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November 26

stereo amplifier output impedance

How do I best go about calculating my amp's output impedance? Should put a given ohm resistor and than just differentiate the voltage drop there? Will this do it?

I need this to figure out its damping factor. — Preceding unsigned comment added by 77.35.20.143 (talk) 03:14, 26 November 2012 (UTC)[reply]


Firstly, be advised that damping factor is a figure often quoted, but in solid state amplifiers, and most valve/tube amplifiers with any pretence of quality, it is quite unimportant and just about meaningless. It was important in valve amplifiers with low or zero negative feedback, especially with loudspeakers designed before the late 1960's as such speakers had low internal damping. Modern speakers are pretty heavily damped internally.
Damping factor is most easily measured this way:-
1. Short one input channel of the amplifier, or if an integrated (ie combines the power and preamp) turn the balance control control to zero that channel.
2. Connect an 8 ohm power resistor (or whatever value is equal to the nominal speaker impedance) between the two "hot" outputs of the two channels (assuming a stereo amplifier - if an mono amplifier, borrow another one)
3. Connect a signal source (preferably an oscillator but an AM radio program will do to get an approximate value) to the unshorted channel, and adjust the level for several volts across the resistor, without overloading the signal channel.
+++ This means you have one channel driving the other channel backwards via the 8 ohm resistor. No harm will come from this provided the signal channel does not overload the driven channel. This will not occur if the two channels of one amp are used or two mono amplifiers are used of the same power output. +++
Measure the voltage with respect to ground of the output terminal of the back-driven amplifier channel. An oscilloscope is best but a multimeter will do. If the signal is not an oscillator, use an analog multimeter - do not attempt to get sensible readings with a digital multimeter.
The damping factor is the ratio of the resistor voltage to the back-driven amplifier output voltage. With modern amplifiers the ratio will be very high - you may have trouble measuring the back-driven voltage with a multiometer due to it being very low. The output impedance is the resistor value divided by the ratio.
IMPORTANT - DO NOT overdrive the amplifiers, as if you do, the readings will be meaningless.
Keit 120.145.149.167 (talk) 03:43, 26 November 2012 (UTC)[reply]

Which caulking should I use?

I have to caulk around a ceiling light fixture, I decide to use silicone caulk for the job, since it is high heat. I took a laser thermometer in the area I need to caulk and its 102° with the light on. I have previously used a particular type and brand of silicone caulk on many jobs and was happy with the results, it is the GE 2 silicone kitchen and bathroom caulk, it has a mold inhibitor and it but when I like most about it is that it has very little smell to it. When drying, some of the caulks have a horrible acetic acid smell to them which I want to avoid. My concern is that if I use this caulking, the mold inhibitor may get released into the air when it gets heated up. I looked into some of the "food safe" silicone caulk but it seems to be the type that smells like acetic acid, I'm concerned with the acidic acid type is that when it gets heated up, it may release that acetic acid smell again. I am going to link the MSDS sheets of both types of caulk I'm considering to use and I would like you to tell me which one I should use.

The Ge 2 kitchen and bathroom 3 hour dry http://www.homedepot.com/catalog/pdfImages/80/803b3b37-83d5-40b0-a182-9e5eba6d304c.pdf

The food safe one DAP Household/Aquarium Adhesive Sealant http://www.homedepot.com/catalog/pdfImages/a1/a12dd342-15df-4ca3-92fc-8204bdf74281.pdf?cm_mmc=seo%7Caltruik%7C100128841 --Wrk678 (talk) 03:37, 26 November 2012 (UTC)[reply]

Let's back up a bit. Why exactly do you want to caulk around a ceiling light fixture ? To keep bugs out ? This isn't normally done. For one thing, won't this be a problem when you need to change the bulbs ? Moisture might also be trapped inside, if it evaporates from the caulk as it cures. And you need to allow the fixture to expand and contract as it heats up and cools down. Another concern is that this may allow heat to build up more than it would if left uncaulked. StuRat (talk) 05:18, 26 November 2012 (UTC)[reply]
The acetic acid is only produced as the caulk cures, and not later. It also becomes soft when heated up a lot. Polypipe Wrangler (talk) 10:53, 28 November 2012 (UTC)[reply]

Power spectral density of speech

What is the energy distribution of human speech like on the frequency spectrum? Is there freely available data of the form

On average, X% of the energy in speech lies below frequency Y Hz.

Thanks. — Preceding unsigned comment added by 98.114.41.201 (talk) 04:41, 26 November 2012 (UTC)[reply]

Unfortunately it's far too variable for any simple statements like that to be useful -- it depends on the voice of the speaker, the language being spoken, and even the specific content of speech. Looie496 (talk) 05:26, 26 November 2012 (UTC)[reply]
If you google power spectral density of speech, you'll get quite a range of papers describing it. As this paper [1] and others show a lot of the information is in the frequency range 1000 to 2500 Hz band, but this band contains very little of the total energy. Most of the energy lies in three distinct bands in the range 100 to 500 Hz (which the actual band centres depending on the speaker's sex and size parameters). While the lower two of these bands sets the voice quality, only the upper of these three bands contains significant information. As is weel known in the telecoms industry, you need a system frequency response essentially flat from about 70 Hz to about 5 kHz to get a natural sounding reproduction of almost all speakers, but the telephone system only transmitts 300 Hz to 3.4 kHz with negligible loss of intelligibility.
You should note that the short term (~ 100 mSec) power spectral density in te upper bands varies quite a bit depending on what is spoken, but the longer term (several seconds) averaged density is quite stable. Keit 120.145.149.167 (talk) 05:53, 26 November 2012 (UTC)[reply]

relativistic kinematics: the solar sail

Most treatments of relativistic kinematics deal with two massive bodies. How about the collision of a photon with a perfectly reflective (at rest) nanosurface of mass m? I'm supposed to calculate the frequency of the reflected photon, since the mirror will acquire momentum. I've tried solving it myself. Currently I've set the K.E. conservation equations and the momentum conservation equations equal to each other and have assumed an elastic collision, where L is the Lorentz factor:

momentum of incoming photon = E1/c = E2/c + L*m*v = momentum of reflected photon + momentum of mirror

then E1 = E2 + L*mvc

but from K.E. conservation

E1 = E2 + L*mc^2 - mc^2 = E2 + (L-1) * mc^2

then

(L-1) * mc^2 = L * mvc

m on both sides drops out. Since L is dependent on v, after distributing the v and c terms in the Lorentz factors on both sides and doing a ton of algebra, I get the cubic equation in x (where x = v^2):

x^3 + 2c^2(x^2) + (c^2 - 3c^4)(x) + c^2(x) = 0

I haven't had access to MATLAB to try to compute the roots in a sane manner yet (c is hard to run through online calculators, but substituting 3 for c says that v only has one real solution), but already the final velocity of the mirror is not dependent on E1 or the mass of the mirror. This can't be right. 207.114.92.89 (talk) 04:55, 26 November 2012 (UTC)[reply]

In your initial equation, are you sure you have the signs right ? It seems to me that the momentum of the mirror and reflected photon should have opposite signs. StuRat (talk) 05:41, 26 November 2012 (UTC)[reply]
v is negative, and the only real root I get for the cubic equation is negative, when I plug in 3 for c. However, this isn't dependent on mirror mass or initial energy. That can't be right. 207.114.92.129 (talk) 06:31, 26 November 2012 (UTC)[reply]
I think you are over-complicating the whole thing. You said the sail is at rest, or, if I am misinterpreting you, then you can put the sail at rest by a change in coordinates. The acceleration from one photon will be very small, so it will always be moving very slowly. That means you don't need to worry about relativistic effects on the sail (so L won't appear). You just need the energy and momentum of a photon in terms of frequency (hint, the formulae involve Planck's constant, h), and then you can apply your conservation laws. --Tango (talk) 09:25, 26 November 2012 (UTC)[reply]
Your problem is with the sign of the momentum. If you take the initial photon to have momentum E1/c in the positive direction, then the reflected photon should have momentum in the negative direction, specifically, -E2/c. Further, it follows that the solar sail moves in the same direction as the initial photon, hence v is positive and not negative as you assert above. If you want to do the relativistic case, then the correct equations are:
It follows therefore that:
Which shows that both the mass of the craft and the energy of the initial photon will enter into the solution. Dragons flight (talk) 11:16, 26 November 2012 (UTC)[reply]


What is the change in wavelength when light of wavelength k hits a perfectly reflective mirror in space? It can't be zero, because the mirror must move to compensate for the light being reflected, but it has an upper bound in order to conserve kinetic energy. I managed to reach this equation too, but v and E1 are both unknown. Yet I started out with two equations with two unknowns. Can I find a specific v or E1? Or is it a spectrum of energies, related to Black body radiation? -- 64.134.100.218 (talk) 16:10, 26 November 2012 (UTC)[reply]
Okay, using the constraint , I managed to solve for v:
However, it seems that both E2 and v are free to vary within a certain range, if E1 is known. Does this mean I have to use laws of Black body radiation? And why don't I notice the huge range of frequency changes and huge light pressures that this equation seems to permit? I have to use relativistic equations because my homework specifically said do not assume v << c. I understand this possibly means the photon might have to be a gazillion times more energetic than a gamma ray for this to matter, but it matters theoretically for my professor. (Plus, the light sail might be moving at relativistic speeds eventually, after enough photons.) 64.134.100.218 (talk) 16:36, 26 November 2012 (UTC)[reply]
You have to eliminate v and L which is a function of v from the pair of equations finding one equation which relates m, E1, and E2. If you don't do it right it will get messy. Hint: square both equations and subtract them finding an expression for the product E1*E2 and right that in terms of (E1 - E2) after using one of the original equations a second time. You will need an expression for L^2*v^2 in terms of L^2 -1 which can be easily obtained by inverting the defining equation for L. Dauto (talk) 16:48, 26 November 2012 (UTC)[reply]

This is why you need to learn about four-vectors and the Lorentz inner product. You can then eliminate the energy and momentum of the particle you are not interested in, in one fell swoop without having to bother about it's velocity and gamma factor.

In natural units, the four component vector (E, p), where E is the energy and p the momentum of a particle, transforms like the space time coordinates under Lorentz transforms, and is called a four-vector for that reason. The Lorentz inner product between two four-vectors defined as:

(E1,p1).(E2,p2) = E1 E2 - p1.p2

where p1.p2 is the ordinary inner product, is an invariant under Lorentz transforms. If p is the four-momentum of a particle of mass m, then we have:

p^2 = m^2 (1)

where the square means Lorentz inner product of the vector with itself. This is because it is Lorentz invariant, so you can evaluate it in the rest frame where p = (m,0).

We can then solve the problem as follows. p1 is the four-momentum of the incoming photon, p2 that of the reflected photon, q1 of the mass at rest and q2 after it has collided with the photon, then we have:

p1 + q1 = p2 + q2

We're not interested in q2, so we move this to one side:

q2 = p1 - p2 + q1


Square both sides:

q2^2 = p1^2 + p2^2 + q1^2 - 2 p1 .p2 + 2 p1.q1 - 2 p2.q1

By (1), we have q2^2 = q1^2 = m^2, and p1^2 = p2^2 = 0, so the equation becomes:

(p1 - p2). q1 = p1. p2

Taking the initial photon to move in the positive x direction, we can substitute here:

p1 - p2 = (E1-E2,E1 + E2),

q1 = (m,0)

p1 = (E1,E1)

p2 = (E2,-E2)

So, the equation becomes:

m (E1 - E2) = 2 E1 E2 ------->

E2 = m E1/(2 E1 + m)

Putting c back, this is:

E2 = m E1/(2 E1/c^2 + m)

Count Iblis (talk) 17:42, 26 November 2012 (UTC)[reply]


Can you walk me through one of the steps? After rearranging q2^2 = p1^2 + p2^2 + q1^2 - 2 p1 p2 + 2 p1 q1 - 2 p2 q1 ,
I get 2 * p2 * q1 = 2 * p1 * q1. After substituting in c and mv for the 3-momentum I get E2 = E1* (1 - mv^2/c^2). 64.134.100.218 (talk) 20:55, 26 November 2012 (UTC)[reply]

So, you have

q2^2 = p1^2 + p2^2 + q1^2 - 2 p1.p2 + 2 p1.q1 - 2 p2.q1

The reason why we brought q2 on one side and squared it, is because q2^2 = m^2, so you get rid of the unknown energy and momentum of the mirror after the collision that we are not interested in. The right hand side does not contain these quantities, only the unkown E2 that we want to solve for. Then q2^2 = q1^2 = m^2, so the m^2 from both sides cancel. And because photons are massless, we have p1^2 = p2^2 = 0. We are thus left with:

0 = - 2 p1 p2 + 2 p1 q1 - 2 p2 q1 --------->

p2.(p1 + q1) = p1.q1

Insert in here: p1 = (E1,E1), p2 = (E2,-E2), q1 = (m,0):

(E2,-E2). (E1+m,E1) = (E1,E1).(m,0)

The Lorentz inner products here are are given by (A,B).(C,D) = AC - B D, so we have:

(2 E1 + m) E2 = m E1 ---------->

E2 = m E1/(2 E1 + m)

We can write this as:

E2/E1 = 1/[1 + 2E1/(m c^2)]

where I've put c back.

