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November 26

Feedback (positive) !!

Guys, believe it or not, I had never visited this page before. The first time I checked the page out was yesterday. I frequent the Help desk considerably, helping out in resolving queries and all that stuff. The general rule out there is that if it doesn't relate to Wikipedia, nuke it. So it was a pleasant surprise seeing the queries here, and even better, seeing the responses, from intergalactic stars to pressure cooker issues. Very interesting, informative and in some cases, absolutely rollicking answers. Just thought I'll leave a note of appreciation. Thanks for making it so interesting. Lourdes 01:49, 26 November 2016 (UTC)[reply]

Comments like this really belong on the talk page, and that page serves to cover the entire ref desk, not just the science one. We'll forgive you this one time, given how great your song "Royals" is. :) μηδείς (talk) 04:27, 26 November 2016 (UTC)[reply]

On behalf of all of us, thank you Loudres for your kind comments! Dr Dima (talk) 04:46, 26 November 2016 (UTC)[reply]

This being the Ref Desk, I can't help but add a link to positive feedback. :-) StuRat (talk) 17:29, 28 November 2016 (UTC)[reply]

puzzled about theta Eridani

Its two components are designated θ¹ Eridani, also named Acamar, and θ² Eridani.

This is the first time I've read of stars with a digit attached to their Greek letter that are not distinct to the naked eye, or of a binary in which a traditional name is attached to one rather than both; usually binaries are distinguished with A/B after the collective name. Have I misunderstood the system? Is θ Eri a special case? —Tamfang (talk) 07:22, 26 November 2016 (UTC)[reply]

Were they further apart in the past? Sagittarian Milky Way (talk) 07:53, 26 November 2016 (UTC)[reply]

They're now 8+ arcsec; how close can a naked-eye double be? —Tamfang (talk) 00:21, 28 November 2016 (UTC)[reply]

Depends on your eyesight but 8 seconds of arc isn't it. The world record is 20/5 vision (previously thought to be sub-threshold) which would be about 15 arcseconds. Sagittarian Milky Way (talk) 03:03, 28 November 2016 (UTC)[reply]

Wikipedia says at Bayer designation#Other Bayer designations that:

A further complication is the use of numeric superscripts to distinguish neighboring stars that Bayer (or a later astronomer) labeled with a common letter. Usually these are double stars (mostly optical doubles rather than true binary stars), but there are some exceptions such as the chain of stars π¹, π², π³, π⁴, π⁵ and π⁶ Orionis.

This is unsourced, but seems plausible. Perhaps someone assumed that Theta Eridani was an optical double. Or perhaps the A/B convention and the 1/2 convention competed for a while and were never made systematic. This article tells something about how chaotic the history of star designations has been. --76.71.5.45 (talk) 13:22, 26 November 2016 (UTC)[reply]

θ Eri seems to have acquired its superscripts rather recently; they're not in Allen's Star-Names of 1899. —Tamfang (talk) 23:58, 27 November 2016 (UTC)[reply]

The definition of amino acid

According to what I understand the difference between peptide and peptide bond is that peptide is two or more amino acids (up to 19 -inc.) which linked together, while the term "peptide bond" refers to the bond between each two amino acids. Is that true? If it is, the definition of protein is that it contains at least 20 amino acids which are linked together. Isn't it? 93.126.88.30 (talk) 11:01, 26 November 2016 (UTC)[reply]

Peptides are biologically occurring chains of amino acid monomers linked by peptide (amide) bonds. An oligopeptide consists of only a few amino acids (between two and twenty). (Poly)peptides that contain less than 20–30 residues are rarely considered to be proteins. Proteins are generally much longer Polyamide chains ranging in size from tens to several thousand amino acids (in a sequence built from 20 standard amino acids) folded into 3-dimensional structures. See the article Amino acid for more information. Blooteuth (talk) 14:45, 26 November 2016 (UTC)[reply]

One small quibble. There are several amino acids that are capable of forming three peptide bonds, so in principle it might be possible to have polypeptides that are tree-shaped rather than linear. These would not be considered proteins: a protein is a linear chain of amino acids linked by peptide bonds. Looie496 (talk) 15:01, 26 November 2016 (UTC)[reply]

After reading the entry of "peptide" now I'm more confused - because I saw so far so many opinion the definition. How could it be that there is not one clear definition for peptide that differs it from the protein? "Peptides are distinguished from proteins on the basis of size, and as an arbitrary benchmark can be understood to contain approximately 50 or fewer amino acids.[1][2] Proteins consist of one or more polypeptides arranged in a biologically functional way, often bound to ligands such as coenzymes and cofactors, or to another protein or other macromolecule (DNA, RNA, etc.), or to complex macromolecular assemblies.[3] Finally, while aspects of the lab techniques applied to peptides versus polypeptides and proteins differ (e.g., the specifics of electrophoresis, chromatography, etc.), the size boundaries that distinguish peptides from polypeptides and proteins are not absolute: long peptides such as amyloid beta have been referred to as proteins, and smaller proteins like insulin have been considered peptides."93.126.88.30 (talk) 04:20, 27 November 2016 (UTC)[reply]

In many fields of science, what seems to outsiders like a fundamental problem of such a basic definition in the field actually isn't important at all. For example, suppose one could define what exactly a peptide is as opposed to a protein. That wouldn't really change the field itself because that level of specificity isn't of interest when those terms are used. Instead, we often focus on the specific length (or range) as a number, or the range of weights where the exact bonding details aren't relevant for separation, or a specific set of polyamides where the relevance to protein (as a function) isn't relevant. Proteins generally are larger (but there is indeed no firm cutoff solely due to length) and may contain more than one chain, and structures other than peptide chains, and have bonding other than peptide bonds among the amino acids. Lots of terminology has just grown over time, and coming from different perspectives, such that eventually it's not consistent and unless there's a scientific problem caused by it...meh. As additional examples, nobody can agree (and most agree that nobody actually cares, and everyone knows that the historical separation is completely bogus) what makes something an Organic compound vs inorganic. DMacks (talk) 04:35, 27 November 2016 (UTC)[reply]

The peptide bond refers to the actual chemical bond between carbon and nitrogen that is formed as a peptide is made. The hydrolysis of a peptide, for example, might be described as breaking a peptide bond, even as it produces two smaller peptides (or a peptide and an amino acid, or in the case of a dipeptide, two amino acids). It is basically a way to 'navigate' while describing the peptide as a molecule. The peptide simply is the molecule, the whole molecule. So if you look at the R chain for any given amino acid, you might say it is not on the peptide bond, but it is on the peptide - it is attached to the carbon adjacent to the peptide bond (i.e. the alpha carbon). Wnt (talk) 14:05, 27 November 2016 (UTC)[reply]

Could a tectonic tsunami from the tropical Pacific enter the Atlantic?

I think not, the lowest routes would be in Nicaragua or Panama and those are at least 50 miles long and tens of yards high. Am I right? There are no fjords I know between the high Andes and Sierras to funnel tsunamis into a 1/3rd mile high flood. Sagittarian Milky Way (talk) 16:34, 26 November 2016 (UTC)[reply]

The Lituya Bay event was that high over a very short distance and caused by a massive landslide into a narrow inlet (albeit triggered by an earthquake). The biggest recorded tsunami that I know of was the 2004 Indian Ocean earthquake and tsunami, with a maximum of about 50 m. The lowest point on the continental divide is the Riva isthmus (about 50 m elevation and the potential location for the western section of the Nicaraguan Canal), which could theoretically be crossed by the largest of tsunamis, but that only takes you to Lake Nicaragua - a long way away from the Atlantic. Tsunamis run out of energy pretty quickly going uphill, which is why it's possible to run away from them with sufficient warning in many cases - not so in Banda Aceh of course, which is very flat. Mikenorton (talk) 22:23, 26 November 2016 (UTC)[reply]

About how small of an asteroid could do it? Maybe an oblique trajectory pointing towards the coast? Sagittarian Milky Way (talk) 22:42, 26 November 2016 (UTC)[reply]

Don't forget about passing S of S America or N of N America. I suspect that all tsunamis do so, to a very minor extent. StuRat (talk) 18:19, 27 November 2016 (UTC)[reply]

Whether a tsunami can pass north of North America depends on the season and whether the ocean is sufficiently ice-free to permit passage of the tsunami. The Drake Passage does not freeze over, and I would expect that a tsunami could pass south of South America. Mariners who sail the Drake Passage, or the extreme southern waters in general (but there is no need to sail that far south except to round Cape Horn) refer to a phenomenon known as the thousandth wave, which is apparently an occasional rogue wave. Given that the so-called Southern Ocean goes all the way around between Cape Horn and Antarctica, once a rogue wave starts circulating from west to east, it may just continue going around the world. Robert McClenon (talk) 03:40, 28 November 2016 (UTC)[reply]

See Antarctic Circumpolar Current. Robert McClenon (talk) 03:44, 28 November 2016 (UTC)[reply]

The NOAA animation at 2011 Tōhoku earthquake and tsunami shows the wave rounding Cape Horn. μηδείς (talk) 23:57, 29 November 2016 (UTC)[reply]

Can two neurons (or networks) have infinite possible connections?

A connection between two neurons could either not exists, or be one point between 0 and the strongest possible connection? They could also be at any distance from each other (given a range of distances between 0 and x). There are infinite values in any range. --Hofhof (talk) 19:48, 26 November 2016 (UTC)[reply]

A direct connection between two neurons is either chemical (a chemical synapse, or "synapse" for short) or electrical (a gap junction). While a maximal signal in either case is limited by the properties of the two cells connected, the number of possibilities is mathematically infinite because you can, if you so desire, define the strength of the connection with an arbitrary high precision. In other words, the number of possibilities is, indeed, as infinite as the number of real numbers between 0 and 1. However, this does not mean that this high precision is either meaningful or necessary. Indeed, operation of a chemical synapse relies on release of neurotransmitter, which is stored in vesicles in the presynaptic terminal. With each activation of the presynaptic terminal, only a few vesicles are released (or sometimes none at all), in a process that is currently thought to be at least partially stochastic. In other words, the chemical synapses are neither perfectly predictable nor perfectly precise. Therefore, if you build a model of a certain neural network, and the model requires you to tune the connection parameters to an unreasonably high accuracy to work properly, then it is likely that you are doing something wrong or at least non-biological. Biological neural networks tune themselves to the right ballpark of parameters by a variety of homeostatic mechanisms, but most networks are robust to perturbations such as a death of an individual neuron. Model networks should also be robust to stochastic imprecision of the synapses, so you may want to include this stochasticity into your model. Dr Dima (talk) 20:12, 26 November 2016 (UTC)[reply]

The "infinity" of values there is probably exaggerated, since the accuracy of biological numbers is not very great. (i.e. if you turn up the subwoofer, the distance your axon travels gets longer and shorter as the low frequency sound waves penetrate the relevant structures. Changes in blood sugar level affect how readily signals are sent, while ionic strength influences ion flow etc.)

