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February 11

Depth of soil

Why in somewhere like England, is the depth of soil so great even on the sides or tops of hills? 92.29.123.158 (talk) 00:48, 11 February 2011 (UTC)[reply]

We have an article on the Geology of England which might be a good place to start. Vespine (talk) 01:13, 11 February 2011 (UTC)[reply]

... and, of course, the soil is extremely thin or even non-existent on the tops of some hills, especially in the Lake District, but peat explains the depth on many hilltops. Dbfirs 07:51, 11 February 2011 (UTC)[reply]

As the geology article mentions, it's partly to do with glaciers, but also why would you expect there not to be soil there? If plants can grow somewhere, which is the case in all of the UK thanks to its relatively warm and damp climate, then eventually soil will begin to build up, even on bare rock. Primary succession and ecological succession aren't particuarly good articles, but do discuss this. SmartSE (talk) 10:28, 11 February 2011 (UTC)[reply]

When was astigmatism first detected and corrected by spectacle lenses

I have invented a new type of spectacle frame and patented it, my reason was to enable spectacle lenses to be fitted which include astigmatism for easy construction in developing countrys. Currently best spherical correction is being used by NGOs globaly but its not providing a 100% accurate solution and 10% of the worlds population cannot work because they are refractivly blind. I will be envolved with a TV program made by the BBC shortly and I need to establish who and when astigmatism was first corrected in spectacle lenses. Any help would be appreciated. John D Snelgrove FBDO 86.167.248.127 (talk) 02:42, 11 February 2011 (UTC)[reply]

This book seems to indicate that astigmatism in the eye was independently discovered by Thomas Young in 1800 and later by George Biddell Airy in 1825. Airy is the one who coined the term, as suggested by William Whewell (who also, incidentally, coined the term "scientist"). The first lens that corrected astigmatism was developed in 1824 at Airy's request. As far as pat histories go, this one seems plausible enough to me. Unfortunately you cannot view the citations on Google Books so it is hard to know what the author is basing it on. --Mr.98 (talk) 03:54, 11 February 2011 (UTC)[reply]

In the Bakerian lecture of Young he describes his experiments on his own astigmatism. --Stone (talk) 07:30, 11 February 2011 (UTC)[reply]

Articel on the history of astigmatism.--Stone (talk) 07:38, 11 February 2011 (UTC)[reply]

The lecture of Airy is also available. --Stone (talk) 08:27, 11 February 2011 (UTC)[reply]

The Sun during solar maximum

What does the Sun look like through a (properly filtered) 13-cm telescope during solar maximum? I became interested in astronomy in my early teens (and bought a 13-cm telescope), when the Sun wasn't terribly exciting to look at because it was near a solar minimum. I'm much older now, but the Sun is STILL in the same minimum, so I'm very curious about what it would look like when it finally becomes active. --140.180.1.16 (talk) 08:40, 11 February 2011 (UTC)[reply]

First, let me slap on the usual DANGER! DO NOT OBSERVE THE SUN THROUGH A TELESCOPE! warning. Now, if you are "doing it right," you should never even look through a filtered telescope either. You should set your telescope up as a heliostat and project the image of the sun onto an imaging plane - do not use your eye as the focal plane! This is significantly less risky than just using a filter, and produces better images.

Usually, during solar maximum, the only thing you'll notice is "more sunspots;" but if you're talented at astrophotography, and heavily invest in specific filters, you can observe the chromosphere and the corona. If you're lucky, you may even catch a coronal mass ejection. CMEs can occur at any time, but are much more likely during solar max. To photograph a CME will be difficult. Historically, scientists waited for a total solar eclipse before attempting to photograph the corona; but modern optical equipment and film emulsions can make it possible to attempt chromosphere and corona photography directly, using a completely opaque sunshade. Here are some photos from NASA: Coronal Mass Ejections... (the photos came from the High Altitude Observatory at Mauna Loa Solar Observatory. Never fear; not everyone has access to such equipment. Here's a good "tutorial" that used a 4-inch refractor and a fairly "amateur" 35mm film camera: Observing and Photographing the Solar Chromosphere Through a Solar Cycle. That paper is a little bit old, so they describe using film (instead of digital) cameras; even with filtering, I'd be very nervous connecting my digital camera up to a heliostat, for fear of permanently burning/damaging my sensor. If you know your optics are safely filtered, and have a way to meter the light intensity, you could use a digital camera too. (Do not use your eyes to test your telescope's filter! Permanent eye injury/damage is a serious risk). Nimur (talk) 11:36, 11 February 2011 (UTC)[reply]

