugation (biochemical) would be a suitable umbrella for all conjugations, protein and otherwise. I don't have time for it now, but eventually... --Seans Potato Business 23:36, 10 November 2007 (UTC)

• I think that should read Conjugation (biochemistry) (so it avoids adjectives in the brackets). - 87.211.75.45 19:29, 11 November 2007 (UTC)

Neurotransmitters

In "Talk:Neurotransmitter#Neurotransmitter effects", I have described a recent experience involving dopamine, norepinephrine, and serotonin, attempting to link symptom clusters with specific neurotransmitter changes and extremes. Does my interpretation appear correct? Which receptors appear to have been preferentially overstimulated or understimulated? 66.218.55.142 15:13, 10 November 2007 (UTC)

DNA sequence around integrated HIV viral DNA attachment site?

Can anyone point me to books, published articles or other currently available research data which describe exactly what is the base pair sequence of integrated HIV viral DNA - I mean after the integrase process is completed, what is the DNA base pair sequence around those "attachment" sites? For example, what is the sequence when DNA in cell is cut in order to integrate viral DNA segment:
--TG...and_here_comes_HIV_DNA...CA--
--AC............................GT--
(This above is just example of what I'm looking for, those TGAC.. are just for example!) And then once the whole thing is integrated, what are those first few base pairs around both attachment sites? For example:
--TG...(U3RU5---HIV-DNA---U3RU5)...CA--
--AC...............................GT--
So, what are U3 and U5 of linear terminal repeat (LTR) attached to once the segment is integrated (the sequence of first few base pairs on both sides) and what does it look like together with U3 and U5 on both sides?? MANY THANKS to anyone who can point me to literature which covers this. --80.95.231.124 16:13, 10 November 2007 (UTC)

Have you seen HIV structure and genome? The literature at the end might be useful. SamuelRiv 17:34, 10 November 2007 (UTC)

Rocket speed

After how long does a space rocket reach 500 km/h and 1000 km/h ? Is there a graph of speed/time for rockets somewhere? I'm not very good with equations and I couldn't find the ones I thought necessary in the article rocket. Keria 17:54, 10 November 2007 (UTC)

Well, obviously, the time dramatically varies depending on the type of rocket. Something like the AIM-9 air-to-air missile accellerates at over 20g's. The space shuttle accellerates at a more pedestrian 3g's - so it piles on speed at roughly 30 meters per second every second. So we convert 500km/h into meters per second (500,000m/3600seconds = 138 meters/second) - takes roughly 138/30 seconds...4.6 seconds! To get to 1000km/h takes only 9.2 seconds. Well, in reality it's a lot more complicated than that - the various rockets start off accellerating fairly slowly because of the weight of all of the fuel - as the fuel burns off, the accelleration builds up - but at some point before the shuttle leaves earth's atmosphere, the speed becomes too high and the engines have to be throttled back to relieve the pressure on the spacecraft. Then as they get higher up and the air is thinner, they can go faster. When the SRB's detach, the spacecraft gradually slows down until (as yet more fuel is consumed), the engines can push the accelleration up again. Finally, the g forces start to get too large and again the engines have to be throttled back. Hence, in reality, the total time is going to be longer than 9.2 seconds - but still, it's piling on speed pretty amazingly fast. However, that AIM-9 rocket adds 200 meters per second every second - so it hits 500km/h within about two thirds of a second and 1000km/h in one and a half second! SteveBaker 18:23, 10 November 2007 (UTC)

You're a hell of a pedestrian. Hope I'll never run into you. —Preceding unsigned comment added by 84.187.90.130 (talk) 00:35, 11 November 2007 (UTC)

Robot Cars

What exactly is an autonomous robot car and how does it work? Could it work for a smaller or toy car? —Preceding unsigned comment added by 68.120.224.217 (talk) 19:27, 10 November 2007 (UTC)

• See DARPA Grand Challenge and links. --Sean 20:35, 10 November 2007 (UTC)

Currently, they take a standard car and add levers to push on the pedals and shift the shifter - also something to turn the steering wheel - those are hooked up to a computer that can therefore drive the car just like a human would. The tough part is that the computer has to be able to see where it's going and have enough intelligence not to do anything stupid like ramming the car into a brick wall. For this they use a combination of digital video cameras, and laser range finders (see Lidar - like Radar but with light instead of radio waves). Each individual technology is quite well known and understood - the tricky part is getting them all to work well together. We could make toys that did this - but the lidar sensors and the amount of computer power needed would be quite significant. But toys like the Aibo robotic dog are fairly sophisticated and can do some quite impressive things. It'll happen sooner or later. SteveBaker 00:24, 11 November 2007 (UTC)

You can do a lot with Lego Mindstorms, too. (I think; I still haven't given in and gotten a set for myself yet.) —Steve Summit (talk) 02:36, 11 November 2007 (UTC)

I have a few of the older sets - the newer ones are simultaneously better and less good for complicated reasons. But yes - you can easily build a "blind" robotic car using Mindstorms - but one that senses it's world and reacts accordingly (such as the ones in the Darpa Challenge) is MUCH harder and isn't really possible with the limited sensors that Mindstorms provides. It's relatively easy to do things like building a car-like robot and having it follow a black line made with electrical tape - or having it seek out light and park itself in a puddle of sunlight on your livingroom floor. But driving around in a complex environment without colliding with things and mowing down pedestrians...there is no way that a Lego Mindstorms machine could do that. It's a lot of fun though - and it's about on the limit of how complex something can be and still be considered a "toy" that kids could build and program. The most fun I ever had with it was making a pair of identical robots that could play "tag" in a darkened room. Each one had a bright light on the top and a 'bump' sensor that told it if it had been run into. One robot was programmed to seek light and the other to avoid it. When the 'bump' sensor detected that it had collided with something, the robot that was seeking light would send an infra-red message to the other robot saying (essentially) "You're It!" and switched from seeking light to avoiding it. When the other robot received a "You're It!" message, it would stop still, beep ten times at one second intervals and then switch from avoiding light to seeking it. The result of two of these little guys scurrying around the room was just hilarious to watch - but (and this is the ENTIRE problem here) the system only worked if neither of them went behind the sofa (killing the light seeking/avoiding behavior) or if they bumped into something else other than the other robot. I was able to add some sophistication to kinda/sorta fix those problems - but it rapidly becomes obvious that you need something better than a simple directional light sensor. Having a camera and lidar would help a lot! Some enterprising people were able to make a very crude lidar-like system using the infrared messaging system to send a message and looking for IR reflections in the light sensor. The strength of the returned reflection enabled some limited range measurement - but it was very crude. SteveBaker 02:51, 11 November 2007 (UTC)

Jupiter (the planet)

How was Jupiter formed and how old is it? —Preceding unsigned comment added by 72.38.227.206 (talk) 22:07, 10 November 2007 (UTC)

You can read about it at Jupiter. —Steve Summit (talk) 22:21, 10 November 2007 (UTC)

...which, I'm sorry, doesn't really answer your question. Anybody have a better reference? (There's a bit of information at Solar System.) —Steve Summit (talk) 22:27, 10 November 2007 (UTC)

Planetary formation (which, for some reason, is currently a redirect to "Nebular hypothesis") may answer this and the following question in more (perhaps too much) detail. In any case, the best one can say about the ages of the planets is that, as we currently understand planet formation, they're all about equally old, and slightly younger than the Sun itself. The reason it's hard to even define an exact age for a planet is that, after the initial aggregation of small planetesimals from the protoplanetary disk, they are believed to have undergone an "oligarchic growth" phase where the proto-planets of various sizes grew by colliding and merging at random. Thus, it's hard to say which collision was the one in which the resulting planet became the one we know today. There's no specific start or end to the oligarchic growth phase either; while the biggest planets eventually settled into more or less stable and non-colliding orbits, the smaller, more numerous bodies continued to collide with them and each other at gradually decreasing frequencies, as they still do to this day.

Also, it's worth noting that, for gas giant planets like Jupiter, there are two competing theories of their formation. The one that, as far as I can tell, seems to be enjoying the greatest popularity at the moment is that they started out like the smaller planets, as solid bodies growing by collisions, but eventually grew big enough that their gravitational pull could directly capture and hold down gas from the surrounding protoplanetary disk, leading to runaway growth as more and more gas fell down upon the initial rocky core. This didn't happen with the inner planets because, by the time they grew big enough, the solar wind had already blown most of the gas away from that part of the solar system. The alternative hypothesis, however, is that the gas giant planets formed directly from the gas disk without any solid "seed"; according to that theory, an eddy in the protoplanetary disk simply pulled enough gas together that its mutual gravitational pull caused it to collapse, much as stars are believed to form. Neither hypothesis is by no means disproved yet — it's even possible that gas giant planets can and do form by both mechanisms. —Ilmari Karonen (talk) 00:07, 11 November 2007 (UTC)

The planet Mars

How old is Mars? —Preceding unsigned comment added by 72.38.227.206 (talk) 22:15, 10 November 2007 (UTC)

You can read about it at Mars. —Steve Summit (talk) 22:21, 10 November 2007 (UTC)

...which, I'm sorry, doesn't really answer your question. Anybody have a better reference? (There's a bit of information at Solar System.) —Steve Summit (talk) 22:27, 10 November 2007 (UTC)

I presume Mars must have been formed at roughly the same time as the Earth - roughly 4.5 billion years ago. (See Formation and evolution of the Solar System and Age of the Earth.) We haven't studied the geology of Mars well enough to know for sure. But the age of the Earth is something of a compromise between the age of the oldest rocks we can find (3.9 billion years) and the age of the solar system (4.6 billion years). It's very likely that even after we have crawled over every inch of Mars looking for old rocks, our answer will be about the same. SteveBaker 00:17, 11 November 2007 (UTC)

Part of the problem with rock ages is that rocks form and reform due to subduction of the Earth's crust and the much different environment of early Earth, so today's rocks are going to be younger than Earth as a whole. Also note that most of our aging techniques won't work on liquids like magma. SamuelRiv 01:27, 11 November 2007 (UTC) Addendum: note there is no plate tectonics on Mars, though there is volcanism. Thus somewhere deep under the surface may be a rock as old as the solar system. SamuelRiv 01:28, 11 November 2007 (UTC)

I have some quibbles with that. Firstly, we know for sure that the young Mars had volcanoes. Olympus Mons - for example - is the largest extinct volcano in the known solar system! If Mars once had a liquid core sufficiently close to the surface to allow volcanism then it seems entirely possible that it once had plate tectonics too. Secondly, despite the ravages of plate tectonics, the date of 3.9 billion years for the minimum age of the earth comes from dating zircon crystals from Western Australia - so despite subduction and volcanism - we still have some pretty old rocks lying around at the surface where we can find them. SteveBaker 02:30, 11 November 2007 (UTC)

I addressed the issue of volcanism, and note the volcanoes on Mars are all shield volcanoes (like Hawaii), and thus are not dependent on plate tectonics. I'm aware that very old rocks have been found on Earth - I'm just addressing why they might be hard to find, and indeed why one as old as the Earth may be impossible to find. SamuelRiv 16:23, 11 November 2007 (UTC)

Cause and Effect

Is it right that not one thing in the Universe hasn’t obeyed the Laws that govern it? If EVERYTHING obeys the rules of Cause and Effect is it not true that everything happens for a reason? If this is true it would seem that nothing can be random and everything is predicatbale. Sorry if this is the wrong place to ask such a question. —Preceding unsigned comment added by 71.150.248.92 (talk) 22:34, 10 November 2007 (UTC)

This topic has dogged philosophy and religion since time out of mind. There's a bunch of stuff on the topic at Causality, Determinism, and Free will. It seems obvious that events are caused by other events. I wouldn't be typing this right now had you not asked the question. But it seems just as obvious that we are not robot-like automatons. I feel as if I am making choices about the words I type. The conflict between these two obvious things is, I think, at the heart of most religions and much of philosophy. Personally, I think that something must be wrong with the question, since people have been asking it for thousands of years and still end up in paradox. Pfly 22:57, 10 November 2007 (UTC)

For many interesting physical systems, predictability is not possible because even small variations in where you are now can result in large changes in where you will be. See Chaos theory. --JWSchmidt 23:29, 10 November 2007 (UTC)

Well, let's be really careful here:

• Is it right that not one thing in the Universe hasn’t obeyed the Laws that govern it? - Yeah - that's right. If some 'law' is ever disobeyed then it isn't a law...so this is true by definition. Is it the case that some of the things we humans believe are 'laws' are wrong? Yes - it's really unlikely that everything we think we know is 100% correct. Newton's "laws" of motion were proven wrong by Einstein's relativity.
• If EVERYTHING obeys the rules of Cause and Effect is it not true that everything happens for a reason? - "Cause and Effect" is not a law. To the contrary, many quantum effects do not follow 'cause and effect' and are fundamentally unpredictable. If you irradiate an atom and stuff some extra neutrons into it - it will eventually decay back down to it's original state by emitting a neutron. When will that happen? Well, we don't know, we cannot know - it's truly, utterly random. Where is the "cause" of that neutron being emitted? There isn't one.
• If this is true it would seem that nothing can be random and everything is predicatbale. - To the contrary, at its heart absolutely everything is completely random and unpredictable. The only thing that makes the universe seem stable and follow nice cause-and-effect rules is the effect of statistics. We don't know when one irradiated atom will decay - but we know with great precision how a kilogram of atoms will decay. We can't know the exact position of an electron - but we can deduce the position of a planet orbiting a star a hundred light years away by measuring the tiny wobble it induces in that star. The universe on the large scale obeys rules - but at the small scale, it's truly random. But the large scale effects are pure statistics. It is perfectly possible (although exceedingly unlikely) for a grand piano to appear out of nowhere in your living room right now. It's only statistics that enables us to say that this "Won't ever happen".

Look at it like this, if you flip a coin, you have no idea whether it'll come up heads or tails. If you flip 100 coins, you can be pretty sure that between 40 and 60 heads will show up - but predicting that you'll get 50% heads is a bit 'iffy'. If you flip a million coins, you can be quite sure that between 499,000 and 501,000 heads will show up - so a 50% prediction is a fairly accurate 'law'. If you flipped as many coins as there are atoms in a grand piano, your prediction of 50% heads would be precise to within one part in a billion billion billion (probably much better than that actually). In effect, you have a cast iron "law" of nature that says "when you flip coins you absolutely always get exactly 50% heads" - but that's not even close to being true for four coins - and it's POSSIBLE to flip a million coins and for them all to come up heads...it's just so unlikely that on a large scale, it's not going to happen. That's how our "large scale" laws operate. They are so accurately true that we can rely on them - even though at their heart, they are relying on completely random events.

• Sorry if this is the wrong place to ask such a question. - This is the perfect place to ask this question!

SteveBaker 23:58, 10 November 2007 (UTC)

Thanks for all the replies. Very interesting answers. —Preceding unsigned comment added by 71.150.248.92 (talk) 02:45, 11 November 2007 (UTC)

I have read alot of extremely well-researched, wonderfully articulated replies on the Science Desk, but SteveBaker's reply to this ultimate of all quandries is truly amazing. The idea that science, even at its finest, boils down to statistics, and that those statistics when viewed from an appropriate macroscopic level may be termed as "laws" is an idea I've been aching to come across. Thank you! Sappysap 04:05, 11 November 2007 (UTC)

Er...wow! Well, thanks! Before we get carried away though - the critical 'take away' point here is that while these macroscopic laws are "only" statistical, the magnifying effect of the sheer quantity of particles on the certainty of the result makes the resulting law quite utterly cast-iron. You cannot and must not take from my explanation the idea that the macroscopic laws are broken routinely because of this statistical stuff. On human scales - they absolutely are not. The probability of anything measurably different from what we expect actually happening is so astronomically small that this makes it impossible for any practical measure whatever. So "certainty" is still present at our scales. But when we deliberately make the small scale visible on the large scale, weird stuff can happen. Listen to the individual clicks of a Geiger counter picking up background radiation (Image:Geiger calm.ogg for example) - each click is the result of the decay of a single atom producing a single neutron. Guess what? It's utterly random - you can clearly hear that - there is fundamentally no way to predict when the next click will happen. SteveBaker 05:43, 11 November 2007 (UTC)

Well, I'm not listening to a Geiger counter per se, but just now I happen to be listening to "Radio-Activity", by Kraftwerk, which starts out with the sound of one. Weird coincidence, or subtle causality? (You be the judge. :-) ) —Steve Summit (talk) 06:01, 11 November 2007 (UTC)

Responding here to Steve's claim that "certainty is still present at our scales" except when "we deliberately make the small scale visible on the large scale". I don't believe that's true. The effect of quantum indeterminacy can easily be blown up to a macroscopic scale by processes not requiring our deliberate intent, by any system that chaotically amplifies small differences with positive feedback. Like weather. So I think, for example, that the question "will it rain in Dallas on the afternoon of July 17, 2063?" is truly non-determined; quantum uncertainty now, on a microscopic scale, will have been amplified to different macroscopic answers by then.

He's certainly correct that there are questions we can ask where the answers are quite deterministic for practical purposes. But those are the questions where small differences tend to cancel out, rather than being amplified. --Trovatore 22:00, 11 November 2007 (UTC)

You're talking about chaos theory. The thing about that is that it produces infinite sensitivity on initial conditions. Chaotic events would be unpredictable as a practical matter (and perhaps as a theoretical matter too) no matter whether quantum-level randomness existed or not. SteveBaker 01:53, 12 November 2007 (UTC)

But the point is that, because of the quantum-level stuff, whether it will rain in Dallas on that afternoon is (I think) unpredictable even in principle, because not enough information exists to determine it. That's different from deterministic chaos, where the only issue is whether you have enough computing power available to run the simulation. --Trovatore 02:00, 12 November 2007 (UTC)

NO! That's not true. It's not about computing power. I can explain this but we need a simpler concrete example. The weather is too complicated to discuss - let's talk about a simpler (but still deterministic and still chaotic) system. This is one of my favorites because it's easy to imagine:

First, the equipment: Take a couple of small, strong magnets and place them about six inches apart in the middle of a large sheet of paper your desk. Now hang a magnetic pendulum bob a few inches above the magnets on the end of a nice long string suspended from the ceiling. You also need a red and a blue crayon. OK - so here is the experiment:

Hold the pendulum over some point on the paper, release it and after it swings around a bit and if the magnets are strong enough, the pendulum will end up hovering over the center of either one or the other magnet. Colour that 'release point' with a small red dot if the pendulum ended up over the right-hand magnet - colour it blue if it ended up over the left-hand magnet. Repeat this experiment for every point on the paper so it's completely covered in red and blue dots.

So what red/blue pattern results? Well, when you release the pendulum near the right magnet the result is always that the pendulum swings immediately over that magnet - so there is obviously an area around the right magnet that ends up red, and an area around the left magnet that ends up blue. Now, if you hold the magnet a bit further off to the right - beyond the righthand magnet, the pendulum will fly right over the righthand magnet and swing over to the lefthand one - by then it's lost enough energy due to air resistance that it'll stop over the lefthand magnet. But you can imagine that from some places the pendulum loops around one magnet then the other crazy swings until it finally loses energy and winds up over one or the other.

So you can imagine a fairly complex pattern of red and blue on the paper.

Now this magnetic pendulum setup is 'chaotic' (in the same way that the weather is). If you crunch the math on this, the actual mathematical pattern you wind up with is a fractal - something like the Mandelbrot set. There are regions of our red/blue pattern that are in big solid patches (like immediately around each of the two magnets) - but there are regions where the red and the blue is all mixed up in whorls and patterns of great complexity. Write a computer program to generate this pattern and you can zoom into these patterns and you see more patterns, you can zoom in deeper and deeper into that map - and in some areas you'll keep on getting more and more red/blue patterns no matter how tightly you zoom. The image is infinitely complex...fractal...chaotic.

