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:A computer simulation may be in order in such cases. If the dynamic loads vary independently of each other, you might vary each independently in the simulation and calculate the moment and stress at each time to determine the max for each (again, the maxes may occur at different times). [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 12:50, 12 April 2014 (UTC)
:A computer simulation may be in order in such cases. If the dynamic loads vary independently of each other, you might vary each independently in the simulation and calculate the moment and stress at each time to determine the max for each (again, the maxes may occur at different times). [[User:StuRat|StuRat]] ([[User talk:StuRat|talk]]) 12:50, 12 April 2014 (UTC)

::Thank you for your answer but I don't quite understand what you mean. The dynamic load factor is a ratio of the deflection as a result of the dynamic load to the equivalent static load deflection. So from this point, I would probably need to do some sort of energy analysis to find the maximum force but I'm not sure what the correct equation is.

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April 7

Induction lamp cooking

As far as I know a powerful enough oscillating magnetic field can light up a lamp. Does it mean that you can make a light bulb glow by putting it on the induction cooking stove? 91.77.167.192 (talk) 08:38, 7 April 2014 (UTC)[reply]

Those induction cookers need a big thick slab of metal to induce currents in. A light bulb filament is not enough. If you irradiated it with millimeter wave or other microwaves, perhaps you could light it. Graeme Bartlett (talk) 09:06, 7 April 2014 (UTC)[reply]
Centimeter wavelengths will certainly light up an incandescent light bulb as generations of schoolboys and numerous You Tube videos can attest. --2A01:E34:EF5E:4640:9D8E:1D42:1F74:5600 (talk) 13:33, 7 April 2014 (UTC)[reply]

And what if you attach a light bulb to this “thick slab of metal”, will it light up? I'm just curious if it is possible to get something glowing with a cooker of this type.91.77.167.192 (talk) 13:25, 7 April 2014 (UTC)[reply]

I don't suggest you do the experiment, but I would think it more likely to melt. What you want is an Induction lamp.--Shantavira|feed me 14:01, 7 April 2014 (UTC)[reply]

Could a machine assemble molecules on demand just starting with the atoms?

That would be easier than synthesizing compounds. If you wanted gasoline (C8H18) or cocaine (C17H21NO4), this hypothetical machine would just pick the right atoms and start putting them together. Or maybe, could a machine generate random molecules from a set of atoms? Maybe putting them under pressure, or pass an electrical current through them and see what gets out of the process. — Preceding unsigned comment added by 88.14.192.119 (talk) 17:58, 7 April 2014 (UTC)[reply]

How do you put them together fast enough? 12 grams of carbon contains 602,000,000,000,000,000,000,000 atoms. Even if a machine could grab and use a million carbon (and equivalent number of hydrogen) atoms per second, that would still take 602,000,000,000,000,000 seconds to make just 16 grams of methane (CH4). That's a bit more than 19,000,000,000 years. The earth has only existed for 4,500,000,000 years. --Jayron32 18:15, 7 April 2014 (UTC)[reply]
(ec, same idea, other numbers) I think you underestimate the number of molecules that make up a useful amount of gasoline. A molecule of octane weights 114 u. Thus a mol of gasoline weights 114 g. That's 6.022×1023 molecules - in other words, one litre has about 3.6×1024 molecules. If your machine makes a million molecules per second, it would have made about 0.01 litres in the time since the beginning of the universe. A big advantage of chemical reactions is that they happen in a massively parallel manner, thus yielding macroscopic amounts of product in realistic time frames. --Stephan Schulz (talk) 18:21, 7 April 2014 (UTC)[reply]
Hmmmm, isn't your body made out of such machines? Count Iblis (talk) 18:42, 7 April 2014 (UTC)[reply]
No, it isn't. You can not program yourself to excrete gasoline or cocaine out of your nose. You (that is, any human) can not even produce all the amino acids the body needs to maintain its structure and function; some of them are essential amino acids, meaning you've got to get them from outside (as food) and can not manufacture them internally (although some other organisms can). There are organisms that can produce hydrocarbons suitable as fuel for internal combustion engines; but there are no "universal" organisms that can produce any chemical compound you want on demand. --Dr Dima (talk) 19:17, 7 April 2014 (UTC)[reply]


That is the intent of Molecular assemblers - an idea to make very, very tiny robots that are capable of assembling things atom by atom. The way such devices would overcome Jayron's objection is by self-replicating themselves - making enough of them to do the needed work in a reasonable amount of time - then disassembling each other again when the work is done. It's not entirely clear whether such things are possible, given the laws of physics as we know them...but the idea is not without merit.
HOWEVER, this won't get you free gasoline. The amount of energy a machine such as this would require to make gasoline would be many, many times more than the energy you'd get back by burning the gasoline subsequently. The general problem is that you need to apply significant amounts of force to (for example) dismantle a CO2 molecule into carbon and oxgen atoms - and yet more energy to force them back together into the form you wanted.
Anyway - check out the Molecular assembler article - it pretty much explains all of this. SteveBaker (talk) 19:12, 7 April 2014 (UTC)[reply]

Are we going out of fertilizer ingredients, in the same way we are going out of oils?

They are the six macronutrients: nitrogen (N), phosphorus (P), potassium (K), calcium (Ca), magnesium (Mg), and sulfur (S); eight micronutrients: boron (B), chlorine (Cl), copper (Cu), iron (Fe), manganese (Mn), molybdenum (Mo), zinc (Zn) and nickel (Ni). I dont mean locally on the soil of a concrete location, but globally. — Preceding unsigned comment added by 88.14.192.119 (talk) 18:06, 7 April 2014 (UTC)[reply]

Peak phosphorus, Peak copper -- Finlay McWalterTalk 18:09, 7 April 2014 (UTC)[reply]
(edit conflict) See Peak phosphorus. This is considered, in some circles, to be a greater problem than Peak Oil, and a major looming catastrophe, specifically because of the need of phosphorus in fertilizers. --Jayron32 18:11, 7 April 2014 (UTC)[reply]
Note that only chemical fertilizers are in danger of running out. There's no shortage of manure (especially around elections), and, if night soil is included, we have quite a dollop of that. StuRat (talk) 18:22, 7 April 2014 (UTC)[reply]
Au contraire, the problem is there isn't enough bullshit in the world, even accounting for politicians, to cover all the fertilizing needs we have. --Jayron32 18:24, 7 April 2014 (UTC)[reply]
If 100% is collected and reused, and no-till farming is employed, along with other methods to prevent soil run-off, then I'd expect very little chemical fertilizer would be needed. StuRat (talk) 18:35, 7 April 2014 (UTC)[reply]
(if 100% is reused, then we have no harvest! ;) Our yield of e.g. corn is approaching ten times the historical values [1]. Some of the increase is due to pesticides, breeding, and other methods. But without cheap, concentrated chemical fertilizers, corn yields would be back in the ~20 bushels an acre across the USA cornbelt. And that would drive up the cost of the many food products using corn and it's derivatives. Sure, you can compost everything and do low-input agriculture in your backyard, but that doesn't change the fact the the majority of the agricultural industry in the USA is based off of cheap and widely-accessible processed fertilizers. It is true that agronomically, we know how to do better, and grow with less energy and inputs. But what matters is what the large agribusinesses actually do. SemanticMantis (talk) 18:53, 7 April 2014 (UTC)[reply]
You'd need to set up a system to collect sewage from sewage processing plants and move it to farms. We'd also need to stop flushing old meds, etc. A dual sewage system could be set up, with things like meds going to a septic tank, instead. StuRat (talk) 19:06, 7 April 2014 (UTC)[reply]
  • Um, phosphorus isn't destroyed. If run-off into the oceans is a problem, then the oceans are also the solution--recovery from salt pans and, as the price increases, use of nuclear energy to extract trace elements from the ocean. It's simply silly to assert we are running out. It's just becoming more expensive to procure. μηδείς (talk) 20:58, 7 April 2014 (UTC)[reply]
I would disagree with that. Recycle before it goes into the sea would help prevent algae blooms that cut off light to the sea-floor, which cause the seabed plants to die off, which in-turn removes a vital food source for the local coastal sea-life which in turn means we reduce the amount sustainable seafood that we enjoy. This is a growing problem around the world and some sizeable lakes too are now dead. The sea is too large a volume for the concentration of Phosphorus to increase above 13mg/kg to point were it becomes economic to recover it.--Aspro (talk) 21:55, 7 April 2014 (UTC)[reply]
I didn't say anything about not recycling or preventing run-off, which will also become increasingly economically urgent. The problem is the question as posed is a typical "what if all the farmland gets used for parking lots?!?!" question that ignores market forces. As phosphorus or farming land or any commodity gets more scarce its value increase, and more money is invested into procuring phosphorus or land values increase, and people sell their sprawling lots and move into efficient high rises high-rises. You are making the same mistake. At some point if phosphorus becomes rare its price will increase and people will invest in new ways to concentrate it and it will be economical to recover it at 13mg/liter.
Hmmm, 160E6 metric tonnes/ yr currently produced = 1.6E11 kg, divided by the 13 mg/kg seawater = 1.2E16 kg of seawater to process. The Nile discharges 2830 m3/s, x365.25x24x60x60x1E6 = 8.9E16 kg/s. So if you desalinate a substantial river of water to meet the needs of dry surface areas, in theory you should get enough potassium phosphorus left over to do the fertilization. If you can separate it somehow. It takes a lot of energy, but so does any solution to the vast deserts predicted to arise from global climate change. Wnt (talk) 23:08, 7 April 2014 (UTC)[reply]
When did we switch to potassium? There's no shortage of potassium. I'm assuming you just mistyped. --Trovatore (talk) 23:36, 7 April 2014 (UTC) [reply]
Burn seaweed in a steam engine (Algae fuel) then process the ashes. Jim.henderson (talk) 23:43, 7 April 2014 (UTC)[reply]
This brings back the old proposal to build a canal to flood the depression in the African dessert from the Med Sea. You could then let it dry out to get your phosphorus, but you'd also have to remove the salt. StuRat (talk) 23:44, 7 April 2014 (UTC)[reply]
Ah - Qattara Depression Project. Wnt (talk) 00:30, 8 April 2014 (UTC)[reply]
Yep, that's the name. Thanks. StuRat (talk) 08:44, 9 April 2014 (UTC)[reply]
Or process brine left over as water evaporates to a low level behind the proposed Red Sea Dam. Since the Red Sea is of fairly recent origin, it's hard for me to feel too bad about destroying a major ecosystem, if the energy goes a substantial way toward ending global warming and provides irrigation for the entire region. Besides, I have a crank notion about it as a prophesied solution to the Israel-Palestine conflict. :) Wnt (talk) 00:18, 8 April 2014 (UTC)[reply]
(Un-indent) What is all this idle talk about extracting phosphorus from seawater, flooding the Sahara, damming the Red Sea, etc., when a much more cost-effective solution has already been devised -- that being to precipitate phosphates out of municipal sewage (at the sewage treatment plant) with, e.g., calcium hydroxide? 24.5.122.13 (talk) 05:13, 8 April 2014 (UTC)[reply]
I think people are assuming that first we manage to use up all the cheap phosphorus, then we say oh no, what are we going to use, when it's already washed out to sea. Which, given our track record, seems not unrealistic. :( Besides, I was thinking of this as a byproduct only - the purpose of the Red Sea Dam is electrical power, and if taken to a perhaps unrealistic extreme, creation of new lands irrigated by water desalinated with that abundant power. Wnt (talk) 19:11, 8 April 2014 (UTC)[reply]
As Wnt correctly stated (I assume it was Wnt, even though the comment appears to be unsigned), the price of phosphorus will begin to rise LONG before all the apatite is mined out, and THAT will make large-scale recovery of phosphorus from sewage the next major source of the stuff even BEFORE the apatite runs out -- and THAT, in turn, will preclude your scenario of all the readily available phosphorus "washing out to sea". 24.5.122.13 (talk) 05:59, 9 April 2014 (UTC)[reply]
I don't see the unsigned comment, but ... while biology may not know theory, economics is outright hostile to logic. The system that gives us Bitcoins and housing bubbles may or may not react sanely to a shortage of phosphorus. Wnt (talk) 11:42, 9 April 2014 (UTC)[reply]
[citation needed] 24.5.122.13 (talk) 01:07, 10 April 2014 (UTC)[reply]
Economics is exceedingly and brutally logical. It's people that aren't. The market makes very logical and unemotional corrections. People don't like it and try to manipulate it to "soften" the reality. Left on it's own, the market will correct for overpopulation. It's people that object to the outcome. (people manipulate the other way, too, like subsidizing fuel from corn/food). --DHeyward (talk) 02:46, 10 April 2014 (UTC)[reply]
Considering that the "market correction" for overpopulation involves millions of people dying, by starvation, war, disease, etc., I can understand why some might object. (It's even possible that it could lead to the extinction of the human race, via nuclear or biological warfare.) StuRat (talk) 03:03, 10 April 2014 (UTC) [reply]

LCD TVs that are not black?

