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August 22

Male autism-spectrum correlations with homosexuality, bisexuality, fetishes and paraphilias?

In males, are autism-spectrum disorders statistically correlated with homosexuality, bisexuality or any fetishes or paraphilias? The one study I found on the subject included only 17 males, and the extent to which it oversampled females leads me to question the representativeness of those males. Neon Merlin 00:54, 22 August 2017 (UTC)[reply]

the article you link should have a link to references, which should include known articles about the same subject. As it seems to me highly politically UNcorrect to try to link ASD to any LGBT behavior, I suspect such study would be rare and statistically unsignificant

Gem fr (talk) 16:15, 22 August 2017 (UTC)[reply]

I think it is a useful topic for study, because some authors think of autism as an "extreme male" phenotype, [1] and so looking at GLBT variations seems like a no-brainer. I've never given that idea much credence (the article I cite doesn't seem to either) but hey, it's biology, and nothing is impossible in biology. That said, the only obvious relevant hit on PubMed for autism homosexuality is this 50-boy study that finds nothing much of note. I actually found three references with autism transsexual, but they amount to anecdotes, something from Med Hypotheses, and this brief report, which takes the idea somewhat seriously and is the closest I have to an answer. But I'll leave it to someone else to pull this off Sci-Hub and decide if there is any meat there. Wnt (talk) 18:52, 22 August 2017 (UTC)[reply]

"extreme male"? Now THAT is a concept! especially when applied to fully functional FEMALE people, by people writing in a Psychology review when we now know that ASD is NOT about psychology (as people though sometime ago)! University job includes producing ideas, most of them utter crap. I already know that Most Published Research Findings Are False and Most Clinical Research Is Not Useful in medicine, i suspect it is even worse in psychology. But that was just my "small" 2 cents Gem fr (talk) 13:16, 23 August 2017 (UTC)[reply]

I wonder if autistic-spectrum people are over-represented in extreme sports… --47.138.161.183 (talk) 17:52, 24 August 2017 (UTC)[reply]

Sunspots

Using premodern equipment, how in the world does one observe sunspots? Even when I was looking through clouds and through a telescope made specifically for eclipse-viewing, the Sun appeared during today's solar eclipse to be a perfect, unblemished shape, aside from the rather big missing chunk of course. I can't imagine how the ancient Chinese, c. 800 BC, or Greeks c. 300 BC (see Solar observation#Early observations, could see sunspots. Nyttend (talk) 02:10, 22 August 2017 (UTC)[reply]

It is fairly easy with premodern equipment to use a pinhole camera to get a projected image of the Sun, on which sunspots are quite readily visible. Double sharp (talk) 02:33, 22 August 2017 (UTC)[reply]

So what, you just take a sheep of papyrus and poke a hole in it, and you're done? Nyttend (talk) 02:36, 22 August 2017 (UTC)[reply]

No poking of sheep is allowed here ! StuRat (talk) 03:09, 22 August 2017 (UTC) [reply]

Poking a piece of parchment would be the most basic approach, I suppose. (Lot's of school children observe an eclipse with two pieces of paper, I don't imagine parchment would be any different.)

But much more advanced pinhole cameras have been in use for a long time. The article Pinhole camera is a bit thin, but it does cover the history. The article Camera obscura also covers the history a bit.

It strikes me as an interesting factoid that cameras were invented thousands of years before film. ApLundell (talk) 14:01, 22 August 2017 (UTC)[reply]

The word camera's etymology runs from Greek kamara "vaulted chamber" --> Latin camera "vaulted room" --> English 1708 "vaulted building" --> early 18c. as a short form of Modern Latin camera obscura "dark chamber" (a black box with a lens or pinhole that could project images of external objects) --> c. 1840 "picture-taking device". Blooteuth (talk) 14:36, 22 August 2017 (UTC)[reply]

Note that sunspots vary dramatically in size and temperature/brightness, and hence visibility. StuRat (talk) 03:10, 22 August 2017 (UTC)[reply]

Any sunspots during the eclipse weren't big as sunspot go and were probably covered by the Moon during the time when there was a rather big missing chunk. The butterfly diagram shows that we're in one of the lowest latitude parts of the sunspot cycle and the Sun's poles are only 7 degrees from Earth's orbit's poles so that couldn't really help bring them to the north or south side of the disc either. Sagittarian Milky Way (talk) 04:23, 22 August 2017 (UTC)[reply]

It would be nice if someone could update the articles with information about where the observations took place. I mean, the Sun has the great advantage of being really bright, so you can project it a long way, but at largest size it is still best viewed in a dark setting. This implies that it would really help to have a monumental roofed building, some manner of cathedral where a pinhole aperture made accidentally or deliberately at one window can project an image into a vast darkened space. (I suppose the right cave can do it but you have to postulate a lot of luck) I imagine the Chinese and Greeks had many such buildings but don't know if those observers were known to use them. Wnt (talk) 05:14, 22 August 2017 (UTC)[reply]

The name and the fact that Galileo and a contemporary worked on it would imply reliance on lenses, which makes that a much higher level of technology than that available to the earlier workers. Wnt (talk) 13:31, 22 August 2017 (UTC)[reply]

Right now, there is just one visible sunspot - a chain on the sun's equator called AR2671. It's possible that you weren't looking in the right place (sunspots are not that big compared to the solar disc, and it's easy to overlook them). That said I find larger sunspots are visible even without magnification, just by looking through eclipse glasses - and if you're very lucky, you can see them in the setting sun. Smurrayinchester 07:59, 22 August 2017 (UTC)[reply]

