The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
I've been checking the weather report lately since the weather has been pretty hot this summer over here, and I've noticed that the heat waves were well-corellated with times of low atmospheric pressure. For example see [1][2]. That seems to be the wrong way around to me, usually we get low pressure with storms, not 24-hour heat. What could cause this? — Preceding unsigned comment added by 78.0.231.47 (talk) 04:25, 18 August 2017 (UTC)[reply]
Cold front) and warm front have nice tab that will help you understand what's going on. wind direction and strength are also important in that matter.
Looking at the temp and pressure diagrams at the first link, I don't see any correlation at all. The 2nd link does seem to show pressure peaking mid-morning each day, and of course temps peak mid-afternoon. Not sure what would cause the pressure to behave like that. StuRat (talk) 07:37, 18 August 2017 (UTC)[reply]
Temperature at high altitude lags behind surface temperature, so you generally get the coldest troposphere in mid-morning. Low temperature means high density, means more weight in that column of air, means more pressure at the surface. So it makes sense. Of course the effect isn't very large. There must be air moving in or out of the column, which happens by high-altitude wind changing with the time of day. PiusImpavidus (talk) 10:39, 19 August 2017 (UTC)[reply]
When the temperature is high, the density of the air is lower. The column of air above you weighs less, so the air pressure drops. If the pressure gets lower than in surrounding areas, the air will rise, clouds will form and the temperature will drop (during the day). But a lot depends on regional effects. When the pressure distribution is such that the wind comes from the continent, the air is dry, which mean less clouds and higher temperatures in summer. PiusImpavidus (talk) 10:04, 18 August 2017 (UTC)[reply]
oh dear. your are basically saying that higher temperature lower pressure. How wrong.
Ideal gas law P =(n/V)RT do not perfectly apply to air, but it does quite well, meaning, when temperature rises, pressure rises, too (opposite of what you said). The higher temperature air do not rise because of lower pressure, it rises because, for a given pressure, if T rises, then density (n/V) must goes down, hence Archimedes' principle applies and the parcel of air must go up. All that without any pressure effect. Gem fr (talk) 12:34, 18 August 2017 (UTC)[reply]
I know the ideal gas law. When the air heats up, it expands. The air isn't locked up in a bottle, it's free to expand and flows away horizontally at the tropopause, driven by a slight increase in pressure up there. As the density drops with rising temperature, the pressure difference between the surface and the tropopause (caused by gravity acting on the column of air) drops, so the surface pressure drops too. That's the pressure we measure. The reduced pressure at the surface causes horizontal inflow, continuing as a vertical flow from the surface to the tropopause. That's how Archimedes' principle drives convection. PiusImpavidus (talk) 10:39, 19 August 2017 (UTC)[reply]
Very complicated way to explain than an isobar process will bring a drop in pressure...which it will not, by definition.
I guess you'll explain likewise that vapor bubbles in a column of boiling water will reduce pressure at the bottom of the column? they won't, either. Actually they would slightly increase the pressure, because slightly rising the whole water surface height (hence pressure, since it depends on height)
Here's a better picture, the last 7 days. The temperature is red, air pressure is dotted gray. You can see how after 14th August temperature is still rising day-to-day, but air pressure has levelled off. I don't think it has much to do with fronts, the first day of that period there was a cold front, but there were no storms/rain on any other day. Even better, check out the relationship from the middle of this graph to the right [3]. (78.0.231.47) — Preceding unsigned comment added by 93.139.110.217 (talk) 04:21, 19 August 2017 (UTC)[reply]
A solar eclipse viewed with a natural camera obscura.During a partial eclipse the eye adapts to the lower light level, but the surface brightness of the sun stays the same. That makes looking at the sun even more dangerous than on normal days. So always use a good filter or project the sun on a screen using a small telescope or a camera obscura. I've always preferred projecting with a small telescope. PiusImpavidus (talk) 10:19, 18 August 2017 (UTC)[reply]
IT IS NOT SAFE. Newscasters have been pointing this out, over and over, and people still don't pay attention. You risk destroying your retina; permanent blindness. To view the partially eclipsed sun safely, go to your eye clinic, or wherever, and get some ISO certified eclipse glasses. Beware of fakes! Or better yet, use one of the methods describe above to project the sun's image. And if you're wearing solar eclipse glasses, do not look through binoculars or telescope - the focusing of the bright light will nullify the effect of the glasses and you'll be blinded. For further details, look for instructions on a TV news website in your area. ←Baseball BugsWhat's up, Doc?carrots→ 14:28, 18 August 2017 (UTC)[reply]
Five seconds is way too long. Consider how long you can tolerate the full sun before you instinctively avert your eyes and (hopefully) suffer no permanent damage. It's like a fraction of a second. ←Baseball BugsWhat's up, Doc?carrots→ 23:24, 18 August 2017 (UTC)[reply]
From the link I gave above: "You have likely heard that looking at a total eclipse can be dangerous. It indeed can be. And this is because the pupil responds to average light. During a total eclipse, the pupil swells to its full size owing to the twilight's low average intensity. Its diameter swells to about 7mm: so the aperture has fifty times the area it has just before the eclipse begins. If you look at the diamond ring just after totality, you can therefore cop a dose of 20mW20mW or so in the eye. This can be enough to cause thermal damage. Evolution didn't kit us out to look at total eclipses". Count Iblis (talk) 20:05, 18 August 2017 (UTC)[reply]
It's not the total eclipse that's the problem, it's when it's not total. The advice I've read is that when the first glint of light appears, avert your gazeimmediately. Then if you want to watch, put your solar eclipse glasses back on. The only time it's safe to not wear those glasses is during totality. ←Baseball BugsWhat's up, Doc?carrots→ 23:23, 18 August 2017 (UTC)[reply]
I heard in the winter time, cloudy days are your friend. That is, clouds trap heat? Keeping the temperature warmer, so on a sunny day with no clouds, temperatures are generally colder?
Clouds radiate infrared radiation back down to Earth, so the maintain higher night time temperature. Absolute humidity is normally higher in summer. But the hot summer day peaks probably have the lowest relative humidity. I don't know about Chicago. Graeme Bartlett (talk) 12:55, 18 August 2017 (UTC)[reply]
lapse rate is your friend on the first matter. Basically, humid air has more energy than dry air (the energy required to turn water into vapor), energy released when vapor turn again into liquid, that is, into clouds. This release of energy warms the air, so it is less cold.
humidity is the answer (or, at least , an entry point for other articles linked in) to your second. You'll notice that it refers to three kind of humidity: absolute, relative and specific.
Clouds make the night warmer and the day cooler, on average. That's for multiple reasons. 1) Clouds are droplets of liquid water, and liquid water has a very high specific heat. That means it can absorb a lot of energy without changes of temperature, so water has a mediating effect on temperature changes; it both cools off slower and heats up slower. That makes for lower highs and higher lows. 2) Clouds tend to reflect radiant heat; that is light energy in the infrared range. That means that clouds during the day tend to reflect sunlight back into space, keeping the ground from getting too hot. At night, when the sun isn't out, the clouds will reflect heat radiating from the ground back towards the ground. 3) Evaporation is an endothermic process, so it tends to remove heat from the environment, while condensation is exothermic, adding heat to the environment. As temperatures rise during the day, clouds can absorb the extra heat of the sun by evaporating (that is, going from tiny droplets to water vapor). At night, condensation takes over, which means the opposite is happening. Wikipedia has an article called Cloud feedback which, ideally, would explain this better, but it's a bit hard to follow (not well written), but this explains it a little bit. --Jayron3213:15, 18 August 2017 (UTC)[reply]
Clouds reduce cooling during the night by radiating heat back the surface and reduce heating during the day by reflecting light back to space. It depends a bit on the type of cloud. Cirrostratus clouds are almost transparent to visible light but not to infrared, so they mostly affect night time temperature.
There's also a second way that clouds are correlated with temperature. You get more clouds when the wind comes from the ocean because of higher water content and this obviously correlates with temperature too: in summer, sea air is cooler than land air, in winter it's warmer. So in winter on a cloudy day, you probably have a (relatively) warm wind from the sea. Whether this is enough to make cloudy days warmer than sunny days in winter, despite lower insolation, depends on where you live. PiusImpavidus (talk) 11:24, 19 August 2017 (UTC)[reply]
Okay looks like the answer is yes and no for winter. Yes in winter because cloudy days means warm air from the sea, no because clouds reflect sunlight and heat back to space. By the way, when you say cirrostratus clouds are most transparent to visible light but not IR, what type of cloud is the most opposite? The most effective to IR? Thanks. 12.130.157.65 (talk) 16:37, 21 August 2017 (UTC).[reply]
Role of the client in large infrastructure projects
In a typical large infrastructure project where most of the accountability is on the designers and construction companies, what is the role of the client? Is it mainly to specify the end product and monitor/regulate the project? 82.132.218.121 (talk) 12:06, 18 August 2017 (UTC)[reply]
it is mainly to chose people that have the known-how to specify the end product and to run the project (may be different people for those two actions), so that he will be happy and builders will be able to do it, and will indeed do it, preferably on schedule.
