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April 2[edit]

Why does the green channel die first and blue last?[edit]

A bar I frequent has a number of flat-panel TVs (I'm not sure whether they're LCD or OLED) that can only display the blue channel. A few weeks ago I saw some that also had a working red channel, but no green. I would've assumed the cause was damaged or loose cables, but tonight I was told the owner -- who's described by the staff I've spoken to as a penny-pincher -- is replacing the screens themselves. Why is it that these dying screens never have a working red or green channel without a working blue, or green without the red? Neon Merlin 06:14, 2 April 2024 (UTC)[reply]

There are too many variables here to consider, including not even knowing if they are LCD (which tend to have a back-lighting and get color through layered color filters) and OLED (which can use white OLED and color filters, or colored OLEDs directly and no color filter). The issue could also be loose or bad cables. In my personal experience, "penny-pinchers" are more interested in what they perceive as the quick and cheap fix rather than actually finding the root cause of a problem. Paradoxically, this can result in taking things become more expensive than necessary as 1) they might have to try multiple solutions to fix a problem when it could have been a single solution if they'd taken the time to look for a cause 2) they find a quick solution that "works" (e.g. new TVs), and so don't bother to see if something actually cheaper might have worked better 3) there solution is just waiting for the next breakdown and will have to be repeated again and again (e.g. buying a cheap TV model with a fast dying color filter that needs replacing ever year instead of a TV costing 1.5 times as much, but lasts 5 years).

Additionally, we would need more information on specific TV models. Are all of the screens at the bar the same make and model, or at least general model (i.e. 55 inch and 65 inch versions of the same TV)? In that case, it could be as simple as that model TV had a defective color filter all on the same color channel, or a defective signal line for that color channel. Or, it could be something else entirely. --OuroborosCobra (talk) 17:45, 5 April 2024 (UTC)[reply]

Why are there no vertical VTOL jetfighters as far as I know?[edit]

Wouldn't it be better to just make a Harrier that can or must take off and land on its tail? It could still thrust vector and fly horizontally just not hover horizontally. How often is hovering horizontally more than just cool but not militarily needed? Over 1 thrust-to-weight ratio is really gas guzzling, and extra nozzles that are only used for extra redirection exhausts seems like unnecessarily weight. Sagittarian Milky Way (talk) 21:50, 2 April 2024 (UTC)[reply]

All VTOLs are vertical; you presumably mean "tail-sitter" aircraft. There have been several, mostly prototypes. The Ryan X-13 Vertijet might be of interest.--136.54.106.120 (talk) 23:59, 2 April 2024 (UTC)[reply]

More vertical than the Harriers that go straight up in horizontal orientation. Sagittarian Milky Way (talk) 01:35, 3 April 2024 (UTC)[reply]

So it seems they tried but too many problems. Sagittarian Milky Way (talk) 01:53, 3 April 2024 (UTC)[reply]

One technical issue it seems with early designs of tiltrotor aircraft was that with more upright designs, like with having the entire wing body tilt up (with the engine bodies fixed to the wing), a big problem is any wind on the runway or during takeoff could cause damage or even knock the thing over. If part of the appeal of VTOL and STOVL aircraft is being able to work from non-optimal infrastructure, then a launching platform for a tail-sitter jet would need both to manage air intake as well as shield the structure from the wind. I don't know whether that's significantly more involved compared to maintaining a small airfield, or a flight deck on an aircraft carrier, say. SamuelRiv (talk) 04:40, 3 April 2024 (UTC)[reply]

In order to be vertical takeoff in any orientation, you need a thrust ratio to weight of over 1:1. That's just the physics of it; if you want to go straight up against gravity, you need to have more thrust pushing up on you than the weight from gravity pulling down. Doesn't matter if you are a Harrier or a Vertijet. Because of this, VTOL capable aircraft that have actually entered service (Harrier, Yak-38, and F-35) have always chosen a configuration that allows for STOVL (short take-off and vertical landing) instead. If your wings are providing at least some of the vertical force, then you do not need to have a 1:1 thrust ratio at takeoff configuration. For the Harrier, that means thrust vectoring where you have them partially angled downwards (but not completely), giving you forward thrust to get airspeed over the wings and vertical thrust to push you up before your wings have a enough lift to do so on their own.