What is the physical meaning of this infinite complexity of red and blue dots? It means that if you start the pendulum over one of those chaotic regions and release it, the magnet it will end up over is very sensitive to where you started it from. Move a millimeter to one side and the answer may be different. Move a millionth of a millimeter to one side and the result will be different, move the width of a hydrogen atom to one side and you'll get a different answer. In fact, move an INFINITELY SMALL distance to one side or the other and the pendulum may end up over the other magnet. The result is that moving the pendulum by (1/infinity) meters can change the answer...but (1/infinity ) is zero (well, kinda - mathematicians might argue it 'approaches' zero - but the result is the same)...so if you move the pendulum by zero distance, the answer can change. It's deterministic - in that there is no randomness in the equations - but you need infinite precision to calculate it - and even if you had that, you still wouldn't get the right answer because displacing the initial position of the pendulum by 1/infinity meters changes the answer.

CONCLUSION: You don't need quantum effects to get a random answer - you don't even need an inaccurate measurement of the initial position because an error of (1/infinity)% is enough to change the answer. The result is independent of computer power or precision.

SteveBaker 15:46, 12 November 2007 (UTC)

Um. The boundary between the red and blue regions may well not be computable. But I think you'll find, in the deterministic idealization of the pendulum problem, that for any ε greater than zero, there's a computer program and a finite amount of information about the initial conditions that -- given sufficient time to run and resources -- would give you the right answer for at least a proportion of 1-ε of the space of possible initial conditions. Which is not true for the quantum version of the same problem.

Computability is beside the point here anyway (or at least, beside my point). In the deterministic case, there exists a function, whether you can compute it or not, that takes the initial conditions and returns the final answer. In the quantum case, no such function even exists,computable or not. --Trovatore 19:47, 12 November 2007 (UTC)

That's true - but you miss the point. For any given starting point, even if you could compute whether to label it red or blue, the problem is that in some areas of the map, the individual red and blue dots are INFINITELY small. So the information from some theoretical computation would be utterly useless because nothing can be positioned infinitely accurately (irrespective of quantum theory). For every red dot in one of these chaotic regions, the distance to the nearest blue dot is zero. But worse, the answer isn't computable because to arrive at the right answer you need infinite precision - that requires an infinite number of bits of storage and (on a finite computer) infinite computation time.

Chaotic systems are deeply weird - even in a 'classical' universe. In a practical quantum universe, there is clearly no way to position anything with anwhere near enough precision and still keep it stationary because of the uncertainty principle. But Newton and Einsteins idea of a 'clockwork universe' where everything is ultimately predictable is blown away by chaos theory every bit as efficiently as by quantum theory. Better actually because you can show the correctness of chaos using pure mathematics. Quantum theory still requires experimental evidence and Einsteins idea that there might be an 'underlying certainty' is really tough to disprove. SteveBaker 21:22, 12 November 2007 (UTC)

SteveBaker 21:22, 12 November 2007 (UTC)

Are you claiming that, in the deterministic version, the intersection of the topological closure of the red set, with the closure of the blue set, has positive Lebesgue measure? I can't refute that off the top of my head but I think it's most unlikely. Whereas if that intersection has measure zero, then the up-to-epsilon computability that I mentioned earlier would be true, and would probably match the "clockwork universe" idea well enough (though admittedly in a completely impractical way). --Trovatore 22:53, 12 November 2007 (UTC)

I don't know whether there is a positive Lebesgue measure - but that's not necessary. We don't know whether the Mandelbrot set has a positive Lebesgue measure either. That doesn't prevent it from being infinitely crinkly which is all that is needed in order to guarantee that for some areas of the diagram, there are areas where we have infinite sensitivity to initial conditions. SteveBaker 00:15, 13 November 2007 (UTC)

Sure, but by itself that's not very interesting. Balance a pencil on its point -- will it fall to your left or to your right? Even in the deterministic version of the problem there's a boundary point, where you have your infinite sensitivity. But for almost all (in the sense of Lebesgue measure) possible positions of the pencil, in the deterministic version, you can predict where the pencil will wind up, given sufficiently accurate measurement. I don't think this would have upset Newton or Einstein too much. --Trovatore 00:28, 13 November 2007 (UTC)

That example isn't quite the same as this case. With the pencil, it will fall in the direction of the error in it's initial position - the consequences of even an infinitesimal error are extremely predictable. With the magnets and pendulum case, you can't predict where it'll end up for an infinitesimal error. SteveBaker 15:25, 13 November 2007 (UTC)

Well, but the point is that you can predict it (in the deterministic version of the problem), except at a boundary point (that is, a point that's neither in the topological interior of the red set nor of the blue set). Of course how small the error has to be depends on how close you are to the boundary. (Note that if you're not on the boundary, then your distance to the boundary is nonzero -- that's because the boundary is closed.)

So the question is, how often are you at a boundary point as opposed to an interior point? My conjecture is that the boundary has measure zero and is nowhere dense. Since the boundary is necessarily closed, the latter condition is the same as saying the boundary has empty interior -- there isn't any region containing an open set such that the infinite sensitivity obtains over the whole region.

But the quantum case is different. If you take these extremely-finely-divided, but still cleanly separated, red and blue regions, and smear them by the tiny uncertainties imposed by quantum mechanics, you see that there are regions with nonempty interior where what you have is genuinely purple. --Trovatore 16:56, 13 November 2007 (UTC)

From the question: To the contrary, many quantum effects do not follow 'cause and effect' and are fundamentally unpredictable. Maybe my understanding was wrong, but this isn't how I interpreted quantum mechanics. Rather, as I learned it, the apparatus for measuring the effect operated on the cause and therefore it was impossible to distinguish the two cause and the measurement. The proccess of observing affects what is being observed. In quantum mechanics it really did matter if there was someone there to hear the tree fall in the forest. Entanglement and the wave/particle duality experiments highlight this. --DHeyward 07:17, 13 November 2007 (UTC)

November 11

Physics and Ancient Greece

I remember my grade school science teacher once telling us that Ancient Greece not only knew the world was round, but more amazingly they also knew its rough diameter. The ancient experiment took place on a wide and very flat plain; on it, two towers spread miles apart. The towers would signal each other in daylight and then quickly measure how long the shadows of each tower fell on the plain. That was all they needed to roughly measure the size of the world. I liked this science teacher quite alot and so I almost hate to ask: is an ancient experiment of this kind roughly possible? A lesser question: without too much research, is there any historical corrobaration for the rough diameter of the world being published more than two thousand years ago? Sappysap 00:16, 11 November 2007 (UTC)

Eratosthenes is the first known to have calculated the circumfrance of the earth. He did conduct an experiment very similar to what you describe to estimate the curvature of the Earth. His estimate was probably a bit off but pretty good for a first shot! You can look at History of geodesy for more discussion of his method and later methods. --24.147.86.187 00:26, 11 November 2007 (UTC)

Yep - the numbers that these techniques could come up with would be very approximate - but they could certainly prove that the earth is round simply by noting that the further North you went, the longer the noon-time shadow is - but the length of the shadow doesn't change when you go East/West. I don't think they used two towers that could see each other though. What I thought they did was to measure the length of the shadow at noon for two locations hundreds of miles apart on the same day. Because they knew (roughly) the North/South distance between those locations, they could do the calculation without requiring communications between the towers. They knew that at noon the shadow was at it's shortest - so precise timekeeping was not needed. Not requiring the two towers to be able to see each other means that they can be a lot further apart - which makes the whole calculation much more accurate. But the observation that the length of the day is more variable at higher latitudes is theoretically enough to allow an observant person to deduce that the earth must be round. SteveBaker 00:42, 11 November 2007 (UTC)

According to the Eratosthenes article: "The exact size of the stadion he used is argued by those who suppose he got it right; but the common Attic stadion was about 185 m, which implies a circumference of 46620 km, i.e. 16.3% too large."

I would consider an accuracy of 1/7 very impressive for someone who did this calculation over 2000 years ago.

Eratosthenes decided to use two cities instead of two towers. The Sun shined directly down a well in one city on the noon of every summer solstice, it was at the zenith at these times. In the other city, he measured the noon sun's elevation during another summer solstice. This would give the cities' latitude difference, and the Earth's circumference can easily be computed if the distance between the cities is known. --Bowlhover 07:24, 11 November 2007 (UTC)

That can't be exactly true. The sun only shines "directly" down a well when you are near the equator. Greece is at roughly 38 degrees north - the earth's axial tilt is 24 degrees - so even at mid-day on the summer solstice, sunlight would shine down the well at an angle of roughly 14 degrees to the vertical. In the end, the problem boils down to the precision with which you can measure the angle of the sun (or the length of a shadow or the depth to which sunlight penetrates a well - all of which amount to the same thing) - and the precision with which you can measure the north/south distance between your two points. Picking two cities far apart makes accurate measurement of the distance between them tough - but reduces your dependence on the accuracy of sun's elevation. Picking two points closer together gives you better precision on the distance between them - but the precision with which you measure the angle of the sun becomes vastly more critical. SteveBaker 14:27, 11 November 2007 (UTC)

Minor correction - you don't have to be exactly on the equator for the sun to be at the zenith at noon on certain days of the year - you only have to be between the Tropic of Cancer and the Tropic of Capricorn. One of the end-points of Eratosthenes baseline was Syene or modern-day Aswan, which is slightly north of the Tropic of Cancer. Our article says:

The latitude of Aswan – 24° 5′ 23″– was an object of great interest to the ancient geographers. They believed that it was seated immediately under the tropic, and that on the day of the summer solstice a vertical staff cast no shadow, and the sun's disc was reflected in a well at noonday. This statement is only approximately correct; the ancients were not acquainted with the exact tropic: yet at the summer-solstice the length of the shadow, or 1/400th of the staff, could scarcely be discerned, and the northern limb of the sun's disc would be nearly vertical.

Eratosthenes presumably knew this bit of "folklore" about Syene, and he took his home town of Alexandria as the other end of his baseline, so he could estimate the size of the Earth from measurements made in his own garden. Gandalf61 16:54, 11 November 2007 (UTC)

Yes, Aswan is in Egypt and not Greece. Using Aswan was logical because if the sunlight illuminated the bottom of a deep well, the Sun's altitude could accurately be determined as 90 degrees. If an accurate data point exists, why not use it? --Bowlhover 17:49, 11 November 2007 (UTC)

One more point: the Earth's axial inclination is not exactly constant, owing to tidal effects. In the time of Eratosthenes, it was a little larger and therefore the Tropic of Cancer was farther north, at 23°43'. Syene still wasn't exactly on it, but it was closer than would be the case today. --Anon, 06:33 UTC, November 12, 2007.

The funny part is that Posidonius repeated Eratosthenes calculation, via slightly different means, and came up with essentially the same result -- and a fairly accurate one at that -- but then revised his calculations such that the circumference was about a third to small. And this figure got incorporated into Ptolemy's Geographia which, along with medieval calculations by Islamic geographers that tended to support the smaller figure, greatly influenced Christopher Columbus's theory of Asia being located approximately where the Americas are. History is funny. Pfly 07:34, 11 November 2007 (UTC)

The bottom line: Eratosthenes and other greeks know perfectly how to estimate a sphere's diameter. The problem for them was that the instruments they had at their disposal were some dracmas and a man's footsteps. But apart from those minor details, they made perfect scientists. --Pallida Mors 76 22:51, 11 November 2007 (UTC)

One point often omitted in connection with Eratosthenes's estimate of the size of the Earth is that the calculation is only correct if the Sun is known to be at a great distance. Say A is Alexandria, B is Syene, S is the Sun, and C is the center of the Earth; at the time of measuring, SBC is known to be a straight line, so points SAC make a triangle with point B on side AC. The angle of 7.2° between the Sun and the vertical, as directly measured at Alexandria, is the supplement of angle SAC. When we say that AB forms an arc of 1/50 of the Earth's circumference, we are assuming that angle ACB is also 7.2°!

But in fact angle ACB = ACS = 180° - SAC - ASC = 180° - SAC - ASB = (supplement of angle SAC) - ASB = 7.2° - ASB. However, since the Sun is 93,000,000 miles away, angle ASB is only about 0.0003° and it is safe to treat ACB as having been measured as 7.2° also. If we believed that the Sun was only 25,000 miles away (and therefore about 230 miles in diameter), then angle ASB would be about 1.1°; we would then compute ACB as 6.1°, producing an estimated figure of 29,500 miles for the Earth's circumference. You could also get a 7.2° change of Sun angle over 500 miles if the Earth was flat and the Sun was about 4,000 miles up!

Fortunately, Eratosthenes did believe that the Sun was at a great distance, although it's not clear exactly how great, so his calculation went the right way on this. (Perhaps someone could work some of this into the article on him; I don't have time now.)

Moving onward from the way Eratosthenes actually did it, let's talk about the situation the original poster remembers the teacher describing: two tall towers in sighting distance across a level plain, with the length of each one's shadow observed at the same time, using light signals to synchronize their observations.

The first point is that towers in ancient times weren't built all that high. The biggest Egyptian pyramids were a few hundred feet. Even if a tower 500 feet high was available, the horizon would be only 28 miles away. Therefore even if we imagine that two towers of that height were used, they would have to be no more than 56 miles apart. This means that the important angle formed between the two towers and the center of the Earth would be about 0.8° instead of 7.2°, and therefore a good deal harder to measure with sufficient precision. If we imagine lower buildings,

Further, the story referred to using the shadow of the tower itself. In ancient times nobody was building vertical-sided buildings anywhere near that high. If a building is a pyramid, or in general any shape without vertical sides, you can't measure the length of its shadow by just starting at the base of one wall; you have to measure from a point below the highest part of the tower, presumably well inside the building. So this would introduce additional error. I don't know what the maximum height for a vertical-sided building was in ancient times, but if it was say 100 feet, then the maximum distance between two such buildings would be about 25 miles, requiring an angle of 0.4° to be measured.

Further, the story refers to a signal passed between the buildings using light, and this would have to be done in the daytime, so lighting a fire wouldn't serve. It is difficult to imagine what could be done fast enough to serve as a good signal that could be plainly seen 25 or 50 miles away without such a thing as a telescope. A mirror reflecting the Sun ought to be visible at that distance, but impossible to aim accurately so it could be seen at the other tower. Perhaps a big sheet of cloth like a sail could be hung over the tower and then let fall, but I don't think it would be practical to make one big enough, which would also have to contrast with the tower. At 25 miles a cloth 100 feet square would appear only 1/10 the width of the full Moon. And if the towers are at the maximum distance for sighting each other, then only the top bit of each tower is visible from the other.

--Anonymous, 06:33 UTC, November 12, 2007.

The OP's story sounds like a strange mix of Eratosthenes, the use of Theodolites in the Principal Triangulation of Great Britain (and of India, which doesn't seem to have a page), and Semaphores... or something. Pfly 09:13, 12 November 2007 (UTC)

Are you looking for Great Trigonometric Survey? —Steve Summit (talk) 14:58, 12 November 2007 (UTC)

Ah ha, yes that's it. If I recall right, the project sometimes involved the building of towers in sight of one another, upon which theodolites were placed to survey via triangulation methods. And interestingly enough, that page says the project was was responsible for the first accurate measurement of a section of an arc of longitude, which relates back, sort of, to the circumference of the Earth. Not quite Ancient Greece though. Pfly 16:37, 12 November 2007 (UTC)

That's quite the overstatement -- I daresay Jean Delambre and Pierre Méchain would have something to say about it! I mean, the Indian survey may well have been more accurate than their earlier work, but still! --Okay, fixed now. (See also "The Measure of All Things" by Ken Alder, and if you read French, fr:Figure de la Terre et méridienne de Delambre et Méchain.) --Anonymous, 06:14 UTC, November 13, 2007.

does inhibition of PKC-β activity decrease pigmentation in vivo?

how? is it permanent? —Preceding unsigned comment added by 81.99.212.22 (talk) 01:03, 11 November 2007 (UTC)

According to this article, PKC-β phosphorylates and activates tyrosinase and "topical application of a selective PKC inhibitor reduces pigmentation and blocks UV-induced tanning in guinea-pig skin". They also discuss regulation at the level of transcription. --JWSchmidt 01:17, 11 November 2007 (UTC)

Feces - taste like what?

I've always wondered what human feces (faeces) taste like.

I'm aware that this might be a strange question. I also realize that I could easily provide my own answer, but I'd prefer to leave it to someone with an interest in coprophagy. —Preceding unsigned comment added by 222.155.51.145 (talk) 07:40, 11 November 2007 (UTC)

Most of the taste sensation is due to the smell. SO I think you could imagine the experience. But you would have the texture and tongue taste as well. The average 1 year old probably can remember the experiencem(but not me)! Graeme Bartlett 10:10, 11 November 2007 (UTC)

Without having performed any OR on the matter, I can nevertheless confidently assert that they taste like shit. —Steve Summit (talk) 14:23, 11 November 2007 (UTC)

After digestion, very little that stimulates taste receptors should remain. Probably the only recognizable taste would be salt, but no saltier than blood. 66.218.55.142 14:40, 11 November 2007 (UTC)

I was under the impression that it had a spongey texture, although I cannot remember where I encountered this gem of information. [14] will perhaps help you on your enquiry. (Really, don't follow that link unless you have a strong stomach). Lanfear's Bane | t 15:08, 11 November 2007 (UTC)

For your sake and for mine, let us hope that that is peanut butter. Can I wash my eyes now? --Russoc4 16:47, 11 November 2007 (UTC)

Knowing the right term might help you: Coprophagia --Mdwyer 20:43, 13 November 2007 (UTC)

commercial radioisotope

What is a commercial radioisotope? —Preceding unsigned comment added by 144.137.98.219 (talk) 10:17, 11 November 2007 (UTC)

It is a radioactive isotope that you can purchase as a standard product, such as ¹³¹Iodine —Preceding unsigned comment added by Graeme Bartlett (talk • contribs) 10:52, 11 November 2007 (UTC) or tritium. Graeme Bartlett

asteroid belt

if some spaceship tries to cross the asteroid belt, then instead of by passing through it, can it go above it and cross the belt?SidSam 10:34, 11 November 2007 (UTC)

A spaceship could go above or below, but asteroids occasionally will orbit at an inclined angle and stray into that space too. The belt is actually very thin and the chances are that a spaceship will fly through and not be impacted. The problem is that more fuel is needed to go outside the eccliptic plane, and then back into it. Space ships always scrape by on the minimum fuel, so such maneuvers are unlikely unless there is a reason to go up and out there. Graeme Bartlett 10:56, 11 November 2007 (UTC)

I heard it's so widely spaced that if you were at one asteroid, you wouldn't even be able to see another asteroid with naked eyes. So passing right through it seems pretty safe. 64.236.121.129 15:08, 12 November 2007 (UTC)

Tbe asteroid belt is more like a flat ring around the sun - like one of Saturns rings - so you could certainly go around it. But the rocks within it are widely spaced - so it's pretty safe to travel through it as though it wasn't there. SteveBaker 14:04, 11 November 2007 (UTC)

The scatter-graphs at asteroid family may suggest how much like Saturn's rings the asteroid belt is not. —Tamfang 22:46, 12 November 2007 (UTC)

Aldol

In the aldol condensation of acetone, mesityl oxide is created. What then, if anything, prevents a further acetone molecule from adding on to the other side of the product, resulting in an endless chain of carbonyls? Or, what prevents the mesityl oxide from adding onto another molecule of acetone, creating a symmetric compound with another isobutene group? What about endless Michael reactions at the double bond? Could't you theoretically just keep making one infinitely larger compound, as long as there are alpha-hydrogens and/or alpha-beta double bonds? --Russoc4 14:56, 11 November 2007 (UTC)

Yes, that's called polymerization. Apparently it can be a serious problem when you don't want to make polymers, because it deactivates the catalyst.[15] [16] That second one says "The active sites for aldol condensation are the same as the active sites for polymer production.". There are even patented methods to inhibit the polymerization.[17] —Keenan Pepper 17:13, 11 November 2007 (UTC)

So then, disregarding any Michael reactions, these [18] would be the possible aldol condensation products? What do you think would happen if I added acetone to a concentrated NaOH solution and heated it? Would that polymer be a likely outcome? --Russoc4 18:01, 11 November 2007 (UTC)

But if you read the page on mesityl oxide, it points out that further condensation yields isophorone, which is a cyclic product. Six-membered rings like that of isophorone are pretty stable and will ressit further attack. Delmlsfan 02:02, 12 November 2007 (UTC)

Heisenberg t-shirt

I am pretty sure I have seen an image containing the Heisenberg uncertainty principle ($\Delta x\Delta p\geq \hbar /2$) and the text "free will". I thought it was a t-shirt or sticker from either xkcd or ThinkGeek, but can't find it. Does anyone know where I might have seen it? —Bromskloss 20:38, 11 November 2007 (UTC)

Not sure, but I'd like a shirt with that inequality and the text "mathematical consequence of a probability theory in the complex field" or maybe "If you get the math, it's obvious". SamuelRiv 22:06, 11 November 2007 (UTC)

Well, it's a trivial consequence of the canonical commutation relation, but is it obvious that the CCR must hold? Algebraist 22:46, 11 November 2007 (UTC)

In the plain math, it comes from the Fourier transform of a waveform, where you find position and momentum space have fundamental uncertainty in deriving one knowing the other. SamuelRiv 22:59, 11 November 2007 (UTC)

So you're claiming it's mathematically obvious that systems are described by wavefunctions? Algebraist 23:16, 11 November 2007 (UTC)

I'm almost positve you can cross XKCD off of the list. As far as I know, they have made a joke of everything nerdy except for the Heisenberg uncertainty priniciple. However, I did find another Heisenberg shirt. Paragon12321 00:16, 12 November 2007 (UTC)

It seems I'm out of luck here. Anyway, you have heard this one, haven't you?:

Police officer: "Have you any idea how FAST you were going back there?"
Werner Heisenberg: "Nope, but I knew EXACTLY where I was!"