A large project will have a complex client team that includes people that specify the requirement in detail, evaluate the proposals, finance people, testers to check that requirements are met, liaison people to answer questions, internal stakeholder relationship management, and possibly a media team. Graeme Bartlett (talk) 12:52, 18 August 2017 (UTC)[reply]
What is the ultimate cause of seasons?
I already know that seasons are caused by the Earth's axial tilt, but how, precisely? Most diagrams depict the Sun's rays shining parallel on Earth to illustrate how the direction it faces matters, but I don't think this is satisfactory. First of all the parallel rays don't explain why some places are hotter than others, since the radiation carries pretty much as much energy reaching the poles (further away from the Sun) than the equator (closer to the Sun), made evident by the fact that the elliptical orbit doesn't affect seasons. Secondly it's an inaccurate illustration. The Sun is larger than the Earth and rays can come in from different directions (which is why solar eclipses occur). So after all this, how does axial tilt lead to seasons? I feel like I've always been told axial tilt = seasons, but something's been skipped in between. Currently I think it is simply down to the centre of the sunlit area being surrounded by other sunlit areas so it doesn't lose heat, whereas towards the edge of the sunlit zone heat is lost to the areas experiencing night time. Is this an adequate explanation? Please correct me if I'm wrong. The Average Wikipedian (talk) 15:24, 18 August 2017 (UTC)[reply]
It's easiest to think about it in terms of what the Sun looks like as viewed from the Earth. Where I live in California, for example, on Dec 22 the Sun is above the horizon for about 10 hours and reaches a maximum angle of about 30 degrees. On June 22 it is above the horizon for 14 hours and reaches a maximum angle of close to 90 degrees. The result is far more solar heating during summer than winter. Looie496 (talk) 15:49, 18 August 2017 (UTC)[reply]
It's true that the equator is closer to the Sun than either of the poles is, but the radius of the Earth 6371 km is so tiny (0.004%) compared with the average distance to the Sun 150 000 000 km (or 1 AU) that this is not enough difference to explain the seasons. The article Season explains how the combination of axis tilt (see diagram) and Earth's elliptical orbit together contribute to the cycle of seasons. Blooteuth (talk) 15:52, 18 August 2017 (UTC)[reply]
The incoming energy is more spread out when it comes in at an oblique angle. You can convince yourself of this with a flashlight. If you want to brightly illuminate a spot on the wall, you do it straight on. If you hold the flashlight the same distance away from the spot, but closer to the edge of the wall it won't be nearly as effective. ApLundell (talk) 15:56, 18 August 2017 (UTC)[reply]
[Edit Conflict] You are wrong.
Firstly, you have ignored the fact that towards higher latitudes the Earth's surface is angled more obliquely to the direction of sunlight, so that (for example) a square metre of sunlight shining vertically on the sub-solar point at noon indeed illuminates one square metre, but the same area/amount of sunlight shining on the tilted surface at high latitudes is spread out over a significantly greater area of surface. This spreading effect varies through the year (because of the constant tilt of the earth's axis): when the tilt effect is at its greatest, the tilt also means that days in that hemisphere are significantly shorter than nights, so the sunlight also shines (and heats) for a shorter time.
Secondly, the illustration has to be "inaccurate" to some degree, because it would be impossible to represent the Sun, Earth and the distance between them all to scale on the same page. (For example, the average Sun–Earth distance is about 13,300 times the diameter of the Earth — I have had to wrestle with this problem in editing illustrated science textbooks!) Because of its distance, the Sun, despite being very large, has an angular diameter of only around half a degree, which is close enough to being a point source to make no significant difference in seasonal terms. {The poster formerly known as 87.81.230.195} 94.12.90.255 (talk) 16:09, 18 August 2017 (UTC)[reply]
In addition to the insolation angle, there's also just more hours of sunlight when your part of the Earth is tilted towards the Sun, and thus more heat accumulates during those times and there are fewer hours at night for the heat to dissipate into space. Note that the difference in number of hours of sunlight is far more dramatic at the poles, hence more difference from summer to winter. StuRat (talk) 16:16, 18 August 2017 (UTC)[reply]
To somewhat avoid technical terms, just think about where the Sun seems to be in the sky. In the far north, it seems far south, and in the far south, it seems far north - in both cases, near the horizon. Now it is TRUE that people in those places see the same Sun, the same size (more or less) as they do at the equator. And it is TRUE that the light and warmth provided if you hold your hand up to the Sun is about the same - just a bit less because the light has to go through extra air when the sun is near the horizon. So why the difference?
Well, consider if you wanted to lay out solar panels on the ground. At the equator, you could lay them out flat and at noon they would all get the absolute maximum energy, intercepting their full area of sunlight. But in Greenland, you would have to tip each one high up to catch the same amount of light and produce the same amount of electricity or hot shower water. Now if you have enough space going east to west you can always put them side by side, tilted or not, at least at noon. But if you don't, and you have to put some north and south of the others -- here's the difference: you can't put them at the same distance apart as you would at the equator, because the shadow of the south solar panel would block most of the one to the north. You have to put them much further apart to collect all the energy. And that reflects the fact that the entire surface of the Earth in the area, like a tilted solar panel, just doesn't block all the light or collect all the energy that one gets when it faces full into the sun. Wnt (talk) 17:46, 18 August 2017 (UTC)[reply]
Does relying on refined cooking oil explain why we cannot get EPA and DHA from plant sources?
It is well known that the conversion rate of Omega-3 to EPA and DHA is too low to supply us with the required amounts. However, it could be that this is only what happens in people who eat a normal diet who then get 50% or more of their calories from refined fats and oils that contain very low amounts of Omega-3 fats. A natural diet that doesn't include any refined fats and oils would yield only about 15% of calories from fat, and most of that would be Omega-3 and Omega-6 fats. This doesn't seem to be relevant to the low conversion rate, however, when sticking to such a diet on the long term, most of the body fat would become the Omega-3 and Omega-6 fats that you would then eat. If in the natural context Omega-3 is present in high concentrations in fat tissue, it seems to me that it's there that the enzymes that convert Omega-3 to EPA and DHA should be present. So, the reason that the reaction rate is low would be because we've filled the reaction chamber with junk. Count Iblis (talk) 22:30, 18 August 2017 (UTC)[reply]
Think I can see were your coming from. It is the ratio of all these fatty acids which is important. So if one's diet is short of one or more of these essential fatty acids (those which humans can not make) then health will suffer. In this case, Junk is getting too many calories from foods which don't don't supply essential fatty acids, vitamins and minerals. P.S. Do you know that the food refiners that remove those very oils you mention, (for the reason is that they go rancid quickly and shorten the shelf-life of modern cooking oils), go on to sell what they have removed, to companies that reformulate them as health supplements. To be bought at exorbitant prices to make good for what one looses in order to be able to buy long shelf-life cooking oils. Madness! Aspro (talk) 23:08, 18 August 2017 (UTC)[reply]
I see this is mentioned in a section at alpha-Linolenic acid. I haven't looked into it as of yet, but the description is actually rather disturbing - there is talk of making linolenic acid free soybeans to make the oil more desirable for cooking so it doesn't "have" to be partially hydrogenated. Note that this runs contrary to the doctrine, frequently espoused in Wikipedia controversies over GMO food, that GMO food is nutritionally equivalent to non-GMO food. Something similar has been attempted by proxy with omega-3 enriched canola oil for use in fish farms, as I recall. After the Holocaust-level casualties involved with humanity's first attempt to tinker with fats for convenience (i.e. partial hydrogenation), I would have thought people would be more careful. Wnt (talk) 11:40, 20 August 2017 (UTC)[reply]
Except that as the very article you linked to says, (I know because I read it before you posted when I linked to it below), this isn't something unique to GMO. It's possible of course to significantly reduce levels levels of omega 3 in soy beans without relying on GMO and this has been done. (Whether this relied in nuking the hell out of the crop and hoping for the best, I don't know.) Not surprising since we humans have been tinkering with our crops and animals for millennia, including with effects on the fats, and often for various reasons of convience, just with a lot less specificity and idea of what we were doing. Nil Einne (talk) 12:55, 20 August 2017 (UTC)[reply]
No, you can't get EPA and DHA from plants because they're not made by most plants. As the articles state, they're only found in large amounts in fish, seaweed, and algae, as well as, interestingly enough, human breast milk. --47.138.161.183 (talk) 06:34, 20 August 2017 (UTC)[reply]
I don't think the OP (no idea about Aspro, I didn't bother to read what they said since I've found it best to ignore anything dietary related coming from them) is suggesting you can get EPA and DHA directly from plant oils. Rather they are suggesting that the low levels of conversion of α-Linolenic acid to EPA and onwards to DHA is somehow because of excessive consumption of fats and oils, particularly refined fats and oils.
