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September 6

Brown bear vs. human

I read that brown bear species (Ursus arctos) is divided into sixteen subspecies, due to morphological/genetical differences and habitat. My question is: from a strict biological perspective, would be ncecessary to divide humans (Homo sapiens) into subspecies, again due to morphological/genetical differences? Notice that I reject the social construct of "race", any kind of racial discrimination, the concept of one "race" being "superior" or "inferior", racial laws and historical definitions such as "Nordic race", Mediterranean race" and so on. I know that dividing humans into three or five or seven "subspecies" is not the same of brown bears, but I read biological differences exist and that's what I ask. Thanks in advance.--Carnby (talk) 07:51, 6 September 2020 (UTC)[reply]

Looking at Subspecies, I don't think it applies to humans. ←Baseball Bugs ^{What's up, Doc?} carrots→ 08:36, 6 September 2020 (UTC)[reply]

[Edit Conflict] The current consensus is that all living humans are members either of the species Homo sapiens which has no recognised subspecies, or of the single subspecies Homo sapiens sapiens, most often contrasted with the now extinct sister subspecies Homo sapiens idaltu.

Neanderthals have sometimes been classified as the subspecies H. sapiens neanderthalensis and sometimes as the species H. neanderthalensis: Denisovans have not yet been given a formal taxonomic status to my knowledge (which may well be outdated), as too little physical material representing them has been identified: one could argue for them being part of H.Neanderthalensis, a subspecies "H. neanderthalensis denisova", "H. sapiens denisova", or something else. The problem is that in biology there is no objective agreement on exact definitions of and distinctions between species and subspecies: different definitions (over 30 for "species") are used in different contexts. At least one prominent paleoanthropologist has expressed the opinion that there has never been more than one human species extant at any one time, given the evident willingness of various named 'varieties' of humans to interbreed when the opportunity arose.

There is also a model currently emerging within paleoanthropology of African Homo sapiens being an amalgam of several African regional populations somewhat physically and genetically differentiated through temporary geographical isolation prior to their subsequent re-merging, before their subsequent interbreeding outside of Africa with Neanderthals, Denisovans and possibly other groups as yet unknown.

Given the potentially fraught outcomes that might result from defining any living human population as a different subspecies from the rest of humanity, I suspect that all responsible scientists are very unlikely to discuss even the theoretical possibility of doing so. {The poster formerly known as 87.81.230.195} 2.122.2.158 (talk) 08:59, 6 September 2020 (UTC)[reply]

I echo the explanation by I.P above. Also you might find our article on Homo floresiensis interesting, who were described as the 'hobbit' people. Populations in the Nazca region and beyond (Easter Island) may have possibly been influenced by the Paracas culture; who are theorized as having naturally elongated skulls that predated and influenced Artificial cranial deformation. That's all a bit fringe though, and isn't well-covered on Wikipedia. Zindor (talk) 09:38, 6 September 2020 (UTC)[reply]

Correct me if I'm wrong but didn't I read somewhere that human races are more genetically similar to each other than different parts of an animal subspecies' range? Cause genetic mixing caused by humans exploring and falling in love or lust. Presumably in low tech eras the skin colors and nose shapes wouldn't mix quite as much as other genes as they were slowly selected by the sunlight and temperature of where they originated. So the races have mixed more than racists think. Also they like to show that Equatorial Guinea has the lowest IQ and don't say it was a small country where the crazy dictator killed anyone who might be smart and even anyone with glasses and they were the Dachau of Africa for awhile with massive population loss and brain drain from purges and people braving the minefields and boat ban to flee. Sagittarian Milky Way (talk) 13:53, 6 September 2020 (UTC)[reply]

Squirrels are able to produce subspecies. Potatoes are able to produce cultivars. Is H. Sapiens indeed the only magical life form totally unable to produce subspecies? I strongly suspect this is a very dangerous question to utter. People have been cancelled for far lesser crimes than asking this question. Tread carefully, as you are walking on politically correct eggshells here. 85.76.78.82 (talk) 16:57, 6 September 2020 (UTC)[reply]

Read what it is that causes subspecies, and it should become clear why humans don't qualify. ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:29, 6 September 2020 (UTC)[reply]

We're all awfully closely related for a species with our numbers. Probably mostly due to recent bottlenecks, only a few thousand generations, in evolutionary times. There's a concept in population biology which I can't remember the name of - being the size of the population you would guess from a given amount of genetic variation. Well, humans have the genetic variability you would expect in a species of 100,000 or so, not billions. If that's missing from our relevant articles, it is probably because they were written by humans. :-)John Z (talk) 06:20, 7 September 2020 (UTC)[reply]

I think the answer to my question is this (from subspecies): "The variation among individuals is noticeable and follows a pattern, but there are no clear dividing lines among separate groups: they fade imperceptibly into one another. Such clinal variation always indicates substantial gene flow among the apparently separate groups that make up the population(s). Populations that have a steady, substantial gene flow among them are likely to represent a monotypic species, even when a fair degree of genetic variation is obvious."--Carnby (talk) 11:25, 7 September 2020 (UTC)[reply]

If a convenient asteroid is modeled as x% gold and platinum atoms and the rest rubbish how high does x need to be to mine it soon?