By doing this, you can carry more fuel and more ordinance into the air than you could in VTOL. You don't need 1:1 thrust ratio on takeoff, so you can carry more weight that makes you heavier than 1:1, while still letting you takeoff from short carrier decks without catapults or hastily/quickly made short field strips. As a bonus, when you've completed your mission, you have a lot less fuel and ordinance on your aircraft, so now you do have a greater than 1:1 thrust to weight ratio. That means you can land on that short carrier or improvised landing strip without needing arresting cables/hook or a brakechute. You can just land vertically with your reduced weight. Every "VTOL" jet fighter that has ever gone into service has actually done almost all operations as STOVL, and not VTOL, for this very reason.

A tail-sitting VTOL aircraft doesn't have this STOVL option. They always have to takeoff vertically (or can only do horizontal with added external equipment and a long takeoff distance that defeats the entire point of having built this VTOL thing for rough conditions in the first place). A tail-sitting aircraft ALWAYS needs a greater than 1:1 thrust to weight ratio (where Harrier and F-35 do not), and therefore cannot carry as much fuel and ordinance.

Lastly, visibility for a pilot landing a tail sitter is terrible. It's very difficult to get things lined up for proper landings, make sure you are completely vertical in force balance (including things like sidewinds) and that your ground speed is actually zero. Think driving in reverse, without a good rear-view mirror, while moving at highway speeds or above, and being subject to crosswinds that you get to ignore in your car, and managing to back into your garage perfectly. Oh yeah, and while you are stopped, instead of nothing happening when you turn the steering wheel, you wildly move away from the garage and probably hit a tree. This has probably become less of a problem today with more advanced computer controls, but for most of the existence of jet fighters, that wasn't an advanced enough tool. Combine that with the limitations of not having STOVL capability, and there's no real good reason for a tail-sitter jet fighter. --OuroborosCobra (talk) 16:30, 3 April 2024 (UTC)[reply]

Of course it has to be >1:1 which is why I was wondering why they didn't skip the belly plumbing probably saving weight. I didn't realize the large area sum of the 4 levitators makes the Harrier levitate more efficient (makes sense though, helicopters levitate with large air fan(s), and they keep trying to raise airliner fan bypass ratio to save fuel).

So there's so many problems, one wonders why they test flew some instead of foreseeing a waste of money. I didn't know the F-35 was designed to VTOL. I knew it had enough thrust but not that it was a designed (probably rarely used) option. Sagittarian Milky Way (talk) 21:44, 3 April 2024 (UTC)[reply]

The Lockheed Martin F-35 Lightning II is a short take-off and vertical landing or STOVL aircraft. The Queen Elizabeth-class aircraft carriers were specifically designed for their vertical landing ability. Alansplodge (talk) 11:21, 4 April 2024 (UTC)[reply]

Only the F-35B variant. The F-35A is for ordinary runways, the F-35C for CATOBAR operations. PiusImpavidus (talk) 12:41, 4 April 2024 (UTC)[reply]

There could be a lot of reasons why pure VTOL "tail-sitter" designs like the Vertijet were test flown. Firstly is simply "did anyone have the idea yet?" VTOL in a horizontal configuration with vectored thrust (or even with dedicated lift engines separate from flight engines) and then transitioning from vertical to normal flight just might not have been much of an idea yet. I think it's pretty telling that we had several designs for pure VTOL before the Harrier and its Kestrel prototype (Bachem Ba 349 Natter, Convair XFY Pogo, Lockheed XFV, Ryan X-13 Vertijet, SNECMA Coléoptère), and basically none after the Harrier/Kestrel (outside of some UAV designs). Once VTOL with STOVL capability was possible, no one bothered with the disadvantages of just pure VTOL.

Second is material. In order to do things like the vectored thrust on the Harrier's Pegasus engine, you need some pretty good metallurgy and mechanics since you now are going to have the hot portion of the exhaust coming out of something that has to move around. The no longer "fixed" nozzles need to handle that, as does any part of the aircraft or fuselage that comes into contact with the hot exhaust. I have a feeling that these technical issues might be why the Soviet Yak-38 contemporary to the Harrier opted for dedicated lift engines rather than vectored thrust (with the disadvantage that outside of takeoff and landing, the dedicated lift engines are dead weight).