:P —Bromskloss 08:26, 12 November 2007 (UTC)

I don't know about a tee-shirt, but as for comics, see Casey and Andy strip 128, and stripts 247 to 249. Related are 89, 67. – b_jonas 11:48, 13 November 2007 (UTC)

November 12

Seriously hurt in the news

We frequently hear in the news that somebody has been "seriously hurt", or even worse "herido muy grave" or the worst "en estado crítico" (I don't know how you would say that in English, normally somebody who is in "estado crítico" dies soon). What do those expressions mean in medicine? --Taraborn 00:35, 12 November 2007 (UTC)

See Medical conditions. These labels are used by hospitals when speaking to the press and are not used among doctors. They vary in meaning from hospital to hospital. --Milkbreath 00:49, 12 November 2007 (UTC)

Proving once again that WHAAOE. (But, dang, Milkbreath, ya beat me to it. *I* was just about to post that link!) —Steve Summit (talk) 00:56, 12 November 2007 (UTC)

Thank you, guys. --Taraborn 01:07, 12 November 2007 (UTC)

I think what you call "en estado critico" is, in english, "in critical condition". Just an FYI. Kuronue | Talk 21:06, 13 November 2007 (UTC)

Jupiter's moon Io

Is there a certain reason Io has volcanoes? —Preceding unsigned comment added by 72.38.227.206 (talk) 00:35, 12 November 2007 (UTC)

Well, I guess the answer is only really stated clearly in the very first paragraph of Io...you couldn't be expected to find something buried that deeply! SteveBaker 01:36, 12 November 2007 (UTC)

tidal stress--( Mulligan's Wake)(t) 01:37, 12 November 2007 (UTC)

I think that prior to IO being surveyed by the Vikings during the 70s, it was just assumed that small remote planets and satellites would all be frozen solid. The incredible phenomenon of IO suggests there could be life-friendly worlds far away from their host stars. Not IO, however. It is estimated that there the entire surface is continually subducting from the heat of tidal friction and is completely renewed every hundred years or so. Myles325a 03:49, 14 November 2007 (UTC)

Jeez! Those Vikings discovered everything! -SandyJax 16:49, 14 November 2007 (UTC)

stpm trial exam question

can ketone react with chlorine?thank you —Preceding unsigned comment added by 218.208.91.249 (talk) 02:04, 12 November 2007 (UTC)

Perhaps haloform reaction may be of some help. (EhJJ) 03:11, 12 November 2007 (UTC)

"brush harder, not smarter"

my dentist told me this. what does he mean? —Preceding unsigned comment added by 172.130.26.74 (talk) 02:08, 12 November 2007 (UTC)

Why don't you ask him?

The usual exhortation is to do something smarter, not harder -- that is, to do it with more care and deliberation, and less brute force. —Steve Summit (talk) 02:16, 12 November 2007 (UTC)

As Steve has pointed out, the usual phrase is "do it smarter, not harder". In this case, your dentist may be using a play on words to imply that you are not brushing hard enough. If you're not sure, you may want to ask him. (EhJJ) 03:14, 12 November 2007 (UTC)

You might be remembering it backwards, or your dentist might have said it backwards accidentally, or your dentist might have said it backwards intentionally. There’s no way we can know which is the case; you’ll really need to ask your dentist. MrRedact 07:50, 12 November 2007 (UTC)

Your dentist wants you to brush gently (not harder), and with a technique (smarter) that gently cleans effectively. If you use a hard brush and scrub, you'll damage your teeth and irritate your gums. If you gently massage your gums with a soft-bristle brush, and hold the brush at the right angles, you will prevent gum disease. --Mdwyer 20:40, 13 November 2007 (UTC)

It means that either you or your dentist are dyslectic. Myles325a 03:51, 14 November 2007 (UTC)

Bamboo nomenclature

72.50.180.145 03:54, 12 November 2007 (UTC)After reviewing your article on bamboo, I'm not sure what you call stalks of bamboo. Are they called stems? I've heard them called that before. Are they called branches? Are they called culms? which according to wikipedia- "originally referred to a stem of any type of plant". What is the appropriate scientific way to name them, because I'm confused.

A bamboo's shoot is like a tree's trunk, whereas a bamboo's stem is like a tree's branch. A culm is a stem of a monocotyledon, which bamboo is an example of. Someguy1221 05:06, 12 November 2007 (UTC)

Laxative agent in apple juice that's not in pear juice?

Here's the strange thing: I get diarrhea when drinking apple juice. Even a single glass causes discomfort, and multiple are certain to cause diarrhea (which passes quickly, however). I thought this could be oversensitivity to sorbitol, but I have no problems drinking even large amounts of pear juice, which likewise contains sorbitol. Ditto for any other fruit juice you'd care to name (though I haven't tried prune juice). I have no problem with raw apples either. It could be an additive of concentrated apple juice (I haven't tried "natural" apple juice either), but I wouldn't know which one.

Though apple juice really is the only thing giving me problems (so it's easy enough to avoid), I'm curious if anyone could guess what could cause diarrhea and is just about only found in (concentrated) apple juice -- a rare chemical or ratio of something? 82.95.254.249 07:57, 12 November 2007 (UTC)

Apples have traditionally "gone through" some people, especially if the apples are the first of the season. I suspect it may be malic acid that causes this, but I stress this is pure speculation. DuncanHill 14:24, 12 November 2007 (UTC)

Well, any hint is welcome, but I don't think it's malic acid. Tart apples are my favorite, actually, and I have no problem with other foodstuffs that contain malic acid (according to the page). 82.95.254.249 20:10, 12 November 2007 (UTC)

I'd take another look at Sorbitol. It has known laxitive effects (many sugars do). As for the difference between pear and apple juice, I can only speculate. Apples and pears are VERY similar. It is possible that the pear juice has been modified in some way? Perhaps it was pasteurized or cooked in some way that would have converted its sugars? --Mdwyer 20:37, 13 November 2007 (UTC)

I get a little stomacheache from eating an apple core, and also from fresh apple juice made from whole apples. I deduce that the core or seeds has something in it that disagrees with me but not the majority of people. I am vaguely aware that apple leaves are supposed to contain some poison also.Polypipe Wrangler 22:10, 13 November 2007 (UTC)

Apple seeds do contain a small amount of a cyanide compound. Snopes has the dirt on it. --Mdwyer 03:44, 14 November 2007 (UTC)

Pseudo-stable isotopes

For isotopes with half-lives longer than the known lifespan of the universe, such as Calcium-48, how do they determine that the isotopes aren't stable? --67.185.172.158 08:06, 12 November 2007 (UTC)

Simple: get a lot if atoms of that isotope and wait for them to decay. Remember that half-life is a statistical property: a half-life of X doesn't mean that no atoms will decay until X has passed. An unstable atom can decay at any time (which is exactly why we talk of "half-life" and not the full life).

The initial suspicion that an isotope is unstable comes from theoretical considerations, and obviously it's hard to determine the half-life for an extremely long-lived isotope with great accuracy. 212.178.108.2 10:55, 12 November 2007 (UTC)

In that case, the article cited in footnote 2 explains it. They thought it was a likely candidate (via theory) for a very rare form of decay, and then spent a lot of time watching it in a test chamber (via experiment) to see if it actually did decay in that way. And it did. Hooray. --24.147.86.187 17:00, 12 November 2007 (UTC)

A radiation counter can pick up a single atomic decay (so long as you shield it from background radiation) - if you have a mole of whatever isotope you have then you have 6x10²³ atoms. A mole is the atomic weight expressed in grams - so a mole of a solid or liquid is a pretty reasonable quantity to measure. The age of the universe is something like 4x10¹⁷ seconds - and a mole is 6x10²³ atoms - so in half the age of the universe, you'd expect 3x10²³ radioactive decay events - so you'd expect about half a million decay events per second. Easily detectable! More to the point of course, if this is a synthetic isotope (made in a nuclear reactor for example) so that it's "fresh", it'll be radiating a lot more than it will be when it's reached it's half-life. SteveBaker 18:51, 12 November 2007 (UTC)

For "a lot more" read "twice as much".

Calcium 48 makes up only 0.187% (1/535) of the atoms in natural calcium, so making a whole mole of the stuff (about 48 grams) would be rather expensive. And then, its half life is not a mere 4x10^17 seconds, but about 100 times that. But all this just means that the number of decay events you'd have to detect is smaller, not that there are too few to detect at all. No doubt the paper by Balysh et al., cited in the article, would have details of how large a sample was used and how long it was observed. --Anon, 23:26 UTC, November 13, 2007.

Penguins in the Arctic? Polar bears in the Antarctic?

Just checked the articles on Penguins and Polar Bears. Just wanted some confirmation, are there any penguins that live in the Arctic? Also are there any Polar bears in the Antarctic? 64.236.121.129 14:20, 12 November 2007 (UTC)

No, and no. DuncanHill 14:22, 12 November 2007 (UTC)

But they could live there, couldn't they? I mean, if you kidnapped some and dumped them there, they'd manage ok, wouldn't they (not that I advocate kidnapping well dressed animals or anything)? Jeffpw 14:44, 12 November 2007 (UTC)

No - Penguins are almost 100% confined to the southern hemisphere (they do creep just above the equator - but not by much) - but not all species live in snow and ice - some are very happy in the tropics. Polar bears are strictly Arctic. Whether you could transplant them depends on an awful lot of things. Bears live on the mainland as well as in the arctic ocean - if you dumped them in the antarctic, they'd have no way to make it do warmer areas in the winter - also the species they are used to hunting would not be present. Some penguins have these very complicated behaviors where one sex gathers together far inland to look after the eggs while the other partner goes off for food. It's not clear how this would work in a continuously moving sea of ice floes - do penguins navigate somehow? I'd be quite surprised if relocating either species would actually work out well. SteveBaker 15:03, 12 November 2007 (UTC)

• The polar bears would probably eat the penguins. Seriously. The kind of penguins that live in Antarctica generally are unable to recognize any land animal as a predator, because the only land animals that have ever attacked them in the last million years or so have been humans or escaped domesticated animals like dogs. If you took a polar bear to the Antarctic, the polar bears would very quickly gravitate towards this huge, meaty, docile food source. If you were to transplant some penguins to the Arctic, then even if they could solve the problem of figuring out which landmass to lay their eggs on, eventually the polar bears (which will eat any kind of animal if they're hungry enough) will figure out that they're easier to catch than their usual prey, and kill them off in short order. --M@rēino 18:54, 12 November 2007 (UTC)

Well, the penguins are prey to Leopard seals and killer whales - so they have seen large predators - just not big, furry white ones. It's hard to know what their reaction might be. Certainly they are generally unafraid of humans. (True story: When my kid was about 5 years old, we went to SeaWorld, saw the Shamu show and then visited the Penguin enclosure for a behind-the-scenes tour. My son (who happened to know a lot about penguins) picked the moment with the most other people around to proudly (and loudly!) announce that "Shamu is a killer whale - he likes to eat penguins!" - the penguin-keepers were torn between their roles as "teaching kids about nature" and "being good corporate employees" - so simultaneously, one of them said "Yes, that's right!" and the other one said "No, Shamu is friendly with EVERYONE!" - then each of them realised their horrible blunder and changed their story - it took them quite a while to decide what story line was the best and lamely trot it out. It was hilarious!) SteveBaker 21:06, 12 November 2007 (UTC)

I will gladly concede that penguins face aquatic predators (and avian predators) -- just not land-based predators. --M@rēino 21:25, 12 November 2007 (UTC)

Wrinkled face

Is there an illness that makes the skin wrinkle excessively? Keria 17:04, 12 November 2007 (UTC)

There is evidence that smoking can do that to you. SteveBaker 18:39, 12 November 2007 (UTC)

I'll point out that our article on Wrinkles sucks, so this question might require some real digging. --M@rēino 18:57, 12 November 2007 (UTC)

As far as I know, wrinkling of the skin is caused by old age, however certain skin disorders may contribute to wrinkles too. Topical Corticosteroids as often used by Eczema sufferers are notorious for thinning the skin and this can definitely cause premature wrinkles. GaryReggae 22:41, 12 November 2007 (UTC)

As the piccie in Wrinkle article demonstrates, masturbation makes wrinkles a lot, lot worse. Myles325a 22:51, 12 November 2007 (UTC)

We must have a few doctors on the science desk, a dermatologist who would like to add a "disease" section on the article wrinkle? Keria 02:55, 13 November 2007 (UTC)

There is a wrinkly skin syndrome, but the description seems to indicate that the wrinkly skin is not on the face. --JWSchmidt 03:55, 13 November 2007 (UTC)

medical ebooks

i am a 2nd year mbbs student of osmania medical college.please give me information about sites to download medical ebooks,so that i can clear evrything. —Preceding unsigned comment added by 117.97.30.240 (talk) 17:13, 12 November 2007 (UTC)

You could try our sister site [Wikibooks] although I don't know if there is anything medical there. Chances are if it's anything too specialist, the best option (apart from spending extortionate amounts of money on textbooks) is a visit to your local library. GaryReggae 22:39, 12 November 2007 (UTC)

how does our organism recognize foreign tissue?

How can our organism distinguish between our own tissue and foreign ones? Of course, the DNA is different, but how our immune system can know it? —Preceding unsigned comment added by 80.58.205.37 (talk) 18:28, 12 November 2007 (UTC)

Antigens on blood cells.

Mainly due to Antigens. These are molecules on the outside of cells, and they are distinctive to each type of cell. For example, there are 2 two types of blood antigen (A and B), so someone with blood type A has "A" antigens on their blood cells. If you give these to someone else whose blood is also A (or AB), the body accepts it, while if you give it to someone of blood type B, the body can't recognise the blood antigens, so it attacks the blood, which causes all kinds of nasty effects inside the body. Laïka 18:50, 12 November 2007 (UTC)

See also, here: "To be able to destroy invaders, the immune system must first recognize them. That is, the immune system must be able to distinguish what is nonself (foreign) from what is self. The immune system can make this distinction because all cells have identification molecules on their surface. Microorganisms are recognized because they have unique, foreign identification molecules on their surface. In people, identification molecules are called human leukocyte antigens (HLA), or the major histocompatibility complex (MHC). HLA molecules are called antigens because they can provoke an immune response in another person (normally, they do not provoke an immune response in the person who has them). Each person has unique human leukocyte antigens. A cell with molecules on its surface that are not identical to those on the body's own cells is identified as being foreign. The immune system then attacks that cell. Such a cell may be a microorganism, a cell from transplanted tissue, or one of the body's cells that has been infected by an invading microorganism." Rockpocket 18:52, 12 November 2007 (UTC)

Potassium chloride and seeds

Is potassium chloride a fertilizer like potash, or a toxin like chlorine? How it will affect the growth of seeds? 199.89.180.65 18:37, 12 November 2007 (UTC)

• The majority of the potassium chloride produced is used for making fertilizer. Just remember that it might kill your livestock if they eat a bag of it. --M@rēino 19:00, 12 November 2007 (UTC)

Mobile phone buzzing

What's the buzzing/beeping that speakers make just before a mobile phone rings? I can't imagine the phone is making an electric/magnetic field strong enough to affect speakers from 3+ metres away. And if it is, is it safe to keep it near my computer? Laïka 18:45, 12 November 2007 (UTC)

Don't worry about it. My cellphone does that to the earbuds on my MP3 player and my FM clock-radio - it's simply the phone acknowledging the cell tower prior to ringing the ringer. The radio signal from the phone is plenty strong to induce a current in a wire of a suitable length to act as an antenna. It should be OK to keep it by your computer - the voltage is enough to hear on a speaker because they are analog devices - also loudspeakers and headphones have long wires leading to them that make great antennas. But no, it won't affect your computer (except that you might hear it on the computer's speakers). SteveBaker 20:55, 12 November 2007 (UTC)

Yes, this is very common although some speakers appear to be sensitive to this and others not - depending on whether the cable is shielded - otherwise the cable will act as an aerial and pick up the electromagnetic waves! It won't do any harm though, other than the obvious interuptions to your audio. GaryReggae 22:37, 12 November 2007 (UTC)

It occurs to me that there is an easier way to explain this convincingly. I should point out that inside the phone is a small computer - you know it survives the most intense radio signals the phone can put out without screwing up. What's more, radio signals decrease in strength as the square of the range. The computer in the phone is at most maybe 2cms from the radio transmitter. So if you put your phone maybe 20cms from your desktop computer, the signal strength will be 100 times less. If the computer in the phone can take the full strength, then for sure your desktop machine will be OK with just 1% as much radio interference. SteveBaker 23:42, 12 November 2007 (UTC)

Also, a desktop computer usually has a metal case (with few holes), so the electromagnetic waves most probably can't get in it. This doesn't apply to notebook computers of course. – b_jonas 11:21, 13 November 2007 (UTC)

Note that this only happens with TDMA (especially current-generation GSM) phones. You aren't hearing the signal itself, so much as the signal being turned on and off something like 50 times per second. This induces currents in some analog circuits -- especially inexpensive amplifiers that lack correct shielding. --Mdwyer 20:34, 13 November 2007 (UTC)

A designer wishes to calculate the thickness of a steel cable for suspending a passenger lift

The total length of cable supporting the lift when it is at the ground floor is 8 m. The mass of the lift when full of passengers is 950 kg. The designer has decided to incorporate a safety factor of 10 into the lift cable: i.e., the cable must be able to withstand 10 times the load it will actually be exposed to in service, before it fails. The steel selected for the cable has a failure stress of 900 MN m– 2 . (a) Using this information, calculate the required diameter of the cable. Show all your working. Assume that the cable is a single piece of steel, with a circular cross-section. (Ignore any effect of the weight of the cable in your calculation.) The downward force F on the cable is calculated by multiplying the total mass m (expressed in kg) by g, the acceleration due to gravity: F = m × g Take the value of g to be 10 m s–2. (In reality there is extra force needed to accelerate the lift upwards, but as this is relatively small, there is no need to consider it here.) Hint: the safety factor means that the cable will fail – the stress will reach its failure stress – when it is loaded to 10 times the intended design load. Use this to calculate the cross-sectional area of the cable, and so its diameter. I found the diameter to be between 11 and 12 mm —Preceding unsigned comment added by Flano1 (talk • contribs) 19:09, 12 November 2007 (UTC)

So do I. 169.230.94.28 19:23, 12 November 2007 (UTC)

This does sound a bit like a homework question to me ("Show your working" is always a dead giveaway!) - please look at the top of this page where you'll see it's our policy not to help people with homework. Sorry. If there is any specific part of the math/physics you can't understand, then feel free to ask specifically about that problem and we'll be more than happy to assist. SteveBaker 20:50, 12 November 2007 (UTC)

It looks like you have all the information you need here to calculate the answer. The most obvious thing to do is ignore the factor of ten for the moment, work out the diameter of cable needed just to support the normal load then multiply it by ten. Our page on Elevators may be of use. GaryReggae 22:34, 12 November 2007 (UTC)

Erm...cough...no, no, NO! Cable strength is proportional to cross-sectional area - not diameter. To get a factor of 10 safety factor, you only need to make the cable sqrt(10) times thicker. About 3.16 times thicker should be fine. SteveBaker 00:01, 13 November 2007 (UTC)

Using the fracture strength of the cable, instead of the yield strength, sounds like a bad idea as well, particularly for applications involving lifting humans... Titoxd^{(?!? - cool stuff)} 00:13, 13 November 2007 (UTC)

Percent solutions

While interviewing undergraduates to be a lab helper, our lab manager was bemoaning how every single candidate could not explain how to make a molar solution. She then asked one how to make a 1% percent agarose solution [weight/volume] and the hapless student suggested filling a measuring cylinder to 1ml with agarose then filling it up to 100ml with water. After we were done laughing at his expense, I noted that doesn't bode well for the future of American science, as it should be simple to work that one out by logic. But this got me thinking that might not actually be the case: a percentage solution [w/v] is considered to be 1g in 100ml. But logically, it should probably be 1mg in 100ml. Are the w/v units different simply by convention, or is there a logical explanation? Rockpocket 19:16, 12 November 2007 (UTC)

In CGS (centimeter, gram, second) units, the base unit of mass is the gram, while the 'natural' unit of volume is the cubic centimeter (equal, of course, to one milliliter). So you shouldn't be deceived by the 'milli' prefix; what you should be thinking is 1% (w/v) is 1 gram in 100 cubic centimeters, which just happens to be 1 gram in 100 mL. TenOfAllTrades(talk) 19:42, 12 November 2007 (UTC)

It’s ultimately only a guess, but I don’t think the name "percentage solution" was originally perceived to be a good name for what it means due to how it works out in CGS units, but rather due to the density of water, which is by far the most common solvent. 1 ml of water weighs almost exactly 1g. (Indeed, the gram was originally defined to be the mass of 1ml of water.) So 1g of solute in a water solution that totals 100ml means that the solute comprises very close to 1% of the solution by weight (w/w). MrRedact 20:15, 12 November 2007 (UTC)

Both suggestions are perfectly logical explanations. Thanks. Rockpocket 23:52, 12 November 2007 (UTC)

Running cool water over frozen drink defrosts. Why?