This isn't something I've looked at that well but the earlier linked articles (both mine and yours) along with Omega-3 fatty acid support the view that strictly speaking neither EPA or DHA are essential fatty acids in humans since even though they are important in human physiology, they can be synthesised from ALA which is an essential fatty acids in humans. However the conversion rates are fairly low. There's some limited disputed evidence that the high ratio of Omega-6 fatty acids compared to omega-3 in modern Western diets may be not ideal. See also Essential fatty acid interactions.
But even if this is true, I think it's fairly misleading to suggest it's just because of refined oils. The seem to be a multitude of reasons for this which relate to modern industrial and agricultural practices and the sources we've chosen for eating and feeding (e.g. grain instead of grass for feedint cattle in many places). This includes the fact that the increased rancidification rate means omega-3 is not desirable when it comes to long term storage. But breeding plants for lower levels of omega-3 isn't directly part of refinement, even if perhaps part of the reason it's done is because of the benefit to refined oils. (In other words, you could probably still have poor levels of omega-3 depending on your diet even if you had low levels of refined oil consumption and you could probably still have decent ratios despite high levels of refined oils.)
And notably I don't think it's clear that the levels of oil consumption, as opposed to the ratio, is a problem. And this seemed to be what CI was suggesting. (Note that this could easily be seperate from whether or not high levels of dietary fat are undesirable.) Note also even if the ratio of omega-3 to 6 in modern western diets is a problem, I don't think it's clear that this is because of the effect on levels of EPA and DHA. Nor of course that EPA or DHA levels either from dietary consumption or conversion are a problem. In other words, I agree with you that CI's premise seems to be basically unsupported by anything (as seems to often be the case when CI brings up dietary issues) but not quite on the reasons for that.
BTW AFAIK and supported by both our article and [4][5] ALA does significantly come from (certain) plant sources.
Another method: charge cap, then discharge into resistor and measure the RC time constant, the time it takes for the voltage to fall to 36.8%. Note that the article implies (by omission) that the fall is linear with respect to time, but that is not the case. The article really needs a graph showing the non-linear fall. Akld guy (talk) 03:35, 19 August 2017 (UTC)[reply]
This capacitance meter circuit comprises two 555 timer ICs. It directly reads capacitance in the range 100pF to 10µF. IC1 and IC2 operate as an astable (with frequency above 80Hz) and as a monostable multivibrator respectively. Time period of the monostable multivibrator is determined by the resistors (R3 to R7 and VR1 to VR5) selected by switch S1 and unknown capacitance Cx. Replace the 1mA moving-coil ammeter with a digital meter if preferred. Blooteuth (talk) 05:45, 19 August 2017 (UTC)[reply]
All of the above. When capacitance meters were a separate thing, they were usually an AC bridge circuit, and even then they needed a variable drive frequency to cope with the wide range of capacitance values encountered. As a crude modern circuit (the cheap, inaccurate, but highly useful $10 "any component testers" found on eBay), they're mostly based on some sort of charge pump where a monostable drives a series of constant current pulses and the voltage rise is measured. This circuit is used because it's cheap: it can be built from low frequency DC circuits attached to the pins of a cheap microcontroller, rather than some AC circuit which would have to be designed and built onto the circuit board with dedicated components. The downside is that it's limited in the range it can measure, and it measures "charge storing ability" (including that charge decaying from leakage over time) rather than an idealised capacitance. Leaky old capacitors in ancient radio circuits highlight just what a difference there can be with those two measurements, over a short or long time period. Andy Dingley (talk) 09:08, 19 August 2017 (UTC)[reply]
The 555 works fine in the µF range. Testing caps in pF range, another circuit is necessary, sometimes another oscillator, followed by a circuit converting the frequency to a voltage. --Hans Haase (有问题吗)15:22, 19 August 2017 (UTC)[reply]
Bridge circuits were mentioned but usually require manual balancing so they do not yield a direct reading. Wheatstone bridge is the basic configuration used to measure resistance. For measuring capacitance the bridge types are Schering Bridge and Wien bridge; the former is independent of frequency while the latter measures capacitance in terms of resistance and known frequency (or can be used to measure frequency when the capacitance is known). Blooteuth (talk) 13:00, 20 August 2017 (UTC)[reply]
Once the microprocessor era was established, there were any number of automatic self-balancing bridges around as test gear. Even before this though, the classic HP 4270 was available from 1970, using digital control circuits and a dozen boards of TTL. It used four oscillator frequencies from 1kHz to 1MHz and measured from sub-pF to μF. Nice article here: "Measuring Capacitance Automatically"(PDF). Hewlett-Packard Journal: 14–20. May 1970. {{cite journal}}: Unknown parameter |authors= ignored (help)Andy Dingley (talk) 20:20, 20 August 2017 (UTC)[reply]
Well, to start, what do you mean by "size"? Height, dry biomass mass, wet volume? Here [6] is a very tall dandelion, here [7] is Guinness'current record holder. Here [8] is a scholarly research article on dandelion demography that can give some context. Here [9] is a study that specifically examines the effects of competition on dandelions. Here [10] is a study specifically about plasticity in a few dandelion species, which is the idea that the same species can grow rather large or rather small, depending on conditions.
Here is an article [11] all about increasing dandelion biomass in agriculture.
I suggest that nobody really knows for sure. You can try some experiments yourself, or you could hire someone to work out some estimates based on models of plant growth, possibly building upon some of the research linked above. But lacking that, the records above are probably pretty good approaches to maximum height. SemanticMantis (talk) 16:39, 19 August 2017 (UTC)[reply]
The first image is not taraxacum officinale. It looks to my fairly experienced eye like either lactuca virosa or lactuca serriola, both of which belong to the lettuce family. Let's get the sheep sorted from the goats before we start comparing. Richard Avery (talk)
Yes, thanks, I had meant to mention that it didn't seem that way but in my haste to find scholarly articles I forgot to. I am a little confused. I didn't think "dandelion" was used for anything other than Taraxacum. Maybe the photo is mixed up? The Guinness bit is perhaps more reliable. SemanticMantis (talk) 16:40, 20 August 2017 (UTC)[reply]
August 20
Umbra speed on Earth's surface
The moon is in a relatively circular orbit, so its shadow inevitably moves across the Earth. However, the speed of the shadow relative to the moving ground depends on several additional factors. (an added comment by a different editor): Not to scale.
During a solar eclipse, why does the speed of the moon's shadow on the surface of the Earth vary significantly?