My father had a question for me last week. One of his customers had told him that LCD TVs would fit much more nicely to their home design if they were white when powered off instead of black. My father asked me if this is possible. I didn't have any idea, as electronics design is not my profession, it's computer software. I know at least that some monochrome LCDs are light gray when powered off, and the lit crystals are black, but is this kind of thing at all possible on a full-colour LCD? JIP | Talk 18:17, 7 April 2014 (UTC)[reply]

I suggest the TV be placed in a cabinet and the doors closed when not in use. (Eventually, I hope the cost of displays will come down and reliability will go up to the point where we can use them as lights, always on, displaying an all-while panel or whichever background we prefer.) StuRat (talk) 18:29, 7 April 2014 (UTC)[reply]
One problem I foresee is that if the error correction is not spot on for every frame, one will notice lots of bright speckles on the picture. Whereas now, when a few pixels at a time are randomly going dark here and there for each and every frame, they are not perceived by the eye. I seem to remember the luminous signal on analogue TV is negative going, for the reason that the radio static is less noticeable when there is a good signal to noise ratio. When the S/N ratio diminishes towards no broadcast signal at all, even the cosmic background radiation can be seen as a mass of speckles often referred to as snow.--Aspro (talk) 22:57, 7 April 2014 (UTC)[reply]
Permit a sceptic to doubt that Aspro's TV is capable of displaying cosmic background radiation. 84.209.89.214 (talk) 12:12, 8 April 2014 (UTC)[reply]
Oh. It is always wise to be a sceptic (especially if the inland revenue says you owe more tax than the 1% pay in ten years) - so count yourself permited. Here is a link (obviously, if the Big Bang theory gets proved wrong – you can demise it) : The CMB is so bright at millimetre-wavelengths that if you de-tune an old analogue TV to show the snow-like static, a few percent of the signal your TV is picking up will have come from the start of the Universe.. Not a lot of people know that.--Aspro (talk) 23:40, 8 April 2014 (UTC)[reply]
And Wikipedia turns out to have a section on the disadvantage of positive modulation (luminance) that I was referring to. 405-line_television_system#Susceptibility_to_impulse_interference.--Aspro (talk) 00:02, 9 April 2014 (UTC)[reply]
Analogue TV's receive at metre and decimetre (VHF and UHF) but not millimetre wavelengths and their no-signal display of electrical noise is dominated by front-end Johnson–Nyquist noise i.e. thermal noise in the input stage transistor. The quoted claim that Cosmic noise forms "a few percent of the signal your TV is picking up" lacks any quantitative values of power or criteria for why CMB is purportedly visible. Does the random speckle display on Aspro's TV look noticeably different when it is moved from a tent to a metal shielded room? Is Aspro in the habit of lecturing to awed visitors that his TV is a viewport to the start of the Universe? And there's no argument about the visibility of impulse interference. 405-line monochrome TV receivers often had an "Interference limiter" control, actually a peak video level clipper or inverter, that had to be carefully adjusted somewhere between white crushing or having white spots bloom on the screen when cars drove by. 84.209.89.214 (talk) 14:17, 9 April 2014 (UTC)[reply]
The CBM 'peaks' at millimetre wave lengths. The gaussian function means there is a few percent of the electromagnetic spectrum in meter wavelength noise still to be received on one's old TV, both visually and backed-up by scientific deduction. As to awe: Here on Ref Desk, we endeavorer to answer the questioners queries. Whether they experience awe or wonderment at our replies is out of our control. Please don't try to colour my motivations with whatever is going on inside you.--Aspro (talk) 23:01, 11 April 2014 (UTC)[reply]
I don't believe that this is easy. Current TVs display white by turning on a backlight, and making the LCD clear so the white backlight can shine through. (roughly speaking) So, unpowered, they're going to be, if not black, then at least dark.
Purely reflective screens can be light-colored when they're powered off. (Think of a digital watch, or a calculator.) But I don't think those would make a very good TV. (Which you want to glow in the dark.) APL (talk) 15:28, 8 April 2014 (UTC)[reply]
From what I understood from what my father's customer wanted, they wanted the screen to be white without any sort of electricity or light output required. It would be practically the inverse of current full-colour LCD technology - instead of having the pixels dark and grow lighter when powered up, they would be light and grow darker when powered up. I understand that this means that when viewed with no outside light source, such as in a darkened room, the whole TV would be pretty much invisible, regardless of what it was showing. I just want to know that even if we accept this, is this kind of thing at all possible, in full colour? JIP | Talk 18:10, 8 April 2014 (UTC)[reply]
3-color reflective LCD display
I think HP used to make some color graphing calculators with that sort of display, but I'm having trouble finding a reference so it may have been a different brand. Obviously the update time and pixel density were nowhere near what you need for a TV, but the basic principle has been implemented. Katie R (talk) 11:52, 9 April 2014 (UTC)[reply]
Ahh, here it is. Casio's "three-color" display did black red green and blue. It's a looong way from a TV display, but no one has had a reason to develop that technology. Katie R (talk) 11:57, 9 April 2014 (UTC)[reply]
Ah, I had almost forgot about those Casio colour graphics calculators. I used to own a couple of models myself, and I think I still have one. From what I can recall, they accomplished the four-colour reflective LCD with two monochrome LCDs laid on top of each other, with different colours. This leads to a new question: Both Casio and Texas Instruments have now developed graphing calculators with 65536-colour displays. Does anyone know if these LCDs also utilise this reflective technology, or are they backlit the same way as LCD TVs and monitors? JIP | Talk 18:23, 9 April 2014 (UTC)[reply]
I have a coworker that just purchased a new TI one. It's a backlit display, and it seems a bit lower quality than the average modern smartphone screen, but incredible compared to the older calculators. Of course, the battery doesn't last for months like the old ones did. Katie R (talk) 19:45, 9 April 2014 (UTC)[reply]
It's certainly possible to make color displays that are reflective instead of backlit. Pixel Qi makes some that will replace the screens on certain laptops, so you could certainly watch a movie on them if you wanted. It'll never get the brilliant colors of a backlit TV on a purely reflective screen though, it's always going to look a bit muted. (But maybe you'd get used to it. Even a glossy magazine cover looks dull and lifeless if you hold it up next to your TV, but they look great on the newsstand next to the other magazines.) APL (talk) 14:15, 9 April 2014 (UTC)[reply]


Hmmm, how's electronic paper doing? (I think it still has problems, and even in concept might not be well suited for TV use) Wnt (talk) 19:08, 8 April 2014 (UTC)[reply]

Is it true that the Pacific Ocean makes the west coast of North America warmer?

My friend told me he wants to move to Vancouver because it's warmer there, and he insists that it's warm because the Pacific Ocean makes the climate warmer there. Is this true? ScienceApe (talk) 19:04, 7 April 2014 (UTC)[reply]

Yes, similar to how the Gulf Stream in the Atlantic makes Western Europe warmer. The water can be warmer near Vancouver, Seattle, etc, than in Los Angeles. Of course, that warm water also brings lots of rain, and global warming may eventually change the ocean currents. StuRat (talk) 19:11, 7 April 2014 (UTC)[reply]
It is Alaska_Current. Ruslik_Zero 19:12, 7 April 2014 (UTC)[reply]
Thanks, StuRat (talk) 08:46, 9 April 2014 (UTC)[reply]
More moderate; both less cold and less hot; see Oceanic climate Jim.henderson (talk) 19:13, 7 April 2014 (UTC)[reply]
See http://cliffmass.blogspot.com/2012/06/why-are-coastal-water-temperatures.html for a pic showing unusual cooler water temps hugging the coast. StuRat (talk) 19:16, 7 April 2014 (UTC)[reply]
(ec)Warmer than where? Large bodies of water have, in general, a mediating influence on peak temperatures, both high and low. So Seattle is warmer in January than e.g. Bismarck, North Dakota. On the other hand, Bismarck is warmer than Seattle in Juli. Also see continental climate vs. oceanic climate. --Stephan Schulz (talk) 19:17, 7 April 2014 (UTC)[reply]
Note that water behaves like land if it freezes solid, but remains mostly liquid near Vancouver year-round. And where the currents go the other way, as in Maine or Korea, it's not particularly warm in winter. StuRat (talk) 19:19, 7 April 2014 (UTC)[reply]
Even in Maine, costal areas are a lot warmer in winter than North Dakota. Similarly, South Korea is warmer in winter than North Dakota. North Korea has a continental climate, with winter temperatures influenced by weather systems coming out of Siberia. But still, its winter is not colder than North Dakota's. --Stephan Schulz (talk) 21:46, 7 April 2014 (UTC)[reply]
Still, if one wants a Canadian city with a mild climate, with little snow, then Vancouver is a better choice than Newfoundland: [2]. StuRat (talk) 00:28, 8 April 2014 (UTC)[reply]
Well, St. John's, Newfoundland and Labrador is about 150 miles further north than Vancouver, so that may have a bit to do with it as well. --Jayron32 01:49, 8 April 2014 (UTC)[reply]
  • "The Pacific Ocean makes the west coast of North America warmer" is only true down to the latitude of Seattle or so. Farther south the ocean has a cooling effect. In San Diego, for example, it is usually very pleasant if you are within a couple of miles of the ocean, but rapidly gets blazing hot if you go a little farther inland. Looie496 (talk) 01:57, 8 April 2014 (UTC)[reply]

If all you care about is the annual average temperature, then the Pacific ocean doesn't have a large effect. Along the Pacific coast, annual average temperatures run from about 1 C below to 2 C above what you would expect based solely on latitude and elevation. That said, the ocean has a very strongly moderating effect on seasonal and weather variation. The difference between winter and summer along the Pacific coast is only 10 C or less, while 20-30 C ranges are typical for most of the US. Similarly, there is much less variation is weather. The Pacific Coast isn't much warmer on average, but it is more predictable, and one can avoid both extreme heat and cold. Dragons flight (talk) 19:42, 8 April 2014 (UTC)[reply]

Like others here, I too suspect Science Ape's friend is referring to the low temperatures (winters) being warmer on the Pacific Coast. This map of the US, for example, shows the average annual extreme minimum temperatures 1976 - 2005. I found it in our article on hardiness zone. ---Sluzzelin talk 09:07, 9 April 2014 (UTC)[reply]
Interesting. I wonder why the Pacific coast has such a wide band which is warmed by the sea, while in Maine it's such a narrow band. Is this due to the prevailing wind direction ? StuRat (talk) 09:55, 9 April 2014 (UTC)[reply]
Yes - see Westerlies. Wnt (talk) 18:24, 9 April 2014 (UTC)[reply]
Thanks. StuRat (talk) 16:16, 10 April 2014 (UTC)[reply]

Inflatable hot tub mystery.

We have an inflatable hot-tub. It's construction is more or less identical to a kiddies paddling pool - three stacked inflatable rings with an inflatable "floor", each with it's own air valve. The "floor" valve is underwater, the other three are on the outside of the tub. The thing is made of some kind of very heavy duty vinyl and is about 5' in diameter and two feet deep and contains around 250 gallons of water at around 38 degC (100F). There is a separate heater/water pump and a gizmo to blow air into the water via a channel around the inside of the bottommost ring..but I don't think that's relevant here. We've been using it off and on for about 6 months - and it's never been completely emptied or deflated in all that time.

The other day, we had occasion to empty the water and deflate all of the air chambers. I was surprised to find that the "floor" had about two gallons of water sloshing around INSIDE the sealed air bladders.

I'm totally mystified as to how the water got in there! The floor is under considerable pressure (2 feet of water resting on it) - so if there were some kind of a leak, you'd expect it to deflate in short order - and it didn't seem to be deflated to any noticable degree. The air must be at about the same pressure as the water when the tub is full. It's conceivable that the air we put into the floor part at the outset was really humid - but even if all of the water vapor condensed out of that air...there was FAR too much water in there to be accounted for in that way.

It's not exactly important that I know the answer - but it's been bothering the heck out of me for the past week...and I can't explain it!

There is a picture of the hot tub here:

 http://www.ebay.com/itm/like/350868160135?lpid=82

...ours has an outer protective cloth cover in bright blue and an inflatable "lid" that covers it when not in use - but that's the thing I'm talking about here.

