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January 3

Cirrus clouds and snow

How can a really thin ciruus cloud produce so much snow?Accdude92 (talk to me!) (sign) 00:45, 3 January 2010 (UTC)[reply]

You might get more responses if you let people know what you are talking about. Dauto (talk) 02:21, 3 January 2010 (UTC)[reply]

The question seems simple enough and clear enough. If you cannot respond to it, no posting is necessary or desirable. How does a large amount of precipitation fall when no large and dense cloud cover is visible? Edison (talk) 02:33, 3 January 2010 (UTC)[reply]

I beg your pardon but I don't think it was clear at all. He seemed to be talking about an specific event without telling us what the event was or giving us some description of what happened. Dauto (talk) 18:15, 3 January 2010 (UTC)[reply]

Our article, and numerous other sources on the internet, agree that cirrus clouds do not produce significant precipitation. However, they can be a precursor of other weather patterns. — Lomn 02:55, 3 January 2010 (UTC)[reply]

Well, I could easily see through the cloud to the blue sky, yet is was snowing lightly.Accdude92 (talk to me!) (sign) 03:05, 3 January 2010 (UTC)[reply]

Your first post said "so much snow" which, to me, would suggest a large amount. Now you're saying it was "snowing lightly". So, which is it? Dismas|^(talk) 03:09, 3 January 2010 (UTC)[reply]

Maybe the OP means a lot of snow considering it was a clear sky. It may have been snow blown in from a nearby cloud of a snowsquall, or perhaps even diamond dust if the humidity was high. Probably wasn't the thin cirrus clouds, even as there were sun pillars at my location (S. Ontario) today. ~A H 1^(TCU) 03:49, 3 January 2010 (UTC)[reply]

Exactly.Accdude92 (talk to me!) (sign) 04:19, 3 January 2010 (UTC)[reply]

Common cold in animals

Other than humans, can other animals (like domestic cats) get the common cold--from the rhinovirus? rʨanaɢ ^talk/_contribs 01:53, 3 January 2010 (UTC)[reply]

Chimpanzees, apparently. Scientists have been able to infect mice with it as well, but it takes some work, it won't happen automatically. The linked-to article discusses some of the reasons why other species can't be infected by it—they lack certain receptors for it. I suppose chimps are close enough to humans for it to work. --Mr.98 (talk) 02:21, 3 January 2010 (UTC)[reply]

Thanks! rʨanaɢ ^talk/_contribs 02:35, 3 January 2010 (UTC)[reply]

The rhinoviruses are normally species-specific, and some animals, especially cats, have their own versions of the common cold. Dbfirs 08:08, 3 January 2010 (UTC)[reply]

Thrust Curve

What is a thrust curve and how can I get one for my motorbike? 86.26.12.165 (talk) 11:41, 3 January 2010 (UTC)[reply]

I've never heard of a "thurst curve" applied to a ground vehicle. Usually, there are power and torque curves, which are plots that show the engine's power and torque output vs. RPM. Presumably, these can be converted in to an effective thrust, but that terminology is more usually used for aircraft. These curves give you an intuitive feeling for the efficiency and acceleration capacity of your engine, and give you an idea of the optimal speed and RPM you should aim for when driving your bike. You can get such curves from enthusiast websites or from the dealer. Here is an introductory article from Total Motorcycle, including charts for Honda Nighthawk bikes and comparable engines, HP and Torque vs. RPM. Nimur (talk) 16:20, 3 January 2010 (UTC)[reply]

(a) it is the thrust v. time graph of a solid-fuel rocket engine, see Solid-fuel rocket#Grain geometry. (b) don't do it unless you are looking to get a Darwin Award for stupidity. SpinningSpark 16:20, 3 January 2010 (UTC)[reply]

I'm surprised Wikipedia has no article on torque curve, power curve, or dyno chart. Power curve directs to power band, but I've never heard that terminology applied to anything (except maybe power chords). We also have engine tuning. Nimur (talk) 16:51, 3 January 2010 (UTC)[reply]