This article does not go back as far as the notations above, but does go back to the early modern period and discusses some fairly simple methods for observing sunspots. Kepler created a simple camera obscura by poking a pinhole in the roof of his house, for example. --Jayron 32 12:07, 22 August 2017 (UTC)[reply]

Note that natural pinholes are quite common. For example, a tree's leaves on a calm day can form many pinholes between the leaves. I often find accidental pinhole lens images. StuRat (talk) 14:39, 22 August 2017 (UTC)[reply]

On rare occasions a sunspot (or sunspot group) can be big enought to be seen directly with the eye when the Sun is dimmed by mist or thin cloud. Over the last 50 years I've read in various astronomy textbooks that ancient Chinese astronomers recorded a number of such observations, although our article on Chinese astronomy doesn't mention them. To give a couple of references:

George O. Abell's Exploration of the Universe; Third edition; Holt, Rinehart & Winston 1975 (in Section 2.1 Earliest Astronomers, on page 11) states, "The Chinese also kept rather accurate records of comets, meteors and fallen meteorites from 700 B.C. Records were made of sunspots visible to the naked eye . . . ."

Patrick Moore's The Guinness Book of Astronomy; Fifth Edition; Guinness Publishing 1995 (in The Solar System ∗ The Sun, on page 6) states, "Naked-eye spots had been previously [to J. Fabricius in 1610–11] recorded, but had not been explained; one given in a Chinese record of 28BC is described as 'a black vapour as large as a coin' and there is a Chinese record of an 'obscuration' in the Sun, which may well have been a spot, as early as 800BC."

{The poster formerly known as 87.81.230.195} 94.12.81.193 (talk) 15:03, 22 August 2017 (UTC)[reply]

What ! Can't see sunspots... reason is below to the tune of La donna è mobile:

Sun spots are not here this year,

Less so than yester year.

Small spots are sometimes seen

Often with H-alpha screen.

Next verse:

Bigger ones are for future years

So get in lots of beers

Get yours before their gone

Come now try one.

See: Real-time solar activity for what I mean.Aspro (talk) 18:32, 22 August 2017 (UTC)[reply]

And here's another and perhaps better ref as to why the ancients saw spots before their eyes but you can't. Sun's Current Solar Activity Cycle Is Weakest in a Century Aspro (talk) 19:01, 22 August 2017 (UTC)[reply]

I highly doubt any "pinhole camera" construction without a lens can help with sky or sun observation. There is no need for any of that to see big sunspots since the sun is big enough to observe with sharp eyes and we have frequent moments to look straight at our central star - in the morning and evening, when the weather conditions are right - without any danger of harming the Retina. Besides that some planets where also known surprisingly early but also there have always been humans with exeptional sharp eyes. --Kharon (talk) 22:57, 22 August 2017 (UTC)[reply]

Ganymede, Uranus, some asteroids and possibly Neptune are visible to good eyesight but their sighting wasn't known to science till 1610, 1789, 1800s, and 1846 respectively. (also a very small amount of galaxies: discovered by telescope, some people can see them) Sagittarian Milky Way (talk) 00:03, 23 August 2017 (UTC)[reply]

Kharon is right about some people having exceptionally sharp eyesight. Anecdata: an acquaintance of mine with comparatively little interest in astronomy once revealed that he could routinely see Jupiter's four Galilean moons with his naked eyes. (They are in themselves bright enough to be visible, having magnitudes ranging from 4.6 to 5.6 at opposition, but are normally lost in the greater glare of Jupiter.) Until a short time before our conversation, he hadn't realised that this was exceptional, so had never mentioned it. Doubtless in earlier history some people had similarly seen previously unknown astronomical phenomena without realising their significance, and so had not recorded the fact. {The poster formerly known as 87.81.230.195} 94.12.81.193 (talk) 16:35, 23 August 2017 (UTC)[reply]

But Kharon is wrong about pinholes being worthless for observing sunspots.

It's a common enough student activity to track sunspots with a pinhole camera.(Example [2]). Ideally you'd want about a meter or more between the pinhole and the viewing surface. This will make the image pretty dim, but with a hood of some kind (or just sticking your head in the cardboard box you're using) your eyes dark-adapt and you can see surprising detail. ApLundell (talk) 17:09, 23 August 2017 (UTC)[reply]

Gluconeogenesis

The gluconeogenesis article says "The process is highly endergonic until it is coupled to the hydrolysis of ATP or GTP, effectively making the process exergonic." What does this mean in layman terms? At what point in a zero carb diet does gluconeogenesis become "coupled to the hydrolysis of ATP or GTP" — Preceding unsigned comment added by 187.1.51.122 (talk) 13:59, 22 August 2017 (UTC)[reply]

Endergonic means energy costly, exergonic means producing energy. An analogous statement to the first part of your question would be that producing gasoline costs a lot of energy (obtaining, transporting, and refining oil) until you burn it in the presence of oxygen. μηδείς (talk) 16:25, 22 August 2017 (UTC)[reply]

it means gluconeogenesis consume energy, that the energy for this is brought in the process by ATP (or GTP), but that not all the energy of ATP is transformed into glucose, the process waste some of it as heat (as is usual for process)

It also means that gluconeogenesis IS coupled to the hydrolysis of ATP or GTP from its very start, it wouldn't happen at all otherwise

Gem fr (talk) 17:54, 22 August 2017 (UTC)[reply]

