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August 19[edit]

Shadow length[edit]

Is it possible to calculate the length of the shadow that a 5 meter tall building would cast at midday in mid summer in Brisbane Australia? The latitude is 27.47 South. Thanks 49.197.49.105 (talk) 05:47, 19 August 2019 (UTC)[reply]

Yes. For any given position on earth, the position of the Sun can be calculated for specific date and time. The solar zenith angle (angle above the horizon) is the angle the top of the shadow makes with the ground (assuming flat ground, including ignoring the curvature of the earth itself), and then just a bit of trigonometry to solve the base of the triangle for the given height. There are online calculators that can do it all for you. Or be sure you cite your source of this information and show your calculations manually if this is a homework problem. DMacks (talk) 06:02, 19 August 2019 (UTC)[reply]

Thanks DMacks, It's not for homework, just for a friend who wants to make a fern garden on the south side of her house. Can you give me a link to an online calculator? 49.197.49.105 (talk) 06:57, 19 August 2019 (UTC)[reply]

A search on "calculate height of sun" gives a number of places that can do this.--Phil Holmes (talk) 07:19, 19 August 2019 (UTC)[reply]

But if you want mid-day in mid-summer, this is the easiest calculation possible - so easy that we can (and should) be doing it in our heads! Earth's axial tilt is about 23 degrees; the latitude is about 27 degrees south; in this case, mid-day during the southern-hemisphere's summer solstice, sun will peak at four degrees (... or, "27.47° - 23.44°"), or, just a bit to the north of of "directly overhead." (If you really need more decimal-digits, we can give you a few; but at some point, we have to start accounting for the apparent size of the sun, which is half a degree - ... we still teach these basic facts in science classes, right? They seem, to me, like prerequisite knowledge for any educated intelligent life-form that plans to inhabit the planet Earth... I mean, we build powerful computers to do the math that's too hard to do in our brains, but ... this one? Easier without the computer - plus, when summer midday rolls around, the shadow presents us with directly-observable evidence to confirm or refute our mathematical model's hypotheses!) Forgive my exuberance, but .. literally the entire body of mathematics that evolved over thousands of years among intelligent Earth primates stems from this specific calculation. Primitive monkey-people had brains that could do this computation, without resorting to the use of digital-logic or semiconductors!

The point, though: Brisbane is particularly close - maybe a couple hundred kilometers - from the Tropic of Capricorn. At midday in mid-summer, the sun appears to be "very nearly overhead," or "so close to apparently directly overhead that the length of any cast shadows will be quite small." I have a gut feeling - nay, more than an intuition, but actual scientific hypothesis informed by my many years of calculating celestial things - that to a mere mortal human who is not equipped with precise and accurate and calibrated scientific instruments, the sun will apparently be "directly overhead" at noon on mid-summer. Any cast shadow will appear "directly below" the object in question. Then, thanks to our friendly specifically-easy-to-approximate tangent equation, the length of the shadow is "zero." Well, I compute thirty-five centimeters of shadow, from a five-meter building - but speaking as a pedigreed physicist, this is "sufficiently close to zero that our other approximations start to break down," notably that the building is an imperfect 5-meter rectangle, sunlight at Earth's surface is not accurately-modeled as a point-source at infinite-distance; and so on - for the record, these computations are much easier on Moon, which has no atmosphere to speak of; but that is regrettably an unsuitable site for most fern gardens).

If you would like to see a user-friendly but extremely-scientifically-accurate simulation, I use a piece of free software called Celestia for puttering around; and if you want "not very user-friendly but much-more-accurate" software tools, I can help find something for any interested reader with more specific needs.

And finally, a recent debate between myself and a more avid gardener resulted in the following conclusion: astronomy-geeks are better at fast calculations about sun-angles than gardeners; but gardeners know practical things about botany that astronomy-geeks may overlook. For example, the sun-angle at midday might be less important than the sunlight in late afternoon, or the temperature or the humidity or any other number of things - at least, this might be important to the fern - so it's a good idea to reach out to local garden experts to see what expertise and domain-knowledge they have to offer!

