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August 31

Winds picking up at night

I know the winds are powered by the sun, therefore the winds can only pick up during the day as the presence of the sun add energy (heat) to the air, causing pressure differences and triggering the wind. But sometimes the winds can pick up in the middle of the night, even though the sun has long been set. How can the winds pick up significantly at night? Maybe the winds must've powered by something other than the sun, but by what? PlanetStar 05:57, 31 August 2017 (UTC)

See Wind#Causes of wind. The Sun heats the atmosphere and ground during the day and the ground temperature sinks during the night as the ground loses more heat to the atmosphere. Large-scale air pressure differences cannot equalize immediately and are the cause of Prevailing winds, including local gusts that arise both by day and by night. All wind is powered by the Sun. Blooteuth (talk) 10:05, 31 August 2017 (UTC)

Bucket-spoon for cement

The masons working with trowels use a tool that is shaped like a small bucket (or a big ladle) and having a handle - for taking the cement close to where they work. In Romanian it's called "cancioc" but I can't find the translation. Can anyone help me to find it's name please? Here's some photos with this tool.

• http://www.alamy.com/stock-photo-old-used-hand-bucket-for-trowel-isolated-on-white-61690157.html
• https://www.emag.ro/cancioc-kubala-profesional-handgrip-1531/pd/D01Y0DBBM/
• https://www.neomathaus.ro/scule-constructii/401-cancioc-maner-lemn-18710567.html

Thanks. —  Ark25  (talk) 08:24, 31 August 2017 (UTC)

Initial googled dictionary searches merely suggest "ladle". In the UK I've not seen such a tool used, and various British online catalogues of bricklaying tools do not include one, so its use may be a non-UK practice, with no English-language term specific to this builder's tool.

British builders sometimes use a different tool for carrying small quantities of mortar/cement, a flat square plate with a vertical handle underneath called a hawk: this is held in one hand/arm with a small quantity of mortar piled on the plate, while the other hand uses the trowel (c.f. Hamlet's remark "I am but mad north-north-west. When the wind is southerly, I know a hawk from a handsaw.") Another name for this tool was a mortar board, but as the link suggests, this name has now mostly transferred to the academic headgear originally so-named as a joke. {The poster formerly known as 87.81.230.195} 90.204.180.96 (talk) 11:19, 31 August 2017 (UTC)

• It's a mortar ladle.[1] More Eastern European than British. Used with wetter mortars than British bricklaying mortar, such as those for rendering. Some have a four foot long wooden handle, so that they can be used as a dipper for lifting mortar up onto staging. In Britain we'd use a bucket, and an assistant to carry it up. Andy Dingley (talk) 12:35, 31 August 2017 (UTC)

The version with the long wooden handle is a brick hod, used for carrying bricks or mortar. Dolphin (t) 13:27, 31 August 2017 (UTC)

A hod is quite a different beast. They aren't waterproof, for one thing (no one wants a rain-filled hod). Andy Dingley (talk) 13:34, 31 August 2017 (UTC)

Thanks for all the answers. Searching for "mortar ladle" on Google images returns images with "Mortar and pestle". Only found one image - named "Construction steel ladle". Does anyone know the name of this tool in any other languages? And maybe there is a Wikipedia article on some other language about this tool? —  Ark25  (talk) 14:05, 31 August 2017 (UTC)

2:56 Have you heard, it's in the stars! See Big Dipper in Ursa Major. [2] [3]. The Western asterism is now known as the "Northern Dipper" (北斗) or the "Seven Stars of the Northern Dipper" (Chinese and Japanese: 北斗七星; pinyin: Běidǒu Qīxīng; Cantonese Yale: Bak¹-dau² Cat¹-sing¹; rōmaji: Hokutō Shichisei; Korean: 북두칠성; romaja: Bukdu Chilseong; Vietnamese: Sao Bắc Đẩu). The personification of the Big Dipper itself is also known as "Doumu" (斗母) in Chinese folk religion and Taoism, and Marici in Buddhism. Blooteuth (talk) 16:47, 31 August 2017 (UTC)

Au contrair User:Blooteuth, surely a dipper is one of these, used for scooping water out of a bigger container such as a barrel. You can get platic ones now. Alansplodge (talk) 17:30, 31 August 2017 (UTC)

Bien sûr, it's what Swedes call a Vannøse and equally useful for water or cement. I looked up "platic" and am no wiser. Blooteuth (talk) 18:41, 31 August 2017 (UTC)

Thanks but "Vannøse" seems to be a word used mostly for frying pan - check the first image in no:Øse - also Google images contain mostly cooking items when searching for "Vannøse". I think I'll ask on russian.stackexchange.com - german.stackexchange.com etc. Too bad there's no Polish SE yet. —  Ark25  (talk) 21:44, 31 August 2017 (UTC)

The water dipper seems to have a longer handle and it's not used for masonry. —  Ark25  (talk) 21:47, 31 August 2017 (UTC)

Feynman Lectures. Exercises. Exercise 12-12 JPG. Lecture 12

. .

...

12-12. An object rests at the base of a frictionless 20° incline 1.00 m long (slant). If the incline is accelerated along the table with an acceleration a = 4.00 m sec ^-2 , how long does it take the object to slide to the top of the slope?

—  R. B. Leighton , Feynman Lectures on Physics. Exercises

I have solved the exercise by means of reference frame connected with the incline . But I would like to know how to solve it in inertial frame of reference, connected with the table. I suppose there is only 2 forces on the object png. But I have no idea, how to find R. We have 2 equations of motion for each axis (3 unknowns) and we cannot use the principle of virtual work because the system is not balanced... Username160611000000 (talk) 15:28, 31 August 2017 (UTC)

I didn't go diving in your bazillion links to see your answer, but I'm thinking that in an inertial frame of reference, the inclined plane is being accelerated upward at 9.80665 m/s^2 and, say, to the right at 4.0 m/s^2. That should give a resultant of 10.59105 m/s^2 that is directed 67.81013 degrees up, 22.18989 degrees right. But the plane itself is tilted right 20 degrees, so the angle is just 2.18989 degrees to the right. Now the object on the plane receives a force that is its mass times the acceleration times the cosine of that; but it receives no force along the plane, so it lags behind according to sin 2.18989 * 10.59105 m/s^2 = 0.4046998 m/s^2. since d = 1/2 (a*t) * t, t = sqrt(2d/a) = 2.223046, I think. Wnt (talk) 03:04, 2 September 2017 (UTC)

I'm thinking that in an inertial frame of reference, the inclined plane is being accelerated upward at 9.80665 m/s^2 and, say, to the right at 4.0 m/s^2 -- Why? According to my image png to the body was at the top of the incline, the incline should be accelerated to the left. In this ex. we have the table (inertial reference frame or lab frame), the incline and the body.Username160611000000 (talk) 10:12, 2 September 2017 (UTC)

Oh, I just went by the text above, and the question didn't say which way the incline inclines. (I have no patience to go through half a dozen links, enable scripts, download what may be pirated textbooks etc. in the process of trying to figure out what the question is) Whichever way, we automatically assume the acceleration is in the direction that would make the object tend to go up it. My mental image was mirror image of yours.