All mixed together homogenously.Sagittarian Milky Way (talk) 13:18, 6 September 2020 (UTC)[reply]

Gold and platinum are native metals so i would imagine they'd be available fairly easily in their elemental form. I'm assuming the 'rubbish' would be carbonaceous chondrite such as found in a C-type asteroid. How to deal with that chemically i don't know, it might be easier to physically mine it. Zindor (talk) 13:34, 6 September 2020 (UTC)[reply]

So when I hear of people mining rocks with invisible gold content it just means the easy stuff is gone and they're scraping the bottom of barrel. So I might find gold nuggets walking around (jetpacking around?) a NEO and just pick them up? Sagittarian Milky Way (talk) 14:04, 6 September 2020 (UTC)[reply]

I doubt they'd exist alone as a sizeable nugget, as space debris would be constantly wearing away at them. As far as i'm aware, these companies mining precious metals out of low% sediment do turn over a profit. I'd suggest that panning in a river for nuggets might be a more viable option that jetpacking around an NEO. Zindor (talk) 14:24, 6 September 2020 (UTC)[reply]

Of course they make profit or they wouldn't do it, I'm wondering how many percent you'd need to scoop up the regolith or whatever and choose which atoms you want to take (maybe vibrating the grains in a centrifuge and taking the dense ones?) and still make profit. No one can be sure an asteroid is profitable without going there probably so it'd be a very expensive risk. Maybe one of our tech billionaires will try one day lol. And one wonders who gets to mine where if other billionaire(s) suddenly want to try the same asteroid if the first one makes high profit. Sagittarian Milky Way (talk) 15:37, 6 September 2020 (UTC)[reply]

Mineral/precious metal deposit speculation is a real business and sometimes money gets thrown at non-profitable mining projects, keeping it afloat, in the hope that they 'strike gold/the motherload' so-to-speak. The key is to invest early, let uninformed investors pile in and raise the share price, then sell your shares just before the project goes bust. It'll be a free-for-all once humanity gets into space; the UN Space Treaty that currently exists will just be ignored for sure. Zindor (talk) 20:19, 6 September 2020 (UTC)[reply]

Per Outer Space Treaty#Key points there is apparently some ambiguity about whether the treaty prohibits space mining, and many space-faring nations have decided it does not. Tigraan^{Click here to contact me} 09:13, 9 September 2020 (UTC)[reply]

Our page asteroid mining cites [1] which does more or less that calculation. It assumes 150ppm platinum (0.015%) and would be deemed too unprofitable for the amount of risk taken. It also assumes between 700 and 70,000 tonnes of platinum are hauled back to Earth; the lower figure is already four times the yearly Earthly production (see [2], table 4) so it seems probable that the market price of platinum would decrease significantly if such a project happened (whereas the article assumes a fixed price). Tigraan^{Click here to contact me} 09:38, 7 September 2020 (UTC)[reply]

All these figures and calculations seem to me to be deeply futile and objectless: The launch alone from the Earth per Atlas V of a one ton vehicle costs at least $110.000 per kilogram, that is already 1.7 times the price of gold and 3.7 times the price of platinum. To bring something back from Mars to Earth would be many many times as expensive per kilogram than that, and the asteroids are on average twice as distant as Mars. So all together to bring one ton of gold or platinum back to Earth will easily cost tens to hundreds times the gold or platinum price here on Earth. So regardless of the content in precious metal of an asteroid, there is no profit in it and there will be predictabily no reason or possibility to mine any asteroid at all (let alone soon). 2003:F5:6F0A:5A00:11D3:AC15:98F8:D5F1 (talk) 10:34, 10 September 2020 (UTC) Marco PB[reply]

It would be much easier (if slower) to mine a distant asteroid than close-by Mars, because said asteroid would requires less delta-v in the return trip, which is the key parameter for the cost of spaceflight (see rocket equation etc.). But yeah, it would still be awfully expensive. Tigraan^{Click here to contact me} 13:18, 10 September 2020 (UTC)[reply]

Some asteroids with similar orbits to Earth need less delta-v than a Moon rock return mission and not that much more than a space shuttle. Still buttloads of fuel of course, but not asteroid belt level buttloads. Sagittarian Milky Way (talk) 01:16, 11 September 2020 (UTC)[reply]

A genetic cause for sex height differences?

Is there a genetic cause for height differences between men and women? Futurist110 (talk) 21:33, 6 September 2020 (UTC)[reply]

See Sexual dimorphism in primates, Sexual dimorphism in non-human primates. Zindor (talk) 21:59, 6 September 2020 (UTC)[reply]

Puberty covers it. The pubertal "growth spurt" is both initiated and terminated by rising estradiol levels. After the onset of puberty, males have lower levels than females, so they generally start the "growth spurt" later, and it lasts longer, resulting in greater average adult height. --47.146.63.87 (talk) 04:47, 8 September 2020 (UTC)[reply]

September 7

Iguanas and snakes: Where was this filmed? What type of iguana and snake?