SteveBaker (talk) 19:31, 7 April 2014 (UTC)[reply]

I'd say an equal volume of air leaked out as the water that leaked in, so the total volume remains the same. The water likely leaked in at the valve, the air might have also leaked out there, but too slowly to notice the bubbles, or perhaps there's a pinhole leak elsewhere. The water pressure might go up instantly when you first sit in the tub, just enough for a few drops of water to drip in each time.
It might actually make more sense to fill the bottom with water than air, as that should make it more stable in the wind. It would also make it heavier, but if you don't plan to move it until you drain it, then that's OK. StuRat (talk) 20:06, 7 April 2014 (UTC)[reply]
My problem with that idea is that when you step into the tub, the weight of your foot on the floor must increase the pressure of the air inside above that of the water on top of it...so I could believe in a bubble being expelled through a tiny hole each time - but the water would have to force it's way in through a positive pressure gradient. It's not like the floor is a rigid container either - it's flexible/stretchable plastic - and you'd think that the tension it adds would keep the air pressure consistently higher than the water pressure. SteveBaker (talk) 20:29, 7 April 2014 (UTC)[reply]
In that case, maybe when you get out, the lowered pressure in the air bladder sucks in some water to replace the gas bubbles which escaped when you got in (escaped from the air bladder, not from you, as those gas bubbles aren't relevant here). StuRat (talk) 00:25, 8 April 2014 (UTC)[reply]
I think a pinhole leak/defect is always possible; it's also conceivable that water could permeate the polymer directly. (Trying to calculate that, based on the exact polymer and plasticizer compositions, would be a decent test of the Refdesk). Wnt (talk) 22:54, 7 April 2014 (UTC)[reply]
Even if there is just a bit of vapor permeability (whether directly through the polymer layer, or through the valves) there will be tendency for liquid water to accumulate on the bottom of the bladder. The liquid in the hot tub (and the polymer layer in contact with it) is likely to often (or always) be at a temperature higher than the ground under the tub. Warm vapor that enters through the top side of the bladder cools and condenses when it reaches the bottom, forming a pool of liquid water...that will just keep growing, for however many months the tub remains in use. TenOfAllTrades(talk) 14:02, 8 April 2014 (UTC)[reply]
Did you inflate it yourself? Is it possible someone deliberately put water in to anchor it before inflation? -- Q Chris (talk) 14:52, 8 April 2014 (UTC)[reply]
I'm going to guess that the floor has one of those valves that is in-only unless you squeeze it a certain way.
If so, and if the water pressure was higher than the floor's air pressure (or became so thanks to temperature changes, or momentary changes in the weight people put on it.) water could flow into the floor without an actual "leak", but air would not be able to flow back out, per its design.
Ok, it's not a perfect theory, and it requires that the cap on the valve is not water-tight, but it could be close. APL (talk) 15:11, 8 April 2014 (UTC)[reply]

April 8

3D Printing using soil

Are there any useful building materials in soil? Like if one day in the future they have such advanced 3D printers that you can just take some soil from your back yard, dump it into the machine, and it's able to build something useful out of it. ScienceApe (talk) 13:06, 8 April 2014 (UTC)[reply]

Depending on what type of soil you have, perhaps:
1) Clay: Can be fired to make ceramics.
2) Sand: Can be melted to form glass.
In both cases, you can't use current 3D printing techniques to make objects directly out of those materials, but you might be able to use a mold or form you create using 3D printing to make an object out of clay or sand. StuRat (talk) 13:13, 8 April 2014 (UTC)[reply]
Yes, there are useful building materials in soil. You don't need to use a 3D printer. Mudcrete is often used in road construction, while Rammed earth has been used for building for many millennia. As you will see from the top of the page, though, we are unable to provide predictions or speculation about future technological developments. RomanSpa (talk) 13:25, 8 April 2014 (UTC)[reply]
Soil has been used as a building material for thousands of years. It is literally the first building material. Besides clay, ceramics, and glass noted above, there's other building materials made essentially out of dirt: brick, adobe, mudbrick, sod, Cob, and dozens more which I am tired of searching for. --Jayron32 14:24, 8 April 2014 (UTC)[reply]
If you're imagining a sci-fi "matter converter" where random unsorted matter is dumped in one end and it's automatically sorted, purified, chemically altered, and used for manufacturing, then we're a long way off.
However, if you're looking for just crudely converting available matter into stuff, you might be interested in this Solar Sinter project, or even more basic, but probably more useful, this open-source brick press from the Global Village Construction Set.
APL (talk) 15:04, 8 April 2014 (UTC)[reply]

What about stuff like silicon to make electronics? ScienceApe (talk) 15:39, 8 April 2014 (UTC)[reply]

Silicon certainly exists in (some) soil. But purifying it would be an ordeal. It's not impossible that in some sci-fi future the silicon is somehow separated for use making chips on demand. That's way in the future though. I don't think anyone even has a good idea how that would work. If we're talking future sci-fi tech, Carbon nanotubes might also be used to make electronics. There's plenty of carbon in most soil.
To be honest, most dirt is probably best used for growing stuff in. Either the old fashioned way, or in some sci-fi way. APL (talk) 16:07, 8 April 2014 (UTC)[reply]
That reminds me of another use for dirt, you can put it in a form (perhaps made with a 3D printer) and grow a fungus inside the form to hold it all together, then seal it in with a vinyl covering, to make a cushion. StuRat (talk) 17:10, 8 April 2014 (UTC)[reply]
A common ingredient in quartz sand is silicon dioxide, so you can purify silicon from that. StuRat (talk) 17:07, 8 April 2014 (UTC)[reply]
Much of the Earth's crust is feldspar, which contains aluminum, silicon, and oxygen. If we had unlimited cheap power, we could make aluminum conductors and structural components, silicon semiconductors, hard corundum pieces out of it. Wnt (talk) 22:10, 8 April 2014 (UTC)[reply]
I'd think circuit boards would be first. Layers of circuit traces, via holes and insulators. A 4 layer PCB would be an awesome 3D printer application for multiple element printers. You would then solder the ICs to it. That would be a huge prototyping breakthrough. --DHeyward (talk) 03:02, 10 April 2014 (UTC)[reply]

Antibiotic resistance arguments applied to vaccines

If it is recommended that usage of antibiotics be limited to discourage antibiotic resistance, why does it not then follow that vaccine usage be limited to discourage vaccine resistance? (Note: I am not a supporter of the anti-vaccination movement, but I have heard this argument against vaccines advanced by some anti-vaccination partisans and would like to hear a counterargument.) —SeekingAnswers (reply) 13:55, 8 April 2014 (UTC)[reply]

Vaccines are far more specific than antibiotics. A typical antibiotic acts against a broad range of bacteria, but a typical vaccine only acts against one very specific type of virus. In fact often the biggest challenge in developing a vaccine to is get it to work against all the different strains of a particular kind of virus. The consequence is that resistance to one type of vaccine does not cause problems in dealing with any type of virus except the specific one that that vaccine is designed for. Looie496 (talk) 14:12, 8 April 2014 (UTC)[reply]
Also, it should be noted that antibiotics act as vaccines for the bacteria they kill. That is, the general idea behind antibiotic resistance is that bacteria are exposed to an agent designed to harm them, and they adapt so the agent isn't harmful anymore. Which is almost exactly what a vaccine does for you. In the first case, you're training bacteria to not be harmed by the antibiotics, and in the second case, you're training yourself to not be harmed by viruses. Furthermore, in science we don't base our understanding of the world by incomplete analogy. We base it by actual data. Antibiotic resistance is an observed phenomenon, and is happening. For most diseases we vaccinate against, we aren't finding any such evidence of that happening. I say most, because there are SOME diseases which show resistance to vaccinations: [3] but as that reference makes clear, most of those are the exception and not the rule. --Jayron32 14:19, 8 April 2014 (UTC)[reply]
Sorry, but "antibiotics act as vaccines for the bacteria they kill" is at best misleading. Vaccines work by activating the body's immune system; antibiotics work by attacking bacteria directly. Vaccine resistance is possible, but the only way for a virus to develop resistance is for it to get better at overcoming the immune response, and there is strong evolutionary pressure in that direction even without any vaccine being present. Looie496 (talk) 14:48, 8 April 2014 (UTC)[reply]
No, vaccines cause the host organism (you) to develop resistance by exposing them to the harmful agent. You know, like antibiotics cause the host organism (bacteria) to develop resistance by exposing them to the harmful agent. The exposure to an agent which causes resistance to that agent is in common to both how vaccines prevent disease in you, and how antibiotic-resistance works in bacteria. --Jayron32 15:00, 8 April 2014 (UTC)[reply]
This sounds like you're thinking of a collection or colony of infecting bacteria as a superorganism or similar. Exposure to antibiotics over time can lead to a colony of resistant bacteria, but any individual either resist, or dies from the antibiotic, right from the start. Over time, only resistant individuals are left. This is rather different than engaging the human immune system. (At least that's the common mode of action... Do you have a ref for acquired resistance to antibiotics within an individual bacterium over time?) SemanticMantis (talk) 16:15, 8 April 2014 (UTC)[reply]
That's exactly correct. --Jayron32 16:37, 8 April 2014 (UTC)[reply]
Well, vaccines do change over time - for example, influenza vaccine is different every year, and the virus keeps evolving, in part to get around current natural immunity and in part to avoid the vaccines. With many other viruses it's not so much of a problem - for example, smallpox and cowpox parted ways a long time ago, so I doubt smallpox will come up with a way to avoid cowpox-based immunity any time soon (not without help anyway). Then there's HIV, which evolves so damn fast it avoids even the natural immune system, and so far vaccine makers are still struggling. Wnt (talk) 16:35, 8 April 2014 (UTC)[reply]
I take your point that influenza quickly evolves, and we make new vaccines every year. But I'm not sure that's "resistance," because the virus is merely scrambling the features that identify it to our immune system, and it would do that just as much even without vaccines, right? Sure, it's a bit of a red queen hypothesis situation, but I don't agree with the semantics of saying "flu resists last year's vaccine." -- It's simply that last year's vaccine loses its effectiveness as the virus mutates on. SemanticMantis (talk) 21:19, 8 April 2014 (UTC)[reply]
p.s. (Do we have a WP:TELEOLOGY warning? No? I'm pretty sure you'd agree that the flu has no goal or end in mind ;)
Well, when H5N1 came out, I remember reading stories that it was spreading quickly due to not being affected by existing vaccines. I don't think it's as dramatic an effect as antibiotic resistance simply because we don't give it time to add up - every year the vaccine can be made differently, while the drug remains the same. Wnt (talk) 11:38, 9 April 2014 (UTC)[reply]
Viruses don't evolve with natural selection the way other living organisms do. Influenza is a classic virus as it moves from species to species picking up pieces as it goes. Influenza moves between pigs, birds and humans. It's survival is mixing different elements of each and is horizontal gene transfer. Whether a particular strain has more human effects or not is not particularly relevant to its survival. Viruses like small pox and polio generally affect only humans so there is no development or adaptation. Because of that, small pox has largely been eradicated. The biggest threat is when a virus picks up a gene sequence that gives it a route to humans. --DHeyward (talk) 03:54, 10 April 2014 (UTC)[reply]
I'm inclined to say influenza has a range of different alleles in a gene pool like any other population, though its sexuality is a bit strange. It expands into a range of different niche environments like many other species. If you think of its spread in humans as an irrelevancy you can say evolution doesn't apply to what you're not looking at, but I wouldn't take that view. Wnt (talk) 14:08, 11 April 2014 (UTC)[reply]
A key distinction between the two is that vaccines don't directly act upon viruses, they act upon the host's immune system, which does all the real "work" in fending off disease. That is why it "doesn't follow." Think about your analogy from the other direction too: there is no analogue to herd immunity in the case of antibiotics and bacteria, because an antibiotic doesn't spread itself from host to host by its own action. SemanticMantis (talk) 22:02, 8 April 2014 (UTC)[reply]
Exactly the opposite. Vaccination promotes the survival of humans resistant to viruses. Antibiotics promotes the survival of bacteria resistant to antibiotics. It is probably true that more rare/severe viruses become higher value targets but only because the more common ones are eradicated. --DHeyward (talk) 00:57, 9 April 2014 (UTC)[reply]
I think I can fairly safely say that the vast majority of arguments I see or hear against vaccination are built on misinformation, logical fallacies and non-sequiturs. This one is no exception. Antibiotics are compounds that act directly upon bacteria to either kill them or prevent their replication through structural disruption or interference with bacterial metabolic or other intracellular biochemical processes. Antibiotics do not modify the individual's immune response to a bacterium. Bacteria develop resistance to antibiotics largely as a result of evolutionary processes that follow on from inappropriate use of antibiotics, creating a milieu that favours the survival and reproduction of bacteria that possess mechanisms which render them impervious to the mode of action of a particular class of antibiotic. Vaccines do not directly kill or otherwise interfere with the normal functioning of the viruses or bacteria they are directed against. Rather, they are generally composed of a weakened form of a particular organism, which is recognised by the immune system and used to develop or augment an individual's cellular immune system response to that organism. Vaccines aren't always 100% effective because they rely on the individual mounting a sufficiently vigorous immune response to the antigens in the vaccine and forming sufficient antibodies that are specific to the infective organism. There's a good vaccine-car analogy that goes like this: If I take a new car, then rip out the engine, rip out the seats and smash the headlights, you'll still be able to identify it if I leave the 3-point star on the hood. If you're really up on your cars you'll even be able to pick the model and year by the body shape. Now, if I take the same car, and then remove all the badges and do a heap of after-market body modifications, you'll struggle to figure out what make of car it was, let alone the year and model. The antibiotic-car analogy goes something like this: take a car, fill it with C4, detonate it, go find all the pieces and chop those pieces up into little bits until there's no sign of the car. Spectacularly effective unless the car manufacturer comes up with a C4-proofing option. Mattopaedia Say G'Day! 01:58, 12 April 2014 (UTC)[reply]