A thrust curve, also known as a tractive force graph, is the linear force at the tyre/road interface and is the force you feel when you are actually riding your bike. Due to most bike's gearing, this tractive force is highest in 1st and decreases with each higher gear, hence you get faster acceleration in lower gears. Some dynos can display this directly on a grpah, or it can be calculated by knowing engine torque and the bike's gear ratios. The maths is a little long to go into here (and i can't write it very well on here), but its fascinating. Alaphent (talk) 16:52, 3 January 2010 (UTC)[reply]

Distance time graph

As the article on distance points out, distance cannot be negative and is not characterized by direction. However, there are several distance time graphs "out there" that do include a sense of direction. For instance, on this site a graph is shown in which the red downward sloping line represents the return journey. But isn't this incorrect? Distance is ignorant of direction, and thus, the return journey should increase the "distance in m" instead of decrease it. In my opinion, this should be called a "displacement time graph". Am I wrong? Lova Falk (talk) 12:12, 3 January 2010 (UTC)[reply]

I would imagine that 'distance' in this case means 'distance from origin' even if that's not explicitly stated, which would decrease on the return journey, rather than 'total distance covered'. Distance cannot be negative but a change in distance can be. Mikenorton (talk) 12:33, 3 January 2010 (UTC)[reply]

That is just my point! "Distance from origin" is displacement. Lova Falk (talk) 12:37, 3 January 2010 (UTC)[reply]

Well, not exactly, as we're only talking about the scalar part of the displacement vector. You could be walking in a circle centred on the origin, your displacement would be constantly changing while your distance remained the same. Mikenorton (talk) 14:54, 3 January 2010 (UTC)[reply]

Clearly, the terminology is used loosely in practice. Abuse of notation is always a problem in scientific/technical presentations - unfortunately, even the best scientists occasionally use the wrong word. Lova Falk seems to understand the intended meaning of the graph, so the actual text-label is irrelevant. Nimur (talk) 16:57, 3 January 2010 (UTC)[reply]

I think 'didtance IS the correct term for that specific graph linked by the OP. That's the term I would use. Dauto (talk) 18:12, 3 January 2010 (UTC)[reply]

Just because it can't be negative doesn't mean it can't decrease (which involves only its slope being negative). You can certainly measure the total distance traveled, as on an odometer, which must clearly be a monotonically increasing function. It's just the integral of speed over time, like the total displacement is the integral of velocity over time. One can also consider the distance from a fixed point to a moving object as a function of time; that function will always be non-negative, but it is free to increase and then decrease, even to 0. --Tardis (talk) 20:19, 4 January 2010 (UTC)[reply]

Plant species mentioned in "Guns, Germs and Steel"

I'm just reading "Guns, Germs and Steel" and cannot correlate some trivial names of plants to species:

Which species of Chenopodium is meant by "Goosefoot"?
Which species of Dioscorea is meant by "African Yams"?
Which plant does the author has in mind when referring to it only as Groundnut?

--Pjacobi (talk) 13:26, 3 January 2010 (UTC)[reply]

There are a number of possibilites for the species of Chenopodium and Dioscorea, and I don't know which ones Diamond is referring to (quite possibly to more than one species, or possibly not to specific species at all). However, the "groundnut" is the peanut Arachis hypogaea.--Eriastrum (talk) 17:06, 3 January 2010 (UTC)[reply]

The goosefoot cultivated in what is now the eastern U.S. was Chenopodium berlandieri.--Cam (talk) 05:16, 4 January 2010 (UTC)[reply]

Diamond is referring to ancient African cultivation of groundnuts, but peanuts are of American origin. He may mean Bambara groundnuts (Voandzeia subterranea).--Cam (talk) 05:33, 4 January 2010 (UTC)[reply]

Many thanks for your answers. Also I now have seen Eastern Agricultural Complex, an article which I should have found earlier by more careful reading. Some quick Google Scholar seem to support the first impression, that the EAC article and the articles about the related plant species, need improvement and update from somebody with access to relevant scientific journals. --Pjacobi (talk) 06:51, 4 January 2010 (UTC)[reply]

percentage illumination of moon's surface by sun light

Could you please indicate what percentage of the total moon's surface is illuminated by the sun at a given time196.209.216.206 (talk) 13:39, 3 January 2010 (UTC)[reply]