A solar filter, which often costs about $50+ USD (full-aperture, NOT eyepiece type, these can crack!!) is usually adequate for viewing the Sun safely by direct optical means, though a small hole or tear could completely compromise safety. A 13-cm telescope is large enough for the Sun's focused rays to do serious damage to the primary and secondary mirrors (assuming reflecting telescope) when using the projection method, and pointing it at the sun is dangerous as well for the finderscope if left uncovered. It is much safer to use both a solar filter and a brightness-reducing eyepiece filter in combination to view the Sun or otherwise leave the outer dust cover on if your telescope has both an outer ring and inner circle objective cover and then apply either projection or aperture-filtered observation, but test it using your camera first to check for any overexposure, although a specially-designed solar telescope is often specifically suited for solar observation but can be far more expensive than the combination of a usual telescope and a solar filter. ~A H 1^(TCU) 22:08, 11 February 2011 (UTC)[reply]

Thanks for the info. One correction: the projection method definitely doesn't produce better images. There's a reason that people buy H-alpha filters even though they cost thousands of dollars. Even with a simple attenuation filter, you get much higher contrast than by projecting onto a piece of paper. --140.180.0.75 (talk) 01:48, 12 February 2011 (UTC)[reply]

By "better", I meant "bigger image." (Here's an example setup with a 3 foot projection from Sky & Telescope). The sun is huge and bright - so it sustains ridiculous magnification factors; and even with a small telescope, you should have enough resolution to capture a lot of detail when you project up to a 1-foot wide solar image. Now, you'd lose that resolution if your camera sensor were directly in the focal plane unless you've got a full-frame imager and tightly packed pixels and you have the optics to precisely control the image projection on your sensor. Depending on the effective field of view of your camera sensor, and the size of your primary, you probably aren't capitalizing on all available optical resolution of your telescope. "Best" results will of course depend on your intent and your available equipment. Nimur (talk) 18:15, 14 February 2011 (UTC)[reply]

Effect of damaging the Earth's core

Not long ago, some classmates presented a plan to dig a tunnel through the Earth's core to provide a high speed train route between New York City and Tokyo. They were unable to answer the question of whether this tunnel would affect the planet's structural stability. Would a tunnel through the core increase the damage that asteroid impacts, etc., could cause? Also, if terrorists managed to detonate a suitcase nuke within the core, what size of earthquake could they cause? (I ask because in 2009, South Korea reported that a North Korean underground nuclear test had created a magnitude 9 earthquake.) Neon Merlin 10:34, 11 February 2011 (UTC)[reply]

There is no conceivable way that a train-sized tunnel would affect the "structural stability" of Earth. What you would need to worry about is whether the tunnel could be structurally sound. But analyzing the hypothetical tunnel's stability is moot - there is no current technology that could be used to dig such a tunnel. Even if the tunnel were not "direct", but instead followed the contour of the earth's surface, we do not have a way to safely excavate a "subway" tunnel underneath the depths of the open ocean across the Pacific. Finally, even if a nuclear weapon could be detonated at great depth (say, even 100 km below the surface, using a hypothetical tunnel or any other proposed mechanism) it is unlikely that it would trigger a catastrophic earthquake. Scientists have been measuring the seismic behavior of nuclear weapons since 1945; the media often grossly misrepresents the energy scales involved. A group at CISAC produced this public report, Technical Analysis of the DPRK Nuclear Test. You may want to re-check where you heard your "magnitude 9" claim; I'm pretty sure that is an inflated number; but it's really irrelevant anyway, because there are so many technical details involved in converting measurements of a bomb into an "equivalent-sized earthquake." The energy released by even the most powerful nuclear bombs is dwarfed in magnitude by even a small earthquake. Nimur (talk) 11:51, 11 February 2011 (UTC)[reply]