About a birds name

I have seen a bird in towns in eastern side of India in West Bengal.the birds are big in size have copper coloured wings and rest of the body is black.But unfortunately I dont know the name of this bird.Please tell me the name of this bird and its habits.Please mention if there is any Wikipedia article about this bird.Another type of bird that has come to my notice is even smaller than a sparrow but I am startled and amazed by its colour.There is a greenish lustre or blueish green lustre.It appears as if the bird's feathers act as a diffraction grating.It has very thin carved beak which I think can prominent feature.Please identify this bird and give its name and corresponding Wikipedia article.Pleasemention the habits of these birds and where you have seen them.117.194.229.124 (talk) 14:01, 8 April 2014 (UTC)[reply]

It sounds like a member of the Starling or Myna family, many of which are known for their iridescent feathers as you describe. --Jayron32 14:09, 8 April 2014 (UTC)[reply]

I see the picture of Starlig but it doesnt match the second type of bird I queried about.Please dont forget to mention about the first bird.117.194.229.124 (talk) 14:14, 8 April 2014 (UTC)[reply]

A straightforward Google search finds the Greater Coucal. Looie496 (talk) 14:17, 8 April 2014 (UTC)[reply]
There are hundreds of different species of birds in the starling family. Your second type of bird could be any number of them, including the Common Starling, which also has a number of subspecies. Don't just look at one picture in the first article you find. Search through pictures of different kinds of starlings and see if you find any matches. --Jayron32 14:21, 8 April 2014 (UTC)[reply]

Searching Google for Greater Coucal I got resuls inn image section purplecoloured sunbird and luckily it was the second type type of bird I asked about.The first type of bird is Greater Coucal. Actually in our locality there is overcrowding of crows a menace with some sparrows so it is a delight to see these wonderful birds.Thanks for your kind help.117.194.229.124 (talk) 14:53, 8 April 2014 (UTC)[reply]

Let me just add a pointer to our Purple Sunbird article. Looie496 (talk) 16:05, 8 April 2014 (UTC)[reply]

Chemical analysis

Suppose a person stumbles upon a substance and he wants to know the chemical composition of the substance.The substance can be quite complex mixture of organic and inorganic compounds.How will the chemist know about what are the compounds present and in what proportion.How this used to be accomplished when there was no NMR spectra or computers available and how is it done nowadays.Isthis type of analysis required in forensic studies and where else this type of analysis required. Can natural substances like fruit skin,fruit juice and products like soaps perfumes creams ointments can be analysed to dtermine composition and what knowledge and expertise and what level of nhemistry knowledge is required to do this.117.194.229.124 (talk) 14:11, 8 April 2014 (UTC)[reply]

The first thing the person would have to do is some form of separation techniques to get the mixture into its component substances. A common method is to combine a separation technique with an analytical technique to then identify the substances as soon as they are separated. Analytical systems like GC-MS or LC-MS (for gas chromatography-mass spectrometry or liquid chromatography-mass spectrometry respectively) are commonly used for that exact purpose. You can also do your separation techniques prior to doing your analysis work; for example using Ion-exchange chromatography and then isolating each product and doing analysis individually. This is commonly done for techniques that don't "marry" well to a gas chromatograph, such as Infrared spectroscopy or Nuclear Magnetic Resonance. --Jayron32 14:35, 8 April 2014 (UTC)[reply]

The methods you mentioned require computers or electronics but without these using these gadgets how can the work be done as was the case in 19th century or early 20th century.Or was such analysis was impossible at that time.117.194.229.124 (talk) 14:58, 8 April 2014 (UTC)[reply]

Basic chromatography techniques such as thin-layer chromatography and paper chromatography and ion-exchange chromatography could be used to separate mixtures without the use of electronics at all. Prior to the use of electronic methods of analysis, the only methods of chemical analysis were called wet chemistry methods, and basically involved a series of chemical tests to elucidate the structure of unknown compounds. For just a few examples, there's the Baeyer's test, universal indicator, the Van Slyke test, Seliwanoff's test, the iodoform test. --Jayron32 15:07, 8 April 2014 (UTC)[reply]

If all matter outside the solar system ceased to exist?

To put the question somewhat more specifically than the header title above puts it, what would be the effects on the solar system if all matter outside a radius of, say, 50 astronomical units from the sun (that is, from approximately outside the Kuiper Belt) spontaneously ceased to exist? This is the only counterfactual; in all else, the laws of physics remain the same. Also to clarify, this is a one-time event; matter that after this event passes across the 50 AU boundary does not cease to exist. (A clarification edit: by "matter", I mean objects with rest mass, so electromagnetic radiation, for instance, would continue to exist outside the boundary.) I'll mention several specific phenomenon, with related questions, in particular.

First, how quickly would scientists realize that this disappearance of matter had occurred, and what pieces of evidence would this conclusion rely upon?

Second, I suspect that without the gravitational effects exerted by matter outside the solar system to counterbalance the gravitational pull of the sun, objects (including planets such as our Earth) would begin to fall toward the sun. Is this supposition correct? And if it is correct, how fast would the fall be (that is, how long from this event before the Earth is swallowed by the sun)?

Third, what would the ultimate fate of the solar system be, given this event? I assume it would be some kind of thermodynamic heat death, but how long would the process take? Would the remaining matter from within the solar system all become packed together forever from gravitational effects, or would some or all of the matter disperse, getting forever further away?

I'd also be interested in hearing about whatever other effects people here can think of.

SeekingAnswers (reply) 15:18, 8 April 2014 (UTC)[reply]

First, 7 hours after the event the Fixed stars disappear from the night sky. It takes about 24 hours for the phenomenon to be confirmed by all earth-bound telescopes but by the 2nd day there are videos on YouTube and TV "expert" commentators explaining whether the End of the World is nigh. A Stock market crash ensues.
Second, the nett pull of the rest of the Universe on our solar system is virtually zero which is why we don't accelerate madly in any particular direction. Gravitational force of the Sun remains the dominant factor keeping the planets in their present orbits unchanged.
Third, if you can explain why the solar system (Address: Local Interstellar Cloud, Local Bubble, Orion–Cygnus Arm, Milky Way) was specially preserved through this singularly destructive event, you probably have the basis for a Belief system that gives a comforting answer to any remaining questions. 84.209.89.214 (talk) 15:46, 8 April 2014 (UTC)[reply]
Good and basically comprehensive answer, 84. But I am not so sure there wouldn't be a huge number of people concluding that it was not the rest of the universe that had been destroyed, but we sinners who had been snatched into the void. μηδείς (talk) 16:26, 8 April 2014 (UTC)[reply]
Regarding the fixed stars disappearing after 7 hours: In my original question, when I mentioned "matter", I was specifically referring to objects with rest mass. So fixed stars wouldn't disappear from our view after 7 hours, since photons from beyond 50 AU that were emitted before the stars ceased to exist would continue to stream toward us. So fixed stars disappearing that quickly couldn't be what tips everyone off that something has happened.
And on the third part, still, what would happen eventually in the end? Would everything be drawn together or disperse?
Also, if I'm understanding your answer, basically, other than human panic, there would be no real negative effects on Earth?
SeekingAnswers (reply) 16:50, 8 April 2014 (UTC)[reply]
IP 84 has already answered your answer correctly, regardless of your further qualification. You seem to know that the light doesn't disappear if the light doesn't disappear, so your question would seem to be resolved. μηδείς (talk) 20:51, 8 April 2014 (UTC)[reply]
cosmic rays would greatly decrease, possibly leading to a decrease in cancer, though the health risk or benefit of low level radiation is unclear; see hormesis. We would also be freer to send out interstellar spacecraft, apart from one small problem. Interest in the Pioneer and Voyager spacecraft would suddenly explode as people became curious if their impact with or passage through the "barrier" could be detected. I think most of us would assume the barrier was some sort of alien quarantine or embargo, or a failure in an existing system meant to simulate an unpopulated universe. Wnt (talk) 16:41, 8 April 2014 (UTC)[reply]
Seems like Mach's principle would be relevant. Katie R (talk) 16:51, 8 April 2014 (UTC)[reply]
If light and other EM radiation kept coming in from the former stars, it would take 4.3 years for the nearest stars to go dark. So, we would know something serious was going on then. I'm not sure if we would detect anything before that, such as the lack of dust clouds just outside the border, or the lack of gravitational perturbation on Kuiper belt objects from nearby stars (this would result in fewer comets, eventually, but this could take thousands of years to become noticeable). The solar system no longer orbiting the galactic center wouldn't be noticeable, as the gravity from the galactic core was exactly counterbalanced by our orbit. StuRat (talk) 17:21, 8 April 2014 (UTC)[reply]
For one thing, the Voyager spacecraft would both stop communicating at the same time. Eris and Sedna would disappear from telescope images. --Bowlhover (talk) 18:57, 8 April 2014 (UTC)[reply]
As other people already pointed out, there wouldn't be any noticeable perturbation in planetary orbits. But there would be an interesting effect when the last light of a given star reaches us: There is a lot of electromagnetic radiation within a star that is normally prevented from escaping quickly by the matter (plasma) of the star. And that would then escape at once, so for the last few seconds before becoming invisible, a star would become a lot brighter (for a timespan of its diameter divided by the speed of light, so e.g. 0.65 seconds for Proxima Centauri and 8 seconds for Sirius A).
And a similar thing must happen with the gravitational field of stars, which should relax and dissipate as gravitational radiation (someone will think of Birkhoff's theorem (relativity): A star is often approximately spherically symmetric, and a spherically symmetric metric is static and thus non-radiating according to General Relativity, but General Relativity isn't really compatible with things suddenly disappearing; and neither is Maxwell's electromagnetism). So gravitational wave detectors will finally detect something a few years after the event.
Icek (talk) 19:28, 8 April 2014 (UTC)[reply]
If we're going with the premise that the light remains but not the matter, note that it takes something like 10,000 to 170,000 years for gamma rays to escape the Sun as visible light. So in that second the star would get something like a trillion times brighter, all in hard radiation. Alpha Centauri starts at magnitude 1.3, so it would come out at something like 3 orders of magnitude brighter than the sun, in X-rays. Not a nice surprise. :) But then again, how much of the lifespan of the gamma rays inside the Sun is counted as "associated with" matter enough that it would be disappeared with the matter, and how much as "free"? (I don't think you can frame the premise meaningfully really??) Wnt (talk) 20:12, 8 April 2014 (UTC)[reply]
If you ask the question of what happens if a charge (like an electron or proton in the star) disappears from the viewpoint of Maxwell's equation, you run into a contradiction: Assuming a spherical charge distribution, and the charge becoming less with time, the field changes according to
Then everywhere tangential magnetic fields should be created according to
where is zero (no current).
But that's not quite possible in a spherically symmetric problem.
The only more fundamental theory is quantum field theory, and I guess renormalization won't be possible anymore if you just take out all the massive particles at a particular point in time.
Icek (talk) 21:08, 9 April 2014 (UTC)[reply]
Could I get an answer on the third part of the original question: in this hypothetical universe, in which the only the only matter remaining is that in the solar system, which of the various theories about the ultimate fate of the universe would likely be correct? —SeekingAnswers (reply) 19:07, 9 April 2014 (UTC)[reply]
The solar system doesn't have enough mass to collapse to anything denser than a white dwarf. And in the long run, as we seem to be in a universe with accelerating expansion and the cosmic background radiation's temperature becomes lower and lower, everything would cool and evaporate.
By the way, the mass density equivalent of the cosmic background radiation is about 4.64*10-31 kg/m3, or about half an electron mass per cubic meter. So within 4.3*1060 m3 of otherwise empty intergalactic space there is about one solar system's mass of about 2*1030 kg. It's only a ball with a bit less than 11000 light years radius. Icek (talk) 21:08, 9 April 2014 (UTC)[reply]
Minor point: the x-ray burst from alpha Centauri wouldn't be a surprise at all. Because of this. - ¡Ouch! (hurt me / more pain) 06:26, 10 April 2014 (UTC)[reply]
I'm pretty that the burst would be significantly different than anything observed so far, and would end up with tons of telescopes pointed at it. And then of course when the burst is over and there's no star left behind we'll probably know something bad happened. Katie R (talk) 18:41, 10 April 2014 (UTC)[reply]
You're pretty? Wow. My real point was that the disappearing Alpha Centauri wouldn't be a surprise because it would be months after Proxima would have disappeared from our sky. Proxima is a red dwarf which is not visible to the unaided eye, but even that would have emitted quite a burst. - ¡Ouch! (hurt me / more pain) 07:49, 11 April 2014 (UTC)[reply]