Why should it not be 50% if the Moon is a sphere ? Even correcting for a non-spherical Moon, it must be very close to 50%. Gandalf61 (talk) 13:51, 3 January 2010 (UTC)[reply]

A lunar eclipse would change that, albeit briefly. Vimescarrot (talk) 14:20, 3 January 2010 (UTC)[reply]

percentage illumination of moon's surface by sun light

The distance between sun and moon is very large and the rays would probably be virtually parallel when arriving at the moon. Given the fact that the sun is much larger, is it possible that this percentage of illumination might be slightly larger than 50%? —Preceding unsigned comment added by 196.209.216.206 (talk) 14:20, 3 January 2010 (UTC)[reply]

If the moon were a perfect sphere, then yes, slightly more that 50% would be illuminated. I think 50% of the ground would be able to see at least half the disc of the sun, but there would be more that can see just a little bit sun sticking up above the horizon. However, the moon isn't a perfect sphere - there are lots of mountains and craters. Mountains near the terminator (the boundary between light and dark) can be illuminated when flat ground wouldn't be and the insides of craters can be dark when flat ground would be illuminated. I don't know which of those factors is bigger or by how much, so I don't know if that would reduce it by enough to take the total illuminated area to less than 50%. --Tango (talk) 15:56, 3 January 2010 (UTC)[reply]

Maybe also some refraction through the moon's (albeit thin) atmosphere to push it up a couple of percent as on Earth? - Jarry1250 ^{[Humorous? Discuss.]} 18:44, 3 January 2010 (UTC)[reply]

Tango I think you are forgetting that the light is not collimated, and that the size of the sun is significant. —Preceding unsigned comment added by 92.22.51.77 (talk) 19:33, 3 January 2010 (UTC)[reply]

Does Earthshine when the moon is more than half full (from Earth) and light filtering through the Earth's atmosphere during a lunar eclipse increase this figure to more than 50%? ~A H 1^(TCU) 20:24, 3 January 2010 (UTC)[reply]

Earthshine will illuminate the dark side of the moon, but not very much - the Earth is far far dimmer than the Sun. Lunar eclipses will reduce the amount of illumination, although light refracting round the Earth will reduce the extent of that - they are rare enough that they won't have a large impact on the average illumination. --Tango (talk) 20:34, 3 January 2010 (UTC)[reply]

I don't think I'm forgetting either of those things. What makes you think so? --Tango (talk) 20:34, 3 January 2010 (UTC)[reply]

The OP is asking a simple question in geometry, don't they teach this stuff any more? The increase in area illuminated for a perfect sphere for small subtended angles is the ratio of the moon-sun distance to the diameter of the sun. This is approximately 1.4x10⁶/1.5x10⁸, or just under 1%. SpinningSpark 21:51, 3 January 2010 (UTC)[reply]

Consider a spherical cow in vacuum. . . Cuddlyable3 (talk) 22:34, 3 January 2010 (UTC)[reply]

That is a bullshit response to a reasonable answer to the OPs question. Of couse it is an approximation, but a valid one. I knew it was an approximation, the OP knows it is an approximation, and there was no need for you to tell us you know it is an approximation as well with a sarcastic response. If you have a more accurate model, I'm sure we would all be delighted to hear about it. SpinningSpark 23:22, 3 January 2010 (UTC)[reply]

The ideal sphere approximation was adequately applied by Tango in the first response. Unlike SpinningSpark, Tango pointed out the secondary effects which make the OP's question not a simple one in geometry. An exact calculation of the area of the Moon that is illuminated directly by the Sun would be a ray tracing exercise based on the actual Moon terrain. This is one of several possible sources of Moon relief maps. From Earth we never see all of the illuminated area of the Moon. Please hold back rhetoric such as "don't they teach this stuff any more?" that shows diminished respect for the preceeding responders. I am sorry that the metaphor of the Spherical cow may have upset SpinningSpark as sarcastic but it is regularly quoted in astronomy sources, see the article. Cuddlyable3 (talk) 00:36, 5 January 2010 (UTC)[reply]

Why is the Large Hadron Collider underground?

