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January 17

Is celiac disease a genetic disorder

Is celiac disease a genetic disorder? It's not listed in List of genetic disorders.Fundon1 (talk) 00:35, 17 January 2014 (UTC)[reply]

Wikipedia has an article you just linked to, which answers that exact question. --Jayron32 03:04, 17 January 2014 (UTC)[reply]

Let me add that Wikipedia's "list" articles are usually not constructed in a very systematic way. Basically editors add whatever they think of. Looie496 (talk) 03:08, 17 January 2014 (UTC)[reply]

Food capsules of the future?

Is it currently possible to compress all the things an adult human needs from food into a capsule to replace eating regular meals? — Preceding unsigned comment added by 174.65.117.118 (talk) 00:38, 17 January 2014 (UTC)[reply]

The closest is Nutraloaf which when fed to prisoners instigates riots.Fundon1 (talk) 00:44, 17 January 2014 (UTC)[reply]

One of the things that humans need from food is the pleasure of eating, often communally, and especially in celebration. Can't imagine a capsule delivering that. HiLo48 (talk) 00:59, 17 January 2014 (UTC)[reply]

If by capsule you mean something very small, the answer is no. The densest form of calories is pure fat, such as vegetable oil, and it takes about 2 cups of vegetable oil to provide as many calories as an ordinary adult needs for a day. If you substitute things that provide nutrients such as protein, vitamins, etc, this will get considerably larger. Looie496 (talk) 02:45, 17 January 2014 (UTC)[reply]

You also need a fair quantity of dietary fiber, and you can't get much of that into a capsule.--Shantavira|^{feed me} 08:27, 17 January 2014 (UTC)[reply]

Possibly in a horse pill size... --Auric talk 17:27, 18 January 2014 (UTC)[reply]

I recall consuming "Space Food Sticks" as a future-astronaut-wannabe; this was supposed to be that sort of thing. ~E:71.20.250.51 (talk) 17:05, 19 January 2014 (UTC)[reply]

Blonde pubic hair and eyebrows

is it possible that a white person have blonde pubic hair and blonde eyebrows? — Preceding unsigned comment added by 70.31.22.79 (talk) 01:52, 17 January 2014 (UTC)[reply]

Sure. --Jayron32 03:03, 17 January 2014 (UTC)[reply]

Sadly, our article Human hair color doesn't mention pubic hair. But Jayron's original research is accurate. HiLo48 (talk) 03:07, 17 January 2014 (UTC)[reply]

There's a well known quote from the film Diamonds are Forever which should give you a clue:

Bond: I tend to notice little things like that - whether a girl is a blonde or a brunette...
Tiffany: And which do you prefer?
Bond: Well, as long as the collar and cuffs match...

Also, if you look at the Albinism page you will notice that the albinistic people in the pictures have white eyebrows - not quite blonde but the principle is the same. Richerman (talk) 10:26, 17 January 2014 (UTC)[reply]

Actually, hair is never blonde, only blond. People can be blonde, if they're female. But I don't suppose that was your point. --Trovatore (talk) 10:53, 17 January 2014 (UTC) [reply]

Not in British English see:[1]. Richerman (talk) 15:20, 17 January 2014 (UTC)[reply]

Since English does not have grammatical gender, only sex (i. e., natural gender), and hair is not a feminine noun, I agree that it does not make sense to say blonde hair, whether the hair belongs to a female or male person. However, it does not make sense to call a woman blonde, either, as blond and brunet would be the only adjectives to exhibit this behaviour, since all others are invariable except for synthetic comparison. As an adjective, blonde is essentially only a variant form – in my POV: a pure misspelling – of blond. The only unarguably correct use of blonde is as a noun (a) blonde for "blond woman", although the obscurity of the use of the male counterpart (a) blond reveals a sexist bias, I suspect. See also my comment at Talk:Declension#English example for an adjective that declines. --Florian Blaschke (talk) 16:55, 18 January 2014 (UTC)[reply]

In British English 'blonde' is correct as a noun or adjective according to the OED and blond is only listed as an alternative spelling. I've never seen the spelling 'brunet' before - it gets a mention in the OED under the entry for 'brunette' but it says 'now chiefly US' Richerman (talk) 19:35, 18 January 2014 (UTC) [reply]

~~"Sexist bias" my ass. I think that's PC nonsense. How is being a "blonde" worse than being a "blond"? --Trovatore (talk) 21:39, 18 January 2014 (UTC)~~ Ah, rereading your comment, I seem to have misinterpreted you. I still think it's a bit hypersensitive though. --Trovatore (talk) 21:42, 18 January 2014 (UTC) [reply]

The Movie Gravity.