A more meaningful range of variation is obtained because neurons can attach to one another in different ways. If one synapse is further out on the dendrite than another, the second synapse might be inhibitory and prevent its signal from getting through. Or the activating signals from the two might sum up - especially if the further one is received a little before the nearer - and put the cell over the potential for an action potential. Or... well, there are a lot of things, and it's been a very long time since I reviewed this topic. But you should look that way. Neurons are covered in synapses; it's an unbelievably complex network. Things like the Cajal stain (hmmm, Golgi stain) work by randomly showing you a tiny fraction of the neurons present because if you could see them all stained the tissue would be just black. Wnt (talk) 13:55, 27 November 2016 (UTC)[reply]

In AI in computing they generally find a byte of 8 bits is quite sufficient for weights though 16-bit floating point values are also common. Great accuracy is not needed and I doubt the bags of mostly water that are our cells depend much on accuracy. Dmcq (talk) 14:20, 27 November 2016 (UTC)[reply]

The lees liquid in the body - the more sugar concentration and decreased freezing point

I saw a citation from a prof. who said the next things (translated): "when you have cold, one of the first things that your body does is to cause to you to get rid off the liquids, such us in urination. This is a simple equation: the less liquids you have, the percentage of the sugar increased and the freezing point decreased.". Well, I don't understand 2 things that he mentioned: 1) why or how the percentage of the sugar depends on the liquids of the body? 2) what is this "freezing point" that decreases?(does it mean that the less water the smaller freezing point"? if it does, it should not be opposite? it's more difficult for a river to be frozen than a drop on the ground.) 93.126.88.30 (talk) 21:00, 26 November 2016 (UTC)[reply]

Read our articles Freezing-point depression and Fluid balance. Graeme Bartlett (talk) 21:31, 26 November 2016 (UTC)[reply]

I am skeptical of this explanation. It is true that eliminating water and making the serum more concentrated will decrease its melting point. However, looking at two totally different formulas for the expected serum osmolality [1] [2], what they have in common is that the concentrations of sodium, glucose, urea and in one potassium are used to come up with a figure that, in the second, is about 280-300 mOsm, (this says 275 to 290 - if I were anywhere near serious I'd need better sources than any of these, but this is strictly back of the envelope/order of magnitude) where apparently 1 Osm = 1.86 K freezing point reduction (that number being the cryoscopic constant). So we're talking about not much more than a degree difference between normal serum and pure spring water. You can't live with pure spring water in your veins, but I'm not sure how much stuff you could tolerate having on the other side - but hypernatremia (and hyperkalemia) is not good, and the sodium apparently accounts for most of the figure. That last link says that under 145 mmol/l is normal and over 160 mmol/l puts you in the emergency room, so that leaves you with basically 10% room for play in manipulating the average serum water content to avoid having your blood freeze, which amounts to not much more than .1 degree Fahrenheit if I understand this correctly. There would seem to be more merit of having the body work as it should to support healthy shivering and sheltering so its tissues aren't freezing! Wnt (talk) 00:38, 27 November 2016 (UTC)[reply]

segway design useful for hovercraft?

Hello, I'm wondering if the way the segway was designed with gyros, microprocessors?, sensors and electric motors could be or has already been used to improve performance and reduce cost of hovercraft--for one thing, my understanding is that sufficient height and sufficient stability compete with each other right now. Thanks.108.252.141.219 (talk) 23:54, 26 November 2016 (UTC)[reply]

The Segway applies an active feedback control system to stabilise its mechanical system of an Inverted pendulum consisting of the rider's Center of mass supported above the wheel axle, with inherent instability in his rotation on that axis. Hovercraft do not have this instability since their center of mass is supported above a large-area cushion of pressurized air. In still air on a level surface, the hovercraft location is stable but it can be moved horizontally by a very small force, almost exclusively that given by Newton's 2nd law:

F=

mass x acceleration. In windy conditions or on sloping ground, keeping the hovercraft stationary requires compensating force e.g. by ducted thrust fan(s), that may be controlled manually or by an automatic Dynamic positioning system whose components may include motion sensors, Gyrocompass, error computer and Servomotor. A fundamental tradeoff in hovercraft design concerns maintaining an unbroken air cushion: the total amount of air needed to lift the craft is a function of the roughness of the surface it travels over. With adequate air cushion height, ground irregularities that displease a segway traveler are imperceptible to a hovercraft passenger. Many hovercraft designs were completed in the 1960s while the Segway patent (US Pat. 6 302 230) was granted in 2001. Blooteuth (talk) 16:56, 27 November 2016 (UTC)[reply]

Some VTOL aircraft are examples of an unstable system which needs active feedback controls. StuRat (talk) 18:23, 27 November 2016 (UTC)[reply]

Yes. But that's rocket science. Blooteuth (talk) 18:58, 27 November 2016 (UTC)[reply]

Not necessarily -- ordinary helicopters (especially the single-rotor type) are unstable in flight and need constant active feedback from the pilot in order to remain under control. 2601:646:8E01:7E0B:4C25:8F4F:2BC7:C702 (talk) 01:22, 29 November 2016 (UTC)[reply]

November 27

Can I say that every protein made by peptides?

amino acids > peptides >protein93.126.88.30 (talk) 04:35, 27 November 2016 (UTC)[reply]

See Protein and Peptide. It's more correct to say that a protein is a large collection of amino acids and a peptide is a small collection - the boundary between the two is arbitrarily set at 50, but there's no real biological basis for this number. Proteins can be artificially formed from peptides (see peptide synthesis), but in nature they're formed from individual amino acid residues. Tevildo (talk) 10:22, 27 November 2016 (UTC)[reply]

My feeling is that proteins and peptides are the same thing, but this is happenstance rather than definition. There was a time when people did not know that proteins were peptides. According to [3], which seems as likely a place to look as any, "protein" was first studied by Antoine Fourcroy and described much later by Gerhardus Johannes Mulder. The first guy was a contemporary of Lavoisier, who named stuff like hydrogen and oxygen, so he didn't have much idea about peptide bonds!

Now where this gets interesting is that tomorrow we could find some bacterium equipped with a trick ribosome that strings boron and phosphorus compounds or fatty acids or any other thing your fervid imagination might desire into the sequence of an otherwise amino acid based protein. Such a thing would be a protein, but not a polypeptide.

For this reason, my call is that properly a "peptide" is any molecule you know is all peptide - you've synthesized it, or you've nailed down every non-H atom on the crystal structure, and you actually know you can make it entirely out of NH2-C(R)-COOH units. (Note many crystal structures have 'disordered segments', though based on the genetic code you can make a very good guess what they are) Until proven otherwise, a new band on a gel is... well, it's very likely a peptide, but let's call it a protein. Technically I could argue that even something like myristoylation disqualifies a protein as a peptide per se, but the way that chemistry works is that you can simply prepend the qualifier, so you can call it a "myristoylated peptide" anyway. That said, you should probably just go with what Tevildo said. Wnt (talk) 14:24, 27 November 2016 (UTC)[reply]

Don't polypeptides have to be folded the correct way to be described as a protein? LongHairedFop (talk) 19:55, 27 November 2016 (UTC)[reply]

Well, a protein can be denatured and sometimes even renatured, and by that standard, I'd say denatured protein is protein. Wnt (talk) 02:56, 28 November 2016 (UTC)[reply]

Could the moon be terraformed?

108.252.141.219 (talk) 05:04, 27 November 2016 (UTC)[reply]

I searched for "terraforming" in the archives of the journal Icarus, and found:

...so it looks like even Icarus hasn't published much on the topic in a while. If reputable and legitimate scientists believed that new information might have developed in the last few years, there'd probably be something in that journal.

A few years ago, at the 2013 meeting of the American Geophysical Union, a special session was devoted to Geoengineering: "deliberately manipulating physical, chemical, or biological aspects of the Earth system." At least a few responsible scientists believe that our species now possesses the technology to actually induce planetary-scale effects; many believe in anthropogenic climate change; and some scientists believe that with adequate planning, we could use the same technology to plan and produce desirable planetary-scale changes - all using existing technology. Perhaps an even smaller number of scientists believe we might use that technology on worlds other-than-Earth.

But for the most part, terraforming - of Earth, of Mars, of our Moon, or of any other world - is still science fiction. We might, some day in the future, develop newer technologies to harness and control massive quantities of matter and energy - but today, the effects of our actions are usually too small to alter an entire world; and when our technology does have world-altering side-effects, those effects are not the result of a deliberate plan. So - no terraforming of the moon is likely, at least not during the next few years.

Here is some excellent further reading: Why We Explore, a series from NASA's office of the Chief Historian; and a book cited therein: The Railroad and the Space Program. An Exploration in Historical Analogy, (1965). See if you can track down a copy at your local library or college.