The Large Hadron Collider was built in the tunnel formerly occupied by the Large Electron–Positron Collider. Is there any particular reason why either of them had to be built underground? I imagine it must be much more expensive than building it above ground, even allowing for the cost of short bridges to carry roads over it. Or railway-like cuttings to keep it level. 78.146.54.230 (talk) 15:10, 3 January 2010 (UTC)[reply]

Particle accelerator says "Very large circular accelerators are invariably built in underground tunnels a few metres wide to minimize the disruption and cost of building such a structure on the surface, and to provide shielding against intense secondary radiations that may occur. These are extremely penetrating at high energies", but does not provide a reference for either claimed reason. Finlay McWalter • Talk 15:15, 3 January 2010 (UTC)[reply]

According to Lyn Evans, the project manager of the Collider: "Cheaper... It would cost a fortune to acquire the land in France and Switzerland to build the racetrack on the surface." See [1].--Fuhghettaboutit (talk) 15:19, 3 January 2010 (UTC)[reply]

Building it underground helps block some of the background radiation from cosmic rays and that way improve the signal to noise ratio. Dauto (talk) 16:09, 3 January 2010 (UTC)[reply]

Is there a reliable source that supports this claim? Super-Kamiokande is 1000 meters down, Sudbury Neutrino Observatory is 2000 meters down. But LHC is only 50m down. -- Finlay McWalter • Talk 17:14, 3 January 2010 (UTC)[reply]

Safety reasons come to mind. The total kinetic energy of the two proton beams is 362 megajoules. That's enough energy to accelerate a Nimitz-class aircraft carrier from rest to more than five knots — or to push a three-ton SUV to nearly twice the speed of sound: [2]. A full-power beam can bore a hole forty meters into a block of solid copper. Safely dumping those beams at the end of experimental run requires controlled beam defocusing and scanning across a custom-designed ten-ton cylinder of graphite, embedded within a thousand tons of concrete and steel shielding. The beam dump hits a peak temperature of 750°C (about 1400°F) and takes a few hours to cool down. [3] If a group of bending magnets fail simultaneously, I'd rather have that uncontrolled beam go through the wall of the facility into nice, solid, Swiss (or French) rock. TenOfAllTrades(talk) 17:03, 3 January 2010 (UTC)[reply]

I guess that's what the article means when it says "to provide shielding against intense secondary radiations that may occur. These are extremely penetrating at high energies" - that the surrounding rock acts as a beam quench in such emergencies. That would explain why linear accelerators, whose failure modes can't send the beam of in any direction, aren't underground. We need to get a clear ref for this for the particle accelerator article. -- Finlay McWalter • Talk 17:19, 3 January 2010 (UTC)[reply]

Does that "secondary radiations" include Synchrotron radiation as the beam is bent? That's another serious concern around the periphery of the ring that is a non-issue in linear configurations. DMacks (talk) 19:19, 3 January 2010 (UTC)[reply]

But if the beam is somehow deflected upward and it can penetrate 40 metres of copper while the tunnel is 50 metres underground, wouldn't there be a slight risk of the beam going above ground? ~A H 1^(TCU) 20:19, 3 January 2010 (UTC)[reply]

Do you have a mechanism in mind that might unintentionally deflect the beam upward? It seems to me that anything that accidentally got in the way of the particle beams would just have holes cut through it. --173.49.16.188 (talk) 22:36, 3 January 2010 (UTC)[reply]

The holes would be microscopic - the beam is very narrow. What if I walked above the beam with a really big magnet? I wonder how big of a magnet I'd need to really mess up the beam. The idea is that left/right the beam is controlled by magnets that might fail. But up/down it needs no control, since it just travels in a straight line. So all it needs is focusing magnets, but not steering magnets. Ariel. (talk) 23:02, 3 January 2010 (UTC)[reply]

The LHC got shut down when a bird dropped a baguette into a cooling unit. Imagine the trouble a bird could cause if the whole machine was on the surface; never mind the havoc a squirrel could create! 75.41.110.200 (talk) 23:51, 3 January 2010 (UTC)[reply]

Here's a link for the bird-baguette story for the curious. -- Scray (talk) 01:08, 4 January 2010 (UTC)[reply]

Things beside baguette fall out of the sky all the time and in Tennessee for instance not only is there danger from hunters chasing dear but from law enforcement faring at anything that might resemble a still, not to mention drunks firing at anything that is metal and has writing on it. 71.100.3.13 (talk) 12:58, 6 January 2010 (UTC)[reply]

Amateur 3D video, also sound editing question.