This YouTube BBC video [3] shows some type of iguanas and snakes, which are not identified. However, there is a longer clip from the original programme, which was broadcast on France 2, and has a complete voice-over narration, in French. I cannot understand spoken French, so I wonder if anyone here can do so, or otherwise identify the species of reptiles being filmed, and the location. I am not sure if it is allowed to post the second YouTube clip here, on WP. It can be found on YouTube by searching for "iguane vs serpents" and choosing the "ZAPPING SAUVAGE" clip, which runs for 5:13 minutes. Absolutely amazing! And if you like wildlife videos, well-worth the five minutes. (A secondary question...would a link to the second video be allowed here? I wished to err on the side of safety, and not post the link.) Thanks, Tribe of Tiger ^{Let's Purrfect!} 21:34, 7 September 2020 (UTC)[reply]

You can turn on English subtitles. However the clip does not say where or exactly what animals these are. Someone in the comments also asked exactly your questions without answer. Graeme Bartlett (talk) 00:29, 8 September 2020 (UTC)[reply]

The BBC site says this is Galapagos Islands.[4] Graeme Bartlett (talk) 01:24, 8 September 2020 (UTC)[reply]

The animals are Amblyrhynchus cristatus and Galapagos racer. Graeme Bartlett (talk) 01:25, 8 September 2020 (UTC)[reply]

@Graeme Bartlett: I did not know about the option for English subtitles, will try this. For some reason, I thought about the Galapagos, based on the volcanic terrain. Lucky guess! I will look at/follow your BBC link, so that in future, I can find such information for myself. Thanks so much for answering my question, and also for describing your method of obtaining it. You are a helpful editor and a good educator. Best wishes, Tribe of Tiger ^{Let's Purrfect!} 03:19, 8 September 2020 (UTC)[reply]

@Graeme Bartlett: I see that our WP article on the Galapagos racer has a reference to this excellent video. Because the terrain seemed so barren, I wondered how the iguanas found food...and WP provides the answer! Thanks again, Tribe of Tiger ^{Let's Purrfect!} 03:34, 8 September 2020 (UTC)[reply]

@Tribe of Tiger: if you mean this video [5] then yes I think linking to it is fine. I'm not entirely sure what Zapping Sauvage is, but the tick is a positive sign. The about page isn't as clear as I would like (at least based on a machine translation) [6] but Matthieu Briere seems to be associated with France TV. I think it's reasonable to assume the channel is in some way associated with France TV and therefore concerns about linking to known or suspected external copyright violation do not apply. Nil Einne (talk) 19:24, 8 September 2020 (UTC)[reply]

@Nil Einne: Yes, this is the video! I did not know about the "tick", as an attribute/attribution, etc,, per copyright, on YT. I did remember, from a year or two ago, some discussions regarding YouTube links, and wished to be careful, per your reference to external copyright violation. Thanks so much, very helpful. I will save your useful cmts for future references to articles. Sincere thanks, Tribe of Tiger ^{Let's Purrfect!} 22:31, 8 September 2020 (UTC)[reply]

September 8

Red squirrels

where are red squirrels native to in North America? — Preceding unsigned comment added by 47.156.108.52 (talk) 22:22, 8 September 2020 (UTC)[reply]

Our article on the American red squirrel has a map with its distribution. --OuroborosCobra (talk) 23:30, 8 September 2020 (UTC)[reply]

Although the OP probably knows this, casual readers should be aware that this is not the same species as the Eurasian red squirrel, which is not native to North America although there may well have been past attempts to introduce it. {The poster formerly known as 87.81.20.195} 2.122.2.158 (talk) 08:56, 9 September 2020 (UTC)[reply]

This reminds me of the problem of the airspeed velocity of an unladen swallow. The two species of red squirrel are even in different genera, although in the same tribe. --Lambiam 11:00, 11 September 2020 (UTC)[reply]

September 9

Normal reading distance

A figure often given for "normal reading distance" is 40 cm (16 in).^[7][8][9] But K.-K. Shieh and D.-S. Lee (2007) (doi:10.1016/j.apergo.2006.06.008) cite the book Human Factors in Engineering and Design (1992 edition) for the statement that "when reading a book or paper-like material, the normal reading distance is usually somewhere between 305 and 406 mm, with a mean of 355 mm." That is awfully precise for something that is quite variable, made vague by "usually". It makes a considerable difference whether the distance of 5 cm between the reported mean and the upper/lower bound is the estimated statistical dispersion, or the 2 sigma associated with the two-sided p = 0.05], or something else. If the latter, 40 cm is somewhat unusual. The uncanny 1 mm precision suggests that this is based on a large number of measurements, but given the general character of the book, covering a wide range of issues, it seems unlikely that the authors performed the measurements themselves. My question is: where did the data come from and how should the spread be interpreted? (Using Google Books I only see unhelpful snippets.) Also, are there other (authoritative) sources that do not just parrot what everyone else is saying? --Lambiam 17:38, 9 September 2020 (UTC)[reply]

Are we looking for a normal in the sense of "the average human" (for any given definition of "average") or are we looking for normal in the sense of "established standard for the purpose of having an agreed-upon value just so we can cut down on variables"? The first will be necessarily fuzzy and imprecise, but closer to actual human usage, the second will be more precise (and thus useful as a standard) but arbitrary and only coincidentally close to a real human average. Which sort of value are you looking for? Because both are useful in different situations. --Jayron 32 17:52, 9 September 2020 (UTC)[reply]

And me with 20/60 vision reads everything at 10 inches cause it degrades after 12 without squinting or optics (I got to 20/15 with optometrist thingy before she stopped, could've done 20/12 probably). Why wouldn't you want your book to be full Quatro HD anyway? 16 inches is wasted resolution. Sagittarian Milky Way (talk) 18:19, 9 September 2020 (UTC)[reply]