Sometimes I put canned soft drinks in the freezer, but I occasionally forget about them. When I finally take them out, they're frozen solid. I found out that I can run cool water (regular temperature) from the kitchen sink over the cans, while holding them at a 45 degree angle and rotating. After a few seconds, they're completely thawed out. If I take too long, they might even be too warm to drink! How does cool running water thaw out a frozen drink in only seconds? Steohawk

The drink you remove from the freezer isn't actually frozen solid — not even close. In fact, the ice that forms is neither contiguous nor very dense, because the sugar and carbonation bubbles in the soda pop interfere with the water molecules forming a contiguous lattice. It feels as though it's frozen solid because it only takes a fairly large chunk of fairly firm slush to provide enough support for the metal can to make it feel solid. Furthermore, water that feels "cool" (or "room temperature") relative to our normal body temperature actually contains a great deal of heat. Figure the soft drink is at 32°F. If you measure the "regular temperature" water coming from your tap, you'll probably find it is between 60°F and 70°F, for a ∂T of roughly 30 to 40 Fahrenheit degrees. The metal drink tin conducts heat very efficiently and uniformly to the contents, so your not-really-very-frozen drink "thaws" in a hurry.

Also, please sign your talk page and reference desk posts properly, not by manually typing in your user name, but rather by ending your post with four tildes ~ ~ ~ ~ (but without the spaces between them). --Scheinwerfermann 20:45, 12 November 2007 (UTC)

The water is cool - but it's above freezing - hence it's warmer than the ice and can melt it. I doubt your drink was frozen solid anyway because the pressure from the liquid expansion would likely split the can. Since the freezer will tend to freeze the liquid towards the outside of the can before the inside, there may not be all that much ice to melt and it's right at the surface where the warmer (but still cool) tap water can melt it most easily. I don't see anything weird going on here. SteveBaker 20:46, 12 November 2007 (UTC)

(edit conflict - GMTA) Who knows? Are we talking about soda (pop) like Coke? A few guesses: The can isn't really frozen solid, it just has a coat of ice all over the inner surface that makes it feel solid. When the water in the drink starts to freeze, it will freeze from the outside in because the heat is being lost through the can, and from the bottom up because of convection. As ice forms, the remaining solution will get more and more concentrated, depressing its freezing point more and more and helping it stay liquid. Running water is terrific at transferring heat, and if it's at all warmer than the can it will transfer heat to it. Do this: next time, take one of your cans outside and saw it in half with a big serrated knife (or get your mom or dad to do it). See if it is really frozen solid. --Milkbreath 20:58, 12 November 2007 (UTC)

Make sure to set up a video camera on a tripod and post the video to YouTube when you're done. It could easily be right up there with Diet Coke and Mentos!  :-) SteveBaker 21:38, 12 November 2007 (UTC)

Thanks for the answers.

BTW, I did sign my last entry with four tildes (no spaces), but it got "converted" when I saved the page. Let me try it again. Steohawk 22:08, 12 November 2007 (UTC)

Yup, that time your signature worked. I'll add that if you are eager to see inside an aluminum pop can, the easiest and safest way to remove the lid is to turn an ordinary can opener (preferably a gear-drive Swing-a-way type) 90° so its wheels are horizontal rather than vertical, clamp the rim of the pop can between the cutter wheels, and turn the can opener's handle to remove the lid and part of the rim. Be careful, for this will create a sharp edge. However, it will not create metallic dust or other contaminants (as the hacksaw method would), so you won't spoil the contents of the can. --Scheinwerfermann 23:02, 12 November 2007 (UTC)

Melted ice contains white particles, even purified water.

Whenever I freeze water and melt it, white particles can be seen floating around in the water. They're never visible prior to freezing. I thought that this might have been caused by some impurity in the water from my faucet, so I froze water from my neighbor's purifier. After melting it, the same white particles were visible. They're even visible in hot water, so I know that they aren't tiny bits of left-over ice. What are these particles and why do they only show up after melting ice? Steohawk 22:14, 12 November 2007 (UTC)

Make sure that the container in which you make your ice is clean and not shedding dust or other bits of material and that you cover it with something similarly clean when you're freezing/melting it. Maybe put the water in a zipper bag during the freeze and thaw? My older plastic ice-cube trays used to flake off over time, and also the paint and other coatings in the freezer itself did. DMacks 22:29, 12 November 2007 (UTC)

Possibilities. 1. Your neighbour’s purifier is full of gunk 2. Your freezer is full of gunk 3. The container you are using is full of gunk 4. The back of your retinas are full of gunk. Myles325a 22:30, 12 November 2007 (UTC)

Is it possible that you live in a 'hard water' area? I could maybe imagine there being a lot of dissolved minerals in the water that could maybe come out of solution when you freeze the water - making these little particles. What we need here is an experiment. (Woohoo! Actual science on the science desk!) The way to prove whether I'm right is to filter the water you get by melting the ice cubes and then re-freeze it in the same freezer/ice-tray that you used the first time. If you melt it again and it has particles in it the second time around then the bits must have come from the freezer/ice-tray. If there are no particles the second time around then DMacks & Myles are wrong and I guessed right. As an alternative experiment, you could try washing your hands with a little soap and the melt-water (wait for it to get to room temperature first) and see if you get lots more lather than you do with tap-water. Soft water produces more bubbles from soap than hard water and if I'm right and you somehow removed minerals by freezing and filtering - then you'd expect to get more lather with the melt-water. SteveBaker 23:57, 12 November 2007 (UTC)

You might enjoy our zone refining article.

Atlant 17:50, 13 November 2007 (UTC)

E = mc2 - how does c relate to the weight and energy units?

Everyone can see why there is so much energy in a given mass, because c, the speed of light is so large, and c2 is enormous. But why couldn’t we just define c as a single unit? How does Einstein define a measure of speed with weights and energy values?

After all, c = sqr(E/m). I have often thought about this, tho obviously I am a layperson in physics, and my brain hurts. Myles325a 22:25, 12 November 2007 (UTC)

We actually have a page about the E=mc2 formula, with some info. One hand-waving reason that "c² is more reasonable than c" here is that we need the units of measurement (dimensional analysis) to fit. A conversion factor for an energy/mass equivalence must be expressed in terms of distance²/time²). Once you have the relationship though, you can do algebra and other manipulations to express t he equation in different forms. Saying "c = sqr(E/m)" is not ideal because it fails if m=0. DMacks 22:37, 12 November 2007 (UTC)

In SI units, mass is in kg, lightspeed is in meters per second and energy is in Joules - which is SI shorthand for: "kilogram meters-squared per second per second" ...which is a bit of a mouthful.

You can work in any scale units you like so long as you are consistent. If you want to define your unit of length as (say) the distance light travels in a second (a "lightsecond") instead of the more usual 'meter' that's OK. The E=mc² equation still works - except that now we've defined c to be 1.0, instead of the result 'E' being in Joules (which is shorthand for kg.m².s^-2) the answer would be in some new unit of energy because we defined our unit of length to be a lightsecond instead of the meter. Let's name it after it's creator and call it the 'Myle'. The Myle is shorthand for kg.lightsecond².s^-2. One kilogram of matter would contain exactly one Myle of energy...very convenient! But the Myle would be a truly phenomenal amount of energy because it's in units that contain lightseconds²!! The Myle simply isn't a very useful energy unit. A 'AA' battery stores about one kJ - but and inconvenient 0.09 attoMyles! On the other hand, you could instead define your time unit as the time light takes to travel in one meter - c is still equal to 1.0 but your unit of time is microscopic - so your unit of energy is still enormous. You can shuffle the unit definitions around however you like - but the result is the same. Dimensional analysis is always very instructive in these cases. SteveBaker 23:31, 12 November 2007 (UTC)

It's also worth noting that when working with problems in relativity, velocities are generally expressed as fractions of c rather than using more common measurements. Donald Hosek 02:00, 13 November 2007 (UTC)

You might want to read up on Maxwell's equations. That's where Einstein got almost all of the theoretical support for the conclusion that E=mcc. --M@rēino 14:27, 14 November 2007 (UTC)

November 13

Technology and Depression

Does the more engrossing technology becomes translate to the more depression society feels as a whole? (Forgive the grammar.)

Living in our parent's house, no post-secondary education or real job experience between us, my brother and I were absolutely entranced with World of Warcraft. "Real" decisions with all their complications were so far removed from the crisply cut and dry work/rewards system in game. I alone spent 7 months of 16-hour grueling days rising to the rank of Grand Marshal of the Alliance and I loved it. We lived every waking hour in WOW for 2 years. Our parents eventually gave us an eye-opening ultimatum and he killed himself with a plastic bag and duct tape about a year ago. I was in a psych unit for about 2 months after. I know now what horror and true self-loathing feel like.

In my heart I know this is a sign of things to come for society as a whole. I'd like to know if depression and technology are correlated. Sappysap 02:13, 13 November 2007 (UTC)

There are arguments that they are, but I don't believe so. Depression is fairly clearly genetic, and has existed in some form for all of history. Suicide, of course, being the worst-case fate of depression, has also been in existence and I don't believe suicide rates have increased or decreased significantly with time, though certainly environmental factors have played a role. I have no significant sources with which to back this, though, other than the fact that "melancholy" was a diagnosis since at least the early 17th century. SamuelRiv 03:14, 13 November 2007 (UTC)

First you need to clear up what exactly do you mean by depression. There are many subtypes of clinical depression, some of which include atypical depression, and dysthymia. Each clinical type will have certain characteristic features; so people with dysthymia may be less prone to committing suicide than people in a major depressive episode.

Correct me if I am wrong, but your hypothesis is that technology allows people who are lonely and/or need human contact but lack it to withdraw into artificial worlds where they get only superficial human interaction. This leads to increasing loneliness and eventually to suicide or total withdrawal from society?

Also when you use the word technology, do you mean only modern technology? If not, then we would need to analyze the impact of earlier technological advances like electricity and the printing press (among many others) to see if they provide support to your hypothesis.

What would be key to supporting your argument is that technology heightens the impact of someone who would be prone to withdrawal from human interact, because I would imagine there is and always has been a certain subset of people who are lonely but still take actions to further isolate themselves.

As to SamuelRiv's claim that depression is "fairly clearly genetic", I do not agree. First it is unclear what that claim even means. Minimally, it would mean there is a genetic component, which does seem to be the case. But the more natural interpretation is that he is claiming it is almost entirely due to genetic factors, which would need some serious evidence to justify. This issue also points to the need to clear up what exactly you are talking about when you use the word "depression", since each clinical subtype is likely to have different roles a genetic component would play.--152.2.62.27 13:00, 13 November 2007 (UTC)

I also disagree that depression is a mainly genetic phenomonen, I have always believed that it is mainly due to the way one has been brought up combined with the life experiences they have been through, ie somebody that was emotionally neglected as a child and/or bullied will take a major battering to their self esteem which is definitely a factor in causing depression. These psychological scars can take years to heal, that is if they ever heal completely. On the flip side, somebody with a more fortunate upbringing is less likely to suffer from these issues BUT there are many exceptions to this and I think there may be a genetic element too...HOWEVER this may be psychological - if a parent suffered from depressive conditions then they may 'rub off' on the child.

Going back to technology, I think it CAN cause depression but it can also be a godsend. For example a depressive person could spend hours "wasting their life" (in the eyes of some people) playing video games but then again this is usually a distraction and definitely doesn't mean that without this activity, the person would be a raving extrovert. On the flip side, I think the internet can be very helpful as people with mental illnesses including depression can find out information about their condition and discuss it with other suffers which is often very cathartic. People may be willing to engage in others online when they wouldn't otherwise.

A third argument is that it is not technology itself that is causing depression but the way society has become. People nowadays are generally more insular than they were say thirty years ago, many people don't know their next door neighbours any more and the general perception is that things are unsafe and there are a lot of people around who cannot be trusted. This could (arguably) be caused partyly by technology, for example people not bothering to talk to each other because they are more interested in watching the TV or gaming while before the advent of these solitary pleasures, most pastimes were more social, take the fact that a lot of families no longer sit around a table for meals, instead choosing to eat in front of the TV. Take also the fact there are a lot more single person households than ever before, this causes isolation but I cannot see any attribution to technology except for perhaps the learning of antisocial trends from television. GaryReggae 15:51, 13 November 2007 (UTC)

Though it is out of date you might fight Oswald Spengler an interesting read—it's not about technology, per se, so much as it is about modernity. But in any case, people have wondered for years (since the 19th century, anyway) if the creation of new technological means of representation and life have "sucked the soul" out of modern human interactions, encouraging depression, vice, etc. I myself am doubtful of a straightforward relationship—first, I don't believe there probably were "good old days" where everybody was chummy and warm and happy (most people are sons-of-bitches no matter what time or place you live—but that's the misanthrope in me talking), and second, I'm unconvinced that technology itself is responsible for the sorts of changes people are talking about. When a family eats its dinners in front of the television rather than together at a table, it isn't the television that has compelled them to do so; it's just as easy not to have the TV on during dinner (even easier as time goes on, with the ability to time-shift shows). As for depression—that's a complicated issue, but I don't see any obvious technological links there. If it wasn't the game, who isn't to say it wouldn't have been some other form of escapism? --24.147.86.187 01:03, 14 November 2007 (UTC)

• Sappysap, I could point you to a ton of Wikipedia articles, such as Uncanny Valley and Escapism. But ultimately, there's still a lot of research to be done before we can say for sure how artificial worlds affect psychology. In a situation like yours, I doubt any answer from us volunteers at Wikipedia will be completely satisfying (especially ones like me who aren't even pre-med students!). You may want to consider the possibility that this is your life calling, enroll in your local college, and get yourself a degree in pyschology. If that's not an option, at least ask your psychologist (or someone you like from when you were at the ward) to refer you to one of their colleagues who specializes in technology, so that you can do some meaningful research. --M@rēino 14:42, 14 November 2007 (UTC)

I don't think it's the technology. If you'd spent two years doing nothing but...I dunno...reading medieval poetry or learning to ride a unicycle - the end result would be kinda similar. MMPORG's are fun - they're fine for a couple of hours of entertainment a few days a week - they help you unwind - but in the end when you finally chuck it in - you haven't achieved anything. Heck you said it was "grueling" - so you weren't even getting the relaxational benefits. But it's pretty clear that making your entire life (16 hours a day, 7 days a week for 2 years) happen in a place some dumb game designer (like me - I'm a game designer) dreamed up is not a healthy thing. I mean, 11,000 hours at minimum wage would buy you a small house or a pretty decent sports car! The problem here isn't the technology - it's the sheer waste of human effort in doing it in the first place. I know where you're coming from - I'm closing in on 10,000 edits over 3 years to Wikipedia. That's an addiction that I recognise - but it's a lot less than 10,000 hours - probably more like 1,000 hours - but still, it's a serious chunk out of my life. The difference is that it doesn't feel like time wasted - there is a positive result - the accumulation of all human knowledge where almost anyone can read it...that's something worth spending your time on. The technology of Wikipedia and that of an MMPORG are not all that dissimilar!

So it's not the technology...it's what you use it for and to what extent you let it take over your life.

I've come across horror stories of MMPORG addiction too (although nothing like on this scale) - we had a neighbour who's smart "straight-A's" teenage son spent 50+ hours straight (no sleep, no food) wrapped up in Everquest. When his parents found out, they took away his computer promising to return it after one week. He then came over to my house, smashed a window, broke in and stole my wife's laptop. Sadly, being a sleep-deprived teenager, he forgot to steal the battery charger and phoned us up the following day to see if we happened to have a laptop charger he could borrow...which certainly raised some suspicions! When we got the laptop back, yep - Everquest had been mysteriously installed on it. But that incident blew away any trust his parents had. He wound up in a downward spiral that put him in military school then downtown Bagdad when he should have been in college.

SteveBaker 19:46, 14 November 2007 (UTC)

Reduction of Fe(III) in FeCl₃(aq) to Fe(II)

How would I reduce the iron in FeCl₃(aq) to Fe(II) in order to use the indicator ferrozine, which only detects Fe(II), not Fe(III)? Can I somehow use ascorbic acid to do so? What would the chemical reaction be? Could this be undergone by using titration? Chickenflicker-♣ 02:48, 13 November 2007 (UTC)

Edit: What would the reaction between Hydrazine (N₂H₂(l)) and FeCl₃(aq) look like? Chickenflicker-♣ 03:03, 13 November 2007 (UTC)

I've seen hydrazine used as a 2-electron reductant (standard way to prepare Pd(0) complexes is from Pd(II) salts, and also the Wolff-Kishner reduction). Not sure about a good 1-electron reduction mechanism. DMacks 14:39, 13 November 2007 (UTC)

Thinking more technically here, hydrazine is N₂H₄. N₂H₂ (diazene) also exists and is a good reducing agent (via hydrogen transfer, such as in hydrogenation of unsaturated compounds), but it's mighty unstable (the pure material decomposes even around –180°C). DMacks 16:50, 13 November 2007 (UTC)

horizontal component of speed

This is a mix between a maths and physics questions...anyway here goes:

An object is being pulled up a slope by a force at a constant speed of say 3m/s. At the top of the slope is a horizontal plane. What is the object's speed at the beginning of its journey along the horizontal plane? Is it 3m/s, or is it just the horizontal component of velocity (ie. 3cos(angle))? Is the vertical component of velocity wiped off or does it become a part of the horizontal velocity?