I have a mid to high-end HD video camera with a hard drive and AVCHD recording format http://www.sony.co.uk/product/hdd-avchd-hard-disk-drive/hdr-xr520ve and wonder if I got a second HD camera (perhaps a cheaper one) or an identical model, would this would be cheaper to film in 3D and somehow use software (assuming there is any) to convert it to an "official" 3D movie format later (assuming there is an "official" 3D video standard yet) or would a dedicated 3D HD camera ever come down in price enough for ordinary consumers? Which option would be the cheapest? Also, when filming Rozi Plain's new song "Wiggling",I spontaneously said "Oh Yeeeaah" in an adenoidal and pervy way. This ruined my recording at the start of the song. If I said the same thing and recorded it repeatedly to create an average signal, is their any software I could buy that could subtract my awful voice and restore the original notes at the beginning? —Preceding unsigned comment added by 80.2.199.128 (talk) 15:33, 3 January 2010 (UTC)[reply]

Per your first question, you can create a 3D movie in that way; the hardest part is the projection technology. If you're willing to force viewers to wear 3d goggles, you just need two video channels to make this happen; a lot of high-end graphics cards (such as those designed for dual monitors) now have hardware acceleration for such two-channel video; NVIDIA 3D Vision is one such example. If you want to project with polarization/Real3D glasses, you need "professional" special purpose equipment. As for your audio trouble, your proposed method may work in theory, but in practice you stand a better chance manually editing the audio track, applying a series of filters. Alternatively, consider re-recording the song and overdubbing the first few seconds - this will be much easier than subtractive audio processing (which is more art than science). Nimur (talk) 15:44, 3 January 2010 (UTC)[reply]

So humans have feet, dogs have paws, horses have hooves...

...and what do ducks have? Are they flippers? Are they feet? It sounds like a bit of a silly question, but I haven't been able to find a clear answer. Thanks! --saxsux (talk) 15:37, 3 January 2010 (UTC)[reply]

Ducks have webbed feet. Nimur (talk) 15:44, 3 January 2010 (UTC)[reply]

Mercury-vapor lamp usage

I understand that streetlights many times use mercury-vapor lamps. What type of areas or towns or cities favor these type of lights, verses say sodium-vapor lamps or other gas-discharge lamps that use an arcing mechanism?--Doug Coldwell ^talk 15:47, 3 January 2010 (UTC)[reply]

Just a gues but somewhere I read that mercury vapor lamps produce a very large amunt of ultraviolet and are therefore valuable in making germicide lamps. 71.100.1.76 (talk) 16:28, 3 January 2010 (UTC) [reply]

I also posted an answer on the humanities reference desk. For the other question, please read the article on street light interference. ~A H 1^(TCU) 19:58, 3 January 2010 (UTC)[reply]

I think it's a matter of color versus cost. Sodium lamps produce more orange-colored light - which is not popular in areas where people are working, eating or shopping - but adequate for driving cars. Mercury lamps produce whiter light - but they use more energy and cost more to make - and when they have to be replaced, the mercury they contain is not good for the environment. SteveBaker (talk) 21:00, 3 January 2010 (UTC)[reply]

What type would you guess are these in Street light interference?--Doug Coldwell ^talk 00:39, 4 January 2010 (UTC)[reply]