It's a little more complicated than that. The thing to bear in mind is that gluconeogenesis is a complicated multi-step process. The change in Gibbs free energy at each step depends on the concentration of all reagents - Le Chatelier's principle is very much in play. Force one reaction with a large energy change, and the next step will have no choice but to go forward ... once a large enough concentration of metabolite builds up. Look at [3] for a sense of the "energy landscape" for glycolysis, which is a similar situation; then see this comparison for some numbers for gluconeogenesis. But the cell can't deal with massive concentrations of some metabolites. So there is a lot of specific engineering and "horse trading" involved, which has been optimized by unfathomable time scales of microbial evolution. But the bottom line is that just a few steps that split ATP are used to drive many others that are neutral or even unfavorable in terms of free energy. Wnt (talk) 19:11, 22 August 2017 (UTC)[reply]

Fair enough, but the OP asked for a translation in "layman terms" of what is written in the article, so i am pretty sure that adding "It's a little more complicated than that" didn't help (even though I can only hat tip) Gem fr (talk) 12:51, 23 August 2017 (UTC)[reply]

From a general overview, it makes sense for gluconeogenesis to consume energy. As noted at nutrient, both protein and carbohydrate provide about 4 kcal/gram to the body in usable energy. Therefore, you gain nothing from an energy point of view by converting protein to carbs; and since you aren't using that protein as an energy source, you're basically burning 4 kcal/gram just to make that glucose. Couple that with energy losses expected due to inefficiencies (a general physics concept is that SOME energy is always lost in any transfer, per second law of thermodynamics, etc) and you can see where the process must be endergonic. --Jayron 32 13:07, 23 August 2017 (UTC)[reply]

Dissociation constant, Ph of water 'n' stuff

The formula is amenable to two different interpretations,

K_w=[OH^-][H⁺]

= 10^-14 x (1000/17.01) x (1000/1.01)

= 5.82 e -10

or

K_w=[OH^-][H₃O⁺]

= 10^-14 x (1000/17.01) x (1000/19.02)

= 3.1 e -11

neither correspond to the given values. This would seem to indicate that my textbook is wrong saying that the value of [H₂O] being "ignored" is 55.5 = (1000 g/L / 18.015 g/mol).

All this leaves me slightly puzzled.

All the best: Rich Farmbrough, 14:52, 22 August 2017 (UTC).[reply]

Why are you getting 17.01? And water's molecular weight is 18.015 g/mol, not 18.5. Yes, I had a chemistry professor that disputed the textbook's usage of H3O+, so, H3O+ and OH- do not make H2O, only with H+ and OH-. 12.130.157.65 (talk) 15:12, 22 August 2017 (UTC).[reply]

18.5 was a typo. 17.01 as indicated below by DMacks. All the best: Rich Farmbrough, 16:08, 22 August 2017 (UTC).[reply]

It might help if you attach units to the numbers. For example, I recognize "17.01", "1.01", and "19.02" as the molar masses of the respective ions, but not what "1000" represents in each case (or why 1000 is in each ionic component but a separate powers-of-ten multiplier that is not part of the literal formula is also present). Are you conflating different substances' "liters" or "grams" or "moles" somewhere? DMacks (talk) 15:18, 22 August 2017 (UTC)[reply]

I don't know because the formula is not clear to me. [] "indicates molar concentration". The figure for water is calculated as shown (55.5) and is predicated on "almost all" of the substance being H2O - I use the same method for the other three species. The ratio of the other two components to the whole is 10e-7 - there are two such factors, hence 10E-14. My text book says "The value of ..Kw at 25 C is 10^-14." which indicates that either the molarity is not part of the equation, or is being subsumed in the constant factor that differentiates Kw, by ignoring the almost-constant denominator.

All the best: Rich Farmbrough, 16:06, 22 August 2017 (UTC).[reply]

do you understand molar concentration, to begin with? It would help you, i think. ~~Please note that it is a dimensionless quantity, and so is any product or ratio of several of them, as Dissociation constant is. That's why we can use a "log" operation on it.~~ (my bad, i stand corrected below). And that's why it is SO important to NEVER, EVER, use raw number without attached unit (for instance, don't use 1.01, use 1.01g/mol, etc.), as DMacks too politely tried to tell you. If you don't you may add up liters with kg, and lose all meaning without seeing it. I suspect this is precisely what just happened to you. Gem fr (talk) 17:30, 22 August 2017 (UTC)[reply]

55.5 M is not dimensionless it is mol/L as is clear from the above.

We would expect the dissociation constant to have the same dimensionality, as it is the product of two concentrations divided by another, and indeed our article confirms this "Kd, which has the dimensions of concentration…" although it later seems to get confused between molality and molarity.

Incidentally it is the cologarithm we (may) take to get Ph, and in that case we have the concentration of H+ ions (or equivalent) over the concentration of H2O, which is trivially dimensionless.

It may be worth re-framing my questions:

When the text book says:

K_w=[OH^-][H⁺]

instead of :

$K={[OH^{-}][H^{+}] \over [H_{2}O^{+}]}$

what is the actual change in value and dimension between K and K_w?
Given that we can treat the ions in water in two different ways, either OH^- and H⁺ or OH^- and H₃O⁺, does this choice affect the value of K or K_w. If so, what does this represent? If not, why not?