    ->>______________________
   -  /
  -  /
    /
   /
  /
 /
/

What about when a ball rolls down a slope and then travels along flat ground? At the beginning of its journey along the flat ground, is its speed just the horizontal component of its previous velocity or is the magnitude of both components of its previous velocity?

                     /
                    /
                   /
                  /
                 /
                /
____________<<-/  

Thanks! D3av 02:53, 13 November 2007 (UTC)

In these types of problems you usually approximate the sharp corner as a tiny rounded corner. Then assuming ideal conditions the ball rolling down the hill will conserve kinetic energy when it hits flat ground. The ball being pulled up a hill will also conserve kinetic energy, but the end result depends on the precise wording of the problem. If you are constrained to be on the surface of the slope, then you move forward with kinetic energy conserved. However, if unconstrained, you fly upwards as a projectile with some initial velocity at some angle. SamuelRiv 03:09, 13 November 2007 (UTC)

So say the one moving down the hill is unconstrained (just freely rolling down). Would you use the horizontal component of velocity or the whole thing?

When you say "with kinetic energy conserved", does that mean that the vertical component of velocity becomes part of the horizontal component?

We had one question in which the object seemed to be constrained as it was being dragged up the slope. The instruction was to use only the horizontal component of velocity when it reached the horizontal plane at the top. Was this a correct instruction? thanks. D3av 03:26, 13 November 2007 (UTC)

As SamuelRiv said, it depends on the precise wording of the problem, in particular how the force is being applied. For example, if there's something pushing on the object from the left (slowly, so that gravity keeps the object from ever leaving the surface), moving at a constant rate to the right,

       ______________________
      /
->|  /
->|*/
->|/
  /
 /
/

then the horizontal component of the velocity will remain constant, and the vertical component will go to zero at the corner. If there's a rope attached to the object that goes up and over the corner (so the corner acts somewhat like a pulley), and the rope is being pulled at a constant rate,

       ______________________->
      /______________________
     //
    //
   */
   /
  /
 /
/

then the component of the velocity parallel to the rope will remain constant, so the horizontal and vertical components will change as the rope's direction changes. If there's something pushing the object in the direction parallel to the slope,

       ______________________
\     /
 \   /
  \*/
   /
  /
 /
/

(...well, I don't know how to make it look like that thing is moving diagonally upwards with ASCII, but anyway) it will actually move faster when it reaches the horizontal surface because the component of the velocity in the direction of the slope will remain constant, but it gets a new perpendicular component when it passes the corner. So it all depends which of these different problems you're trying to answer. —Keenan Pepper 06:27, 13 November 2007 (UTC)

The vertical component won't magically go to zero, regardless of whether the object is being pushed or pulled. Unless the object is held on some sort of track, it will rise into the air above the horizontal plane until gravity brings it back down. Clarityfiend 08:00, 13 November 2007 (UTC)

Unless, as I said, it goes "slowly, so that gravity keeps the object from ever leaving the surface". This is a quasi-static approximation, which is actually very good for everyday objects. If you pull a rubber ball up a ramp with a string, slowly, it doesn't fly into the air when it reaches the top. —Keenan Pepper 23:04, 13 November 2007 (UTC)

Thankyou very much for those answers. They are very comprehensive and clear. I'm still unsure though (as I wrote above) about what will happen to a ball that is rolling down a slope and then continues along a flat surface. (The only forces acting on it are the force of gravity and the normal). Thanks D3av 09:45, 13 November 2007 (UTC)

It depends on the nature of the ball. When the ball hits the bottom of the slope, there is a collision. What happens next depends on what the ball and the ground are made out of. A softer ball might bounce a few times - eventually rolling off along the new direction at roughly the speed it was moving down the slope - much conserving kinetic energy (assuming we ignore losses). What happened was that the kinetic energy temporarily turned into elastic energy from the compression of the ball's material - then that elastic energy was turned back into kinetic energy directed in a different direction. That little energy interchange permitted the velocity vector to change but the kinetic energy to be conserved. A harder ball might simply give up the vertical component of its energy in the collision (so the kinetic energy turns to heat) - continuing to move along the horizontal region at a speed equal to the horizontal component of it's former motion. Without more information, you can't say which it will be.

When the ball is travelling up the slope, the same thing happens. At the top of the slope the ball will continue upwards (because of conservation of momentum) - travel on a free parabolic trajectory and eventually collide with the top of the slope. Once again, the ball will either bounce - causing it's new velocity to be horizontal - with the same speed as it was travelling up the slope - or it'll impact without a bounce - causing it to lose the vertical component of it's former velocity.

With real balls, the behavior is somewhere between the two cases - energy is lost in the bouncing and the actual horizontal speed will be a blend between the old horizontal speed and the old net speed.

SteveBaker 12:26, 13 November 2007 (UTC)

To original poster, remember kinetic energy is NOT a vector. It has no horizontal and vertical components. When kinetic energy is conserved, all velocities must be taken into account. SamuelRiv 14:56, 13 November 2007 (UTC)

But also remember that kinetic energy is NOT necessarily conserved -- it's only conserved if the collision between the ball and the horizontal surface, described by Steve, is elastic. To make this clear, imagine rolling a cannonball down a steeply sloped, almost vertical wall into a lawn. It'll probably stop dead at the bottom and embed itself in the grass. However, the same ball coming down a gentler slope, one that's almost horizontal, will probably keep rolling and may well do it without much loss of kinetic energy. --Anon, 23:32 UTC, Nov. 13.

Thanks for all your help. I see that my maths course has oversimplified the issue and ignores bouncing/heat/other forces etc. I'll just try to follow the guidance of the question! D3av 00:03, 14 November 2007 (UTC)

That's pretty typical of a math course - they aren't really interested in the physics - because it's a math course! However, in this case, it's not the same thing as simply ignoring friction and air resistance for the sake of getting an answer. In this case the outcome will be quite utterly different for an elastic versus an inelastic collision. The issue of whether kinetic energy is conserved or whether conservation of momentum applies is not an insignificant matter. SteveBaker 18:46, 14 November 2007 (UTC)

Well, momentum (unlike kinetic energy) is always conserved, but that's not useful to consider in this problem. Whether or not the ball's speed changes, its velocity certainly does, since it's moving in a different direction; therefore there is a transfer of momentum between the ball and the Earth. --Anonymous, 23:01 UTC, November 14, 2007.

stars come down

hi my question: there is a place in space that stars come down(and it is a holy place...god says in quran)Have you heard anything about it? —Preceding unsigned comment added by 213.207.252.64 (talk) 10:20, 13 November 2007 (UTC)

Could you provide a reference to a specific passage in the Quran? Scripture often contains sections that need to be read very carefully to determine what they refer to exactly. My first guess would be a black hole. (EhJJ) 11:41, 13 November 2007 (UTC)

It's possible that the book refers to meteorites as stars. -- JSBillings 11:43, 13 November 2007 (UTC)

If "come down" means "move deeper into a gravity well", then I suppose you could say that "stars come down" as they fall into a supermassive black hole. Those are found at the centers of many (and perhaps all) galaxies. But there is no one, single place where this happens in the universe - it's likely that there are at least as many of these objects as there are galaxies - and there are billions of those. The nearest one of these things is at the center of our own Milky Way galaxy.

I find it amusing that a black hole might be considered a holy place. Good and useful things (like stars and planets) go in - nothing (including light or information) comes out! But really, any sufficiently vague and ancient statement (with likely translation errors and a wide laxity of interpretation) can be made to fit some real thing. I don't think I'd read anything into this statement beyond that whoever wrote it didn't know what a star really was and thought that the idea of the seemingly immutable stars 'falling' was a dramatic image. Spotting an occasional meteor and (incorrectly) assuming that was a 'falling star' might serve to reinforce that. Sadly, with what we now know, it really doesn't make any sense.

SteveBaker 12:07, 13 November 2007 (UTC)

Stars falling can refer to the yearly Leonid meteor shower which offered a particularly vivid display on November 13, 1833 bringing quite a few people to thinking it was the end of the world. Not that this particular event would be mentioned in the Qur'an but maybe a very early occurence of this same phenomenon. Keria 14:26, 13 November 2007 (UTC)

Meteor showers such as the Leonids and Perseids are not localised to one place though. Every place on the earth and the moon is peppered by them at one time or another. So this couldn't be a specific holy place - except by the very broadest interpretation. SteveBaker 14:56, 13 November 2007 (UTC)

The representations of the meteor shower often show a sort of fountain of stars as if there was an origin to these moving lights. I'm not sure what the questioner means by "that stars come down" but if it's from I could definitely see a 7th century scholar/mystic (or Muhhamad!) seeing a holy fixed source of stars in space in the apparent origin of these showers. Keria 15:55, 13 November 2007 (UTC)

Yes - both the Leonids and the Perseids appear to come from one place in the sky - in fact their names come from the constellations that they appear to radiate from (that point is called "the radiant"). The reason is that these showers each come from tiny bits of an ancient broken-up comet that have spread out into a gigantic loop around the sun. The earth moves through each of those streams once a year - which is why the debris appears to come from a fixed spot in the sky year after year - over the exact same couple of days each year. But the problem with this theory is that this 'radiant' effect means that the 'shooting stars' radiate outwards - they are all heading off to different places - not falling to the same place. SteveBaker 23:51, 13 November 2007 (UTC)

• The Qur'an also says that Mohammed flew from Mecca to Jerusalem on a magical flying horse, so you should probably take the unrealistic parts as metaphor. --Sean 19:06, 13 November 2007 (UTC)

The authors of the quran can't possibly be intending to refer to black holes, since its authors lived long before anyone knew anything about black holes!

I did a search on "stars" within English translations of the quran.[19] There are only about 15 hits, although that number varies by translation. There are a couple passages that look like what you might be referring to:

Al-Waqi’a 56:75 refers to what is translated in the Yusuf Ali translation as "the setting of the stars". The Muhammad Asad translation, however, uses the phrase "coming-down" instead of "setting". (The Muhammad Asad translation also uses "parts" instead of "stars", so the meaning of that passage is particularly unclear.) If this is the passage you are referring to, then no, there is no place where stars literally come down. They merely appear to come down, to the west of wherever you happen to be located, due to the rotation of the Earth.

Al-Hajj 22:18 talks about the stars "bowing down" before god, or "prostrating" themselves, depending on the translation. And Al-Hajj is about the pilgrimage to Mecca, which is certainly considered a very holy place among Muslims. If this is the passage you are referring to, then no, no such bowing motion is observed in stars, even when they appear to be over Mecca. MrRedact 19:30, 13 November 2007 (UTC)

Thanks! Excellent work! That's pretty much what I thought - vague translations of an uncertain source...but it's nice to see it confirmed. SteveBaker 23:51, 13 November 2007 (UTC)

Alabama? --Trovatore 00:02, 14 November 2007 (UTC)

What snake a specific book refered to

I'm looking for a king of snake who may havg from a tree over the Amazon river or one of its tributaries, would look like a vine to someone in a canoo going under that tree, and can swim away if grabbed off the tree. I looked in Category:Reptiles of South America, but couldn't find it - the Emerald tree boa looked promissing, but I couldn't find evidence of it being able to swim. Od Mishehu 11:29, 13 November 2007 (UTC)

Perhaps an anaconda? (EhJJ) 11:38, 13 November 2007 (UTC)

I don't think it hangs from trees. In the book in question (Magic Tree House book called Afternoon on the Amazon), Jack tries to get to the river bank by pulling on a "vine" that turns out to be a snake. Od Mishehu 11:47, 13 November 2007 (UTC)

It doesn't look like the author actually used real animals in her books, but just made up a convenient fictional snake as a plot device. -- JSBillings 13:27, 13 November 2007 (UTC)

I'm afraid I don't understand the question. It was a fictional snake. --Milkbreath 12:01, 13 November 2007 (UTC)

Indeed, it's difficult to see why a snake would hang above a river unless it could swim. But why would it do that? It's not as though tree snakes normally go fishing.--Shantavira|^{feed me} 13:52, 13 November 2007 (UTC)

I thought all snakes could swim, no? I know hedgehogs can. Lanfear's Bane | t 16:46, 13 November 2007 (UTC)

Condensation and Deposition

I need to measure the rate at which Condensation and Deposition (physics) occurs at standard temperature and pressure from a particular inorganic vapor with a melting point in the 1000s K produced by combustion. I have had no luck so far. How do physicists and chemists measure condensation and deposition? As a biologist without a whole lot of physical chemistry, I haven't even been able to figure out where to begin. I am hoping that there is a formula but I fear that these rates are determined only empirically. Thanks for any help. Spc303 12:59, 13 November 2007 (UTC)

You'll want to look up the phase diagram for your substance and then calculate the vapor pressure, I believe (not an engineer). Deposition is a much harder problem, and I'm afraid I don't know the physics of that phase transition, but for condensation you may also need to look up the constants a and b in the Van der Waals equation, but not for these formulas. Here's the relevant equations: for the change in the number of particles going from liquid to gas phase, we have

$\Delta N={\frac {16\pi \gamma ^{3}}{3n_{l}^{2}}}(k_{B}T\log(p/p_{0}))^{3}$

where $\gamma$ is the surface tension (in erg/cm³), $n_{l}$ is the number of molecules per volume of the liquid, and $p$ is the vapor pressure defined by

$p=p_{0}\exp(-L_{0}/RT)$

and $L_{0}$ is the last thing to look up is the latent heat of vaporization per molecule. Note T is temperature, kB is Boltzmann's constant, and R is the gas constant. SamuelRiv 14:53, 13 November 2007 (UTC)

Biofuels

Hi there, I have a question about Biofuels which I can't find a meaningful definitive answer to online. Where I work, we have a well used old 2002 reg Citroen Berlingo van (diesel) and I have been toying with the idea of using waste cooking oil as fuel as we have tons of the stuff available. According to some sources I have seen (for example [20]), engines need to have conversion kits fitted, such as adding an extra fuel tank or heating the fuel, however I have read many cases of people just putting used cooking oil (once it has been filtered) straight into their fuel tank - not neat but making sure they have half a tank of standard mineral diesel as well as half a tank of veg oil, ie 1 50:50 mix. Would this work or damage the engine? It's located in the South of the UK so unlikely to be susceptible to really cold temperatures. It's probably too much hassle to get the engine modified but if we can use it as it is then great! The van is a bit of a 'rattler' so it's not exactly the end of the world if it needs repairing after any 'experiments'.

Disclaimer: I know we need to pay fuel duty tax if we use more than 2,500 litres a year and obviously we would monitor this and ensure that we do not go over this amount so this is NOT a legal question, just a technical one. GaryReggae 15:36, 13 November 2007 (UTC)

There are two problems you need to deal with:

• Filtering
• Cold weather

Typical cooking oil has lots of bits of food in it - and sometimes some dissolved water - and it absolutely must be carefully filtered before you put it into the tank of your car. I'm told that coffee filters work - but they fall apart and clog before you get enough oil through them. You need something more robust if you are going to be doing this for any length of time.

Cold weather is the more serious matter. Veggy oil is not very runny at cold temperatures - that can strain your fuel pump and clog the system up pretty badly. A proper conversion kit uses a secondary tank of straight diesel to start the car - then picks up waste heat from either the exhaust or the hot water entering the radiator to heat up the fuel lines coming from the veggy oil tank so you can switch over to veggy oil a few minutes after you've started the car when everything is nice and toasty. If you lived here in Texas, you could probably not bother and just stick filtered oil straight into your car, switching to straight diesel for a few weeks in the depths of winter - but in the South of England - no, you're going to need a heating system of some kind at least 6 months of the year. Certainly you can experiment. I don't see the harm in mixing diesel and oil - although I suppose they might settle out if not mixed leaving the oil being the first thing that enters the system on a cold weather start. If I were you, I'd start in the summer when things should work OK even if you're not really getting it right - and as the weather gets colder, and the thing gets harder to start - or just won't run at all - then you need to switch to diesel (at least in some mixture). But before you do any of those things, you need to figure out a good way to filter your oil.

SteveBaker 16:37, 13 November 2007 (UTC)

This is something you need to ask a sharp mechanic (like the Car Talk guys) if your mind is really set on it. A car has internal computer regulators of the octane level of your fuel, among everything else including the integrity of the fuel-air mixture that is injected into the engine. It may be that this regulation will cause serious consequences to your exhaust or fuel efficiency, and will certainly damage the engine if you're using just any old cooking oil. The engine is designed to self-clean and lubricate, and it only can do this if the nature of the exhaust of the fuel you put into it is predictable. Cooking oil almost certainly burns different and probably is dirtier than gasoline, causing premature wear in your pistons. I would definitely recommend against it for all these reasons. But please, if you feel you must do it, ask an actual mechanic. SamuelRiv 20:37, 13 November 2007 (UTC)

Whaaat??? Your answer is almost 100% wrong! We're not talking about gasoline. This is a diesel engined vehicle. The whole business of octane levels are relevent to gasoline powered cars because you need to avoid knocking - but diesel engines are quite different in that regard because they rely on the fuel detonating without a spark. The idea of running diesel cars on left-over cooking oil is extremely well understood (see Vegetable oil used as fuel for example). Thousands of people have done it - it works (although without engine modifications - it only works well in warm climates and with older vehicles). The issues are the ones I've described. The issues you bring up are not a problem.

The onboard computer (the ECU) doesn't "regulate the octane level" - how the heck could it possibly do that? There isn't some separate 'octane supply'! What it generally consists of is a small microphone that detects the onset of 'knocking' in the engine due to insufficient octane in the fuel. The computer then alters the fuel/air mixture to avoid the damage that might result. But that's all irrelevent here because this isn't a gasoline engine!

SteveBaker 23:45, 13 November 2007 (UTC)

A computer that can arbitrarily alter hydrocarbon composition of a mixture would be impressive, indeed... -- 128.61.20.199 13:35, 14 November 2007 (UTC)

Thanks for all your ideas. I will do some more research on conversion kits and suchlike. It seems another option is to add methanol to the veg oil to thin it down but I don't know how well that works. GaryReggae 12:36, 14 November 2007 (UTC)

A big problem with methanol is that it is an organic solvent, so you'd want to talk to someone who knows organic chemistry before subjecting the various fuel lines and seals in your engine to that. It may also react with or give the fuel undesirable properties, but I don't know enough chemistry to say. Anyway, running diesels on veg oil works beautifully from my experience, just make sure you take the precautions that Steve mentioned. -- 128.61.20.199 13:32, 14 November 2007 (UTC)

Yeah - absolutely. Even the 10% ethanol that they are putting in gasoline these days can seriously mess up the engine of an older car. I don't know about methanol - but the results could be similar. Ethanol dissolves rubber and some kinds of plastics. On modern cars they've been aware that ethanol was likely to be added to gasoline for many years - so modern cars have seals made of plastics that don't dissolve - but I've noticed all sorts of seal failures in my 1963 Mini since we switched over to E10 a year or two ago here in Texas. There are other considerations too - ethanol produces acidic residues from combustion that can destroy the viscosity of non-synthetic motor oils, it also conducts electricity so E25 ethanol (coming soon to a gas station near you!) can short out the fuel gauge sending unit or your fuel pump on some older cars.