I would believe that the whole business of Street light interference is just so much bullshit. Hence it doesn't matter a damn what kinds of light they are. Between people who are so desperate to feel more important than everyone else (who'll make up any story to make that happen) and observer bias (if a street light just happens to turn on or off as you walk past...) it's all just complete nonsense. We don't need to explain anything. I'm sure that if you (or someone you know) can reliably do this then James Randi will give you a million dollars when you demonstrate it under reliable scientific conditions. SteveBaker (talk) 01:32, 4 January 2010 (UTC)[reply]

Jeezz Steve take it easy... Here is a perfect example of where the Wikipedia WP:NOR policy has failed. The validity of the subject matter is not ours to question or refute so long as that content can be found in prior publication. Live with it. 71.100.3.13 (talk) 12:35, 6 January 2010 (UTC) [reply]

Results of oppositely charged or poled bodies in outer space

Normally when considering the dynamics of force in outer space one thinks of gravity only. But what about the magnetic/electric force which is likewise limited in intensity by distance like the force of gravity but which intensity can be increased or decreased irrespective of distance. My question is what if the force responsible for propelling the universe to expand is opposite electric charges and/or opposite poles? Has an experiment ever been done using either or both electric charge or magnetism to see what would happen to identical objects with opposite charges or poles over a long period of time while orbiting some body in space... and to see if their distance apart can be controlled by varying the intensity of the field? 71.100.1.76 (talk) 16:25, 3 January 2010 (UTC) [reply]

Metric expansion of space is not due to electromagnetic interaction. However, electromagnetic effects are non-negligible in certain situations of orbital mechanics. Dynamics of Dust near the Sun describes electromagnetic interaction and radiative forcing for very tiny grains of dust. The consequence of this is an effectively "reduced gravitational force" - there's a name for this type of electrostatic drag (which eludes me now, I will look it up later). There is also Poynting–Robertson effect drag, due to electromagnetic radiation pressure, not net electric charge. Around planets such as Earth (because we have a geomagnetic field), there are electromagnetically trapped particles (ions and electrons), which we commonly call the van Allen belt. Those magnetic field alignment and field-trapping effects are only relevant for single-atom-sized objects; electrostatic effects can have consequences for objects as large as dust grains. As objects become larger than micrometeorite size, the effect of electromagnetic and radiative forcing becomes more negligible, and purely gravitational effects dominate the orbital mechanics. Experiments to quantify the total effective net charge of astrophysical objects like planets have yielded ambiguous results. We discussed the experimental and theoretical implications of electric charge on Earth, on May 9, 2009, "Net charge of earth". Nimur (talk) 16:31, 3 January 2010 (UTC)[reply]

Battery life

I have a bag full of AA batteries and a volt meter. I would like to discard the bad ones. Is there a minimum voltage level below which I should toss out a battery? What about using the low-voltage ones in a device that requires a low current, such as a portable clock? Hemoroid Agastordoff (talk) 17:07, 3 January 2010 (UTC)[reply]

The trouble with non-rechargeable batteries is that they retain the same voltage until just before they die. So you can't tell whether they are nearly dead. Worse still, a battery that went 'dead' while you were using it will recover a little after a matter of a minute or two and give a few seconds more life before dying again - so measuring the voltage with a meter might not tell you anything. Once the voltage has dropped significantly - you should toss it out. Using them in low-current devices like clocks might work for a short time - but it's hardly worth the effort. That said, non-rechargeable AA batteries are supposed to produce 1.5 volts - anything much lower than that means that the battery is probably dead. SteveBaker (talk) 19:46, 3 January 2010 (UTC)[reply]

I rarely disagree with Steve, but in this case I do. (Most) rechargeable batteries keep their voltage until shortly before they die - hence they are recommended for high-current devices like cameras. Non-rechargeables, on the other hand, do show a slow drop in voltage as they are depleted (as anyone will be able to attest who has ever uses a flashlight that went dimmer and dimmer and dimmer but kept going). --Stephan Schulz (talk) 21:40, 3 January 2010 (UTC)[reply]

A high impedance load such as a voltmeter would continue to show about 1.5v even with a somewhat worn out AA cell so, what would be a reasonable load resistor value in parallel with a voltmeter to better access its true condition? hydnjo (talk) 22:26, 3 January 2010 (UTC)[reply]