(EC) The structural damage is probably negligible - a tunnel is tiny compared to the size of the earth (or the core). For the same reason, a nuclear explosion in the core would not or hardly be noticeable up here. Aside from the engineering difficulties of protecting the tunnel from the stresses and temperatures prevalent in the core, are you aware of the kind of slope the train has to go down and up again when it is near the surface? I'm too lazy to calculate it but it'll probably be more than 45 degrees, much more than any normal train can handle. --Wrongfilter (talk) 11:54, 11 February 2011 (UTC)[reply]

49 degrees, according to my calculations assuming a spherical ~~cow~~ Earth. A straight-line tunnel would reach a depth of about 2200 km, penetrating far into the lower mantle, but missing the core by some 500 km. –Henning Makholm (talk) 13:06, 11 February 2011 (UTC)[reply]

And, as we all know (I assume ;-), a spherical train in a frictionless vacuum moving under the influence of gravity only would need 49 (IIRC) minutes for the trip, regardless of the distance covered. --Stephan Schulz (talk) 14:30, 11 February 2011 (UTC)[reply]

Schults, that's only true if the shape of the tunnel is a cycloid and the density of the earth is uniform, which it isn't.Dauto (talk) 16:01, 11 February 2011 (UTC)[reply]

Uniform density and non-rotating Earth are, of course, part of the spherical cow universe! --Stephan Schulz (talk) 19:55, 11 February 2011 (UTC)[reply]

There's one, erm, small technical obstacle to this plan. The inner core rotates relative to the Earth's surface, something like once every 400 years if I remember correctly. As for nukes, they can be detected on a seismograph, but they are very small. Unfortunately Comprehensive Nuclear-Test-Ban Treaty Organization doesn't get into the specifics of the amplitude, but it's something like 3. There's a formula for the size of the bubbles of vaporized rock from a nuke at Underground nuclear testing - but it's something like hundreds of feet near the surface, and in the core I bet the pressure would reduce the size further. Wnt (talk) 18:07, 11 February 2011 (UTC)[reply]

Someone mentioned above that this hypothetical tunnel is supposed to miss the core by some 500 kilometers so the rotation of the inner core wouldn't be a problem, if that happens to be correct. Dauto (talk) 00:38, 12 February 2011 (UTC)[reply]

Something else to mention is that plate tectonics all occurs in the Earth's crust, so an explosion far below the crust is unlikely to have much effect. I would suspect that the right-sized nuke, placed at the hypocenter (focus) of the potential earthquake, a few miles underground, typically, could trigger a quake. However, this would only work if strain had built to the point where a quake would occur soon, anyway, and the quake would be no bigger than the normal range for that fault (it might actually be smaller, if it occurs somewhat earlier, with less strain accumulated). Consider an analogy with cloud-seeding to make rain. It doesn't work when there's no moisture in the air, it has to be almost ready to rain anyway. StuRat (talk) 01:20, 12 February 2011 (UTC)[reply]

Actually, the base of the lithosphere (and all plates) is in the uppermost mantle, the asthenosphere, but you're right that you can't cause an earthquake in that way, only trigger it if it's already near a critical point in the earthquake cycle. There are often foreshocks before an earthquake, but opinion is divided as to whether they actually trigger the main event or are just a manifestation of the increasing stress levels in the area where the mainshock happens. Current thinking is leaning towards the latter explanation suggesting that it's actually difficult to trigger an earthquake before it's ready to go. Mikenorton (talk) 10:13, 12 February 2011 (UTC)[reply]

Castrating cats

Is it right to castrate domestic cats to make them stay at home? — Preceding unsigned comment added by Sina-chemo (talk • contribs) 11:57, 11 February 2011 (UTC)[reply]

The reference desk doesn't give moral judgements; it only answers factual questions. The article Neutering has information on the positive and negative effects of neutering a cat (but please consult a vet if you have specific questions as we can't give veterinary advice). --Colapeninsula (talk) 12:31, 11 February 2011 (UTC)[reply]

I don't know if people do it to make it stay at home. I always thought it was to avoid it getting nuts. Wikiweek (talk) 12:42, 11 February 2011 (UTC)[reply]

It helps keep the cat from getting testy. Googlemeister (talk) 14:19, 11 February 2011 (UTC)[reply]