Ok, but then the incline can't be accelerated upward, because the table acts on it with reaction force equal to the weight. Username160611000000 (talk) 03:51, 3 September 2017 (UTC)

dozen links, enable scripts, download what may be pirated textbooks -- When I type jpg or png, these are direct links to the images , no need scripts or to download something (actually all images in browser windows are downloaded. If you will turn off javascripts globally, it's safe). Username160611000000 (talk) 03:55, 3 September 2017 (UTC)

Why was Hurricane Adolph added to the name list?

Sagittarian Milky Way (talk) 16:35, 31 August 2017 (UTC)

Do the articles Tropical cyclone naming and List of retired Pacific hurricane names help? Blooteuth (talk) 16:55, 31 August 2017 (UTC)

I can see why it was retired. Why was it on the name list and for so long? Sagittarian Milky Way (talk) 17:07, 31 August 2017 (UTC)

They try to use a variety of names to avoid picking from one cultural group. Adolph is a common name. Once a storm is given a name, the name is commonly retired to avoid confusion by having a second Hurricane Adolph. Similarly, Katrina has been retired. So has Hugo, and many others. 209.149.113.5 (talk) 17:14, 31 August 2017 (UTC)

It's unreasonable to blame Adolf for Adolph. Blooteuth (talk) 18:48, 31 August 2017 (UTC)

This is more for the language desk, but I note in passing that it seems to be more acceptable in British publications, than in American ones, to anglicize names, for example to write "Adolph Hitler". The one that particularly jangles my nerves is when people refer to "George" Cantor. On the other side, I think Americans are more likely to write "Josef" Stalin, which can be seen as a hyperforeignism, given that "Josef" is not particularly closer to "Ioseb" than "Joseph" is. --Trovatore (talk) 19:28, 31 August 2017 (UTC)

I thought that the question may be wondering why the name of a bad person was used. I assumed that the question was based in intelligence, even though I know that the United States is currently in a wildly fascist frenzy of censoring history. That is very new. Not long ago, liberal groups were fighting all the way to the Supreme Court to allow idiots to say and do things that any reasonable person would find offensive. Therefore, I expect the current spasm to pass as soon as the media has trouble selling advertising with it. 209.149.113.5 (talk) 19:53, 31 August 2017 (UTC)

It's perhaps worth noting what actually happened here. While this isn't explicitly noted anywhere that I noticed if you look at our articles and the sources [4] and understand how things wrong, the name Adolph was I guess used since naming started in 1960. None of the retired ones would suggest Adolph was a replacement. Then in 1995 Ismael was retired due to the damage and deaths and replaced with Israel. When people noticed this (I'm not sure if the name Israel was only decided in 2001, or it had been decided after 1995 but no one paid attention) some people raised concerns over Israel especially since there was a minor risk of "Israel destroys ....." headlines. So Israel was never actually used and replace instead with Ivo which didn't actually do anything. People also noticed Adolph at the same time as it was in the same season and so Adolph was retired after that season [5]. In other words, if Ismael hadn't done all that damage or Israel hadn't be chosen to replace it, there's a fair chance we'd still have Adolph. Nil Einne (talk) 20:21, 31 August 2017 (UTC)

Adolph was retired because it was a major hurricane. All major hurricanes have their names retired. 71.85.51.150 (talk) 00:37, 1 September 2017 (UTC)

Except that all sources presented thus far disagree with you. All including NOAA say that the reason it was retired was due to political reasons, and NOAA and IIRC some other sources (I think this includes at least one of our articles) also say hhurricanes which don't make landfall and don't cause any notable damage nor deaths or other significant impact to human populations (the worst seems to have been closure of two ports to small ships) despite some warnings and fears it would, are not necessarily retired even if they were very strong. If you have other sources which contradict these sources, please present them. To be fair, Adolph was new record for the basin in May since recording begun in the mid 1960s which in itself is not necessarily that significant since there could have been a lot of those, but was also the first category 4 in May since recording begun, so there are reasons it may have been retired even without making landfall (although I'm not sure if records 'in May' are really enough to push them into retiring in the absence of something else). So perhaps there was discussion over whether to do so and the political reasons are what pushed them over the edge. Or maybe they actually decided to retire the name due to the significance of the last Adolph, but then said they did it for political sensitivity reasons. Maybe there are blogs or records which talk about these issues. But none of them have been presented this far and they don't seem to be in our articles so we can only go by what the sources actually say namely that it was retired for political reasons, and with a strong hint the reason this came about was because the plan to name a source Israel drew attention to the issue. Nil Einne (talk) 09:10, 1 September 2017 (UTC)

BTW, I'm not sure what you meant by major hurricane, but if if you simply meant it was a category 4, you may want to take a look at List of Category 4 Pacific hurricanes. It lists 118 hurricanes. Unfortunately it doesn't separate between the eastern and central Pacific although I believe eastern Pacific hurricanes may be more common and in any case I'm fairly sure you'll find other category 4sin the eastern Pacific which weren't retired because they too didn't have any significant impact. As the earlier list shows, the retired names is a lot smaller and not all of these are even category 4 or higher. In fact I'm fairly sure that means you'll find some List of Category 5 Pacific hurricanes which weren't retired. Edit: Actually I just found Hurricane Dora (1999) which was category 4, if you look at Tropical Storm Dora, you'll see the name is still being used.Nil Einne (talk) 10:13, 1 September 2017 (UTC)

Are foot orgasms real?