Nimur (talk) 06:49, 27 November 2016 (UTC)[reply]

Terraforming of the Moon is our redirect on the subject, but the target fails to mention the possibility or otherwise. Graeme Bartlett (talk) 10:39, 27 November 2016 (UTC)[reply]

It doesn't seem practical, since the gravity there is too little to keep an atmosphere from escaping into space, due to the solar wind and tidal forces from the Earth, even if you could manage to create one. (You could create habitable domed or underground areas, but that's not really terraforming.) StuRat (talk) 18:26, 27 November 2016 (UTC)[reply]

I read once that if the Moon were somehow given an Earthlike atmosphere it would lose one-tenth in 1e4 years. —Tamfang (talk) 00:33, 28 November 2016 (UTC)[reply]

That process is Jeans' escape - and in many planetary physics classes, it's a common homework problem to compute the time-constant for many familiar moons and planets. The equation governing this time scale is an equilibrium (or non-equilibrium) between gas diffusion and gravity. More complicated versions account for thermal effects, solar wind, creation of new gas via geological processes, and so on. Nimur (talk) 02:36, 28 November 2016 (UTC)[reply]

Keep in mind that terraforming is expected to take thousands of years, so even a slow bleed-off rate is problematic over such a long period. The option of building domes or tunnels to contain the atmosphere might be a good first step, with the idea of building up atmosphere in those places for eventual release into the atmosphere, once we have a solution to the bleed-off. StuRat (talk) 17:22, 28 November 2016 (UTC)[reply]

No, you can't terraform the Moon as such, it is too small to hold an earthlike atmosphere with its gravity (ec with Tamfang). Colliding Venus and then Ganymede with Mars will give a body with a mass slightly less than the Earth's, and a lot of water, but sneaking Venus past Earth would be a delicate operation. You'd also have to wait perhaps tens of millions of years for the surface to cool and stabilize. In the meantime, you could tow in a body like Titan to make a cool moon, since tides are vital in maintaining productive coastal ecosystems. Finding an Earthlike exoplanet would probably be a lot easier and quicker. μηδείς (talk) 00:42, 28 November 2016 (UTC)[reply]

^{[citation needed]}? Did you just make this stuff up, or can you attribute those ideas and concepts to an actual scientific reference? Nimur (talk) 02:38, 28 November 2016 (UTC)[reply]

I dunno, Nimur. I took physics for science majors and passed, and looked up the masses of the bodies I mentioned. If you have a specific question, ask it, and identify which of the many people here who have said that the moon won't hold an atmosphere whom you are addressing. μηδείς (talk) 04:17, 28 November 2016 (UTC)[reply]

Yes you can, given sufficiently advanced technology. However, if you want it to last more than 10,000 years or so, you will need to actively maintain your atmosphere, which will otherwise gradually bleed into space. I think you could reasonably expect to find a technical fix for this little problem during that 10,000 year window, but I suspect your goals will have changed by then. -Arch dude (talk) 00:58, 28 November 2016 (UTC)[reply]

Might the relatively low amount of geological activity within the Moon also be a problem? ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:23, 28 November 2016 (UTC)[reply]

Is any Moon soil like ground glass? You might want to avoid running barefoot through such areas. Sagittarian Milky Way (talk) 02:47, 28 November 2016 (UTC)[reply]

If you're going barefoot on the airless Moon, getting cuts on your feet might be the least of your worries. ←Baseball Bugs ^{What's up, Doc?} carrots→ 15:20, 28 November 2016 (UTC)[reply]

I mean after there's air and an ozone layer. Sagittarian Milky Way (talk) 16:31, 28 November 2016 (UTC)[reply]

Some obvious challenges include:

Getting suitable atmosphere material to the Moon - see Interplanetary Transport Network, perhaps.
Keeping the atmosphere in long-term - wrapping it in plastic seems like an option, but maybe you can think of something fancier.
Spinning up the rotation rate to deal with hot and cold extremes - requires a lot of energy/propellant/time, you'd think.
Dealing with tectonic activity from the spin-up and renewed tides - time and earthquake prediction, I suppose.
Given the time needed for some other steps, I think the beaches will churn themselves to proper sand in the meanwhile.

Benefits include the having the best orbit around the best star, the best geological materials a planet can have, not to mention the human powered sport flying and the truly massive surfing opportunities. Wnt (talk) 03:18, 28 November 2016 (UTC)[reply]

I just read our entire article on the Interplanetary Transport Network... it's not outright wrong, but it presents a very over-simplified description that might lead the unskilled astrophysicist into some very very wrong conceptual conclusions. Although lines of equipotential energy connect many places in our solar system, it is not valid to suggest we can use "zero fuel" or "zero net change in energy" to get from point A to point B along these lines. The key insight that is missing is that, in orbital mechanics, energy takes the form of stored gravitational potential energy and also kinetic energy. In one simple example: your space ship might get to a distant point, but at the wrong velocity to stay at that point! The only way to get to, and then stay at, the point where you want to be, when you're orbiting, is to change your kinetic energy - delta v - which means, using the only technology we have today, to expend fuel. If we forget about this "small detail," when reading our Interplanetary Transport Network article, we reach the wrong conclusion based on a very superficial understanding of the physics. We might conclude that we can use these complicated orbits to "cheaply" and "efficiently" navigate around, following equipotential lines and following unstable orbits without firing the rockets. That's the wrong conclusion: we can pass by a lot of points, but we won't stay there... unless we pursue massive rockets for very large delta v; or pursue direct entry and slam into the target at orbital velocities - slowing from orbit by colliding with the destination! In both cases, we suffer the same issue: today's technology does not enable us to apply this orbital approach to move non-trivial quantities of cargo. We can't build giant rockets cheaply and reliably enough to provide that much delta-v at the destination; and we don't have great technology options for atmospheric entry when the entry-velocity is extraordinarily large!

A great book, for the interested reader, is To Rise From Earth, available from many book resellers. It presents orbital mechanics in a format that a non-mathematician can appreciate, and lays a great foundation for the future physicist who will (eventually) study the equations and mathematical formalisms that describe spaceflight. Another fantastic book is the much-harder-to-find Realities of Space Travel. Our technology has changed since it was authored, but the harsh laws of gravity have not - and despite our best efforts, we really have not learned much to change our understanding of the basic physics of gravity in over a century.

Nimur (talk) 03:53, 28 November 2016 (UTC)[reply]

One does need a lot of energy to get out of the gravity well around planets, but otherwise Gravity assist does enable travel around with expending much energy. That's how Voyager manged to get out of the Suns gravity well. One might have to wait for a fruitful conjunction but the energy is got from speeding or slowing the planets by imperceptible amounts. Dmcq (talk) 13:28, 28 November 2016 (UTC)[reply]

Gravity assist is great and useful, but to use the technical terminology, it is only able to perform conservative work. If you actually want to change your orbit, you need delta v, which requires non-conservative work. To use non-technical terminology, that means either hitting a planet (crushing on its surface or aerobraking in its atmosphere); or, expending rocket fuel. There are no other options: every option is a variation on these themes. Using gravity assist, one can expend fuel or impart a collision at the best possible time - but there is no way to steal energy from the planet's orbit unless you collide with the planet. Complex equations and rotating reference frames do not invalidate conservation of energy.

Don't let the complications fool you - orbital mechanics is really built on very basic physical law: gravity is a conservative force, so it requires exactly the same amount of energy to get to your destination no matter what path you use. Gravity-assisted orbit boost, or even orbit-capture, does not change this fundamental principle. All it does is permit the engineers to move around when the spacecraft's stored energy gets spent, so that it gets spent when it is the best and most efficient time and place. Gravity-assisted direct-entry (implying a collision with the planet's atmosphere or hard surface) is another really neat option - but with today's materials technology, our spacecrafts cannot survive entry, descent, and landing, from interplanetary orbit velocity. In non-technical terms, the spacecraft will break up or burn up in the planetary atmosphere; or crush on the planet's hard surface. If we can engineer materials and structures that could withstand that event - say, some amazing foamed-titanium truss structure that could crush and dissipate the energy from an impact at 20 kilometers-per-second - as recommended in this 1960 NASA technical report, Landing Energy Dissipation for Manned Reentry Vehicles, we might be able to land somewhere without firing rockets. Unfortunately, NASA scientists realized that no available material in the 1960s could yield a survivable landing - which is exactly why our Apollo missions used giant rockets to perform orbital insertion and landing on the moon; and more giant rockets to perform Earth orbit reentry, followed by aerobraking, followed by parachutes, followed by a just-barely-survivable-by-human water landing. Aerobraking is not a great option on many worlds we might want to visit, especially in places where the atmosphere is sparse - so it's either crushable landing gear, or giant rockets.

Nimur (talk) 15:35, 28 November 2016 (UTC)[reply]

What you say is wrong an has been proven wrong in practice. It would have required a huge rocket to get Voyager away otherwise. If one only had a satellite and a single planet then one would not be able to steal energy just from the planet it set off from, but it is a Three-body problem effect going around various planets. Getting off a planet or down to one is not helped by gravity assist, only moving around once one can get off enough to visit another body. Dmcq (talk) 15:49, 28 November 2016 (UTC)[reply]

Voyager did not stay, and could not stay, at the planets it visited. It passed beyond them. This is a very important detail. The Voyagers were launched on some of the most powerful rockets ever built; a lot of kinetic energy was imparted to the space probes when they left Earth. The probes traveled to the outer planets along an orbital trajectory. For Voyager to stay at any of the planets it passed by would have required a change in its orbital trajectory, which means an even bigger rocket. Notice that neither Voyager ever entered any planetary orbit - it was never captured (and could never be captured) during any of its gravitational slingshot maneuvers. Compare that spacecraft to, say, the Cassini mission, and look how different the spacecrafts looked. Cassini entered planetary orbit with a huge expenditure of fuel from a massive chemical rocket. Nimur (talk) 16:37, 28 November 2016 (UTC)[reply]

As I said Gravity assist doesn't help with taking off or landing on a planet, there is no intermediary body to gain or lose energy from when in its gravitational well. But once one can reach other bodies it is most certainly possible to gain or lose energy. What helped Voyager leave the solar system was gravity assist, it did not have enough speed on launch to escape the sun. The rockets were large for Voyager to take it to Jupiter directly instead of using gravity assist from the inner planets. New Horizons, the fastest rocket ever made, is still slower than Voyager. When going to Mercury gravity assist is used to slow down satellites so they approach it much slower than they would otherwise - the speed is slowed such that if they crashed on the planet it would be with something close to the escape velocity for Mercury. Dmcq (talk) 17:44, 28 November 2016 (UTC)[reply]

@Nimur: See [4] - "a low-energy transfer to a lunar-libration orbit saves 400 meters per second (m/s) of deltaV and often more. This is a significant savings..." I understand this network isn't quite magic, but the three body problem notoriously interjects some rocket science into rocket science, I mean, you can't just calculate everything in a neat little equation, it takes modelling and clever thinking - cleverer than I claim to bring to the table, to be sure. Wnt (talk) 00:42, 29 November 2016 (UTC)[reply]

Right - 400 m/s less than a less-efficient orbital transfer! The objective, in orbital trajectory planning, is usually to work towards the theoretical best orbital trajectory, or minimum possible delta-v. There's no way to change your orbit with zero delta-v; but if you're not at the theoretical minimum, you can find ways to use less delta-v. Nimur (talk) 19:45, 29 November 2016 (UTC)[reply]

I don't want to defend a strawman here - I realize you can't get off the mountaintop with clever orbital engineering. But see [5] page 3:

In general, a low-energy transfer is a nearly ballistic transfer between the Earth and the Moon that takes advantage of the Sun’s gravity to reduce the spacecraft’s fuel requirements. The only maneuvers required are typical statistical maneuvers needed to clean up launch vehicle injection errors and small deterministic maneuvers to target specific mission features. A spacecraft launched on a low-energy lunar transfer travels beyond the orbit of the Moon, far enough from the Earth and Moon to permit the gravity of the Sun to significantly raise the spacecraft’s energy. The spacecraft remains beyond the Moon’s orbit for 2–4 months while its perigee radius rises. The spacecraft’s perigee radius typically rises as high as the Moon’s orbit, permitting the spacecraft to encounter the Moon on a nearly tangential trajectory. This trajectory has a very low velocity relative to the Moon: in some cases the spacecraft’s two-body energy will even be negative as it approaches the Moon, without having performed any maneuver whatsoever.