1. When George Clooney untethers himself from sandra bullocks, why does he drift away from earth ? Shouldn't he be revolving around the earth just like a satellite ?

2. Opening the hatch from outside - will not drain all the air from the escape vessel ? — Preceding unsigned comment added by 115.113.11.188 (talk) 10:22, 17 January 2014 (UTC)[reply]

I haven't seen the movie, but you might be interested to read Gravity (film)#Scientific accuracy.--Shantavira|^{feed me} 10:36, 17 January 2014 (UTC)[reply]

SPOILERS: I have seen the movie, and yes, Clooney would not drift away in a real situation like that. As for the second, if you saw the film, you would hopefully have noted that the hatch never actually opened. Mingmingla (talk) 17:09, 17 January 2014 (UTC)[reply]

1. If Clooney was tethered to Bullocks but still accelerating (and stretching the tether) then breaking the link would make Clooney drift away.

2. he didn't do that, it was all Bullocks' dream. But no it wouldn't drain all the air that is still compressed in some sort of cylinder. Bullocks could open and close the oxygen supply, so there was air somewhere.

For watching films you need Suspension of disbelief. Gravity is not a documentary about how things work in space. The whole story is full of holes to make the story look more appealing. Traveling from the Hubble telescope to one space station and then to another, showing Bullocks in sexy hot-pants (instead of a more appropriate thermal suit below the space suit), and much more, makes the story progress. OsmanRF34 (talk) 18:05, 17 January 2014 (UTC)[reply]

Even the actress seems to have progressed to a plural person. :) -- Jack of Oz ^{[pleasantries]} 23:29, 17 January 2014 (UTC) [reply]

Thanks Jack - that load of bullocks was annoying me too. :) Richerman (talk) 00:00, 18 January 2014 (UTC)[reply]

For suspension of disbelief you need a fiction which has its own rules and which abides by them consistently. The core rule undercutting most fiction is the world is the same unless otherwise stated, and if it makes it not the same without justification, the suspension breaks, the immersion breaks, and the enjoyment is reduced.

For (1) it makes sense if you assume that they are spinning slightly so there is some centrifugal force pulling him away from the tethers. They did such a good job with the rest of the science that I bet the writers had this much better, but it got lost somewhere between storyboarding and the edit room floor. 109.70.142.60 (talk) 23:44, 17 January 2014 (UTC)[reply]

Neil Degrasse Tyson Tweeted several criticisms of the movie (although apparently he really liked it on the whole). Your first question is one of the issues he raises. In an interview with Huffington Post, director Alfonso Cuaron explained it this way:

What happens is she's grabbing the tethers and he comes with momentum. His momentum pulls her. They're moving together. There's a wide shot that shows they keep moving and you can see the background keeps on moving. What happens is, if he lets go, his force stops and the force of the tether takes over. And, look, by saying that, this is not a documentary. We took certain liberties. Part of the liberties we took were in the sense of we would stretch the possibilities of certain things.

--— Rhododendrites ^talk | 19:52, 19 January 2014 (UTC)[reply]

Eating when not hungry

People say eating when not hungry can lead to becoming overweight but is this true even if the recomnended daily calorie intake is not exceeded? 82.40.46.182 (talk) 12:44, 17 January 2014 (UTC)[reply]

Yes. Simply because you may not be using all the calories you're taking in. Recommended daily intake is vague at best. Many other factors, such as age, gender, level of activity, genetics, etc., can affect what YOUR RDI is. See Weight_gain 196.214.78.114 (talk) 13:27, 17 January 2014 (UTC)[reply]

Only in general though. Any number of external factors could be affecting the person's appetite. (Making them not feel hungry when they really do need calories, or vice-versa.)