The no-load voltage from a cell (originally the word "battery" was reserved for multiple cells) doesn't show much difference between a new and a 50% depleted cell. I would discard as "bad" any non-rechargeable cell that gives 0.9V or less. Saving such cells, such as keeping them to run an undemanding clock, is risking leakage that can damage internal contacts, especially from cheap carbon/zinc cells. Cuddlyable3 (talk) 22:30, 3 January 2010 (UTC)[reply]

My recommended procedure for cheap cells is "remove from package; discard immediately". That way you don't find yourself wondering why the remote control for your new video player that came with no-name batteries is sitting in a lake of white goo on the coffee table. </off-topic rant> Franamax (talk) 23:36, 3 January 2010 (UTC)[reply]

Whilst I agree to some extent with Franamax above, I don't follow this advice. I hate to throw out cells that might have some remaining charge, so I waste time comparing voltages under both no-load and a heavy load (just an ohm or two) to see how much use they have left. I even have a multimeter that includes this function. I then throw out only the worst, and keep some "used" cells for non-critical use later (battery clocks are a good example where a significantly-reduced voltage often makes little difference to operation). It is probably not worth the effort, but I can't stop myself from doing this! Rechargeable batteries can also be compared by this method. Dbfirs 23:52, 3 January 2010 (UTC)[reply]

I hear ya, I hate throwing stuff away too (especially batteries, which should always be kept out of the waste stream). I just hate throwing away the entire device more than I hate taking a poorly-constructed battery to a hardware store for disposal. And why would you keep battery clocks anyway? I saw a TV show about that, it's so cruel keeping the clocks in a tiny cage. :) Franamax (talk) 00:42, 4 January 2010 (UTC)[reply]

They have more time to lay eggs that way. Unfortunately, they don't hatch into either new cells or new clocks, but I suppose one could use them to go back to a fully-charged time? :) Dbfirs 10:06, 4 January 2010 (UTC)[reply]

You might consider capturing the decision parameters of a battery tester (needle in red= bad/needle in green=good). For a 1.5 volt cell, such a tester would load the battery with a resistor which put some well chosen drain on it, and noted whether the voltage dropped below a criterion. The criterion might be 10% below nominal or 1.35 volts, and the load might be 300 ma, implying a load resistance of 5 ohms. If your device will function at a lower voltage, or if it has a different load current, the cutoff voltage and the load resistor might be chosen differently, per Ohm's Law. The "no-load" voltage as measured by a high quality voltmeter, is largely uninformative, except that any cell with a very low no-load voltage is shot. I have little faith in AA batteries with 1.3 volts or less with no load. An appreciable but small load will collapse the voltage of a battery in poor condition, but a good battery will sustain near-no-load voltage. Never test a battery with an ammeter, only with a load resistor and a voltmeter. The actual device to be operated, or its electrical equivalent, can be a good test circuit. Edison (talk) 01:22, 4 January 2010 (UTC)[reply]

I agree that the output voltage from a depleted cell at zero current can be surprisingly close to the nominal voltage. Therefore using a voltmeter to measure the voltage at zero current is not a suitable method to determine the state of the cell. The output voltage must be measured while the cell is supplying an output current. If the cell is is a good state the voltage drop will be small, but if the cell has little life left the voltage drop will be substantial. I have an inexpensive instrument called a battery tester that contains a galvanometer and internal loads. If you don't have such an instrument, any resistor of suitable resistance to serve as a load will do the job. As Edison suggested above, the actual device to be operated by the cell will make a good load while you measure the output voltage. Dolphin51 (talk) 02:56, 4 January 2010 (UTC)[reply]

A few years ago AA cells had a built-in tester which when activated (pressed) indicated the cells condition (depletion?). This feature has been since eliminated (cost vs benefit?) but at the time it showed some indication of remaining energy. I'm not the OP here but am asking: what would be the appropriate load (ohms) to read a voltage which would give some approximation of the remaining energy of an AA cell. hydnjo (talk) 03:17, 4 January 2010 (UTC)[reply]

My reading of the above suggests the answer to your question is: the load for which you propose to use the cell. That test strip was pretty cool, but ultimately just tested the cells ability to operate the test strip. Franamax (talk) 04:45, 4 January 2010 (UTC)[reply]