There is an interesting map of Jacobi constants that determine "forbidden regions" by velocity on page 38. I think but certainly will not swear that provides some hard limits. I am not sure if this is related to the statement on page 13 that "Sweetser computed that the theoretical minimum ΔV that a space craft would require to travel from a 167-km altitude circular orbit at the Earth to a 100-km altitude circular orbit at the Moon, just passing through L1, is approximately 3.721 km/s" It is fairly clear from the text and the complicated shape of these bands that there are some transfers possible within a "network" that require no energy, once a spacecraft has gotten to some location. Wnt (talk) 12:48, 30 November 2016 (UTC)[reply]

An interesting alternative: [ https://blogs.scientificamerican.com/guest-blog/lets-colonize-titan/ ] --Guy Macon (talk) 17:15, 28 November 2016 (UTC)[reply]

Oh, nonsense. "Titan is the only place..." I mean, look up from that frigid wasteland to Saturn. It's got the perfect gravity, liquid water, with a few percent oxygen added you can make hydreliox mix to live at 10 atm where that is, and there are bands of relatively calm, upwelling atmosphere. True, you have to fly all the time and a shaker of salt is probably more precious than gold on Earth, and yeah the next layer of atmosphere is industrial grade stink bomb, but relatively speaking, doable. Venus is probably better, though. Wnt (talk) 00:48, 29 November 2016 (UTC)[reply]

Feynman Lectures. Lecture 39. Ch. 39–4. Temperature and kinetic energy [6]

...Ultimately, what will be the distribution? Answer: It will be equally likely to find any pair moving in any direction in space. After that further collisions could not change the distribution.
They are equally likely to go in all directions, but how do we say that? There is of course no likelihood that they will go in any specific direction, because a specific direction is too exact, so we have to talk about per unit “something.” The idea is that any area on a sphere centered at a collision point will have just as many molecules going through it as go through any other equal area on the sphere. So the result of the collisions will be to distribute the directions so that equal areas on a sphere will have equal probabilities.
Incidentally, if we just want to discuss the original direction and some other direction an angle θ from it, it is an interesting property that the differential area of a sphere of unit radius is sinθdθ times 2π (see Fig. 32–1). And sinθdθ is the same as the differential of −cosθ. So what it means is that the cosine of the angle θ between any two directions is equally likely to be anything from −1 to +1.
— Feynman • Leighton • Sands, The Feynman Lectures on Physics, Volume I

Feynman first explains that directions of molecules are random, but probability that some molecule go through some area on the imaginary sphere is equal for equal areas. I agree. Last paragraph is unclear.

According to Fig. 32–1 $dA=2\pi r^{2}\sin \theta \,d\theta$ . dA is a change of the area between angle θ and (θ+dθ). But this area is the area of spherical segment limited by 2 conical surfaces, not 2 directions, and not 2 equal areas.

Feynman suggests next manipulation $dA=2\pi r^{2}\,d(-\cos \theta )$
and for r=1 $dA\propto -d\cos \theta$ ;

$A_{12}\propto \int _{\theta _{1}}^{\theta _{2}}-1d\cos \theta =\cos {\theta _{1}}-\cos {\theta _{2}}$

If e.g. $\cos {\theta _{1}}-\cos {\theta _{2}}=\cos {\theta _{5}}-\cos {\theta _{6}}$ , then $A_{12}=A_{56}$ and then these areas have equal probabilities. But I can't see it's correct for cosine itself. Besides different dA's are different because ΔCosθ is different for different angles (maximum at θ = π/2).

Can you show how Feynman derived that cosine must be from −1 to +1?

Username160611000000 (talk) 08:08, 27 November 2016 (UTC)[reply]

This seems like a long and tortuous way of saying the result by Archimedes that the area of a slice of a sphere is the same as the area sliced on a cylinder round the sphere but I suppose trigonometry and calculus are the way to do that now. Anyway see [7] Dmcq (talk) 11:16, 27 November 2016 (UTC)[reply]

Maybe Archimedes' Hat-Box Theorem is correct for equal h (or dh), but we use equal dθ. Username160611000000 (talk) 11:25, 27 November 2016 (UTC)[reply]

Even if ΔCosθ would be equal with increasing θ on equal steps, I still don't understand the logic. Why does cosine of the angle between 2 directions of 2 pairs of molecules vary from -1 to 1? This statement is absolutely not correlated with previous one about differential area . Username160611000000 (talk) 11:35, 27 November 2016 (UTC)[reply]

Feynman next uses this statement to prove that average cosine is zero. So I need to understand this paragraph by all means. — Preceding unsigned comment added by Username160611000000 (talk • contribs) 11:39, 27 November 2016 (UTC)[reply]

I assume that a conic surface is associated with some direction of the spread of pair of molecules (the spread of 2 molecules occurs in opposite directions). So molecules can go in any direction inside surface. Then this conic surface is associated with little area (in fact the section of surface and sphere (circle) must be in the middle of the area , but with infinitesimal values it's not important). Is it correct? Username160611000000 (talk) 11:58, 27 November 2016 (UTC)[reply]

The cosine of θ is −1 in the straight downwards direction and +1 in the straight upwards direction. Dmcq (talk) 12:53, 27 November 2016 (UTC)[reply]

I agree. But it doesn't mean that every direction is equiprobable, because it does not connected with equal areas. Username160611000000 (talk) 13:09, 27 November 2016 (UTC)[reply]

They said equal areas were equiprobable. They never said that equal changes in θ gave equal areas. Dmcq (talk) 13:24, 27 November 2016 (UTC)[reply]

dθ is used to calculate differential area , every dθ is equal. Why is this differential area needed at all ?

To show that angle can get any value between 0 and π with equal probability, we must show that there is some area associated with each direction in some way and that these areas are equal. Feynman did not show that, he just said that there is some differential area. So what? Differential area exists. It is not connected with direction, it is not connected with cosine. Username160611000000 (talk) 13:54, 27 November 2016 (UTC)[reply]

To get equal differential areas of segments you must have equal d(cos θ) not dθ. Nobody put in any requirement for every differential of θ to be the same. When you have dy/dx you get the ratio of changes out and you do not expect a constant. Dmcq (talk) 14:15, 27 November 2016 (UTC)[reply]

When you have dy/dx you get the ratio of changes out and you do not expect a constant. When we build a graph of g(x)=dy/dx we take some point y(1) and find Δy during some Δx, then we take next point y(2) and again find Δy during SAME Δx. 1st and 2nd Δx are equal, 1st and 2nd Δy are not equal. Username160611000000 (talk) 14:27, 27 November 2016 (UTC)[reply]

To get equal differential areas of segments you must have equal d(cos θ) Agree. What are the next manipulations with d(cos θ)?

No the Δx do not have to be equal. d(cos θ) is equal for equal displacements in the vertical direction. Dmcq (talk) 14:41, 27 November 2016 (UTC)[reply]

d(cos θ) is equal for equal displacements in the vertical direction. Ok. Let displacements be equal. What is next?Username160611000000 (talk) 15:04, 27 November 2016 (UTC)[reply]

Let the vertical diameter of the sphere consist of ten parts Δh₁, Δh₂ ..., Δh₁₀. Consider some vertical displacement Δh₃. What angle θ is associated with Δh₃? θ ∈ (53°;66°], probability P(θ ∈ (53°;66°])= 10%. Consider Δh₂, θ ∈ (36°;53°], probability P( θ ∈ (36°;53°])= 10%. What is probability θ ∈ (53°;54°]? It is

P(\theta \in (53^{\circ };54^{\circ }])={\tfrac {P(\theta \in (53^{\circ };66^{\circ }])}{13}}={\tfrac {10\%}{13}}=0.8\%

.
By analogy

P(\theta \in (36^{\circ };37^{\circ }])={\tfrac {P(\theta \in (36^{\circ };53^{\circ }])}{17}}={\tfrac {10\%}{17}}=0.6\%

;

P(\theta \in [0^{\circ };1^{\circ }])={\tfrac {P(\theta \in [0^{\circ };36^{\circ }])}{36}}={\tfrac {10\%}{36}}=0.3\%

.
So we get increasing probabilities in equator direction [png].Username160611000000 (talk) 15:12, 27 November 2016 (UTC)[reply]

Diapason is not a term normally applied in anything like this. It is a musical term and only applied in other contexts to mean a range in a literary sense. Range would be a better name and range does not apply to a single angle. Dmcq (talk) 13:08, 28 November 2016 (UTC)[reply]

Also formula for A is A(θ)=2πR²(1-Cosθ). Letting R=1 and expressing θ we get : θ(A)=arccos(1-A/2π).