Obviously, if the question is motivated by something personal you'd want to talk to an actual doctor. APL (talk) 22:57, 17 January 2014 (UTC)[reply]

I've heard two pieces of advice: "Never eat when not hungry" and "Always eat breakfast". Unfortunately, I'm never hungry when I first wake up, so those two are in conflict for me. I decided to go with the first bit of advice, and not eat breakfast. StuRat (talk) 23:59, 17 January 2014 (UTC)[reply]

Eating when bored may lead to distracted eating or panic eating.--Auric talk 17:21, 18 January 2014 (UTC)[reply]

There was recently, an article in a UK newspaper that stated the obvious. Families who serve dinner at the table have slimmer children - because they learn when they are full. When I go State-side I always put on weight. One can’t help it. Don't get me wrong – I do like a big beef stake -once in a while- but the ginormous meals one that get served up with is about three time my daily calorie intake – and that is just in one meal. It is gluttony taken to extremes. Does the recommended daily calorie your talking of, excluded meals eaten outside the family home?--Aspro (talk) 21:20, 21 January 2014 (UTC)[reply]

Second-order differential equation in Simple Harmonic Motion

I am reading a physics book in which the following equation is given:

-kx(t)=m{\frac {\mathrm {d} ^{2}}{\mathrm {d} {t}^{2}}}x\left(t\right).

i.e., second-derivative of position (x) w.r.t., time (t) is equal to negative of position (x) times spring constant (k) divided by mass (m). My book just says "it is a second-order differential equation and after solving this equation you get

x\left(t\right)=A\cos \left({\sqrt {k \over m}}t\right).

". But it didn't mention how to solve this and only give the answer. If possible, I want someone to show me steps how to solve second-order differential equation. Britannica User (talk) 13:57, 17 January 2014 (UTC)[reply]

It sounds like you're reading a physics book that's a notch ahead of your mathematical preparation. By the time you read that book, its author assumes that you should have the equation for simple harmonic motion etched indelibly into your memory, and recognize its solution by inspection; maybe you skipped a few books ahead of your level. But how is the solution derived?

The solution is predicated on using Euler's formula to represent the cosine in terms of a complex exponential. Solve the differential equation by the separation of variables method, and recognize that the exponential function is a good ansatz because it equals its own integral (rather, it is linearly related to its own integral... and in the sloppy math we do in physics, that means it's equal, because we don't worry about a scale factor or a constant offset - those are just coordinate changes). Apply the physical constraints (which are stated implicitly, because no initial condition was provided), and drop the sinusoidal term (representing an arbitrary phase lag). This sort of thing is covered in gorey detail in a book on methods of solution for ordinary differential equations.

So, this is just another case where "thinking like a physicist" means "knowing the math so well that you do all of the above by inspection." Most people need two or three years of mathematical training beyond their initial integral calculus work before they can just see this sort of thing. Some people, no matter how much time they pour in, never get there. And every few centuries, we get a Newton or a Fourier and they just get it. Nimur (talk) 14:23, 17 January 2014 (UTC)[reply]

Oh, and if you don't like Euler's formula, or if you're a skeptic who doesn't want to use e^iωt as your trial function, then you can re-express your ansatz as a Taylor expansion-style infinite sum of polynomial terms, and it can be proved that the series will converge by solving explicitly for the polynomial coefficients. You will also prove that the exponential would have been a good choice. So, you can solve this equation numerically, too. In fact, there are many ways to find solutions to this very common equation. Nimur (talk) 14:30, 17 January 2014 (UTC)[reply]

I am a 10th grade student, but grade doesn't matter. I know about complex numbers, Euler's formula, differentiation (ordinary as well as partial), integration, and how to solve first-order ordinary differential equations; however, I don't how to solve second-order differential equations. A book on Calculus says this "There is no general analytic method for finding solutions to second-order and higher differential equations. Solutions of such differential equations are often obtained by educated guessing, or numerical methods". If the solutions are obtained by educated guessing, then I noticed that there can be many possible solutions of the second-order DE I mentioned above. Therefore, why did the author of my physics book just stick to a single solution? How did he know only that solution would fit to that equation? Britannica User (talk) 15:47, 17 January 2014 (UTC)[reply]

Not a problem... just don't panic if there are some complex parts in the math. With enough time and effort, you'll get it.