Third is flexibility. I've already gone over the advantages of having STOVL ability in your VTOL capable aircraft. Now, for some applications, those advantages might not matter as much (except for the extremely difficult landing). A dedicated point-defense interceptor aircraft would be a great application for dedicated VTOL, and that's where we saw designs like the SNECMA Coléoptère and the Bachem Ba 349. The thing is... that type of design basically can only do that one thing. It can be a point-defense interceptor. It can go up really fast, have some dedicated air-to-air weapons, and (hopefully) return to the ground with a living pilot. It can't do much else. It can't be easily adapted for interdiction or fighter-bomber roles or anything else. If you look at the history of Cold War and post-Cold War aircraft, you will see a lot of designs (often on paper only) for highly dedicated single mission aircraft, often interceptors... and you won't see a lot of them enter service. There are outliers, like the F-106, but for the most part, the military isn't spending its money on dedicated unitaskers. The F-104, for all of its problems, ended up often in roles where it was used for ground attack and not point-defense. The F-4, developed as a pure air-superiority fighter, ended up being a fighter-bomber with a greater bombload than a World War II B-17 heavy bomber. The F-16 was designed to be so pure an air-superiority fighter that much of the driving force behind it was "not a pound for air to ground," and has become one of the best strike, ground attack, and suppression of air defense aircraft of all time. Even the F-14 could carry bombs, and even the F-102 saw use with its IRST looking for campfires in Vietnam and attacking with rockets.

Pure VTOL aircraft, especially tailsitters, just can't be flexible like that. They are probably going to be good for only one thing, and that's not something the military usually wants to spend money on. The only aircraft exception to that is, well, helicopters. But their entirely different design allows them that flexibility, even as dedicated VTOL, that fixed wing VTOL does not. --OuroborosCobra (talk) 16:44, 4 April 2024 (UTC)[reply]

April 3[edit]

Identification of tree[edit]

What is the tree at the right side of the image? Link

It is located in Ladera Alta, close to Las Palmas, Gran Canaria. It there a Wikipedia article about it?

--2001:4650:869A:0:E44B:AD5A:F990:FD8D (talk) 13:33, 3 April 2024 (UTC)[reply]

Hard to say at that distance but it looks to me like a Norfolk pine. (EDIT) Morphology doesn't mean much but a quick google (images) of Norfolk pine shows similar trees. 41.23.55.195 (talk) 14:05, 3 April 2024 (UTC)[reply]

28°07′29″N 15°28′54″W / 28.12461°N 15.481581°W, if you want to have a close-up in Street View. -- Verbarson ^talk_edits 14:32, 3 April 2024 (UTC)[reply]

Thank you! I think this tree, and other similar trees that I saw in Teror, are indeed Norfolk Island Pines. --2001:4650:869A:0:E44B:AD5A:F990:FD8D (talk) 16:26, 3 April 2024 (UTC)[reply]

April 5[edit]

What is the star nearest to TRAPPIST-1?[edit]

I have this question for a while now. So, what is the closest star to TRAPPIST-1, or in another terms, what is the star that has the shortest distance to TRAPPIST-1? I tried to search online for an answer to no avail. CactiStaccingCrane (talk) 17:49, 5 April 2024 (UTC)[reply]

To be clear, it's not TRAPPIST-1. CactiStaccingCrane (talk) 17:50, 5 April 2024 (UTC)[reply]

Judging by the chart in the infobox, something like Phi Aquarii (φ Aquarii) (see List of stars in Aquarius) may be close, but I'm no Martin Rees. MinorProphet (talk) 22:57, 5 April 2024 (UTC)[reply]

According to WolframAlpha, Trappist-1 and Phi Aquarii are 182.6 light years (56 parsecs) apart, and TRAPPIST-1 is 40.66 light years from Sol according to our article, so Phi Aquarii is not the answer.-Gadfium (talk) 00:02, 6 April 2024 (UTC)[reply]

I don't know how to find the answer directly, but Gliese 898 is 11.6 ly from TRAPPIST-1 [1], while HR 8866 and HR 8544 are about 29.3 light years from it. Gl 849 and Gliese 884 are 13.66 and 16.26 ly away. Everything else I've tried is further away, but that's no guarantee there isn't a closer star.-Gadfium (talk) 00:15, 6 April 2024 (UTC)[reply]

Using the data from TRAPPIST-1 and List of stars in Aquarius I get the same closest three, but different distances:

10.26 ly: Gliese 898
14.44 ly: Gliese 849
16.82 ly: Gliese 884
25.68 ly: Gliese 876
27.47 ly: 53 Aquarii

--Lambiam 11:13, 6 April 2024 (UTC)[reply]

As TRAPPIST-1 is only about 40 ly away, any star less than 10 ly from it can be up to 15 degrees away as seen from Earth, so such a star may not be in Aquarius. We should check Pisces and Pegasus too. And calculate proper error bars; I'm pretty sure that using those the results given by Gadfium and Lambiam will overlap. I won't do that today. The proper way to do it is to download all objects in the Gaia catalogue between right ascension 21h45m and 0h25m or thereabouts, declination between -22⁰ and +12⁰, parallax between 58 mas and 128 mas (there shouldn't be that many stars in that search volume) and write a little computer program to calculate the distances, incuding proper error bars. PiusImpavidus (talk) 18:58, 6 April 2024 (UTC)[reply]