{\tfrac {d\theta }{dA}}={\tfrac {d}{dA}}\arccos {(1-{\tfrac {A}{2\pi }})}={\tfrac {1}{2\pi {\sqrt {1-{(1-{\tfrac {A}{2\pi }})}^{2}}}}}

. graph . We see the angle changes nonlinearly.Username160611000000 (talk) 17:04, 28 November 2016 (UTC)[reply]

The relevance is fairly marginal, but I went through the same kind of confusion looking at the distribution of the sine of an angle (which is not uniform) in this previous discussion here. Wnt (talk) 14:39, 27 November 2016 (UTC)[reply]

what are proteins, lipids, nucleic acids and carbohydrate classified?

what are proteins, lipids, nucleic acids and carbohydrate classified? I mean to ask if they are groups of or families of biochemistry. For example I want to say that we have 4 families in biochemistry: proteins, lipids, carbohydrates, and nucleic acid. Is it true to refer them like that? 93.126.88.30 (talk) 15:46, 27 November 2016 (UTC)[reply]

Generally speaking, chemicals involved in biology ("biochemistry") are described as belonging to four families, amino acids (including proteins), lipids, carbohydrates, and nucleic acids (including nucleobases, nucleosides, DNA, RNA, etc.). That can lead to some amount of oversimplification, however, as it doesn't necessarily account for some more complex compounds, such as chlorophyll or heme porphyrins with metal centers, some vitamins, etc. --OuroborosCobra (talk) 19:36, 27 November 2016 (UTC)[reply]

I took the question as asking for a category that includes the named groups but excludes others. You can generalize, using such terms as organic molecules, but that includes things like octane and methyl alcohol, both dangerous toxins. You could mention nutrients, but not all carbohydrates, for example, are nutritious, and we don't require nucleic acids as such, since we can synthesize them. This sounds like a homework question. Is it? I have a suggestion, but the poster should tell us what he thinks before we supply suggestions. μηδείς (talk) 00:29, 28 November 2016 (UTC)[reply]

One clue is to see how Wikipedia classifies them. If you read the articles, down the bottom will be categories. You can click on these categories and see what else is in them, or what their parent categories are. Graeme Bartlett (talk) 05:43, 28 November 2016 (UTC)[reply]

Commons also has a category tree eg Commons:Category:Lipids. Graeme Bartlett (talk) 11:42, 28 November 2016 (UTC)[reply]

This page and These lecture notes basically agree with the above, regarding the four major classes of biomolecules. Wikipedia's article titled Biomolecule doesn't directly cover the 4-class structure, but does also go into other types of biomolecules that cannot be classified such as terpenes, steroids, and lignin, etc. --Jayron 32 14:11, 28 November 2016 (UTC)[reply]

Biochemistry#Biomolecules gives the same four major types, cited to a published textbook. DMacks (talk) 14:16, 28 November 2016 (UTC)[reply]

November 28

video

What's happening in this video[8]? Is it for pipeline construction or something? ECS LIVA Z (talk) 04:28, 28 November 2016 (UTC)[reply]

It looks like an enormous mine-clearing line charge test; but without more context, it's hard to be certain. Do you have any further information, like where you found the video? Nimur (talk) 04:44, 28 November 2016 (UTC)[reply]

It was just randomly on the internet without any context. ECS LIVA Z (talk) 04:52, 28 November 2016 (UTC)[reply]

It appears to be an extract from Dykon Blasting Corporation's promotional video for the ISEE Conference a few years ago. Here is their 2016 video; there is an archive of older videos on their website. This company sells explosives services for construction and military purposes. Nimur (talk) 04:58, 28 November 2016 (UTC)[reply]

Here is the entry form for next year (2017): Give us Your Best Shot: 24th Annual Photo Contest. Be aware that the contest is intended for explosives professionals who have appropriate permissions, exercise proper procedures, and exhaustively enforce safety controls; it's not intended for home-brew bomb-makers. Don't play with explosives. Nimur (talk) 05:04, 28 November 2016 (UTC)[reply]

These seem like the kind of people who consider *only* triple checking something to be dangerous.Naraht (talk) 16:06, 28 November 2016 (UTC)[reply]

Thank a lot!

What are these blasted trenches used for? In the video I see some pipelines, so pipelining is probably one of the uses. Anything else? Would a road/railway need blasted trenches like these? ECS LIVA Z (talk) 00:21, 29 November 2016 (UTC)[reply]

Possibly, if the road has to e.g. cut across a ridgeline. 2601:646:8E01:7E0B:4C25:8F4F:2BC7:C702 (talk) 01:17, 29 November 2016 (UTC)[reply]

Removing red-28 dye from a carpet - and Red-28 in children's toothpaste.

So over Thanksgiving, two of our grandkids decided to have a "tea party" in their bedroom...which had a new carpet laid just a couple of weeks before.

They started out playing with cups of water - but then decided they needed "food"...and (by some peculiar twist of a 5 year-old's brain) they decided to use toothpaste - a Crest childrens paste in watermelon flavor.

Inevitably, this got dropped onto the new carpet - and in their efforts to clean it up before any parent/grandparent could find out - got it rubbed deeply into the pile over a large area.

OK - so we wash it, we try every carpet spot-remover we have, we Google "getting pink stain out of carpet" and try things like hydrogen peroxide and heat from an electric iron.

No luck whatever...which is weird because these techniques seem to work well for EVERYONE else.

In desperation, I look on the ingredients list - and discover that the toothpaste's coloring is "RED 28". This is not listed in Food dyes - and only after much cunning Googling do I discover that this particular chemical is banned for foodstuffs in the USA! It's allowed only for certain cosmetics - and for "drugs".

This paper: https://ntp.niehs.nih.gov/ntp/htdocs/chem_background/exsumpdf/reddyes_508.pdf seems to suggest that the stuff is pretty nasty - but my knowledge of the chemistry and the vocabulary of these kinds of reports is sadly lacking, so maybe I'm misunderstanding it.

Anyway: Two questions:

How do you get RED-28 out of a carpet?
How the heck can they put a not-allowed-in-foods chemical into toothpaste that's recommended for "2 years and up" kids - who are pretty much certain to swallow the stuff? Can they really classify toothpaste as "A Drug"? Also why would they do this? There are plenty of other red colorants they could have used instead.

TIA SteveBaker (talk) 19:01, 28 November 2016 (UTC)[reply]

How to clean the carpet. The linked report seems to imply that D&C Red No. 27 (insoluble) changes to D&C Red No. 28 in a high-pH environment. I would therefore try flushing with dilute sodium hydroxide (caustic soda). Wear rubber gloves, protect eyes and rinse thoroughly with water.
How could a potentially phototoxic colorant be allowed in toothpaste? According to the linked report, D&C Red Nos. 27 and 28 were approved as additives for some drugs and cosmetics in 1982. A toothpaste manufacturer's lawyers might see that as adequate and even argue that toothpaste is not intended to contact the eye nor be exposed long to light. Crest (toothpaste), possibly like this, is made by Procter & Gamble and there is consumer support at https://pgconsumersupport.secure.force.com/ContactUs/emailus?brand=Crest&country=United+States+of+America&language=English-US Blooteuth (talk) 20:51, 28 November 2016 (UTC)[reply]

Thanks...

I'm concerned that over-aggressive treatment might bleach out the other dyes in the (oatmeal-colored) carpet. So how dilute is "dilute" in this case?
The product you linked to seems to be green, and to use two food-legal colorings...it's from Colgate. The one we're concerned about is "Crest Stages Pro-health kids": LIKE THIS. I already contacted them twice - but so far, no response. SteveBaker (talk) 21:42, 28 November 2016 (UTC)[reply]

Obviously, you should test a series of bleach dilutions on a portion of the carpet that will be covered by a couch. Or test on a large region of the carpet, if you dislike it to begin with. Or just bleach the whole damn thing, to keep it even :) Someguy1221 (talk) 22:44, 28 November 2016 (UTC)[reply]

Lye will damage the rug and there are better alternatives, if you wanted high pH or alkalinity (such as washing soda). Page 5 does say No 28 is more soluble, but also says "The transition from D&C Red No. 27 to D&C Red No. 28 occurs between pH 3.4 and pH 5.0 (CRC, 1986)." So it appears you want an acid. So something like a mixture of baking soda (a little bit to raise the pH), lemon juice (pH 2) and water would be better. Since the dye was a toothpaste ingredient you might also try scrubbing with the brand's white toothpaste version in the hope that the dye is soluble in that. The iron may have set the dye into the fibers, so you may have to resort to experimenting with diluted chlorine bleach (perhaps testing solutions on a rag with the pink toothpaste rubbed into it). Four or five parts water to one part laundry bleach should suffice. Ideally the dye will vanish within 20 seconds or less (longer times would indicate it's ineffective and further risks fading and/or damaging the fibers). Ventilate the room and keep pets away before you begin: opening windows and using fans to direct any fumes away is recommended. Having put on gloves first, sponge the carpet with the bleach and then blot it with paper towels to remove the excess bleach as fast as you can and immediately flood with hydrogen peroxide. The peroxide will react with the remaining bleach and immediately stops further unwanted action on the rug. Apply only to small areas because noxious gases (chlorimides) can form from any nitrogenous compounds that happen to be in the rug. With a new rug this should not be a problem, but an old rug can accumulate these. --Modocc (talk) 22:52, 28 November 2016 (UTC)[reply]

I don't know. If I understand this correctly, the whole damn thing is an aromatic system (9 double bonds = 18 electrons, following Huckel's rule. Every position is oxygenated or halogenated. I don't know what bleach is going to do to that, though I'm not good enough at organic chemistry to be sure. If it works, great... The photoactivation might mean that room light has bound some of it to the carpet - I don't know that for sure either. Would some sort of nucleophile/reducing agent break apart ether linkages? No idea, I'm talking out my ass. If I had some beta-mercaptoethanol handy I'd be sorely tempted to dilute it way out and see what happened, but that would only get me in a lot more trouble... Wnt (talk) 01:16, 29 November 2016 (UTC)[reply]

I suppose the real answer here is "contact the manufacturer". They might have a clue, and if not, well, it would be nice if they couldn't play a "well, we would have just said..." game should you ever end up suing their asses. Wnt (talk) 01:26, 29 November 2016 (UTC)[reply]

Yes, the dye is a fluorescein derivative and it tends to be a very stable structure because its highly halogenated. I don't know either if bleach would help. When my friend bleaches his white dress shirts, if a touch of dilute bleach and an adequate quantity of hydrogen peroxide fails on a stain, he'll apply the dilute bleach then a drop of lemon juice or vinegar and then the peroxide. Unfortunately, this combination of bleach and acid liberates toxic chlorine gas which can be lethal in quantity, why you should not mix bleach and vinegar, until its stopped by the H₂O₂ . In fact, it is this unpleasant gas that you smell when CO₂ in the atmosphere reacts with the bleach solution. -Modocc (talk) 02:34, 29 November 2016 (UTC)[reply]