Count Iblis (talk) 23:27, 26 November 2012 (UTC)[reply]

What's the momentum of the mirror after it begins moving? 128.143.1.41 (talk) 19:59, 29 November 2012 (UTC)[reply]

By conservation of momentum, this is minus the change of the momentum of the photon. The initial momentum of the photon was E1 in the x-direction, the final momentum is -E2 in the x-direction, so the change is -E2 - E1, the momentum of the mirror will thus have increased from zero to E1+E2. Putting back c, this is (E1 + E2)/c. Count Iblis (talk) 17:27, 30 November 2012 (UTC)[reply]

HIGGS BOSON

I HAVE A QUESTION IN MY MIND IT GOES LIKE THIS WHAT IS THE MASS HIGGS BOSON CONVERT ENERGY INTO ? [LIKE METAL, NON-METAL ETC] — Preceding unsigned comment added by 115.240.26.191 (talk) 06:22, 26 November 2012 (UTC)[reply]

I'm afraid your question doesn't make sense... The Higgs Boson doesn't convert energy into anything - it is involved in creating the effects of mass. --Tango (talk) 09:29, 26 November 2012 (UTC)[reply]
That's not entirely true either - it doesn't create the effects of mass, the effects already exist. Plasmic Physics (talk) 11:54, 26 November 2012 (UTC)[reply]
The Higgs (within the standard model framework) gives mass to (some) fundamental particles including the W-boson, the Z-boson, and all of the fundamental fermions - that is quarks and leptons, though it is not clear how exactly neutrinos get their masses. Dauto (talk) 15:51, 26 November 2012 (UTC)[reply]
I'm not sure I understand the question properly, but maybe you are looking for a simple unit conversion. The previously linked article on the Higgs Boson says its mass is ~126 GeV/c2, which is equivalent to ~2.25×10−25 kg. Annihilation of the particle would yield ~126 GeV of energy, which is equivalent to ~20×10−9 joules. Astronaut (talk) 20:46, 26 November 2012 (UTC)[reply]

the found or posthypnotic Higgs boson has not be as thought one which give mass to all particles .it is only boson such as others , the nature dose not obey from which we think it works really as it is.--Akbarmohammadzade (talk) 16:05, 27 November 2012 (UTC)[reply]

Sorry, I don't understand. Please write in your native language and we will translate it. --Tango (talk) 23:43, 27 November 2012 (UTC)[reply]
you never heard of the post hypnotic Higgs boson with the nature dose not obeying which we think..?GeeBIGS (talk) 00:20, 28 November 2012 (UTC)[reply]

برای بوزون هیگز قبل از آشکارسازی آن نقش های خاصی قائل بئدند و جرم پذیری ذرات رابه آن نسبت می دادند. این نقش تا حد جایگاه فراطبیعی پیش رفته بود اشکار سازی و به عبارتی حدس نزدیک به کشف بوزون هیگز که کاملا هم به یقین تبدیل نشده ،در سرن نشان می دهد که این بوزون نیز مانند سایر بوزونهاست ،منظور اینکه طبیعت هر آنچه در واقعیت ذاتی خود داراست نشان می دهد و تابع تصورات یادتئوریهای ما نیست.--Akbarmohammadzade (talk) 06:39, 28 November 2012 (UTC) با عرض پوزش مکان جغرافیایی ما با اروپا و آمریکا اختلاف ساعت دارد بنابراین اوقاتی که شما در اینجا حضور دارید و بحث می کنید احتمالا بنده در اینترنت نیستم ووقفه حاصل می گردد. با این حال بنده بسیار خرسندم که ارتباط با دوستان فرهیخته و پزوهشگر دارم.این برای من باعث افتخار است و بهره می برم از این مباحث علمی.--Akbarmohammadzade (talk) 06:44, 28 November 2012 (UTC)[reply]

V-1 flying bomb

What kinds of metal-processing/assembly machinery (other than a lot of slave labor) were used in the manufacture of the V-1 flying bomb? Also, where else were V-1s being made besides the main factory at Nordhausen-Mittelwerk? Thanks in advance! 24.23.196.85 (talk) 07:20, 26 November 2012 (UTC)[reply]

Read V-1 flying bomb facilities. Page 31 of the manual describes the Geräte und Sonderwerkzeuge (special tools and equipment) they used. Trio The Punch (talk) 13:10, 26 November 2012 (UTC)[reply]
Danke! The info on the location of the production plants and storage depots was quite valuable. The location of the storage depots, in particular, is just right for the thriller novel I'm working on. As far as the Gerate und Sonderwerkzeuge, though, these appear to describe mainly the equipment needed for the V-1s transportation and final assembly before launch -- I was hoping for info about the equipment used to actually manufacture the major assemblies prior to shipping. 24.23.196.85 (talk) 03:24, 27 November 2012 (UTC)[reply]
Germans like making lists, but I haven't found one that lists all the machinery involved in manufacturing the V1's, and I am not sure if such a list exists because they used a lot of pre-existing steel/car/airplane factories. For example, 336,600 people paid in advance for the Kdf-Wagen but when Germany invaded Poland on 1 September 1939 the automotive production was replaced by war production and none of buyers received their car. I would recommend contacting something like the Stiftung Gedenkstätten Buchenwald und Mittelbau-Dora, they know a lot about Mittelwerk. Page 37 of this PDF mentions straigthening rolls, spot welders, guillotines, press brakes and forming presses and may give some inspiration, but the Combined Intelligence Objectives Subcommittee was more interested in destroying the bunkers than documenting all the equipment used, because they had the same stuff back home. Trio The Punch (talk) 04:44, 27 November 2012 (UTC) p.s. Category:V-weapons[reply]
There must be exact copies of all procedures in some US archives. The US took everything from Mittelwerk. They tested the V-1 extensively, they even launched one from a aircraft carrier. Wernher had for sure all the plans with him. --Stone (talk) 20:31, 27 November 2012 (UTC)[reply]
They guesstimated that the Mittelwerk and Nordwerk tunnel system (which was used to produce and assemble the components for both the V1 and the V2, the Taifun, Schildrote and the engines for the Junkers) contained about 5000 machine tools. I copypasted some of the more interesting stuff. Trio The Punch (talk) 00:48, 28 November 2012 (UTC)[reply]
Think you'll find the smaller US V1 clones where developed for Landing Ship, Tank's and it was the V2 clone that was launch from the carrier.--Aspro (talk) 20:51, 27 November 2012 (UTC)[reply]
See Republic-Ford JB-2. Trio The Punch (talk) 21:13, 27 November 2012 (UTC)[reply]
This second PDF has EXACTLY the kind of info I was looking for! (I was trying to get a feel for the production process and in particular for any critical bottlenecks in it, so as to know which machines the Maquis should smash.  :-) ) Looks to me that blowing up the main transformers would have been very effective, or else destroying the overhead cranes. Danke schoen! 24.23.196.85 (talk) 05:47, 28 November 2012 (UTC)[reply]
Page 7: "In no case was any standby or local [electricity] generating plant provided". According to page 18 there was only one main high tension intake station with a number of transformer sub-stations. If you destroy it it would be difficult to fix, and the electric hoists mentioned on page 20 would be useless. The lights would turn off too, which may aid or hinder your escape. Trio The Punch (talk) 06:35, 28 November 2012 (UTC)[reply]
Clarification: the Maquis attack takes place not at Nordhausen-Mittelwerk, because it's too deep within Germany proper; rather, it takes place at another (fictional) combined V-1 factory and V-weapons storage depot in the Lille-Calais area (in real life, that area had several storage depots but no V-weapons factories). In fact, this is the reason why I asked specifically about the V-1 and not the V-2 -- the latter were all made at Mittelwerk. 24.23.196.85 (talk) 05:09, 29 November 2012 (UTC)[reply]

Insect

Unable to place this insect

What is this insect? I have based the name on a guess. How can I identify the species? The photo was taken at Idukki Didtrict, Kerala, India. — Preceding unsigned comment added by Drajay1976 (talkcontribs) 08:33, 26 November 2012 (UTC)[reply]

I found a website that helps you identify 39 different species of grasshoppers found in the region of the Western Ghats and a local expert. Trio The Punch (talk) 13:15, 26 November 2012 (UTC)[reply]
Do you have any other pictures of it from another angle? It definitely looks like a tettigoniid though from the antennae length, or at least ensiferan. Male as well from the lack of an ovipositor. Also, are the wings damaged? They look chewed off.-- OBSIDIANSOUL 21:42, 29 November 2012 (UTC)[reply]
Thanks. I sadly couldn't get pictures from other angles. I could not get closer to it because of the terrain. Couldnt get a better angle as well. The insect was wet from the rain and was just sitting there. I dont think that the wing was chewed off. --Drajay1976 (talk) 09:41, 1 December 2012 (UTC)[reply]

Reason for type of emotional response to music and poetry

It came out on this thread on the Miscellaneous reference desk that "The thing with Limericks is their metrical structure alone is funny, regardless of meaning", though that was also disputed. But what is it about that metrical structure that makes a limerick sound funny to most people? And why do certain tunes sound funny, or light-hearted, or sad, or pensive, or foreboding to most people? I'm not looking for a description of what strikes people in this way or that way; I'm wondering if it is known why the brain reacts the way it does. Is it that that sound pattern somehow gets processed at a spot in the brain that is physically near the spot that processes that emotion? Duoduoduo (talk) 15:54, 26 November 2012 (UTC)[reply]

I'd say (without references) that it's just a matter of expectation and normalization (psychologists have a better word for that). Say you've heard 100 limericks, and 50 of these (the figures are made up) were funny, 1 was an odd one, and 49 were just bad. Then you'd have the expectation during the next limerick that "it's gonna be funny (or maybe it's just gonna suck)" and that makes you think limericks are funny.
And if you read it, the voice in your head will probably be a funny one. Which i why it works so well with limericks - the rhythm and rhyme are easy to remember.
Once limericks were expected to be funny, more funny limericks were made, since after all, the audience was already expecting something funny.
I think that's all, no nerve tickling its neighbors or stuff like that... - ¡Ouch! (hurt me / more pain) 16:24, 26 November 2012 (UTC)[reply]
p.s. about the voices in my head. I'm not mad. I asked the voices, and four of them said I'm not. - ¡Ouch! (hurt me / more pain) 09:10, 28 November 2012 (UTC) [reply]
See "Music and emotion" and "Psychoacoustics".
Wavelength (talk) 17:23, 26 November 2012 (UTC)[reply]
I can't find a physiological answer in those cites. Also, as for Ouch's comment that "if you read it, the voice in your head will probably be a funny one", this extends the question to include this: for what physiological reason do certain "voices in your head", or more generally certain sentence intonations, come across as funny? Or is that too known to be purely learned? Duoduoduo (talk) 18:45, 26 November 2012 (UTC)[reply]
One reason for a song's personal emotional appeal can have to do with some past association. For example, if you hear an "oldie" on the radio, it might remind you of where you were and what you were doing, when you first heard it. ←Baseball Bugs What's up, Doc? carrots12:05, 27 November 2012 (UTC)[reply]
But I think I see what the OP is getting at. Why does a minor scale sound "sad" compared to a major scale? I've wondered about that myself; a couple of sound waves of different frequences; change the frequency of one of them a little bit, and for some reason you whole emotional reaction changes. What's up with that? Where in evolution does that confer a survival advantage? Amazing. Gzuckier (talk) 05:10, 30 November 2012 (UTC)[reply]

Common emotional response to specific colors

Similar question to the above. Red is commonly interpreted as an aggressive color, a certain shade of pink as a soothing color (so for example I've read that some prisons have pink walls to lower inmate aggression), pastel colors as gentle, hot pink as attention-getting, highly saturated colors as attention-getting, etc. Do we know why the human brain interprets certain colors in certain ways? Is it that, for instance, red is processed in a location in the brain that is near where aggression is processed?

Related question: What is the physical reason that it is widely perceived that certain colors "clash" -- that is, viewing them together leads to an unpleasant feeling? Duoduoduo (talk) 16:04, 26 November 2012 (UTC)[reply]

I think the difference between pink and red is just an example that stronger stimuli provoke a stronger response in our brains. Some colors and emotions are linked through (sub)culture, for example certain medieval European queens used white to mourn instead of black. In certain parts of Africa red is a color of mourning, representing death. In both Thailand and Brazil they prefer purple. Many people who see bullfighting on TV believe that the color red causes the bull to be aggressive. In reality they torture the bull beforehand (offcamera). Cattle is red-green color-blind. There have been some attempts to make lists, but they are incomplete and unreliable. The brain area that deals with face recognition sits right next to the color area, which is problematic for achromats. More interesting stuff. Trio The Punch (talk) 16:39, 26 November 2012 (UTC) p.s. Color_theory#Color_harmony_and_color_meaning[reply]
Color in Chinese culture tells us that red is a lucky colour in China. HiLo48 (talk) 18:33, 26 November 2012 (UTC)[reply]
Right, maybe I overstated the premise. So let me back up: (1) Is there some commonality across cultures, such as more saturated colors being viewed as "stronger" and eliciting a stronger response? And (2) if so, is there any known physiological reason for this? E.g., why are unsaturated colors not viewed as "strong"? (I don't see that item (2) is addressed in any of the references given.) Duoduoduo (talk) 18:40, 26 November 2012 (UTC)[reply]
The article Color psychology makes some interesting claims, but doesn't really answer most of your questions. There might be a more germane article somewhere? Dbfirs 19:32, 26 November 2012 (UTC)[reply]
Isnt' that like asking why are dense clouds of odor molecules considered "strong" and diffuse clouds of odor considered "weak"?165.212.189.187 (talk) 21:12, 26 November 2012 (UTC)[reply]
Nope. Saturated colors don't have more photons; they simply have a lower variance of wavelengths. Duoduoduo (talk) 22:00, 26 November 2012 (UTC)[reply]
Point? The purest (most saturated) color is achieved by using just one wavelength at a high intensity (dense cloud), such as in laser light. To desaturate a color of given intensity in a subtractive system (such as watercolor), one can add white (diffuse cloud), black, gray, or the hue's complement. Density across the spectrum. Maybe I should not have used the term molecules but just gas.165.212.189.187 (talk) 13:54, 27 November 2012 (UTC)[reply]
Aposematism extends this topic to other species --Digrpat (talk) 22:06, 26 November 2012 (UTC)[reply]

100% electric spacecraft engines

I've looked around a bit, and do want to know if there is any real, practical, lightweight spacecraft engine that uses no fuel but electricity? I've looked at ion engines and others, but they just use heated materials as propellant. 3 bit (talk) 17:51, 26 November 2012 (UTC)[reply]