I don't know which (if any) of these things also applies to methanol - but unless you know for sure that methanol has been sucessfully tested by a lot of people - I wouldn't use it. Mix your veggy oil with regular diesel fuel by all means - but getting creative is not advised! SteveBaker 18:39, 14 November 2007 (UTC)

"From the Fryer to the Fuel Tank", by Joshua Ticknell is the standard book I read on the subject. (try to get the second or any later edition). Also, ReNew magazine published in Aust has articles occasionally. Lots of considerations. The price of old chip oil in rural victoria has gone up in the last 5 years as a result of conversions.Polypipe Wrangler 20:14, 14 November 2007 (UTC)

PMTs, APDs, DUV, and other exciting acronyms

Anybody happen to know roughly what the window of time for detection of a Cherenkov radiation event in a particle detector is? This statement in the first paragraph of the photomultiplier tube article struck me as possibly outdated and unsubstantiated:

"Semiconductor devices like avalanche photodiodes have replaced photomultipliers in some applications, but photomultipliers are still used in most cases."

Modern APDs (like the ones my group produces for UV wavelengths) approach the gain afforded by PMTs and enable single-photon detection with enormously lower cost, power requirements and physical size, but higher reliability, bandwidth, speed, quantum efficiency, etc. The advantages of APDs for this application seem to be somewhat confirmed by a presentation given by a KEK scientist about a next-generation Cherenkov radiation based neutrino detector, though his avalanche gain figure is outdated (gain over 10⁴ is quite doable with APDs these days).

The only area I can think of where PMTs might still hold an advantage regards the window of event detection. I have no idea how much time you have to detect the photons emitted from a Cherenkov event, nor exactly how PMTs are operated (are they run continuously or sampled?). Typically high-gain APDs are operated in Geiger mode--they are biased below breakdown and then pulsed into breakdown wherein they operate at very high gain for the duration of the pulse. While this pulsed/sampled mode of operation can be run reasonably fast (several MHz with pulses on the order of tens of nanoseconds with a good device), I don't believe it's possible to operate very high gain APDs in breakdown continuously for risk of various destructive breakdown mechanisms. The higher quantum efficiency of APDs may make up for the possible deficiency of Geiger mode measurement, but I have no figures for PMTs to compare with.

So enough rambling, here are my burning questions: How long do you have to detect Cherenkov radiation? Presumably the data coming from these neutrino detectors is sampled for digital processing, but how quickly? Are the photomultiplier tubes operated continuously? Does the whole system operate in a sort of "triggered" mode, operating the photomultipliers continuously and waiting for a detection? What percentage of "missed" events is considered acceptable? Does anyone happen to know what the quantum efficiency of a PMT typically is? (I'd guess it's mostly limited by the front-end optics and the quantum efficiency of the photoelectric effect in the photocathode) Have I lost more than 90% of the readers of this page by now? Would I be better off learning Japanese and calling up the KEK to ask my questions? :)

Thanks for any insight you can provide.

Bonus: Impressive picture of the innards of the Super-Kamiokande detector. Notice the two technicians and the size of each photomultiplier tube (enhanced by the big lens assembly). Now consider the infamous Super-K PMT implosion accident which took out over half of these in a chain-reaction, and the motivations for using micron-sized APDs rather than enormous evacuated tubes become apparent!

-- mattb 16:42, 13 November 2007 (UTC)

IceCube operates continuously with triggered events. Trigger timestamping is at <5 ns, and the resulting current pulses are digitized at 250 MHz. The size of pulses allows one to discriminate events running from 1 to ~200 photons per 15 ns. Quantum efficiency for those PMTs is typically 20%, with practical efficiencies with a single photon closer to 10%. Many events involve more than 1 photon. For IceCube like environments, you also need to operate at -40 ° C. Dragons flight 17:44, 13 November 2007 (UTC)

Ah ha. So the necessary detection window is around 15 ns, and the detector needs to be able to distinguish ~1-200 photons in that time? That's a lot shorter window than I thought. Continuous linear operation with high gain is indeed a big boon.. You can operate APDs continuously in their linear (low gain) region and outpace the PMTs, but high-gain Geiger mode operation is necessary for single-photon detection and isn't nearly as fast... With a 15 nanosecond window for counting photons, I don't think current APDs alone could cut it, since the shortest Geiger pulse I've seen an APD used with is a bit under 1 ns. Even with good active quenching, there's a necessary 1 or 2 ns delay before the APD can be reset for another count, so if you're right there's no way they could distinguish between 200 photons in 15 ns using Geiger mode... Realistically, I don't think Geiger mode is even practical for periods in the tens-of-picoseconds range; generating timing pulses that short is difficult, and quenching fast enough is even more difficult. That explains the interest in hybrid detection schemes using an APD in linear mode... Thanks for the info. -- mattb 18:51, 13 November 2007 (UTC)

This page indicates a maximum 30% quantum efficiency and overall gain up to about 10⁸. DMacks 17:47, 13 November 2007 (UTC)

Free-falling

When we fall, we fall at a rate of approximately 33 feet per second, per second.

Is this an 'exponential' rate? Do we fall at a speed that is increasing 'exponentially?'

Peter Lamont —Preceding unsigned comment added by 72.39.249.249 (talk) 17:09, 13 November 2007 (UTC)

Free-fall due to gravity on earth is at an acceleration of about 32 feet per second, per second. Note it's "per second per second", so it's not a simple speed of falling. What it is, is how fast the speed increases: every second of falling, one is falling at a speed 32 feet/second faster. It's a linear increase in speed because gravity is an approximately constant force in this situation. DMacks 17:15, 13 November 2007 (UTC)

Neglecting air resistance and the changes in gravity due to distance and (eventually) relativistic effects - then over time:

• Your accelleration is constant (at roughly 9.8 ms^-2 - which is about 32 feet per second per second).
• The speed with which you are falling is increasing linearly.
• The distance from your starting point is increasing exponentially (but not in the mathematical sense of e^x).

SteveBaker 17:38, 13 November 2007 (UTC)

Actually, that's not an exponential increase. For the speed to increase, then we need to have $v(t)=ab^{t}+c$. The velocity, is, in fact, given by $v(t)=ct+v_{0}$ where c is a gravitational constant (32 ft/s² or 9.8 m/s²) and v₀ is the starting velocity. Donald Hosek 17:43, 13 November 2007 (UTC)

The technical term for this type of increase is "increasng polynomially." That's faster than "increasing linearly" but slower than "increasing exponentially." in this case the ploynomial is of order 2 (the square,) so we can also say is is O(n²) -Arch dude 17:51, 13 November 2007 (UTC)

• And therefore we can alternatively say "increasing quadratically". --Anon, 23:35 UTC, Nov. 13.

(ec) Let's run successive integrals: constant acceleration, linear speed, quadratic position. DMacks 17:52, 13 November 2007 (UTC)

see also Big O notation. -Arch dude 17:53, 13 November 2007 (UTC)

Also note that terminal velocity will tend to put a cap on acceleration if you give it enough time (and aren't in a vacuum) GeeJo ^(t)⁄_(c) • 18:18, 13 November 2007 (UTC)

Length of US State Coastlines

I saw a VERY cool entry called "List of Countries by Length of Coastlines"

And naturally wanted to extend that to our 50 States and was hoping to find that already computed somewhere .....

Has it been?

Many thanks! Drew

(Email address removed to prevent every spammer sending you 'special offers'... Skittle 18:45, 13 November 2007 (UTC))—Preceding unsigned comment added by 71.160.50.240 (talk) 17:58, 13 November 2007 (UTC)

I did some digging, but I have yet to find such a list. In fact, I'm having trouble finding the info state-by-state. We could compile a list and create a Wikipedia page with it if we ever find the facts. --Milkbreath 20:12, 13 November 2007 (UTC)

You could start here - just the Atlantic and Gulf coasts. Cheers Geologyguy 20:17, 13 November 2007 (UTC)

Here are the rest. Cheers Geologyguy 20:18, 13 November 2007 (UTC)

It's also traditional to bring up How Long Is the Coast of Britain? with these questions, emphasizing that states like Maryland can see a particularly large coastline versus what might be expected. — Lomn 20:19, 13 November 2007 (UTC)

One source (see the section "Coastline and Shoreline" and note the disclaimer about being "generalized" and made from "small scale maps" -- meaning the figures probably err on the side of being too small)

NOAA has been creating high-quality shoreline datasets recently. These would probably provide the most accurate info on coastline length currently available. I'm not sure if they have finished building the datasets yet, or whether one would need to use something like GIS to get a list of coastline lengths by state. The main webpage for the project is here (NOAA's Coastal Geospatial Data Project). The project includes far more than just coastline data. A page specifically on shoreline data is here (NOAA Medium Resolution Shoreline, GIS Data). More info here. There's tons of pages and info on these NOAA pages, so it might take some effort to figure it out. Still, the best coastline length data would be from there. I looked at the two PDFs linked above and it looks like they use NOAA data, but not quite the most recent data, athough I could be wrong. Pfly 21:10, 13 November 2007 (UTC)

This is a famous introductory problem in fractals. Basically, there is no such thing as an "accurate" estimate of coastline length, as the coastline is self-similar to multiple scales. Any measure of coastline must state what the length of the "meter stick" used to measure is. Otherwise, you get a length that approaches infinity with increasing precision. Two separate books will quote two wildly different figures because of this. SamuelRiv 23:47, 13 November 2007 (UTC)

You could give the length as a function of how close you looked, couldn't you? Also, what if you took the difference in area covered in high tide and low tide, and divided that by the average distance between the high- and low-tide shores? — Daniel 00:34, 14 November 2007 (UTC)

Coastlines are fractal-like and self-similar among some scales. But when measuring at a scale fine enough to be affected by tides, for example, the analogy of coastlines as fractals breaks down. The problem then is one of defining what to measure -- low tide line, high tide? Also, rivers often widen into estuaries and merge with the sea without an obvious demarcation. At the kind of measuring scales common to things like the NOAA projects, these are the problems with coastline measurement, not fractals. Plus, when measuring at this fine scale, differences of coastline definition (where in the tide zone to measure, where to delineate estuary vs. sea, etc) are not going to result in the kind of wildly divergent figures one finds at the larger scales where fractal issues arise. In short, the idea that coastlines are fractals is only true under certain conditions and scales -- coastlines are not fractals in the strict, mathematical, absolute at all scales notion. Like most real world phenomena, mathematical analogies break down at some point, once you look close enough. I suspect the fractal coastline thing has been a little too much of a good introductory example for explaining fractals. But remember the difference between pure mathematics and the real world. Coastline lengths do not "approach infinity" as measurement precision increases. Just go to the beach with a centimeter-long ruler and take a stab at measuring the coastline and you'll find the problem isn't so much that you are measuring something fractal but more that you keep getting your feet wet as the waves wash up; and what time is low tide today anyway? (sorry for being overly critical, the fractal coastline thing is a minor pet peeve of mine!) Pfly 07:35, 14 November 2007 (UTC)

It is fractal to many orders of magnitude. Obviously it is not a perfect fractal. But let's say you were on a rocky coastline, not a sandy one, so tides wouldn't affect things too much. How do you measure into the crevasses of each rock where water gets in with a centimeter ruler? The point is that there's practically no such thing as an "true" coastline measurement. SamuelRiv 13:14, 14 November 2007 (UTC)

I absolutely agree - there is no correct answer to this question - it depends ENTIRELY on how precise your ruler is. The difference between measuring with a 10m ruler and a 10cm ruler is huge - it's not going to be a 1% error or even a 10% error - the length you end up with could easily double when you measure around every little inlet and rock rather than taking 'short cuts' across them. Truly, we should not be encouraging people to look at the results of this or that survey - the correct answer to this question is that the length of a coastline is unknowable. SteveBaker 18:30, 14 November 2007 (UTC)

I understand the inherent problems, but I still don't see the harm in making a list of US states ranked by coastline length based on a measurement system standardized for the nation. Just make sure to point out the uncertainty of the figures and the vagueness of what a coastline is in the first place. That's what is done with lists of rivers by length (or ought to be done!), even though measuring river lengths has the same problems as measuring coastlines. Pfly 22:44, 14 November 2007 (UTC)

User:Milkbreath made the page yesterday List of U.S. states by coastline. Perhaps the participants in this thread may want to add the appropriate caveats and links to fractals. Cheers Geologyguy 22:48, 14 November 2007 (UTC)

Squinty eyes focus better

I notice my vision has changed since I originally got my glasses (I'm near sighted). However, I notice that if I slightly squint my eyes while wearing glasses, my view is much clearer. What's the science behind this? Am I slightly reshaping my cornea by squiting or is less light passing through causing my iris to have a higher "f stop" or something entirely different? --24.249.108.133 18:13, 13 November 2007 (UTC)

First, we have to get this out of the way... This is not a medical diagnosis. Some Wikipedia users feel that any description of the human body is a diagnosis and delete both the question and answer out of fear of someone suing Wikipedia.

It is very common to have extremely clear vision when you first get a new pair of glasses (or contacts) and then it blurs. Being nearsighted, your eye is basically too long for the lens. When you get new glasses, the muscles learn to relax and make it just a tad longer. By squinting, you do two things - you put pressure on the lens and you shorten the length of the eye. Combined, this corrects the nearsightedness problem. Of course, you can't squint hard enough to correct any significant level of nearsightedness - so walking around squinting all the time is not an option to wearing proper corrective lenses. -- kainaw™ 18:19, 13 November 2007 (UTC)

Kainaw, could you please avoid dragging meta-discussions onto the desks? Particularly argumentative misrepresentations of people's motivations.

On another note, that wearing glasses causes vision to worsen is certainly not proved, according to my (actually trained) optician. It could be, but the evidence isn't there yet. (Also, I'd always heard that squinting was more to do with creating a smaller aperture than reshaping the eye, but I'm not so sure about that one). Skittle 18:44, 13 November 2007 (UTC)

I haven't considered it a "meta-discussion". There is no discussion. Some users delete questions and answers without any discussion at all. If you know of a better way to stop the problem, feel free to make a suggestion. I should note that this isn't an issue with one user and one question. I've seen many questions deleted by many users and I've seen no means of controlling the problem. -- kainaw™ 20:26, 13 November 2007 (UTC)

The relevant discussions have been had on the talk page many, many times. When people consider a removal contentious, they tend to discuss it on the talk page and determine consensus. The reasons for deleting medical questions and advice are explained at some length all over the talk page archives, as well as at a user subpage (I forget which one). If you have a problem with a particular deletion, please bring it up on the talk page and explain why you feel that particular deletion was inappropriate. If necessary, you could also place a message on the talkpage of the person who carried out the deletion. If you have further to say on this topic, please take it to the talkpage. Skittle 15:23, 14 November 2007 (UTC)

Squinting also has the effect of making your depth of focus deeper, similar to how a pinhole camera doesn't need a lens. --Mdwyer 20:28, 13 November 2007 (UTC)

I think that's a big part of it, and another part may be that "stopping down" the lens (using only the central part of it) reduces aberrations. Photographic lenses certainly benefit from being stopped down (until diffraction becomes important), and I would expect the relatively simple optical design of the eye to likewise benefit. -- Coneslayer 20:39, 13 November 2007 (UTC)

So in short, the answer is yes, squinting is like increasing your eyes' f-stop (i.e. narrowing their aperture). --Anonymous, 23:37 UTC, November 13, 2007.

AND the change in eye shape, as kainaw mentioned. I think they are both significant. --Mdwyer 03:42, 14 November 2007 (UTC)

So it's a bit of both, eh? Interesting. (BTW, I wasn't seeking medical advice. I was just curious about this phenomenon.) --24.249.108.133 18:14, 14 November 2007 (UTC)

Hard Coupled Forces Problem

Imagine a train consisting of N frictionless cars following a locomotive that is accelerating the whole train forward with acceleration of magnitude a. Assume that the first car behind the locomotive has mass M. If the tension in the coupling at the rear of each car is 10% smaller than the tension in the coupling in the front of the car, what is the mass of each sucessive car as a fraction of M? What is the total mass of the cars in terms of M? [Hint: You might find it helpful to know that 1+x^2+x^3+...+x^n=(1-x^n+1)/(1-x).] —Preceding unsigned comment added by 164.107.244.238 (talk) 18:59, 13 November 2007 (UTC)

How is it that you're in a position to give us a hint, yet unable to solve the problem yourself? -- Coneslayer 19:45, 13 November 2007 (UTC)

There's a good chance the hint was provided with the original homework question. — Lomn 20:14, 13 November 2007 (UTC)

It was in the original homework problem and you don't need to be a jerk coneslayer

Nor do you, and Coneslayer posed a fair question. Anyway, as noted above, the Reference Desk will not do your homework for you. If a particular part of the problem is causing difficulty, then perhaps we can assist or advise. — Lomn 20:29, 13 November 2007 (UTC)

(To anyone else who has solved this: it's quite neat, isn't it?) To the OP:We could certainly help you better if you told us what you have attempted and what you are having trouble with. No-one's likely to just post a full solution, and we can't give specific advice without knowing the problems you're having. For example, are you familiar with Newton's second and third laws of motion? Algebraist 20:42, 13 November 2007 (UTC)

When setting the accelerations equal

A1=A2

you can say

F1/M1=F2/M2

(T1-T2)/M=(T2-T3)/(M*x)

T2=.9T1

T3=.9T2=.81T1

so

(1-.9)=(.9-.81)/x

so x=.9

thus each car is ninety percent the mass of the car in front of it.

The total mass

Mt = M(1+x+x^2+...+x^(N-1)) = 10M*(1-.9^N) —Preceding unsigned comment added by 164.107.244.238 (talk) 21:44, 13 November 2007 (UTC)

Well done! Unfortunately, you have made a slight error: your arguments do not apply for the Nth (ie rearmost) car, which is not being pulled from behind. You need to consider this case separately. Algebraist 21:49, 13 November 2007 (UTC)

The condition in the question(T(K+1)=.9T(K)) will never apply to the last car. If it did, every tension would have to be zero. —Preceding unsigned comment added by 164.107.244.238 (talk) 22:33, 13 November 2007 (UTC)

Indeed, that condition must fail at the bacl. Thus the last car experiences a forward pull of TN=0.9^(N-1)T1, and no rearward pull at all. Algebraist 22:37, 13 November 2007 (UTC)

Is there a landmass under Antartica and the Arctic?

If the ice and snow melted, would you see dirt/rock? 64.236.121.129 20:09, 13 November 2007 (UTC)

Antarctica is a continent and thus has a landmass. The Arctic has no such continent. — Lomn 20:14, 13 November 2007 (UTC)

Antartica, yes. Arctic, no. —Preceding unsigned comment added by 134.84.156.80 (talk) 20:15, 13 November 2007 (UTC)

By the way, not even all of Antarctica is "under" any ice; there's a certain amount of bare dirt. Not a lot of it, but some. See Dry Valleys. --Trovatore 20:18, 13 November 2007 (UTC)

So is the Arctic just a big slab(s) of ice floating on water? 64.236.121.129 20:20, 13 November 2007 (UTC)

Generally speaking yes. See Arctic. And just for fun, you might want to read up on USS Nautilus (SSN-571). --LarryMac | Talk 20:27, 13 November 2007 (UTC)

If you look on Google Earth, there is actually nothing at all shown at the North Pole (the Arctic) while there is a large icy landmass at the South Pole, the overall earth imagery must have been taken during the Northern hemisphere summer months as the Arctic virtually all melts away...of course while it's summer at the North Pole, it's winter at the South, hence why the Arctic is not there but there is loads of ice at the Antarctic. GaryReggae 20:42, 13 November 2007 (UTC)

Antarctica is actually (at least) two landmasses. --Carnildo 23:42, 13 November 2007 (UTC)

Geography of Antarctica doesn't say that. On the other hand, Geography of Greenland does say that Greenland, the largest Arctic island, may really be three islands. The thing is that the ice sheet (in both places) is so thick that it's hard to measure what's under it. --Anon, 23:49 UTC, November 13, 2007.