When laundering, a few minutes before washing I'll often use toilet bowl cleaner on the worst stains because the soil here has a lot of iron oxides in it. Sometimes only pure soup and a lot of heavy brushing gets whatever I happened to get into out. --Modocc (talk) 05:21, 29 November 2016 (UTC)[reply]

Ah - yes fluorescein - from the little chemistry I know, that sounds like bad news. :-( With a dye that's as stable as that, it's likely to be as stable as the dyes that we don't want to destroy in the original carpet. Having a white patch on an oatmeal carpet would be at least as bad as a pink patch! SteveBaker (talk) 14:47, 29 November 2016 (UTC)[reply]

A white patch or any fading or damage to the original carpet would be bad. My friend and I have used this peroxide stopping technique on various carpets without causing that problem. The key factor is to not let the bleach react with the carpet for any length of time (on clothes after a few minutes it will turn it white and even eat holes in it too, also, undiluted bleach will do this in mere seconds so you have to slow the reaction down with dilution). To gain confidence and get the hang of it, you could try this on old clothes and carpets that you don't care about so you can estimate what to expect. Since you have to be quick with the peroxide, it helps to limit the size of area that you are targeting to something manageable. --Modocc (talk) 15:23, 29 November 2016 (UTC)[reply]

I suspect that even if you succeed in removing the colour of the stain without bleaching the carpet's colour, the area involved will always show a noticeably changed texture. I suppose you've considered just putting a rug over the stain? {The poster formerly known as 87.81.1230.195} 176.248.159.54 (talk) 08:20, 29 November 2016 (UTC)[reply]

Sadly, the "put a rug over the stain" doesn't really work here because it's right by the door as you step into the room...but something of that nature seems necessary here. SteveBaker (talk) 14:47, 29 November 2016 (UTC)[reply]

I suggested to my wife that we smear kids toothpaste all over the carpet and dye it all the same shade of pink!

This did not go over well! :-)

New rule: Anything with Red 28 in it does NOT get into our house anymore! We'll be switching to kids' toothpaste with regular food dyes that are easy to get off in case of another "tea party" related disaster! (And also putting toothpaste into a cabinet that's out of kid's reach!)

Thanks for all the help! SteveBaker (talk) 14:47, 29 November 2016 (UTC)[reply]

You may have already found this, but one of the primary uses of Red 28 is to dye fabrics. In fact it is somewhat famous as one of the first modern dyes that didn't require a secondary fixative. All of which suggests you are pretty screwed. There are industrial systems for degrading Red 28, especially in the waste water from dye operations, but most of those are either going to be inappropriate or inaccessible. The only thing that comes to mind that you might consider is that prolonged exposure (hours) to UV light will degrade the dye especially is combined with activation agent like H2O2 or an aqueous suspension of powdered TiO2. However, I would suspect that anything strong enough to destroy Red 28 is going to fade the other dyes in your carpet. Depending on personal preference a larger whitish stain might be preferable to a small pink one. If you do try UV, at least there is the potential to illuminate it narrowly on the stain, which may allow for better targeting than bleach. If your carpet is a uniform color (rather than patterned) you might be able to excise the affected region and apply an inconspicuous patch. However, be warned that it takes a lot of skill to patch carpet in a way that looks good. Poor to middling efforts tend to be very conspicuous. Dragons flight (talk) 15:32, 29 November 2016 (UTC)[reply]

I realized after reading Modocc's comments below that I was apparently thinking of the wrong "Red 28". I had in mind "Direct Red 28" (aka Congo red) [9], which is apparently different from "FD&C Red 28". Sorry for any confusion. Dragons flight (talk) 09:05, 30 November 2016 (UTC)[reply]

Another option: a professional carpet cleaner outfit may have what's needed to clean it. The dye is soluble so there is hope, and the newer carpet fabrics are made to be more stain resistant, so its possible the dye has not yet fixed such that its indelible, but after already using various solvents and a hot iron on it... -Modocc (talk) 16:23, 29 November 2016 (UTC)[reply]

This data sheet of the compound list different manufacture names for the compound: [10] which should help us find out more about it. It has two table entries for color fastness against light: 1 and fading from soaping: 3. But I haven't found what these AATCC values mean yet. OK this source states that "Lightfastness is judged on a scale of 1 to 8, where 8 is most fade-resistant. Washfastness is judged on a scale of 1 to 5, where 5 is most resistant to washing out." I can't find mention of the AATCC and it does not list this dye (you would think it would be, errrrr). But assuming these are the scales being used having respective values one and three, it is very susceptible to light as Dragons flight pointed out above and will fade anyway with repeated washings. --Modocc (talk) 17:54, 29 November 2016 (UTC)[reply]

Turns out we do have an article on this dye, its called Phloxine. It states "For humans, the Food and Drug Administration deems phloxine B to be safe up to a daily dosage of 1.25 mg/kg." This Handbook of Biological Dyes and Stains: Synthesis and Industrial Applications lists it as Phloxine B. --Modocc (talk) 20:29, 29 November 2016 (UTC)[reply]

Good news, the Dictionary of the Coal Tar Colours states that this dye is not colorfast; confirming what I found above in this source. In other words, a moderate amount of light and washings should make a noticeable difference without affecting the carpet's more durable dyes. --Modocc (talk) 22:23, 29 November 2016 (UTC)[reply]

What is the common denominator of all the lipids?

I've read the article here, but I couldn't understand the answer and I'll explain why. In biochemistry we have monomers and polymers while the monomer is the basic unit molecule for the polymere. In protein for example, we have amino acid that this is the basic unit of the proteins. Amino acid is characterized as structure of 4 groups: Alpha carbon, amine group, carboxylic group, hydrogen and R chain. this is the basic of all the amino acids. But when we are talking about lipid as a main family of molecules (like what we have we have with protein) then I can not understand what is "amino acids" (parallel) of them, i.e. what is the basic unit of the lipids and what is the structure. Well, I could understand that the fatty acids are the analogous for the amino acids, but the problem is that fatty acids according the article (and other schemes that I saw on google pictures), they are considered as subgroup of the lipids, and that says that not all of the lipids are made of them. So I don't have clue for the answer to my question. 93.126.88.30 (talk) 21:07, 28 November 2016 (UTC)[reply]

Unlike amino acids, lipids as a group don't have a common structural definition. They are defined by certain properties rather than by structure, in particular the properties of being insoluble in water but soluble in certain nonpolar organic liquids. Looie496 (talk) 21:31, 28 November 2016 (UTC)[reply]

Yes, there are some odd "lipids" like cholesterol. But most "lipids" in the body are triglycerides, phospholipids and glycosylated derivatives of those, which have a fairly consistent structure. Blythwood (talk) 23:02, 28 November 2016 (UTC)[reply]

The definition of "lipides" apparently traces back to this article [11] from 1925. Needless to say, as always with ACS, they're demanding credentials to access the content. Aaron Swartz could have suggested something here. In general note that biological terms are typically defined historically and procedurally (though not infrequently they are redefined, which can add to the fun) - an enzyme is defined as an activity that converts one chemical to another in an assay, an allele is defined in terms of how it affects an organism. It came as a pretty big surprise over the years that so many of these things turned out to have a straightforward physical basis, most of the time. Wnt (talk) 01:00, 29 November 2016 (UTC)[reply]

@Wnt: They aren't really demanding credentials so much as payment, either in the individual case or through some academic affiliation. Subtle distinction, and yes, it is bullshit, but let's at least recognize that un-credentialed are welcome to read if they pay enough (for an article whose author, as well as original editors and publishing staff are all dead).

Anyway, the definition of 'lipides' given there is:

“

Substances having the following characteristics : as ether, chloroform, benzene. potential. a. Insolubility in water and solubility in the fat solvents, such b. Relationship to the fatty acids as esters, either actual or c. Utilization by living organisms. Fats-esters of the fatty acids with glycerol. Waxes-esters of the fatty acids with alcohols other than glycerol. Compound lipides. Esters of the fatty acids containing groups in addition to an alcohol and fatty acid. Phospholipides-substituted fats containing phosphoric acid and nitrogen-lecithin, cephalin, spingomyelin. Glycolipides-compounds of the fatty acids with a carbohy- drate and containing nitrogen but no phosphoric acid- cerebrosides. Aminolipides, s’ulfolipides, etc.-groups which are at present : not sufficiently well characterized for classification. Derived lipides. Substances derived from the above groups by hydrolysis. Simple lipides. Esters of the fatty acids with various alcohols. Fatty acids of various series. Sterols-mostly large molecular alcohols, found in nature combined with the fatty acids and which are soluble in the fat solvents-cholesterol ( C2,H4,0H), myricil alcohol (C30He10H), cetyl alcohol (CXH~~OH), etc.

”

Apologies for formatting; I'm happy to provide a pdf of the article to anyone who is interested. SemanticMantis (talk) 16:16, 30 November 2016 (UTC)[reply]

Stubborn fat???

Is some fat stubborn? Like visceral fat being more difficult to get rid off than subcutaneous fat?? — Preceding unsigned comment added by 31.4.136.234 (talk) 23:28, 28 November 2016 (UTC)[reply]

A detailed reading of our articles Adipose tissue and Abdominal obesity does not give a clear cut answer, but suggests that it is difficult to target specific areas or types of fat rather than the body's fat overall (see also Spot reduction).