The OP's added clarification that incoming photons from distant stars would continue after the stars themselves are magically deleted suggests that the event would for many years go undetected here. However astronomers would not long miss the steadily accumulating number of position hops by the fixed stars; this is due to disappearances of gravity lensing by intervening stars, although they might be puzzled about the cause. I am already puzzled to think how the OP's destructive phenomenon might interact with a Black hole and wonder if anything in the proposition rules out the possibility of its arbitrary border slicing a black hole in half. The "laws of physics remain the same" condition seems to call for the Impedance of free space to be unpeturbed even at the concave destruction limit but if this cannot be guaranteed, we would see at least some clouding or distortion of all the fixed stars at time E+7 hours (due to partial reflection and refraction) followed by a sudden great brightening of the whole night sky at E+14 hours due to internal reflection of our own Sun. Such a scenario takes away the luxury we have enjoyed of having time to ponder over a likely Ultimate fate of the universe and we would find ourselves more than usually lonely and ill prepared to predict how such a cosmic vacuum metastability event plays out on our doorstep. 84.209.89.214 (talk) 23:26, 10 April 2014 (UTC)[reply]

Do not see. I'd think the bursts would come in at the very moment the grav lensing ceases; thus we'd not get an early warning unless there's a massive non-radiating body (which could be anything from brown dwarf to black hole) within 4.something light years. A rogue planet might qualify, too, but it'd have to be really close to the line of sight to cause observable lensing.
Black hole slicing: since the mass of a black hole is at a single point, it's hit or miss. Even if a black hole is a fuzzball (string theory), I'd rule out that it can be cut in half if photons are unaffected.
I wonder how refraction and internal reflection mix with it. The refractive index difference would be so tiny that we probably wouldn't detect it, given that we'd look at the surface at about 90 degrees, where refraction is generally at its weakest. Maybe the pulse frequency of pulsars would change. - ¡Ouch! (hurt me / more pain) 07:49, 11 April 2014 (UTC)[reply]
The non-indent of my post is because I haven't much to say about Proxima Centauri's flares that you brought up in the thread flirting with pretty Katie. The size of a Black hole to us is the finite, optically observable diameter of its event horizon, not the abstraction of a point singularity. The OP's proposition effectively partitions the Universe into our local factual part and a momentarily counterfactual surrounding in which mass spontaneously vanished. It supposes a massive superluminal violation of the physical law of Mass–energy equivalence and leaves us clinging with whatever sanity remains to an easy assurance that all physical laws are thereafter re-established. But devoid of mass there can be no Gravitational constant which at a stroke invalidates what Isaac (1687) and Albert (1915) laboured to elucidate. My post speculates a visible Schlieren at the factual-counterfactual interface; actually anything might happen there including the Dark matter (more than 1/4 of our Universe) turbulently boiling away into pre-existent nothingness. The alarming implication of your belief that we would get no early warning of this is that it could already have started. ¡Hush!. 84.209.89.214 (talk) 14:58, 11 April 2014 (UTC)[reply]

Is the technic of the suction cup - base on surface tension?

I've read our article about, but it's not clear for my question. 5.28.177.131 (talk) 18:58, 8 April 2014 (UTC)[reply]

No, a suction cup works on the principle of differential pressure. There's lower pressure inside, higher pressure outside, so the difference in pressure keeps it pressed against whatever it is sticking to. --Jayron32 19:01, 8 April 2014 (UTC)[reply]
What Jayron32 said. When the suction cup is not stuck on something (i.e. when it's like most other objects you interact with) it's surrounded by air, so the atmospheric pressure we experience here on earth kind of balances out. When you remove all of the air from one side, however, the only pressure is what's pushing down on it from the other side. The strength of the suction is basically a measure of how easily air can sneak back in. This is also why suction cups can't work in space, where there is no atmospheric pressure. --— Rhododendrites talk20:44, 8 April 2014 (UTC)[reply]
Perhaps that air not sneaking back in quickly is what this Q is about, as that is the interesting part, at least to me. There's obviously a pressure differential, yet the vacuum doesn't immediately suck the air back in along the edges. Adhesion of the suction cup material to the material on which it's mounted gets the credit there, I believe. StuRat (talk) 00:52, 9 April 2014 (UTC)[reply]
There is definitely something going on that we haven't explained yet, but I don't think we need to invoke adhesion or any other chemical or electrostatic forces. I think it's all to do with the elasticity of the cup. Imagine the situation with a rigid vessel, like the Magdeburg hemispheres. It's obvious what's happening there, because somebody used a vacuum pump to suck out the air. Now imagine that the copper hemispheres are replaced with rubber ones. You don't need a vacuum pump any more: you just squash the two hemispheres together, expelling the air, and then let them expand by their own elasticity. They could have called them the "Magdeburg sink plungers". The suction cup works like that, only with one hemisphere replaced by a flat surface. A suction cup is its own vacuum pump, using its elasticity to expand a sealed volume against atmospheric pressure. Thus it causes the atmosphere to exert a force on it, and that force pushes the rubber against the wall to create the seal (and incidentally, to create the friction that stops it from sliding down the wall). --Heron (talk) 18:30, 9 April 2014 (UTC)[reply]
Well, the grease was needed in the hemisphere case, and water helps to keep a suction cup stuck, so I believe those liquids are an important part of the equation, with adhesion serving to plug the gaps with those fluids. StuRat (talk) 23:11, 9 April 2014 (UTC)[reply]
Every kid with the suction cup darts know they work better when darts are licked. My guess is the liquid on the rubber has an adhesion to the rubber and the pressure on the liquid pulls the rubber tight on the surface through its adhesion to the liquid. --DHeyward (talk) 09:52, 10 April 2014 (UTC)[reply]

April 9

Chemical Test

How is a chemical test developed and/or discovered.Please explain with an example.117.194.234.95 (talk) 03:22, 9 April 2014 (UTC)[reply]

Please do your own homework.
Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. StuRat (talk) 04:51, 9 April 2014 (UTC)[reply]

I am not a student of school college university.I asked you question that has nothing to do with homework.I disapprove prejudiced notions.I dont think it will be a good thing to go to school/college in thirties.I soloicit your answer to my query and not these malicious conjenctures.117.194.249.163 (talk) 13:32, 9 April 2014 (UTC)[reply]

What chemical test did you have in mind? Wikipedia has an article titled Chemical test that discusses many wet chemistry chemical tests. There's a long list of them. If you read articles on any one of them, it may discuss how that test came to be. --Jayron32 14:14, 9 April 2014 (UTC)[reply]
I think the OP is thinking about a "chemical test" like the pH paper that you find so often in Chemistry labs. In that case, we may have to look up the history of the pH paper. But it all depends on whatever "chemical test" you're talking about. 140.254.227.100 (talk) 22:02, 9 April 2014 (UTC)[reply]

Skin cream gun control study

Hello, maybe you've heard of the study in which participants had to assess the results of a (fake) study about the efficacy of either skin cream or g*n c*ntr*l laws (with exactly same numbers), which purportedly showed that peoples' math reasoning ability declined when the math problem in question had to do with a politicized issue. I'm interested in the (fake) problem itself:

rash got better rash got worse
patients who used the cream 223 75
patients who didn't use the cream 107 21

After that the participants had to "indicate whether the experiment shows that using the new cream is likely to make the skin condition better or worse." It says on the internet that it's not enough to compare the numbers, one has to do the ratios. But what ratios? I see two ways of going about solving this:

  • Of those who used the cream, 223/(223+75)=75% got better, vs 107/(107+21)=84% of those who didn't => ergo, cream sucks
  • Of those whose rash got better, 223/(223+107)=68% used the cream, vs 75/(75+21)=78% of those, whose rash got worse =>cream still sucks

Which way is the correct one and why? (If I'm ever recruited for a similar study:) And is it incidental that in this example they devised, the conclusion is the same either way or is it mathematically inevitable?

Asmrulz (talk) 05:37, 9 April 2014 (UTC)[reply]

I'm not an expert, but it seems to me that the 1st method is correct. The 2nd group is your control group, you are comparing the effectiveness of the treatment vs the control group, which can be "no treatment" as in this case, or can be the current best available treatment (especially where withholding treatment all together would be unethical). So yes, put simply with the treatment 75% got better, without it 84% got better, so the treatment actually performed WORSE then no treatment at all. Vespine (talk) 06:02, 9 April 2014 (UTC)[reply]
You can only use the second method if you extrapolate the numbers ("normalize"?) to make the population sizes equal. The first group has 298, the second only 128. Multiply the second row by 2.328 and you get 249 and 49. That makes it clearer that using the cream has no particular effect. In fact, it appears that letting the body's own immune system handle it seems to be better than using the cream. But as to whether it "sucks" or not, you would have to do some kind of statistical analysis to see whether the 249 and the 223 are statistically different across the given population. ←Baseball Bugs What's up, Doc? carrots08:25, 9 April 2014 (UTC)[reply]
Why the asterisks on "gun control" ? Is it considered an obscene term in your part of the world ? StuRat (talk) 08:33, 9 April 2014 (UTC) [reply]
It's from the Jargon file of hacker slang which I read long ago (link) It's funnier this way :) Asmrulz (talk) 10:16, 9 April 2014 (UTC)[reply]
The first method where one comparess the ratios of improvement in two unrelated populations is a rational test of the cream. The second method conflates the ratio of interest with the ratio of test and control populations, which is arbitrary, and cannot justify its conclusion. In this data the cream demonstrated less improvement 75% than whatever else (unspecified) other people did that improved 84% of them. The cream really sucks but the lesson here will not be welcomed by g*n r*ghts l*bby*sts with kn**-j*rk th*nk*ng st*ck *n 1791. 84.209.89.214 (talk) 13:08, 9 April 2014 (UTC)[reply]
  • I agree with others that the first method is the correct one. But let me just add a note that the differences are not actually statistically significant (p = 0.0574 using Fisher's exact test, calculated using this tool, so the conclusion in a published paper would be that the data don't establish that the cream has any effect. Looie496 (talk) 14:02, 9 April 2014 (UTC)[reply]

Thanks to all of you for your answers. Asmrulz (talk) 14:43, 9 April 2014 (UTC)[reply]

Heat and magnetic fields.

As I understand it, heat destroys magnetism. So why then, does the magnetic field of the earth survive when the core of our planet which is hot enough to create a molten core? — Preceding unsigned comment added by 86.6.96.72 (talk) 09:59, 9 April 2014 (UTC)[reply]

Your understanding is wrong. --DHeyward (talk) 10:12, 9 April 2014 (UTC)[reply]

Explain?

Heat can stop a magnet being magnetic, this is true. Read the following to understand why we have a magnetic field around the Earth. http://en.wikipedia.org/wiki/Earth's_magnetic_field#Physical_origin 217.158.236.14 (talk) 11:29, 9 April 2014 (UTC)[reply]

A Magnetic field is produced by electric charges in motion i.e. Electric current. The way that works was explained by Maxwell in his elegant set of differential equations which hold true at all temperatures that we experience. 84.209.89.214 (talk) 12:09, 9 April 2014 (UTC)[reply]
Just to clarify, I am aware of the above and I am not debating it, however it is also true that ferromagnets can lose magnetic properties when heated to above the Curie temperature.217.158.236.14 (talk) 12:23, 9 April 2014 (UTC)[reply]
Yes, but that isn't because "heat destroys magnetism". It's because heat alters the crystal structure of the iron so it no longer has ferromagnetism. Different ideas. --Jayron32 12:31, 9 April 2014 (UTC)[reply]
Agreed, which is why I said "Heat can stop a magnet being magnetic", and not "destroys magnetism". — Preceding unsigned comment added by 217.158.236.14 (talk) 13:14, 9 April 2014 (UTC)[reply]
  • To add detail to the answers above: ultimately magnetism arises from the movement of large numbers of electric charges in a synchronized way. In a ferromagnet (an ordinary solid iron magnet), the synchrony comes from the fact that large numbers of atoms are frozen into alignment with each other. If you heat a ferromagnet to near the melting point, the atoms lose their alignment and the magnetic field disappears. But the Earth is not a ferromagnet. Its magnetic field arises from electric currents flowing coherently through the magma on a vastly larger scale. The reason why that happens, incidentally, is not very well understood -- but clearly it does happen. Looie496 (talk) 13:52, 9 April 2014 (UTC)[reply]
Vast energies are stored in the Earth's magnetic field and evidence in the Geologic record that it has periodically reversed polarity is evidence that it is part of a resonant system. As yet no simple global model of Earth's resonant modes, that may be linked with resonances in other bodies, has emerged because the short time span of collected data is plagued by Heteroscedasticity of local anomalies. 84.209.89.214 (talk) 16:02, 9 April 2014 (UTC)[reply]
(Hey, you seem to be giving some good responses/refs here recently... or at least I think so, but it's hard to keep track of IPs... any chance we can convince you to register? It is less anonymous and more pseudonymous than signing with an IP, but it may actually increase your privacy too :) SemanticMantis (talk) 22:58, 9 April 2014 (UTC)[reply]
I'm pretty sure the editor behind the IP has had multiple accounts and are technically WP:block evading. People are just turning a blind eye because they aren't harping on about its/it's and other grammar and spelling issues. However registering another account is still probably not a good idea, instead requesting an unblock under one of their blocked accounts, preferably the main one. Nil Einne (talk) 05:01, 10 April 2014 (UTC)[reply]
(Thank you for those kind comments. Nil Einne can identify me easily. "Its" and "it's" mean different things. A registered user pointed that out a while ago.) 84.209.89.214 (talk) 12:52, 10 April 2014 (UTC)[reply]
Many factors establish the earth's magnetic field. It's not constant and rotation and heat affect it. If you want an extreme case, look at the sun. It has no iron core. It's created by heat and rotation. It flips N/S polarity every 11 years and often has a weak quadropole during maximum sunspot activity. The resultant magnetic field is very strong at sunspot minima. The earth has a different mechanism, but it's not correct to simply associate heat with the destruction of the magnetic field. There are solid magnetic properties as well as rotating, conducting fluids. See Dynamo theory to see how the Earth's magnetic field is created and changes. A ferromagnetic solid may reflect the magnetic field imprinted on it, but it's not necessarily the force that is responsible for the magnetic field. --DHeyward (talk) 05:42, 10 April 2014 (UTC)[reply]

