Science desk
< June 5	<< May \| June \| Jul >>	June 7 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

June 6[edit]

Uranus & references to its original name[edit]

In the 'Naming' section of Uranus's page it states that Georgium Sidus was the name selected by Hershel. Are there any scanned textbooks or documents online from the time that show this being used in a list of the names of the planets? --Anonimasimio (talk) 09:02, 6 June 2012 (UTC)[reply]

Here are two Herschel documents about the planet,[1][2] neither including a list of names, but the latter includes in passing the names of both Jupiter and Saturn. Thincat (talk) 09:27, 6 June 2012 (UTC)[reply]

Interesting to see that Hershel called it The Georgium Sidus in writing. I guess what I'm most curious to find is the kind of list that would be given to school children to memorize. As in, here an the eight planets: Mercury... Saturn and Georgium Sidus. --Anonimasimio (talk) 09:31, 6 June 2012 (UTC)[reply]

Here is the Nautical Almanac for 1820 where the name is in a list and it is called "Georgian" (page 34, for example, of the PDF). The preface refers to "the Planet Herschel, called the Georgian Planet by us" (page 25). Hardly appropriate for school children though. Thincat (talk) 10:55, 6 June 2012 (UTC)[reply]

Thanks, Thincat. That's the first contemporary, non-Hershel reference I've seen. I'm digging this up because there is a modern story told by many of how Uranus was almost named 'George' but I've never been able to find anything from the time that actually uses that name. If anyone can find a that reference, or any other contemporary usage of the planets' name before we settled on Uranus, I'd be grateful. --Anonimasimio (talk) 11:56, 6 June 2012 (UTC)[reply]

I have seen (in the library of a National Trust property, but I won't say which one, because I don't think we were supposed to be looking at the books) a school atlas that lists the planet as "Hershall" (It was a long time ago, but I'm pretty sure of the spelling). --ColinFine (talk) 15:38, 6 June 2012 (UTC)[reply]

Anonimasimio, I've found a publication on Google Books by Tiberius Cavallo in 1803 that refers to Uranus as "the Georgian Planet" in several places -- [3] He does, in a footnote, mention that the planet is also called Uranus or Herschel by some. There are a number of other titles from the 19th Century in Google Books that also refer to it as "the Georgian Planet", but I thought Cavallo was of sufficient stature to be an important one to mention. I didn't hunt for "Georgium Sidus", your original question, but you may have luck searching for it as well. Best of luck with it! Jwrosenzweig (talk) 06:02, 7 June 2012 (UTC)[reply]

Another citation perhaps worth mentioning is this book by Jacques Ozanam [4], which I point out only because it's an example of a non-British scientist who is clearly familiar with the name (and appears to be using it as the planet's standard name, although I don't know if this book is translated from a French original) -- anyway, most of the citations I'm finding are from scholars born in or working in England and Scotland, and I thought it was worth pointing out that "Georgium Sidus" had traveled across the Channel as at least one of the planet's names, if not its standard name. I'll also note that I'm seeing citations as late as the 1850s that seem to be using "the Georgian Planet" as Uranus's standard name, though it might be worth investigating whether these are just reprints of earlier editions, or if the name persisted in use that long. Jwrosenzweig (talk) 06:10, 7 June 2012 (UTC)[reply]

To clarify, Sidus is merely Latin for "star" (which in the old sense included planets). And Georgium is the accusative singular of Georgius, a proper name from the ancient Greek Γεώργιος, a name based on γεωργός (farmer) = γῆ (earth) + ἔργον (work). We recognize Γῆ to this day as Gaea, and thus, oddly enough, Uranus was very nearly named after the Earth! Wnt (talk) 17:20, 7 June 2012 (UTC)[reply]

What's the distinction between sidus and stella? ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:58, 7 June 2012 (UTC)[reply]

That's a question we should give serious consideration to. -- ♬ Jack of Oz ♬ ^{[your turn]} 05:03, 9 June 2012 (UTC)[reply]