Should we look forward to your forthcoming novel about interstellar trade routes in the Aquarian Sector :-)? {The poster formerly known as 87.81.230.195} 151.227.130.213 (talk) 20:52, 6 April 2024 (UTC)[reply]

lol, I'm just curious. I'm finding data about the TRAPPIST-1 system and wanting to make a {{Infobox planetary system}} for it, similar to the Solar System. CactiStaccingCrane (talk) 02:24, 7 April 2024 (UTC)[reply]

April 7[edit]

Can light move in a quantum vacuum at the maximal velocity C?[edit]

If it cannot, then why doesn't our article speed of light indicate that the concept of "speed of light in vacuum" is only a theoretical limit and does not exist as a real velocity? 147.235.221.47 (talk) 20:04, 7 April 2024 (UTC)[reply]

Propagation through a state, including a quantum vacuum state, is nonsensical. It's like asking "how fast can you run through the number 2?" or "how fast can a car drive through the color yellow?" It does propagate in a vacuum at the velocity of c, obeying all other concerns. --OuroborosCobra (talk) 20:34, 7 April 2024 (UTC)[reply]

Despite the link, I didn't refer to a quantum vacuum state, but rather to a vacuum with the lowest possible energy. This vacuum contains no matter; Yet it may contain other aspects of quantum vacuum, e.g. photons, virtual particles, electromagnetic fields, dark energy, and the like.

So my question is: When measuring the speed of light that moves in such a vacuum, is the result expected to be the maximal velocity C? In other words, does the refractive index in such a vacuum equal exactly 1? More generally and more importantly: Is there any real medium/vacuum whose refractive index equals exactly 1? 147.235.221.47 (talk) 20:49, 7 April 2024 (UTC)[reply]

"Quantum vacuum" is short for "quantum vacuum state". It is a description in terms of quantum theory of the current understanding of a vacuum when viewed as a quantum system. The vacuum-viewed-as-a-quantum-system is not a different thing than the vacuum of "empty" space, but a description of the same thing in terms of a different theoretical framework. --Lambiam 08:21, 8 April 2024 (UTC)[reply]

Your remark is correct, and that's why I have already clarified, that despite the link I've supplied, I didn't refer to a quantum vacuum state, but rather to a vacuum with the lowest possible energy.

Anyway, I suspect I didn't receive a direct answer to my original question, so let me be more specific now:

Let V be a vacuum, containing no matter, yet it may contain other aspects of quantum vacuum, e.g. photons, virtual particles, electromagnetic fields, dark energy, and the like. So, when measuring the speed of light that moves in the vacuum V, is the result expected to be the maximal velocity C? In other words, does the refractive index of the vacuum V equal exactly 1?

Additionally, is the vacuum V real or theoretical only? 147.235.221.47 (talk) 10:25, 8 April 2024 (UTC)[reply]

I think the answer is you're right, light would be slowed down by a tiny amount by the virtual particles. Really tiny. I wonder if one could make it go slightly faster by sending it between two plates as in the Casimir effect. I'm pretty certain if one cold measure that it would make Nature :-) NadVolum (talk) 12:02, 8 April 2024 (UTC)[reply]

In other words, the refractive index can never be exactly 1 in the real world (rather than the theoretical one), right? In other words, light can never reach the maximal velocity C in the real world, as far as measured velocities are concerned, right? 147.235.221.47 (talk) 18:52, 8 April 2024 (UTC)[reply]

Vacuum as defined is invariant under Lorentz transformations. It will look exactly the same in any frame of reference. Therefore the speed of light will be the same in any reference frame as well. If this is the case, it must be equal to

c

. If the vacuum were not invariant, then some privileged frame of reference would exist, i.e. there would be aether. Ruslik_Zero 19:54, 8 April 2024 (UTC)[reply]

Everyone having basic knowledge in mechancis knows that any particle moving at the maximal velocity C moves at this velocity in any reference frame, so this was not what I was asking about. Actually my question is another one: It's about whether light moves at the maximal velocity C in a theoretical vacuum only, i.e. only in a vacuum which contains no energy and actually nothing, or also in a vacuum which indeed contains no matter but still contains other aspects of quantum vacuum, e.g. photons, virtual particles, electromagnetic fields, dark energy, and the like.