In general, if a solution space is spanning the problem space, we can guarantee the solution is unique. In methods for solving differential equations, we construct an orthonormal set of basis functions from which we construct the solution. Even if we guess the answer - that is, if we use an ansatz function - we can still prove it is correct. Purist mathematicians have lots of tricks to help us make good guesses; but your book is correct, sometimes there is no analytic method when you've got very hard equations. (Sloppy physicists just learn how to guess right, and prove it later).

This probably sounds jargon-y, but what it means is that we have a formal, mathematical way to prove that our solution is complete and that it is the only possible solution for this problem. Here, we would use A cos (ωt) + B sin (ωt) and we know B=0, and ω²=k/m because we solve the equations explicitly. B could be nonzero if initial conditions were specified. As physicists, we don't care, because the function is periodic, so we can slide around the time we define for "t=0" until B is also zero. Nimur (talk) 15:52, 17 January 2014 (UTC)[reply]

If you want to learn the technique that applies here, the key thing to know is that the equation is linear. There is a general method for solving linear differential equations of all orders. Our article unfortunately is written at a level of abstraction that you probably won't be able to handle, but any textbook that covers differential equations will have a section on this class of problem. Looie496 (talk) 16:07, 17 January 2014 (UTC)[reply]

Indeed. The simple harmonic oscillator equation is especially easy to solve because it is both linear and homogeneous. Your book which said "There is no general analytic method for finding solutions to second-order and higher differential equations" probably had in mind more general *non-linear* differential equations, which are much more difficult to solve explicitly. Gandalf61 (talk) 16:27, 17 January 2014 (UTC)[reply]

And for the simple harmonic oscillator, there are many methods to analytically find the solution; separation of variables, for example. You can also take the Laplace transform (or any other differential transform) and solve algebraically in the transform domain, and then apply the inverse transform to find the result. This specific equation, x’’ = ±Cx has been analytically solved for hundreds of years in thousands of different ways. Nimur (talk) 16:38, 17 January 2014 (UTC)[reply]

We have an article specifically on the Harmonic_oscillator that may help. It is often easiest to find the Characteristic_equation_(calculus), which reduces the problem to simple algebra. The solution behavior depends on the roots of this equation: each root leads to a solution, and the general solution is a linear combination of these solutions with arbitrary real coefficients. You might also look at some other calculus books: both of mine discuss harmonic oscillators in a chapter on elementary differential equations. Hope this helps. OldTimeNESter (talk) 17:15, 17 January 2014 (UTC)[reply]
You can do the following:
1. Introduce a new function: $y(t)={\frac {dx}{dt}}$
2. Now we have: ${\frac {d^{2}x}{dt^{2}}}={\frac {dy}{dt}}={\frac {dy}{dx}}{\frac {dx}{dt}}=y{\frac {dy}{dx}}={\frac {1}{2}}{\frac {d(y^{2})}{dx}}=-kx$
3. The above equation can be solved by simple integration: $y^{2}=C-kx^{2}$ , where $C$ is a constant
4. Taking into account the definition of $y$ we have: ${\frac {dx}{dt}}={\sqrt {C-kx^{2}(t)}}$
5. Or: ${\frac {dx}{\sqrt {C-kx^{2}(t)}}}=dt$
6. The latter equation can be again solved by simple integration: $t=t_{0}+\int \limits _{x_{0}}^{x}{\frac {dx'}{\sqrt {C-kx'^{2}(t)}}}$
7. The latter integral you can calculate yourself as an exercise.

Ruslik_Zero 19:49, 17 January 2014 (UTC)[reply]

How do people have sex in "impossible sex positions"?

It's a short story, written by a well-regarded author, William H. Coles, here. The short story only mentioned it briefly, yet it got me thinking what type of character Denise was like and what she could have placed in her story. How do people know if a particular sex position is "impossible"? When there is no penetration or something? 140.254.227.125 (talk) 15:28, 17 January 2014 (UTC)[reply]

I think the "back-to-back" position would fall into this class! -- Q Chris (talk) 15:38, 17 January 2014 (UTC)[reply]

Please explain the qualifications that allowed you to make your judgment. 140.254.227.125 (talk) 16:12, 17 January 2014 (UTC)[reply]