Sidus is a constellation, but it can also refer to a single star or at the other extreme the entire night sky.
Stella means a star or a planet.
Sidus gives us words like 'sidereal' and 'consider'. Stella gives us 'stellar', 'constellation' etc. And the words 'star', 'Astarte' (= Venus), 'Ishtar' (= Venus), 'asterisk', 'asteroid', 'astro-' words (like 'astrolabe', 'astrology', 'astronaut', 'astronomy', 'astrophysics', etc), 'disaster' and 'catastrophe' are all related. It's written in the stars. -- ♬ Jack of Oz ♬ ^{[your turn]} 05:26, 9 June 2012 (UTC)[reply]

Battery charging time[edit]

I have installed a custom made 12 V 750W UPS with 170Ah battery. The charging current is to 14A, so how long will it take to charge the battery.

2. If I use a 24V 1000W UPS with 2 170Ah battery what will be the charging time. — Preceding unsigned comment added by 182.185.144.194 (talk) 11:12, 6 June 2012 (UTC)[reply]

To properly answer this question, you need to tell us more about the battery and the load - I'll say more about this but first the simple, and very inaccurate answer:

The battery capacity rating Ampere-hours (AH) is defined as the product of current and hours from full charge to a defined flat condition under conditions meant to be typical for the battery in question. The charge capacity is roughly about the same as the discharge capacity. Thus, in your first case, time to battery "flat" = 170 Ah / 14 A = 12 hours. Since the maximum output of a 750 W 12 V power supply is 750/12 ie 62.5 A, I assume your load is equal to output - charge ie 62.5 - 14 = 48.5A. Your second question cannot realy be answered as you have not given the charging current. If it is assumed to remain 14 A then the charge time is unaltered.

In real batteries, the product of discharge (or charge) and current is a function of current. The capacity increases as current is reduced. Eg for typical lead acid batteries, halving the current will not double the discharge (or charge) time, it will as much as treble it.

In real batteries, the ampere-hour capacity is strongly temperature dependent. Capacity increases with temperature.

How long a charge takes or lasts is a function of how flat you can tolerate. There are industry standard cutoff voltages for each type of battery, so that you can make comparisons, but in practice your load may cease to work properly at a different voltage. Lead acid batteries in particular are damaged if flattened. Types intended for standby service will give a short life if routinely flattened more than 50 to 70 % of nominal capacity.

All this means that in order to meaningfully calculate a charge or discharge time, you need more comprehensive data about your battery, and you need to know more about your load. Battery manufactuers usually provide graphs of performance over a range of charge and discharge currents.

You mention a "custom made 12V UPS". Note that if you are using a nominal 12V battery, and your power supply has a constant output voltage of exactly 12V, it will not fully charge the battery. For instance typical lead-acid batteries require to be charged to 2.3V per cell (13.8V for a nominal 12V battery) at 25 C. Other types are similar.

If you are installing your own system, you should ensure you are thoroughly conversant with battery safety - particularly with ventilation requirments if using lead acid, and charging requirements if using lithium-iron types. Get it wrong and you can have explosions, posionous gas, or both.

Wickwack58.164.238.58 (talk) 12:00, 6 June 2012 (UTC)[reply]

The load will like this 2 120W fans and 1 65 W laptop. All rated at 220V AC.

the cutoff is set at 10V.