All the best: Rich Farmbrough, 18:49, 22 August 2017 (UTC).[reply]

Rich: to start out with, there's no H2O+ -- you want H2O there. But to answer your question, [H2O] = 55 mol/l, so the dimensions of the constant of the other side change accordingly. Kw is mol^2/l^2 and Keq is mol/l. Wnt (talk) 16:54, 23 August 2017 (UTC)[reply]

To start out with, bear in mind that K_w is equal to the product of two concentrations; therefore it has units of (mol/L)^2. This is relevant if someone asks you to do this in the gas phase and your concentrations turn into partial pressures in atm; then you need to bear in mind that the ideal gas constant R = 0.082 L atm / mol K --> then you can multiply your mol/L by (RT)^2, where T is the temperature of the gas, and get something in atm^2 that you can use to figure out the relevant partial pressures ... or at least, could if the dissociation constant didn't depend on temperature and phase a lot - as you see in self-ionization of water the values for steam and supercritical water are ridiculously high since the charges aren't stabilized the same way as in polar aggregates. But you couldn't compare them at all without having units on your equilibrium constants to guide you!

Now, it should be clear that the K_eq for [H+][OH-]/[H2O] will be reduced by a factor of 55.5 M (by which I mean 55.5 mol/l; honestly, I despise the italic M notation). And so the units of that will be in mol/l only. You get that number by noting that 1 g H2O ~= 1 cm^3 H2O and 1 mol H2O = 18.01 g H2O and 1 cm^3 (cubic centimeter) = 1 ml = 1/1000 L. You can multiply anything in chemistry by the ratio of two things that are equal, because that is always 1, so putting these together like dominoes you get (1 g H2O / 1 ml H2O) * ( 1 mol H2O / 18.01 g H2O) * (1000 ml / 1 L ). By cancelling out all the units on both top and bottom and multiplying or dividing numbers as given, that's a net result of 1 mol H2O * 1000 / 18.01 L = 55.5 mol/L.

As for your other numbers, I'm not sure where they come from. At room temperature K_w is around 10^-14 (mol/L)^2, i.e. the pK_w = 14 + 2 log L - 2 log mol. But nobody else on Earth but me would try to keep track of units once they take a log; they say oogledyboo and the units go away and they just try to remember not to take a log of anything but mol/l in the same context. Wnt (talk) 19:42, 22 August 2017 (UTC)[reply]

Thanks! I think you can say "the units go away" in a fairly legitimate manner, if everything is normalised, or you are otherwise Very Careful™. You answered Q1, and also made me feel that my questions weren't somehow stupid, which is the vibe I was starting to get from the other answers. So thanks again.

All the best: Rich Farmbrough, 19:59, 22 August 2017 (UTC).[reply]

(Actually my initial question was stupid, because it didn't explain itself enough.) All the best: Rich Farmbrough, 20:01, 22 August 2017 (UTC).[reply]

I think you should read self-ionization of water. Please, remember that the basic quantity is activity, which is dimensionless by definition. Only in case when concentrations are low it is proportional to molality. Ruslik_Zero 20:04, 22 August 2017 (UTC)[reply]

It is indeed true that the equation is really for activity. That said, introductory students and biochemists alike rarely are in a position to figure out the activity rather than the simple concentration. As explained at Thermodynamic activity, it is still debatable whether the activity of H+ ion is even a measurable quantity! To be sure, pH is a very deep rabbit hole and the more you look into it the more confusing it gets. I don't pretend to have understood it the last time I went there. But to the basic student it is all very simple. ;) In any case, for equilibria it is convenient to use activity coefficients, which are dimensionless and typically close to 1, which are multiplied by the concentrations. Thus, the equilibrium constants retain the same units as if true concentrations are used - so the activities are, for most practical purposes, empirical corrections used to make the formula match observed behavior.

Rich: so far as units are concerned, there are different schools of thought on the matter. There are some who love natural units and would like to do everything dimensionless. Personally I hate dropping units and like having the check as I go along. The key thing with units is that as long as you watch your math, they'll tell you if you're doing the wrong calculation. You rarely get an answer in the right units if you used the wrong formula. Wnt (talk) 22:59, 22 August 2017 (UTC)[reply]

The thing I always dislike, pedagogically, about thermodynamic activity (which you get at) is that it is both literally true and yet it is fully tautological in practice. By that I mean that a) Strictly speaking, equilibrium constant calculations are only absolutely true for ratios of activities and b) The only way to find activities is to know the equilibrium constant first; that is to "back calculate" from a known EQ constant. Since I can't actually use activity to calculate K, it's a pretty useless concept, even if true. --Jayron 32 11:15, 23 August 2017 (UTC)[reply]

I don't feel comfortable saying that - the article describes a lot of different things that the activity effects, like melting point depression and vapor pressure if I recall correctly. The main thing that hinders me in applying the concept comfortably is that I could use a better mental picture of how you inactivate a substance - I would think in some cases there are dimers or even little crystals formed or something? But I don't actually know. Wnt (talk) 16:58, 23 August 2017 (UTC)[reply]

The origin of the activity concept lies in a specific parameterization of the chemical potential of component i:

\mu ^{i}(T,p,n)=\mu _{0}^{i}(T,p)+RT\ln a^{i}(T,p,n)

(n - concentration). The second term corresponds to some additional entropy and

\lim _{n\to 0}a=0

. In equilibrium the thermodynamic potential should be the same before and after the reaction. So, one can obtain the equilibrium conditions. Ruslik_Zero 19:24, 23 August 2017 (UTC)[reply]

That math seems informative - I didn't realize the concentration-dependent term would tend to zero at low temperature (even if that isn't generally achievable). But I still don't understand how the extra entropy would come about. I mean, one of the examples given is sodium chloride - I know the sodium ion doesn't vibrate into some weird conformation where it can't react, right? Wnt (talk) 20:57, 23 August 2017 (UTC)[reply]

This is similar to the chemical potential of ideal gas. Ruslik_Zero 18:06, 24 August 2017 (UTC)[reply]

If anyone is interested the fundamental error I was making was using the same formula for concentration of ions other than H20. In all cases (in this example) the solute is essentially H20, therefore the availability/molarity/molality of the ions are, respectively, identical. The remaining difference from 10^-14 is a simple constant, which can be mopped up at the same time we ignore the constant denominator. (Unless anyone knows better?)