Who is the "we" who is looking for a normal in some sense? Personally I'm satisfied with my customary reading distance without comparing it to a supposed norm. But it would be nice to have an authoritative source for linking to in articles that mention the concept (Naked eye, Optics, Visual system). If the source is any good, they'll surely also define what they mean by "normal" in the context. I'm not happy with citing just any odd source; although 40 cm; is frequently mentioned, equally "reliable" sources give sometimes considerably different values. I also think the concept is sufficiently notable that it should actually deserve an article of its own – or perhaps a dedicated section of an article with a wider scope and less normative title, such as "Viewing distance". --Lambiam 19:42, 9 September 2020 (UTC)[reply]

There are smartphone and tablet distance standards which are further for tablet or maybe I'm misremembering something where a corporation really pushes "typical distance" to make their minutes of arc look better. There's also a desktop computer rule of thumb that I can't focus at and it's too close for myopia glasses without slow eye damage. 19 to 24 inches according to this probably trustworthy image I've downloaded from god-knows where. Sagittarian Milky Way (talk) 22:03, 9 September 2020 (UTC)[reply]

If you can touch the screen with your arm without moving your head and shoulders closer, it's too close.^{[citation needed]} 93.136.121.193 (talk) 01:58, 10 September 2020 (UTC)[reply]

^{If I can't touch it it's at least 3 times too far. Sagittarian Milky Way (talk) 03:06, 10 September 2020 (UTC)}[reply]

The 305, 406 numbers would have been derived from 1 foot and 1 foot 4 inches, which does not have so many significant figures. But if a scientist or serious medical practitioner has been using feet, they must have been doing this measurement long-long ago, and so is likely seriously out of date, from the pre-computer/mobile phone era. Graeme Bartlett (talk) 22:08, 9 September 2020 (UTC)[reply]

That is a perfect explanation of the uncanny (and apparently totally unwarranted) precision. Presumably they just copied what someone wrote, something like "normally between 12 and 16 inches", and went metric and overboard on that. And the arithmetic mean of these two possibly guesstimated measures, converted to mm while rounding, is 355.5 mm, which, with a regrettable loss of precision :), was rounded down to 355 mm. It appears that even if the book cites a source, we won't learn anything from it. --Lambiam 09:32, 10 September 2020 (UTC)[reply]

Reminds me obliquely of the time I read news of an artifact that was expected to fetch between [two dollar amounts] at auction, and accurately inferred the then current exchange rate of the pound. —Tamfang (talk) 01:27, 11 September 2020 (UTC)[reply]

Considering the near point of a healthy eye is at 25 cm, I'd say choosing to read at 30 cm is a little far-fetched unless the study wasn't limited to people with healthy eyes. 93.136.121.193 (talk) 01:56, 10 September 2020 (UTC)[reply]

I could focus 4 inches from my eyeball when I was 18. Now it's more like 5. God I'm getting old. Sagittarian Milky Way (talk) 03:04, 10 September 2020 (UTC)[reply]

Presbyopia --47.146.63.87 (talk) 09:22, 10 September 2020 (UTC)[reply]

Perhaps the "normal reading distance" is the reading distance normal people with normal eyesight normally prefer for normal lettering. My preferred reading distance for letters with a corpus size of one millimetre is not the same as for letters that are one metre tall, and neither is 40 cm. There seems to be a lack of thorough and comprehensive treatments of the topic in the literature, which I think could easily fill a monograph. This is a bit surprising given the huge practical importance. Something that is relevant for the topic is Technical Note 1180 of the National Bureau of Standards, Size of Letters Required for Visibility as a Function of Viewing Distance and Observer Visual Acuity by Gerald L. Howett (1983). But this does not deal with the issue of accommodation. A comprehensive treatment should deal with the combined effects of (not always perfect) accommodation and (always limited) visual acuity. --Lambiam 09:29, 10 September 2020 (UTC)[reply]

September 10

What is the over/under for the closest a planet has been to Earth in the Phanerozoic? What about any planet pair?

2. Is the probabilistically most likely value for "what was the biggest Phanerozoic perturbation?" enough to cause detectable climate change if it magically happened now?

Planet means the kind we have 8 of. Sagittarian Milky Way (talk) 03:41, 10 September 2020 (UTC)[reply]

It is generally thought that the Solar System as a whole is unstable (see the article on Stability of the Solar System) and that the system of equations is is chaotic in the technical sense of mathematical chaos theory. On a time scale of a few million years nothing dramatic is to be expected, and the effects are too small to have been observed since Homo sapiens gazed up to the skies. But even the most precise long-term models for the orbital motion of the Solar System are not valid over more than a few tens of millions of years, let lone over the 500+ million years the Phanerozoic has lasted. Numerical integration of the differential equations on supercomputers suggests the possibility of collisions of Mercury, Mars or Venus with the Earth some 3.34 billion years into the future, but not sooner than within one billion years. Since the equations are time-symmetric, the models can equally be used to postdict the past planetary motions. I don't know if anyone has actually done that. It seems that for a time scale below a billion years into the past the result is likewise not likely to be dramatically different from the current orbits. But the lack of major upheaval for a long future time span is no guarantee of the same for the past. Until someone runs that simulation backwards the best answer to the question in the heading is that we don't know. As to the second question, current climate models have no contingency for a direct collision of Mars with Earth, but I think it is a fair bet the climate change would be detectable.^[10] --Lambiam 08:05, 10 September 2020 (UTC)[reply]