So, do you claim that light moves at the maximal velocity C, also in a vacuum which indeed contains no matter but still contains some low energy, including photons, virtual particles, electromagnetic fields, dark energy, and the like? 147.235.221.47 (talk) 21:30, 8 April 2024 (UTC)[reply]

It seems Wikipedia has an article about this! Scharnhorst effect. Not that it says much. NadVolum (talk) 22:39, 8 April 2024 (UTC)[reply]

Oh, quite interesting. Thank you. So, any answer to my question is still speculative, just as the Scharnhorst effect is still "hypothetical". 147.235.221.47 (talk) 23:44, 8 April 2024 (UTC)[reply]

It is well known that in a metallic waveguide the phase speed of waves exceeds c. No problem here. The phase speed of X-ray radiation in matter exceeds the speed of light as well. Again no problem here. Ruslik_Zero 20:59, 9 April 2024 (UTC)[reply]

But the Scharnhorst effect refers to the front speed, whereas the measured value of the front speed - of light moving in vacuum - is considered to be the maximal measured value of any front speed of light (because only the front of waves carries information, whereas no information can be transmiited faster than the maximal speed possible in nature). Anyway, you are right, no problem here, although the Scharnhorst effect refers to the front speed. The reason for there being no problem is found in the last sentence of the article Scharnhorst effect. 147.235.221.47 (talk) 06:33, 10 April 2024 (UTC)[reply]

The article does not say anything about the "front", only about the index of refraction. Ruslik_Zero 20:53, 10 April 2024 (UTC)[reply]

If the article had referred to the phase speed only, the last sentence of the article would have been needless. 147.235.221.47 (talk) 20:58, 10 April 2024 (UTC)[reply]

I only want to make clear that there is just one vacuum with all virtual particles etc and it is invariant under any reference frame transformation meaning that the speed of light is invariant as well. You basically asking whether the luminiferous aether exists and whether the physical vacuum can be considered as such. The answer is no. Ruslik_Zero 20:56, 9 April 2024 (UTC)[reply]

"I only want to make clear that there is just one vacuum with all virtual particles etc and it is invariant under any reference frame".

"With all virtual particles etc"? Have you got any source for this? Maybe this "just one vacuum" must contain no energy at all? 147.235.221.47 (talk) 06:37, 10 April 2024 (UTC)[reply]

I admit that most of the details of the above long discussion are over my head, but to the extent that I understand the issues, surely "a vacuum that contains no energy at all" is not something that can exist in our Universe? {The poster formerly known as 87.81.230.195} 151.227.145.123 (talk) 20:32, 11 April 2024 (UTC)[reply]

Yes, it is, and this is exactly the origin of my question: Since (as you correctly say) "a vacuum that contains no energy at all" is not something that can exist in our Universe, so the only question remaining is whether the ~~maximal velocity~~ [upper limit of all velocities that are] possible in our Universe is the speed of light moving in a vacuum that contains no energy.

In my view, this is really an important question, because If the answer it positive, i.e. if the ~~maximal velocity~~ [upper limit of all velocities that are] possible in our Universe is the speed of light moving - in a vacuum that contains no energy i.e. a vacuum that cannot exist in our Universe - then also the ~~maximal velocity~~ [upper limit of all velocities that are] possible in our Universe is not something that can exist in our Universe (although all velocities that can exist in our Universe must be slower than this upper limit which does not exist in our Universe, just as all natural numbers must be smaller than "infinity" which actually does not exist among natural numbers). 147.235.221.47 (talk) 22:08, 11 April 2024 (UTC)[reply]

By the usual meaning of the word possible, the maximal velocity possible in our Universe can exist in our Universe. (In the sense used in "possible world", there is no upper limit.) In any case, c refers to the speed of light in the kind of vacuum that can exist, that is, a vacuum with vacuum energy. --Lambiam 13:34, 12 April 2024 (UTC)[reply]

As for what "possible" means, you're right, so due to your first remark, I've just struck out the words "maximal velocity" in my previous contribution, and replaced them by the words "upper limit of all velocities that are" [which I've put in brackets].