All the best: Rich Farmbrough, 18:04, 25 August 2017 (UTC).[reply]

STDs that spread despite use of condoms

What STDs can still be caught even if people use condoms correctly and systematically for vaginal/anal, but not for oral sex? Assuming a non-monogamous person, with regular sexual contact (including kissing) with other promiscuous people.--Hofhof (talk) 16:27, 22 August 2017 (UTC)[reply]

You're question need clarification for me: are you talking about the effect of condom on sexually transmitted infection, or about STI that also transmit NON sexually, through kisses and oral contact?

Gem fr (talk) 17:08, 22 August 2017 (UTC)[reply]

Oral sex counts as sex.B8-tome (talk) 18:38, 22 August 2017 (UTC)[reply]

You can get papilomavirus infection, syphilis, gonorrhea and possibly chlamydia by oral route (in both directions). In rare cases you can get HIV. Ruslik_Zero 18:33, 22 August 2017 (UTC)[reply]

Herpes (HSV-1, or oral herpes and HSV-2, or genital herpes) and HPV. You can also catch gonorrhea in the throat, or transmit it to someone else’s genitalia. — Preceding unsigned comment added by B8-tome (talk • contribs) 18:35, 22 August 2017 (UTC)[reply]

It's important to note as well that people can still use condoms correctly and the condom can have a fault in it. Nothing is perfect. So the answer to "What STDs can people still caught even if they use condoms correctly and systematically" is every STD. Assuming "can" means "it could happen at all to at least one person in the entire history of humanity" The world is a big enough place that even low percentage events occur in large absolute numbers. This source from the CDC is probably a good place to start your research, and it leads to additional sources. --Jayron 32 19:37, 22 August 2017 (UTC)[reply]

You are technically right. I should have asked which STDs are common in the scenario above, without disregarding low risks. Hofhof (talk) 20:07, 22 August 2017 (UTC)[reply]

The only perfect STD barrier is sexual abstinence, which is a form of isolation. If you isolate yourself, then you will not catch a pathogen. Even if you do catch a pathogen, the likelihood of falling ill really depends on the health of your immune system. If you are immunocompromised, then you will be susceptible to all sorts of diseases. If you are relatively healthy, then you will be unaware that you carry the pathogen. 140.254.70.33 (talk) 22:18, 22 August 2017 (UTC)[reply]

That's a dangerously broad statement. To begin with, the "healthiness" of the immune system can only be tested by putting it to the test, which makes this a tautology -- if you shake off a virus, you had a healthy immune system! It would be different if you could stick your finger in a machine and get back a message that thanks to eating your vegetables you're AIDS-proof, but that won't happen. (There are folks with mutations like CCR5 that make them resistant to HIV, but they are actually more vulnerable to other pathogens for the same reason) The other thing is that the virulence of pathogens varies - there are some, like HIV, that a person can be exposed to a dozen times and get lucky every time, and others like pneumonic plague where exposure pretty much guarantees a gruesome death without artificial treatment. Wnt (talk) 23:05, 22 August 2017 (UTC)[reply]

One's strong immune system can also contribute to organ damage and death when an infection occurs, so I agree that "if you have a strong immune system" is a misleading overgeneralization. —Paleo Neonate – 06:03, 23 August 2017 (UTC)[reply]

It's also simply wrong. The effectiveness of "sexual abstinence", like all other forms of protection from pregnancy or sexually transmitted disease, is evaluated in terms of both theoretical effectiveness and use effectiveness. While the theoretical effectiveness of sexual abstinence may be zero, the use effectiveness isn't. So, for example, the theoretical effectiveness of male condoms is about 97%. Use effectiveness is about 86%. As an estimate of the use effectiveness of "abstinence," about 60% of people who took virginity pledges in high school were sexually active in their college years. This is probably the biggest "gap" of any method between theoretical and use effectiveness. When you recommend sexual abstinence, you recommend a method with a 60% failure rate. - Nunh-huh 05:19, 23 August 2017 (UTC)[reply]

... though I would expect that more than 99% of individuals would know, looking back at the past, whether or not they have practised abstinence consistently ... Dbfirs 06:23, 23 August 2017 (UTC)[reply]

But then you need to define common and low risk. Depending on the community you're in and other factors, your risk of being infected by HIV is fairly low even if you regularly engage in unprotected vaginal sex in non-monogamous relationships. It's still generally considered a bad idea though considering the consequences are quite high and the risk isn't so low that you can assume it isn't going to happen. Nil Einne (talk) 08:50, 23 August 2017 (UTC)[reply]