I assume the time of day should be a factor. Think of a shadow cast on a globe -- if the Sun is just rising, a small movement of a shadow across it will go a long distance on the ground. Likewise the latitude should have an effect. Also, the rate at which the ground moves due to the rotation of the Earth will be different by latitude. The Moon's orbital speed should not vary by much, so the shadow itself (ignoring the terrain and curvature of the Earth in its way) should not change speed much. Wnt (talk) 02:51, 20 August 2017 (UTC)[reply]
I don't think time of day should matter. Essentially, the shadow itself doesn't move (relatively speaking) -- the Earth rotates within the umbra; the time as determined by the position of the sun in the sky (not "clock time") is the same for all observers (for same eclipse phase). 2606:A000:4C0C:E200:A90E:D475:2878:2440 (talk) 03:04, 20 August 2017 (UTC)[reply]
To excerpt from that source "Because of the geometry of the Earth’s shape, the shadow will travel faster across its surface and the ends of the eclipse path, and slowest right in the middle." (It should say "at the ends". Must be a typo.) Bus stop (talk) 03:07, 20 August 2017 (UTC)[reply]
I guess the relative speed of the moon (~one orbit per month) does matter. Looking at the top image from that source, the moon's shadow always remains in the "center" of the Earth's surface, since the Sun's direction (vector) must be normal to the surface, with the moon intersecting the vector. 2606:A000:4C0C:E200:A90E:D475:2878:2440 (talk) 03:29, 20 August 2017 (UTC)[reply]
I can't see the image you refer to, but (speaking as an ex-astronomer) I think you must have misinterpreted it, if not the entire physical situation. During a solar eclipse, the angle to the Earth's surface of the direction to the eclipsed sun can vary from 0° (if the Sun & Moon happen to be at the at the Zenith, which would actually be a rare occurance) to 90° (if they're on the horizon, which happens not infrequently). {The poster formerly known as 87.81.230.195} 94.12.90.255 (talk) 05:52, 20 August 2017 (UTC)[reply]
I can see now that the alignment need not be a normal vector to the Earth; but, I reiterate my argument that the reason the shadow moves is due to the rotation of the Earth, and the extraordinary change in the velocity of the shadow is curious. -OP:2606:A000:4C0C:E200:5560:5BBD:5A94:28E2 (talk) 08:00, 20 August 2017 (UTC)[reply]
Umbras start at sunrise and end at sunset. If it goes over the pole before hitting the Earth it can go the "wrong way" (sunset to sunrise). The shadow cannot only move because of the rotation of the Earth, if that were the case there'd be no way to get off the Earth and there'd be a total solar eclipse somewhere on Earth 24/7, even at Full Moon. Sagittarian Milky Way (talk) 09:13, 20 August 2017 (UTC)[reply]
The rotation of the Earth contributes to the movement of the umbra over the Earth's surface, but is by no means the dominant contributor. Also important are the movement of the Earth due to its revolution round the Sun, the movement of the Moon due to its revolution around the Earth – both of which alter the relative alignment of the syzygy (astronomy) – and the sphericality of the Earth which means the umbra moves faster towards the Earth's limbs (from the Moon's point of view) limbs due to the umbra's increasingly oblique angle to the Earth's surface. Consider setting up a rough 3D model with balls and a flashlight so that you can demonstrate these factors to yourself, rather than trying to visualise them from descriptions and 2D diagrams. {The poster formerly known as 87.81.230.195} 90.204.183.114 (talk) 19:50, 20 August 2017 (UTC)[reply]
Viewed from an observer at Polaris at this time of year, the Earth is a blue dot (formerly white...) and according to Orbit of the Moon the Moon goes around it counterclockwise. The Earth is going around the Sun counterclockwise, or if you want to be parochial about it, the Sun goes around the Earth counterclockwise also, in a non-inertial frame of course; but either way this is so slow we can almost ignore it. And since the Moon is able to extract angular momentum from the spinning Earth, it's a good guess the Earth is also spinning counterclockwise, if your scope can make out those details. So the chief competitors in this race are the Moon, moving an average 1.022 km/s = 2290 mph, and the surface of the Earth, moving 24000 mi/24 = 1000 mph at the equator, times the cosine of your latitude. (The North Pole at 90 N, of course, does not move when rotating). So the Moon's speed is definitely faster than the land's speed, and they're both going counterclockwise, so it's going to catch up. So the East Coast, racing toward sunset, will not outrun the approaching eclipse... unless, that is, it actually reaches sunset first. But when the Moon's shadow is nearing that side of the Earth, then it is extending further and further back along the side at a faster and faster rate, so it is entitled to a last burst of speed near the end of the eclipse (as near its beginning) to cover a large amount of twilit territory. Wnt (talk) 00:51, 21 August 2017 (UTC)[reply]
Hmmmm.... The Moon is thought to have moved from 5 Earth radii away to the present 60. The orbital speed is proportional to the square root of this radius. So I'm thinking when it was something like 60/(2.3)^2 = 11 Earth radii away, the orbital speed should have been slow enough that the equator could match its speed. And of course the umbra back then would have been much larger. Does that mean there was a time early on when eclipses could last for hours, or you could even have a double eclipse where the Sun could peek past the Moon, then return behind it? Dang that doesn't seem to make sense - the Earth's surface isn't in orbit - yet I rechecked, Earth's rotation is 1040 km/h, orbit of the Moon is 2290 km/h. Is the orbital speed formula approximation that wrong?? Wnt (talk) 01:54, 21 August 2017 (UTC)[reply]
No, I was just being stupid. The orbital speed is proportional to the inverse square root of radius, i.e. something a billion light-years out has an orbital speed of basically zero, while closer orbits are faster. So the Moon doesn't slow down to current rotational speed at the equator until it is 6 times further out, if it ever gets there, at which point the rotational speed has also decreased, and there are no total eclipses anyway. This also reflects that the rotational speed would have to be very much higher than lunar orbital speed to reach orbital velocity near the surface. Now, I still don't actually know that the faster rotation of the early Earth was never faster than the faster orbital speed of the early Moon, and I can't rule it out logically; I'd need formulas for how both have chanced over time. Wnt (talk) 03:33, 21 August 2017 (UTC)[reply]
Yes, they're in there to limit the current. Unless you use a very weak source like a penlight battery which has enough internal resistance anyway. Some batteries can have very low resistance so unlike the article I wouldn't leave out the resistance even with 3V coin cells. See LED where it talks a little bit about this in the section on considerations for use. Dmcq (talk) 15:18, 20 August 2017 (UTC)[reply]
LED bulbs need something that limits the current, but that something does not have to be a resistor. Unless the product is really cheap, it is far more common to use a tiny Switched-mode power supply because the batteries last a lot longer. --Guy Macon (talk) 17:33, 20 August 2017 (UTC)[reply]
There are two aspects to this: controlled current or controlled voltage. Unlike an incandescent lamp, LEDs have a strongly non-linear relation between current and their junction voltage. So a simple resistor isn't a great solution. It's important to control (or at least, safely limit) both aspects.
For decades, LEDs were low-powered dim indicators, not powerful illuminators. So their power consumption was low and they were fed from a broadly constant voltage supply. So the simple series resistor was adequate. Also the LED forward voltage (typically 1.7V for a red) was low (compared to the supply), so the resistor was essential. This had both advantages and a disadvantage, also it's somewhat unclear whether the resistor is there to control voltage or current. Because the supply voltage was so much higher than the LED forward voltage, there was a large voltage dropped across the resistor making the circuit relatively insensitive, i.e. the LED current changes little if the supply voltage changed (a good thing). The other advantage of the resistor is that it's cheap and simple, the disadvantage is that the power is mostly dissipated uselessly in the resistor, not the LED.
It's necessary to control both current and voltage - or at least, to not exceed limits on either, by controlling one of them. For these early LEDs, controlling the current was enough, and was easy.
Then the 1990s changed everything - LEDs became powerful sources of illumination. Also their forward voltages increased, to maybe 3.6V. We were now trying to make powerful LED lighting, also using LEDs as torches, from low battery voltages (maybe just a single cell at 1.2V) and trying to cope with battery voltages that changed as the cells discharged.
The series resistor was no longer an adequate control mechanism. As the supply voltage approaches the LED voltage, the circuit no longer regulates so well. But the supply voltage was also reducing at this time, both because the junction voltage was increasing and approaching battery voltages, and because powerful LEDs (especially with batteries) could no longer afford to waste power across the ballast resistor, as they had before.
Constant current supplies are now popular for lighting, because they are easy to control and they're also usable with varying numbers of LED chips in a series circuit. However they need a supply voltage that's close to the LED junction voltages (or the difference ends up as waste somewhere), so they might require an embedded switch-modebuck converter (a downward voltage converter) to efficiently match these.
Constant voltage supplies are also used, mostly when driven by batteries. These are boost converters, which can raise the voltage, thus allowing batteries to be used as their voltage falls under discharge. However voltage regulation is difficult, as the LEDs are sensitive to their voltage, and a small excess voltage can cause such a great over-current that the LED overheats and is destroyed. Andy Dingley (talk) 18:39, 20 August 2017 (UTC)[reply]
2 watts, 12 to 16 LEDs feed by LED switching mode driver from 12 V AC or DC >90% efficiency
Internal AC adapter driving a series circuit of LEDs >80% efficiency
Full insulated single device passive filament LED driver >90% efficiency
This question leads to a broader question: is it a good idea to save animals from natural threats? Shouldn't the evolution just do its work? That is, I wonder why save the whales at all in this situation, shouldn't we better let thoughtless stranded whales die?--Hofhof (talk) 16:07, 20 August 2017 (UTC)[reply]
"Is it a good idea...?" and "Shouldn't" both implicitly rest upon some value system. Maybe you don't want to "interfere", but surely other people do think there is good to be done by saving a beached whale. Could be something about how humans are pretty much 100% responsible for endangering wales, and maybe some people think we owe them a little help when we can. Nobody can tell you what you think is right or wrong, but we can give you references that discuss what experts in biology and ethics have said about it.