Do they truly exist? — Preceding unsigned comment added by Uncle dan is home (talk • contribs) 21:53, 31 August 2017 (UTC)

Well, I only found this reference. It appears that foot orgasm does exist for one woman in the world. 50.4.236.254 (talk) 23:31, 31 August 2017 (UTC)

Did'nt Coca Cola build an world empire on that? When we are heated up and very thirsty the drinking of cold sweet water with a fruity taste can get so intense that we may even be stunned and have "involuntary muscular contractions". That looks pretty close to "orgasms". --Kharon (talk) 02:36, 1 September 2017 (UTC)

I'd like whatever Kharon is getting, please. This study describes how to induce in animals an itch paroxysm comparable to that which occurs in man. Blooteuth (talk) 12:33, 1 September 2017 (UTC)

• Guinea worm. μηδείς (talk) 21:36, 1 September 2017 (UTC)

You're definitely not one of the Knights who say Ni in organism. Dmcq (talk) 21:46, 1 September 2017 (UTC)

I just bought reading glasses today. μηδείς (talk) 18:29, 2 September 2017 (UTC)

Whatever floats your boat, some have spontaneous orgasms for neurologic reasons or through intense fantasy ideas, some have them while dancing or kissing... some even have them from pain or the sensation of submission (and others via sadism). I remember some tabloid news mentioning someone having orgasms from being thrown mellons at. For more information: orgasm, BDSM, fetishism. Feet are also quite sensitive, it's possible that for some they are like an erogenous zone. —PaleoNeonate – 21:48, 1 September 2017 (UTC)

• Look at the sensory homunculus. The big toes are next to the genitals, as far as the brain is concerned. Have your lover suck your big toes while stimulating you genitally. Any good hetaira knows this. μηδείς (talk) 18:27, 2 September 2017 (UTC)

Rocket consumption and energy

A curiosity: but the rocket consumption (assuming that the weigth of propellent is very small compared to the weight of the accelerated object and an efficiency of 100%) is proportional to the impulse given to the object or to the kinetic energy given to the object? Because in the first case, for example, with a dobule velocity given the consumption would be double, but in the second case the consumption wolud be four times higher. The more logical solution would be the first (because according to the relativity an object can be considered ever stationary in his frame of reference, so the consumption can only be a function of the time multiplied for the force and even because a force produces a consumption even if doesn't move an object) , but how could this conciliate with the kinetic energy theory? Thanks for answering, I know that this isn't an easy question, Francis. — Preceding unsigned comment added by 82.48.80.146 (talk) 22:03, 31 August 2017 (UTC)

You may be interested in our article on the Oberth effect. -- ToE 22:20, 31 August 2017 (UTC)

By conservation of energy, we can say that the initial potential energy of the rocket (the fuel) equals the final kinetic energy of the payload and propellant, plus the gravitational potential energy, plus energy lost to friction and heat. Typically, the kinetic energy is much greater than the gravitational potential or heat because rockets go very fast. So in a simplified analysis, you need four times the fuel to make your payload go twice as fast. This is true for cars as well: it takes four times as much energy to accelerate a car to 20 m/s compared to 10 m/s (ignoring friction).

The theory of relativity says that the laws of physics are the same in all stationary reference frames. It does not apply to the payload being accelerated in a rocket. If the velocity of an object changes, then the object's reference frame is not stationary. You might be interested in the twin paradox, which is resolved in a similar way.

For rockets going into space, our quadratic approximation is very optimistic, because propellant takes up 85% or more of the mass of real rockets. That means that adding more propellant increases the weight of the rocket significantly, and you need to add more propellant to propel that extra weight, and so on. Your velocity increase from adding propellant diminishes, or phrased another way, you need an exponential amount of propellant to keep increasing the velocity of a rocket. NASA has a good explanation with real numbers. C0617470r (talk) 00:05, 1 September 2017 (UTC)

Our article Tsiolkovsky rocket equation may be of interest. {The poster formerly known as 87.81.230.95} 90.204.180.96 (talk) 16:45, 1 September 2017 (UTC)

Thanks to all of you for answering. But there is one thing that I don't realize. I give you an example: the Earth moves respect to the Sun at a speed of 35 km/s. If I launch a 2 kg object on the Earth in the same direction of orbitation movement at the speed of 5 m/s the object gains respect to the Sun an accretion of kinetic energy of 350.025 Joules (35.005^ equals to 1.225.350.025 respect to the quadratic of 35.000 that is 1.225.000.000) with a consumption of energy of only 25 Joules. And in other, if for example is given to a 100 kg object a thrust (in the opposite direction of gravitational force) of only 50 kg of force, there is howewer a propellent consumption even if there isn't any movement. How colud be this explanable without thinking to the more logical solution: the consumption is proportional to the force applied, not to the kinetic energy? Francis — Preceding unsigned comment added by 87.3.17.123 (talk) 22:33, 1 September 2017 (UTC)

What may be bothering you is that the change in kinetic energy of an individual object is not invariant under Galilean relativity, whereas change in momentum is invariant. Consider a 2 kg object R (for Rocket) which changes its velocity by 1 m/s. In the reference frame where the object was initially at rest, its momentum changed from 0 kg⋅m/s to 2 kg⋅m/s, and it's kinetic energy changed from 0 J to 1 J. But in the reference frame where it was initially traveling at 1000 m/s and finally traveling at 1001 m/s, its momentum changed from 2000 kg⋅m/s to 2002 kg⋅m/s (the same Δ of 2 kg⋅m/s), and its kinetic energy changed from 1,000,000 J to 1,002,001 J. At fist glance it may not make sense that in one frame the object gained only 1 J, but in another it gained 2001 J. How can that be consistent with conservation of energy? Part of the answer is Newton's first law. Your object wouldn't have changed velocity unless it was acted upon by a force, and you must consider the entire system, including those other objects from which the force originated, in order to apply conservation of energy. So, lets give that object a reason for changing velocity.

Consider a second object with an attached spring, together called F (for Fuel), total mass (object and spring) 2 kg, such that if F is connected to R with the spring compressed between them, and then the connection is broken, the two objects will fly apart at a relative velocity of 2 m/s. In the frame where the objects were initially at rest, after the catch is released each is traveling at 1 m/s, so the kinetic energy of each object went from 0 J to 1 J, and that 2 J total increase came from the potential energy of the spring. Now consider the frame where they are initially traveling at 1000 m/s, with final velocities of 1001 m/s and 999 m/s for R and F, respectively. R's kinetic energy changed from 1,000,000 J to 1,002,001 J, the same 2001 J increase we calculated earlier, and F's kinetic energy change from 1,000,000 J to 998,001, a loss of 1999 J. Combining the two, we have that same 2 J increase in total kinetic energy. Energy is conserved.

Now look at it just a little differently and consider two of these 2 kg rockets which started from rest, one loaded with only one unit of fuel and the other loaded with a huge amount of fuel so that after expending all but one unit of fuel it is traveling at 1000 m/s. Then they both expend their last unit of fuel, and the first rocket accelerates from 0 m/s to 1 m/s gaining 1 J, and the second accelerates from 1000 m/s to 1001 m/s gaining 2001 J. If you think if unfair that the same unit of fuel did so much more work on the moving rocket than on the stationary one, remember that half of the huge amount of fuel loaded onto the second rocket was spent getting that last unit of fuel up to speed, and it is giving up some of its kinetic energy which was gained at that fuel's expense. You could even say that the last bit of fuel was supercharged beyond its 2 J of potential energy with the 1,000,000 J of kinetic it had as well, though it only manages to give up 0.2% of that. But change your reference frame and that supercharging seems to disappear. -- ToE 23:44, 1 September 2017 (UTC)

Is global sea level drop due to heavy rainfall events measurable?