The sub-ice configuration of both Antarctica and Greenland is fairly well known, and has been published in National Geographic atlases for years. One version for Antarctica is here. Cheers Geologyguy 23:58, 13 November 2007 (UTC)

The problem is that the land is pushed down under that mass of ice. If the ice ever melted, the land would tend to 'spring back' - increasing the height of all of the terrain. On the other hand, if that amount of ice were ever to melt, the sea level rise would inundate more of the antarctic continent. Between those two counteracting effects and our relatively poor view through all of that ice, I think it's hard to guess the precise result. SteveBaker 01:45, 14 November 2007 (UTC)

Uplift totally wins. The isostatic rebound at equilibrium would be roughly 35% of the ice thickness, which at places means more than 1 km of uplift. The sea level rise due to melting is only ~70 m. Discounting small islands, both Antarctica and Greendand would be a single land mass after accounting for the uplift. Dragons flight 16:06, 14 November 2007 (UTC)

Wow! That's a lot! I'd have guessed the rebound was only tens of meters...but a whole kilometer of uplift! Amazing! SteveBaker 18:24, 14 November 2007 (UTC)

Values around 300-350 meters of rebound are well documented in the Hudson Bay and Scandinavian Shield areas [21] - I couldn't find a ref quickly but I agree with Dragons Flight that values more than 1000 m are known. Cheers Geologyguy 21:08, 14 November 2007 (UTC)

But that uplift takes place over thousands of years (the Hudson Bay / Scandinavian Shield uplift mentioned is still ongoing since the last ice age), whereas the sea level rise could be much faster. Short term, the sea level rise might be the much more important effect. --169.230.94.28 21:57, 14 November 2007 (UTC)

Anyone have a 'sub ice' map of Greenland? I'd be interested in seeing that... --Kurt Shaped Box 02:10, 14 November 2007 (UTC)

Guess what! asked and answered right here! To a degree, at least. Cheers Geologyguy 15:41, 14 November 2007 (UTC)

Hard disk

Is there an associated change of mass, even at the atomic level, when writing data to a hard disk? In other words, does a full disk drive weight more than an empty one? Rockpocket 20:37, 13 November 2007 (UTC)

I doubt it, an HDD is a sealed unit so nothing can 'get in' to increase the weight. All the read/write heads do is rearrange magnetic particles but those particles have to be there in the first place. GaryReggae 20:39, 13 November 2007 (UTC)

(Off on a tangent...) Hard drives aren't actually sealed; there's always some path that allows a slight flow of air in and out of the head/disk assembly, if only to easily accommodate changing barometric pressure.

Atlant 02:33, 14 November 2007 (UTC)

No, you aren't adding or removing particles. You are flipping existing particles upside down. --Mdwyer 20:48, 13 November 2007 (UTC)

Doubtful, even on a subatomic level. Writing to a hard drive does not involve adding or removing matter, and as far as I'm aware, the magnetic potential energy of a full hard drive does not differ from that of a blank one. --Carnildo 23:45, 13 November 2007 (UTC)

Well (blatantly armchairy speculation ahead, beware), when you write to a disk, you're changing its entropy. A disk with all 0's or all 1's on it has less entropy, and is therefore (I think) somewhat less energetically favorable, than a disk with random patterns of 1's and 0's. A disk with highly ordered patterns of 1's and 0's obviously has less entropy, too. It takes energy to create order, and (if what little I know about thermodynamics is accurate) there's a certain energy cost in creating order / decreasing entropy above and beyond (or perhaps I should say below and beneath) the energy required to actually (in the case of a magnetic disk) induce the tightly-polarized little magnetic fields and stuff.

Now, as we all know, devices with potential energy stored in them weigh slightly more, due to e=mc². So does a device that contains an infinitesimal amount of anti-entropic energy in it therefore weigh an even more infinitesimal (less tesimal?) amount more? (I have no idea, and actually I kind of doubt it, because I think I'm mixing apples and apples here, because I'm not at all certain that entropic randomization of an unstably ordered state is an exothermic reaction; rather, it's an inevitable result of other, exothermic reactions. But anyway. Perhaps someone who knows more about this stuff than I do can set me straight.) —Steve Summit (talk) 03:28, 14 November 2007 (UTC)

I think Information entropy and Thermodynamic entropy are different. --antilived^{T | C | G} 05:07, 14 November 2007 (UTC)

I'm surprised that nobody has mentioned the actual effect utilized to store data in most modern hard disks, giant magnetoresistance. GMR does not involve rearranging particle position per se, and I'd be hesitant to foster weird notions about physics by comparing particle spin with macroscopic orientation. Basically GMR enables a large resistance change in material stacks with differing magnetic orientations because it takes more energy to cause electrons to flow through a magnetic domain if their spin is aligned differently than the magnetic moment ordering of the domain. (see ferromagnetism) Back to the original question, any mass gained or lost is totally negligible; macroscopic object masses fluxuate many orders of magnitude more when a dust particle settles on them. (if that's unsatisfying, then the random accumulation of electrical charge may provide compelling reason not to care much about these infinitesimal mass changes) -- 128.61.20.199 13:19, 14 November 2007 (UTC)

Entropy is not a form of energy. Sure, you need non-thermal energy to reduce entropy, but this energy is converted to heat (which can easily leave the hard drive).

The energy, and thus the mass, of the hard drive is probably higher when there are long runs of all 0's or all 1's, compared to alternating 0's and 1's, due to magnetic repulsion between domains of the same orientation. Icek 18:07, 14 November 2007 (UTC)

Antenna length

The radio station WUMB has a page here that suggests a way to make a dipole antenna. They say each branch of the antenna should be 30.5" long to optimize your reception of their 91.9 MHz signal. I've read Dipole antenna, which shows a formula of 468/f(MHz) feet, which leads to an antenna length of 0.5223 inches. Can anybody explain how WUMB arrived at their figure, and how this might be generalized to any radio frequency? Thanks, jeffjon 20:39, 13 November 2007 (UTC)

I think you've made a math error; 468/91.9 = 5.09 feet, which is 61 inches (or 30.5 inches for each half). -- Coneslayer 20:43, 13 November 2007 (UTC)

Hmm. The only thing that's better than a math error, is a math error that's so obvious I can't fathom how I might have made it. Thanks. jeffjon 21:36, 13 November 2007 (UTC)

If you're wondering where this seemingly arbitrary number comes from, it corresponds to a little less than half the wavelength (in air). Exactly half the wavelength gives two quarter wave rods, which provides the best resonant properties. However, the antenna input impedance is slightly reactive if its length is exactly a half wavelength, and adjusting the length down a bit causes the reactive part to vanish. Purely real impedances are very easy to match, and conveniently the real impedance of a slightly-less-than-half-wave dipole is fairly close to 75 Ohms. You can, of course, match an antenna with complex-valued input impedance, but it's additional trouble, could involve this beauty, and easier for hobbyists to just cut the antenna to appropriate length. The main reason you want the antenna impedance matched to the line is so the maximum amount of power can be delivered to whatever is at the other end (a LNA, perhaps) instead of being reflected back. As I hinted at above, the dipole configuration mentioned has a real impedance that results in a very low amount of reflected signal due to antenna mismatch when connected to a 75 Ohm line, less than 2%, actually. -- 68.158.1.192 06:53, 14 November 2007 (UTC)

Simplified Electromagnet Math

I'm doing the grade-school experiement of making electromagnets in my kitchen. I'd like to maximize the strength of the magnet while minimizing the copper. It seems like I should be able to plug in all the variables and get an answer. The math is pretty daunting, however! Is there a simplified way to handle this math:

$H\approx 787\ {\mbox{ampere.turns/meter}}$

For sxample, say I've got a .25" thick nail, and I'm going to run 12VDC through it. Is there a way I can draw a graph of turns vs ohms and figure out where the sweet-spot is? In the above math, is 'meter' actually the square-meter cross-section of the nail?? --Mdwyer 21:08, 13 November 2007 (UTC)

Yeah, this is a pretty gnarly optimization problem. It wouldn't be so difficult if you assumed an ideal voltage source (capable of supplying unlimited current) and a linear magnetic material (that reacts the same to an arbitrarily strong magnetic field and never saturates), but that's totally unrealistic. In practice, you're going to be limited by the internal resistance of the power supply (which limits the amount of current you can draw from it) and the magnetic saturation of the ferromagnetic iron core (which limits the magnetic field, at about the amount you quote). Frankly, I think trial and error would be easier to do than a thorough theoretical analysis, and guaranteed to give realistic results. —Keenan Pepper 23:18, 13 November 2007 (UTC)

The theory is straightforward: magnetic inductance is linearly proportional to the number of turns. Increase turns, and you increase inductance. The resistance encountered per length of wire depends on the wire's gauge (look it up), but will be negligible until you approach many hundreds of turns. Basically, from what I recall, we got about 8 ohms per meter of very thin copper wire in one experiment. Magnetic inductance is also directly proportional to current, so at constant voltage, doubling the number of turns doubles the length but reduces current by half, so your net gain is zero. Basically, in the end resistance is not going to be your problem. SamuelRiv 23:54, 13 November 2007 (UTC)

On this page there's a chart showing that, given a fixed voltage, they got a stronger magnet by using fewer turns of thicker wire, but the magnet had heat problems. That just seems backwards to me, for some reason. I always see magnets use lots of turns of really fine wire, but the page seems to suggest that, given a big enough power supply and a way to handle the heat, you should actually be using fewer turns of thicker wire. Am I still missing something? --Mdwyer 03:55, 14 November 2007 (UTC)

Yes. The magnetic field is proportional to the current flowing through the wire, and the current is limited by the internal resistance of the power supply and the resistance of the wire itself. The way to make a strong magnet is not only to have many turns, but also thick, low-resistance wire, and something to pump a heck of a lot of current through it. The resistive cores of the most powerful magnets in the world look like this: Florida-Bitter coils. See how much metal there is? The only holes are for cooling, because they have to pump dionized water through the magnet incredibly fast to keep it from overheating. —Keenan Pepper 06:13, 14 November 2007 (UTC)

Geez... Given the insane power and cooling requirements for those 35 T magnets, I'd think a high-temperature superconducting coil would be more effective. -- 128.61.20.199 12:47, 14 November 2007 (UTC)

They do use superconducting magnets as well, but the problem with that is that superconductors tend to expel magnetic fields (the Meissner effect), and if the field becomes too strong, the superconductor changes phase into a normal, resistive state. So there is a fundamental limit to the strength of a superconducting magnet, and the strongest magnets are in fact hybrids, with a resistive core inside a larger superconducting magnet. —Keenan Pepper 15:50, 14 November 2007 (UTC)

If I were faced with the design problem you stated, I would want to know the characteristics of the 12 volt DC supply. Is it like a car battery, which could supply hundreds of amps for a short while, or is it like a lab power supply, which might be current limited to a few amps or a fraction of an amp? As for heat problems, how long does the magnet have to stay energized?Many electromagnetic devices are designed for temporary or intermittent operation. As for the wire, magnet wire is made with a thin coating of insulation, so that more turns can be placed closer to the magnetic core than would be the case for, say doorbell wire or wire intended for 120v or 240 v AC service. Joseph Henry in the 1830's created the first truly powerful electromagnets (able to lift thousands of pounds) by using many turns of insulated wire. One thing he did was to place several windings on the same core, in parallel, so that each winding had a low resistance and therefore could carry relatively high current, but the multiple windings added their magnetizing effects. A nail makes a poor electromagnet compared to a u shaped or horseshoe shaped core, where the pole pieces are in proximity. Edison 20:27, 14 November 2007 (UTC)

BINDEEZ

Hi, I need a FULL ingredient list for Bindeez/ aqua dots —Preceding unsigned comment added by 68.193.21.179 (talk) 21:30, 13 November 2007 (UTC)

Since they're not considered food, there's little chance of a publicly-available breakdown of what goes into the product. — Lomn 21:33, 13 November 2007 (UTC)

Pretty sure it would be a trade secret. bibliomaniac15 A straw poll on straw polls 23:21, 13 November 2007 (UTC)

Unless it was patented. But I don't see any obvious patent claims, so maybe it's just a trade secret. --24.147.86.187 01:13, 14 November 2007 (UTC)

If you're interested in the "date rape drug" aspect, see our article on it Bindeez, which describes what chemical was substituted for the one that was supposed to be there. --24.147.86.187 01:13, 14 November 2007 (UTC)

Two identical humans

Two identical human both "weights" 70 kilograms.

They both jump off a high place (simultaneously). One of them is attached to a frictionless pulley and to a 65 kilograms counterweight. Assuming no air resistance, will both of them hit the ground at the same time? Will the one with the counterweight survive the fall if the height is deadly for the other human? 202.168.50.40 23:21, 13 November 2007 (UTC)

More homework, is it? We don't do homework. --Anon, 23:51 UTC, Nov. 13, 2007.

I'd be a bit surprised if none of us do homework. :P —Tamfang 23:59, 13 November 2007 (UTC)

Some of us don't even have homework :-p Someguy1221 00:20, 14 November 2007 (UTC)

But for the OP, this is a very straightforward problem; draw out your force diagrams, and calculate the acceleration (including direction) of everything involved. Someguy1221 00:22, 14 November 2007 (UTC)

And they 'll both die since no air resistance implies no air so they'll asphyxiate. Donald Hosek 00:24, 14 November 2007 (UTC)

When I was designing problem sets (for the one class I taught that required them) I really wanted to kill off a few of our hypothetical astronauts and things like that, just to see if the kids were paying attention. --24.147.86.187 01:16, 14 November 2007 (UTC)

I think they both hit the ground at the same time because for the human with counterweight, his/her mass is effectively only 5 kilograms (70kg - 65kg = 5 kg). But Gallieo says that both masses will fall at the same rate in Vacuum, so they must both hit the ground at the same time regardless of their different mass. 202.168.50.40 03:46, 14 November 2007 (UTC)

Hmm, interesting no one has picked up on the important principle here. Try a reductio ad absurdum thought experiment. Suppose that the second man weighs 65 TONS instead of kilograms. Now would they both hit the ground at the same time, because Galileo said that in a vacuum objects will fall at the same speed? Hmmm…ponder ponder….

The answer is that the second man will fall much more slowly because of the inertia of the 65 kg weight he has to move. Inertia of massive objects is very important in these considerations. For example people think that because objects are weightless in space, you can just pick up a 1 ton block and throw it around like a beach ball. Wrong. The block doesn’t want to move. You would still have to exert a lot of effort to get it going. Try this next time you are down at an ice rink. Get your skates on and put some castor type skates on a grand piano. Now try moving the piano around the rink (get permission from the manager first). You will find that it’s very difficult to start and stop the piano, and to steer it. It will feel like the piano is trying to waltz YOU around the rink. Similarly, while heavy objects will fall at the same speed in a gravitational field (let’s just forget about air resistance, it’s negligible for massive objects), they don’t behave the same way if something is trying to stop them. That’s why a large asteroid will make a huge hole, while a little meteorite won’t. The second man can be said to weigh only 5 kg, and he would feel like he only weighs 5 kgs when he is wearing his counterbalance on the ground. But does that mean he will fall with the same speed as the 70 kg man, because different weights fall at the same speed (a la Galileo)? No, because the second man is, to use the old-fashioned terminology, expending most of his “potential energy” doing the work of lifting the 65 kg weight upwards, while the first man can covert ALL of that potential energy into kinetic energy. And I eat my homework. Myles325a 05:28, 14 November 2007 (UTC)

Blade metallurgy chart

I've been searching for a chart that shows the properties of differant metals found in blades including iron and titanium alloys. I can't seem to get anything showing many metals or numarical properties that I'm searching for. I would like to know strength hardness and efficiency of hardening on the alloys, not just this can hold an edge better then that. —Preceding unsigned comment added by 71.102.37.149 (talk) 23:41, 13 November 2007 (UTC)

November 14

What is a Dye Front?

I'm working on a PCR and I need to divide the distance of the bands that I have by the Dye Front, but our teacher never explained what this looks like. Can someone give an explanation or a diagram? Thanks. M@$+@ Ju ~ ♠ 00:29, 14 November 2007 (UTC)

It is the front edge of the dye as it travels across the gel. this page has a good picture with notation. If you can't see the dye front, it may have left the gel - which means you can't do analysis on it. -- kainaw™ 03:32, 14 November 2007 (UTC)

Thanks very much! M@$+@ Ju ~ ♠ 03:58, 14 November 2007 (UTC)

labrador coat

Is it normal for a labrador puppy to have a thickest fur along its "back"? (My dog-related English speaking is not the best, I hope you understand it anyway). Thanks in advance. 217.129.241.186 00:30, 14 November 2007 (UTC)

My girlfriend worked a couple years in the kennels of a large guide dog school that uses labradors almost exclusively, so she’s something of an expert on labs. According to her, yes, it’s normal. She also says that a lab’s coat changes substantially during its first year (it becomes coarser, sometimes wavier, and is two layers), so what a puppy’s fur looks like is a lot different from what it will look like as an adult. MrRedact 16:41, 14 November 2007 (UTC)

Our Lab/Retriever has much coarser hair down her back than elsewhere - I think it's typical of the breed. I somehow got the idea that it was an adaptation for swimming since the soft fur is what ends up getting wet and the coarse fur stays pretty much above water...but I can't imagine why that's an advantage! I have a book about labs at home - I'll see what it says. SteveBaker 18:19, 14 November 2007 (UTC)

Chemtrail Crazies, why is the bandwagon so big?

In my opinion the whole chemtrail phenomenon is a load of bollox. From what I've seen from online research is that those who believe in chemtrails piece together unrelated elements onto a big illogical ball of *cough*. I don't know of anyone on MCAGCC 29 Palms running around with black lung desise and their hedaches are from working with the Marine Corps or PTSD. What I want to know is how people hear these illogical reffrences to contrails being poison and then they fall for it the second they hear about it? Is it that people in general are just that gullible or the bandwagon effect at it's best, because thats all I can think of. This may seem like a nonsensical rant but I have a friend who claims the government is out to get us all and everyone seems to believe him when he mentions chemtrails. —Preceding unsigned comment added by 71.102.37.149 (talk) 01:52, 14 November 2007 (UTC)

Perhaps the chemtrails page can help you. Delmlsfan 02:27, 14 November 2007 (UTC)

(ec) This comic may provide some insight. I agree that the chemtrails stuff is one of the dumbest conspiracy theories around (on par with the "we never landed on the moon" thing). I can't give you any kind of definite explanation, but I do have some ideas about this kind of behavior. I think the whole thing has to do with Occam's razor. This tells us that a good strategy for finding the best explanation for some observation is to pick the simplest explanation that fits the facts. While this usually works, it does leave the person using this strategy very exploitable (and the human mind is shaped by eons of lying and exploitation). For instance, if my best friend is sleeping with my wife, all he has to do to stop me from figuring it out, is make sure that all clues I might find have a simpler explanation than him sleeping with my wife. Humans seem to have evolved an ability to second guess Occam's razor (which may actually be what we call paranoia) in order to save ourselves from being exploited this way. Part of this happens when an explanation that is more complex involves someone benefiting from us using Occam's razor. Once I realize that there is an alternative hypothesis where my best friend wants me to accept the simplest explanation, that hypothesis gets a lot of attention, and I reject Occam's razor.