What you're really after here are engines without reaction mass. Examples include solar sails (have been demonstrated in space) or theoretical things such as magsails and Bussard scoops. Note however, that in all of these there is still reaction mass in a way, just not carried on board. Solar sails and magsails use the momentum carried by photons and charced particles from the sun. The Bussard scoop uses light atoms encountered along the way. Full reactionless drives are thermodynamically impossible, as there still needs to be a transfer of momentum onto the spacecraft in some form. Fgf10 (talk) 18:22, 26 November 2012 (UTC)[reply]
(edit conflict)Do your mean batteries? (Otherwise there must be a source of fuel to generate the electricity.) The problem with using an electric "engine" is that there is nothing to "push against" so an electric motor cannot be used to propel the spacecraft. Electricity can, of course, be used to accelerate particles to eject in an ion engine, but I cannot think of any other usage for electricity in a spacecraft, other than the obvious application of powering computers and equipment. Perhaps someone else can think up a novel way to drive a spacecraft by onboard electricity (assuming that there is a way to generate it)? Dbfirs 18:26, 26 November 2012 (UTC)[reply]
The Bussard ramjet mentioned by User:Fgf10 above would be one such application. Dbfirs 18:32, 26 November 2012 (UTC)[reply]
Hmm, generating electricity without fuel or batteries? If only there was a nearly limitless supply of energy we could use.... ;) Fgf10 (talk) 19:06, 26 November 2012 (UTC)[reply]
Well, obviously, except in deep-space, but I class the sun as a source of fuel. The OP asked about "no fuel but electricity". Dbfirs 19:24, 26 November 2012 (UTC)[reply]
Newton neatly sums up why it's not possible to have an engine with no reaction mass: "A body will preserve its velocity and direction so long as no force in its motion's direction acts on it." In a (very nearly) frictionless vacuum (like outer space) you need to have some sort of mass to carry that force. A light sail and laser engines use photons to carry that mass and bounce it off of an ablation surface or a sail. A conventional rocket engine uses molecules of hot gas to carry the momentum. An ion engine (surprise!) uses ions. It could be anything thanks to the fact for every action there is an equal and opposite reaction. Hence the name "reaction engine" and "reaction mass". You could be throwing bowling balls out the back or letting off compressed gas (as in a positioning thruster) but you have to have SOME kind of reaction mass in order to carry away the momentum in a controlled direction. HominidMachinae (talk) 22:06, 26 November 2012 (UTC)[reply]
EmDrive--Stone (talk) 20:20, 27 November 2012 (UTC)[reply]
That reminds me of Eric Laithwaite and his magic gyroscope. Dbfirs 09:22, 28 November 2012 (UTC)[reply]

Length contraction

If I have a long, thin, dense rod with radius less than its Schwarzschild radius and make it go at a relativistic speed in the direction of its axis, will it collapse into a black hole if it's going fast enough? --168.7.228.96 (talk) 21:09, 26 November 2012 (UTC)[reply]

Fast enough relative to what? AndyTheGrump (talk) 21:11, 26 November 2012 (UTC)[reply]
That doesn't matter, the answer is no. In its own reference frame the object is motionless and its length is unchanged. Motion can generate an apparent reduction in length from the viewpoint of an outside observer, but it doesn't generate a gravitational field. Looie496 (talk) 21:18, 26 November 2012 (UTC)[reply]
I'm not certain the answer is that simple. Doesn't velocity imply energy, and doesn't energy inherently effect a gravitational field? And doesn't the generalization of gravity tell us that neither reference frame is canonical? Both the "stationary" observer and the "moving rod" view a relativistic motion, as measured in the respective reference frame. The question presents a legitimate conundrum. It seems the resolution is, of course, that an event horizon must be viewed from some reference frame. Not all reference frames see the same horizon; nor do they see a singularity in the same location. By symmetry, whichever reference frame the observer happens to be in, they will see (or rather, not see) the other reference frame as on the opposite side of an event horizon. To define that apparent horizon in any kind of useful way, we need to solve a very ugly and difficult set of "elegant equations." With luck, one of our regulars who are better-versed in relativistic geometry will be along shortly to provide a more thorough explanation, and perhaps even a good book. (Unfortunately, I know of no text or web-page that makes this topic any more accessible). Nimur (talk) 22:35, 26 November 2012 (UTC)[reply]
It's worth noting that the equations that point out which bodies will become black holes (basically the various forms of the Schwarzschild metric) are solutions to Einstein's field equations in rather trivial situtations, ones which require no linear momentum. Once you change the system to one in which your potential black hole is moving at extreme velocity, you will need new equations to describe its gravitational field; you can't simply plug in the relativistic mass to the Schwarzschild metric. Someguy1221 (talk) 03:42, 27 November 2012 (UTC)[reply]
No. Whether you consider something to be moving or not is merely a change of coordinates, namely, a Lorentz boost. The geometry of spacetime isn't affected by what coordinate system(s) you choose to describe it in. I.e., what events, four-vectors, tensors and geodesics exist, and what the geometric relationships are among those geometric objects, is independent of any coordinate system. In particular, what absolute horizons exist, and what set of events they consist of, is invariant with respect to any change in coordinates. If a rod doesn't produce an absolute horizon as expressed in one coordinate system (atlas, actually), then it doesn't produce an absolute horizon as expressed in any other coordinate system. Red Act (talk) 08:04, 27 November 2012 (UTC)[reply]
Looie496 and Red Act are correct. It is a common misconception to think that space contraction and relativistic mass makes it possible for a fast moving object to turn into a black hole. That's not correct. Dauto (talk) 17:03, 27 November 2012 (UTC)[reply]
... otherwise the Earth would turn into a black hole every time an observer passed by at the speed of light! Dbfirs 22:21, 28 November 2012 (UTC)[reply]
The Earth just happens to inexplicably turn into a wicked time-warped pancake every time an observer passes by near the speed of light... oh wait. ;) -Modocc (talk) 04:13, 30 November 2012 (UTC) [reply]

average intensity of three vibrations on a string of constant frequency and individual amplitude, but random phase of pi or 0

This is a homework problem I'm trying to make sense of. The amplitude A and frequency f of each of the vibrations is constant, but phase varies "randomly and independently" in values of pi or 0. It's the "randomly and independently" part that confuses me. I think it's related to quantum wavefunctions but I am not sure. Anyway, I deduced there are 6 probability states for the string, 2 with a peak amplitude of 3A where all of the vibrations are in phase with each other, and 4 where one of the vibrations is out of phase with the other two.

I used the formula for power of a wave on a string I got from physicsforums. Then P_avg = A^2 * 2pi^2 * f * v * mu, where mu is linear mass density, v is wave velocity and A is peak amplitude. This is for an individual wave though.

If the vibrations' phases vary "independently and randomly" does that mean that mean that I_avg = (2 * 3^2 + 4*1^2)/6 *(A^2)*(2pi^2)*f*v*mu, that is, 6(A^2)*(2pi^2)*f*v*mu? 64.134.100.218 (talk) 21:38, 26 November 2012 (UTC)[reply]

The reasoning is basically correct, I think, but you are miscounting the number of states. If you have three random variables, each with two possible values, the number of global states is 2*2*2 = 8. Looie496 (talk) 21:57, 26 November 2012 (UTC)[reply]
Also, the words "individual amplitude" makes me think that the waves amplitudes are not all equal to each other. Dauto (talk) 17:00, 27 November 2012 (UTC)[reply]
The problem specifically says they are equal. 128.143.1.41 (talk) 21:39, 29 November 2012 (UTC)[reply]


November 27

forest extent in very ancient India

Hi, I`m interested in finding out the extent of forest cover in India ideally around 7,000 BCE (but willing to follow leads for a few thousand years either side of that) A map would be great but I can`t remember the name for this kind of map so am having trouble googling it. All pointers appreciated. Thanks. 184.147.123.169 (talk) 01:33, 27 November 2012 (UTC)[reply]

I doubt that the data exist to make comprehensive forest zone maps, but lake levels show that the monsoon was much stronger at that time than it is now, so you can expect that the forest zone covered most of the subcontinent. Even the Thar desert region had permanent lakes then. Searching Google Scholar for "holocene climate India" will get you into the literature, for what it's worth. Looie496 (talk) 04:12, 27 November 2012 (UTC)[reply]
Thanks very much, those search terms do help. 184.147.123.169 (talk) 01:27, 28 November 2012 (UTC)[reply]
You might also go about it from the other direction, look at these results for deforestation grazing india. μηδείς (talk) 17:41, 28 November 2012 (UTC)[reply]

Problems about determine habitable zone in 7.5 billion year future sun

I understand determining habitable zone in 7.5 billion year tip of RGB if difficult but I still have trouble understand why is it so difficult. The reasons is we don't know exactly how big the sun will be, the second reason is we don't know how bright the sun be. I thought calculate the luminosity is quite easy to determine It tell us to use the square root/square power rule. So 1000-3000 luminosity if you square root, the HZ is 45-60 AU. It is that easy. I do it is 3000 luminosity/45 AU square, the intensity is 1.xx just to play a wildcard. I still don't get the table What does Saturn (?) Uranus(?) or Jupiter (???) means. Is there anything else we need to know besides the luminosity and how big the sun will be in order to know the HZ. Do we know exactly how hot the sun will be and how much heat will the sun produce because I thought luminosity is the key idea to determine how much heat will star produce. When sun becomes a giant do me know exactly how hot the sun going to be, or what other issues about knowing it? --69.226.43.162 (talk) 02:09, 27 November 2012 (UTC)[reply]

We just had this exact question asked from Los Angeles in September: http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Science/2012_September_18#how_to_figure_out_the_habitable_zones. Was that answer no good? Please let us know if the prior response was not helpful. μηδείς (talk) 05:50, 27 November 2012 (UTC)[reply]
Is just when a respondant said The main reason for uncertainty in the Solar System's habitable zone 7.6 Ga (billion years) from now is that we're not sure how much the Sun will expand when it reaches RGB, and exactly how bright it will be. he pointed out Jupiter and Saturn's could still be in HZ at that time. I tried to calculated it on my calculator I assume sun's expansion in 7.6 Ga is at least where Venus orbit is now, to at most where Mars orbit is now. The reason I think that is I never heard in my lifetime that Mercury will survive at tip of RGB, I didn't see the expansion of the sun's uncertainty make that much difference. I definitely know just because planets fit in HZ it doesn't mean the planet itself will be habitable that is quite obvious, I mainly describe the HZ by if the planets fit in habitable zone, not if the planets is actually habitable (it is not really that important).--69.226.43.162 (talk) 00:45, 28 November 2012 (UTC)[reply]

Neutron binding

Can two or more neutrons bind together via the strong nuclear force and become stable? From what I understand, a neutron is stable in an atomic nucleus because of the strong nuclear force generated by the protons and neutrons, but neutrons alone should also generate this force as well. So shouldn't they also be able to bind with each other and become stable? ScienceApe (talk) 03:48, 27 November 2012 (UTC)[reply]

That would be Neutronium. Our article describes observed and hypothetical properties of the various isotopes. In brief, two neutrons can form a strong attraction to one another, but cannot truly form a bound state. Someguy1221 (talk) 03:56, 27 November 2012 (UTC)[reply]
I thought neutronium is unstable outside of a neutron star's immense gravity field? ScienceApe (talk) 07:06, 27 November 2012 (UTC)[reply]
You're thinking of neutron degenerate matter. Neutronium can also mean a hypothetical element on the periodic table that would have atomic number zero, i.e. would be composed solely of neutrons. Double sharp (talk) 07:20, 27 November 2012 (UTC)[reply]
(edit conflict)Precisely, which is why no one has ever observed it directly in a laboratory. Although, you still have to decide what "neutronium" means – does it refer only to neutron-degenerate matter, or does it include "atoms" with an atomic number of zero? Regardless of what you call it, a system of just two neutrons will not form stable bonds through any means. And based on currently accepted models, the same goes for three, four, five, etc. What have been observed in the lab are very transient interactions between two otherwise free neutrons. Someguy1221 (talk) 07:20, 27 November 2012 (UTC)[reply]
For four neutrons (not just two), see Tetraneutron for more info. Clusters of three and five neutrons should not be bound, and should not exist for even a short time. Double sharp (talk) 05:05, 27 November 2012 (UTC)[reply]
Your two neutrons together is called a dineutron. Graeme Bartlett (talk) 11:36, 27 November 2012 (UTC)[reply]
But it appears it's not stable even though they are within the range of the strong force. Why is that? ScienceApe (talk) 17:25, 27 November 2012 (UTC)[reply]
Read Deuterium#Spin and energy for an explanation. Dauto (talk) 18:33, 27 November 2012 (UTC)[reply]

Diamond (thought experiments)

Theoretically predicted phase diagram of carbon

To keep it cheap and simple: If I have a pure diamond of the size of a soccer ball (DSB) and