We can read here:

"Harvey has unloaded 24.5 trillion gallons of water on Texas and Louisiana"

If all that water is on land at the same time, then that amounts to roughly 0.3 mm sea level drop. Now it may be that a lot of the water has flowed back to sea by the time the last part of the 24.5 trillion gallons fell from the sky, so the sea level drop may be a bit less. Count Iblis (talk) 22:35, 31 August 2017 (UTC)

I concur with your calculation: 24.5E12 us_gal / 360E6 km^2 ^Ocean = 0.26 mm. -- ToE 23:16, 31 August 2017 (UTC)

Do you mean measurable as in within the level of significant figures or greater than noise level? Or do you mean measurable as in "I can see that distance on a ruler"? The answer is different depending on which you are asking about. --Jayron32 01:26, 1 September 2017 (UTC)

No, you have to include all water in the atmosphere, all water in the underground, in rivers and inland seas, all water bound by Biomass and all water bound by ice, because all that is part of the Water cycle, as is the Harvey rain volume and the oceans you picked out. Additionally water is not a static mass and therefor you should include Ocean currents, moon position, local Atmospheric pressure(weather) and last not least the local Physical geodesy to conclude a sea level rise or drop. --Kharon (talk) 02:11, 1 September 2017 (UTC)

It might be worth considering that 'Harvey' has not been the only catastrophic event in the world recently. Richard Avery (talk) 07:27, 1 September 2017 (UTC)

Just to be explicit, that is not a discernible change. The precision of annual, global average sea level is 5-10 mm. Local sea level can be discerned more precisely when the weather is very calm, but most of the time it is not. Dragons flight (talk) 08:44, 1 September 2017 (UTC)

The first bit sounds reasonable (not to say unimpressive!), any ref for that? SemanticMantis (talk) 14:17, 1 September 2017 (UTC)

It would seem trivial to construct a pseudoharbour to emulate "Very calm" weather. All the best: Rich Farmbrough, 12:52, 2 September 2017 (UTC).

If something useful seems trivial, but is not widely done, there are is usually at least one explanation. It is not as trivial as it seems ;-). In this case, you would need a body of water that communicates with the sea (so as to reflect the actual sea level), but which excludes not only surface waves, but also dynamic changes in pressure (e.g. those caused by waves colliding against an underwater inlet). --Stephan Schulz (talk) 13:55, 2 September 2017 (UTC)

It's a big assumption that the amount of water that fell was an addition to the amount in the air anyway or all came from the sea. And shouldn't the sea level actually rise nearby if the land is heavier with water on it though it might fall elsewhere in the world? If so that makes the concept of sea-level just a little murkier.Dmcq (talk) 17:17, 1 September 2017 (UTC)

I see, so an order of magnitude below what can be detected. Count Iblis (talk) 05:28, 2 September 2017 (UTC)

• Also, consider that the land itself is plastic, so when it is depressed in one area, it may rise in another. Think about what happens when you press into dough or clay with your hand. The handprint is depressed, but the rest of the soft matter rises around the depression. If it were sea-water causing this to happen, the effect would pretty much balance out. μηδείς (talk) 18:23, 2 September 2017 (UTC)

September 1

Affective "clipping"

In signal processing, clipping is the phenomenon where a signal goes undetected because it is in phase with another signal that already drives the detector to its limit. What is the equivalent phenomenon in affective psychology called? An example of it would be that a controlled study exposes the experimental subjects to stimuli x+y and the control subjects to only y, but their reaction to y saturates all the affective dimensions that x affects, and thus both groups respond the same, and the researchers incorrectly conclude that the subject population is insensitive to x (especially likely when that population's sensitivity to y has been underestimated, or when the exposure to y was inadvertent). What is this called? Also, is it known whether such a phenomenon is relevant to the management of autism-spectrum disorders? NeonMerlin 06:23, 1 September 2017 (UTC)

The term 'Clipping' is normally reserved for analogue systems. Yet, this sound like Information overload. Something that people diagnosed has having asperger's appear to suffer from. It is not so much slow mental processing, rather the reverse, where they are aware of all the stimulus from their senses (and past memories) all at once. A gift (and sometimes burden) that us lesser mortals don't posses.Aspro (talk) 11:58, 1 September 2017 (UTC)

In my experience this sort of thing is usually referred to as a ceiling effect (statistics). Looie496 (talk) 13:37, 1 September 2017 (UTC)

I would say in your example that the signal of the response to x is "swamped" by a larger response in "y". Here [6] is a book that discusses signal swamping in regards to human hearing. Hannington_transmitting_station uses "swamped" in the same context for radio signals. I can't find any good coverage of this concept on WP. Saturation has some leads, but nothing quite like what you're describing of "I can't see the response this one signal because there is another stronger signal that is masking it". The ceiling effect mentioned by Looie can also cause this kind of swamping, but in my scientific experience "the signal was swamped" is a more general phrase that is widely understood. SemanticMantis (talk) 14:15, 1 September 2017 (UTC)

They are different things. A ceiling effect occurs when a measured parameter "hits the ceiling" in some sense and therefore is unable to rise any more. It doesn't really have anything directly to do with the size of the signal. Looie496 (talk) 14:40, 1 September 2017 (UTC)

Clipping is a phenomenon in an analog electronic amplifier or signal processing system where the amplitude of the input signal reaches the threshold of what one of the system's devices can handle before gross distortion sets in. All systems distort input signals in that they don't faithfully follow the variations in the input signal, but here we are talking about a situation where the device (transistor, tube) saturates so that it can no longer produce output that increases with the increasing level of signal and cuts off the highest values, grossly distorting the signal. The device literally clips off the tops of the waveform, hence the name "clipping". The phenomenon has nothing to do with two signals being in phase, and therefore the analogy used by the OP is not appropriate. Akld guy (talk) 20:56, 1 September 2017 (UTC)

I think it does. If we add a small square wave that is in phase to a square wave that is only just clipped we will see no change If it is perfectly out of phase we will see maximal change.

All the best: Rich Farmbrough, 12:56, 2 September 2017 (UTC).