I think this is the mechanism that generates conspiracy theories. Al-Qaeda explains 9-11 without leaving anything open. The Apollo program is a perfect, flawless explanation for the moonlanding footage. All the simplest possible explanations work perfectly. Yet the suggestion that somebody wants us to believe the easiest explanation seems impossible to reject. We are so tempted by these theories, because if we weren't we'd be easy to take advantage of. And then there's the confirmation bias. Once the paranoia kicks in, subtlety seems to go out the window. All we're concerned with is finding confirmation, rather than verifying the hypothesis. I'm not sure how to explain that, but I think it has something to do with defense mechanisms lacking the subtlety of everyday mechanisms, since they're only activated situations that warrant defense. risk 02:52, 14 November 2007 (UTC)

Well, there there are a number of reasons why people believe weird things like "chemtrails":

• People don't like to believe that they are easily fooled. So, when they think that they've seen through a "trick" or "conspiracy" it makes them feel smart. Or, if they're being told that they're being tricked or fooled, they may prefer to accept a belief that makes them feel more clever than the people they believe are attempting to trick them. The problem is, they can fool themselves this way, and after that admitting that they have made a mistake by believing foolish claims is even worse than saying that they used to be tricked by someone quite clever.
• People like to feel that they are special. This is related to the previous item. Part of the reason why people believe some weird things is that it makes them feel like one of the "brave few", or some such thing, in that they have some special knowledge, power, purpose, or experience that most others do not have.
• People don't like things they don't understand. Things like vapor trails and epidemiology are difficult because they require an understanding of science, however "planes spread sick juice" is simple. This leads me to the next point.
• It's easy to make claims that are hard to refute, even when those claims are entirely false. Some people tend to accept any claim that is not easily disproved if they like it, and thus fall into the trap of believing things simply because they don't understand why it's probably wrong. For example, "you're an adulterer" is an easy claim to make, but can be impossible to refute even if 100% false.
• People prefer beliefs that reinforce their worldview. If they are the kind of people that want to believe that the government is evil, then "chemtrails" or "9/11 was an inside job" is an appealing idea to them. This appeal becomes more important to them than evidence for or against those ideas. You see this in debates where you show evidence against the conspiracy you're discussing, and rather than show evidence for their conspiracy, they start talking about other conspiracies as though they are evidence that their conspiracy is true.
• People like the idea that someone powerful controls things. Thinking that things occur due to intelligent intervention allows people to believe that things don't just happen randomly, but that there is someone who has control of things, and thus they might also be able to control things. Random is scary, thus it's comforting to think that someone is in charge because then you can feel that things happen for a reason, and perhaps you can understand that reason.
• We tend to see patterns, even in entirely random events. This is related to the previous item, but it also means that people may see all sorts of data, and assume that they're related. They often see the "hits" that fit the "pattern" they're looking for, while ignoring most or all of the misses. Even in entirely random data, patterns will appear if you have enough data. The problem is that some people assume that "there are no coincidences" or "it's too coincidental to be random", therefore believe that any patterns must have some cause.

These reasons, among others, are why people tend to believe weird things for very poor reasons. You can take a look at most weird beliefs and see some combination of those factors at play. This is why we need to be skeptical of claims that are not supported by good science.

That being said, there really aren't that many people on the "chemtrail bandwagon", but the vocal few make it seem like it's a more popular belief than it really is. If you go to a place where people promote the idea, you'll see a disproportional percentage of the population that believe the idea. But start asking around randomly and you'll see how few people actually believe it. Hope that helps! -- HiEv 11:33, 14 November 2007 (UTC)

I would add:

• The truth is neither as interesting nor exciting as the fiction.
• People would love to find an 'easy way out' of whatever bothers them.

This is why things like water fuelled cars continue to be bandied about. It's interesting to hear that some guy with almost zero science training (they always have "almost zero science training") was able to figure this out when all of those smarty-pants scientists couldn't. People like to think that there is a quick fix for global warming just around the corner - and the only thing that's stopping it is 'big oil' suppressing the technology - that makes the whole problem "somebody elses fault".

It's really the same with this one: You're feeling tired and stressed out? Well no wonder! The government are spraying us with all of those tranquilizers! It's much easier to blame "them" than to start taking exercise and eating right. The truth, by comparison, is very boring. Water fuelled cars simply cannot work because of the first law of thermodynamics...jeez - what a boring answer! You scientists are just too dumb to figure out how to get around it like that guy in the garage did...Or: Contrails look different at different times because of weather conditions in the upper atmosphere - and besides, there are contrails all over the USA - how would the politicians doing this manage to avoid breathing them in themselves - and how would the enormous chemical industry needed to produce these chemicals manage to keep it so secret? All of those explanations are just tedious compared to a good old conspiracy theory.

SteveBaker 18:16, 14 November 2007 (UTC)

Polystyrene cups

What are the effects from drinking hot beverages out of Polystyrene cups? Zain Ebrahim 09:13, 14 November 2007 (UTC)

Could you please explain what you mean by the question? Do you mean the effects on the person drinking from the cups or the environmental impact from the practice of using these cups?

In the first case, I would have thought there was no difference to the person drinking whether they were using a polystyrene cup or a ceramic one. Polystyrene is generally inert at the temperatures it would be subjected to by this use (if it melted due to the heat or reacted with the fluid then it would not be a suitable material for that use).

The environmental effect on the other hand is more serious, recycling of plastic cups is very limited at best and due to the fact that they are so inert, they sit in landfill sites for hundreds of years and biodegrade incredibly slowly (if at all). Producing plastic cups utilises precious hydrocarbons which could (arguably) be put to better uses than disposable cups and once they are landfilled or incinerated, the hydrocarbons are gone.

There are also several different types of polystyrene cups; the rigid plastic ones you commonly get by water coolers and the foamy (expanded polystyrene) ones hot drinks come in. I assume you mean the latter?:GaryReggae 12:47, 14 November 2007 (UTC)

On the other hand, polystyrene undergoes depolymerization pretty easily. (Note - we need a separate page on depolymerization). So it can be recycled into styrene monomer and reused. Options such as wax-covered paper cups are not nearly so recyclable. Delmlsfan 13:10, 14 November 2007 (UTC)

The critical thing with plastics is not whether they are "recyclable" - it's whether they are "recycled". It's no use having a substance that's easy to recycle only to find that people toss them into landfill anyway. That's the problem with plastic cups of all kinds - they are very rarely collected up and sent to the special recycling center that can turn them back into something useful. Paper cups are not recyclable - but at least when you toss them away, they behave much like a pile of leaves would - and decay down to normal soil nutrients in a matter of months.

Thanks a lot. I was referring to the effects on the person drinking from the cups. And yes, I meant the expanded polystyrene ones. Zain Ebrahim 13:38, 14 November 2007 (UTC)

The polystyrene is pretty inert - it's not going to harm you - the only thing that might be a problem is if the manufacturers use some kind of mold-release agent to make it easier to get the cups out of the injection molding machine. But I seriously doubt they'd use anything problematic. SteveBaker 16:57, 14 November 2007 (UTC)

chemicel contents of ferro titanium

i would like to know the chemical contents of ferro titanium.if the compound has silica 3.64%,iron (fe2o3) 35.25%,titanium(tio2) 48%,mangansese 2.55%,tungsten(w)6.09%,molybdenum 2%,carbon 0.79%,alumina 1.93% and its bulk density is 2.476% what is this material called. —Preceding unsigned comment added by Sammy 203 (talk • contribs) 10:03, 14 November 2007 (UTC)

Help on the Antlion

Please help?

I'd like to know how to go about answering the below mentioned challenge copy 1 vs copy 2 presented to 3-12 year olds

Copy 1.I also challenge you to find a sticky substance in an antlions hole. The antlion is the larvae. The struggles of the ant to get out of the hole cause the dry sand to slide down trapping the ant. The antlion flicks sand up from below with it's pincers until the ant is within reach and it can grab the ant. Anything sticky would cause the sand granules to stick together making the antlions hole useless.

Copy 2. The Antlion Antlions think ants are so delicious, they spend their whole lives building sneaky traps in dry sunny soil for ants to fall into. They even live in these traps - and can dig one in just 15 minutes! They then prepare a gooey, sticky substance called Larvae, and wait for some unlucky ant to fall in! When the gooey Larvae traps the ants in the hole the antlion get very happy and say "yummy, a fine breakfast is served!"

Any help in this regard would be much appreciated

Confused!÷ —Preceding unsigned comment added by 196.36.13.113 (talk) 13:51, 14 November 2007 (UTC)

I can't make heads or tails of this. The whole thing is in non-standard English and makes little sense. Where did it come from? There is nothing sticky in the hole, especially not the larva (larvae is plural) which is the ant lion itself. They don't spend their whole lives catching ants; they only do that as larvae. See Antlion. --Milkbreath 14:55, 14 November 2007 (UTC)

Plasma rockets

If the rocket shoots out plasma instead of fire, will it go faster? 64.236.121.129 18:04, 14 November 2007 (UTC)

Thrust is not dependent on "fire" versus "plasma", but rather the mass and velocity of the expelled material. — Lomn 18:45, 14 November 2007 (UTC)

Plasma would come out faster I think since it's hotter and it's trying to escape to an area of lower pressure. 64.236.121.129 19:59, 14 November 2007 (UTC)

Fire is plasma, you will find many ions in a flame. Graeme Bartlett 20:12, 14 November 2007 (UTC)

Neither criteria you mention (heat, pressure) is meaningful. Heat does not affect mass or velocity, and high-to-low pressure is a function of all rockets. A more useful consideration is the combustion characteristics of your fuels: what provides the most thrust per fuel mass? What byproducts are acceptably non-toxic? What fuels are feasible to store and handle? These are the sort of concerns rocket engineers have. — Lomn 20:38, 14 November 2007 (UTC)

Gold Detecting

Would gold show up on a metal detector? What kinds of metals show up? —Preceding unsigned comment added by 209.203.103.2 (talk) 18:24, 14 November 2007 (UTC)

Yes. See prospecting and metal detector.--Shantavira|^{feed me} 18:31, 14 November 2007 (UTC)

Dried Banana Chips

Do dried banana chips still contain high level of potassium similar to fresh bananas?84.66.175.242 19:23, 14 November 2007 (UTC)

Speaking out of pure speculation, I would expect the majority of the potassium to go with the water (i.e, out of the bananas) during the drying process, being that potassium ions, K⁺, dissolve very well in water. shoy (^words _words) 19:59, 14 November 2007 (UTC)

When banana is dried evaporation is used. Only water and volatile substances leave it. Potassium is not volatile at the low temperatures used, so it stays in the banana. But be aware some dried banana has sugar and oil added, so it may not be so healthy as you think. Graeme Bartlett 20:05, 14 November 2007 (UTC)

Why don't they make bullets out of steel instead of lead/tungstun?

What advantages does lead have over steel bullets? 64.236.121.129 20:01, 14 November 2007 (UTC)

Off the top of my head, bullets are swaged when fired. This means softer metal is better. If the bullets were as hard as the barrel, I imagine the barrels wouldn't last very long. Friday (talk) 20:03, 14 November 2007 (UTC)

Yea but, if they were made of steel, you could shoot into water with them. I saw on Myth Busters, they fired bullets into water to see if water was "bullet proof", and the bullets fired from high powered rifles had a tendency to shatter when they hit the water. If they were made out of steel, they wouldn't shatter. 64.236.121.129 20:08, 14 November 2007 (UTC)

Tungsten is not soft, however like lead it is dense, so it will carry more kinetic energy for a given velocity and size. This means it will travel further through air before slowing down. Graeme Bartlett 20:11, 14 November 2007 (UTC)

Some bullets are meant to expand on impact, some are not. However it's possible when hitting something like water than even a full metal jacket bullet might come apart. I imagine this would happen more with smaller ones than bigger ones, but I'm just speculating. Bullets are designed for traveling through air rather than water. You might want something less bullet-like if you're intending to shoot through water. Friday (talk) 20:13, 14 November 2007 (UTC)

Yea, full metal jackets were used, and they shattered on the water. I think steel bullets would have greater armor penetration than lead. The tungsten sounds promising, but I read on the tungsten page that steel with small amounts of tungsten greatly increases its strength. So I think tungsten/steel bullets would be best. Yea bullets are designed at shooting through the air, but if you are trying to kill someone under water, it would be best if your gun's bullets can penetrate that water. 64.236.121.129 20:19, 14 November 2007 (UTC)

Even if the bullet stays together, I can't imagine you'd get much penetration. Water is dense. It's probably not as bad as shooting into sand, but it's not going to be easy to get through it. Friday (talk) 20:24, 14 November 2007 (UTC)

Yea, water is dense, but if the bullet can stay together, it means that at higher velocities, you will indeed get greater penetration. This contrasts with the fact that at higher velocities, a lead bullet will simply be more likely to shatter, which ironically means, a handgun has more penetration into water than a machine gun. 64.236.121.129 20:29, 14 November 2007 (UTC)

Even the handgun bullet doesn't penetrate very well through water. Are you imagining some military application? I can't think of why you'd want your gun to be able to shoot someone underwater. Wait til they come up, and shoot them then. Or, if they're in a submarine, they have torpedoes for that. Friday (talk) 20:33, 14 November 2007 (UTC)

I didn't say handgun bullets penetrate very well through water did I? I said handguns ironically penetrate through water better than high powered rifles because the bullet doesn't shatter. You are arguing strawmen, please don't do that, it's annoying. Yes, I am imagining a military application. I'm talking about guns and bullets, DUH! What application would we be using this for? Lol. You shoot at someone in the water to KILL them! DUH!!! A gun that can shoot at targets in the water AND the air is superior to a gun that is limited to only targets in the air. You can't figure that out? Lol. 64.236.121.129 21:41, 14 November 2007 (UTC)

Not exactly bullets, but they use steel shot in shotguns for waterfowl. The lead shot was poisoning the wetlands. The shot has a sleeve around it to keep it from wrecking the barrel. It doesn't hit as hard or go as far, but who wants lead in their duck dinner? --Milkbreath 20:24, 14 November 2007 (UTC)

Wait that makes no sense dude. Steel should hit harder than lead. 64.236.121.129 20:29, 14 November 2007 (UTC)

No, lead is denser. Steel is harder but when you're looking at retained velocity of the shot, density helps. Friday (talk) 20:31, 14 November 2007 (UTC)

No, mass helps. Newton's 2nd law. Hmm, I do agree that lead would retain velocity better than steel. But I think a steel/tungsten alloy would be best, as is indicated in the tungsten article. 64.236.121.129 21:41, 14 November 2007 (UTC)

For a bullet of a given size, density is effectively mass. A lead bullet masses more than a steel bullet because lead is denser than steel. — Lomn 22:07, 14 November 2007 (UTC)

Supposing steel is superior to lead for firing into water -- why do you want to fire into water? That's a really really really specific consideration with no practical relevance that I can see. Even if you want to shoot a guy who's underwater, doesn't he have to surface? Waiting to shoot him then (with a standard bullet) not only solves the problem of bullet composition but also energy dissipation, aiming despite refraction, and a host of other things. Ultimately, the answer to this "why don't we change..." is the same as several others -- it's a solution in search of a problem.

(ec) I see Friday has raised the same question above. — Lomn 20:34, 14 November 2007 (UTC)

You can't figure out that a gun that can shoot at targets in both the water and the air is better than a gun that can shoot at targets only in the air? You can't figure that out? 64.236.121.129 21:41, 14 November 2007 (UTC)

Again, it's a solution in search of a problem. BTW the people here are trying to be helpful and answer your questions. There's no need to get testy. It's less obvious that a shooting-through-water gun is useful than a shooting-through-air gun. You still need to be able to see your target and aim at it, for example. Water makes this harder. Friday (talk) 21:45, 14 November 2007 (UTC)

BTW, the article sucks but there are Spearguns meant to shoot through water. But, it's quite different than a bullet. Friday (talk) 20:35, 14 November 2007 (UTC)

Yea but a speargun isn't that good at killing targets in the air! Versatility is always better than a gun with limitations. 64.236.121.129 21:41, 14 November 2007 (UTC)

Shooting through walls, for example, is arguably even more useful than shooting through water, depending on the application (in some applications, excessive penetration is specifically not wanted.) Mass of the projectile (as well as hardness) generally help penetration- but it means less ammo that an individual is able to carry. Also note that copper-jacketed lead bullets are cheap to produce. Do you want ammo that penetrates more, if it's significantly more expensive? Like most practical questions, it's about making trade-offs. Friday (talk) 21:52, 14 November 2007 (UTC)

(ec) And so by your argument everyone should be driving SUV's, or even better, tanks around, just in case it needs to go off road into the mud or cross trenches, even though all you do is dropping the kids to and from school and grocery runs? Can you think of actual military or civilian needs to be able to shoot guns in water? Why sacrifice capabilities for its usual purpose for some obscure, not commonly used purpose? --antilived^{T | C | G} 21:56, 14 November 2007 (UTC)

Y'know, thinking this through, I realized that at no point has the OP submitted evidence that steel bullets would be preferable for shooting into water. With no evidence of my own, I would surmise that steel's hardness (relative to lead) would in fact make catastrophic fracturing even more likely than that of lead bullets, since shattering is a characteristic of hard things. Anyway, I think it's quite clear that Friday and I aren't idiots. If a gun/bullet system existed that fired at submerged targets just as well as at normal ones, with no meaningful disadvantages, it would be... nice? But let's face it -- there are disadvantages. Given that the last time soldiers were forced to fire rifles at submerged targets as a best-case option was... let's see... never, I'm quite confident in my opinion that this is a dead-end. Defending against frogmen is a possibility, but the concussive force of a small depth charge (that is, a hand grenade) works just as well and is already widely deployed. Our opinions are most certainly not the result of failing to imagine that a magic gun would be cool. — Lomn 22:04, 14 November 2007 (UTC)

Mystery of the freezing water

A few years ago in a trip to Norway early one morning, my friends and I were about to get on the road with our car. The temperature was about 0 degrees celsius or maybe below, and we found that we'd left two glass bottles with some water in the car overnight. Picking up the (sealed) bottles for taking a sip, we saw the water freeze and turn into ice in just an instant. Can anyone please explain this? Gil_mo 21:25, 14 November 2007 (UTC)

Your water was supercooled. Picking it up jarred it and started nucleation. Friday (talk) 21:30, 14 November 2007 (UTC)

Levitating at the center of the world

There's a well in my backyard. It's lined with a material that makes it almost impervious to heat and pressure. (Pretty amazing stuff!) The well is so deep it actually runs straight through the center of the world and connects up both Antipodes. Yesterday my poor cat Mr. Cuddlypants accidently fell into the well. Am I safe to assume Mr. Cuddlypants is now levitating at the center of the world? Sappysap 22:04, 14 November 2007 (UTC)

Eventually, yes, due to atmospheric drag. Negate that and he'll cycle back and forth from endpoint to endpoint. — Lomn 22:14, 14 November 2007 (UTC)

Should I continue to throw food down the well for him? —Preceding unsigned comment added by Sappysap (talk • contribs) 22:43, 14 November 2007 (UTC)

If you throw food down and hit him, it will kill him. Of course, if there's any water levitating at the center of the well, your cat would have died when he hit it. If there isn't any water, what was the point of the well? — Daniel 22:45, 14 November 2007 (UTC)

What if there's water at the other end - i.e. the antipodal point is somewhere in the ocean? Man... --Kurt Shaped Box 22:53, 14 November 2007 (UTC)

Why do raw carrots taste like soap?

I've recently began eating raw carrots, peeled, they seem to leave a soapy aftertaste. I've tried fresh carrots from other store and they have the same soapy taste. Is this because of chemicals used by farmers, should I be worried? Or perhaps certain chemicals, fatty acids... gives the taste buds a similar sensation as when one smells a bar of soap? Thanks for any info! --Taktser 22:08, 14 November 2007 (UTC)

I don't get that taste, myself. This may be a stupid question but have you washed them? Friday (talk) 22:23, 14 November 2007 (UTC)

Well, if he/she peeled them him/herself, then washing isn't really a major issue. I don't taste anything soapy though I do think they have an odd texture and smell. It might be one of those things that some people are more sensitive to than others (i.e. like those people who really can't enjoy raw tomatos because they taste more acidic to them than to others). --24.147.86.187 23:02, 14 November 2007 (UTC)

Potassium/Rubidium-filled bullets?

How effective would bullets containing small pieces of metallic potassium or rubidium be against human targets? Would it be possible to seal small fragments of the pure alkali metal inside lead bullets which were designed to shatter upon penetration - with the end result being that the target/victim takes a round to the chest, then almost immediately begins to be incinerated from the inside out as the alkali metals react with his own bodily fluids...

I'm mainly thinking of the possible applications with regard to sniper rifles and near-certain one-shot kills. I'm aware that pure alkali metals are expensive. --Kurt Shaped Box 22:44, 14 November 2007 (UTC)

Sodium bicarb and ethanol turning black

What might react with baking soda to turn a solution of water and ethanol black? Solution probably also contains ethyl acetate, fusel oils and trace amounts of methanol. Lowerarchy 23:05, 14 November 2007 (UTC)