Avoiding the British winter

Where would be the nearest place to go to avoid the British winter and the oppressive long dark winter evenings? Going all the way to Australia or New Zealand would be one solution, but its a long uncomfortable plane trip. Are there any nearer solutions? Thanks 78.147.11.181 (talk) 21:20, 3 January 2010 (UTC)[reply]

Spain or the south of France? In Madrid, for example, the shortest day was about 9 hrs 20 mins this Winter (sunrise to sunset - sunset being 1751hrs). That compares with 7hrs 50mins for London (sunset 1553hrs local time). Some of the difference in sunset times is due to timezones - Madrid is pretty much due south of London but is an hour ahead (mainly to be the same same as France, I think). --Tango (talk) 21:33, 3 January 2010 (UTC)[reply]

On any particular date during the northern hemisphere winter, the farther south you go, the more daylight you get each day (until you hit 24-hour daylight somewhere in the Antarctic). So this is a matter how much more sunlight you would need to see in order to consider that you have "avoided the long dark evenings". If you go south and slightly west from England until you run out of Europe, you reach the Costa del Sol in Spain. At Malaga for today's date you have about 1 hour 45 minutes more sunlight than in London (even more if you live farther north in England). As Tango says, the time zone in Spain shifts the sunlight to later in the day, so Malaga today had sunrise at 8:30 am and sunset at 6:12 pm. Will that do? If not, you need to go further south. --Anonymous, 21:37 UTC, January 3, 2010.

The Canary Islands are a popular destination. Egypt is traditional. If you really want to hit it, go to Réunion, Mauritius or anywhere in Southern Africa. As opposed to South America or Australia, these are (a bit to much) closer, and they have much less difference in time zone, and hence less jet lag. --Stephan Schulz (talk) 21:59, 3 January 2010 (UTC)[reply]

IDoes anyone know if there is an article about the very large piece of rock in the Canary Isles that is prediced to slide into the sea and cause a large tsunami? I cannot find any. 78.147.11.181 (talk) 23:19, 3 January 2010 (UTC)[reply]

Mega-tsunami#Canary_Islands Franamax (talk) 23:38, 3 January 2010 (UTC)[reply]

La Palma has some on that subject. 75.41.110.200 (talk) 23:45, 3 January 2010 (UTC)[reply]

If you want 24 hours of daylight you need to go at least as far south as latitude 66.6° South where the penguins live. Alternatively move to Northern Norway and enjoy Scandinavian instead of British winter. There is less pollution. Cuddlyable3 (talk) 22:10, 3 January 2010 (UTC)[reply]

You could go to Chicago and gain more than a hour of daylight over London. Of course, Chicago was expecting -18 Fahrenheit windchill this morning so you have to consider more than simply hours of daylight. 75.41.110.200 (talk) 23:45, 3 January 2010 (UTC)[reply]

In Chicago, you could avoid the balmy British sort of winter where it rarely gets very cold, and experience true bone-chilling cold. Farther north from there could let you experience far colder weather. Edison (talk) 19:46, 4 January 2010 (UTC)[reply]

The nearest place to go to avoid the British winter would be the nearest place that is not Britain. Then you would be dealing with the Irish or French winter or maybe Isle of Man. The nearest place to go to escape oppressive long dark winter evenings is somewhere (a pub on trivia night, your friend's place, a concert, Wikipedia) where you are enjoying yourself and are simply not thinking about whether a long evening is oppressive or not. Changes in day length may (or may not) be associated with serious psychological effects and these can sometimes be countered with light therapy (or not). But many people do get the "winter blues", IMO there's not much point in wishing the Sun would appear more often, 'cause it never listens. Franamax (talk) 00:28, 4 January 2010 (UTC)[reply]

If avoiding Winter is the goal here then you are talking about avoiding only two flights a year. How about a slow comfortable ride on a boat to the British Caribbean with excursions to Brazil as part of your itinerary? 71.100.3.13 (talk) 12:45, 6 January 2010 (UTC) [reply]