No kitten; it really works. Matt Deres (talk) 14:52, 11 February 2011 (UTC)[reply]

Purrr-fect answers. 10draftsdeep (talk) 16:05, 11 February 2011 (UTC)[reply]

I get the feline you're treating this like some kind of joke... Vimescarrot (talk) 18:01, 11 February 2011 (UTC)[reply]

Please, don't make catty comments on the ref desk. It pusses me off. Tinfoilcat (talk) 19:07, 11 February 2011 (UTC)[reply]

Surely a better reason to castrate cats is to avoid getting too many more cats. HiLo48 (talk) 21:46, 11 February 2011 (UTC)[reply]

See feral cat. ~A H 1^(TCU) 22:01, 11 February 2011 (UTC)[reply]

Uncastrated toms are rather objectionable, doing things like spraying their "scent" on the drapes. If I had such a cat, the only choice would be between castration and putting it to sleep. If it could make the choice, I think it would prefer to skip the nuts. StuRat (talk) 01:03, 12 February 2011 (UTC)[reply]

It is a good idea to castrate them. It doesn't make them stay at home, but it makes them roam, fight and mate less. And, of course, they will not produce unwanted offspring. You should do this to your cats unless you are breeding them. Zzubnik (talk) 14:57, 14 February 2011 (UTC)[reply]

Castration would seem an in effective method of birth control. Unless every single tom in a neighborhood is castrated, a queen in heat will find a willing an capable mate. Spay (ovario-hysterectomy) 50% of the stray queens and you cut the birth rate in half. Castrate 95% of the toms and you have no effect at all. -- 119.31.126.67 (talk) 00:34, 16 February 2011 (UTC)[reply]

differentiation by first principles

how to obtain derivative of e to the power root x by using limits? —Preceding unsigned comment added by 180.215.34.14 (talk) 12:28, 11 February 2011 (UTC)[reply]

This is a duplicate of a recent mathdesk question, probably by the same anonymous user: WP:RD/MA#differentiation by first principles. Please don't ask the same question on several desks. –Henning Makholm (talk) 12:44, 11 February 2011 (UTC)[reply]

If you read closely, you'll see that this is actually a different question, in that it adds the specification "...by using limits". The answer given on the math ref desk used various rules about differentials, rather than using just the definition of the derivative as a limit, and expressing a derivation of the result by using expressions involving limits. Red Act (talk) 22:40, 11 February 2011 (UTC)[reply]

 Please solve USING LIMITS.  —Preceding unsigned comment added by 180.215.44.196 (talk) 07:55, 12 February 2011 (UTC)[reply]

It follows by definition if you choose characterization 4 from Characterizations of the exponential function! Anyway the maths reference desk is the right place. Dmcq (talk) 12:13, 12 February 2011 (UTC)[reply]

Extra 4 minutes

If a day is 23 hours and 56 minutes long instead of exactly 24 hours, how do the extra 4 minutes fit in? jc iindyysgvxc (my contributions) 12:52, 11 February 2011 (UTC)[reply]

It's the difference between the solar day and the sidereal day. The Earth rotates once every 23 hours and 56 minutes with respect to the stars. But it also circles the sun (or, geometrically equivalently, the sun circles the Earth) once per year. So the Earth has to do 4 minutes worth of catching up to the relative movement of the sun every day. Note that 365*4 minutes is nearly exactly one day, i.e. the extra rotation due to the motion around the sun. --Stephan Schulz (talk) 13:00, 11 February 2011 (UTC)[reply]

In other words, if we measure the day by tracking the location of the sun, we would have to wait an extra 4 minutes before we turn a full 360 degrees. And that is the source of the discrepancy. That's because the earth has moved along a small distance in its orbit of the sun between today and yesterday. --Jayron32 13:46, 11 February 2011 (UTC)[reply]

Thanks. I've heard the term "sidereal" and read of the difference but never had it explained so clearly in terms of "one day per year." Edison (talk)

Supposing for the sake of simplicity that the year was exactly 365 days long, the sidereal day would be exactly 365/366 of the solar day, and the number quoted above as "4 minutes" would turn out to be be exactly 1/366 of a year. --Anonymous, 03:40 UTC, February 12, 2011.