Institutional dependence

Hi there. I am interested in a phenomenon whereas an individual who spends many years in a restrictive environment, e.g. prison, becomes dependent on the institution and develops unhealthy reactions sabotaging normal release process for instance. I am also interested if the phenomenon like this has ever been coded, if there is a corresponding ICD-9 or ICD-10 code describing it. Thank you --AboutFace 22 (talk) 20:01, 9 April 2014 (UTC)[reply]

Stockholm syndrome came to my mind. Brandmeistertalk 20:07, 9 April 2014 (UTC)[reply]

Oh lordy! I just was about to get back and add that in my understanding the phenomenon I am looking for or the code thereof is NOT the same as Stockholm syndrome. It should be distinct from it. Thanks.--AboutFace 22 (talk) 20:10, 9 April 2014 (UTC)[reply]

I think Learned_helplessness is relevant, since it describes the effect that a punitive environment has on individual volition. OldTimeNESter (talk) 20:17, 9 April 2014 (UTC)[reply]

So, you are asking about individuals who become institutionalised? 86.146.28.229 (talk) 21:21, 9 April 2014 (UTC)[reply]

Thank you. The last two suggestions are quite valuable. I have to think about it, especially the institutional syndrome. --AboutFace 22 (talk) 01:50, 11 April 2014 (UTC)[reply]

Quadruple axels and beyond

What actually prevents a figure skater from making quadruple or even quintuple axels and similar feats? And what capabilities should (s)he theoretically have to perform them and similar stuff? Is it just a muscular force of legs during the jump-off, adequate jump height giving more time and a good sense of balance? --93.174.25.12 (talk) 20:04, 9 April 2014 (UTC)[reply]

The article on Jumping mentions the force-velocity ratio for muscles, which establishes the biomechanical limits on jumping height. OldTimeNESter (talk) 20:29, 9 April 2014 (UTC)[reply]
Basically you have to do it "just right", or you'll be either over or under rotated and will land with the skate blade not pointing in the right direction, and will probably fall. You also have to keep your center of gravity "just so", or you'll land at the wrong angle... and will probably fall. The more spins you have in a given jump, the harder it is to do. But it will happen eventually. The double axel was once considered tough to do. And lots of skaters fall on triple axels. As the commentators said at the Olympics recently, "It's really hard!" But someone will eventually succeed in doing the quad axel (4 1/2 times around) and that will set the bar higher. ←Baseball Bugs What's up, Doc? carrots20:54, 9 April 2014 (UTC)[reply]
Maybe we'll see it done some day. But there are limits to the torque a skater can apply to himself, as well as (perhaps softer) limits to the ability to control. I'm confident there's some n for which humanity will never see an n-tuple axel, but what that n is is probably best investigated by watching competitions and waiting, rather than trying to model it from first principles. In other words, I think the answer to your last question is "yes"; it's "just" those three things that would have to improve. Maybe replace "sense of balance" with kinesthetic intelligence. SemanticMantis (talk)
It probably varies per-person, given the above comments. But the basic thing is that you must have enough angular velocity to complete those axels in the time you remain in the air. A quantitative analysis would require the maximum impulse the skater could deliver using the muscles in question, and how quickly (so the time for spinning can be calculated). The axis of rotation should be close to vertical for balance purposes, you do not want any torque components other than the vertical component to deal with (torque is a pseudovector).--Jasper Deng (talk) 07:48, 10 April 2014 (UTC)[reply]
One technical point: If you intend "axels" to mean "rotations", note that the jump known as the "Axel" is named for a guy, Axel Paulsen or some such. It's the only one of the jumps that starts facing forward. Hence a single Axel is 1 1/2 revolutions, a double Axel is 2 1/2, and so on. ←Baseball Bugs What's up, Doc? carrots05:07, 11 April 2014 (UTC)[reply]

April 10

Nature of carbon atom

Why do carbon atom form bonds tetrahedral in shape? Why do two carbon atoms not form a quadruple bond? 117.242.108.105 (talk) 05:53, 10 April 2014 (UTC)[reply]

Only diamonds are tetrahedral, I believe. Graphite is hexagonal and in sheets. The 3 double bonds are actually shared with 6 carbon atoms. Bond order and quantum mechanics are the reason why quadruple bonds aren't common. Basically, the higher the order the less stable so mixtures with lots of carbon will reform into lower order bonds. Acetylene has a triple bond and the amount of energy released when the bond is broken is tremendous. It would be difficult or impossible to have a carbon substance with quadruple bonds, and the diatomic distance associated with it, without it reducing to paired covalent bonds at farther spacings and less potential energy. --DHeyward (talk) 06:40, 10 April 2014 (UTC)[reply]
To answer question 1) All atoms that form 4 bonds to 4 different atoms will assume a tetrahedral geometry, because that geometry maximizes the angular distance between electron pairs on the valence level of that atom. Because electron pairs will tend to repel each other, the geometry that is most stable will be the one that maximizes the distance between said pairs of electrons. See VSEPR for more details. The reason why Carbon doesn't form quadruple bonds is pretty complex, you'd need to have some background understanding of molecular orbital theory. You can read a general overview of such bonds at Quadruple bond if you wish. --Jayron32 11:08, 10 April 2014 (UTC)[reply]
Dicarbon (C2) exists, but it doesn't have quadruple bonds as simple valence bond theory would predict. Looking at it with MO theory shows that there are two sets of paired electrons in the sigma system (one bonding and one antibonding) and two sets of pairs in a degenerate pi bonding set. In total the bond order is 2, so we have a double bond instead: and indeed the MO diagram shows two pi bonds and no sigma bonds. Confirmation of this prediction can be seen from B2, C2, and N2 having increasing bond dissociation energy across period 2, indicating single, double, and triple bonds respectively.
In any case, you are not likely to find dicarbon samples lying around, as it is a very strong acid and only persists in dilution or as an adduct. Double sharp (talk) 15:15, 10 April 2014 (UTC)[reply]
Jayron's VSEPR bond-counting explanation omits an important detail (though not relevant to 4-bonded carbon specifically): it's not solely the number of bonds, because lone-pairs also count as a "direction". An atom with 4 bonds and a non-bonded pair is the geometry similar to 5 bonds (AX4E1) not tetrahedral. Going further, especially into the transition metals, lots of molecules have square planar molecular geometry. DMacks (talk) 15:19, 10 April 2014 (UTC)[reply]

Deep water pressure

Why Titanic and other deep water debris is not crushed flat by water pressure?--93.174.25.12 (talk) 07:34, 10 April 2014 (UTC)[reply]

It is because the pressure is basically uniform over the entire surface of the object, including the part on the seabed (yes, the part on the seabed gets more force because that force must be the sum of the debris weight in addition to the water weight, but that extra force has always been there regardless of whether there was water), so the forces could only ever act to compress the object. It would be natural to think the metal panels have compressed already, so they will compress no further.--Jasper Deng (talk) 07:43, 10 April 2014 (UTC)[reply]
(edit conflict)Because the water is on the inside as well as the outside. The solid pieces of steel are strong enough to withstand very high pressure, though they will compress very slightly so that they have internal stress to match the external pressure. Dbfirs 07:46, 10 April 2014 (UTC)[reply]
For clarity, my comment applies to solid objects, not hollow objects which have no internal stress.--Jasper Deng (talk) 08:26, 10 April 2014 (UTC)[reply]
Most solid objects don't compress much at those pressures, just like rocks on the sea floor. Styrofoam cups, however, do compress to a small fraction of their original size, under such pressures, because they contain gas bubbles inside each cell. StuRat (talk) 04:20, 12 April 2014 (UTC)[reply]

Pretty lakes

Why are supposedly "polluted" lakes always so full off vegetation? It seems to me that unnatural chemicals would stop plants from growing! — Preceding unsigned comment added by 2001:708:110:1004:CACB:B8FF:FE24:8A97 (talk) 08:51, 10 April 2014 (UTC)[reply]

"Unnatural" and "natural" aren't very specific words. If the waste in question contains a lot of nitrogen and/or phosphate compounds, then I'm sure it could lead to an algae bloom, for example.--Jasper Deng (talk) 09:04, 10 April 2014 (UTC)[reply]
Maybe start with eutrophication. 86.146.28.229 (talk) 09:05, 10 April 2014 (UTC)[reply]
Yes, many of the pollutants are "natural", or organic. Sewage spill-over and agricultural runoff are two common sources of such pollution. So, while turds floating in the lake are entirely "natural", and algae just loves it, we find it rather unpleasant. StuRat (talk) 16:14, 10 April 2014 (UTC)[reply]
  • The growth of algae in lakes is naturally limited by the level of the least abundant limiting resource, which is usually nitrogen. The removal of this limit by adding a little fertilizer has the same effect as removing predators from a system of herbivores like rabbits or deer; a population boom followed by starvation. In this case, the decay of dead algae will consume the oxygen in the water, and kill most of the animals in the water. μηδείς (talk) 17:21, 10 April 2014 (UTC)[reply]
See Liebig's law, which expands on the issue of limiting factors. μηδείς (talk) 20:07, 10 April 2014 (UTC)[reply]

Underwater "pings" in the Indian Ocean

Recently, searchers have been hearing several pings in the Indian Ocean, as these officials search for that missing airplane, Malaysian Flight 370. I don't understand: why are they not saying "definitely" that the pings are coming from a plane's black box? What other things in the ocean might account for these pings, other than a plane's black box? Are there any other possibilities? What are they? What other items in the ocean might make such a ping? Thanks. Joseph A. Spadaro (talk) 14:36, 10 April 2014 (UTC)[reply]

False data, as appears to have been the case in the earlier Chinese reporting, is the most likely alternative. While the particular frequency characteristics of the ping functionally assure that what is being heard is man-made, there's a whole lot of man-made technology clustered around the search area (namely, all of the searchers). It is not inconceivable that something else is generating a similar audio characteristic, even if it's unlikely. Also, from a PR standpoint, it's much easier to walk back a "probably" statement than a "definitely" statement. — Lomn 15:43, 10 April 2014 (UTC)[reply]
One report said that whales might make similar sounds. I don't know if it's true, but that's what they said. StuRat (talk) 16:12, 10 April 2014 (UTC)[reply]
Whales, dolphins, or other sea life will not make a sound at a single specific frequency with a single specific repetitive interval for a sustained period of time -- something to the effect of 37 kHz for 10 ms every second. Whales might make "similar" sounds in that they make high-frequency sounds and sometimes repeat them, but that's a use of the word "similar" that is completely misleading in terms of the actual science and technology at work. Black box pings are most certainly not natural, and the reported multi-hour contacts are completely inconsistent with sea life as a possible explanation. — Lomn 18:10, 10 April 2014 (UTC)[reply]
I've heard (but cannot confirm) that a number of undersea cables have locater beacons attached to them to make them easier to find to repair. Undersea acoustics can do strange things at times, so it's possible that they're picking up long-distance transmission of a cable's beacons. --Carnildo (talk) 01:16, 11 April 2014 (UTC)[reply]

What is the relation between death rates and life expectancy?

"In the nineteenth century twenty-two out of every thousand people died each year, more than 2% of the population– today only 5 out of a thousand people die each year. In the nineteenth century the average lifespan was thirty-six years – today it’s about eighty. The biggest killer in the nineteenth century? Tuberculosis. Also known as consumption, this disease was rampant, believed to be hereditary, and in spite of numerous claims of ways to cure the disease, no cure or effective treatment was available." - From a review of The Remedy on Goodreads by Watchingthewords.