  1. "Close to absolute zero" (i.e. space temperature) I increase the pressure on DSB very, very, very slowly (to avoid heating) and uniformly until I reach pressures which are found in the center of suns. What will happen to the diamond structure?
  2. I repeat the above experiment (with a second DSB) without taking care of heat dissipation (lets say, we reach 5000-10000 °C). What will happen to the diamond structure? Thanks for answers. GEEZERnil nisi bene 08:27, 27 November 2012 (UTC)[reply]
at 1 TP (10000000 atmospheres) it is predicted to form bcc-carbon and at high pressures metallic carbon could also exist. Graeme Bartlett (talk) 11:45, 27 November 2012 (UTC)[reply]
Slow compression: If the diamond does not change into bcc-carbon or metallic carbon, it will only be compressed. However, this does assume that the compression is exactly uniform (which it may as well be if it is very slow). If there are irregularities, they might shatter the crystalline mesh, as even the structural strength of diamond is no match to that kind of pressure. If the diamond changes its structure, say to bcc carbon, it may or may not form one new crystal, depending on the time it takes to form the bcc crystal.
The article on allotropes of carbon states that "There is an evidence that white dwarf stars have a core of crystallized carbon and oxygen nuclei." However, "it is more likely an exotic form of carbon." I take it as possibly degenerate.
Fast compression: There will probably be not much of a difference between different amorphous phases at that kind of pressure. Whether it is a liquid, a gas, or a plasma will not matter much. The carbon will be in a degenerated state of extremely high (relativistic) thermal motion and mean free path close to zero. However, if the compression is extremely fast, from zero to full pressure without any time in-between, chances are that you'd set off a carbon catastrophe. You should find a "volunteer" to do that for you, and be somewhere else when the experiment starts. Not "in a different room" or "in a different building." In a different city.
- ¡Ouch! (hurt me / more pain) 13:14, 27 November 2012 (UTC)[reply]
What exactly is a 'carbon catastrophe'? AndyTheGrump (talk) 15:06, 27 November 2012 (UTC)[reply]
Going from context (extremely fast compression of carbon), I'm assuming the meaning is setting off fusion in the carbon, and the 'carbon catastrophe' refers to the sudden start of carbon fusion in a white dwarf: carbon detonation. But I could be wrong of course. Fgf10 (talk) 18:18, 27 November 2012 (UTC)[reply]
Thanks Fgf10, that's the one. I think I had an iron catastrophe in the back of my head. - ¡Ouch! (hurt me / more pain) 09:12, 28 November 2012 (UTC)[reply]
Ah, I suspected as much, after thinking about it a bit more. Given that even a soccer-ball sized supernova might be a little, um, 'unfriendly', I can see why Ouch advises getting someone else to try it... AndyTheGrump (talk) 18:50, 27 November 2012 (UTC)[reply]
Being bored, I decided to calculate just how 'unfriendly' this would be! A regulation football has a circumference of 69 cm, translating to a volume of 5.5 l. A diamond this size, given a density of 3.5 g/cm2, would weigh 19.4 kg. At a formula mass of 12.01, this means 1616.7 moles/football, or 9.7x1025 atoms/football. Assuming the ideal case of all of these atoms fusing with their next-door neighbour, this leads to half that number of fusions. Carbon-carbon fusion has a number of options, but a quick look suggests that at least in stars, the most prominent reaction is the fusion to Mg, liberating 13.93 MeV each. All of these fusions in a diamond football, again assuming ideal case, will then liberate 6.8x1026 MeV. So the bottom line? In a ideal case, your completely fusing diamond football will give you 25.9 kt of TNT equivalent. Call it a Fat Man with change (and a rapidly expanding Mg plasma and a load of gamma radiation), so yes quite unfriendly.... But a lot less than I was actually expecting! Fgf10 (talk) 21:47, 27 November 2012 (UTC)[reply]
Interesting calculation. So Ouch's 'different city' to be safe is probably about right for his football-sized diamond experiment. Since the yield is presumably proportional to the mass, the average engagement ring will only be capable of more local damage if detonated this way. I could probably do the maths myself, but maybe in the interests of romance, I'd best not... AndyTheGrump (talk) 05:19, 28 November 2012 (UTC)[reply]
Geezer: This is going O/T (for you asked about fast compression but not about the fast limit case) but meh. For some "real" yield, one could always add some crystalline deuterium. Pocket Tsar.
Andy: And you thought a divorce could end up messy... ;)
Since the next topic goes towards silicon compounds, we could investigate the yield of breast implants.
"Oooo, what a nice pair of WMDs..." - ¡Ouch! (hurt me / more pain) 09:12, 28 November 2012 (UTC)[reply]
Hehe, yes it's OT, but it's fun! An engagement ring? Say 2 carats? About 0.5 kt, still enough to ruin your day!
And Ouch? Silicones? That's going to be one hell of a calculation; silicon, carbon, oxygen and hydrogen? Way over my head! I'll stick to my dayjob, regenerative neuroscience is so much easier.... Fgf10 (talk) 09:50, 28 November 2012 (UTC)[reply]
No, no, not O/T ... I enjoyed reading the comments. They added color to thoughts about a tiny object in an otherwise cold universe... May even serve as excuse one day, if a Female requires attention in form of a diamond ring...  ;-) GEEZERnil nisi bene 15:23, 28 November 2012 (UTC)[reply]

Hassocene

In 2008 there were plans at the GSI to synthesize the compound hassocene (Hs(C5H5)2), the heavier homologue of ruthenocene and osmocene (see this presentation by Cristoph E. Düllmann). Does anyone have access to the results of the experiment? Was the experiment even conducted in the end? I haven't found any info on this. Double sharp (talk) 13:42, 27 November 2012 (UTC)[reply]

Nothing in SciFinder by that name or chemical formula. DMacks (talk) 19:24, 27 November 2012 (UTC)[reply]
I am fairly certain the Hassocene is the geological period from 1982 until 2000. μηδείς (talk) 03:34, 28 November 2012 (UTC)[reply]
The older archaelogical strata have a pretty solid Foundation layer, but towards the end there are some slowly-undulating areas of a weird silicone-like material. DMacks (talk) 04:04, 28 November 2012 (UTC)[reply]
<chortles> We're currently in the middle of the Hassocene-Beiberanian extinction event -- OBSIDIANSOUL 21:38, 29 November 2012 (UTC)[reply]
LOL. Is that related to the Bieberian Law of thermodynamics, that "in a closet system, gaiety can only increase with time"? - ¡Ouch! (hurt me / more pain) 07:12, 30 November 2012 (UTC)[reply]

Best ... reference desk jokes ... ever. μηδείς (talk) 21:57, 1 December 2012 (UTC)[reply]

walking vs running

which burns more calories? — Preceding unsigned comment added by PlanetEditor (talkcontribs) 14:46, 27 November 2012 (UTC)[reply]

Running. See [2] and [3]. Dismas|(talk) 15:09, 27 November 2012 (UTC)[reply]
if they are done for the same duration. eg, walk for a day, run for a minute, then walking burns more.165.212.189.187 (talk) 18:31, 27 November 2012 (UTC)[reply]
Uh, that's not the same duration. Same duration would mean walk for a minute, run for a minute. Looie496 (talk) 19:17, 27 November 2012 (UTC)[reply]
That's a typo, I suppose. Anyway, does running/distance or running/time burn more calories than walking/distance or walking/time respectively? OsmanRF34 (talk) 20:18, 27 November 2012 (UTC)[reply]
Walking is simply a controlled fall, running requires lifting the body off the ground. The first is more efficient, and so burns fewer calories per distance. If you want to burn calories, rowing is the best low-impact all body exercise and the stair-master is the best low-impact calorie burning exercise. One gym I went to had to put a 20-minute limit on the stair master since some of the ladies were so skinny they were getting blown of their feet by the stage fans they used to cool the place. This does not constitute medical advice. μηδείς (talk) 03:29, 28 November 2012 (UTC)[reply]
Running for an hour definitely uses more energy than walking for an hour. Now running for a mile versus walking for a mile is a bit trickier, and might depend on the individual. The basal metabolic rate, which varies by the individual, must be considered, since the walker will take longer, and thus burn more calories in this way. So, someone with a high BMR might do better to go faster, to minimize calories burned in this way, while someone with a low BMR might want to go slower, to take advantage of the more efficient movement method. I suspect that, for the average person, a slow jog might be the most efficient method per mile. StuRat (talk) 08:31, 28 November 2012 (UTC)[reply]
That's a bit of special pleading StuRat. The person with the high BMR is not going to turn his off once he stops running--the BMR calorie consumption will continue over the same period in both exercisers. It's the difference in calories burned between running and walking the same distance that that matters. The BMR calories are going to be burnt no matter what at a rate of 24 hours per 24 hours. μηδείς (talk) 17:34, 28 November 2012 (UTC)[reply]
Power walking may expend more calories. ~AH1 (discuss!) 20:50, 29 November 2012 (UTC)[reply]

Mattress pad science

This might be the most ridiculous question so far posted but it's bugging me...

The wife and I have a thick (a couple inches) down pad that sits between the mattress and the fitted sheet on our bed. It is constantly sliding down the length of the bed so that after about a week, the pad is protruding off the foot of the bed by at least 5-6". And therefore, the pillows are also in a bit of a trench since there isn't any pad to push them up slightly. What causes the pad to slide down like this and so quickly? All we have to do is sleep on it and it moves down over the course of the week. Thanks, Dismas|(talk) 15:01, 27 November 2012 (UTC)[reply]

If we first assume that your nocturnal motions induce small movements in the pad, we would expect the probability of finding the pad at a given position would follow a normal distribution. But you're reporting a systematic bias in one direct (footwards). I believe this is caused (in part) by the random walk the pad takes being constrained at one end but not the other. Random motions you induce in a headward direction are sometimes contained by the pillows and headboard, but not in the footward direct. So this induces a bias, and the pad's distribution is pushed footward. I confess I've always thought this process was called a drunk walk, because of the way an inebriate winds his way along the road - the walls of buildings constrain him, so rather than his position at any time being gaussian, random offsets tend to bounce him along one wall or the other. But we don't seem to have a drunk walk article, so I don't know what it's really called. -- Finlay McWalterTalk 16:02, 27 November 2012 (UTC)[reply]
Drunkard's Walk, which redirects to random walk. I've never heard it called "drunk walk" myself, but another redirect won't do any harm. Tevildo (talk) 23:24, 27 November 2012 (UTC)[reply]
Try this experiment: turn the mattress so the foot end is now the head end and see if it still happens, and then with just the pad. This will rule out (or not)surface features of the two materials like a lint brush goes easy in one direction but rough in the other.165.212.189.187 (talk) 18:35, 27 November 2012 (UTC)[reply]
I can't tell you why, but to stop it get some sticky velcro and stick it to the top end of the bed and pad, and the pad should stay put. Now if I could only find a way to stop my duvet from working its way down the duvet cover in much the same way as you describe! --TammyMoet (talk) 19:04, 27 November 2012 (UTC)[reply]
The shifting bedding can be simply due to the bed being slightly lower at one end, such that gravity is pulling you and your bedding downward whenever the lateral forces on the pad from your movements overcome the static friction between the pad and bed. It doesn't take a steep incline for this sort of thing to happen, and often times floors that appear level are actually not. I would try raising the apparent low end with some shims and see if that helps put a stop to the creeping pad. -Modocc (talk) 21:43, 27 November 2012 (UTC)[reply]
  • This is very common, I have the problem all the time. The bedclothes ride down when you sit on the edge or side of the bed. You don't sit at the headboard. At night you can push away from the headboard, but you can't reall push effectively toward it. Hence the drunken walk away from the headboard. (Oh, and BTW, this doesn't even make it to the lowest third of ref desk question, let alone close to the stupidest.) μηδείς (talk) 03:20, 28 November 2012 (UTC)[reply]

Can separate universes be causally connected?

This is perhaps more of an issue of definitions, but I was arguing with my roommate last night about the idea of a multiverse. My position was that you could have two x-dimensional universes contained in a y-dimensional multiverse, and that these universes could theoretically exchange gravitons and be detected in this way (the manner of detection and interaction here isn't important), and that it makes sense to call each of these objects a "universe". But he was saying that if we can detect them at all then it's not another universe, it's just another part of our universe. In this case, what I'm calling a multiverse, he's just calling the universe. To him, a multiverse necessarily connotes entirely separate universes. Who has the more accurate definition, from a physicist's standpoint? Goodbye Galaxy (talk) 15:14, 27 November 2012 (UTC)[reply]

I am a physicist (at least, I carry the degree of one, which I obtained by battling other physicists in close quarters in the halls of a fine, accredited university). So, in principle, anything I say is "from a physicist's standpoint," even when I spout nonsense. It is my opinion that "multiverse" is a bunch of pop-science drivel. The term is ill-defined, and is primarily used to permit non-physical descriptions of hypothetical scenarios. For the sake of citing a reference, (as I've been drawn to the habit of spouting my opinions quite frequently this week), I'll recommend Richard Feynman's Six Easy Pieces, which I re-read recreationally during the Thanksgiving holiday. This book excerpts several of the best lectures from his course at Caltech. Feynman will fall back on this definition: physics is guided by experiment. Experiment trumps everything else. If you hypothesize a multiverse, or a Wakalixes, or whatever other ridiculous word you choose, you will have to describe what consequences you hypothesize. And then you will have to test for that consequence. In the absence of predictive hypothesis, corrected by experimental result, science degrades into Aristotelian drivel. So, ask yourself the more important question: if, as you hypothesize, there are some set of distinct regions of interacting matter, how would we test for that? And then, construct the test, and draw a conclusion, and let the language-experts fight over the proper name for your result. Nimur (talk) 16:01, 27 November 2012 (UTC)[reply]
That's a pretty non-answer, you have to admit. It is recognized as completely valid within the scientific community to spend some time thinking about abstract, not-yet-measurable things, with the idea that such things are what give you the ideas as to what kinds of experiments might be interesting to conduct. There are set technical definitions of universes within the multiverse theories, which is what the OP is asking about. He or she is not proposing a new theory of multiverses and doesn't probably need to be told that experimental results are important for determining which of many competing theories is actually true. Nobody disputes that experiments are necessary for determining validity between theories, but experiments are not the only source of inspiration for theories, and indeed many theories are in fact the source of inspiration for experiments. --Mr.98 (talk) 16:09, 27 November 2012 (UTC)[reply]
Most "multiverse theories" do not make scientific predictions. There is a difference between thinking abstractly, and thinking without any contraint whatsoever; this is what distinguishes physicists from religious teleologists. Nimur (talk) 16:12, 27 November 2012 (UTC)[reply]
Apparently it doesn't distinguish them, under your model, since many physicists do find it worthwhile to think about such things. What distinguishes them from religious teleologists is, I would argue, something else. There is a grand tradition in theoretical physics of taking big leaps that don't as of yet correspond with any experimental outcomes. Some of them work, some of them don't. There's no need to take your professional angst out on an OP. --Mr.98 (talk) 16:16, 27 November 2012 (UTC)[reply]
My very non-physicists' understanding is that it really depends on which flavor of multiverse theory you're talking about. In some of them, I believe, the non-interactiveness is key to the definition of what a universe is. In some I'm not sure it matters. My conclusion would be, generally, that if you're not being extremely precise about which theoretical framework you're talking about, it doesn't make sense to try and be precise about the linguistic terms. The linguistic differences only matter if you are actually reducing them to technical definitions. --Mr.98 (talk) 16:15, 27 November 2012 (UTC)[reply]


If you start from the perspective of real observers like we are, who have access to only a finite amount of information (due to finite brain capacity), you get a totally different picture. There are then identical copies of us located in different universes. Because the copies are exactly identical, you can't say that you exist in either one or another universe, you are in all of them. The identical copies are the same w.r.t. to their memomories, their bodies can be in different states. So, if you are not bald, then the amount of hair you have is not exactly determined unless you have measured this.