It's what we call Desi UPS http://www.wiredpakistan.com/forum/30-engineering-corner/

I have also attached a multimeter to the battery, the max value reached while charging is 13.8 — Preceding unsigned comment added by 182.178.210.45 (talk) 13:08, 6 June 2012 (UTC)[reply]

Something's not right somewhere, or information is missing. Your stated load totals 305 W. Allowing for typical invertor efficiency that means a DC current of ~28A on a 13.8V bus. For a 750W system, 54 A is available. For a continuous bus system, charging current will equal 54 - 28 ie 26 A. But you siad charging current is 14 A. This is possible if the UPS is of the switched bus type (battery on a separate charger circuit, and switched to feed the invertor when required), but you haven't told us about the 1000W UPS. Is it continuous bus or switched bus? If switched bus, what is the charging current? Possibly the link you gave was meant to tell us more, but it is not functional. Wickwack58.164.238.58 (talk) 14:52, 6 June 2012 (UTC)[reply]

When I had the job of testing some early UPS units a few years ago, I noticed that the discharge rate was a brutal one in UPS mode, and much higher than the charge rate. This made sense, because the device is not really designed to operate on ,say, a daily basis, but rather to operate in the occasional emergency. The battery is called upon to discharge at an extremely high rate, at the cost of decreasing the battery life, in order to keep initial cost down, but still provide a relatively large amount of AC power to the protected computer or device. Protecting the continued operation of the computer (don't lose the term paper you have just finished typing, or don't lose all the pizza orders held in the computer, or the financial transactions, or the control of the elevators in a highrise, or the air traffic control, or the power system operations) is judged more important than the well-being of the battery. Fewer ampere-hours could be obtained from the battery in this mode than if the discharge rate was lower. Lead acid batteries have very low internal resistance, so they can be called on to do heroic discharge rates for a short while, like cranking a car engine (hundreds of amps from a 70 amp-hour battery for a few seconds). Power interruptions are often a few seconds if automatic reclosing of distribution feeders clears the fault, or a few minutes if supervisory switching can solve the problem. If a PC is operating on UPS with no generator backup, the typical practice is to save your work and do an orderly shutdown rather than keeping on for an hour or whatever until the battery dies abruptly. Ten volts as the battery voltage when the discharge is ended,, as 182.178.210.45 stated, seem an extreme state of discharge, and likely detrimental to the battery life (number of charge discharge cycles) which can be expected. Wasn't a ten hour rate a classic rule of thumb (capacity divided by ten = charge rate and discharge rate?) Even so, a typical charger tapers off the charge rather than continuing at the initial high rate, which would require raising the charging voltage over time. The website of the battery manufacturer might offer specific recommendations for charging protocal. Edison (talk) 14:31, 6 June 2012 (UTC)[reply]

You have it pretty right, Edison. However, the 10 hour rule used to be commonly quoted for small batteries intended for powering portable equipment such as cameras, tape recorders, and the like. It has largely gone by the wayside now, and was never used for larger sizes, nor used for vehicle, industrial, or UPS batteries. Also, constant-current chargers became common when switchmode AC/DC conversion became economic. In continuous bus UPS designs (common because they're cheap, reliable, and simple), the current available for changing the battery is simply the AC/DC convertor output, which must be capable of sustaining the full rated UPS system load, minus the current drawn by the DC/AC convertor in supplying the load. Hence if for some reason the load is minimal, the full AC/DC convertor output is available to charge the battery. Wickwack58.164.238.58 (talk) 14:52, 6 June 2012 (UTC)[reply]

Do you have a reliable source for charge rates way higher than capacity/10 being the present day common practice? I agree that for occasional emergency use (UPS, emergency stairway lighting) a rate which discharges the battery in an hour or so at the cost of battery health and at the cost of ampere hours is common, but I question charging it really fast, since I expect damage to result. The AC/DC converter is ONLY used to charge the battery in many UPS systems, which run the load off the mains until mains failure, then quickly switch to inverter from the battery, fast enough that the PC stays online. Better and more expensive UPS systems indeed run from the inverter off the battery all the time, perhaps with a generator offline to start when the mains fail. Edison (talk) 04:29, 7 June 2012 (UTC)[reply]