2% of the population is the same as one person out of 50. So if 2% of the population dies every year, then that means that, in any given year, on average, one person out every 50 people will die. If 50 people die every year, then it will take 40 years to kill 1,000 people, which means that your average lifespan is going to be about 40 years. 40 is not that much different from 36, so maybe we are on the right track.
But then we look at today's numbers and we get something that doesn't make any sense. If only 5 people out of 1,000 die every year, then it will 200 years to kill 1,000 people, and nobody lives 200 years.
I suspect there is something wrong with this picture, but I don't know what. I suppose an expanding population could skew the numbers somewhat, but I don't see how they can skew it that much. Pergelator. — Preceding unsigned comment added by 50.43.12.61 (talk) 17:11, 10 April 2014 (UTC)[reply]
The current world's population is young, and therefore has low death rate (5 per 1000). It means that it is growing. Your conclusion that "it will [take] 200 years to kill 1,000 people, and nobody lives 200 years" is wrong because you incorrectly assumed that the population is static. Ruslik_Zero 19:45, 10 April 2014 (UTC)[reply]
And presumably we will see a much higher death rate when the Baby Boom generation starts to die off in quantity the US, and even more so when all those born before the One Child Policy in China start to die en masse. StuRat (talk) 13:10, 11 April 2014 (UTC)[reply]
Why only the US? Baby boomers lurk in lots of un-American places, Stu. -- Jack of Oz [pleasantries] 20:33, 11 April 2014 (UTC)[reply]
The article I linked to only mentions the US, Romania, and Africa, and I'm not sure if the magnitude of those other two is sufficient to warrant a mention. PS: I'm aware that Australia had a baby rabbit boom; did it have one with humans, too, or do you still need to import criminals/prostitutes to supplement your population ? :-) StuRat (talk) 23:41, 11 April 2014 (UTC)[reply]
You're not sure if a baby boom across the entire continent of Africa would be of sufficient magnitude to mention, compared to a baby boom in the United States? 85.255.232.229 (talk) 23:46, 11 April 2014 (UTC)[reply]
Nope, and the article didn't say it was across all of Africa, either. From the description, it sounded like it excluded the Arab part, and perhaps more. StuRat (talk) 23:49, 11 April 2014 (UTC)[reply]
Death rates from birth are relatively difficult to compare. Really, the best way to look at life expectancy is when a person reaches a certain age and average remaining years. Infant and child mortality dominate stats 50-100 years ago as well as 3rd world countries. But if you look at the CDC tables they report average life remaining at each year attained. Make it to 5 and your remaining life span is nearly the same as when you were born, reflecting higher death rates for young children. Make it to 80 and your remaining life is very nearly the same as it was 50-100 years ago. Whence the huge gains in life expectancy was reducing infant mortality and childhood diseases. Those lucky enough to reach 100 have only seen an improvement of a month or two. Notice how "the oldest person in the world" doesn't appear to be getting any higher. --DHeyward (talk) 04:30, 12 April 2014 (UTC)[reply]

Effective normal stress in geotechnical engineering

Normally, effective normal stress is calculated by subtracting the pore water pressure from the normal stress, in natural conditions. However, why is it that in the untrained triaxial compression test, the pore water pressure is negative, meaning it is added to the normal stress to obtain the effective normal stress? Is it because the compressing effect, creates a force which pushes in the opposite direction, thereby adding to the normal stress? 2.221.71.15 (talk) 21:19, 10 April 2014 (UTC)[reply]

The pore water pressure in an unconfined sample is negative due to the capillary effect. When a confining pressure is applied the pore water becomes positive, see for instance here. Mikenorton (talk) 22:34, 10 April 2014 (UTC)[reply]

April 11

Longest rotational period

Which object that orbits the Sun and does not orbit any other object orbiting the Sun, has the longest rotational period (prograde or retrograde, doesn't matter)? --Atethnekos (DiscussionContributions) 04:49, 11 April 2014 (UTC)[reply]

Longer orbital periods are associated with greater distances from the Sun (see Kepler's second law), so your question is basically asking about the objects that are the furthest from the Sun while still within the Sun's gravitational potential well. I'm not sure which one specific object would be the answer to your question, but I think it'd be something within the Oort cloud. There are long-period comets (which are thought to originate from the Oort cloud) which have orbital periods in the thousands of years. For example, Comet McNaught has an orbital period of about 92,600 years. However, your criterion of an "object that orbits the Sun" is a bit fuzzy, because when you're talking about the objects with the longest orbital periods, that are the most weakly gravitationally bound to the Sun, there's fuzziness as to how long the object has to remain in the vicinity of the Sun before it gets lost as an interstellar comet for it to still fit your definition of an "object that orbits the Sun". Red Act (talk) 05:38, 11 April 2014 (UTC)[reply]
The OP asked about rotational period, not orbital period. --140.180.250.141 (talk) 05:57, 11 April 2014 (UTC)[reply]
Yes, I seem to have misread the question, so I have retracted my answer. Red Act (talk) 06:17, 11 April 2014 (UTC)[reply]
One could say it was redacted! Har! Dismas|(talk) 15:36, 11 April 2014 (UTC)[reply]
Our article on Rotation period includes these figures for major bodies in the Solar System. Extreme rotation periods for some smaller bodies can be found here. Of the bodies known to Wikipedia, it looks like Venus is your answer. RomanSpa (talk) 06:56, 11 April 2014 (UTC)[reply]
Since you didn't specify a size, there are an almost infinite number of objects orbiting the Sun directly, when one includes the asteroid belt, Kuiper Belt, and Oort Cloud and allows for sizes down to a pebble. Of all those, I'd expect some have essentially no rotation at all. StuRat (talk) 13:03, 11 April 2014 (UTC)[reply]
It's infinite, there are neutral hydrogen atoms that orbit the Sun and they can have a total angular momentum of zero (electron in the 1s state and the electron-proton system in the total spin zero singlet state). Count Iblis (talk) 23:18, 11 April 2014 (UTC)[reply]
Probably Mercury (planet) is the correct homework answer as it is tide locked with the sun and Mercurians only get one "day" every 2 "years". (but I didn't check venus) --DHeyward (talk) 04:40, 12 April 2014 (UTC)[reply]
Please do so. —Tamfang (talk) 05:41, 12 April 2014 (UTC)[reply]

Gas pipeline

How heavy must a piece of debris be (assuming it's travelling at the speed of a typical avalanche) to rupture a major aboveground gas pipeline? For example, would a good-sized boulder do the job? How about a snowmobile? Or a typical four-door sedan? Or a 3-ton truck? Thanks in advance! 24.5.122.13 (talk) 06:57, 11 April 2014 (UTC)[reply]

I guess we've won the War on Terror if Al Qaida is reduced to this. — Preceding unsigned comment added by 96.227.210.243 (talk) 12:47, 11 April 2014 (UTC)[reply]
Not a question about Al-Qaida, but a research question for a disaster novel. 24.5.122.13 (talk) 23:11, 11 April 2014 (UTC)[reply]
The shape and way it hits would make a huge difference. Imagine a block of masonry hitting it. If it hit corner first, it might well poke a hole, while a glancing blow on the side of the block would not. The difference is the pressure exerted at a given point. However, enough total force, even if evenly distributed, could still rip the pipeline off it's supports and rupture it. StuRat (talk) 12:53, 11 April 2014 (UTC)[reply]
So, a boulder could do it if it was big enough and/or it hit the "right" way, right? This is just what I hoped to hear -- the rupture of a gas pipeline (which is promptly ignited by a downed 39,000 volt power distribution line, incinerating an automobile with its occupants and threatening to do so to another in which a mother and her daughter are trapped and badly hurt) is an important plot element, and I'd be VERY disappointed if it turned out to be implausible. 24.5.122.13 (talk) 23:11, 11 April 2014 (UTC)[reply]
Possible, sure. Make sure a particularly sharp protrusion from the boulder hits the pipeline first. Also, you'd need to explain why they would build a pipeline where there's an obvious avalanche danger. My suggestion, explain it with acoustic lubrication. This is a situation where the sound/vibrations of the landslide hit just the right frequency, providing a type of lubrication that allows the landslide to go much further than predicted. When you see film of such a landslide, it looks like rushing water, due to the greatly reduced friction. To explain it to the readers, the sound frequency makes the boulders vibrate, so they are only in contact with the ground and each other part of the time, significantly lowering the average friction versus if they were inconstant contact. The larger the objects, the lower frequency is needed for this type of resonance, so boulders require quite a low frequency sound. StuRat (talk) 23:30, 11 April 2014 (UTC)[reply]
In the Canadian Rockies, there's often no choice but to build stuff despite a known avalanche hazard. A more important question is why that part of the pipeline wasn't built underground -- perhaps it would be to save on construction costs, or maybe it was supposed to be built underground but they decide to turn on the gas first and only then cover the trench because the project is running badly behind schedule? 24.5.122.13 (talk) 23:45, 11 April 2014 (UTC)[reply]
It would be nice to actually cite references when stating that something is "possible, sure". Here's a South African report that talks (on page 5) about how impacts are a common cause of pipeline ruptures, although it's talking about underground pipelines. Also note the following sections where it talks about fireballs and such, which may be useful for the novel. Here are US NTSB reports on pipeline incidents, but the table does not classify them by cause, so you might have to read a number of them to see to find ones that are relevant to your scenario. --50.100.193.30 (talk) 00:53, 12 April 2014 (UTC)[reply]

Bug ID

What is this? Should I be worried that this dropped down onto my desk? -- Zanimum (talk) 12:10, 11 April 2014 (UTC)[reply]

Wood louse (Isopoda) suborder, pill bug (Armadillidiidae) family. Very common here. I've never known them to be harmful. In fact, they are kind of fun. Poke them on the back and they will roll up into a little ball like an armadillo, to defend themselves. You can then play marbles with them, until they get tired of it and walk away. :-) StuRat (talk) 12:30, 11 April 2014 (UTC)[reply]
A solitary one is not a problem, but; "When large numbers of woodlice are found indoors, perhaps clustered in wall crevices or under skirting boards etc., it is always worth checking for excessive dampness in these places - just in case there is a structural problem with the damp proofing or damp course".[4] Alansplodge (talk) 13:03, 11 April 2014 (UTC)[reply]
Right, but they still aren't a problem (other than a cosmetic one), but rather they just indicate that a problem exists. StuRat (talk) 13:56, 11 April 2014 (UTC)[reply]
Well, the question was "should I be worried". Referenced answer: "Maybe". Alansplodge (talk) 22:43, 11 April 2014 (UTC)[reply]
Please do not worry unduly because it looks like Armadillidium vulgare has survived the fall. Let's not be negative about his lifestyle. 84.209.89.214 (talk) 12:44, 11 April 2014 (UTC)[reply]
'Round these parts, we call those roly polys. Justin15w (talk) 14:38, 11 April 2014 (UTC)[reply]
I had an interesting experience with one once. I wasn't in the mood to play with it when it crawled into my house, so I flushed it. At the moment it disappeared down the toilet, everything went black and silent, so I thought "Damn, that pill bug must have been God, and now the universe has ended" ... at least until the power came back on a minute later. :-) StuRat (talk) 12:48, 11 April 2014 (UTC)[reply]
:) -- Zanimum (talk) 14:44, 11 April 2014 (UTC)[reply]
In Kent, according to a former colleague, pill woodlice are known colloquially as "monkey peas",[5] presumably because monkeys are supposed eat them like peas? The origin of this seems to be completely unknown - there are no wild monkeys in the south east of England. Alansplodge (talk) 12:56, 11 April 2014 (UTC)[reply]
:) -- Zanimum (talk) 14:44, 11 April 2014 (UTC)[reply]
In Norse areas of northern England they were known as "thuslice" /θʊslaɪs/ (originally Thor's lice). Dbfirs 21:16, 11 April 2014 (UTC)[reply]

Thanks all! I've seen pillbugs a lot outdoors, never in. I work in a 148-year-old building, so we frequently have surprises. -- Zanimum (talk) 14:44, 11 April 2014 (UTC)[reply]

Simple beam calculations

I have a beam with a simple support at one end and a roller support at the other end. The beam has a 30kN load 1m from the left support, a 50kN load 2m from the left and a uniform load 3m from the left, which spans 2m. The overall span of the beam is 6m. I've simplified this to a simply supported beam, which I'm assuming is correct. In order to find the reactions at either end I've done the following. R1+R2=w1+w2+(w3*length of uniform load/2), and for the moments R2*L=w1*x1+w2*x2+w*length of uniform load*distance to centre of uniform load from left of beam. I've basically done this by simplifying the uniform load to a concentrated load acting at the centre of the length of the uniform load. But apparently this is incorrect as my instructor has used W*l instead of w*l/2 to find the contribution of the uniform load to shear force equilibrium, and wl^2/2 instead of wl^2/8 for uniform load contribution to moment equilibrium, which in my opinion treats the uniform load section as a cantilever. Where have I gone wrong in my calculations? Why is that section a cantilever? Clover345 (talk) 13:32, 11 April 2014 (UTC)[reply]