Leap years

then why do we leap 1 day every 4 years? —Preceding unsigned comment added by 165.212.189.187 (talk) 16:52, 11 February 2011 (UTC)[reply]

That's another matter entirely. We leap a day every so often because a year is actually slightly longer than 365 days. Dauto (talk) 17:05, 11 February 2011 (UTC)[reply]

In other words, it takes about 365 1/4 turns on the earth's axis to arive at the same point in the earth's orbit. That extra 1/4th of a day adds up and needs to get tacked on every 4 years. --Jayron32 18:09, 11 February 2011 (UTC)[reply]

Specifically, 97 times every 400 years because it takes 365.24 turns. :) —Preceding unsigned comment added by 205.193.96.10 (talk) 20:45, 11 February 2011 (UTC)[reply]

The "one day per year" is because 364 days are due to the Earth turning, and the last one is because we are going around the sun. If the Earth didn't rotate at all, we would still get one day per year due to us going around the sun. Leap days, on the other hand are caused because it takes an extra 0.24 days to get around the sun and start the seasons over; it has nothing to to with the Earth's own rotation. —Preceding unsigned comment added by 205.193.96.10 (talk) 20:49, 11 February 2011 (UTC)[reply]

Actually we have 366 (plus about a quarter) days due to the earth's rotation and (-1) days due to the earth's motion around the sun for a total of 365 (plus about a quarter). Dauto (talk) 00:11, 12 February 2011 (UTC)[reply]

See leap year and leap second. The sideral motion of the Moon also varies in this matter as it catches up to the Earth on one side of its orbit while remaining gravitationally locked to face one side to Earth. ~A H 1^(TCU) 21:59, 11 February 2011 (UTC)[reply]

Radio power

Hi, does anyone know how much radio signal power is needed in order to pick up a station on an ordinary household radio (with no special aerial or other fancy equipment)? I'm talking about the power "in the ether" at the place where the radio is located, not the total power of the transmitter, which is at an unspecified distance. I think I'm looking for an answer in watts/m^2 (if not, please correct me). Does the answer differ much between FM and AM? 86.179.4.118 (talk) 14:39, 11 February 2011 (UTC)[reply]

The integrated power at the aerial terminal is given (albeit without source) in Orders of magnitude (power) as on the order of femtowatts for an FM signal and a good-quality radio receiver, but I'm not sure what the appropriate 'effective area' would be for the aerial. A similar figure (roughly 10 fW) is given for a minimum receivable spread-spectrum cellular telephone signal. As a very rough (order of magnitude) estimate, a detectable FM signal flux is going to be on the order of femtowatts to picowatts per square meter, depending on the size and configuration of your antenna. TenOfAllTrades(talk) 15:24, 11 February 2011 (UTC)[reply]

See "Field strength." The "signal power" of a broadcast signal at the receiver location is typically given in units of microvolts per meter, rather than in watts per square meter. Then the signal at the radio's antenna terminal or first RF input stage is affected by the type and orientation of the antenna, or the "gain" of the antenna. Modern radios often use a little ferrite loopstick in the radio, while older ones used a flat coil on the back panel, and older ones still, with less RF amplification, required a long overhead antenna and an Earth ground. Naturally a weaker signal or stronger signal will produce an audio output with varying signal to noise ratio, and with varying clarity compared to atmospheric static, interference from electrical equipment, thermal noise in the receiver circuitry and interference from other stations on the same or adjacent frequencies. The bandwidth affects the signal to noise ratio. A signal to noise ratio of 3 deciBels is sometimes described as the minumum usable signal, but it would sound horrible. A 10 dB SNR is often used in describing the sensitivity of a receiver, per [1]. Someone with recent skills or training in communications calculations might be able to come up with the microvolts per meter for a 10 dB Signal plus noise to noise ratio on a typical (say AM) receiver in the broadcast band. I once saw a map of Italy showing the signal strength at varying locations in that country, regardless of what the transmitter was, but I haven't seen that for the US. Says that the sensitivity of modern radio receivers ranges from several microvolts to several millivolts. I suppose that to convert this to "power" you would need to know the input impedance of the radio at the antenna connection, which for most ordinary AM radios is somewhere internally where the loopstick connects to the first RF stage. Edison (talk) 16:00, 11 February 2011 (UTC)[reply]