For the medium to large scale UPS's (10's to 100's of kW) I had experience with, the battery charging rate was roughly of the same order as the discharge by application design. This is because they were continuous buss systems, so the charge current is essentially the rated output minus the actual load, as I said. This should be taken into account when choosing the battery. Switched bus systems can deliver a fixed charge current independent of actual load and are thus kinder to the battery as you have said. And since the switching can be from raw AC input to invertor output, the AC/DC convertor can be much smaller and cheaper. However, continous bus systems can be prefered for their reliability and inherently stable and spike-free output. Lead acid batteries in particular will not take kindly to rapid charge. Generally, the slower the charging, the longer the life and the greater the capacity. As far as I am aware, batteries designed for rapid charge are used for small portable applications like personal electronics, phones, small tools, and the like. For performance, and service life, for any charge rate, consult manufacturer's data. Batteries are made for a wide range of cycle conditions - even within the same basic chemistry (gell lead acid, nickel-iron, whatever). So I don't think a source with one simple rule of thumb or formula, beyond what I gave above, can be cited. Incidentally, where diesel backup is employed, it usually because an outage cannot be tolerated, such as telecomms, business critical server farms, or hospital operating theaters. This is quite different to a PC or minicomputer application, where you only need enough time to shut down in an orderly way (a few minutes). Diesel backup is usually provided with a UPS battery capacity for 3 hours or so. This is so that, should the mains not came back on in that time (an unlikely occurance) and the diesel fail to start, there is enough time to call out the mechanic, plus time for him to fix it immediately if he can, or request delivery of a portable genset. In such cases, the UPS battery will be very large and inherently tolerate a lot of charging current. Wickwack120.145.54.86 (talk) 05:28, 7 June 2012 (UTC)[reply]

Which will system will be better, 24V or 12 V ?

It depends on many factors. But, as a general rule, if the load power is the same, and the battery capacity in Ampere-Hours (Ah) is kept the same, then 24V will be better as the discharge current (and potentially the charge current) will/can be halved. I stress that this is only a very rough guide. You would normally expect to halve the battery Ah size as well on 24V. Of perhaps more reliable to state is that AC/DC conversion and especially DC/AC conversion is slightly more efficient at 24V. Wickwack58.167.249.78 (talk) 12:05, 7 June 2012 (UTC)[reply]

What about charging time ? Will it be reduced in 24V system ? — Preceding unsigned comment added by 182.185.220.86 (talk) 11:25, 8 June 2012 (UTC)[reply]

I've already answered that to the limitted general degree possible given that you have not supplied sufficient information. Is the 24V sytem to be switched bus or continuous bus? If switched bus what is the charging current? 2 x 170 Ah batteries, or 2 x 85 Ah batteties, or what?? And, as I siad, battery capacity varies with operating conditions (cutoff voltage, charge current, temperature, etc etc). You need to look at manufactuers data for batteries you are considering. Your questions are like asking us "If I now eat 2kg of food each day and propose to eat 2.5 kg, how much will I weigh?" How should I know, it depends on so many factors that haven't been given. About all I can say is you will get fatter, if all other factors are kept constant. Similarly, I can't say what the charging time will be, beyond the rough guide I've already given. Wickwack121.215.41.248 (talk) 15:31, 8 June 2012 (UTC)[reply]

Charging current is 14A, Battery 2 x 170 Ah, cutoff voltage 10V, room temperature is around 36C . Manufactures usually don't provide such extensive data. — Preceding unsigned comment added by 182.185.171.47 (talk) 09:04, 9 June 2012 (UTC)[reply]

In that case, since you have neither changed the ampere-hour capacity, nor the chanrge current, then the charge time is unchanged. However the discharge time for the same load power will increase somewhere around 2.5 to 3 times, ignoring the 35 C temperature. Battery manufacturers certainly DO provide such "extensive data" - it is essential for the reasons I have explained. You just haven't asked in the right place. I have used such data myself. A 35 C temperature is excessive for an office environment. Wickwack124.178.139.104 (talk) 12:47, 9 June 2012 (UTC)[reply]

Source of uncertainty of the Earth's age ?[edit]

On the page Age of the Earth it is stated that the age is 4.54 ± 0.05 billion years. What are the main sources of the 1% uncertainty, and what is most dominant?