The closest planet to earth, on average, is Mercury (no really: [11], [12]). The closest planet to earth at the closest point on either of their orbits is Venus. The gravitational effect of Venus at its closest point is easy enough to calculate; it is neither close enough nor large enough for General Relativity to matter, so use F = G (M1 * M2)/(d^2). You should also do the calculation for the gravitational effect of the Sun on the earth too; I'd imagine the effect of Venus is well out of the range of the significant figures of our measurements. You may also want to do Jupiter, given the difference in masses, Jupiter may have a larger overall gravitational effect on Earth than Venus, but again, these numbers are going to be several orders of magnitude into the uncertainty range for the effect of both the Sun and the Moon, and at that level, basically meaningless as they would get lost in the normal expected variations of those bodies (basically just part of the "noise"). --Jayron 32 10:50, 10 September 2020 (UTC)[reply]

I've heard the Mercury thing before, apparently the math makes it not tied with Venus. Sagittarian Milky Way (talk) 01:09, 11 September 2020 (UTC)[reply]

One can estimate an upper bound on any perturbations of Earth's orbit by considering the hypothesis that the largest perturbations in the Earth's climate with unknown cause, was caused by a perturbations of the Earth's orbit. Now, going a bit beyond the Phanerozoic period, we can consider the snowball Earth period, and for that you actually need quite small perturbation of earth's orbit, way smaller than what can happen due to the effects mentioned by Lambian above. Count Iblis (talk) 15:03, 10 September 2020 (UTC)[reply]

September 11

Time it would take for Earth and Mars to collide

A friend brought up the question of how long it would take for the Earth and Mars to collide. There were several diversions but the core of the question is, if Earth and Mars were the only objects in space, how long would it take for them to collide. We're assuming the gravitational constant remains what it currently is. And no other bodies are affecting them. The distance would be, let's say a light year. And they start out stationary. I thought it would be fairly straightforward. Something along the lines of just using the two masses, gravity, and distance. But I'm not sure how that math problem would be set up. And my Google skills are failing me for what seems like a simple equation. Any help?

I know that school is just starting here in the US but I can assure you this isn't for a school project. I'm a dude in my 40s with no homework due any time soon. Thanks, †dismas†|^(talk) 02:28, 11 September 2020 (UTC)[reply]

Per Newton (not Einstein) and his law of universal gravitation

F=G{\frac {m_{1}m_{2}}{r^{2}}},

gives the force F pulling two objects together. Solve for

G = 6.674×10⁻¹¹ m³⋅kg⁻¹⋅s⁻².^[1], m_earth = 5.97237×10²⁴ kg, m_mars =  6.4171×10²³ kg

Then Newton's 2nd Law of Motion a = F / m

gives the accelerations a_earth and a_mars of each planet towards the other. We can arbitrarily consider one of the planets as a stationary reference. Then the separation distance

s = 1 Light-year - radius_earth - radius_mars

 = 9.46 x 10¹² - 6371.0 - 3.39 km

will be closed by the other body accelerating at a = a_earth + a_mars

Remembering the equation of motion s = ut + ½ a t ² the actual calculation of

t=2sqrt(s/a)

is left to the reader. 84.209.119.241 (talk) 08:25, 11 September 2020 (UTC)[reply]

^ "2022 CODATA Value: Newtonian constant of gravitation". The NIST Reference on Constants, Units, and Uncertainty. NIST. May 2024. Retrieved 2024-05-18.

I'm afraid that's wrong. Crucially, the gravitational force and hence the acceleration does not stay constant as the two bodies approach each other. I'm not going to attempt to solve the equation of motion (Newton's 2nd law), but merely point to the article free-fall time, which gives the solution. --Wrongfilter (talk) 08:34, 11 September 2020 (UTC)[reply]

(ec) To start, I'd make the problem more general. We start out with a Newtonian universe containing two stationary bodies, being point masses of magnitudes

m_{1}

and

m_{2}

at a given distance

d_{0}

. Obeying Newton's laws of motion they start moving toward each other at time

t=0

and their distance

d_{t}

will decrease as a function of

t

. Solve

d_{t}=0

for

t

. The centre of mass of the two-body system is stationary at time

t=0

, and Newton's laws (specifically the conservation of momentum) imply it will remain so, so that the bodies will meet there. Likewise, the conservation of angular momentum implies they will move along the straight line connecting them and passing through the centre, which we will take to be the origin. (Even without appeal to conservation of angular momentum, this already follows from elementary considerations of symmetry.) We can therefore think of the two bodies as being located at the positions

d_{1,t}={\tfrac {m_{2}}{m_{1}+m_{2}}}d_{t}

and

d_{2,t}=-{\tfrac {m_{1}}{m_{1}+m_{2}}}d_{t}

. To simplify the notation, we define

s=s_{t}=d_{1,t}

and

\mu ={\tfrac {m_{2}}{m_{1}+m_{2}}}

, so that

s=\mu d_{t}

(and therefore

d=d_{t}=\mu ^{{-}1}s

), and solve

s=0

. The forces attracting the bodies to each other are opposite in sign but equal in magnitude, which is given by

F=Gm_{1}m_{2}d^{{-}2}=

G\mu ^{2}m_{1}m_{2}s^{{-}2}=

\gamma m_{1}s^{{-}2}

, where

\gamma =G\mu ^{2}m_{2}

.

The acceleration on the first body, taking the proper sign so that the acceleration is towards the origin, is given by

{\ddot {s}}=a=-F/m_{1}=\gamma s^{{-}2}

.