You can approximately convert electric field strength (volts/meter) to radiant intensity (watts per square meter) by squaring and dividing by the antenna impedance; in practice, you need to account for the antenna matching properties, impedance of free air, antenna directivity, and so on. A comprehensive theoretical treatment and several useful conversion tables are provided by Antarctic Impulse Transient Antenna website at University of Hawaii: Field Intensity and Power Density, which is an excerpt from the Navy Electronic Warfare & RADAR Systems Handbook. Nimur (talk) 22:14, 11 February 2011 (UTC)[reply]

James Watt steam engine patent

I am looking for the patent that James Watt filed in 1784. This patent is mentioned in [article "History of Rail Transport"], but I cannot locate an electronic copy of the patent itself. Thanks. 128.223.222.16 (talk) 16:30, 11 February 2011 (UTC)[reply]

I don't see it cited there or at James Watt#Early experiments with steam or even anything specifically about the 1780s at all in Watt steam engine. That's bad. DMacks (talk) 19:25, 11 February 2011 (UTC)[reply]

I looked briefly at some old Google Books entries about Watt's 1784 patent; it seems to be a real thing, but none of them gave patent numbers. The European patent office search doesn't seem to let you search very old patents very easily, even though they may be on there (like the 1769 Watt patent linked in our article). Seems like it is not going to be a very easy thing to dig up online. --Mr.98 (talk) 21:54, 11 February 2011 (UTC)[reply]

Not sure whether Google Patents goes that far back, but this link does mention the invention and existence of a patent. ~A H 1^(TCU) 21:57, 11 February 2011 (UTC)[reply]

Google Patents only contains US patents. The US didn't issue its first patent until 1790. I've assumed the 1784 patent must be a British one, in part because his 1769 one is. --Mr.98 (talk) 00:00, 12 February 2011 (UTC)[reply]

Your local library may be the only source at the moment. According to Birmingham City Council the European Patent Office is currently embarked on the "digitisation of UK Patents 1870 - 1920", although this espacenet publication contradicts that and suggests that there's not much hope for anything before 1890.

~~By the way the number of the 1874 Watt patent is GB1432. --Heron (talk) 11:48, 12 February 2011 (UTC) Ignore all that. I misread the dates. Sorry. --Heron (talk) 12:15, 12 February 2011 (UTC)[reply]~~

Got it in three ! Project Gutenberg has Kinematics of Mechanisms from the Time of Watt by Eugene S. Ferguson which has a very detailed history of Watt's production, competition, and patents, including a diagram and description of "Watt's mechanisms for guiding the upper end of the piston rod of a double-acting engine (British Patent 1432, April 28, 1784)". SamuelRiv (talk) 20:19, 12 February 2011 (UTC)[reply]
Bengal fire
[edit]

~~Resolved~~
What exactly is Bengal fire (a.k.a. Bengal light), and how did it get its name? All I found on the web so far is this – and many indirect uses, such as flowers and tea carrying that name. — Sebastian 21:21, 11 February 2011 (UTC)[reply]
This Everything2 post explains that it is a firework, mostly potassium nitrate, adulterated with copper and/or barium to add a blue-green color. Is is so named because these fireworks/explosives were originally originally manufactured and supplied from Bengal to England. The Everything2 node isn't quite a reliable source; and though it cites a few references, they are also of dubious quality. But this journal article, Pyrotechnics in Fireworks (2004), seems to corroborate the general claims, though. It seems that other "brightly colored" things use the term "bengal fire" because it sounds neat. Nimur (talk) 21:41, 11 February 2011 (UTC)[reply]
Thank you, Nimur! — Sebastian 22:20, 11 February 2011 (UTC)[reply]
Actually, as I'm trying to incorporate that information, I noticed that we used to have a redirect from Bengal fire to Bengal fire stick. I restored the latter in my user space so I can ask here if there's anything to it; it is at least not an obviously fictitious page, as the deleters assumed. — Sebastian 22:45, 11 February 2011 (UTC)[reply]
~~I recall purchasing "Bengal matches" around November 5th many years ago. Were they an early alternative to sparklers? Dbfirs 00:20, 12 February 2011 (UTC)[reply]~~