Is it the accretion time (100 million years is indeed 2% of the 5 billion years age) itself?
Is it the uncertainty of the accretion time?
Or is it the uncertainty of the dating methods?

Wolfsson (talk) 13:17, 6 June 2012 (UTC)[reply]

From the first citation on that page, the uncertainty seems to be from the dating methods. The time appears to refer to the end of the accretion of solid bodies such as Earth, which should have occurred at roughly the same time throughout the inner solar system.-RunningOnBrains^(talk) 13:39, 6 June 2012 (UTC)[reply]

Also, I bet there is uncertainty into just what qualified as Earth. Did a molten ball of magma qualify ? StuRat (talk) 15:13, 6 June 2012 (UTC)[reply]

Given that the Earth is a molten ball of magma (except for quite insignificant parts), I'd say yes. --Stephan Schulz (talk) 17:48, 6 June 2012 (UTC)[reply]

Seems unfair on the inner core. Sean.hoyland - talk 17:54, 6 June 2012 (UTC)[reply]

The Earth doesn't currently contain much molten magma. Even the vast majority of the mantle is a rheid, not molten magma. Red Act (talk) 19:03, 6 June 2012 (UTC)[reply]

There was an interesting article a few weeks back that one of the longer lasting isoptopes had been found to have a different half-life than believed, and that the earth's age might need to be recalibrated . Anyone recall the article or isotope? μηδείς (talk)

No, and such a change at this point given the progress of science in this field seems highly unlikely. Uranium-lead dating is extremely solid in both theory and practice. Interestingly, our article on Calcium-aluminium-rich inclusion suggests that this age of the Earth may be a lower limit on the age of earth rather than the actual age.-RunningOnBrains^(talk) 06:52, 7 June 2012 (UTC)[reply]

Yes, that kind of radioactive dating tells you when that particular rock formed. What we know is that the Earth must be at least as old as its oldest rock (everything melted during the formation, so any older rocks won't have survived intact). It is unlikely that the Earth is significantly older than its oldest surviving rocks. Our estimates of the age of the Earth are consistent with the ages of meteorites, which theory tells us are left over from the formation of the planets. It's also consistent with our estimates of the age of the Sun. It's very unlikely that our age estimates are completely wrong. --Tango (talk) 16:12, 7 June 2012 (UTC)[reply]

To expound further, I realize my original point was unclear: I'm not saying there couldn't be new evidence (or new theories consistent with the current evidence) that pointed to an older earth, but the idea that we have a half-life wrong is about as close to impossible as I'm comfortable to say. -RunningOnBrains^(talk) 17:12, 7 June 2012 (UTC)[reply]

bond energy of ionic compounds[edit]

how will we define the bond dissociation energy or bond dissociation enthalpy for ionic compounds ? i have read the definition of bond dissociation energy as " The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form product (gaseous atoms) in gas phase ." the problem is that ---firstly the ionic compounds don't occur in gas phase , secondly if we heat it we will get ions (& the energy involved is lattice energy) & not the atoms, moreover the definition tells only about the covalent compounds .ncert class 11 chemistry text book part one pg 171 http://ncert.nic.in/NCERTS/textbook/textbook.htm?kech1=0-7 117.225.240.240 (talk) 14:03, 6 June 2012 (UTC)[reply]

See Lattice energy. Bond dissociation energy refers only to covalent bonding. --Jayron32 18:05, 6 June 2012 (UTC)[reply]

As a side note: boiling sodium chloride, produces discrete molecular sodium chloride or 'chloridosodium'. Further heating to the point of ionization leads to diassociation into a gas plasma. This consists of a menagerie of ions, including sodium(1+) and chloride(1-), as well as polyatomic ions. Plasmic Physics (talk) 23:30, 6 June 2012 (UTC)[reply]

I've converted the OP's footnote into a "small" comment; the refdesk pages don't handle footnotes properly.Matt Deres (talk) 22:54, 9 June 2012 (UTC)[reply]

distribution of angular momentum[edit]

Most of the angular momentum of the Solar System, I seem to remember reading, is that of the orbit of Jupiter. But of course every rotating or revolving body has its own bit of a.m., and their vectors are not all parallel. Is the variance of these vectors (mass-weighted, of course) a meaningful concept? If so, is it known (or estimated)? If so, how big is it in radians?