Different people live in different sectors of the multiverse. E.g. given what I know, Nimur could have a big beard or not. In my sector, those two different Nimurs exist. But from Nimur's perspective, these two sectors are different, he is a different person in the two sectors. Count Iblis (talk) 18:18, 27 November 2012 (UTC)[reply]

@ Nimur, your answer was perfect, if maybe a bit too charitable. (Not only do they not make testable predictions, there's no evidence for them, and there's not even a coherent description of one.) But otherwise perfect. μηδείς (talk) 22:11, 27 November 2012 (UTC)[reply]

There is evidence. E.g., in quantum mechanics, time evolution is unitary, no deviations have ever been detected, so there is no experimental evidence for models that postulate a non-unitary collapse of the wavefunction. Count Iblis (talk) 22:30, 27 November 2012 (UTC)[reply]
Doesn't that mean there is no evidence? -- Jack of Oz [Talk] 23:14, 27 November 2012 (UTC)[reply]
You need a real collapse of the wavefunction to get rid of the quantum multiverse, so far there is no evidence for this. Quantum mechanics has so far survived every experiment designed to falsify it. Take e.g. the double slit experiment, you can verify that electrons don't go through either one or the other slit, both possiblities are equally real (we refer to this as a "superposition" of the two possible classical possibilities, it is often said that the electron goes to both slits as once, but it's misleading to put it that way).
The question is then why we don't see such strange things happening in the macroworld. You could postulate that there is a real wavefunction collapse that causes superpositions of different classical states collpase into one or the other classical state. But you don't need to make such assumptions, so-called decoherence leads to the emergence of a classical world at the macro level. Also, there is no evidence that a real wavefunction collapse happens (this is something that could be unambigously detected in experiments). Count Iblis (talk) 00:09, 28 November 2012 (UTC)[reply]
That's not really evidence, it's based on taking an interpretation as fact and then insisting something that we can neither formulate coherently nor detect must be going on because we are sure of our interpretation. That compares almost exactly to the argument for phlogiston, although that theory at least made testable predictions. μηδείς (talk) 03:15, 28 November 2012 (UTC)[reply]
And what happened in the past can just as well be considered as events that are not real but exist in some other realm with which we cannot intereact anymore. So, by your logic, we shoud prefer the interpetation that says that the past actually is fictitious. The alternative interpretation that it isn't is, after all, just an interpretation with zero "real" evidence for it (you can't intereact with the past). Count Iblis (talk) 00:22, 29 November 2012 (UTC)[reply]
That's silly, Iblis. My grandmother was real and her effects and remains are real, she is just not alive now, which is no different from saying her corpse is not buried on Mars. You know this. μηδείς (talk) 21:54, 1 December 2012 (UTC)[reply]
What's the point of postulating a completely separated universe that cannot ever have a connection to ours? Isn't that like the flying spaghetti monster or your favorite God? And why would you need a word like multiverse? The concept of universe encompasses everything. Maybe at some parts of it the natural rules are completely different than what we know, and not yet discovered, but there is no "outside." OsmanRF34 (talk) 23:06, 27 November 2012 (UTC)[reply]
Right on. "In my Father's house (= the Universe) there are many mansions (= sub-universes)". -- Jack of Oz [Talk] 23:14, 27 November 2012 (UTC)[reply]

This has already happened. We can never send a message to the galaxies seen in the Hubble deep field. They now live in realities totally separate from our own, even though we see them clearly as they once were. Hcobb (talk) 23:18, 27 November 2012 (UTC)[reply]

there you go. Math being incorrectly interpreted. Like the traveling professor marvel in the wizard of oz.GeeBIGS (talk) 00:26, 28 November 2012 (UTC)[reply]
  • The whole "multiverse" thing may or may not be hokum, but it's really just renaming the limits. If the Universe is "everything", how do you have more "everythings"? You don't, just just redefine what "everything" means: you can either give it a new name, or you can redefine the old terms. It is important to remember in this context that less than 100 years ago, it was widely assumed in the scientific world that the entire universe consisted of the Milky Way galaxy. In less than a century we've shown how completely ridiculously small the Milky Way is. In terms of scale, it's a bigger redefinition of the "universe" than anything Copernicus or Galileo came up with. And there are people alive today who lived in a world where the prevailing scientific thought was that everything in creation was in the Milky Way. So, whether or not there are, or are not, something like a "multiverse", if it were found that there was, it wouldn't be an unprecedented redefinition of creation. It wouldn't even be the first such major redefinition in the past century. --Jayron32 05:52, 28 November 2012 (UTC)[reply]
But here we have a clear case of different bits of one universe that mingled in chaos and void until some old dude turned on the lights and are now so far apart that they can never interact in any way again. Hcobb (talk) 12:32, 28 November 2012 (UTC)[reply]
As fun as this is to say, "ah, these guys must just be idiots!," the people who work on this aren't idiots, and it isn't just about linguistic semantics. Hugh Everett's many-worlds interpretation, for example, is an attempt to make sense out of quantum mechanics on a very deep level. Not everyone agrees with it, but a startling number of theoretical physicists do find it the most useful way to understand quantum phenomena. Those other "universes" are not simply areas of one bigger universe; they are alternate realities. I fear there is a lot of ignorance on exhibit in this particular thread. I don't claim to be an expert here but I know that the pat little "ah, misunderstood math!" and "mansions without houses" sorts of answers are just plain uninformed. I suppose I prefer Nimur's, "why even talk about it if we don't know their real" sort of response, though it is also unhelpful to the OP. It's a pity that one can't ask a question about a controversial theory on here without provoking a debate as to whether the theory is valid or not, especially from a lot of people who clearly don't know much about the theory. It is a bigger shame that we can't get a SteveBaker-like answer to this sort of thing, rather than a Yahoo! Answers chorus. --Mr.98 (talk) 23:56, 28 November 2012 (UTC)[reply]
Quantum entanglement + wormhole = ??? ~AH1 (discuss!) 20:46, 29 November 2012 (UTC) Shared noise. Hcobb (talk) 17:57, 30 November 2012 (UTC)[reply]

Is a covalent bond a cooper pair or a magnetic link?

Is a Covalent bond more like a Cooper pair or is it a magnetic link between two atoms as the electrons drop into reversed spins to cancel out their two magnetic fields? If so, isn't the photon emitted when a covalent bond is established simply the result of Magnetic reconnection? Hcobb (talk) 16:35, 27 November 2012 (UTC)[reply]

No, covalent bonding isn't anything exotic like that. It's just good old electrostatic interaction. Nothing fancier than Coulomb's law is needed to explain it. In a covalent bond, the electrons exist in a molecular orbital that has more electron density between the nuclei than around them. To put in in simplest terms, the negative electrons hang out in the space between the positive nuclei, and electrostatics keeps the whole thing together. There's a whole lot of Shroedinger's equations and wave functions and Hamiltonians and whatnot describing how the electrons hang out in that space; but basically they do, and that's what keeps it all together. It's the same forces that keep electrons around the nucleus of an atom, or stick the balloon to the wall after you rub it in your hair. It isn't magnets or anything like that. Cooper pairs are exotic low-temperature behavior kind of stuff. --Jayron32 05:43, 28 November 2012 (UTC)[reply]
An Ionic bond is held together by electrostatic interaction. The most famous covalent bond would be something like carbon-carbon, which is entirely electrically neutral. It is the magnetism of unpaired electrons, not their charges that is "neutralized" therein. Hcobb (talk) 12:29, 28 November 2012 (UTC)[reply]
What? That's a rather odd perspective. A non-polar molecule has a balanced charge, from an external point of view, but internally it is the electrostatic interactions between the nucleus and the electrons that holds it together. It isn't "magnetism", it's the localized concentration of electrons between the nuclei. Even in a C-C bond, the two Cs are held together by the electrons concentrated between them. You seriously need to reconsider what loony is teaching you basic chemistry and physics here if you think there's "magnetism of unpaired electrons" which is holding a C-C bond together. The forces holding ions to each other are identical to the forces holding a molecule together internally; it's just a difference in organization not in fundemental forces. --Jayron32 17:51, 28 November 2012 (UTC)[reply]
Agree with Jayron. All forms of chemical bonding are based ultimately on a combination of electrostatic interactions and exchange interactions. To the extent that magnetism enters into it, it is only as a secondary consideration. Dragons flight (talk) 21:40, 28 November 2012 (UTC)[reply]

Direct Evidence Against Sodium Channel Inactivation in Spike Frequency Adaptation

Hello. Is there any evidence against sodium channel inactivation in spike frequency adaptation (e.g. the failure to stop adaptation with the administration of a pharmacological agent that interferes with this inactivation)? Thanks in advance. --Mayfare (talk) 18:49, 27 November 2012 (UTC)[reply]

On the contrary, it is clear that multiple mechanisms contribute to SFA, and there is pretty strong evidence that sodium channel inactivation is one of them -- see for example http://jp.physoc.org/content/566/2/519.full. If you want to get into the literature on this topic, I recommend doing a Google Scholar search for "sodium channel inactivation spike frequency adaption" -- this will give you a number of relevant papers. Looie496 (talk)!~

Bombarding proton with neutron

What will happen if we bombard proton with a high speed neutron ? I shall be thankful if someone suggest me some articles related to my question in Wikipedia or anywhere else. Sunny Singh (DAV) (talk) 18:54, 27 November 2012 (UTC)[reply]

Many things might happen, but you're like to produce a bunch of hadrons. At high enough energies these hadrons are likely to form jets. Dauto (talk) 19:21, 27 November 2012 (UTC)[reply]

Can you mention some of those things that might happen ? Sunny Singh (DAV) (talk) 05:35, 28 November 2012 (UTC)[reply]

Stability of neutron and proton

I know the half-life of neutron is about 10 minutes, 11 seconds, when it is outside the nucleus- hence unstable. I want to know the half-life of proton when it is outside nucleus. Sunny Singh (DAV) (talk) 20:49, 27 November 2012 (UTC)[reply]

No one knows. Proton decay has never been observed, which implies a half-life of more than 1033 years. Dragons flight (talk) 20:57, 27 November 2012 (UTC)[reply]
It has been estimated to be years, see e.g. here. Count Iblis (talk) 21:05, 27 November 2012 (UTC)[reply]
In the Standard Model the proton is stable. In some Beyond the Standard Model theories like GUTs the proton is unstable but there is no exprimental evedence that protons decay, or hint that the theories that predict proton decay should be favoured over theories that pridect protons are stable. Dja1979 (talk) 22:52, 27 November 2012 (UTC)[reply]
Protons should decay even within the Standard model because of non-perturbative sphaleron effects. Dauto (talk) 23:56, 27 November 2012 (UTC)[reply]
The deuteron can decay this way to a positron and an anti-muon neutrino or to an anti-muon and an anti-electron neutrino. The deuteron lifetime due to this process is approximately 10^(218) years. Count Iblis (talk) 21:51, 28 November 2012 (UTC)[reply]
10^218 years? That's long enough that not a single decay would be expected to have occurred to any particle in the entire history of observable universe thus far. Dragons flight (talk) 00:12, 29 November 2012 (UTC)[reply]
For an interesting tangent in theoretical particle physics, see tetraneutron. ~AH1 (discuss!) 20:43, 29 November 2012 (UTC)[reply]

Why poisoning with 210Polonium?

It's difficult to obtain and expensive, which limits the circle of suspects. Radioactive elements are also easier to detected and it also decays rapidly, so you have to hurry up once you obtain it. There are other more efficient poisons, with a lethal dose of less than 1g. Why not use traditional poisons? OsmanRF34 (talk) 23:41, 27 November 2012 (UTC)[reply]

To send a clear unambiguous message to others: "That's what will happen to you if you cross us". Dauto (talk) 23:50, 27 November 2012 (UTC)[reply]
The perpetrator might want two things: to be unambiguous or to keep everything under wraps. If it was the case of Arafat's death, couldn't an F16 be a better choice to deliver the same message unambiguously? Or to be as discreet as possible, use an easily available poison? I still see Polonium as a total overkill (excuse the pun). OsmanRF34 (talk) 00:14, 28 November 2012 (UTC)[reply]
An F16 lives no deniability. Dauto (talk) 00:21, 28 November 2012 (UTC)[reply]
Deniability leaves no chance for your unambiguous message. If Arafat was killed with Polonium, people will know that a state was behind it (which can only be the Israeli state, for obvious reasons). If they wanted it to look like a health issue, then, why not another poison? OsmanRF34 (talk) 00:31, 28 November 2012 (UTC)[reply]
It could also be a rival in the Palestinian movement, as was suggested at the time he died. ←Baseball Bugs What's up, Doc? carrots02:00, 28 November 2012 (UTC)[reply]
I don't know about Arafat. I was thinking of Alexander Litvinenko. But the point is how to send a clear message AND still preserve deniability? That is a balance act that requires a specific choice of murder weapon, hence the choice of polonium-210. Dauto (talk) 00:48, 28 November 2012 (UTC)[reply]
"which can only be the Israeli state, for obvious reasons" — I don't think it's so obvious. Arafat was, at the very least, a known quantity. It strikes me as not at all clear that the Israelis would consider his death in 2004 to be a desirable thing (at an earlier point in time it would be more obvious). There are lots of potential candidates of people who would have benefited from his death either directly or indirectly by the confusion it caused within the Israeli-Palestinian relations. I'm not trying to start a debate, but I think you should expand your analysis as to whom "the state" behind the Polonium might or might not have been. There are lots of states (even some private companies) with the technical wherewithal to create Polonium-210. More difficult than the technology is the experience with handling it, but even that is probably wider than is commonly thought. --Mr.98 (talk) 16:01, 28 November 2012 (UTC)[reply]
It is also worth noting that very small amounts of Polonium-210 (i.e. 1/100000th of a lethal dose) can be easily obtained for industrial and research purposes for roughly $100. Given that it has been years since he died, it isn't that hard to imagine that someone could deliberately introduce traces of polonium after the fact to intentionally mislead an investigation. That is especially true when talking about the traces reportedly found on Arafat's effects, probably less so if they actually find it in his bones, which is after all the point of exhuming him now. Finding very small amounts of polonium now isn't really a guarantee that it was used to kill him, especially since many of his symptoms weren't generally consistent with what you expect from radiation poisoning (e.g. no hair loss). By the way, we have an article of the death of Yasser Arafat. Dragons flight (talk) 20:55, 28 November 2012 (UTC)[reply]
You only need about 50 nanograms of Polonium 210 to kill someone. The alpha radiation emitted by the decays don't exit your body, so it isn't easily detected. Count Iblis (talk) 23:52, 27 November 2012 (UTC)[reply]
Polonium poisoning is also difficult, if not impossible, to treat. Plasmic Physics (talk) 05:09, 28 November 2012 (UTC)[reply]