That is a special case of clipping in which the clipping (of the larger signal) has already occurred. So it's not a definition as claimed by the OP. Akld guy (talk) 13:24, 2 September 2017 (UTC)

September 2

Spider identification

Is there a way to identify this small spider I took some shots of? I live in southern California, and a small tan and black hairy spider has been living on my fig tree for at least a month, and I've seen it with a fly in its fangs on a couple of occasions when I caught it while it was feeding. The best shot of it is this one which has a ladybug on the same leaf for perspective of its size, and you can see a tan cross/plus sign on its abdomen. The front of it near the fangs is also a bluish green color. And you can see from this close up how hairy it is.--十八 00:50, 2 September 2017 (UTC)

This is a Jumping spider, family Salticidae. Looks familiar but can't remember the species off the top of my head, will look it up. Dr Dima (talk) 04:02, 2 September 2017 (UTC)

I thought Phidippus californicus but it doesn't look quite right. It's almost certainly Phidippus, but it's up to the BugGuide to figure out what the species is. Sorry. --Dr Dima (talk) 04:21, 2 September 2017 (UTC)

Phidippus putnami looks pretty close, so this may be it. Still, you better ask the BugGuide people to be sure. Dr Dima (talk) 04:26, 2 September 2017 (UTC)

Chicken feet

Chicken feet does not contain muscles. So how does 100 grams of chicken feet contains 17 g protein— Preceding unsigned comment added by RabbitDog (talk • contribs) 03:42, 2 September 2017 (UTC)

"Type I collagen, is present in many forms of connective tissue, and makes up about 25% of the total protein content of the mammalian body". - Fibrous connective tissue. —PaleoNeonate – 03:56, 2 September 2017 (UTC)

Also, don't forget that some protein exits in bones. Aspro (talk) 11:28, 2 September 2017 (UTC)

aircraft airframe material

What is "LA"[7] in the context of aircraft airframe material? Mũeller (talk) 04:59, 2 September 2017 (UTC)

• My (educated) guess is "light alloy" of aluminium. NorthBySouthBaranof (talk) 05:03, 2 September 2017 (UTC)

Adding small amounts of Lanthanum (I have seen 0.9% specified) improves the machinability of Titantium Ti-64 alloy. Adding 1% Lanthanum to A390 Aluminum alloy improves the ductility from 0.7 to 1.8% and increases the tensile strength from 100 to 150 MPa. I have seen these alloys referred to as "A390 La" and "Ti-64 La". But in this case, I think our article on AgustaWestland AW101 has the answer: "The fuselage structure is modular and comprises an aluminium-lithium alloy, designed to be both light and damage-resistant". --Guy Macon (talk) 07:45, 2 September 2017 (UTC)

If it would describe Aluminium-lithium alloy it would read Al-Li or alike. Certainly not "LA". The "LA" honeycomb part is very special tho and searching for that i found the company http://www.lacomposite.com/, which also is an official subcontractor (parts supplier, see SoA)for the company AgustaWestland which manufactures and sells these helicopters. Thus i bet the "LA" only means these parts are from lacomposite. --Kharon (talk) 20:10, 2 September 2017 (UTC)

Interesting. I have never heard of anyone making stringers out of honeycomb, but it could work. --Guy Macon (talk) 11:19, 3 September 2017 (UTC)

Martial arts question

How does breaking power correlate with knockout power? For example, if you can break 4 boards with one strike, how big of an attacker can you knock out with the same strike? How about if you can break 6? 2601:646:8E01:7E0B:4DFA:5B5D:E0B4:4FDF (talk) 10:10, 2 September 2017 (UTC)

Boards are not standardized so as we know how much force/power is needed to break them. But it also does not matter, since even extremely weak people have enough force to break some bones, which kind of knocks any opponent out. --123abcnewnoob (talk) 10:57, 2 September 2017 (UTC)

See Martial arts and Breaking (martial arts). An ability to break things does not correlate well with one-on-one Combat sport where there can be many more effective ways of disabling an opponent than trying to break his bones. See Self-defense#Unarmed. Traditional Japanese martial art schools place little, if any, emphasis on board-breaking. Blooteuth (talk) 12:41, 2 September 2017 (UTC)

Also people "playing on" with broken bones is the stuff of sporting legend. All the best: Rich Farmbrough, 13:38, 2 September 2017 (UTC).

Be aware that most board breaking is with the grain, boards will break fairly easily in this orientation, and are rarely used that way structurally.

A knockout punch - that renders one's opponent unconscious, may be more a combination of placement and timing than extreme force. All the best: Rich Farmbrough, 13:38, 2 September 2017 (UTC).

OK, in this case, let me break this question into parts: (1) Is there a correlation between a person's size and the amount of force needed to knock him/her out? (2) About how much force is needed to knock someone out? (3) How much force is needed to perform a power break on an unpegged stack of 4 standard 1-inch-thick boards, and how much more force is needed if the number of boards is increased to 6? 2601:646:8E01:7E0B:4E8:FDC:7D20:30AB (talk) 08:56, 3 September 2017 (UTC)

Oak or pine? balsa? Maple? Ash? Plywood? Particle board? Sagittarian Milky Way (talk) 09:30, 3 September 2017 (UTC)

1) No. Unfortunately, our relevant article at Chin (combat sports) is pretty useless. Size may play a small role, if only because a larger person has a larger head, but such differences are overwhelmed by technique, timing, etc. In combat sports it's generally accepted that it's extremely difficult to knock someone out who is expecting it and surprisingly easy to knock out someone who is not expecting it. Matt Deres (talk) 02:27, 4 September 2017 (UTC)

Problem solving abilities of scientifically trained/mathematically literate

If you have, for example, extensive training in some science or math, but over the years do not integrate it in your day to day job, would that still make your general problem solving abilities better than average joe's (without science/math training)? Is there some scientific research about how science/math educated people perform outside their direct field of expertise? --123abcnewnoob (talk) 10:55, 2 September 2017 (UTC)

Problem solving requires two distinct types of mental skill, analytical and creative. This article contrasts analytical thinking that predominates in solving closed problems with creative thinking characterized by fluency, flexibility and elaboration in using the imagination in solving open problems. Science/math education purportedly develops left-brain thinking, is more logical and analytical, and is predominantly verbal, possibly with neglect of right-brain thinking that is more holistic and is concerned with feelings and impressionistic relationships. Problem solving skills may be trained as a distinct discipline, see Problem solving. Blooteuth (talk)

Knowing the scientific method would help to solve problems. For example, say you notice you get sick after eating X. Some would just stop there, and assume that X is the cause. But a better way would be to keep a food journal, so you can look back at what you ate and correlate that to when you got sick. You might then find out that you often eat X and Y together, and get sick after. Then, eating them separately would help to determine which, if either, is the true cause. StuRat (talk) 13:09, 2 September 2017 (UTC)

We certainly have Life itself as "benchmark". Have you ever wondered why for example not even one of all our very famous Nobel Memorial Prize in Economic Sciences Laureates managed to make a fortune at the stockmarket or anywere else in financial biz for himself? Economic Science is mostly based on math btw. To the contrary surprisingly many of the richest people actually did not even finish their education, not to speak of becoming an respected scientist afterwards. So no, it does not look like a huge help. --Kharon (talk) 20:29, 2 September 2017 (UTC)