And should it be expected to decrease over the eons? —Tamfang (talk) 16:48, 6 June 2012 (UTC)[reply]

It does seem like an interesting concept, although I doubt if it would be measured in radians. If we assume that the initial rotating cloud of gas was uniform in it's rotation (not sure if this is true), then any variance must be caused by outside influences (passing comets, etc.). However, once there is even the slightest variation, it's possible for two objects to give each other more variance by passing near each other and knocking each other out of position (perhaps a bit more with each orbit). Pluto seems to be the (dwarf) planet with the most deviation of the inner planets, so something interesting must have happened there. Objects out in the scattered disc and Oort cloud are far more random in their vectors than those in the inner solar system and, to a lesser extent, the Kuiper belt (with exceptions in the inner solar system allowed on for smaller objects, easily knocked out of the plane of the ecliptic, like the centaurs). The following chart may also be relevant. StuRat (talk) 19:07, 6 June 2012 (UTC)[reply]

"Inner planet" usually refers to Mercury, Venus, the Earth and Mars. Pluto isn't an inner planet. It's orbital inclination is normal for Kuiper belt objects, as your chart shows. --Tango (talk) 19:39, 6 June 2012 (UTC)[reply]

OK, then how does one refer to the region inside of Pluto's orbit ? StuRat (talk) 22:56, 6 June 2012 (UTC)[reply]

Cis-Plutonian. But as the orbit of Neptune, rather than of eccentric Pluto, is taken as the delimiter between the outer solar system and the next broad category of solar system objects, people would be more apt to talk about Cis-Neptunian objects. List of trans-Neptunian objects shows there are about 40 known TNOs with perihelions closer than Neptune's, but none with an average orbital radius smaller than Neptune's Tautologically; they wouldn't be TNOs if they were. -- Finlay McWalterჷTalk 09:19, 7 June 2012 (UTC)[reply]

Let's start by checking if what you remember reading is correct, since it's a simple enough calculation. I would guess the major contributors are going to be Jupiter's orbit and the Sun's rotation, since the Sun and Jupiter make up essentially the entire mass of the solar system. Jupiter's orbit gives an angular momentum of 779,000,000 km (orbital radius) * 13 km/s (orbital speed) * 1.9*10²⁷ kg (mass) = 1.9*10³⁷ km²kg/s. The Sun (assuming uniform density, which will overstate its angular momentum since a lot of the mass is concentrate in the core) has an angular momentum of 2/5*m*r²*2*pi/T (see List of moments of inertia, m, r and T are the mass, radius and rotation period of the sun, respectively). That gives us 2/5*2.0*10³⁰kg*(0.7*10⁶km)²*2*3.1/25 days=1.1*10³⁶ km²kg/s. So yes, Jupiter is the main contributor. I'll just quickly check Neptune, since it is so far away it might be higher than I expect: 4.5*10⁹ km * 5.4km/s * 1.0*10²⁶ kg = 2.4*10³⁶ km²kg/s. That is higher than I expected, but not as high as Jupiter's. --Tango (talk) 19:39, 6 June 2012 (UTC)[reply]

Awesome Science things in Europe?[edit]

So, not exactly a science question, but i'm hoping you can help. It's my dad's 60th soon and we want to take him somewhere in western europe to see something awesomely sciencey. CERN is an obvious one but it seems like most of their stuff for tourists is museumy rather than seeing anything real. Trying to pull some strings to get into ATLAS but i'm not hopeful. Something like the JET Tokamak comes to mind but we live in the UK and it's not exactly an exotic location to spend more than a day. Any thoughts would be appreciated or reprimands for putting this here also welcome! 137.108.145.21 (talk) 18:50, 6 June 2012 (UTC)[reply]