We need a diagram. Is this correct (not quite drawn to scale) ?
                          UNIFORM LOAD
         30kN       50kN /            \
          |          |  /              \
          |          | /                \
<-- 1m -->V<-- 1m -->V/<------ 2m ------>\<------ 2m ------> 
============================================================
o                                                          ^
I assumed the uniform load is centered 3m from the left. Also, is the magnitude of the uniform load specified ? StuRat (talk) 13:51, 11 April 2014 (UTC)[reply]
The uniform load is centred 4m from the left and it's value is 10kN/m. Also the simple support is at the left and the roller support is at the right. Clover345 (talk) 14:46, 11 April 2014 (UTC)[reply]
Like this ?
                                    UNIFORM LOAD
         30kN       50kN           /   10kN/m   \
          |          |            /              \
<-- 1m -->|<-- 1m -->|<-- 1m --> /<----- 2m ----->\ <-- 1m --> 
          V          V          vvvvvvvvvvvvvvvvvvvv
==============================================================
^                                                            o
StuRat (talk) 16:31, 11 April 2014 (UTC)[reply]

Ye, that's it exactly. That is a simply supported beam right? According to my instructors calculations that's a cantilever but I can't understand why. Clover345 (talk) 16:48, 11 April 2014 (UTC)[reply]

Unfortunately, it's been too many years since I've done such a problem to answer reliably, but now that the Q is clear and we have a diagram, hopefully others can answer. (Your approach of dealing with it as two simple supports seems right to me, too, so either I'm missing something or the instructor made a mistake.) StuRat (talk) 17:08, 11 April 2014 (UTC)[reply]
If the uniform load is specified as 10kN/m then its (not spelled "it's") total force is 10k/m times 2m. This is what the instructor means by W*I and the OP has no reason to divide by 2 in the expression for R1+R2 which comes out to 100 kN. The sum of the clockwise moments about the left end is 30 + 50*2 + 10*2*4 kN m = 210 kN m which is in equilibrium with 35 kN force on the support at the right end. That leaves 65 kN force at the left end. The beam is not a cantilever unless it extends (not shown) beyond the right hand support. It's commendable that contributor StuRat shows correct spelling here. 84.209.89.214 (talk) 19:38, 11 April 2014 (UTC)[reply]
Thanks but why is this? I thought that in a simple beam with a uniform load, the moment is given by wx/2*(1-x), whilst the shear force is given by w(l/2-x). Clover345 (talk) 20:07, 11 April 2014 (UTC)[reply]
This may help. 84.209.89.214 (talk) 21:42, 11 April 2014 (UTC)[reply]

Weather/travel

I have heard there is a possibility of a storm on the east coast early next week what is the chance that incoming flights into airports in the northeast will be affected — Preceding unsigned comment added by 75.69.246.118 (talk) 14:55, 11 April 2014 (UTC)[reply]

OP geolocates to Massachusetts, so that's probably the coast he's interested in. Rojomoke (talk) 16:24, 11 April 2014 (UTC)[reply]
Looking up the weather for Boston, I get rain and winds up to 25 mph on Tuesday, which doesn't seem like enough to cancel flights, to me. If you can be more specific as to the exact arrival date, time and location, we can give a better answer. StuRat (talk) 16:27, 11 April 2014 (UTC)[reply]
The OP should call his airline. We don't have some secret knowledge they don't have. μηδείς (talk) 20:11, 11 April 2014 (UTC)[reply]
The National Weather Service's area forecast for the Boston region, for April 11, 1745 UTC, includes notification for an AIRMET (AIRMET Sierra) with IFR conditions, mountain obscuration, overcast skies, low ceilings, thunderstorms, severe or greater turbulence, severe ice, low level wind shear, anf IFR conditions. You can check the Terminal Aerodrome Forecast for most major commercial airports to get specific forecast probabilities for specific types of weather. It is at the discretion of the airline whether they ought to delay or cancel flights ahead of time when such weather is forecast. Nimur (talk) 23:41, 11 April 2014 (UTC)[reply]

April 12

Please identify yellow flowering plant

Can anyone identify this beautiful yellow-blossomed flowering plant? Thanks. μηδείς (talk) 01:23, 12 April 2014 (UTC)[reply]

Hard to tell. Could be an Alstroemeria (Peruvian Lily) or a Daylily or a True Lily. --Jayron32 02:19, 12 April 2014 (UTC)[reply]
The image is way too fuzzy for me to have any hope, but the leaves look more dicot-ish than monocot-ish. Looie496 (talk) 02:34, 12 April 2014 (UTC)[reply]
Yes, it really does look like a lily, but the flowers remind me of Jimsonweed, the leaves do look dicotish (you can see they don't have parallel veins in the original), and I am fairly sure it is a member of Solanales, if not Solanaceae, and perhaps a Cestrum, whose leaves and size are similar. Not this seems to be one inflorescence from one shoot. The screenshot is from 30:09 of episode 2 season 1 of Boston Legal. The plant looks like a spike cut from a much larger bush. μηδείς (talk) 02:45, 12 April 2014 (UTC)[reply]
I left a note on the talk page of @Cullen328:, I seem to have some recollection that he might know something about flowers. --Jayron32 02:59, 12 April 2014 (UTC)[reply]
It sure looks like some type of lily to me, but my wife who knows more than me says that the photo is not clear enough to narrow it down beyond that. Very similar flowers are growing by my front door right now, planted over ten years ago by the home builder. A slight correction, Jayron32: You are right that I love flowers and post at least one of my flower photos on Facebook every day, along with its location, as I travel around Northern California on business and pleasure. This is my way of letting my friends know where I travel, and sharing a bit of beauty. But I am not an expert and usually rely on my Facebook friends for help identifying species. Thanks for remembering my interest in flowers, though. Cullen328 Let's discuss it 03:11, 12 April 2014 (UTC)[reply]

Is it true that some Prebiotic fibers could actually make us fatter?

I've heard that some particular types of Prebiotic fibers (whether they'll be Soluble, Insoluble but Fermentable, or Resistant starch), could actually give us the opposite effect and make us fatter, because unlike most of the Prebiotic fibers that supposedly make our Probiotic gut flora differentiate and (basically that's it), these particular aforementioned one's could also resemble the extra effect of breaking into notable amounts of available energy\handable sugars, in the gut, that will then enter the bloodstream directly.

What are the particular types of such "human-fattening prebiotic fibers", if any of them are actually known to Science except from general theoretical thought...? Much thanks!!! — Preceding unsigned comment added by 109.67.164.204 (talk) 08:55, 12 April 2014 (UTC)[reply]

See Dietary fiber#Fiber and calories. According to http://www.menshealth.com/mhlists/facts_about_nutritional_fiber/printer.php:
"Fiber is essentially composed of a bundle of sugar molecules. These molecules are held together by chemical bonds that your body has trouble breaking. In fact, your small intestine—can't break down soluble or insoluble fiber; both types just go right through you. That's why some experts say fiber doesn't provide any calories. However, this claim isn't entirely accurate. In your large intestine, soluble fiber's molecules are converted to short-chain fatty acids, which do provide a few calories. A gram of regular carbohydrates has about 4 calories, as does a gram of soluble fiber, according to the FDA. (Insoluble fiber has essentially zero calories.)"
"Fiber's few calories are more than offset by its weight-control benefits. The conclusion of a review published in the journal Nutrition is clear: People who add fiber to their diets lose more weight than those who don't. Fiber requires extra chewing and slows the absorption of nutrients in your gut, so your body is tricked into thinking you've eaten enough, says review author Joanne Slavin, Ph.D., R.D. And some fibers may also stimulate CCK, an appetite-suppressing hormone in the gut."
So note that the benefits of soluble dietary fiber, such as slowing the absorption of sugars and thus preventing blood sugar spikes followed by insulin spikes and sugar crashes, outweigh the negatives. StuRat (talk) 12:41, 12 April 2014 (UTC)[reply]
I'm intrigued by the OP's words "could actually give us the opposite effect". Opposite effect to what? HiLo48 (talk) 12:50, 12 April 2014 (UTC)[reply]
It is a bit tricky to figure out what they meant, but I took them to mean that fiber is normally good for weight loss because it fills you up and contains no digestible calories. Here the "opposite effect" would be if it causes weight gain because it does contain digestible calories. StuRat (talk) 12:57, 12 April 2014 (UTC)[reply]

Using aggregated data for statistics

If you want to do medical research, one way is to ask for instance GPs to sent all their medical records. This method has obviously huge implications regarding privacy. Even if data like Social Security numbers and names are omitted, sometimes it's not too hard to find out which record is about which patient. Another problem is that the GP doesn't have a guarantee about what the research is used for and by whom.

I'm wondering if it's possible that every GP collects his data (about, say, 2500 patients) and turns it into a file that contains facts like "5% of patients aged 10-15 have diabetes, of which 3% need insulin", etc. So part of aggregating the data takes place at the computer of the GP, and (more) anonymous results are sent to the research institute. Of course, if the practice has only a few people suffering from a particular disease, you might deduct some more information about them, but at least it's better than just sending separate records.

I was told that this method would make some research impossible, or at least very difficult. If the right set of questions can be asked and answered correctly by the system of the GP (like "of all patients using medication A, older then B, how many are also prescribed medication C"), is it true that for some research you actually do need the individual records? Joepnl (talk) 11:16, 12 April 2014 (UTC)[reply]

I don't see that approach working. For example, the portion of patients who "need insulin" will vary by doctor, as some may rely more on other methods to control diabetes. StuRat (talk) 12:26, 12 April 2014 (UTC)[reply]
The article Medical ethics notes the value placed on the patient-physician privilege. This is upheld because patients might withold sensitive information if unsure whether telling it can have consequences beyond their medical treatment, but it is not an absolute privilege because various laws require physicians to report STD's, gunshot wounds, child abusers and vehicle drivers impaired e.g. by epilepsy. Medical research involving both medical ethics and Research ethics raises questions like the OP's question. Shall results of Human subject research be published for study even if their collection was abusive and unethical (the WW II examples are well known)? Is research without knowing consent of the subjects justifiable? (Ethical questions are raised over Stanley Milgram's 1961 experiment of misleading subjects into applying an increasing scale of shocks to people ostensibly to "improve" them, yet today Wikipedia administrators believe themselves empowered to apply electronic blocks on an identical increasing scale that ends in indefinite dehumanisation of participants.) The problem with the OP's proposed procedure is that a General practitioner is already hard pressed by lack of resources and pharmaceutical company salespeople and has a skill set focussed on giving care to individual patients that is different from being an objective data-collection functionary. The GP might argue in modern terms to have "sworn by Apollo" never to "cut for stone" (take on the role of a surgeon/researcher ?) and "keep secret" (the patient's identity). These considerations detract from the end-to-end transparency, verifiability and accountability that medical research demands. (Readers may be interested to see how voluntary reports on side effects of medicines are collected in Norway, a country with well developed Publicly funded health care. On this site in Norwegian anyone living in the country is invited to report side effects they or others experience from medications. The pink button links to the reporting form; this appears however only for residents.) 84.209.89.214 (talk) 13:12, 12 April 2014 (UTC)[reply]
Well, actually the problem is that many GPs in The Netherlands don't seem to realize what kind of data they're sending out for research (or many other purposes). As a tech person, who does know the kind and volume of data leaving the practice without the doctor or patient knowing exactly what's happening, that bothers me. A lot of really important research can't be done without data from their practices, however. So I'm looking for a way that would be good enough for the researchers and at the same time doesn't imply that the most intimate things people share with their doctors can't be traced back to a single patient by them. Joepnl (talk) 13:32, 12 April 2014 (UTC)[reply]
Yes, some research couldn't be done this way. You would have trouble doing a longitudinal study for example. Furthermore, if you need to know information like diet, which GPs don't routinely collect, you would not be able to do the research. But the important thing is that this approach would open up a new avenue for research. It would be difficult to get it through ethics committees, so it might have to be handled through private institutions, which are often subject to different ethical processes. IBE (talk) 13:29, 12 April 2014 (UTC)[reply]

Dynamic load.

How do you find the maximum load, as well as max moment and stress on a cantilever beam if you have the dynamic load factor? I calculated the DLF using FL^3/3EI but I'm lost on where to go from here. — Preceding unsigned comment added by 82.132.236.40 (talk) 12:37, 12 April 2014 (UTC)[reply]

If you only have one dynamic load on the beam, it seems to me that the max moment and stress would both occur when that dynamic load is at the maximum. With multiple dynamic loads, the situation is more complex. There the max moment and stress could occur when one of the dynamic loads is at it's max, but this isn't always true, and even if it is, it isn't necessarily the largest dynamic load that's at max at that time.
A computer simulation may be in order in such cases. If the dynamic loads vary independently of each other, you might vary each independently in the simulation and calculate the moment and stress at each time to determine the max for each (again, the maxes may occur at different times). StuRat (talk) 12:50, 12 April 2014 (UTC)[reply]
Thank you for your answer but I don't quite understand what you mean. The dynamic load factor is a ratio of the deflection as a result of the dynamic load to the equivalent static load deflection. So from this point, I would probably need to do some sort of energy analysis to find the maximum force but I'm not sure what the correct equation is.