@Kharon: Economists do not get assessed by speculating in the stock market. Some are not even allowed to do that (all those who work for central banks, for example). And it's completely false that famous economic laureates have never become rich in the financial industry. Your answer has so many issues that it's funny. Hofhof (talk) 06:13, 3 September 2017 (UTC)

Hydride generation

I need some help to design an experiment. I want to synthesise mercury(II) hydride and isolate it so that I can study it. I'm considering using a metathesis reaction between mercury(II) chloride and lithium tetrahydroaluminate. I think that this can be accomplished by using liquid dimethyl ether as a solvent, flushing the system with nitrogen, and conducting the experiment in a dark room. What equipment would I need for this experiment, and how would I liquify the solvent, and control the temperature of the reaction so that it does not exceed -125°C. How would I isolate the product, while minimising decomposition? Plasmic Physics (talk) 11:06, 2 September 2017 (UTC)

Postscriptum: Does the cold version of a heating mantle exist, or can I only use a cold-finger or reflux condenser cooled with liquid isobutane? Plasmic Physics (talk) 00:59, 3 September 2017 (UTC)

Regarding some usefull properties of mercury(II) dihydride, it is a nonpolar solute with relativistically supressed electron deficiency, meaning it does not readily form stable solvate adducts. I believe that its solubility should be close to the chloride reagent, hence I believe that I must use the LTHA in excess to ensure that chloride reagent contamination can be minimised in the final product. Perhaps I could attempt crystalisation by evaporation the solvent by vacuum. Plasmic Physics (talk) 13:22, 2 September 2017 (UTC)

I'm not sure we can field a team of chemists good enough to answer this one. Certainly I am not on that team. I'm not really supposed to be giving advice here, but my feeling is that when you ask about basic equipment for cryogenic reactions, you might want to steer clear of messing around with an unstable compound of a toxic metal, especially if by lithium tetrahydroaluminate you mean lithium aluminum hydride, which tends to spontaneously explode and hurt chemists, then dissolving that in an ether. And, well, how do you keep the mercury vapor safely contained while allowing liquid nitrogen (which I assume you need somewhere in this apparatus) to safely vent? I don't have any idea if HgH2, if released in an uncontrolled way, decomposes rapidly enough to have a toxicity no more than ordinary mercury vapor or if it can interact with organic compounds to create things with more methylmercury like effects.

If you're serious about this at all though, I think you should look up the people who have actually published about doing this in an argon or krypton matrix and ask them why they did things the way they did. I'm not sure if there's anyone else who could begin to answer this question properly. Wnt (talk) 01:48, 3 September 2017 (UTC)

I could negate the danger of the LTHA by using a sparged solution instead of the pure compound. I can destroy any mercury hydride vapour present in the flue gas by UV-photolysing it to mercury vapour, and running it through a rudamentary wet scrubber containing dilute nitric acid. See, you've already helped. Thanks. I do hope that there is a practical chemist here somewhere though. Plasmic Physics (talk) 03:39, 3 September 2017 (UTC)

If other users are abstaining from giving advice because of safety concerns, then they are actually doing me a disfavour, as come 2018, I will go ahead with this experiment, with or without their advice. If the safety issues are as severve as you suggest then, I think we can both agree that I would be better off prepared than not. In the very least, abstaining users could give advice on what research to conduct to gain a better understanding of mitigating the risks involved, or on proper technique; perhaps I coul practise on less risky experiments to gain experience, etc. Plasmic Physics (talk) 04:46, 3 September 2017 (UTC)

As someone who has been here as long as you have, you should well know that we cannot give actual health advice, and you are 100% on your own regarding any risk in the real world regarding what is said here. The premise and tone of this above comment is highly inappropriate. DMacks (talk) 13:12, 3 September 2017 (UTC)

Calm your nerves, I was just being frank. I did not ask for health advice, I made a suggestion on how to give 'health advice' without actually directly giving health advice. Furthermore, I actually acknowledged that I am completely fine with singly taking on any associated risks. Plasmic Physics (talk) 05:33, 4 September 2017 (UTC)

Let me make it absolutely clear to any interested party - the pripmary aim of this query does not include obtaining health & safety information, but does not exclude its provision. Plasmic Physics (talk) 08:22, 4 September 2017 (UTC)

Chemistry questions.

1. Water has a higher density as a liquid than it is as a solid. What other substances are like that?

Some answers that I've heard are paraffin, bismuth, and hydrogen peroxide, but can't verify. On Wikipedia, paraffin is a solid that is what candles are made of, but I couldn't find the density as a liquid. Same with density of bismuth as a liquid. And can't find the density of hydrogen peroxide as a solid either.

2. When magnesium sulfate is put in water, water turn hot. When ammonium nitrate is put in water, water turns cold. What's the mechanism of how you can tell whether reaction will turn hot or cold? For ionic compounds at least. Is the answer also the same for covalent compounds?

3. What's a better drainer to use acid-based or base-based?

Acids dissolve greases, proteins, and carbohydrates.

Bases dissolves greases, proteins, and fats and oils.

Most drainers like Drano are base-based, I guess, because bases dissolve a wilder range of substances. Thanks. 12.130.157.65 (talk) 12:21, 2 September 2017 (UTC).

Didn't you answer your own question just now for no. 1? Regarding no. 2, whether ther overall reaction is endo- or exo-thermic depends on the bond-energies involved. Electron-deficient cations tend to readily undergo hydration during dissolution. Hydration typically releases energy. In your example, the magnesium cation is very much electron deficient, being able to accept four water molecules to form the tetraaquo complex, whereas the ammonium cation is electronically satieted, not being able to accept any water molecules. Ammonium can however, form hydrogen bonds with water, but that does not release energy in the same way as hydration does. Regarding no. 3, in my opinion, more often than not, a block will contain more fat/oil than other carbohydrates, so I would recommend a base-based drain cleaning agent for general use. Plasmic Physics (talk) 13:10, 2 September 2017 (UTC)

Well, all cations are electron-deficient. Okay, so what I'm absorbing here is that it has to do with whether the cation is a single atom cation or a polyatomic cation? Thanks. 12.130.157.65 (talk) 13:53, 2 September 2017 (UTC).

I'm using the term 'electron deficient' in the valence-theory sense, not in the general sense. In ammonium, nitrogen has a full octet, and so does bond to other electron rich species, it is just attracted to the anion. So, it has nothing to do with being monoatomic or polyatomic. Point in case: the polyatomic trichlorocobalt(3+) ion is electron deficient, and bonds to ammonia like the monoatomic magnesium(2+) ion bonds to water. Plasmic Physics (talk) 21:34, 2 September 2017 (UTC)

Regarding drains containing more fat/oil than carbs, but that is not necessarily a complete list, of what acids and bases dissolve. There could be more stuff that acids dissolve that bases do not, that I did not include, that could give acids an overall edge than bases. Shrug. 12.130.157.65 (talk) 14:14, 2 September 2017 (UTC).