Exactly. I don't know the quality of the sources but these say the risk of HIV transmission from unprotected insertive oral sex [4] [5] [6] may be 0.04% per act between serodiscordant people, which is equal to the "estimated risk for insertive anal sex between men, with use of a condom, with a partner whose HIV status is unknown". (Although I would imagine this isn't perfect use for the condom.) Nil Einne (talk) 08:50, 23 August 2017 (UTC)[reply]

I know Nil knows this, but most people are surprised that the risk of infection depends on the rate of prevalence in the population. Yet it's true, and you can't really give a meaningful estimate of risk without knowing prevalence. - Nunh-huh 23:34, 24 August 2017 (UTC)[reply]

Also note that condoms won't prevent the spread of pubic lice. StuRat (talk) 20:04, 23 August 2017 (UTC)[reply]

I'm curious how a person who practices abstinence can get an STD. Some of the above advice is suggesting that abstinence is a dangerous practice. Akld guy (talk) 01:37, 24 August 2017 (UTC)[reply]

Some STDs can also be spread by non-sexual contact, like some forms of herpes from kissing. StuRat (talk) 02:04, 24 August 2017 (UTC)[reply]

Herpes spread in that way is a form of virus, not an STD. Genital herpes spread by sexual intercourse is indeed an STD. So again, I ask. Akld guy (talk) 03:10, 24 August 2017 (UTC)[reply]

Another example is AIDS, which can also be spread by blood transfusions and sharing needles. It's still a predominantly sexually transmitted disease, no matter how a given person gets it. StuRat (talk) 14:23, 24 August 2017 (UTC)[reply]

I guess the trouble is that too many person who "practices abstinence" ... really don't, and when they do have sex, is is not in the safest way for obvious reason. Gem fr (talk) 08:24, 24 August 2017 (UTC)[reply]

Exactly. If you're "practicing abstinence" as a way of avoiding disease, and nonetheless have sex and get that disease, that's a failure of the abstinence method, in the same way that forgetting to use a condom is a failure of the condom method. - Nunh-huh 23:34, 24 August 2017 (UTC)[reply]

StuRat, AIDS is not a sexually transmitted disease, it is the end process of HIV infection (which can be a sexually transmitted disease). I am sure you are well aware that people can't catch AIDS from having sex, but we must strive to be specific about such matters as there is still a lot of ignorance about HIV and AIDS. RichYPE (talk) 11:25, 25 August 2017 (UTC)[reply]

That's a rather unusefully false precision. HIV is not a disease. It is a virus. AIDS is the disease stage, which not all people infected with HIV reach. It was known that AIDS was sexually transmitted before the virus vector was conclusively identified. The proper place for extended technical specificity is the article, not in a simplified list. μηδείς (talk) 16:19, 25 August 2017 (UTC)[reply]

What causes red/green coloration in eclipse images?

Here [7] are a few images of the recent eclipse. The first is taken through a telescope with a cell phone camera. Note how the sun/sky border is limned with green tones, while the sun/moon border has red. I also saw the same pattern of colors using different imaging methods. I could see it with the naked eye looking at natural camera obscura e.g. through leaf shadow like the second image in the album. The third photo is of a projection by binoculars on to a white paper, red/green was also visible in live viewing. N.B. you can't quite see the colors in the leaf shadow photo but I could see it as it happened, and several nearby watchers said they could see the same color pattern.

Q: What causes this red/green coloration on the different boundaries? Perhaps some sort of diffraction? Since I see it with three fairly different methods, it seems unlikely to be a property of any given camera or lens.

Thanks! SemanticMantis (talk) 16:39, 22 August 2017 (UTC)[reply]

red appears for the very same reason the sun look red at sunset or sunrise

I cannot explain the green, someone else may help

Gem fr (talk) 17:16, 22 August 2017 (UTC)[reply]

I believe you are seeing various forms of chromatic aberration, which is fundamentally an optical artifact that is aggravated by the very high light/dark contrast ratio that is a characteristic of this type of photograph of a solar eclipse - whether it is an image of a projection, or a direct view of the sun.

To make matters worse, complicated engineering systems - carefully-designed complications of optical glass lenses, and carefully-designed software and hardware inside the camera - may attempt to correct for chromatic aberration. Those corrections only work up to a point - but for the extreme case of a direct solar image or a high-contrast projection of the sun - these corrections might actually aggravate the artifact.

First, the lenses - there are achromats, apochromats, super-apochromats, multi-element super-apochromats, ~~superheterodynes~~, ... I sure wish this book wasn't so darned expensive!

Next, the algorithms - there are radial correctors, chromatic gradient detectors, desaturation filters, nonlinear edge estimators, ... heck, it's 2017... somebody's probably trained a recurrent convolutional neural network to detect-and-correct chromatic aberration.

The core cause of the visual artifact we call chromatic aberration is optical refraction occurring at the exact same part of the image as optical diffraction - places where a hard edge in the image corresponds to an extreme, high-contrast change in the illumination.

Here's an open-source algorithm - Fix-CA - that is commonly distributed with the GIMP. It's pretty simple, and it works in RGB-colorspace, so it should be "easy" for an ordinary programmer to follow.

Here's a much more famous and complex/sophisticated algorithm - Radial, part of PanoTools. Here's a blog documenting how it works, and here's another one. The algorithm is harder to read, because it's split across the tools library, but the meat of it is in correct.c.

Those tools are meant to allow a pro-user to fix up an artifact in a digital image after they've captured it. A modern-vintage camera probably implements a built-in, probably-proprietary, possibly-similar algorithm to fix up the image before you ever see it.