I'll leave it to the two of you to solve this simple ODE and plug in the values for the variables in the model, but don’t be shy to tell us if you need more help. BTW, the planets you want to collide, in their regular orbits, are never more than 4.25 × 10^-5 light-year or 23 light-minutes apart, so putting them a light-year apart is a far stretch~~; in the universe as we currently understand it, you'd not be able to ignore relativistic effects~~. --Lambiam 10:41, 11 September 2020 (UTC)[reply]

From the energy conservation law it follows that

{\frac {\mu }{2}}\left({\frac {dr}{dt}}\right)-G{\frac {m_{1}m_{2}}{r}}=-G{\frac {m_{1}m_{2}}{r_{0}}}

,

where

r_{0}

is the initial distance,

\mu =m_{1}m_{2}/(m_{1}+m_{2})

. Then the time

\tau

to collision will be

\tau ={\sqrt {\frac {\mu }{2Gm_{1}m_{2}}}}\int \limits _{r_{E}+r_{M}}^{r_{0}}{\frac {dr}{\sqrt {{\frac {1}{r}}-{\frac {1}{r_{0}}}}}}={\sqrt {\frac {\mu r_{0}^{3}}{2Gm_{1}m_{2}}}}\int \limits _{(r_{E}+r_{M})/r_{0}}^{1}{\frac {{\sqrt {t}}dt}{\sqrt {1-t}}}

.

You can calculate the integral yourself. Ruslik_Zero 20:41, 11 September 2020 (UTC)[reply]

Does milk slosh in a sealed tetrapak?

When I shake a new sealed tetrapak of milk, it always seemed like the milk doesn't slosh. Once I open it and remove the seal it does. Is something about the sealing process preventing the milk from sloshing (even though surely it's still got space to move around)? Or is it because the milk is sloshing, but the sound can't travel through the vacuum? (Which would be very cool, like having a tiny microcosm of space in your milk carton.) --2A01:4C8:64:898F:1:2:A027:D730 (talk) 16:46, 11 September 2020 (UTC)[reply]

Milk only sloshes at a phase boundary. If there is no air in the tetrapak, there is no phase boundary against which the milk can slosh. Once opened, air gets in, and now you have the ability to slosh. --Jayron 32 17:52, 11 September 2020 (UTC)[reply]

So a question that remains, for me, is whether the milk gets mixed when it is shaken, or does one need to open and seal it again before shaking in order to mix the milk?--Shantavira|^{feed me} 18:23, 11 September 2020 (UTC)[reply]

If there is no air inside – whether a tetrapack, bottle or can – shaking has no or very little effect. Turning the pack on different sides and slowly keeping turning may help to mix if separated components (e.g. milk and cream) have different densities; this requires no vigorous anything but just patience to let gravity do its thing. --Lambiam 23:10, 11 September 2020 (UTC)[reply]

Tetrapaks aren't used here, so I have no familiarity with opening them, but if there was no air inside, I can't see how there wouldn't be milk spilling out whenever you do. However, there might be just a small bubble of air that could move around without "sloshing". --174.88.168.23 (talk) 01:19, 12 September 2020 (UTC)[reply]

Assuming the container isn't rigid, it can slosh even without air by deforming the containiner itself. I just tried a "tall rectangular half-gallon" and shaking it up and down caused it to bulge alternately on the bottom vs top of the sides. That means the liquid is able to move inside even if there is no non-liquid space. DMacks (talk) 20:28, 12 September 2020 (UTC)[reply]

An additional consideration is that cow milk froths pretty readily. Depending on how much air is in there (assuming there is some amount), you might not hear much sloshing because the foam is a) deadening the sound and b) filling the previous air gap with a structure of milk foam. Matt Deres (talk) 12:40, 12 September 2020 (UTC)[reply]

LED Lightbulb Problem

I bought a new LED light bulb for a 3-way lamp, labeled 50-100-150W, which means that its output is equivalent to that of an old 50-100-150W incandescent three-way lightbulb. When I put the bulb into the lamp and cycled the switch, it went 0-50-100-50. I couldn't get it to produce what was supposed to be full illumination at full power. So then I thought that maybe there was something wrong with the lamp, so I switched this new bulb and another existing bulb between two similar lamps. The bulb still switched 0-50-100-50, and did not go up to 150. My question is: What is wrong? It isn't that one of the two sets of light-emitting diodes doesn't work. Each of the sets of diodes works. They just don't work at the same time. Why doesn't the higher-power set of diodes work when I have switched the lamp to energize both sets of diodes? Robert McClenon (talk) 17:50, 11 September 2020 (UTC)[reply]

This page talks about incompatibility of using incandescent lamp dimmers for LEDS. It might help, i don't really know much about the topic tbh. Zindor (talk) 18:02, 11 September 2020 (UTC)[reply]

Dimmers are irrelevant. The traditional 3-way light is a bulb with two separate filaments powered independently, and does not involve a dimmer. I don't know about Robert's problem but I would suggest contacting the manufacturer. --174.88.168.23 (talk) 01:22, 12 September 2020 (UTC)[reply]

Yes, that is a little bit weird. Robert, if you have a replacement incandescent bulb, maybe you can try that to see if the 3-way switch works properly. Also, examine the contacts on your LED bulb to see that they are in the right places. A 3-way LED bulb is kind of a hack, though. LED's are dimmable, though the dimming electronics are different from old-fashioned incandescent (triac) dimmers. Places like homedepot.com will have tons of stuff like that. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 02:25, 12 September 2020 (UTC)[reply]