Basic drain cleaners work better than acidic ones because drains get a lot more acidic solutions run down them under ordinary working conditions. Therefore the buildup will be resistant to acids, but not so much to bases. Abductive (reasoning) 18:30, 2 September 2017 (UTC)

Do acids damage metal pipes more? Sagittarian Milky Way (talk) 15:56, 3 September 2017 (UTC)

2) See endothermic and exothermic reactions.

3) Aren't greases essentially the same as fats and oils ? StuRat (talk) 13:13, 2 September 2017 (UTC)

"Wax expands considerably when it melts and this allows its use in wax element thermostats for industrial, domestic and, particularly, automobile purposes." Paraffin wax All the best: Rich Farmbrough, 13:47, 2 September 2017 (UTC).

For (1), this happens when some part of the structure of the solid is broken to allow denser packing as a liquid. Among the elements, this happens for Ga, Ge, Sb, and Bi, for instance. (Source: Greenwood and Earnshaw 2nd edition, p. 222.) Double sharp (talk) 15:43, 2 September 2017 (UTC)

Heliosphere Moved from Math Desk.

(Moved from Math Desk.)

I don't quite understand why the solar wind interacts with the interstellar medium at all. Aren't the particles such low density that they would just pass by each other without any effect on the other ? StuRat (talk) 02:00, 2 September 2017 (UTC)

The net effect of the charged particles supposedly constitutes a significant enough magnetic field to cause large-scale interactions even though direct collisions between particles may be "relatively" rare. See Interplanetary magnetic field. — Preceding unsigned comment added by 73.232.241.1 (talk) 02:35, 2 September 2017 (UTC)

Perhaps what's missing are the elements of time and distance. That is, that while they don't interact on a human scale, over the course of a billion years and one light year, they do ? StuRat (talk) 13:18, 2 September 2017 (UTC)

I guess it's a matter of scale. The density is low but it's not zero. Consequently the mean free path of particles is large, but not infinite. The mean free path sets the "thickness" of the transition or interaction layer. I'm not familiar with the conditions around the heliopause, but we might take the value for "extreme vacuum" from the table in article mean free path as indicative (for charged particles in a plasma, the interaction should be stronger, hence the mean free path shorter). Even increasing generously from 10⁵ km to 10⁶ km or more, that is still small compared to the radius of the heliopause, which is some 10¹⁰ km. That is a rather sharp transition. --Wrongfilter (talk) 14:13, 2 September 2017 (UTC)

No, at densities prevalent in the solar wind and in interstellar medium the collisions are so rare that they can be neglected (at least at the scales of about 100 au). So, the plasma is effectively collisionless. However, because charged particles can interact over long distances electromagnetically, plasma behaves more like a liquid than like a cloud of particles. You can look here. Ruslik_Zero 16:59, 2 September 2017 (UTC)

Thrown off by the previous question into thinking that this was WP:RDS? -- ToE 02:49, 2 September 2017 (UTC)

Lol, got so caught up with the whole alien discussion, I hadn't even noticed myself!73.232.241.1 (talk) 03:05, 2 September 2017 (UTC)

Yep, that's exactly what happened. StuRat (talk) 13:16, 2 September 2017 (UTC)

Why do the solar wind and interstellar medium remain charged ?

That is, if they contain positive and negative ions and particles, why don't they attract each other and form neutral particles, given billions of years ? StuRat (talk) 17:06, 2 September 2017 (UTC)

Take a look at Interstellar medium, in particular at the section on heating and cooling. To a first approximation, much of the ISM is very hot (and tenuous), so it forms a plasma. For a nucleus to catch an electron, not only do they need to meet, there needs to be a way for the pair to to get rid of the extra kinetic energy. In the very low pressure environment, there are not many ways to do that. --Stephan Schulz (talk) 18:18, 2 September 2017 (UTC)

Solar wind ions move hundreds (thousands?) of km/s (at least around Earth), they don't have billions of years. Sagittarian Milky Way (talk) 18:32, 2 September 2017 (UTC)

According to Heliosphere, they are emitted from the sun at 400 km/s, and at the termination shock slow to below the speed of sound, which is, under these circumstances, 100 km/s (And boy, Speed of sound really goes over my head ;-). So it takes them on the order of magnitude of one year to go from the sun to the heliopause, unless I dropped a zero or five somewhere... --Stephan Schulz (talk) 18:44, 2 September 2017 (UTC)

How about just considering solar wind particles with opposing charges, then. Presumably they are moving at much the same velocity, so that their relative velocity is low. Why don't they all "hook up" ? StuRat (talk) 01:31, 3 September 2017 (UTC)

They are still at high temperature, and at high temperatures, gases are mostly ionised. The kinetic energy of the electrons is high compared to the binding energy of the outer electrons, so chances for combination are low and chances for knocking off electrons are high. See Plasma (physics), in particular Plasma_(physics)#Temperatures. --Stephan Schulz (talk) 08:23, 3 September 2017 (UTC)

Speed of the solar wind is about 500 km/s, temperature is about 100 kK. This means that the thermal velocity of the protons is about 40 km/s, much less than the bulk motion, and that of the electrons about 1700 km/s, much more than the bulk motion. The protons mostly move with the flow, the electrons have a random motion on top of that, constrained by the magnetic field. Of course they can recombine and they will, emitting a photon, in some fraction of the cases when a proton and an electron meet. However, when one of the few neutral atoms collides with an electron or a proton, there's a large probability it will get ionised again.

Not mentioned by anyone yet is the ultraviolet radiation. The UV from the sun and other stars (and energetic non-stellar objects) is quite intense and responsible for keeping almost the entire universe ionised. It's the UV that keeps the ISM hot. PiusImpavidus (talk) 09:07, 3 September 2017 (UTC)

Why do the solar wind and interstellar medium repel each other ?