Nimur (talk) 15:39, 19 August 2019 (UTC)[reply]

People should of course remember that solar noon often does not correspond exactly to noon local time in areas of modern time zones (potentially including daylight savings). --47.146.63.87 (talk) 22:18, 19 August 2019 (UTC)[reply]

Sure, but the OP's purpose seems to be to find the smallest shadow of the year, not the shadow at 12h00 AEST. —Tamfang (talk) 23:19, 19 August 2019 (UTC)[reply]

Point of trivia - the smallest shadow is going to be "no shadow," because (in the southern hemisphere) the ecliptic swings south of due-east- and due-west during summer time; but this will never occur at mid-day. That means that if there is no obstruction on the horizon, the southern wall can receive full sun - no shadow - in the early morning and late afternoon during summer time (rather, this varies throughout the year, with its most extreme apparent excursion at each solstice). At mid-day (solar noon), every single day of the year, this side of the building will be shaded; but it can see the direct sun during some portions of the year. This is perhaps non-obvious, but the sun's most northern excursion is not coincident with its greatest elevation. The reciprocal, of course, applies in the Northern hemisphere. This is a fun fact known to our most well-educated astronomers and gardeners alike.

Although I've very dramatically made the case for how easy it is to approximate the mid-day shadow, if we care about the shadow at other times of day, things get a lot more tricky. The time-variation of the shadow is a difficult calculation of trigonometry, requiring many coefficients and numerical constants (the solar ephemerides), and we might actually want to refer to an almanac or a digital computer software tool that effectively computes an almanac. Nimur (talk) 02:45, 20 August 2019 (UTC)[reply]

Right, just a reminder, if they want to actually test this they'll want to look up the solar noon time for the day at that location. --47.146.63.87 (talk) 06:16, 20 August 2019 (UTC)[reply]

Note that the actual area always in shadow and partially in shadow is difficult to calculate, because it depends on the shape of the roof line, the edges of the building, placement/size/shape of trees, etc. And the Sun doesn't go straight across the sky, either, but follows an arc. So, you might do better to rely on direct observation or the presence of indicators, like moss, which only grow in shadows. SinisterLefty (talk) 11:21, 19 August 2019 (UTC)[reply]

(OP) Thanks Nimur and Sinisterlefty, for those great answers!! I appreciate the points you make about the possible origin of mathematics, the different viewpoints of gardeners versus astronomy geeks, and the position of the ecliptic at sunrise and sunset. 49.197.49.105 (talk) 07:17, 20 August 2019 (UTC)[reply]

If the purpose of the fern house is to provide shady growing conditions (and most ferns prefer some shade, though it isn't an absolute rule - it depends on the species and latitude), the simplest solution here would be to install shading as part of the construction, either through blinds, slats or netting, or liquid products applied to the glass. For any shade loving plant, it's important that it receives protection from the sun at its fiercest - what happens at sunrise and sunset doesn't really matter, as the radiation received is too weak. At the latitude of Brisbane, the sun in midsummer would be so close to overhead at midday that a 5m-high building wouldn't cast much shade, and bear in mind that the foliage of the ferns inside the house wouldn't be at ground level - they might be grown on benches, and even if planted in the ground, their foliage would grow upwards away from it, making them more exposed. PaleCloudedWhite (talk) 07:07, 20 August 2019 (UTC)[reply]

Oops, sorry, I've just re-read the opening post, and the intention was to create a fern garden, not a fern house. Well, the same principle applies - the owner will need to supply extra shading, either by constructing something overhead, or growing taller plants to provide shade. PaleCloudedWhite (talk) 07:13, 20 August 2019 (UTC)[reply]

Top 18 Shade Structure Designs. Alansplodge (talk) 20:53, 24 August 2019 (UTC)[reply]

Radon chemistry[edit]

Has there been much recent work on this? I have found some good reviews (10.1021/bk-1987-0331.ch018 and 10.1524/ract.1983.32.13.163), but both are from the 1980s. Double sharp (talk) 16:28, 19 August 2019 (UTC)[reply]

Radon#Chemical properties has numerous sources and is fairly extensive. Perhaps that will give you some good leads. --Jayron 32 17:57, 19 August 2019 (UTC)[reply]

I probably ought to have clarified that I wrote a significant amount of that section, so what I really meant is "has there been work more recent than that I found and cited in the article, because the latest review I found and cited is from 1998". I found some good information in Felice Grandinetti's Noble Gas Chemistry: Structure, Bonding, and Gas-Phase Chemistry (2018), but it seems to be theoretical (the section on Rn chemistry is part of a chapter with the title "Chemistry in Silico"). Double sharp (talk) 02:52, 20 August 2019 (UTC)[reply]