You enjoy reprimands ? How about spankings ? :-) StuRat (talk) 18:57, 6 June 2012 (UTC) [reply]

I presume you've been to the National Space Centre in Leicester, or Jodrell Bank? --TammyMoet (talk) 19:33, 6 June 2012 (UTC)[reply]

The Deutsches Museum in Munich has a massive collection of all manner of science and industry, but it's not an active site of research, which seems more like what you're going for. --Mr.98 (talk) 01:00, 7 June 2012 (UTC)[reply]

You might try one of those bigger-on-the-inside-than-the-outside timey-wimey boxes. μηδείς (talk) 01:43, 7 June 2012 (UTC)[reply]

I know you say you want to visit western europe but a few locations in England stick out too much not to mention, even though they are probably more of a day trip rather then a nice stay away somewhere. Maybe if not this time, you can at least put them on a list for "next time". I'm in Australia and just over a year ago I stayed for a month with my brother who was living in London at the time. I did a day trip to Down House, i'm a big fan of Darwin so that was very high on my list of things to do, I thought it was set up really well. The other two places I'd love to see would be blatchley park and the cavendish laboratories, those places obviously hold legendary status in the history of science, not having seen them I can't say how "interesting" their exhibits would be to people not already interested in their history. Vespine (talk) 04:09, 7 June 2012 (UTC)[reply]

I know it's not quite what you're asking, but I'm 62 and on the list of things I'd like to do before I die is to to go on one of the northrn lights cruises and hope to see the Aurora Borealis. Richerman (talk) 09:00, 7 June 2012 (UTC)[reply]

Have you considered Iceland? There are various geothermal energy experiences to appeal to the sciencey-minded, the opportunity to see active geysers, lava fields and other awesome volcanic things, and of course the possibility either of 24-hour daylight or the aurora, depending on when you travel. We're approaching solar maximum, so aurora-spotting has an improved chance of success next winter, particularly in higher latitudes. Ka renjc 18:16, 7 June 2012 (UTC)[reply]

Some more ideas: 184.147.126.249 (talk) 18:39, 7 June 2012 (UTC)[reply]

UK (but probably not close to where you live!) - Callanish, at the right date to check the solar alignments for himself

UK (but really cool anyway) - Kew Bridge Steam Museum for the giant engines

France (or Sweden, or Germany, or Switzerland) - any built-to-scale Solar System models. The Swedish one is the biggest

Austria - Ars Electronica Center (fusion of science and art?)

Belgium - Euro Space Center

The Max Planck Institute of Plasma Physics's Wendelstein 7-X in Greifswald is pretty big and flash and sciency. Last time I was there they were running public tours. It is a bit out of the way though (4 hours drive north of Berlin). 101.171.127.244 (talk) 19:26, 7 June 2012 (UTC)[reply]

Locating the South Pole[edit]

In the days of Scott and Amundsen, how would an explorer identify his position as being exactly at the South Pole, and how accurate would this be? --rossb (talk) 22:52, 6 June 2012 (UTC).[reply]

I'd say the easiest way would be to use the motions of astronomical bodies. Can't say how accurate that would be. Plasmic Physics (talk) 23:11, 6 June 2012 (UTC)[reply]

Not so much "motions" as "positions" -- we note at history of longitude that "determining latitude was relatively easy in that it could be found from the altitude of the sun at noon with the aid of a table giving the sun's declination for the day." For the poles, that's enough right there -- latitude 90, longitude irrelevant (also, time of day functionally irrelevant). This site confirms that a sextant reading of the sun was used, and notes that Amundsen was within 200m of the true South Pole. — Lomn 23:25, 6 June 2012 (UTC)[reply]