As for your last remark, do you think you've got any sources for it? Actually, my original question has been about whether the well known constant c refers to a vacuum with vacuum energy - in which case c can exist as an actual velocity in our universe, or c refers to the absolute vacuum containing no energy at all - in which case c is only an upper limit of all velocities that are possible in our Universe - but c itself cannot exist as an actual velocity in the universe. 147.235.221.47 (talk) 10:34, 14 April 2024 (UTC)[reply]

April 9[edit]

Mixing Ozone and Carbon Monoxide[edit]

I am having difficulty finding out what happens if you combine O3 and CO. As far as I can tell, O3 wants to be O2, and CO has no qualms about combining with other molecules. If I injected ozone into a airstream containing CO would the ozone seek it out to become O2 and CO2? ^Thanks,L3X1 ◊distænt write◊ 02:53, 9 April 2024 (UTC)[reply]

It would not be seeking it out, but rather bumping into each other at random. The activation energy would have to be exceeded for a reaction to occur. This means that the gases would have to be warm enough. But a catalyst could make this happen faster.^[1] Graeme Bartlett (talk) 03:31, 9 April 2024 (UTC)[reply]

How to find a constellation...[edit]

...that a given right ascension and declination is in? Jo-Jo Eumerus (talk) 13:56, 9 April 2024 (UTC)[reply]

You could look here. For higher resolution I'd look in my copy of A Field Guide to the Stars and Planets, but I have the first edition and don't know how the later ones show this information. --142.112.220.50 (talk) 22:31, 9 April 2024 (UTC)[reply]

If the celestial coordinates are too close for comfort to a boundary in this image on the Commons, you can check the constellation boundaries defined by the IAU here. It is more work for some constellations (Draco!) then some others (Crux), but doable. --Lambiam 09:16, 10 April 2024 (UTC)[reply]

Note the following complication: The constellation boundaries were originally defined (for equinox B1875) such that the edges were ligns of constant right ascension or declination, so that the question could have been answered naively by a (in some cases lengthy) sequence of "less than" and "larger than" conditions. For J2000, the equinox that we're using now, this is no longer the case and a somewhat more sophisticated algorithm is needed. --Wrongfilter (talk) 10:00, 10 April 2024 (UTC)[reply]

How does the equinox impact the suitability of the (somewhat arbitrarily drawn) constellation boundaries? --Lambiam 14:53, 11 April 2024 (UTC)[reply]

What do you mean by "suitability"? The boundaries are still the same with respect to the stars, but the coordinates have changed due to precession. For Crux both eastern vertices (and the entire eastern edge) were at R.A. 12^h50^m, the western edges at 11^h50^m in B1875 coordinates (I think I have "east" and "west" correctly). In J2000 coordinates they are the R.A. values from the file you linked to; the values are of course different, and the edges are now also tilted with respect to a constant R.A. line. --Wrongfilter (talk) 19:33, 11 April 2024 (UTC)[reply]

As long as the coordinates of a spot on the celeastial sphere to be assigned to one of the 88 constellations are given with respect to the same epoch as those of the boundary list, everything should be fine. It seems simpler to me to adjust the coordinates of the spot than those of the list. --Lambiam 01:01, 12 April 2024 (UTC)[reply]

If you know and use Python, astropy has a function for that. --Wrongfilter (talk) 22:36, 9 April 2024 (UTC)[reply]

The celestial coordinates of constellations change owing to the precession of the equinoxes - the reference year shows the date of calibration. 2A00:23D0:443:BB01:EC7D:7F9E:B865:32A8 (talk) 15:08, 10 April 2024 (UTC)[reply]

The function uses the algorithm described in this paper by Nancy G. Roman. If you want to do it by hand, the tricky bit is to precess the coordinates you have (presumably in J2000) to 1875. This online tool can do that. --Wrongfilter (talk) 19:41, 11 April 2024 (UTC)[reply]

The algorithm in that paper is most suitable for being executed by a human computer. For an electronic computer, a version of the winding number algorithm may be simpler. --Lambiam 01:13, 12 April 2024 (UTC)[reply]

The astropy function that I referenced above implements the Roman algorithm for an electronic computer, the actual algorithm only needs a handful of lines, it uses boundary data that are a rearranged, but not precessed, version of the original definition. And it works. --Wrongfilter (talk) 08:47, 12 April 2024 (UTC)[reply]

April 10[edit]

Economic independence of future space colonies[edit]

Assuming that at some point during future space colonization Earth will eventually become uninhabitable for some reason, have there been estimates in what time colonies on Moon, Mars, etc. may become fully independent of Earth supplies and deliveries? Brandmeister^talk 21:37, 10 April 2024 (UTC)[reply]