I switched the bulb in question with a working 3-way incandescent bulb. The incandescent bulb worked fine. The questionable LED bulb continued to work 0-50-100-50 in another similar lamp. One comment I got is that maybe the complicated electronics in the bulb is trying to control the current to the diode assemblies, and that there is some error in the "intelligence". That is, if you put artificial intelligence in something, you risk putting artificial madness in it. So I have a bulb that doesn't work the way it is supposed to work. Robert McClenon (talk) 04:53, 12 September 2020 (UTC)[reply]

"if you put artificial intelligence in something, you risk putting artificial madness in it" sounds like the tagline to a great cyber-dystopia film. Flesh it out a bit with computer screens that cast glowing text on people's faces and maybe some boobs and you're got a hit on your hands. Let me know how to invest in this. Matt Deres (talk) 12:47, 12 September 2020 (UTC) [reply]

The phrase should be voiced by Jeff Goldblum, hopefully in a more pithy formulation. --Lambiam 20:26, 12 September 2020 (UTC)[reply]

Though this be artificial madness, yet there is an algorithm in 't. --Lambiam 20:30, 12 September 2020 (UTC)[reply]

It's worth closely reading the packaging to check whether it actually promises that it lights all the LEDs with both live contacts powered, or if it just promises to be "compatible" with 3-way lamp fixtures. Every standard bulb is compatible with a 3-way fixture because the fixture was designed to be "backwards-compatible"; there's just a second hot contact for the second filament in an incandescent. In a "3-way LED lamp", the power controller has to monitor both hot contacts and switch on all the LEDs when they're both powered. A chintzy bulb that was designed to be super-cheap while fooling you could shave off components. (Electronics people: would the bulb work as described if it just connected both hots to a common input? That would explain why it doesn't have a "150W" mode.) --47.146.63.87 (talk) 23:25, 12 September 2020 (UTC)[reply]

Reduction potential at mercury electrode

Hydrogen has high over-potential at mercury electrode. But I could not find the value when I searched Google. What are the reduction potentials of hydrogen, sodium, water and oxygen at mercury electrode in alkaline media? Thanks. Horus1927 20:21, 11 September 2020 (UTC) — Preceding unsigned comment added by Horus1927 (talk • contribs)

September 12

Ventilation fan

I'm wondering why typical household ventilation fans are as small as they are. My reasoning is as follows (tell me if I'm wrong somewhere). Ventilation requirements are basically in air changes per hour which means for a given room size you need a given cfm (cubic feet per minute) of air, and fans are rated in CFM. 1000 cfm would be a typical small indoor fan. Basically the fan pushes out a cylinder of air the same diameter as the fan blades, and its velocity (per minute) is the cfm divided by the area swept by the blades. From the molar weight of air and the ideal gas law, you can find the mass of 1000 cfm of air. The kinetic energy per minute is 1/2·m·v² which after some unit conversions gives you the electric power in watts needed to run the fan (there is a small loss factor but this wattage really does seem to be reflected in the fan specs). The thing here is, that the cfm stays constant but the airspeed (and therefore power) decreases with the square of the diameter. So to minimize your electric bill, you really should want the largest diameter fan you can get.

This shows up in big ceiling fans, but right now I'm using a 20 inch box fan, it is pretty noisy and power hungry, and I'm wondering why I couldn't get one that was say 4 feet in diameter. Any idea? Am I missing something? Thanks. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 02:31, 12 September 2020 (UTC)[reply]

Additional question: I have the 20" box fan blowing through a furnace filter to help clean the air coming out.[13] The filter slows down the airflow somewhat, so that would be one advantage of a smaller (higher velocity) fan. Is there a reasonable low-tech way to gauge the airspeed change? Say, by seeing how much a piece of paper (maybe weighted by a penny) hung in front of it gets deflected? I could try to concoct something like this. I have a voltmeter so in principle I could measure the electrical power going into the fan, but I'm not really concerned about the power bill right now. I mostly want to compute the cfm getting through the filter. Thanks. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 02:38, 12 September 2020 (UTC)[reply]

The something that you're missing is practicality. A large fan on the ceiling is acceptable, but consumers are concerned with factors in addition to efficiency, such as: How would a 4-foot fan fit with my decor; and where can I find space for such a thing? -- They don't make 4 ft. residential fans because nobody would buy them (with one exception?); but, if you really want to, you can buy one: [14] —2606:A000:1126:28D:FDD2:BED1:91BF:2ACB (talk) 05:32, 12 September 2020 (UTC)[reply]

That fan you linked is a huge, expensive, noisy, 1.5hp (1.1kw) industrial fan. I was imagining something that ran at much lower speed. very quiet, using little energy. If not 48", then 30" or so seems practical for many settings such as in a wide window. Right now my 20" fan is sitting on a chair in the middle of the room, which is terrible decor by any stretch, but it's helping the air and a 30" would fit about as well. I also have to wonder why anyone uses 12" or 16" fans when they can use 20". Finally I'm wondering whether this relationship between diameter and power efficiency (watts per cfm) is well known--or alternatively, whether it is mistaken. It does make me more interested in getting a large ceiling fan for hot days. Home Depot has some residential ones up to 7 feet across, and industrial ones (appropriately called "big ass fan"[15]) up to 14 feet (750 watts, they don't even tell you the cfm but I think I can calculate it from the diameter and power). Anyway, thanks. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 08:15, 12 September 2020 (UTC)[reply]