You'll have to tell me if these have anything new: Review of high-sensitivity Radon studies (2017), Radon: A Tracer for Geological, Geophysical and Geochemical Studies (2016), The use of radon as tracer in environmental sciences (2013). Someguy1221 (talk) 03:41, 20 August 2019 (UTC)[reply]

These seem mostly focused on other aspects, but at least do have some chemical information; thank you! Double sharp (talk) 04:01, 20 August 2019 (UTC)[reply]

Russian nuclear rocket explosion, part II[edit]

Russian government has been trying to cast a veil on the magnitude of the disaster in the North region. Two geographical landmarks over there are two cities: Severodvinsk and Archangelsk. At the same time they shut down four monitoring stations, first in Dubno and Kirov, then 3 days later Bilibino and Zalesovo. Google maps give us a perfect way to get geographical estimate of how far the radiation has spread.

Dubno is about only 50 miles or less north of Moscow, it lies to the West of the Severodvinsk meridian, that is the radiation kept spreading west. It is difficult to measure on google maps but it seems to be about 2 thousand miles down south from the explosion. Kirov, a large city is exactly on Severodvinsk meridian about a thousand miles down south.

It is clear why only those two stations were first shut down and then the other two were added. Bilibino is in the Chuckcha region, thousand of miles to the east from the explosion. When the radiation came in there, which took exactly 3 days, as it seems, they shut it down. Zalesovo is in Altay Mountains, to the east and south of the explosion, perhaps 3-4 thousand miles from Severodvinsk.

Thus it is a full blown Chernobyl, not a mini Chernobyl as one of the Russian source has said. People in Alaska may expect the deadly Russian guest soon. AboutFace 22 (talk) 20:02, 19 August 2019 (UTC)[reply]

Is there a question in this?

Is there a source for any of this? Andy Dingley (talk) 20:22, 19 August 2019 (UTC)[reply]

The Source is WSJ of today, Aug 19, 2019. [1] AboutFace 22 (talk) 20:31, 19 August 2019 (UTC)[reply]

You might be applying Western standards of transparency, which would be to share info unless it is so devastating as to panic the nation. But in totalitarian nations, they tend to hide all bad news, however minor. Thus, I suspect there is a detectable increase in radiation at those stations, and that is enough for them to hide it, even though it is not dangerous. SinisterLefty (talk) 21:51, 19 August 2019 (UTC)[reply]

There is just no way this can be a "full blown Chernobyl", as far as radioactivity is concerned. Chernobyl_disaster#Radioactive_release indicates an estimated 6 tons of fuel spread around into the atmosphere, with a lot of fuel remaining in situ exposed. The nuclear engine of the failed rocket would use some kg (maybe tens?), and Russians for sure treat the info on this as top secret: this is reason enough to shut down all info, since radiation level would help other nation guess how much fuel the rocket use. Gem fr (talk) 23:40, 19 August 2019 (UTC)[reply]

Worth noting, I think, that radiation alarms were going off in Sweden only two days after Chernobyl. With all of the initial Soviet denials, other countries may not have known exactly what happened, but they knew something had happened. Right now, the winds in the general region of the crash would be taking any radioactive clouds north, into the arctic. Whatever there is to detect, the Russians can't hide that. If there is extreme local contamination, that's another matter. Someguy1221 (talk) 23:59, 19 August 2019 (UTC)[reply]

For what it's worth, Norway might have detected radioactive iodine from the explosion, but if they did, it was an incredibly small amount, and apparently those small amounts are detected in Norway from Russia several times a year. There's no good way of knowing if it had anything to do with this specific explosion. Clearly, other countries are monitoring, and so far haven't found anything conclusively related. This was a very small radiation release, several orders of magnitude smaller than "full blown Chernobyl." --OuroborosCobra (talk) 23:50, 20 August 2019 (UTC)[reply]

Obviously, it means Russia is crashing nuclear test rockets several times a year. Someguy1221 (talk) 02:08, 21 August 2019 (UTC)[reply]

Not great; not terrible. --47.146.63.87 (talk) 20:32, 23 August 2019 (UTC)[reply]