The best research on this can be found in the Mars trilogy by Kim Stanley Robinson. Abductive (reasoning) 22:55, 10 April 2024 (UTC)[reply]
Let's be careful calling works of fiction "research." Certainly, authors can conduct research while writing their fiction, but there is zero guarantee that they have done so, that they have done rigorous research if it was attempted, nor are they beholden to the results of said research. --OuroborosCobra (talk) 13:51, 11 April 2024 (UTC)[reply]
Do you seriously think I wasn't sure that Kim Stanely Robinson did a ton of research for his books? Robinson is known for that. Abductive (reasoning) 17:14, 11 April 2024 (UTC)[reply]
*sigh* Please be careful to not make this personal. I'm commenting on fiction authors, not you or even Robinson specifically. However, that is why I included the clauses "that they have done rigorous research if it was attempted" and "nor are they beholden to the results of said research." In the case of the former, rigorous research usually requires an academic or other functional background relevant to the field. While Robinson has a PhD in English, well deserving of respect, that's not necessarily a background needed for rigorous research (including interpreting/analyzing what they have found) in fields like economics, space travel, aerospace engineering, physics, atmospheric science, terraforming, etc. Nor, as in the latter clause, is an author of fiction, are they beholden to the results of said research should they decide that the narrative needs of their story require some deviation, otherwise known as artistic license. --OuroborosCobra (talk) 19:03, 11 April 2024 (UTC)[reply]

Robinson assumed vast amounts of subsurface ice, for which (afaik) there is not good evidence. Also he was caught in an engineering blunder (the windmills, whose secret purpose he retconned in the next volume). —Tamfang (talk) 19:39, 11 April 2024 (UTC)[reply]
...and still it was the best research on the subject. Abductive (reasoning) 08:13, 12 April 2024 (UTC)[reply]
What are you basing that belief upon? —OuroborosCobra (talk) 12:00, 12 April 2024 (UTC)[reply]

I doubt that it has been rigourously established that full independence is reachable with currently available technology, not requiring completely untested ideas or hoped-for future inventions. --Lambiam 14:48, 11 April 2024 (UTC)[reply]

Note that for the scenario to make economic sense, Earth would have to become more uninhabitable than the Moon or Mars, which both are indeed completely uninhabitable with technology prospects in the medium term. The only way this makes sense to me offhand is if Earth becomes completely barren of all life and all water in any form and all convertible energy sources. SamuelRiv (talk) 18:01, 11 April 2024 (UTC)[reply]

April 11[edit]

Gender benders[edit]

Re this comment in an earlier discussion:

...As I understand it, Rishi agreed to ban, for example, mechanical restraints that prevent natural development and cripple young women, and hormone treatments that oppose it. That is because girls of this age are immature and in many cases young women who were crippled in this way later said that their lives had been ruined by what was done to them. So is Stonewall saying that Rishi agreed to the opposite of what he agreed to? 92.10.96.64 (talk) 11:07, 1 November 2023 (UTC)[reply]

Last night's Evening Standard carried this harrowing report:

The most high-profile case of "detransitioning" is that of Keira Bell, who began taking puberty blockers aged 16 before transitioning to a male. She later decided to reverse treatment, claiming she never should have been given the medication and that the Tavistock clinic should have challenged her more.

I'm no medical expert, so can someone explain what physical (as opposed to intellectual) aspects of adulthood have not been reached by age sixteen? The process begins at age seven. 92.28.114.98 (talk) 10:56, 11 April 2024 (UTC)[reply]

We have an article on Bell v Tavistock. According to this article, the complainant had a double mastectomy at the age of 20, which to me seems a mature age in terms of both physical and mental development, so this would hardly be a relevant case concerning gender-conforming care for children.

An intended effect of the most commonly used puberty blockers is diminished sex-specific physical characteristics, which should no longer be an issue at age sixteen. A potential risk for patients having a change of heart at a later age is compromised fertility; I have not investigated for which age class this may be an issue. --Lambiam 14:40, 11 April 2024 (UTC)[reply]

Luminosity of air plasma[edit]

Is there some empirical formula or theory that describes the expected (optical) luminosity of a plasma? Say depending on ionization rate and/or ion density. JoJo Eumerus mobile (main talk) 13:43, 11 April 2024 (UTC)[reply]

Are you asking about the black body radiation, atomic emission (or ionic, as it may be), or the combination of both? --OuroborosCobra (talk) 13:48, 11 April 2024 (UTC)[reply]

Both, really - I wonder less about the mechanisms and more about the quantities. JoJo Eumerus mobile (main talk) 15:41, 11 April 2024 (UTC)[reply]

