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August 21

The Octaeteris and the Cycles of Venus

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]

Which ancients? There are thousands of cultural groups around the world. --Jayron 32 10:35, 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]

If you don't find a clue in Egyptian astronomy, Babylonian astronomy or Babylonian astrology, we can only try guess

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

Gem fr (talk) 16:37, 22 August 2017 (UTC)[reply]

Yes, I am aware, and, more than that, the following number, 21, also appears there, twice, once in conjunction with a 13. — 79.113.240.75 (talk) 11:34, 24 August 2017 (UTC)[reply]

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. --Jayron 32 19:25, 22 August 2017 (UTC)[reply]

Okay you're right, inferior planets do seem to go around their main epicycle once per real year in inertial space (this is simplified from small epicycles on main epicycles and maybe more layers right?) but from the observer their years are ~1/3rd and ~8/5ths years or the model wouldn't work. And I wonder if 8/13ths years would be more important to geocentrists than 8/5ths years when the center of the epicycle is glued to the Earth-Sun line like a rod. When it comes down to astrology (which was what astronomy mostly was at the time and paid the astronomologers' bills) who cares what sign Venus is in and what stars it conjuncts as seen from the center of it's epicycle, what sign is Venus in and what stars it conjuncts from Earth? And what's it's angular distance from the Sun as seen from Earth so we can see if it's conjunct or semisextile or semisquare and makes good or bad sex juju? And when's a good time to see Venus in the morning (early bird)/evening (everyone else) so I can contribute to planet location records so the theorists can add epicycles to improve the ephemerides? (which all repeat once every ~8/5ths or 8 years). Sagittarian Milky Way (talk) 21:49, 22 August 2017 (UTC)[reply]

Thanks, guys ! Problem solved ! Quoting from Ptolemy's Tetrabiblos: “Venus, taking in charge the third age, that of youth, for the next eight years, corresponding in number to her own period...” — 86.125.198.169 (talk) 23:27, 23 August 2017 (UTC)[reply]

Resolved

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. --Jayron 32 12:14, 21 August 2017 (UTC)[reply]

Oh, and it doesn't get overpowered because he hooks up a ballast resistor. You can clearly see him doing this. --Jayron 32 12:20, 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.

If you want some real electrical knowledge from YouTube, try Big Clive instead. Andy Dingley (talk) 16:01, 21 August 2017 (UTC)[reply]

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]

Also he uses the finest of safety equipment, such as the Explosion Containment Pie Dish Andy Dingley (talk) 16:42, 21 August 2017 (UTC)[reply]

Interesting, but my favorite is still David L. Jones at [EEVblog]. --Guy Macon (talk) 18:19, 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]

He's using LEDs from a bag labelled "Red". In the bulk shots you can see the internal frame for a single LED. Andy Dingley (talk) 10:28, 22 August 2017 (UTC)[reply]

Does Wikipedia have an article on the territorial behavior of various animals, including humans?

^ 140.254.70.33 (talk) 13:03, 21 August 2017 (UTC)[reply]

Territory (animal) and Territoriality (nonverbal communication). Rojomoke (talk) 13:51, 21 August 2017 (UTC)[reply]

Is it possible to have a total solar eclipse on a Full Moon instead of a New Moon?

Is it possible to have a total solar eclipse on a Full Moon instead of a New Moon? 96.66.16.169 (talk) 16:21, 21 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]

For 'around' read 'at'. —Tamfang (talk) 19:12, 25 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]

[ec] -In other words, in order to be in the moon's shadow, the sun must be directly behind it, which means the moon's dark side is facing you. 2606:A000:4C0C:E200:F9E3:2F35:FA19:8314 (talk) 16:33, 21 August 2017 (UTC)[reply]

You certainly can have one if you're on the Moon! ;) But to Earthly eyes that's a lunar eclipse. Wnt (talk) 19:39, 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]

Well, technically, a bulky object rather than a massive object. —Tamfang (talk) 19:12, 25 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]

"Here's a total solar eclipse on Earth as seen from the Moon". CNET. — 2606:A000:4C0C:E200:F9E3:2F35:FA19:8314 (talk) 19:31, 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_Zero 20:40, 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?

What's considered a shallow one? Are there any other terms like this? (moderate?) Presumably this would be in eclipse magnitude or obscuration. (width and area covered) Sagittarian Milky Way (talk) 23:42, 21 August 2017 (UTC)[reply]

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. --Jayron 32 19:20, 22 August 2017 (UTC)[reply]

Solar eclipse: why did the sun still look round in direct images?

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? Neon Merlin 23:52, 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]

The difference between 100 feet and 93 million miles is irrelevant, as focus is not at all linear, and both those would fall under the infinite focus position. For example, this camera considers anything beyond 7 feet to be infinite focus: [1]. There are cameras which can distinguish longer focal distances than that, but eventually they all go to "∞". StuRat (talk) 16:49, 23 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]