For whales specifically, see this report [12] on whale welfare and ethics, this [13] position piece on beached cetaceans from the WDC, including decision making.
From [14] it appears that this might be a matter of mission creep -- NOAA apparently more often rescues whales tangled in manmade objects, and I would assume the implication there is that they don't expect the sea to be full of crap for the next 10 to 20 whale generations to make it a useful thing to select against. Whether that's because the crap gets cleaned up or the whales die out before then is yours to decide... Of course, the literal answer to your question is that it's a great idea: a million dollars in suckers' money waiting to be spent, and in the process you make yourself a star "real life hero" in two weeks of paid work. Wnt (talk) 00:36, 21 August 2017 (UTC)[reply]
Another possible argument for saving a stranded animal is if it's species is endangered. Of course, it could also be argued that it has defective genes, and should be eliminated for the good of the species. But this must be weighed against the reduction in genetic diversity that will occur if that individual dies. StuRat (talk) 00:54, 21 August 2017 (UTC)[reply]
Laymen typically highly overestimate the effects of selection pressure in driving evolution. Typically, most deaths are entirely accidental in regards to any specific gene. Rather, even if only .01% of deaths are differentially caused by a gene per generation (that is, a surplus of only 1/10,000 deaths) it is enough to drive evolution in geologic time, even though very few excess deaths can be attributed to that gene.
Consider the extinction of the non-avian dinosaurs. There was no specific gene that the Chicxulub Meteor selected for that day. Likewise, very few people who die who have blue eyes die because they have blue eyes. Also, people who die in old age have already reproduced. Selection pressure doesn't apply at that point, hence the plethora of diseases of old age.
Finally, there's the example Stephen Jay Gould gave of a single dog which killed over 500 kiwis in 1988, about half the local population. There was no single gene these birds shared that the massacre survivors didn't.
This is Gould's hypothesis that much of evolution is driven by radical contingency (i.e., blind luck), not gene selection. See Gould's Wonderful Life. If most whale strandings are due to blind misfortune, attempting to save the whales is unlikely to contribute to their long-term extinction.
There are, of course, counter-examples. Should world civilization fall, the bloodlines of type I diabetics, hemophiliacs, and women whose hip structure allows only Caesarian births, for example, would quickly plummet to almost negligent numbers. That is because we address the effects of these genetic issues without eliminating the genes themselves from our own population. μηδείς (talk) 16:24, 21 August 2017 (UTC)[reply]
A small nitpick. In the Māori language, the plural of kiwi is kiwi. One kiwi, 200 kiwi. One Māori (person), 20,000 Māori. One waka, 30 waka. By convention, this is carried over into New Zealand English when using Māori words. There are many New Zealanders who uncaringly add the 's', but the preference is to follow the Māori style. Akld guy (talk) 20:29, 21 August 2017 (UTC)[reply]
I bet the Chicxulub Meteor selected pretty heavily for some hibernation/estivation variants, and blue eyes could be the difference between life and death under the Nazis. But in order to be effective, selection has to be sustained - the meteor of course could wipe out whole species, but if it left a few copies of a given allele within a species, that allele might eventually return to its original equilibrium. Despite all these things, evidence of selective sweeps is not hard to find. Even if a given gene only accounts for 1/10000 deaths ... there are 30,000 of them. One gene usually matters little, but some gene usually matters. Wnt (talk) 19:52, 21 August 2017 (UTC)[reply]
The birds are normally called kiwis (pl) in American English, this is the term Gould uses repeatedly in Bully for Brontosaurus recounting the story I referred to, and I am sticking to it. You can call moose "elk"; rabbits "coneys"; and opossums "possums" as much as you like.
As for selection for estivation or hibernation as a means of surviving K/T level events, either life at the time would have had to be small due past such events, or have become small afterwards to survive regular future such events. Neither was the case, so this was one mass culling, not the evolution of small size by (repeated) natural selection. Dinosaurs did not evolve to become extinct and mammals did not evolve to stay small to avoid future mass extinctions.
Don't see any evidence, Akld guy was "nitpicking for points", not sure Wnt was doing so either. Most logical conclusion would be that the person who assumes nitpicks are for points probably believes so because they aren't being quite honest in what their cup of cricket is hence adding non existent motivations just because it's what they normally do. Incidentally, in modern contexts it's not very common to call opossums as possums in NZ. Mostly of course we don't talk about opossums much since they don't really concern us, but possums do. Our article Opossum suggests it's an American thing. There is some history of the possum being called opossum in NZ, including in legislation [15][16] but from my experience in modern contexts very few Kiwis actually do that. (I don't think I've actually read or heard any example in the years I've been here except for those I read when researching this. I.E. whenever someone has mentioned opossum they're been talking about opossums although as said, this isn't really that common in NZ. But I have read and heard about possums. A lot. And although it isn't something I have that much experience with, a Google Scholar search supports my view that this applies even to scientific papers in modern contexts. Most references to opossum relating to New Zealand since 2000 seem to be titles of older citations. The modern papers themselves use possum. This source suggests it was probably about in the late 1970s and the 1980s [17] that the scientific usage caught up with the vernacular usage.) Nil Einne (talk) 08:52, 22 August 2017 (UTC)[reply]
Basically TLDR, Nil, but both those editors understood my point exactly, and I said they could call opossums "possums" if they liked. All that matters is they were making points irrelevant to the actual science. Mass extinction does not equal evolution, although it obviously opens up niches that other animals can evolve to fill--but mass extinction and evolution by natural selection are separate concepts. As for PC nomenclature, they are still starfish and jellyfish no matter how much the mermen complain. μηδείς (talk) 16:36, 22 August 2017 (UTC)[reply]
How aware were the ancients that the Octaeteris coincides with cycles of Venus ? (I am asking this because these two numbers, eight and thirteen, appear conjoined in at least two instances in this ancient monument, dating from the first two centuries BC, and I want to avoid anachronistic interpretations). — 79.118.174.82 (talk) 10:30, 21 August 2017 (UTC)[reply]
Europeans (such as the ancient Greeks, for instance), and the civilizations (directly or indirectly) known to them in ancient times (i.e., Egyptians and Babylonians, but not Mayans). — 79.118.174.82 (talk) 12:11, 21 August 2017 (UTC)[reply]
however, be aware that 8-13 is also part of the Fibonacci sequence, with ubiquitous presence in nature, so you WILL find it pretty everywhere whether people meant it or not. And hence in numerology
Plus you can cannot even know it's 13 Venus years if you think the Earth is the center of the Solar System. Aristarchus the Ancient Greek thought the Sun was the center but that didn't catch on in the West till a few centuries ago. The more relevant numbers for Earthbound observers are 5 and 8 (5 Venus apparition pairs per 8 years) and 1.6 years (AKA 8/5ths years, 1 and 3/5ths years). (5 is coincidentally a Fibonacci number) The octaresis also sucks. It's wrong by a whole tenth lunation after only 2 cycles (16 years) and the Metonic cycle's only wrong by a tenth lunation in ~a millennium (dozens of cycles)) so that's a much better accuracy to length ratio. Sagittarian Milky Way (talk) 17:38, 22 August 2017 (UTC)[reply]
No, the ancients knew how long the planetary years were even in their geocentric model because they could still track the patterns in the repeating motions of the planets. That is, they could calculate the time periods for the patterns they saw, and could arrive at the numbers even if their explanations were crap. See Deferent and epicycle for a rough overview. Pre-modern astronomers had strikingly accurate calculations of the movements of objects. They didn't know what they were actually watching, but they could describe and predict it just fine. --Jayron3219:25, 22 August 2017 (UTC)[reply]
LED on mains power how is it possible
In this video some guy solders what looks like a ~1v DC Light Emitting Diode onto an AC mains power socket and somehow it works. Can someone explain to me how this is possible? I thought AC travels in both directions so shouldn't the LED not light up? How come it doesn't get hot and / or explode by being overpowered by 239v? — Preceding unsigned comment added by 49.49.98.175 (talk) 11:08, 21 August 2017 (UTC)[reply]
Devices called rectifiers are used to convert AC to DC. Any simple diode can work as a rectifier (with some extra components, IIRC. I am not an electrician, but I did take a few electronics classes many decades ago, and some of it is coming back to me) including a light emitting diode. I suspect the LED acts as its own rectifier in some way for this circuit. --Jayron3212:14, 21 August 2017 (UTC)[reply]
LEDs are diodes, so they only pass current in one direction. When they do that, they light up. When reversed, they don't conduct. However they're not very good diodes, so they have only a low reverse breakdown voltage. LEDs are available where two chips are placed in one package, connected in reverse parallel and usually of two different colours. They were used as a two-wire connected two colour indicator, or as a simple polarity indicator.