You can't have the quantities without an understanding of the mechanisms. Our article on black body radiation does a good job explaining the necessary mathematics for that aspect. As for atomic emission, that is far less trivial (to the extent that black body radiation can be considered "trivial" to calculate to a given quantitative accuracy). As for atomic emission, that gets far more difficult. For a hydrogen-like atom, i.e. an atom with only one electron, this isn't terribly difficult to solve using the Schrödinger equation. However, once you start having more than one electron, you now have a many-body problem. We do not have exact solutions to the Schrödinger equation for many-body systems. These require approximation methods, such as the Hartree–Fock method, density functional theory, coupled cluster method, etc. That's not even getting into the issue of things like spin–orbit interaction, or relativistic effects that come into play once you start getting to large numbers of electrons around a single nucleus. If we go to molecules, then we may need to consider other effects and make other approximations, such as the Born–Oppenheimer approximation. So... it kind of matters what you are looking at and what you want out of it. --OuroborosCobra (talk) 19:23, 11 April 2024 (UTC)[reply]

The degree to which these issues are relevant depend on the composition of the plasma, density, temperature and the degree of ionisation. Even for astrophysical plasmas (that I am somewhat familiar with) the conditions vary quite a bit. As the intro to bremsstrahlung explains, the emission mechanisms can be classified into free-free, free-bound and bound-bound processes. A very hot plasma, such as the intracluster medium in clusters of galaxies, the free-free process of thermal bremsstrahlung dominates: an electron is scattered, i.e. accelerated, in the electric field of an ion and emits radiation. The resulting spectrum is a continuous spectrum, but not a black-body spectrum (electrons and ions are in thermodynamic equilibrium but the radiation is not). In cooler plasmas, free-bound radiation becomes more important; this is emitted when an ion captures an electron (this is called recombination) and again results in a continuous spectrum. The recombined electron often lands in an excited energy level of the resulting atom (or ion) and can then cascade to lower levels or the ground level. This then leads to bound-bound radiation, which happens in distinct emission lines (for example the reddish glow of an HII region around a hot star is from the Hα line of neutral hydrogen, following recombination of a proton (H⁺) with an electron). There are computer programmes that can compute the full spectral energy density for a specified plasma; I could dig out some formulas for the luminosity of a cluster of galaxies but I suspect that is not what you're actually after. --Wrongfilter (talk) 21:59, 11 April 2024 (UTC)[reply]

@Wrongfilter and OuroborosCobra: I am mostly looking at air plasma and gas-discharge lamp, not astrophysical things. The plasmas I am interested in have a low ionization fraction and low temperature, and I want to know how bright they would be to the naked eye (or not, if the luminosity is way too low) for a given energy input/ionization rate. I guess that recombination radiation is the most important component. Jo-Jo Eumerus (talk) 07:05, 12 April 2024 (UTC)[reply]

April 13[edit]

Temperature at which dogs stick out tongue[edit]

At what air temperature on average dogs start to stick out their tongues and at what they start to keep them inside mouth? 212.180.235.46 (talk) 18:21, 13 April 2024 (UTC)[reply]

According to this source, the air temperature surrounding the animal must not rise above 29.5°C (85°F). The same article mentions 31°C (87.8°F) as the skin temperature; at higher ambient temperatures panting becomes the only available means by which the dog can cool itself. Normal human skin temperature is a few degrees higher, so you may still be comfortable while your hot dog is not. --Lambiam 21:15, 13 April 2024 (UTC)[reply]

Long experience tells me it varies a lot by breed and even by individual dogs. I'd be fascinated if anyone has ever worked out an average. HiLo48 (talk) 23:33, 13 April 2024 (UTC)[reply]

I would think it would also depend on the dog's activity level, i.e. whether running around or just setting. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:40, 14 April 2024 (UTC)[reply]

For sure. HiLo48 (talk) 01:21, 14 April 2024 (UTC)[reply]

Hm, I've seen dogs panting during walkout well below 29.5°C which looks quite high (at around 20°C or so). 212.180.235.46 (talk) 11:09, 14 April 2024 (UTC)[reply]

I imagine there are local factors and a lot of diversity. The dog that scammed his way into my house is quite heat intolerant apparently having forgotten his young wild days in a hot eucalyptus forest, whereas the semi-feral street dogs seem unbothered by temperatures like 34C while bathed in IR from concrete etc. Sean.hoyland (talk) 11:30, 14 April 2024 (UTC)[reply]

April 16[edit]

How can some birds have lifelong high core temperatures?[edit]

Understandably they need to be hot to fly well but something must be different. Protein differences? Sagittarian Milky Way (talk) 18:58, 16 April 2024 (UTC)[reply]

Some possible answers here. Mikenorton (talk) 19:21, 16 April 2024 (UTC)[reply]

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