Recall Hamlet's words after killing Polonius: "Indeed this counsellor is now most still, most secret, and most grave, who was in life a foolish prating knave." In the Litvinenko case, I think this is precisely the message someone was sending through their choice of poison. By the way, the verb "prate" can here be read as having a double meaning: "to prattle" but also "to tell tales or inform against someone"; Shakespeare used it in the latter sense in Macbeth.--Rallette (talk) 06:43, 28 November 2012 (UTC)[reply]

What was the message from Litvinenko? If you are no longer or friend we go and kill you. We will do it any country. Going to a journalist or a foreign secret service will not help. We will do it in a way that everybody knows that it was murder. We will not be prosecuted or even questioned. We do not care if everybody knows that we have done it. Even if the dieing is on youtube and every other media we will not care.--Stone (talk) 07:45, 28 November 2012 (UTC)[reply]
Are you suggesting, Rallette, that the use of Polonium was an actual intended allusion to Polonius? μηδείς (talk) 19:22, 28 November 2012 (UTC)[reply]
Yes, actually, that was what I thought it must be, back in 2006. Of course I can't be sure, but I think it fits rather neatly.--Rallette (talk) 07:15, 29 November 2012 (UTC)[reply]
I suppose it's conceivable, but what it doesn't fit very neatly with is my image of the Cheka. Subtle allusions to a foreign writer that hardly anyone will even think of? Perpetrated by those humorless thugs? Kinda doubt it. But who knows; maybe they've been taking night classes. --Trovatore (talk) 07:47, 29 November 2012 (UTC)[reply]
On the contrary, the Russians seem to be very big on old-fashioned education, including the literary canon, and indeed take some pride in being able to insert apt references and quotes into their conversation. Take for instance the movies "Peculiarities of the National Hunt" and "Peculiarities of the National Fishing", which are extremely low-brow comedies, yet their names are an allusion to the opening paragraph of The Brothers Karamazov.--Rallette (talk) 08:17, 29 November 2012 (UTC)[reply]
Also, it has been pointed out that "Polonius" and "Polonium" are spelt identically in Russian: Полоний. --Rallette (talk) 08:26, 29 November 2012 (UTC)[reply]
I didn't say Russians. I said Cheka. The second point is mildly interesting, though. --Trovatore (talk) 08:27, 29 November 2012 (UTC)[reply]
Yes, I see your point! Although I can’t say I personally know any Chekists, still my image of Russian spies is rather different. The typical KGB man in my mind is the Soviet-era “cultural attaché” in a foreign capital, who unofficially represents the real power of the Kremlin, the Party, and the national security apparatus, as opposed to the regular diplomatic bureaucrats who work for the Foreign Ministry, far down the food chain. There used to be lots of them here in Helsinki, and they were most cultured and charming people, or so I’ve heard... But let me admit for the record that a major reason I believe in the Hamlet connection is, of course, that it makes me feel clever! But this is getting a bit OR, and I suppose belongs more in the Humanities desk.--Rallette (talk) 08:46, 29 November 2012 (UTC)[reply]
Hmm, that's not the image I ever had. The stereotype, here, is men in badly fitting suits, with comical accents, trying to appear Western. But I'm happy to admit that may just be a successful propaganda effort but on whose part? Let's see, whom would it benefit? :-) --Trovatore (talk) 10:25, 29 November 2012 (UTC)[reply]

November 28

Bathing

100 years ago people bathed anywhere from once a month to once a year. 200-300 years ago it was once a year. Have there been any studies done on bathing frequency and skin infection (staph infection, boils, fungal infection ect) and also UTI's. Most people bath 1-7 times a week now and I'm curious if skin infection and UTI's have gone down or remain the same. --Wrk678 (talk) 06:31, 28 November 2012 (UTC)[reply]

I don't think you can generalize on bathing frequency like that. In tropical areas on the beach, people probably bathed far more often than in cold climates inland, due to the difficulty of obtaining and heating the necessary water. Then there were those who didn't bathe because they thought it unhealthy, which it was, if they didn't have the sense to keep warm until they dried off. Also, lye soap can be rough on the skin, if used daily. StuRat (talk) 06:40, 28 November 2012 (UTC)[reply]
It also depends a lot on place. During the Roman Empire, people frequented the Thermae several times per week. --Jayron32 06:53, 28 November 2012 (UTC)[reply]
Nearly always, if you go to the doctor and tell him about some skin rash you've got, unless it is certain specific diseases, he'll tell you to stop using normal soap and instead use a bag of oats or some dermatitis "soap" that isn't actually soap. And quite often the rash will then clear up, or at least get better. Trouble is, without soap, you don't feel clean, and you smell a bit. Also, in hot climates (eg Australia), a daily wash with soap seems to help you keep cool and not sweat so much. Ratbone 58.169.233.62 (talk) 07:54, 28 November 2012 (UTC)[reply]
What's so unhealthy about being cold? I hope you're not referring to this unfortunate misconception. Evanh2008 (talk|contribs) 09:57, 28 November 2012 (UTC)[reply]
Your link doesn't say it's a misconception. Dry, cold air leads to cracked, chapped lips and an irritated throat and nasal passages, both of which allow viruses in. The rest of your body is normally protected, being covered with clothes and oil, but, after a bath, you have neither, if you used lye soap and no moisturizer. StuRat (talk) 23:57, 28 November 2012 (UTC)[reply]
Maybe, but if you are low on calories (most people would be 200 years ago or earlier) cooling off your body in winter is a bad idea for your health in general, because calories will be used to rewarm it, intead of being invested in building up your imune system. Any germ can take advantage of a slightly weaker imune system. --Lgriot (talk) 10:13, 28 November 2012 (UTC)[reply]


can anyone answer my original question? In the usa 100 years ago on the east coast they didnt bathe much. --Wrk678 (talk) 23:51, 28 November 2012 (UTC)[reply]

I doubt anyone can answer your question for the simple reason that information about the frequency of skin infections and urinary tract infections occurring 200 ago is not widely available (if available at all). Richard Avery (talk) 08:23, 29 November 2012 (UTC)[reply]
Lots of US families a hundred years ago had the Saturday night bath as a standard. I question the OPs assertion that they only bathed once a month back then. Maybe people who liked being filthy, or those who vnever did enough work to get dirt on their bodies. 75.34.30.62 (talk) 20:02, 29 November 2012 (UTC)[reply]
Or those who lived in places where water was very scarce. 24.23.196.85 (talk) 09:18, 30 November 2012 (UTC)[reply]

Fluoroauric acid

As gold(V) fluoride is a stronger fluoride accepter than antimony pentafluoride, I'd imagine it would create a stronger acid than fluoroantimonic acid based purely on fluoride affinity if mixed with hydrogen fluoride (as with fluoroantimonic acid). Am I right, or are there other factors affecting the strength of such an acid?--Jasper Deng (talk) 06:55, 28 November 2012 (UTC)[reply]

Variants of this have been asked several times over the past year or two. Executive summary has been IIRC "no matter what you think, you're missing something (or something isn't fully known period) if you think you're making something stronger than fluoroantimonic". Likely places to search are the archive of WP:RD/S, the talkpage of the acid, and maybe my talk-page.DMacks (talk) 07:06, 28 November 2012 (UTC)[reply]

Did Carl Sagan know the intended use of his calculations and still willingly participate?

From this story: "According to the report in The Sun, the US would have used an atom bomb, because a hydrogen bomb would have been too heavy. The planning reportedly included calculations by astronomer Carl Sagan, who was then a young graduate." 20.137.2.50 (talk) 15:02, 28 November 2012 (UTC)[reply]

The author of one of Sagan's biographies suggested that he may have committed a security breach in 1959 after revealing the classified project in an academic fellowship application. Reiffel concurred. Read Project_A119#Consequences Trio The Punch (talk) 15:15, 28 November 2012 (UTC)[reply]
He certainly knew about it, and he willingly participated. It's one of those ideas that seems silly or crazy in retrospect, but from a scientific standpoint, a nuclear blast would not have done much more to the moon than creating another crater. The craziest part of the project is the idea that the American people — much less American allies internationally — would have considering testing nuclear weapons on the moon to be a positive public relations stunt. --Mr.98 (talk) 15:51, 28 November 2012 (UTC)[reply]
I have to agree with Mr.98. Those crazy American people should have known bombing a little country no one likes would have been a much better idea than exploding a device on the moon, where it would harm no one (what's the point in not harming anyone?) and where everyone could actually see what they were up to! μηδείς (talk) 17:25, 28 November 2012 (UTC)[reply]
It would certainly show how powerful the US was, without contaminating the Earth. This seems a good way of working on your self-marketing during the cold war. I don't see anything crazy here. OsmanRF34 (talk) 18:58, 28 November 2012 (UTC)[reply]
Mr98 says "from a scientific standpoint, a nuclear blast would not have done much more to the moon than creating another crater." But our article says
Another factor [in the cancellation of the project], cited by project leader Leonard Reiffel, was the possible implications of the nuclear fallout for future lunar research projects and colonization.
Duoduoduo (talk) 18:52, 28 November 2012 (UTC)[reply]
Umm... what fallout? There's something known as solar wind which would very quickly sweep all fallout off the Moon and into deep space. The only radioactive material left on the surface of the Moon would be the debris welded into the Moon's surface by the heat of the blast, and even that would be slowly broken up by continual bombardment by meteoroids of all sizes and itself swept off the Moon by the solar wind. Whoop whoop pull up Bitching Betty | Averted crashes 20:51, 28 November 2012 (UTC)[reply]
Haha, the solar wind is not that strong, man! The moon is covered in dust. I agree the issue would be of no concern at all because the moon is huge and you'd ask scientists where they were least likely to go for the next few years, because cosmic rays would be worse than the radioactivity, and for a host of other reasons. But the solar wind doesn't sweep the moon clean. μηδείς (talk) 20:58, 28 November 2012 (UTC)[reply]
Every sample of lunar soil and ice will be contaminated from that point on. We could possibly never find out the exact process of lunar formation thanks to that. Consider how it's no longer possible to manufacture steel that's free from Cobalt-60 contaminants on Earth.Dncsky (talk) 21:28, 28 November 2012 (UTC)[reply]
Yes it is - just make steel without cobalt. Whoop whoop pull up Bitching Betty | Averted crashes 22:27, 28 November 2012 (UTC)[reply]
If it were that easy we wouldn't be digging up battleships for steel.Dncsky (talk) 22:47, 28 November 2012 (UTC)[reply]
The escape velocity on the moon is 2.4 km/s (5400 miles / hr), and the local gravity is 1.6 m/s2. A substantial fraction of the dust and rock fragments kicked up by a nuclear explosion on the moon will fall back onto the moon. I'm not sure where you get the idea that the solar wind would scour the moon, but for anything heavier than gas molecules it is basically not true. The solar wind moves ~400 km/s but has a density of only about 6 protons / cm3 at the Earth's orbit. That implies a solar wind pressure of around 10-9 Pascals. Unless an object has an extremely low mass in relation to its cross-sectional area, the force imparted by solar wind bombardment won't be large enough to move it away from the moon's gravity. Dust particles on the moon can't be dislodged by the solar wind. Dragons flight (talk) 21:20, 28 November 2012 (UTC)[reply]
Practically all of the radiation which is not welded into the Moon's surface would be atomised gas. Unlike on Earth, there is zilch blast wave to generate dust, so anything which is not directly vaporised by the radiation from the detonation would stay firmly attached to the Moon's surface. Whoop whoop pull up Bitching Betty | Averted crashes 22:27, 28 November 2012 (UTC)[reply]
If it is at the surface of the moon, the mere act of vaporizing rock will create a blast wave and excavate a crater. It may be smaller than on Earth, but it wouldn't be inconsequential. They could diminish the effects by placing the explosion well above the surface, but if you just wanted to set off a bomb in space then why carry it all the way to the moon? Dragons flight (talk) 22:51, 28 November 2012 (UTC)[reply]
I can see the contamination being an issue from a scientific standpoint — it could make it hard to do certain types of future measurements on lunar soil, and this is from a pre-Apollo understanding of the lunar surface anyway. I wasn't trying to imply it was a good idea, just that the "blow up the moon" stuff is a bit hyperbolic. It's an awful idea from a P.R. standpoint, as the attention given to this (decades old) story makes somewhat clear: people have at best mixed feelings about setting off nukes anyway, but to do it to something as timeless, visible, and apparently "pristine" as the moon ought to have been an absolute no-go. (I wonder how people would feel, incidentally, if they knew that we left several small plutonium power generators on the moon as a result of Apollo...) --Mr.98 (talk) 23:44, 28 November 2012 (UTC)[reply]
My question is not about any of the technicalities of the consequences of trying to blow up the moon, but how in the hell a man like Carl Sagan, who seemingly placed great value on natural celestial objects and not destroying them, was convinced to contribute to the project. 67.163.109.173 (talk) 22:23, 28 November 2012 (UTC)[reply]
That amounts to a request for speculation and opinion. Unless on the off chance he explained himself, say in The Collected Letters of Carl Sagan, we can't help you with that, we do not make windows into men's souls here. μηδείς (talk) 22:54, 28 November 2012 (UTC)[reply]
Of course I'm only interested in what he himself said of it. "Off chance"? Seems like something people would definitely have asked him or pressed his response on if it came to light before he died. 67.163.109.173 (talk) 23:12, 28 November 2012 (UTC)[reply]
Based on the information in Project A119, it would appear that the project was mostly secret until after Carl Sagan's death. Dragons flight (talk) 00:07, 29 November 2012 (UTC)[reply]
The ostensible goal of the project, aside from the P.R. aspects, were to find out more about the constituents parts of the moon. They wouldn't really "blow up the moon" — that's just journalistic hyperbole. For a comparable sort of thing, see also Operation Argus. --Mr.98 (talk) 23:39, 28 November 2012 (UTC)[reply]
The OP seems to be starting with the premise that nuclear devices are inherently evil (the Evilonium-235, no doubt) and that any use of them is therefor unimaginable. But many scientists supported notions like Project Orion in principle. And if a demonstration of the destructive power of the bomb were visible worldwide, not to mention would give plenty of scientific data, is it impossible that a man like Sagan might consider the possible experiment? Questions framed in the form of "how in the hell" seem to imply a very strong desire to engage in soapboxing, rather than research. μηδείς (talk) 01:56, 29 November 2012 (UTC)[reply]
If you or others transfer the blowing up of the moon part to the journalists or append obtained knowledge about the true state of mind of the project runners that they meant "but not really blow it up, just scare 'em" (just talking desired intent, not at all thinking the ability were ever there) then I retract my soapbox-impression-inducing use of "hell." 67.163.109.173 (talk) 02:18, 29 November 2012 (UTC)[reply]
Okay, hehe. I thought it was obvious the notion was not actually to blow up (think Death Star) the moon. 03:18, 29 November 2012 (UTC)
It's worth remembering a fair bit of Carl Sagan's fame and quite a few of his writings came later in his life. People's opinions of things often change over time and they may not even always remember the change in their opinions. In other words, even if it's true the personality you understood Carl Sagan had from his writings, interviews, television shows etc, would not have liked the concept, and even if your understanding isn't that far off the mark from Carl Sagan's real personality at some point in his life (perhaps a melding over time), it doesn't necessarily mean he would always have felt that way. In fact, this project is probably a decent example of that. I suspect if you ever ask any people involved who are still living, most would say it was a dumb idea even though obviously a fair few did not think this at the time. Nil Einne (talk) 15:11, 29 November 2012 (UTC)[reply]