There are two problems here. He solves the excess forward voltage and current issue by using a resistor (see further up this page too). But he leaves it with a high reverse voltage, so the LED is likely to fail fairly soon. Andy Dingley (talk) 13:21, 21 August 2017 (UTC)[reply]
Looking at a few more of this guy's videos, he's past idiot and into dangerous idiot. Many of the things here are not just pointless, but so pointlessly hazardous that they will try to kill you - in a way that isn't even educational or amusing.
Agreed. BC has a bit of an obsession with pyrotechnics, but at least he *warns* you when he is doing something that is not 100% sensible. Gandalf61 (talk) 16:32, 21 August 2017 (UTC)[reply]
Nobody has done minimum calculations for efficiency or life span of this circuit. There are BiLEDs or double core LEDs avail, containing two "anti parallel" green/red or yellow/red LEDs chips in a single LED device case. For a red LED, the one in the video looks little "orange". The second LED is a reverse to the first one. Most LEDs allow a max reverse voltage of five volts. As even blue LEDs have the highest voltage drop in operation mode from 3 to 5 volts, the dual LEDs protect each other form reverse voltage. But nobody has seen if his multiple socket is plugged to the wall. There's also no flicker or interference from camera to the red LED visible which makes me being in doubt if the LED is powered with AC. --Hans Haase (有问题吗)09:52, 22 August 2017 (UTC)[reply]
No. You get lunar eclipses around the time of a full moon, solar eclipses around a new moon. The ordering has to be either S/E/M or S/M/E for each, and accurately aligned. Andy Dingley (talk) 16:27, 21 August 2017 (UTC)[reply]
Technically no, because a full moon occurs when the moon is on the far side of the earth from the sun, rather than between the earth and sun. But a full moon does occur half a month before and after an eclipse, and there could be back-to-back solar and lunar eclipses 15 days appart if conditions are right. μηδείς (talk) 16:28, 21 August 2017 (UTC)[reply]
No. By definition, the moon would be on the exact wrong side of Earth during a full moon.
The only way it could happen is if some other massive object showed up to block the sun. At that point, arguing over whether or not that "technically" still counted as an eclipse would probably be the least of our worries. ApLundell (talk) 14:07, 22 August 2017 (UTC)[reply]
Astronomers don't have to wait for a total eclipse to study the solar corona. They can place a circular mask at the end of their telescopes. A calendar year can have a maximum of seven eclipses (four of the sun and three of the moon if I remember rightly) or a minimum of two (both solar). Eclipses happen when the orbits of the earth and the moon intersect (which happens every six months). So for the maximum seven the orbits would intersect at the beginning of January, in June and at the end of December. 82.14.24.95 (talk) 16:02, 22 August 2017 (UTC)[reply]
Would a solar eclipse be visible from the moon?
Another eclipse question: I was reading Total eclipses on the Moon, and wondered if the reverse could occur. Would a solar eclipse (as seen from earth) be visible from the moon? I.e. if you were standing on the moon, looking at the earth, would you see the moon's shadow passing across (umbra and penumbra), or would it be too small to see with the naked eye? Optimist on the run (talk) 19:20, 21 August 2017 (UTC)[reply]
The moon's shadow on the earth would be the same size as the moon looks from earth, about a half a degree. Seems like it would be highly visible if you were on the moon looking at the earth. It would look like a small dark spot moving across the earth. 69.243.146.7 (talk) 19:26, 21 August 2017 (UTC)[reply]
You are wrong: moon's shadow on the Earth is very small - about 250 km. From the the Moon the angular size of this shadow will be about 2 angular minutes. In another words it will be barely visible. Ruslik_Zero20:40, 21 August 2017 (UTC)[reply]
Binoculars would definitely help, and it would be more visible if it (or rather the lit area surrounding it) was hitting a highly reflective surface, like clouds or ice. StuRat (talk) 22:02, 21 August 2017 (UTC)[reply]
Approximately what is considered a deep partial solar eclipse?
I can find no formal definition of the term, which means that deep means "a lot" and shallow means "only a little" without any formal % as the dividing line between the two; it's probably like the differences between "some" "a few" "several" "a lot" "many", etc. These are imprecise relative terms and are not defined strictly. At least, that's how I interpret the absence of evidence on this one; if it had a formal definition, we'd find it. If it lacks a formal definition, we're stuck to falling back to how language works, which is as fuzzy as always. --Jayron3219:20, 22 August 2017 (UTC)[reply]
Solar eclipse: why did the sun still look round in direct images?
2017 solar eclipse, captured at close to maximum eclipse in a reflection off the edge of the lens(?), showing that the sun still appeared round in unfiltered direct images but that reflections made the eclipse visible.
In my camera-phone photos of today's solar eclipse, taken where it was not total but peaked above 75%, the sun still appeared round or slightly elliptical in the direct image. Only a reflection, which I'm guessing came off the edge of the lens (since it persisted when I removed the clear plastic case), made the crescent shape of a partially-eclipsed sun visible. Why does this happen? NeonMerlin23:52, 21 August 2017 (UTC)[reply]
I saw something similar today, but with no camera involved (just using a lens pointing the image at paper). My guess is that the direct image was badly out of focus, since any image sufficiently out of focus will look like that. The reflected image just worked out to be perfectly in focus. StuRat (talk) 23:58, 21 August 2017 (UTC)[reply]
Here in the Northeast the sun appeared crescent both in my pinhole camera and when directly visible through heavy cloud cover. My dad's camera made an hourglass shape, until I asked him how he made the pinhole, which was with a blunt paperclip that tore the foil rather than puncturing it. When he made a new, round pinhole the image appeared quite clearly crescent-shaped. My sister is not too far from Vermont, I'm planning on seeing totality there in 2024. μηδείς (talk) 00:48, 22 August 2017 (UTC)[reply]
A pinhole lens has the feature that everything is always in focus, so that would explain why you wouldn't get a big blob. The downside, of course, is that the image is quite dim. Normally this is a showstopper, but with something as bright as the Sun, some dimming is a good thing. StuRat (talk) 01:09, 22 August 2017 (UTC)[reply]
Here in Virginia, cellphone-carrying people with whom I was talking typically had one of two images: a vaguely round really bright figure through thin or no clouds, or a crescent-shaped figure through not-as-thin clouds. I was left suspecting that cellphones simply can't capture the solar figure properly unless it's dimmed by clouds or something else. Nyttend (talk) 02:06, 22 August 2017 (UTC)[reply]
I had a looooooong discussion about this yesterday before the eclipse. A cell phone (and any cheap auto-focus camera) is designed to focus on objects that are designed to focus on objects that are between 6 inches and 100 feet away. This depends greatly on the exact camera and the type of auto focus. So, lets assume yours is a very nice autofocus and it can focus on objects that are 100 miles away, not 100 feet. The sun is almost 100 million miles away. As a result, your camera cannot focus on it and you get a blurry bright dot. That isn't all. The brightness of the sun is so much that it floods the photo sensors and bleeds into nearby sensors, causing more fuzziness. If you take the photo through eclipse viewing glasses, you will reduce the light by about 99%. But, it will still be greatly out of focus. Even during totality, when you can look directly at the sun, your phone cannot focus on an object that far away. So, instead of a pretty ring, you get a fuzzy circle. The fix is to use manual focus. Not all phones have manual focus. Even then, you need to be able to focus on something very far away. An alternative that I saw was to use two cameras. Each had eclipse filters to block 99% of the light. Each got an out-of-focus photo from a slightly different angle (they were about 4 feet apart). Then, the image was fed into a computer that used the two photos to reconstruct an in-focus photo. It looked acceptable, but nothing near the quality of a photo from a filtered telescope. 209.149.113.5 (talk) 11:52, 22 August 2017 (UTC)[reply]
I think the scale seems like an important factor here. I mean, in the photo above the reflection is very clearly a small crescent. That implies to me that the real image is of the same size, buried in the middle of that immense white blob. I'm tempted to think that if the reflection is in focus, the original image is also in focus, but I don't know that - depends on how that camera produced it, especially since it might even have auto-focused the reflection because it provides a better contrast... Wnt (talk) 13:28, 22 August 2017 (UTC)[reply]
August 22
Male autism-spectrum correlations with homosexuality, bisexuality, fetishes and paraphilias?