If I recall Huntford's book The Last Place on Earth correctly, in those few days of at-pole peregrination, Helmer Hanssen was probably the one who got closest to the geographic pole. It was my recollection that Scott carried a theodolite, which would have allowed him to make much more accurate measurements of the position of the Sun than Amundsen could with his sailor's instruments (and that the additional weight thereof was one of many contributory factors to Scott's failure). But, while there was a theodolite on the expedition, it seems it wasn't taken to the pole (picture and info). -- Finlay McWalterჷTalk 20:10, 7 June 2012 (UTC)[reply]

At night, they could track the celestial South pole to give a rough direction. Somehow I don't think that option was available, you'd be pretty foolish to attempt such an expidition during the polar winter. Plasmic Physics (talk) 23:34, 6 June 2012 (UTC)[reply]

For the next three days the men worked to fix the exact position of the pole; after the conflicting and disputed claims of Cook and Peary in the north, Amundsen wanted to leave unmistakable markers for Scott. After taking several sextant readings at different times of day, Bjaaland, Wisting and Hassel skied out in different directions to "box" the pole; Amundsen reasoned that at least one of them would cross the exact point, from Amundsen's South Pole expedition. If my visualisation of the trigonometry is right, when you're at the South Pole the sun stays at the same altitude all through the day (+- a little depending on if it's rising or setting). These days of course it's much easier; you know you're there when you trip over this thing. FiggyBee (talk) 23:31, 6 June 20126 (UTC)

Not really, that thing is not really at the pole, the pole drifts around. Yes, incase you're wondering, both the magnetic and actual poles drift. That is just an arbitrary point that someone chose when the pole actually used to be in the vicinity. LOL, it never even passed through that particular point either. Plasmic Physics (talk) 23:38, 6 June 2012 (UTC)[reply]

Most of the change comes from the ice underneath that barber pole drifting, to the tune of 10 meters per year. --Carnildo (talk) 00:17, 7 June 2012 (UTC)[reply]

Yes, but the Earth also moves irregularly on its axis, e.g. the Japan earthquak moved it quite a bit. Plasmic Physics (talk) 01:07, 7 June 2012 (UTC)[reply]

4 to 10 inches. Rmhermen (talk) 04:57, 7 June 2012 (UTC)[reply]

Wow, that's alot! I guess I should have checked the data first. Plasmic Physics (talk) 05:08, 7 June 2012 (UTC)[reply]

I'd think we could tell that much change all over the world, if it happened during the quake. If not, by what mechanism does the Earth's axis slowly change as a result of the quake ? StuRat (talk) 06:25, 8 June 2012 (UTC)[reply]

Does anyone know what the flag to the left of the Union Jack is? ElMa-sa (talk) 19:50, 7 June 2012 (UTC)[reply]

It's either the flag of Belgium or the flag of Romania. Per this page, the flags at the ceremonial pole are those of the signatories of the Antarctic Treaty System. WP pictures of the site suggest that the flag arrangement is not constant. — Lomn 20:14, 7 June 2012 (UTC)[reply]

unidentified monocot in South Jersey[edit]

Can anyone identify or suggest an identification for this plant? It is one foot tall. I will be able to recognize the flowers if someone suggests the correct identification. A friend had one such plant in his yard in South Jersey last year, and now it has taken over the entire property. The blooms, when they come, are white and look like bluebells Hyacinthoides non-scripta, although the stalk is more erec,t or wild orchids of some type. Thanks. μηδείς (talk) 23:29, 6 June 2012 (UTC)[reply]

As a drive by shout I'm thinking some kind of Helleborine. Richard Avery (talk) 06:14, 8 June 2012 (UTC)[reply]

Looking at the links and pictures I can find on google that is close but I haven't seen anything that looks like an exact match. I'll have to see if I can get a picture sent to me after the flowers mature. μηδείς (talk) 15:52, 8 June 2012 (UTC)[reply]