See if you can spot where and why such an algorithm could fail - and imagine what a proprietary camera algorithm might do differently - and imagine how that might fail too! The great bane of digital image processing algorithms is that in some corner-case conditions, instead of solving a visual artifact problem, the algorithms can make the problem worse!

Nimur (talk) 19:34, 22 August 2017 (UTC)[reply]

Couldn't this be caused in the same way as the Green flash?B8-tome (talk) 20:22, 22 August 2017 (UTC)[reply]

Thanks @Nimur:. That's helpful but two things I'm still not clear on: 1) I can accept that my neighbor's binoculars and my friends' telescope may have the same pattern of aberration, but why would they consistently show red on one the sun/moon border while showing green on the sun/sky boundary? 2) I suppose the same aberration can occur in my own naked eye?

Put another way, if I accept this to be chromatic aberration in the telescope, the binoculars, and my naked eye, I'm still left with the question of what is different between the sun/moon boundary and the sun/sky boundary. I have yet to see any photo that shows both sides with the same color aberration, or with a reversed color scheme. If you search google images for /eclipse through telescope/ (e.g. [8]) or /eclipse projection/ (e.g. [9]), you will see many photos with the same pattern that I describe, and I feel like there should be a good reason why none of the putative aberrations have the same color on both sides, or a reversed pattern, etc. SemanticMantis (talk) 21:02, 22 August 2017 (UTC)[reply]

Well, don't those two sides have very different things on them? On one side you are seeing light that comes from the center of the Sun and grazes the edge of the Moon; on the other side you are seeing light that comes from the edge of the Sun's globe. Looie496 (talk) 21:41, 22 August 2017 (UTC)[reply]

The root cause of red/green coloration of images is dispersion which is the change in all optical materials of their refractive index with the frequency (color) of light. Dispersion causes a glass prism to spread a beam of white light into spectral colors that Newton observed in 1666, gives rise to the Rainbow, and causes a single glass lens to have too much chromatic aberration for camera use, though wearers of spectacles have to get used to it. Compound lenses of greater complexity and cost have been designed to achieve compromises between various optical aberrations (spherical, coma, and chromatic) but a perfectly achromatic or even Apochromatic lens will not be found in a cell phone. Incidentally, the human eye has only a single lens that presumably exhibits all the optical aberrations which must thereafter be corrected by retinal/neural processes. These processes are not normally trained to work with extreme contrasts and may be saturating during SM's eclipse observations. Blooteuth (talk) 22:09, 22 August 2017 (UTC)[reply]

At a more concrete level, the reddish tinge of the inner edge probably comes from bending of light by the Earth's atmosphere, the same process that causes the sky to be red at sunrise and sunset. The greenish tinge of the outer edge probably results from the fact that the light has to pass through a long stretch of the Sun's atmosphere, which filters out reddish hues more than greenish hues (this second part is speculation). Looie496 (talk) 14:25, 23 August 2017 (UTC)[reply]

Physical mechanism of leafhopper song

This article has a pretty good description of how leafhopper communicate and the purpose of their vibrations. But how do they make the vibration? Is it by moving their wings back and forth, or using their legs to strike drum-like tymbals or something else? http://www.wired.com/2014/09/lustful-leafhoppers-locate-good-vibrations/ --Captain Breakfast (talk) 18:40, 22 August 2017 (UTC)[reply]

I'm also curious if these sounds are easily detectable by human ears, or if they are only heard by other leafhoppers.--Captain Breakfast (talk) 18:46, 22 August 2017 (UTC)[reply]

Wikipedia has a brief overview article on Leafhoppers, unfortunately it doesn't say anything about the song. Leafhopper is a pretty large family of insects, which is part of the cicada superfamily, and THAT article describes how they produce their song. --Jayron 32 19:14, 22 August 2017 (UTC)[reply]

As described in our Article Tymbal. Generally its easy to generate vibrations. If you use one of your fingernails to bend another fingernail and then let it flip by gliding off the edge you can make a loud sound yourself like if you had an exoskelet. Do the fingernail flip right next to your ear for a try to notice how loud it actually is. If you could do that very rapidly and you had a resonating shell on top, you could make similar impressive loud sounds too. --Kharon (talk) 23:15, 22 August 2017 (UTC)[reply]

I think you are looking for stridulation and Cicada#Song

Gem fr (talk) 12:29, 23 August 2017 (UTC)[reply]

The Multiplication of Food Grains

Let's say I have one of each seeds of typical wheat, rice, barley, maize, millet ... and many other common food grains. They are all properly planted in well-prepared fields under favorable climate. They are all successfully harvested.

How many grains can I get from each plant?

How many grains can I get if I plant one seed of typical wheat, rice, barley, maize, or millet ... -- Toytoy (talk) 21:27, 22 August 2017 (UTC)[reply]

There are many varieties of each of these grains. The number of grain could easily be double between varieties of one species, and it should be noted that not all grain is fertile seed, such as in two-row barley.Of 19 (talk) 22:14, 22 August 2017 (UTC)[reply]

One seed of maize (sweetcorn or field corn) typically produces one or two ears of harvested corn if it sprouts and grows to maturity without suffering from insects, drought, blight, etc. (Maize intended to be harvested pre-maturely for "baby corn" may have more ears and thus more immature harvested kernels). [10]. Each ear has 500 to 1000 kernels, with 800 being typical. [11]. That could imply 500 to 2000 for each healthy stalk. In the Genesis 26:12 in the Bible, a farmer harvested 100 fold, so he might get 100 seeds harvested from each seed planted. In Mattheh 13:8 seed sown into "good ground" produced a yield of 30 to 100 fold. Not sure what grain the Biblical farmers planted. Edison (talk) 22:16, 22 August 2017 (UTC)[reply]