the article you link should have a link to references, which should include known articles about the same subject. As it seems to me highly politically UNcorrect to try to link ASD to any LGBT behavior, I suspect such study would be rare and statistically unsignificant
I think it is a useful topic for study, because some authors think of autism as an "extreme male" phenotype, [18] and so looking at GLBT variations seems like a no-brainer. I've never given that idea much credence (the article I cite doesn't seem to either) but hey, it's biology, and nothing is impossible in biology. That said, the only obvious relevant hit on PubMed for autism homosexuality is this 50-boy study that finds nothing much of note. I actually found three references with autism transsexual, but they amount to anecdotes, something from Med Hypotheses, and this brief report, which takes the idea somewhat seriously and is the closest I have to an answer. But I'll leave it to someone else to pull this off Sci-Hub and decide if there is any meat there. Wnt (talk) 18:52, 22 August 2017 (UTC)[reply]
Sunspots
Using premodern equipment, how in the world does one observe sunspots? Even when I was looking through clouds and through a telescope made specifically for eclipse-viewing, the Sun appeared during today's solar eclipse to be a perfect, unblemished shape, aside from the rather big missing chunk of course. I can't imagine how the ancient Chinese, c. 800 BC, or Greeks c. 300 BC (see Solar observation#Early observations, could see sunspots. Nyttend (talk) 02:10, 22 August 2017 (UTC)[reply]
It is fairly easy with premodern equipment to use a pinhole camera to get a projected image of the Sun, on which sunspots are quite readily visible. Double sharp (talk) 02:33, 22 August 2017 (UTC)[reply]
Poking a piece of parchment would be the most basic approach, I suppose. (Lot's of school children observe an eclipse with two pieces of paper, I don't imagine parchment would be any different.)
But much more advanced pinhole cameras have been in use for a long time. The article Pinhole camera is a bit thin, but it does cover the history. The article Camera obscura also covers the history a bit.
Any sunspots during the eclipse weren't big as sunspot go and were probably covered by the Moon during the time when there was a rather big missing chunk. The butterfly diagram shows that we're in one of the lowest latitude parts of the sunspot cycle and the Sun's poles are only 7 degrees from Earth's orbit's poles so that couldn't really help bring them to the north or south side of the disc either. Sagittarian Milky Way (talk) 04:23, 22 August 2017 (UTC)[reply]
It would be nice if someone could update the articles with information about where the observations took place. I mean, the Sun has the great advantage of being really bright, so you can project it a long way, but at largest size it is still best viewed in a dark setting. This implies that it would really help to have a monumental roofed building, some manner of cathedral where a pinhole aperture made accidentally or deliberately at one window can project an image into a vast darkened space. (I suppose the right cave can do it but you have to postulate a lot of luck) I imagine the Chinese and Greeks had many such buildings but don't know if those observers were known to use them. Wnt (talk) 05:14, 22 August 2017 (UTC)[reply]
The name and the fact that Galileo and a contemporary worked on it would imply reliance on lenses, which makes that a much higher level of technology than that available to the earlier workers. Wnt (talk) 13:31, 22 August 2017 (UTC)[reply]
Note that natural pinholes are quite common. For example, a tree's leaves on a calm day can form many pinholes between the leaves. I often find accidental pinhole lens images. StuRat (talk) 14:39, 22 August 2017 (UTC)[reply]
On rare occasions a sunspot (or sunspot group) can be big enought to be seen directly with the eye when the Sun is dimmed by mist or thin cloud. Over the last 50 years I've read in various astronomy textbooks that ancient Chinese astronomers recorded a number of such observations, although our article on Chinese astronomy doesn't mention them. To give a couple of references:
George O. Abell's Exploration of the Universe; Third edition; Holt, Rinehart & Winston 1975 (in Section 2.1 Earliest Astronomers, on page 11) states, "The Chinese also kept rather accurate records of comets, meteors and fallen meteorites from 700 B.C. Records were made of sunspots visible to the naked eye . . . ."
Patrick Moore's The Guinness Book of Astronomy; Fifth Edition; Guinness Publishing 1995 (in The Solar System ∗ The Sun, on page 6) states, "Naked-eye spots had been previously [to J. Fabricius in 1610–11] recorded, but had not been explained; one given in a Chinese record of 28BC is described as 'a black vapour as large as a coin' and there is a Chinese record of an 'obscuration' in the Sun, which may well have been a spot, as early as 800BC."
The gluconeogenesis article says "The process is highly endergonic until it is coupled to the hydrolysis of ATP or GTP, effectively making the process exergonic." What does this mean in layman terms? At what point in a zero carb diet does gluconeogenesis become "coupled to the hydrolysis of ATP or GTP" — Preceding unsigned comment added by 187.1.51.122 (talk) 13:59, 22 August 2017 (UTC)[reply]
Endergonic means energy costly, exergonic means producing energy. An analogous statement to the first part of your question would be that producing gasoline costs a lot of energy (obtaining, transporting, and refining oil) until you burn it in the presence of oxygen. μηδείς (talk) 16:25, 22 August 2017 (UTC)[reply]
it means gluconeogenesis consume energy, that the energy for this is brought in the process by ATP (or GTP), but that not all the energy of ATP is transformed into glucose, the process waste some of it as heat (as is usual for process)
It also means that gluconeogenesis IS coupled to the hydrolysis of ATP or GTP from its very start, it wouldn't happen at all otherwise
It's a little more complicated than that. The thing to bear in mind is that gluconeogenesis is a complicated multi-step process. The change in Gibbs free energy at each step depends on the concentration of all reagents - Le Chatelier's principle is very much in play. Force one reaction with a large energy change, and the next step will have no choice but to go forward ... once a large enough concentration of metabolite builds up. Look at [19] for a sense of the "energy landscape" for glycolysis, which is a similar situation; then see this comparison for some numbers for gluconeogenesis. But the cell can't deal with massive concentrations of some metabolites. So there is a lot of specific engineering and "horse trading" involved, which has been optimized by unfathomable time scales of microbial evolution. But the bottom line is that just a few steps that split ATP are used to drive many others that are neutral or even unfavorable in terms of free energy. Wnt (talk) 19:11, 22 August 2017 (UTC)[reply]
The formula is amenable to two different interpretations,
Kw=[OH-][H+]
= 10-14 x (1000/17.01) x (1000/1.01)
= 5.82 e -10
or
Kw=[OH-][H3O+]
= 10-14 x (1000/17.01) x (1000/19.02)
= 3.1 e -11
neither correspond to the given values. This would seem to indicate that my textbook is wrong saying that the value of [H2O] being "ignored" is 55.5 = (1000 g/L / 18.015 g/mol).
Why are you getting 17.01? And water's molecular weight is 18.015 g/mol, not 18.5. Yes, I had a chemistry professor that disputed the textbook's usage of H3O+, so, H3O+ and OH- do not make H2O, only with H+ and OH-. 12.130.157.65 (talk) 15:12, 22 August 2017 (UTC).[reply]
It might help if you attach units to the numbers. For example, I recognize "17.01", "1.01", and "19.02" as the molar masses of the respective ions, but not what "1000" represents in each case (or why 1000 is in each ionic component but a separate powers-of-ten multiplier that is not part of the literal formula is also present). Are you conflating different substances' "liters" or "grams" or "moles" somewhere? DMacks (talk) 15:18, 22 August 2017 (UTC)[reply]
I don't know because the formula is not clear to me. [] "indicates molar concentration". The figure for water is calculated as shown (55.5) and is predicated on "almost all" of the substance being H2O - I use the same method for the other three species. The ratio of the other two components to the whole is 10e-7 - there are two such factors, hence 10E-14. My text book says "The value of ..Kw at 25 C is 10-14." which indicates that either the molarity is not part of the equation, or is being subsumed in the constant factor that differentiates Kw, by ignoring the almost-constant denominator.
do you understand molar concentration, to begin with? It would help you, i think. Please note that it is a dimensionless quantity, and so is any product or ratio of several of them, as Dissociation constant is. That's why we can use a "log" operation on it. And that's why it is SO important to NEVER, EVER, use raw number without attached unit (for instance, don't use 1.01, use 1.01g/mol, etc.), as DMacks too politely tried to tell you. If you don't you may add up liters with kg, and lose all meaning without seeing it. I suspect this is precisely what just happened to you. Gem fr (talk) 17:30, 22 August 2017 (UTC)[reply]
55.5 M is not dimensionless it is mol/L as is clear from the above.
We would expect the dissociation constant to have the same dimensionality, as it is the product of two concentrations divided by another, and indeed our article confirms this "Kd, which has the dimensions of concentration…" although it later seems to get confused between molality and molarity.