Optically hide a big fan in new construction with a dropped ceiling, pitch black tallest open-ended honeycomb that's worth it ceiling grate for the fake lower ceiling, pitch black entire ventilation system and largest gap between top of grate and bottom of fan that's worth it. You can paint the bottom of the grate any color you want, in fact a light color might camo the fan more. Put as many lights as you can that still seems homey, all under the grate with caps to prevent direct grate illumination and evenly distributed in area and brightness, summing to a typical total room illumination level in all. Well that's the best I could come up with. Sagittarian Milky Way (talk) 13:16, 12 September 2020 (UTC)[reply]

Or slowly and slightly turn the top half or more of a very tall but otherwise standard square subway grate-type grate, you can make it geometrically impossible to see the fan no matter where you put your eyeball. The top of the grate cells should be pointing up and to the left or right or front or back, whichever is downwind in the exhaust system. Sagittarian Milky Way (talk) 17:32, 12 September 2020 (UTC)[reply]

If I got a ceiling fan I wouldn't want it to be hidden. They look cool as hell imho. But, part of that is they look a bit retro since the modern era generally uses a/c instead of fans. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 20:12, 12 September 2020 (UTC)[reply]

I may be mistaken, but I think most of the energy expenditure goes into the mechanical friction of the motor plus electrical losses because electrical motors are not 100% efficient even disregarding friction. The energy saved by using a fan with a large diameter may be disappointing. Noise production costs energy, so go for the less noisy fans. Commercial ceiling fans have anywhere from two to ten blades, sometimes curved like the wooden propellors of early planes, sometimes plain and flat. I think the cuved ones are that way because it looks slick, and the flat ones because it is cheaper to manufacture. I doubt the areodynamic properties are a serious factor entering the considerations of the manufacturers. --Lambiam 20:22, 12 September 2020 (UTC)[reply]

I haven't done the math for very large or small fans, but for midsized ones (and for vacuum cleaners), the energy really does seem to mostly go into moving the air. See air watt. I believe my 20" box fan uses around 100W (haven't measured) and it definitely makes noise, but think of how loud a 100W stereo would be. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 20:45, 12 September 2020 (UTC)[reply]

What makes radar so valuable?

I read about radars, and it seems that except for special beyond-horizon radars; most radars can detect only objects which are in a sight line with the object. Meaning that theoretically, can been seen with a naked eye(with binoculars I guess?). So why then the radar makes such a different? why nobody uses a visible light for example?--Exx8 (talk) 18:47, 12 September 2020 (UTC)[reply]

A fighter jet at radar limit is too small for binoculars. In daytime. Sagittarian Milky Way (talk) 19:16, 12 September 2020 (UTC)[reply]

See Chain Home for the first early warning radar, which had a detection range of 160km (100 miles) back in 1936. Mikenorton (talk) 19:29, 12 September 2020 (UTC)[reply]

Another factor is that radar can "see through" clouds and other meteorological phenomena. See Radar#Principles. AndrewWTaylor (talk) 19:34, 12 September 2020 (UTC)[reply]

In addition to the points above, even in fair weather you'd need to keep scanning around in a 360° sweep to notice vehicles, possibly several, approaching from unpredictable directions. With radar you need to look merely at the screen. --Lambiam 19:57, 12 September 2020 (UTC)[reply]

@Sagittarian Milky Way, why? what makes radar able to detect thing in a distance which can't be observed with visible light?--Exx8 (talk) 20:25, 12 September 2020 (UTC)[reply]

Well, at night it is dark, so you can't observe anything with visible light unless it has lights on or you light it up somehow, in which case black paint would do a good job of stopping reflections. During the day your sensors will be flooded with sunlight because it is daytime so you have to somehow resolve a tiny dark speck. Maybe you could do that with a high resolution wide angle camera these days but that takes fairly modern tech. With radar, you illuminate the target and it's enough to notice a very faint reflection since there is not much background noise at radar frequencies. It's like being able to see a distant star at night because of the faint light reaching you. Finally, you can measure the doppler shift in the reflected radar signal and that tells you the velocity of a moving target. 2602:24A:DE47:BB20:50DE:F402:42A6:A17D (talk) 20:51, 12 September 2020 (UTC)[reply]

At night, if you don't have too much light pollution, you can look for artificial satellites passing overhead. Except, you don't get to look at any one for very long, because they'll pass through your line of sight in tens of seconds. Things in orbit are moving quite a bit faster than aircraft, but still, this gives you an idea. Now imagine trying to pinpoint something in the daytime moving fast at 50,000 feet, and keep your bearing on it while it moves, and trying to identify it. And then do that for possibly dozens of aircraft at a time. --47.146.63.87 (talk) 23:31, 12 September 2020 (UTC)[reply]

RADAR = "RAdio Detection And Ranging". In addition to the detection and determination of azimuth, which radar does better than visual sighting as described above, it also gives accurate ranging to aid in tracking or targeting. -- ToE 00:16, 13 September 2020 (UTC)[reply]

September 13

The speed of electricity

How fast does electricity move? I've heard it said that electricity moves at the speed of light. If this is the case then surely there is something else, other than light, that moves at the speed of light. Does this not contravene the rules of physics? Thank you

[physconst-G-1] "2022 CODATA Value: Newtonian constant of gravitation". The NIST Reference on Constants, Units, and Uncertainty. NIST. May 2024. Retrieved 2024-05-18.

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