After reading those articles, you might want to investigate some of the further links in their "See also" sections. {The poster formerly known as 87.81.230.195} 176.248.159.54 (talk) 17:23, 30 November 2016 (UTC)[reply]

November 29

Causes of sudden death in dogs

Is there a list of causes of sudden death in dogs? One of my two dogs just died about an hour ago out of the blue.Uncle dan is home (talk) 02:39, 29 November 2016 (UTC)[reply]

It's going to be essentially the same as in humans: Cardiac arrest, arrythmia, heart attack, stroke, aneurysm, poison and pulmonary embolism. There are other, more rare circumstances, but they'll generally be some major defect in in breathing or circulation. Someguy1221 (talk) 02:48, 29 November 2016 (UTC)[reply]

My condolences. I looked at your user page and saw that your dog was 15 years old, so it was likely something related to old age. Our pet cat developed cancer at around the same age, and we had to have him euthanized. If you're in the Northern Hemisphere, it's possible the cold weather may have exacerbated an underlying health issue. That's why I went to your user page, to see if you stated your location. Let me clarify that I'm certainly not accusing you of neglect; no matter how warm and cozy you are, your body has to work harder to maintain its temperature in cold weather. Also as most people know infectious diseases spread more in cold weather. --47.138.163.230 (talk) 06:25, 29 November 2016 (UTC)[reply]

I'm sorry to hear it. We have an article on Dog health which lists common diseases of dogs as well as the foods and chemicals that commonly cause poisoning, and one on Aging in dogs. Neither makes for very pleasant reading though. Smurrayinchester 09:22, 29 November 2016 (UTC)[reply]

When I lost my last dog, it turned out that she had cancer throughout her body and in many vital organs. She'd shown no sign of it until a couple of days before she died. Our vet remarked that dogs are incredibly tough animals and can keep going despite amazing amounts of disease and injury - but when it finally overwhelms them, they go quickly. There is much to be said for that, I think.

FWIW: The only known cure for the loss you feel from losing an old friend is a new puppy...they have a way of providing the ideal distraction. SteveBaker (talk) 14:52, 29 November 2016 (UTC)[reply]

A new dog -- not necessarily a new puppy. We adopted a sweet little terrier who was maybe 10-12 years old at the time. Already housebroken, and old enough to be calm and not chew up the furniture. There's something special about older dogs, just like there's something special about puppies. Shock Brigade Harvester Boris (talk) 03:13, 30 November 2016 (UTC)[reply]

And by the way, her date of birth is September 11,2002, not 2001.Uncle dan is home (talk) 00:15, 30 November 2016 (UTC)[reply]

Water temperature at the bottom of a kettle

Suppose you heat the bottom of the kettle to a temperature of 200 degrees Celsius, what will be the water temperature right near the metal layer? Will there be a sharp temperature difference or gradient? Gil_mo (talk) 09:59, 29 November 2016 (UTC)[reply]

When you boil a kettle the temperature inside eventually rises to 212 degrees F. As the water reaches that temperature it boils away. I guess this would happen here - so don't leave your kettle in contact with the heat source after the water is all gone. 86.145.54.170 (talk) 10:15, 29 November 2016 (UTC)[reply]

What would happen is this: the outside side of the bottom of the pan would be at 200°C while the inside side of the bottom of the pan would be at 100°C as long as the water lasted. If your pan has poor conductivity (thick stainless steel) you will need less energy to maintain that 100 degree difference. If your pan has good conductivity (thin copper) you will need a lot more energy -- likely more than you have available to you in any kitchen or even industrial setting, and a steady supply of water to replace what boils off. I doubt that any flame would be enough; you would likely need something like liquid sodium at 200°C (boiling point 882.8°C). --Guy Macon (talk) 10:31, 29 November 2016 (UTC)[reply]

...while the inside side of the bottom of the pan would be at 100°C as long as the water lasted... - not true, even under the (possibly wrong) assumption that there is no film of gas causing Leidenfrost effect here. The layer of fluid next to the solid will have a finite temperature difference with the solid just next to the fluid; in other words, such an interface is described by a (finite) heat transfer coefficient rather than by a thermal diffusivity coefficient. Tigraan^{Click here to contact me} 12:31, 29 November 2016 (UTC)[reply]

So to answer Gilmo's question, there will be a "sharp temperature difference" (i.e. a discontinuity in temperature across the interface), but how much exactly (and whether it is negligible compared to conduction inside the kettle, convective effects etc.) cannot be said without more precisions. As Jayron32 mentioned, this is a complex subject. Tigraan^{Click here to contact me} 12:31, 29 November 2016 (UTC)[reply]

What you are asking to describe is a complex physical phenomenon; local temperature will vary throughout a substance due to various effects, including specific heat capacity, temperature differential (i.e. Newton's law of cooling), heat flux, the effect of convection in fluids, etc. There are entire higher-university level courses that are dedicated to teaching the complex relationships and mathematics involved in calculating what you are asking. You might want to read up on things like Non-equilibrium thermodynamics as well. --Jayron 32 12:08, 29 November 2016 (UTC)[reply]

In short, will or will not the water just at the hot solid surface be above 100 Celsius, or not? Should we also take into account that the water at the very bottom is under pressure of the water above, thus raising the boiling point? Gil_mo (talk) 13:49, 29 November 2016 (UTC)[reply]

This image helps to visualize the water in contact with the kettle bottom. The lowest layer of molecules are water vapour i.e. Steam, superheated over 100 °C (212 °F). Water vapour is a gas and a thermal insulator so there is a steep thermal gradient through this layer. However it is usually turbulent and its depth is unstable, except in the special case of the Leidenfrost effect which can occur where there are only small water drop(s) on a very hot surface: a stable cushion of vapour forms under the droplet and insulates it from rapid evaporation. Most of the temperature difference 200°C to 100 °C is is across the vapour layer. Above this is a relatively thick layer of liquid water at 100 °C which absorbs conducted heat from below, the latent heat of vapourization that transforms it to gas. Whether there is a further temperature variation higher in the water depends on how recently the kettle was filled; the liquid may soon all be at 100 °C with streams of vapour bubbling to the surface. Blooteuth (talk) 20:15, 29 November 2016 (UTC)[reply]

If this layer of gas existed at the bottom of an ordinary boiling pot, we would see it as being silvery at certain angles. We don't see the silvery effect, therefore the water is still wetting the metal. Of course once the conditions are right for Leidenfrost effect (as happens when you pour water into an overheated cast iron pan), the physics change radically. --Guy Macon (talk) 22:01, 29 November 2016 (UTC)[reply]

The water will be at 100°C. under normal conditions, water above 100°C isn't water -- it is steam. The water at the very bottom will indeed be under pressure from the water above, thus raising the boiling point a bit, but for a normal-sized pot the effect will be negligible, and smaller than that caused by changes in altitude or barometric pressure. (Note that I said 100°C, not 100.0000°C.) And as long as that water is still wetting the metal, the top layer of atoms in the metal will be at 100°C. --Guy Macon (talk) 21:54, 29 November 2016 (UTC)[reply]

Let's denote the cross section of the bottom of the kettle by $A$ , the temperature on the outside $T_{1}$ , the inside temperature by $T_{2}$ , and the thickness of the kettle by $d$ . Then the heat flux into the kettle is:

J_{1}=\lambda {\frac {T_{1}-T_{2}}{d}}A

where $\lambda$ is the thermal conduction coefficient of the metal. This must be balanced by the heat escaping from the kettle which will be primarily due to the latent heat of vaporization. If the cross section of the opening is $B$ , then this is:

J_{2}=L\rho vB

where $L$ is the latent heat of vaporization, $\rho$ the density of the escaping steam and $v$ the flow velocity of the steam moving through the opening. Equating $J_{1}$ to $J_{2}$ allows you to solve for the mass flux $\rho v$ (note that the latent heat does not depend strongly on the temperature). Conservation of mass implies that the product of the mass flux times the cross section this flux moves through, is conserved. So, this yields the flux of escaping steam from the surface of the water. The next step is to estimate the required overpressure in the kettle that would support this flux, this is not so straightforward it depends on the frictional losses particularly at the opening. This overpressure then yields the temperature via the Clausius–Clapeyron relation. The higher the overpressure, the higher the temperature will be above the normal boiling point of 100°C. Count Iblis (talk) 22:53, 29 November 2016 (UTC)[reply]

Thank you all for your enlightening answers! Gil_mo (talk) 06:41, 30 November 2016 (UTC)[reply]

At a microscopic scale, the temperature of water/steam in bubbles can be surprising at times - see sonoluminescence. Wnt (talk) 12:13, 30 November 2016 (UTC)[reply]

Helicopter (in)stability

Is it true that twin-rotor helicopters are more stable than single-rotor ones? Of all the commonly used rotor systems -- Sikorsky system (conventional single-rotor), NOTAR (if different from Sikorsky system), Piasecki system (tandem rotors), Kamov system (coaxial rotors), and Flettner system (synchropter), how do they rank in the order from least stable to most stable (or if you prefer, from most stable to least stable)? 2601:646:8E01:7E0B:4C25:8F4F:2BC7:C702 (talk) 10:56, 29 November 2016 (UTC)[reply]

Helicopter yaw control

How is yaw controlled in helicopters with tip jets? 2601:646:8E01:7E0B:4C25:8F4F:2BC7:C702 (talk) 10:57, 29 November 2016 (UTC)[reply]

If they don't have a tail rotor (like the NHI H-3 Kolibrie and the American Helicopter XH-26 Jet Jeep), I would expect that they need a rudder (like the Sud-Ouest Djinn), which would depend on the forward motion. However, I don't see one in the image of the Hiller YH-32 Hornet (unless it uses its two tail planes for that), so I don't see how that would work.

A brake could allow turning in one direction, but I don't think that would be used. Rmvandijk (talk) 14:12, 29 November 2016 (UTC)[reply]

Seems to me that because there is no engine inside the body of the helicopter, the torque that normally tries to spin the body of the helicopter in the opposite direction of the rotors is vastly less (presuming the friction in the shaft is kept low). So it might not take very much at all to control the direction of the helicopter. In forward flight, a small rudder would be all that you'd need...it might also be possible to route some of the thrust that goes to the tip jets to a couple of small jets on the tail of the helicopter. However, it's not clear what they *actually* do...these kinds of craft are rare and exotic beasts! SteveBaker (talk) 14:57, 29 November 2016 (UTC)[reply]

Indeed, they are experimental aircraft. This means that they're "non-standard" and can be modified irregularly. Among the few famous tip-jet helicopters, some used bleed exhaust or a drive shaft from the main powerplant to power the tail rotor; some used a redundant smaller powerplant to drive the tail rotor; some did not have any tail rotor at all, like the Hiller Hornet. The point is, these were "skunk works" projects, metaphorically and literally. The designs changed any time the mission planners demanded they needed to change, sometimes bypassing regulatory testing, paperwork, and documentation.