For this to happen, they would need to have the same electric charge, but aren't any two particles passing near each other just as likely to have opposing charges and attract each other ? StuRat (talk) 17:08, 2 September 2017 (UTC)

Well, your assumption about required charges is wrong - two baseballs that hit each other in there are quite able to bounce back, even if both are uncharged. But I think the misunderstanding goes deeper. Why should the solar wind and the ISM repel each other? The solar wind eventually merges into the interstellar medium. --Stephan Schulz (talk) 18:26, 2 September 2017 (UTC)

Our heliosphere article states: "The solar wind flows outward from the Sun until encountering the termination shock, where motion slows abruptly". That sure sounds like the two are repelling each other. If they simply passed through each other, why does the solar wind slow so much ? Could it be that the charged particles in the solar wind do merge with those in the interstellar medium, and take on the resultant of the two motions ? StuRat (talk) 19:08, 2 September 2017 (UTC)

What is "interstellar medium"? There is maybe a single atom per cubic mile in the space inbetween star systems. Want to calculate how many atoms you could send flying thu that cubic mile befor a single one might have a chance to come close to the one waiting there? --Kharon (talk) 20:43, 2 September 2017 (UTC)

Interstellar medium says densities don't go below 1 ion per 10 liters. Sagittarian Milky Way (talk) 20:49, 2 September 2017 (UTC)

An interstellar medium is a medium who took an astral plane beyond menopause and all the way to the heliopause. StuRat (talk) 20:58, 3 September 2017 (UTC)

They could claim it has 2 little unicorns per 10 liters. Its all assumptions. Maybe educated calculations, maybe bluntly made up to fit another assumption or theory. Noone and nothing was or is out there to measure the factual. Same with the Dark matter. They could claim its from the apples the unicorns drop. If they would stick together, agreeing to that, noone could prove they bullshit us all. --Kharon (talk) 21:06, 2 September 2017 (UTC)

And your atom per cubic mile was somehow based on better information? --Trovatore (talk) 21:19, 2 September 2017 (UTC)

The fundamental forces of individual particles generate interference patterns which form tensor fields which can most certainly cause relatively strong collisionless reactions. And geometrically speaking, it's fairly easy to work out the average density of particle distributions, so your argument that it is somehow "all assumptions" is a little weak. And just take a look at the densities that are directly measured here within the Earth's orbital radius[8], 150 million kilometers from the Sun. Particle speeds of over 500 kilometers per second and densities of at least 1 to 10 particles per cubic centimeter. So yes, the resulting force fields can be substantial.73.232.241.1 (talk) 23:32, 2 September 2017 (UTC)

I can't answer Stu's question, per se, but I can provide a timely xkcd. --Trovatore (talk) 21:33, 2 September 2017 (UTC)

Also, the force of attraction must exceed the relative momentum difference between the particles in order for them to interact directly to any appreciable degree. Consider a hypothetical situation where only two particles exist in the universe, separated by billions of light-years of distance. If they are initially moving away from eachother at a separation velocity above some threshold, then the shared gravitation between them is overcome and they will remain separated until the gravitational force finally has time to catch up (and in this case, it eventually will, because gravity is an acceleration (quadratic) while velocity is linear).73.232.241.1 (talk) 22:58, 2 September 2017 (UTC)

I don't think that the interstellar medium should be able to repel particles with a given charge inward from all directions. This isn't one of Maxwell's equations per se, but a charge on all sides should cancel out somehow. (shell theorem? There must be something more general) But it should be possible for opposite charges to stop moving charged particles - I mean, proton meets electron, they settle down, make beautiful babies together. Cathode rays travel in vacuum, but not in air - charged particles can interact with neutral atoms. As for the specifics, there's some general prose here but I can't say I really understand the situation as yet. Many astronomical objects, Sun included, are tremendously hot and are surrounded by big bubbles of hot outward moving gas, whose pressure gets weaker as it expands, but why that slows so abruptly to form a termination shock is another question. Wnt (talk) 01:17, 3 September 2017 (UTC)

The termination shock is just an acoustic shock, where speed suddenly goes down and pressure up. Compare it to a jet engine in a supersonic flow. Right ahead of the engine inlet, the air flows in at supersonic speed. In the inlet you get a shock, where the flow suddenly slows to subsonic, whilst the pressure increases. (Actually, jet engines use a number of oblique shocks.) It's not much different with the termination shock of the solar wind. Given the Mach number of the solar wind, there is a distance from the sun where pressures and densities are just right to form the shock, and that's where the shock forms. The only difference is that the solar wind is a magnetised plasma, so part of the pressure is gas pressure, part is magnetic pressure. PiusImpavidus (talk) 09:49, 3 September 2017 (UTC)

See http://en.wiki.x.io/wiki/Heliosphere#Termination_shock. According to this lecture the termination shock is expected to be a fast mode magnetohydrodynamic shock wave. A "reverse shock" directed toward the sun. Oh, I also found this paper titled The Heliospheric Termination Shock on arxiv.org. I haven't looked at this article yet. It was later published here in a book titled Physics of Collisionless Shocks. Also, this article on collisionless shock waves seems accessible (I've only skimmed it). --Modocc (talk) 15:12, 3 September 2017 (UTC)

September 3

Human hair color

Reading human hair color, a question popped into my head. Is black eumelanin biochemically unstable or incomplete dominance? I mean, one parent may have only black hair, and that black hair is bred true because of ancestry. Another parent may have brown hair, but carries the blonde allele; and the person may have other genes that control the amount of brown eumelanin throughout one's age. When they cross, the offspring may have some shade of brown or even blonde hair, maybe even black hair. I am not sure if the black eumelanin is actually more biochemically unstable so the brown version takes over or if there is some kind of incomplete dominance going on. 50.4.236.254 (talk) 15:32, 3 September 2017 (UTC)

What? μηδείς (talk) 16:38, 3 September 2017 (UTC)

Articles: Human hair color#Genetics and biochemistry of hair color, [9], Genetics of Blonde Hair.

Hair color is determined by the amount of eumelanin (which is dark brown) and pheomelanin (which is reddish). There are numerous examples of two brown-haired parents having red-haired offspring, which would suggest that it is determined by a recessive allele, but but Neel (1943) found many offspring of two red-haired parents to have non-red hair. Reed (1952) reviewed the various hypotheses that had been proposed, including that red was recessive; that red was dominant; that red was dominant, but could be masked by brown; or that red was usually recessive but could sometimes be dominant. This report concludes that there are a large number of different MC1R (Melanocortin 1 Receptor) alleles, and other genes affecting the amount of brown pigment that plays a major role in determining hair color. Blooteuth (talk) 17:38, 3 September 2017 (UTC)

What is this plant called?

Mystery plant

Anybody know the name? I would really like to add it to my garden!Wikigirl97 (talk) 20:35, 3 September 2017 (UTC)

Kniphofia, I think. {The poster formerly known as 87.81.230.195} 90.204.180.96 (talk) 20:44, 3 September 2017 (UTC)

I would have called it the "burning match plant". But, much like a burnt match, it looks ugly when used up (after the petals fall off), so be sure you know what you're getting into. StuRat (talk) 20:53, 3 September 2017 (UTC)

Another common name is "Red Hot Poker". Dbfirs 23:49, 3 September 2017 (UTC)