Harold White / NASA research on warp drive

Going off of this article here -- so I'm not a scientist by any stretch of the means, and most of the stuff in the Alcubierre drive article makes my head hurt. Main thing is - even if White did legitimately find a way to reduce the energy requirements - doesn't this still all depend on the existence of the same theoretical exotic matter Alcubierre's original idea talked about? If that's the case, what are White and his people at NASA even researching? --Brasswatchman (talk) 16:41, 28 November 2012 (UTC)[reply]

NASA is a huge organization; it directly and indirectly supports an absolutely vast amount of research. Here's a very useful and informative website maintained by NASA's Glenn Research Center: Warp Drive, When?. It provides links to numerous research avenues that have been explored, and helps explain in plain language exactly why this problem is so difficult. I find that NASA web-page far more insightful than most pop-science publications on the topic. And, for the enthusiastic reader, there are links to more technical research papers. The NASA "Breakthrough Propulsion Physics" laboratory is also very helpful if you're not familiar with the topic. You might find this summary enlightening, as well: All NASA support to sustain cognizance on these possibilities has been withdrawn as of October 1, 2008.[4] In other words, it is the official position of the current administration that this research does not warrant NASA support: irrespective of whether the physics is interesting or impossible or even completely legitimate, the expected returns on the investment of time, talent, and money, are not seen as particularly worthwhile. Nimur (talk) 17:31, 28 November 2012 (UTC)[reply]
Thanks for the idea of searching at NASA - can't believe I didn't think of that. Wound up finding these papers, if anyone is curious: Warp Field Mechanics 101 and Eagleworks Laboratories: Advanced Propulsion Physics Research. It looks like White and his team are specifically focusing on trying to measure the expansion of spacetime to see if they can measure any possible preexisting "warps" -- which I suppose would be a precursor for trying to hunt down any kind of the exotic matter Alcubierre's ideas require. Scanning the papers, I get the impression that White and his cohorts may perhaps be motivated more by enthusiasm and a certain amount of optimism rather than solid experimental fact... still, I suppose this is something that's worth looking into, just in case. --Brasswatchman (talk) 19:03, 29 November 2012 (UTC)[reply]
It may be worth looking into but not by these guys, who are pretty clueless. In the second document they talk about a "quantum vacuum plasma thruster", which is apparently a reactionless drive that pushes against vacuum fluctuations. They think not only that this is possible but that models have actually been built and work, and they seem to think there's nothing controversial about the concept. I conclude from this that they've never talked to a real particle physicist. -- BenRG (talk) 06:39, 30 November 2012 (UTC)[reply]
Yes, it still needs exotic matter. I don't know how White used/uses his grant money, but according to the article one thing he did was build a modified Michelson–Morley interferometer and a ring of ceramic capacitors. Don't ask me why. -- BenRG (talk) 17:47, 29 November 2012 (UTC)[reply]

Size of bubbles in weightlessness

What determines the size of bubbles in weightlessness and do all liquids have the same bubble diameter in weightlessness? Thanks--93.174.25.12 (talk) 20:36, 28 November 2012 (UTC)[reply]

Surface tension determines the size of bubbles. As not all liquids have the same surface tension, different liquids have different bubble diameters even in weightlessness. Whoop whoop pull up Bitching Betty | Averted crashes 20:45, 28 November 2012 (UTC)[reply]
Unless you are saying that density actually varies in weightlessness, Whoop, all that matters is volume, i.e., mass divided by density. Surface tension holds the bubble together and makes it tend toward spherical shape when not perturbed or sticking to something else. But surface tension is not affected by gravity. μηδείς (talk) 20:55, 28 November 2012 (UTC)[reply]
Yes, that is why gravity or lack thereof does not and cannot make different kinds of liquid have the same sizes of bubble. Whoop whoop pull up Bitching Betty | Averted crashes 21:05, 28 November 2012 (UTC)[reply]
Moving this down to first indent for clarity: per Medeis, volume is a function of mass and density, and the basic sizes are unchanged between microgravity and any Earthbound laboratory. If liquid A fills less of a beaker than liquid B here on Earth, it will likewise form a smaller blob in space (assuming otherwise equal conditions like temperature and pressure). — Lomn 21:29, 28 November 2012 (UTC)[reply]
Now, for the counterpoint: if you're instead asking "how big of an air bubble can I create within a blob of liquid in microgravity?" then yes, surface tension is a significant factor. Again, you'll observe a basic correlation between behavior on Earth and behavior in space, though I would expect a substantial scaling factor. You might also have to quibble over definitions of what constitutes "a bubble of air within a blob of liquid". — Lomn 21:36, 28 November 2012 (UTC)[reply]
That would make sense of Whoop's statements--although gravity would still not matter, and only certain mixtures will allow you to make air bubbles. μηδείς (talk) 22:48, 28 November 2012 (UTC)[reply]

Cooling of Air-cooled engines

I'm trying to track down the original source of this claim from Air-cooled_engine: "In all combustion engines, a great percentage of the heat generated (around 44%) escapes through the exhaust, not through either a liquid cooling system nor through the metal fins of an air-cooled engine (12%). About 8% of the heat energy finds its way into the oil, which although primarily meant for lubrication, also plays a role in heat dissipation via a cooler.". The given source [5] does not contain any references. I'm interested in the breakdown of the remaining 36% of the waste heat.

Googling "engine heat 44% 12% 8%"[6] only produces sites with a verbatim copy of the erroneous quote from WP.

Alternatively if anyone knows where I can similar figures like this that would be great too; basically any ballpark break-down of how much waste heat is exhausted via each route in an ICE.Dncsky (talk) 21:17, 28 November 2012 (UTC)[reply]

I notice that only adds up to 64%. I wonder where the remaining 36% goes. Radiating off the rest of the engine block and heating the passenger compartment when the heat is on ? In any case, those numbers don't seem that unreasonable, to me. StuRat (talk) 00:10, 29 November 2012 (UTC)[reply]
The left over 36% is the mechanical power output to the pistons. Radiation of the engine surfaces is much much lower. Ratbone 124.178.55.2 (talk) 16:40, 29 November 2012 (UTC)[reply]
Have a llook in Heywood "Internal Combustion Engine" Greglocock (talk) 00:19, 29 November 2012 (UTC)[reply]
You, sir, are a Godsend. Page 675[7] had exactly what I was looking for. Do you have a script that checks this page for engine related questions or something? ;)Dncsky (talk) 00:42, 29 November 2012 (UTC)[reply]
For a good discussion, with typical values, in an easy to read format (without math or proof), of where the fuel energy goes, see later editions of Harry Ricardo's classic The Hight Speed Internal Combustion Engine (Blackie pub). A very good treatment, theory and data, is Charles Fayette Taylor's 1960's era textbook. The amount of heat lost in the exhaust is easily calculated approximately by applying the otto formula, with some corrections for the finite combustion time and for dissociation, and the heat lost to the coolant. As the heat lost to coolant via cylinders and head is quite small, you need only make a rough approximation. (Note that the heat rejected to ambient by the radiator is considerably greater than the heat lost from combustion gasses within the cylinder, as it includes heat generated by friction and heat lost from exhaust gasses to the exhaust passages in the head and manifold. Plenty of measured data is available to support such approximation. Manufacturers data for sales and application purposes of industrial engines usually includes the fractions of fuel energy converted to mechanical output at the flywheel, heat lost to coolant, and heat lost to exhaust. There is always a gap between the total of the three and the energy value of the fuel, this gap being the heat lost from engine surfaces by radiation and convection, being typically (in large engines) about 5 to 8% of the engine maximum output fuel energy. Ratbone 121.221.30.167 (talk) 01:09, 29 November 2012 (UTC)[reply]
So, are the numbers in our article correct ? StuRat (talk) 01:49, 29 November 2012 (UTC)[reply]
I seem to remember answering a similar question from you on Ref Desk before. The numbers look roughly right. I don't have any measured data immediately to hand, and I'm not familiar with aircooled engines, however I have a computer software package that gives accurate (+,- 2%) calculations for large diesel engines (after you enter a large amount of engine physical measurements). For a typical large truck size engine, (15 litres, 4 valves per cyl, compression ratio 15:1): The data (naturally aspirated at full output, 1500 RPM) from this package is :-
Brake Power Output:........265 kW (39%)
Total Loss to Coolant:....164 kW (24%)
Loss to exhaust:..............233 kW (34%)
Loss from engine surfaces..19 kW ( 3%)
The sum of the heat lost to coolant and to air from engine surfaces (164 + 19 kW) is made up of:-
Cylinder wall loss to coolant/air:.........57 kW (includes combustion heat loss to oil via pistons)
Exhaust port loss to coolant/air..85 kW
Friction loss to coolant/air:.......41 kW
These figures look somewhat different to the data cited by the OP. However, it should be noted that : 1) diesel engines are substantially more eficient than gasoline engines, due to higher compression ratio, no air throttling, and lower combustion temperatures; 2) aircooled engines are most commonly small engines, so the surface area compared to power is relatively high. So the data cited by the OP is not unreasonable.
This of course is something you have to take on trust, as you don't have my software, but you can check in C F Taylor's book. Ratbone 121.221.30.167 (talk) 02:53, 29 November 2012 (UTC)[reply]
The first number is how much energy is used, not excess heat, which is what we're discussing here. The total energy produced by an engine, of course, being mechanical, heat, exhaust pressure, and a tiny bit as sound and engine wear. The "loss to exhaust" might include both heat and pressure. StuRat (talk) 03:11, 29 November 2012 (UTC)[reply]
I can't see what you are trying to say, Stu - your first sentence is too cryptic. The OP said 44% (of the combustion heat) is lost to exhaust, compared to my calculated estimete of 34%. The discepancy is accounted for by the higher thermodymanic efficiency of the diesel engine used in my example, as I said. The figure "Loss to exhaust" in both cases is the energy lost out the exhaust pipe - yes, this lost energy is manifested in the temperature (most of the energy) and pressure (some of the energy) of the exhaust gasses, both temp and press being above that of the ambient air. For the 15 litre diesel example, the acoustic power generated from engine surfaces is of the order of a few tens of watts at most and is always ignored in engine thermodynamic efficiency calculations. If this surprises you, consider a domestic stereo system - typical electrical output 50 W RMS per channel, Loudspeaker efficiency 3%, music mean to peak ratio 0.1, mean acoustic power output 150 milliwatt. And, while it's not like a big engine in full cry, you can certainly deafen people with a stereo. Most of the acoustic energy comes from the exhaust, which is already accounted for. In theory, the wearing of bearing surfaces represents an energy loss. However, the energy lost in this way, which can be calculated by the science of tribology, is really minute, and nobody would bother. In any case, it is already accounted for in the figure for friction loss. Ratbone 121.221.7.194 (talk) 03:52, 29 November 2012 (UTC)[reply]
My interpretation of the OP is that it asked what percentage of the waste heat was dissipated by various means. That is, he is excluding the actual work done by the engine by the moving cylinders. Thus, the percentage of waste heat going to the exhaust is 44%, but the percentage of the total energy going to the exhaust would be less. StuRat (talk) 04:01, 29 November 2012 (UTC)[reply]
Well, only the OP can really say what he had in mind, however he /she accurately quoted from the Wikipedia article. The OP and the article says "..percentage of heat generated..", not percentage of waste heat generated - so the article is talking about the combustion heat, not the waste heat. The article could have been better worded, but the meaning is never the less clear - to an engine man at any rate. However, the OP used the phrase "waste heat" in the last sentence of his question, and that implies he misunderstood the data. Using the waste heat as a base, for my example above, the figures are:-
Combustion heat:........686 kW
Brake power output:....265 kW
"Wasted heat" is 686 - 265 = 421 kW
Exhaust heat:............233 kW i.e., 55% of wasted heat. (34% of combustion heat as given in my previous post)
55% is not 34% and for a typical air cooled gasoline engine the descrepancy will be about as bad. I can't think why you would want to calculate in this way. In any case, the reference for the Wiki article on this aspect, http://www.avweb.com/news/maint/182883-1.html, makes it very clear they meant 44% of the combustion heat, not the wasted heat. The answer to the OP's question, "where does the remaining 36% go", is of course the indicated power output (the mechanical effort to the pistons). Most of this ends up as the mechanical (i.e., brake) power output (typically about 25% of the total heat in an ordinary gasoline engine), the rest is lost in bearing and piston ring friction, valve gear and oil pumping loses. Ratbone 121.215.61.79 (talk) 12:26, 29 November 2012 (UTC)[reply]

November 29

Where do drug names come from?

Back in August I had a minor heart attack, and was prescribed an anticoagulant drug called Ticagrelor. Apparently it's an updated version of something called Clopidogrel. I'm also prescribed Ramipril and Bisoprolol. I'm just wondering where these names come from.

I know there are IUPAC names (Ticagelor is, God help us, (1S,2S,3R,5S)-3-[7-[(1R,2S)-2-(3,4-Difluorophenyl)cyclopropylamino]-5-(propylthio)- 3H-[1,2,3]triazolo[4,5-d]pyrimidin-3-yl]-5-(2-hydroxyethoxy)cyclopentane-1,2-diol, for instance) and trade names (ticagrelor is "Brilique" in some parts of the world), but why ticagrelor? Why not boffinokpum or gluponifen? Why four syllables? Why is ticagrelor not called "numptum" or "bonktyrumptimbul"?