If you get a chance to visit San Carlos, California, the Hiller Aviation Museum has tons of neat stuff. How did these weird aircrafts really work? We shall know soon - it's been nearly fifty years since most of them flew, and the details have begun to emerge... Some say that a band of rowdy helicopter pilots meet on Monday nights to dispel rotorcraft myths and share experiences.

Last week's discussion was on LAX00FA306 ([12], [13]), which has incredible explanatory power for the mandatory Special Awareness Training in 14 CFR 61. Certain helicopters - particularly a few conventional models like R-22 and the UH-1 - do not behave well in zero-G condition. If the pilot attempts to correct this unusual condition by application of cyclic or adverse yaw, the rotors flutter and flap and just ... fly away. We will probably never know if tipjet rotorcraft suffered that problem... but if you visit the CIA Memorial Wall, you can imagine whatever you want about those unmarked stars.

Here's an Army Agency for Aviation Safety video hosted on YouTube: Mast Bumping - Causes and Prevention.

Nimur (talk) 15:27, 29 November 2016 (UTC)[reply]

So, how did yaw control work in the Hornet? 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 21:14, 29 November 2016 (UTC)[reply]

"Barely." From the exhibit page of the Smithsonian's National Air and Space Museum, you can read about the various modifications: in the first incarnation, a rudder (only) was used, and it was controlled by the collective. (I imagine the yaw control wires were bungeed to the stick like some early Bonanzas and the Ercoupe). Later, a (very unusual asymmetric) tail rotor was added to the aircraft, driven by the starter motor (...a battery-powered electric propulsion system). Though its performance is not explicitly described, if you have any familiarity with aircraft design, you can imagine that this probably didn't result in great yaw authority, nor great duration of flight. The Air and Space Museum lists fuel consumption rates, and computes a maximal flight time of 30 minutes; in actual truth, the helicopter probably could not fly at all. (In the United States, it is generally illegal to begin a flight in an aircraft that is carrying less than 30 minutes of fuel remaining on board, 14 CFR 91.151 - well, twenty minutes for rotorcraft, so ...).

"The flames coming out of the ramjets produced an incredibly bright white halo when the HOE-1 was flown at night. This was a considerable disadvantage in the military environment, and the effect prompted a number of UFO sightings when operated in the vicinity of populated areas.... The noise generated by the ramjets was also quite considerable, and did not endear the United Helicopter's Palo Alto, California facility to its neighbors." (Smithsonian Institution).

The interested reader will be happy to know that the Palo Alto test facility, declassified by the National Reconnaissance Office in year 2006, was not a Hiller Helicopter facility, as it was loudly and widely advertised and publicly known; but the location was in fact a CIA "General Participant" development facility for Project CORONA, America's first observation and surveillance satellite. You can find some great historical photographs of the facility during the 1940s and 1950s in this book, Palo Alto Over Time, featured in the Palo Alto Daily News. Interestingly, the location of this facility was on Willow Road, which is today the mailing address of a more modern ~~surveillance program~~ corporation.

When one wants to keep a secret project quiet, one has to find some excuse for moving so many engineers into and out of a building each day.

Nimur (talk) 21:40, 29 November 2016 (UTC)[reply]

Mostly it doesn't need to be. There's no major torque reaction to deal with, so no need to counteract it. Yaw is only needed for literally yawing the helicopter. As helicopters also have a cyclic pitch which can translate them sideways, then there's not even as much need for that.

Tip jets can be hot or cold. Cold tip jets are powered from a compressor bleed on a gas turbine, implying that there's a supply of pressurised air available and so puffer jets could be used (as worked for the Hawker Harrier), although I can't think offhand of a tipjet helicopter which used them. A few helicopters, particularly the French designs, but also the Fairey Ultra-light used a tailplane with conventional rudder and their gas turbine exhaust over this was enough to give very good yaw control, even in the hover.

Hot jets, such as the British fuel-burning jets supplied with air from a central compressor, or the US designs with a self-contained jet on each rotor, have a harder time of it. The Hiller Hornet was tested as a gunship (claimed to be the first, although the Germans had tried it during WWII) but its poor yaw control put paid to that. Various Gyrodynes, such as the Fairey Rotodyne and its precursors, had good yaw control at speed, by controlling their propulsion props, but were limited in the hover. A few, such as the Jet Jeep and the (un)Flying Ocarina, took a geared drive from the main rotor hub and used this to drive a conventional helicopter tail rotor, although much undersized from the usual. Andy Dingley (talk) 16:36, 29 November 2016 (UTC)[reply]

Do you know if the control surfaces at the tail of the Hiller Hornet (Are those inverted ruddervators?) gave some yaw control even while hovering by deflecting downwash? -- ToE 13:15, 30 November 2016 (UTC)[reply]

November 30

Set in the stars....

Whether your favorite iconic image is Trump, Castro, or Obama, if it can be drawn with simple, heavy lines, it seems like in theory, a field of stars might line up with that artwork in a suggestive way. As a very large number of stars have been mapped by now, I'm thinking that for each of these folks there may be a constellation waiting to be promulgated, even if it might be a very small one. But has anyone designed a publicly available interface by which a piece of submitted art can be searched among all the stars at all possible scales until something abnormally suggestive is uncovered? Is there a statistical analysis possible for how good or bad a match that might turn out to be on average, given available data? Wnt (talk) 02:43, 30 November 2016 (UTC)[reply]

Just for reference, this is a rendering of every star visible to the naked eye, in a light-pollution free area, from...somewhere, not really sure. Now, this is a bit less than you'd be concerned with, since this is everything you'd be able to see standing still, not counting what stars might be behind you, etc. Someguy1221 (talk) 02:50, 30 November 2016 (UTC)[reply]

Sorry, to be clear, I didn't just mean a search for small constellations of the desired shape, but also faint ones - even if a constellation is only visible through a powerful telescope, there is still substantial amusement value to be had if someone found it and posted an image, where people could say "oh yeah, that looks like him" or have fun peering at adjacent stars and saying "you know, I think there's somebody over there behind Kennedy on that starry knoll..." Wnt (talk) 12:08, 30 November 2016 (UTC)[reply]

A bit of terminological pedantry: what you are describing would be asterisms, not constellations. {The poster formerly known as 87.81.230.195) 176.248.159.54 (talk) 14:14, 30 November 2016 (UTC)[reply]

You may be interested in the Coathanger. Sagittarian Milky Way (talk) 15:00, 30 November 2016 (UTC)[reply]

If you mean that anyone looking at that set of stars would say "Wow, that's person X !", I doubt that is the case. Even most of the current constellations really don't objectively look like what they are said to represent, with a few possible exceptions, like the Big Dipper, and that's only 7 stars. To make a convincing representation of a PARTICULAR person, you would need far more stars than that, and without extraneous stars visible in the same field. On the other hand, for a generic "smiley face", perhaps as few as 5 stars could work, with 2 for eyes and 3 to form the mouth. StuRat (talk) 15:17, 30 November 2016 (UTC)[reply]

Borked Latin taxonomic names

It seems that ungrammatical Latin taxonomic names (like Baracktrema obamai instead of correct Latin genitive obamae) have been around for at least several years. Presumably scientists are supposed to know basic Latin to give correct names. Is it centuries-old or a relatively recent phenomenon? Brandmeister^talk 11:50, 30 November 2016 (UTC)[reply]

It's been a complaint since at least the 1890s. So, depends on your definition of "recent". --Jayron 32 11:56, 30 November 2016 (UTC)[reply]

Thanks, didn't know it goes back to the 19th century. Weird. Brandmeister^talk 12:01, 30 November 2016 (UTC)[reply]

Note also that the name "Obama" is not Latin: it is from a Kenyan dialect, Dholuo. In order to say that obamae is the "correct Latin genitive", you need to assume that the name is taken into Latin with the same spelling and assigned to the first declension. It would be equally reasonable to Latinize it as some other declension. If you choose to use the second declension, Latinizing by adding -us to the name to make the nominative Obamaus, then the actual form obamai is correct. And this becomes more a Language question than a Science question. --76.71.5.45 (talk) 13:19, 30 November 2016 (UTC)[reply]

Only tangentially related, but also check out This common mistake which also involves bad latin. --Jayron 32 13:42, 30 November 2016 (UTC)[reply]

Recall that, generally speaking, binomial nomenclature doesn't really have anything to do with Latin per se, people can and do put in any kind of nonsense they want. Binomial_nomenclature#Derivation_of_binomial_names lists some non-latin, non-greek examples. Also, while some taxonomists may be a bit sharper on their classics, modern scientists in the USA generally receive no mandatory formal training in Latin or Greek, though some of us did choose to pursue it. SemanticMantis (talk) 16:05, 30 November 2016 (UTC)[reply]

Elements 119-120

Now that all of the first 118 elements have been named, are scientists working some on trying to create atoms of elements 119-120?? If not, please explain what they're waiting for. Georgia guy (talk) 14:40, 30 November 2016 (UTC)[reply]

They are! But new technology is needed and 119 and 120 are stretching the limits of current technology. We should be able to get to these elements in the next five years (though it could be as soon as next year or as late as in ten years). Double sharp (talk) 15:00, 30 November 2016 (UTC)[reply]

See the Wikipedia articles titled Ununennium and Unbinilium. --Jayron 32 17:25, 30 November 2016 (UTC)[reply]

@@ Line 317: / Line 317: @@
 Is some fat stubborn? Like visceral fat being more difficult to get rid off than subcutaneous fat??  <!-- Template:Unsigned IP --><small class="autosigned">—&nbsp;Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/31.4.136.234|31.4.136.234]] ([[User talk:31.4.136.234#top|talk]]) 23:28, 28 November 2016 (UTC)</small> <!--Autosigned by SineBot-->
-:A detailed reading of our articles [[Adipose tissue]] and [[Abdominal obesity]] does not give a clear cut answer, but suggests that it is difficult to target specific areas or types of fat rather than the body's fat overall.
+:A detailed reading of our articles [[Adipose tissue]] and [[Abdominal obesity]] does not give a clear cut answer, but suggests that it is difficult to target specific areas or types of fat rather than the body's fat overall (see also [[Spot reduction]]).
 :After reading those articles, you might want to investigate some of the further links in their "See also" sections. {The poster formerly known as 87.81.230.195} [[Special:Contributions/176.248.159.54|176.248.159.54]] ([[User talk:176.248.159.54|talk]]) 17:23, 30 November 2016 (UTC)