See hyperbole. Were a student to regularly describe lovers copulating while hanging from chandeliers or standing on their hands, her creative writing instructor may well describe her as being "fond of [writing about] intercourse in impossible positions" even thought such positions may be possible with sufficiently strong and well hung lighting fixtures and acrobatic participants. --ToE 05:39, 18 January 2014 (UTC)[reply]

(Was your use of the term "well hung" really necessary?) HiLo48 (talk) 22:12, 18 January 2014 (UTC)[reply]

Generally we consider gravity in determining impossibility. Most of the impossible positions require zero gravity, a series of straps, or an underwater environment to be feasible.--Auric talk 17:18, 18 January 2014 (UTC)[reply]

How to calculate probability of a complex system

It's clear that if you have a die or a bag full of red and black balls you can see how probable are one result or the other. But what if you cannot run several tests (like in the case of nuclear reactor or a space shuttle), and you have a extreme complex system, how can you calculate the probability of failure? OsmanRF34 (talk) 17:58, 17 January 2014 (UTC)[reply]

You have to look at all the possible failure modes, figure out the probability of each, and then combine those. For example, failure mode and effects analysis of an electrical system involves looking at a break in every wire, to determine which breaks will cause serious problems, and which will not. In many cases multiple failures are required to cause a disaster, so then those probabilities much be combined. StuRat (talk) 18:11, 17 January 2014 (UTC)[reply]

If you want a simplified example, let's say there are two ways in which a system can fail, and the first requires 2 things to go wrong, while the 2nd requires 3, and for convenience, we will say each component has a 1% chance of failure per year:

P₁ = 0.01 × 0.01 = .0001

P₂ = 0.01 × 0.01 × 0.01 = .000001

P_total = P₁ or P₂ = 1 - (1-P₁)×(1-P₂) = 0.0001009999

Note that P_total is very close to just being the sum of P₁, P₂, etc., for small values, so sometimes that approximation is used. Here P_total was the chance of a disaster in one year, but you could get the chances for a decade or a century just by combining them in the same way as the last step above, where P₁ would then be the probability of any failure in year 1, P₂ in year 2, etc., and you can just keep adding more terms like that to get the total probability over the full time period.

One limitation of this type of analysis is that it assumes that each component failure is independent of the rest. However, this isn't always the case. You can have a cascade failure, where one failure causes others, like an electrical grid where one wire going down puts too much power on the rest, and then they fail, too. You can also have a common outside cause for multiple failures, like an ice storm causing many electrical lines to go down. For a real example, the Fukushima Daiichi Nuclear Power Plants lost primary reactor power, backup electrical power from the grid, and the backup generators, all due to the same earthquake and tsunami. Not having any power then made it difficult and dangerous for them to open and close valves manually, which made it impossible to control the reactors.

With this is mind, the historic failure records of a complex system may be more useful in figuring out the overall risk. However, this isn't possible for a new system, unless similar to existing systems. And, even then, quite a minor change can have a huge effect on the overall risk. For example, had the backup generators been stored at the highest elevation in the complex, they would have been above the waterline and thus would have functioned to prevent the disaster.

I should probably also mention that human factors figure in to risk, as well. Specifically, once a system has worked well for many iterations, everyone becomes complacent, and safety standards drop, until a disaster happens, which causes them to tighten up on safety, until many more safe iterations, when safety standards start to drop again. This can lead to a cycle of disasters, relatively evenly spaced out, over years. The Space Shuttle disasters are an example of this. StuRat (talk) 18:34, 17 January 2014 (UTC)[reply]

To give an example of historic failure rates, there have been three major incidents in the nuclear power industry (Three Mile Island, Chernobyl, and Fukushima) out of 14500 reactor-years of operation. There are currently about 450 reactors in operations, which would suggest that a major incident might be expected to occur on average every 11 years. Of course, one might hope that each major failure leads to improvements that make future failures less likely. The fact that Fukushima occurred 25 years after Chernobyl might support that conclusion (or it might just have been good luck). From a practical perspective, one makes the systems that one understands as safe as possible, and hopes doing that is good enough to protect against problems that no one planned for. Dragons flight (talk) 22:36, 17 January 2014 (UTC)[reply]

I don't think I'd include Three Mile Island as a major incident. It was a minor incident blown out of proportion by the media. Nobody died. StuRat (talk) 00:05, 18 January 2014 (UTC) [reply]