Incidentally it is the cologarithm we (may) take to get Ph, and in that case we have the concentration of H+ ions (or equivalent) over the concentration of H2O, which is trivially dimensionless.
It may be worth re-framing my questions:
When the text book says:
Kw=[OH-][H+]
instead of :
what is the actual change in value and dimension between K and Kw?
Given that we can treat the ions in water in two different ways, either OH- and H+ or OH- and H3O+, does this choice affect the value of K or Kw. If so, what does this represent? If not, why not?
To start out with, bear in mind that Kw is equal to the product of two concentrations; therefore it has units of (mol/L)^2. This is relevant if someone asks you to do this in the gas phase and your concentrations turn into partial pressures in atm; then you need to bear in mind that the ideal gas constant R = 0.082 L atm / mol K --> then you can multiply your mol/L by (RT)^2, where T is the temperature of the gas, and get something in atm^2 that you can use to figure out the relevant partial pressures ... or at least, could if the dissociation constant didn't depend on temperature and phase a lot - as you see in self-ionization of water the values for steam and supercritical water are ridiculously high since the charges aren't stabilized the same way as in polar aggregates. But you couldn't compare them at all without having units on your equilibrium constants to guide you!
Now, it should be clear that the Keq for [H+][OH-]/[H2O] will be reduced by a factor of 55.5 M (by which I mean 55.5 mol/l; honestly, I despise the italic M notation). And so the units of that will be in mol/l only. You get that number by noting that 1 g H2O ~= 1 cm^3 H2O and 1 mol H2O = 18.01 g H2O and 1 cm^3 (cubic centimeter) = 1 ml = 1/1000 L. You can multiply anything in chemistry by the ratio of two things that are equal, because that is always 1, so putting these together like dominoes you get (1 g H2O / 1 ml H2O) * ( 1 mol H2O / 18.01 g H2O) * (1000 ml / 1 L ). By cancelling out all the units on both top and bottom and multiplying or dividing numbers as given, that's a net result of 1 mol H2O * 1000 / 18.01 L = 55.5 mol/L.
As for your other numbers, I'm not sure where they come from. At room temperature Kw is around 10-14 (mol/L)^2, i.e. the pKw = 14 + 2 log L - 2 log mol. But nobody else on Earth but me would try to keep track of units once they take a log; they say oogledyboo and the units go away and they just try to remember not to take a log of anything but mol/l in the same context. Wnt (talk) 19:42, 22 August 2017 (UTC)[reply]
Thanks! I think you can say "the units go away" in a fairly legitimate manner, if everything is normalised, or you are otherwise Very Careful™. You answered Q1, and also made me feel that my questions weren't somehow stupid, which is the vibe I was starting to get from the other answers. So thanks again.
What STDs can still be caught even if people use condoms correctly and systematically for vaginal/anal, but not for oral sex? Assuming a non-monogamous person, with regular sexual contact (including kissing) with other promiscuous people.--Hofhof (talk) 16:27, 22 August 2017 (UTC)[reply]
You're question need clarification for me: are you talking about the effect of condom on sexually transmitted infection, or about STI that also transmit NON sexually, through kisses and oral contact?
It's important to note as well that people can still use condoms correctly and the condom can have a fault in it. Nothing is perfect. So the answer to "What STDs can people still caught even if they use condoms correctly and systematically" is every STD. Assuming "can" means "it could happen at all to at least one person in the entire history of humanity" The world is a big enough place that even low percentage events occur in large absolute numbers. This source from the CDC is probably a good place to start your research, and it leads to additional sources. --Jayron3219:37, 22 August 2017 (UTC)[reply]
You are technically right. I should have asked which STDs are common in the scenario above, without disregarding low risks. Hofhof (talk) 20:07, 22 August 2017 (UTC)[reply]
What causes red/green coloration in eclipse images?
Here [20] are a few images of the recent eclipse. The first is taken through a telescope with a cell phone camera. Note how the sun/sky border is limned with green tones, while the sun/moon border has red. I also saw the same pattern of colors using different imaging methods. I could see it with the naked eye looking at natural camera obscura e.g. through leaf shadow like the second image in the album. The third photo is of a projection by binoculars on to a white paper, red/green was also visible in live viewing. N.B. you can't quite see the colors in the leaf shadow photo but I could see it as it happened, and several nearby watchers said they could see the same color pattern.
Q: What causes this red/green coloration on the different boundaries? Perhaps some sort of diffraction? Since I see it with three fairly different methods, it seems unlikely to be a property of any given camera or lens.
I believe you are seeing various forms of chromatic aberration, which is fundamentally an optical artifact that is aggravated by the very high light/dark contrast ratio that is a characteristic of this type of photograph of a solar eclipse - whether it is an image of a projection, or a direct view of the sun.
To make matters worse, complicated engineering systems - carefully-designed complications of optical glass lenses, and carefully-designed software and hardware inside the camera - may attempt to correct for chromatic aberration. Those corrections only work up to a point - but for the extreme case of a direct solar image or a high-contrast projection of the sun - these corrections might actually aggravate the artifact.
Next, the algorithms - there are radial correctors, chromatic gradient detectors, desaturation filters, nonlinear edge estimators, ... heck, it's 2017... somebody's probably trained a recurrent convolutional neural network to detect-and-correct chromatic aberration.
The core cause of the visual artifact we call chromatic aberration is optical refraction occurring at the exact same part of the image as optical diffraction - places where a hard edge in the image corresponds to an extreme, high-contrast change in the illumination.
Here's an open-source algorithm - Fix-CA - that is commonly distributed with the GIMP. It's pretty simple, and it works in RGB-colorspace, so it should be "easy" for an ordinary programmer to follow.
Here's a much more famous and complex/sophisticated algorithm - Radial, part of PanoTools. Here's a blog documenting how it works, and here's another one. The algorithm is harder to read, because it's split across the tools library, but the meat of it is in correct.c.
Those tools are meant to allow a pro-user to fix up an artifact in a digital image after they've captured it. A modern-vintage camera probably implements a built-in, probably-proprietary, possibly-similar algorithm to fix up the image before you ever see it.
See if you can spot where and why such an algorithm could fail - and imagine what a proprietary camera algorithm might do differently - and imagine how that might fail too! The great bane of digital image processing algorithms is that in some corner-case conditions, instead of solving a visual artifact problem, the algorithms can make the problem worse!
Thanks @Nimur:. That's helpful but two things I'm still not clear on: 1) I can accept that my neighbor's binoculars and my friends' telescope may have the same pattern of aberration, but why would they consistently show red on one the sun/moon border while showing green on the sun/sky boundary? 2) I suppose the same aberration can occur in my own naked eye?
Put another way, if I accept this to be chromatic aberration in the telescope, the binoculars, and my naked eye, I'm still left with the question of what is different between the sun/moon boundary and the sun/sky boundary. I have yet to see any photo that shows both sides with the same color aberration, or with a reversed color scheme. If you search google images for /eclipse through telescope/ (e.g. [21]) or /eclipse projection/ (e.g. [22]), you will see many photos with the same pattern that I describe, and I feel like there should be a good reason why none of the putative aberrations have the same color on both sides, or a reversed pattern, etc. SemanticMantis (talk) 21:02, 22 August 2017 (UTC)[reply]
Wikipedia has a brief overview article on Leafhoppers, unfortunately it doesn't say anything about the song. Leafhopper is a pretty large family of insects, which is part of the cicada superfamily, and THAT article describes how they produce their song. --Jayron3219:14, 22 August 2017 (UTC)[reply]
The Multiplication of Food Grains
Let's say I have one of each seeds of typical wheat, rice, barley, maize, millet ... and many other common food grains. They are all properly planted in well-prepared fields under favorable climate. They are all successfully harvested.
Resolved
2 watts, 12 to 16 LEDs feed by LED switching mode driver from 12 V AC or DC >90% efficiency
Internal AC adapter driving a series circuit of LEDs >80% efficiency
Full insulated single device passive filament LED driver >90% efficiency
![{\displaystyle K={[OH^{-}][H^{+}] \over [H_{2}O^{+}]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b9a11a4e90d45ad4eec70b98e873b340db1b8b1)
![[19]](http://isite.lps.org/sputnam/AdvancedChem/Metabolism/ch15_E-landscape_glycolysis.jpg){kind=link}
![[21]](http://francetrip.wp.urbanschool.org/files/2014/10/solar-eclipse-through-telescope-Chloe.jpg){kind=link}
![[22]](http://www.mikesastrophotos.com/wp-content/uploads/2012/06/eclipse-projection.jpg){kind=link}