[double edit conflict; I wrote this before Of 19] Two further considerations: (1) are they sheltered from birds, and (2) do you have a second maize plant nearby? We tried growing a few wheat and oat plants in our vegetable garden (just to see what would happen) when I was growing up, and they all produced good grain, but between the fewness of the plants and the hunger of nearby birds, the "entire" yield was eaten before it ripened. Most of your grains are self-pollinating, but it looks like maize isn't, so exactly one maize plant will produce nothing. Nyttend (talk) 22:21, 22 August 2017 (UTC)[reply]

Certainly they must be sheltered from birds and bugs. If I were on Mars without potatoes, I need to grow something for my breakfast cereals and Saturday night beers. I need to have grains that multiply A LOT!!!! -- Toytoy (talk) 03:55, 23 August 2017 (UTC)[reply]

Corn is monoecious; the tassel holding the male organs, the ear holding the female organs. The two organs however may not emerge at the same time so for a small garden staggering planting by a week or two will help with pollination. Also Self-incompatibility in plants can be an issue for about half of flowering plants but I'm not sure for these grains. Of 19 (talk) 22:53, 22 August 2017 (UTC)[reply]

It's not clear whether the OP is literally planting just one seed of each item. In any case, you're right that maize can self-pollinate (or be induced to do so) but any significant time gap between emergence of the male parts (tassel) and emergence of the female parts (silk) can result in little or no yield. ←Baseball Bugs ^{What's up, Doc?} carrots→ 08:00, 23 August 2017 (UTC)[reply]

It is well known among gardeners that maize has to be planted in substantial patches (say 2 meters across at minimum) in order to pollinate reliably. If you just plant one or two plants, you probably won't get any ears. Looie496 (talk) 14:19, 23 August 2017 (UTC)[reply]

So the question remains, how many maize seeds does the OP intend to plant? ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:08, 23 August 2017 (UTC)[reply]

To answer this, i would just divide the maximum yield by the standard weight of seeds needed

wheat for instance can yield more than 1O t for a seed dose of 125 kg, meaning you can expect ~100 grains par seed with good but not exceptional earth condition (i guess not all seed contribute to yield, some die in the process)

I leave it to you to find both maximum yield and standard seed dose for different plants using your favorite search tool.

Gem fr (talk) 12:14, 23 August 2017 (UTC)[reply]

Biological maladaptions of living too far away from homeland

When light-skinned people migrate to Africa near the equator, their skin is probably going to be overwhelmed by the sun and get skin cancer. Likewise, when dark-skinned people migrate to the polar region, they may develop rickets. African sickle cell heterozygotes are protected from malaria. Are there others? With the advent of fast pace movement from one continent to another, how can humans adapt to the environment? If a population of light-skinned settlers move to the equator, will their descendants develop dark skin? 140.254.70.33 (talk) 22:56, 22 August 2017 (UTC)[reply]

Unless the cancer or rickets reduces a group's ability to have otherwise healthy children who will also go on to breed at a rate similar to the rest of the local population there is no pressure on the group (forgetting silly race based violence) to lose their seemingly negative traits. So with modern medicine, the settler's offspring will continue to have light coloured skin.

Humans can adapt to these challenges with clothing and sun-screen and vitamin fortified foods. In the distant future we may be able to edit our DNA to change our skin colour or disease resistance, or anything really so long as the change doesn't kill us. Of 19 (talk) 23:13, 22 August 2017 (UTC)[reply]

"How can humans adapt to the environment?"

We don't have to. In general, we adapt the natural environment to us. There is some cold and heat adaptations and high-altitude adaptation, that tunes our body to the environment though.--B8-tome (talk) 23:31, 22 August 2017 (UTC)[reply]

Genetic studies have shown that skin color does adapt over the course of thousands of years. Dark skin is the base state that evolved in our African ancestors, but in populations that move far from the equator, skin color gradually lightens over generations. There are also other interesting features relating to this. For example, people who live in Africa experience minimal changes in the length of the light and dark parts of the day, even at the extreme north and south ends of the continent, but people who live on other continents experience much larger changes. It would be interesting to know what effect this has had on the genes that underlie biological rhythms, particularly those involved in sleep. Looie496 (talk) 14:15, 23 August 2017 (UTC)[reply]

As for other "geographic" traits:

There are high altitude adaptations, such as being short with large lungs, more red blood cells, etc. I suspect they would be fine at sea level, though those adapted to live at sea level would not survive at high altitudes.

There are adaptations to extreme cold, such as having a short, pudgy body. Handling extreme heat might be difficult for such people. Conversely, a tall, thin body is best in hot climates, but could be deadly in the cold. See ectomorph and endomorph. StuRat (talk) 20:31, 23 August 2017 (UTC)[reply]

You do know that the article you cited is about a well-debunked bit of pseudoscience, right? --Jayron 32 16:26, 25 August 2017 (UTC)[reply]

Lactose-tolerance has developed in areas where dairy farming is most practical. I can't see this being a problem to those who move away, though. Lactose-intolerance, on the other hand, could be a problem for those moving into a dairy farming area, at least before we knew how to deal with it (lactose-free milk, etc.). StuRat (talk) 20:14, 23 August 2017 (UTC)[reply]