I'm aware that some drug families have similar names (-pril indicates ACE inhibitor and -olol means Beta blocker) but why (for instance) "bisoprolol" rather than "bogolol", why "ramipril" rather than "nobbipril"?

Same goes for statins. They seem to have names (and again, I'm not talking trade names like "Lipitor") of the form thingvastatin, such as Atorvastatin (ironic, given its similarity to Atora), Simvastatin and Lovastatin.

I can see why systematic names such as IUPAC designations are assigned, trade names come from some completely barmy marketing outfit (and I'd like to meet the chap who decided that "Brilique" was just such a super name for a platelet inhibitor to be prescribed following heart attacks) but where do half-way houses, neither systematic nor marketing, such as "ticagrelor" come from? Tonywalton Talk 00:12, 29 November 2012 (UTC)[reply]

The only rules (in the US) are officially described here. If it doesn't break a rule, they can call it whatever the hell they want. Someguy1221 (talk) 00:33, 29 November 2012 (UTC)[reply]
International Nonproprietary Name. DMacks (talk) 01:47, 29 November 2012 (UTC)[reply]
  • There are lots of different considerations. First there's the difference between the chemical itself and the brand name it's marketed under. Minoxidil, 6-piperidin-1-ylpyrimidine-2,4-diamine 3-oxide, is the active ingredient of Rogaine. The chemical name is ultimately arbitrary, but suggests diamine oxide, with the NO group suspected of being the active site. The name Rogaine was originally suggested as Regaine, for obvious reasons, but was rejected by regulators as too suggestive. Atorvastatin, the chemical in the brand Lipitor, is a statin class drug with cardiovascular benefits. 'Lipitor' is obviously something powerful (think "terminator") that has to do with fighting lipids.
Beyond such associations, drug companies don't want to pick a name that is offensive or has bad connotations in other languages, such as the case of the Chevy Itdoesn'tgo. They want the name to be different enough from other names that the two are not confused, but not so different that it is hard to pronounce or remember. People have been killed by drug name confusions, see the ref below. Names with C, P, V, X, and Z are popular, frankly because they sound powerful and a little magical. Our article Brand name didn't lead anywhere useful, but this very good article gives all the basics: http://www.medicinenet.com/script/main/art.asp?articlekey=53208. Sometimes I have fun with the pharmacist, and ask if they filled my prescription for fratastatin, or generic dammitol. But I wouldn't do something like that if she were busy or we weren't on a first name basis. μηδείς (talk) 01:48, 29 November 2012 (UTC)[reply]
You mean Chevy Nova#Urban legend? Nil Einne (talk) 11:45, 29 November 2012 (UTC)[reply]
Yes. I assumed the main article would link to it, but thanks for the direct link. I should also mention there have been many deaths linked to accidental dispensation of a drug confused with the one actually prescribed due to similarities in names. The medicinenet link I gave above documents some. μηδείς (talk) 16:37, 29 November 2012 (UTC)[reply]
Oh, I did mention it. Forgot to take my GleeMONEX. μηδείς (talk) 21:29, 29 November 2012 (UTC)[reply]
I think you (or possibly I) have missed the point. I am not talking about trade names like "Brilique". This is the registered trade name for (see IUPAC name above). Trade names can be anything.
I'm wondering about is why it's also called "ticagrelor". Why, come to that, is 3-piperidin (and so on) called "Minoxidil". Why not "Boggyplonk"? There would appear to be three names to any given (prescription) drug. A IUPAC name which is, far from what μηδείς suggests, particularly exact, not "arbitrary". The chemical name is an exact description of the compound. Then there is a name whose nature this question is trying to ascertain (such as Ticagrelor or Clopidogrel). Then, finally. there is a trade name such as Brliique or Plavix. Why three, and what generates the "non trade" name? Tonywalton Talk 23:56, 29 November 2012 (UTC)[reply]
I seem to have been misunderstood. By chemical name I meant the one assigned to the item patented, not the IUPAC formula. I referred to minoxidil as the chemical name above which suggests the amine and oxide of the IUPAC formula. That one is indeed arbitrary within the obvious parameters of pronounceability and the ones I gave, a unique, memorable, non-offensive name that may suggest the class of drug, etc. The company that patents the drug names it, then usually markets it under a trademark name that is more customer than industry oriented. Both are made up. There are no set rules. It's no different than inventing something with specific dimensions, say a 10"-diameter pressed aluminum cake pan, patenting it under the industry name 10DPACP, and marketing it under the trademark name Frisbee. What you are calling the trade and non-trade names are the trademark and patent names. Both are used in trade. μηδείς (talk) 01:29, 30 November 2012 (UTC)[reply]
As I mentioned earlier (and others also stated in various ways), the generic drug-name, which you identify as the "non trade" name, is made up, but does follow some set patterns based on various biochemical and/or structural properties (as you also noted). The parts that are patterned follow published rules (INN, BAN, USAN, etc.), but the exact combination and details that are not part of the pattern rules are arbitrary (but tend to follow patterns again based on structure or activity). To answer one of your specific cases ("bisoprolol" rather than "bogolol"), bisoprolol appears to have two isopropyl groups and bi/bis is a common chemical prefix for "two". And "3–4 syllables" is one of the published standards (I'm sure the standards-bodies have reasons for making that declaration...ease of remembering and writing, I guess?). So if you say "we have this new beta-blocker that has as one of its distinguishing structural characteristics a 'bis-isopropyl'", contract that down to a two-syllable prefix for the '-olol' class. DMacks (talk) 04:59, 30 November 2012 (UTC)[reply]
Aha. Yes, I'd misunderstood your use of the term "chemical name", μηδείς. I can't see "clopidogrel" or "amlodipine" being "chemical names" in any meaningful sense, to be honest. "Generic name", fine. And there are published standards for these names, per DMacks, who also says they're sometimes contractions of a more precise chemical name (Ibuprofen is a good example of this, the name coming from iso-butyl-propanoic-phenolic acid). Thanks, all, for your contributions (but I still think "ticagrelor" sounds like one of Dr Who's more unpleasant foes). Tonywalton Talk 09:31, 30 November 2012 (UTC)[reply]
Yes, chemical name is obviously ambiguous. One thing that isn't clear is whether the various regulatory bodies actually propose generic names for newly approved drugs, or if the just approve those suggested by the patent holder as I asserted. See United States Adopted Name which DMacks mentions as USAN. That article also has a link to this excellent list of several hundred stems used in naming new medications], which is what I assume you have really been looking for. μηδείς (talk) 17:59, 30 November 2012 (UTC)[reply]
Generic names are proposed by the company/organization that files the New Drug Application (or equivalent in other jurisdictions). The name has to fit within the guidelines noted above, and then must be approved by the relevant regulatory agencies. -- Ed (Edgar181) 18:10, 30 November 2012 (UTC)[reply]

The Mechanism of Coldness and sore throat ?

what is happening in coldness (from feeling the right amount and time of cold, windy or not), and why our body's tend to develop sore throat (a very annoying situation) as for it. why not something else? — Preceding unsigned comment added by 79.182.153.70 (talk) 02:51, 29 November 2012 (UTC)[reply]

Cold air can't hold much humidity, and that dry air dries out the throat. Add to this the thermal shock of inhaling cold air, and damage is caused to the lining of the throat, making it sore and vulnerable to infection. StuRat (talk) 02:54, 29 November 2012 (UTC)[reply]
wow thanks ! enlightening.. ! what did you mean by "Lining" of the throat?, and why is soreness (redness?) accumulates infection?,, Blessings. 79.182.153.70 (talk) 04:25, 29 November 2012 (UTC)[reply]
The throat (both esophagus and trachea) are lined with mucus, which keeps them moist. This mucus provides a barrier to bacteria and viruses, since, in the time it would take for them to migrate through, they might be coughed up or swallowed and killed by stomach acid. We don't have skin inside, to protect the tissue, but this serves a similar function. Also, when we swallow things which would normally irritate the esophagus, like acidic sodas, the mucus keeps it away (the stomach lining also has mucus to protect it). When the mucus layer is reduced, then bacteria and viruses find it easier to infect our tissue, where they cause inflammation, apparent as redness and soreness. StuRat (talk) 06:49, 29 November 2012 (UTC)[reply]
There's another possibility. When it's cold, the temperature of the body surface can drop dramatically, and immune responses are slowed considerably in cold tissue. Looie496 (talk) 18:21, 29 November 2012 (UTC)[reply]
Which has to do with vasoconstriction in the extremities of the body. This happens especially when the person in question is not only cold but also wet, because the water takes away a huge amount of heat as it evaporates. 24.23.196.85 (talk) 09:24, 30 November 2012 (UTC)[reply]

Computer in Matrix movies

I am currently writing a paper on virtual reality, and referencing Bishop George Berkeley's work on idealism and how his views today can be seen clearly exemplified in movies such as "The Matrix". I saw all three movies in the series when they came out. Now, I am wondering if the massive computer that keeps the Matrix in operation is, in the movie, itself conscious. Because if it is not, then it would mean that you need more than just enormous complexity to attain consciousness. This makes it a very different computer to the one in sci fi movie "The Thirteenth Floor", in which conscious entities exist but are not bound to bodies. They are, in that movie, fully conscious beings which exist as avatars of the computer in which they reside. This is a far more complex situation than the one in Matrix, I believe.

I myself do not agree with the strong AI position on this, as I think that while complexity is a necessary condition for consciousness, it is not a sufficient one. But then, I do not know what WOULD comprise such sufficient conditions. But I am not alone in this. As long as we don't know what makes us conscious, we cannot say with finality what conditions are necessary to achieve that state. And the "hard problem" of philosophy remains. Myles325a (talk) 06:04, 29 November 2012 (UTC)[reply]

I'm not sure what your question is, but if you're looking for resources, the "conscious computer" trope is a well tapped field for science fiction. Two that jump to mind as particularly notable and famous is "Mike" from Robert Heinlein's The Moon Is a Harsh Mistress and "HAL 9000" from the 2001: A Space Odyssey film, book, and various sequels thereof. The idea of consciousness being unbound from a body is also a well-used trope, the character of "The Dixie Flatline" from Neuromancer is a particularly seminal use of it. More recently, John Scalzi's Old Man's War makes use of the idea of mind transfer to transfer the consciousness from one body to another. If you're looking for the philosophical tradition that underpins the Big Questions regarding consciousness in general, Wikipedia's article consciousness is a good start, as is the Mind–body problem and if you want to get really nuts, dig into the Phenomenology philosophy of Husserl and Heidegger. --Jayron32 06:16, 29 November 2012 (UTC)[reply]
The Terminator (franchise) also comes to mind. StuRat (talk) 06:41, 29 November 2012 (UTC)[reply]
Sufficiently advanced computers may be indistinguishable from a human in their responses, but that doesn't automatically mean they are conscious. StuRat (talk) 06:41, 29 November 2012 (UTC)[reply]
If you can't 'distinguish' conciousness, what is it? AndyTheGrump (talk) 06:47, 29 November 2012 (UTC)[reply]
Well, then we get into philosophy, and the idea that I really don't know that anyone else in the universe in conscious. StuRat (talk) 06:51, 29 November 2012 (UTC)[reply]
See also p-zombie and hard problem of consciousness. --Trovatore (talk) 07:52, 29 November 2012 (UTC)[reply]
We don't have a good definition for consciousness. But imagine this thought experiment: Program a computer to simulate a bunch of neurons. This is clearly technologically possible for small-ish groups of neurons even now. A program/computer that did this would be dumb as rocks, certainly neither intelligent nor conscious. Now, suppose we could somehow scan a human brain, tracing every connection and measuring everything. That's difficult with present technology - but it doesn't seem to be impossible from a fundamental science standpoint. Now dump all of that data into our neuron simulator and start it running. If we have a sufficiently powerful computer to run all of that software in realtime - and if we connect the simulated brain to microphones and cameras via simulated ears and eyes - and to an audio jack via a simulated larynx - then what should happen is that the brain should behave pretty much exactly like the person you scanned. Every possible external test for consciousness would succeed. We ask the neural simulator "Are you conscious?" and it would reply "Yes!" - if you asked "Is it OK for me to shut down the software now?" and it would presumably reply "No!! Please don't kill me!"...and so forth. Should we say that this is "consciousness"? Notice that we didn't change the "dumb as rocks" software or the computer to do it - we just gave it more data. So is the consciousness somehow "contained" in that long stream of numbers? It's not a scientific question, it's a philosophical one.
That's one of the reasons why we don't have a good definition of "consciousness". SteveBaker (talk) 14:31, 29 November 2012 (UTC)[reply]
If we presume for argument's sake that we are conscious, then would it follow that animals are also? They would not say yes to the Q "are you conscious"165.212.189.187 (talk) 15:43, 29 November 2012 (UTC)[reply]
The question is whether consciousness is just piling on more connectivity or not. We really don't have a good sense as of yet how the human brain works except in the most basic terms. Going from a basic neurological description to a qualitative description of consciousness is still pretty far out there. I'd put our understanding of cognition in the single digits percentage-wise at this point (and I take that assessment from neurologists/cognitive scientists). Whether just simulating the neural wiring in and of itself will produce anything that acts like a human brain is still unclear to me; there's more than just wiring in there. There's still a lot of "scientific questions" there beyond the philosophical ones. I'm not saying its unanswerable — I think the consciousness must ultimately have a material basis, obviously, and presumably will yield to systematic analysis upon continued investigation — but I don't think we're really in a position to say, "oh, just pile on more complexity" and have that be compelling, yet. --Mr.98 (talk) 16:35, 29 November 2012 (UTC)[reply]
Too much thinking, overanalysis. "Consciousness" is simply "awareness of your own existence" on some level. ←Baseball Bugs What's up, Doc? carrots16:10, 29 November 2012 (UTC)[reply]
"Awareness"? "On some level"? What does that mean, precisely? Is my laptop "aware of its existence on some level"? (About This Mac --> I self-refer, therefore, I am?) The reason that philosophy spends so much of its time doing "too much thinking" is because flip answers yield no results upon attempts to actually use them for anything. Once you start trying to use said results to answer the questions people are actually interested in (much less the questions philosophers are interested in), you start having to refine them down, "overanalyze" them, "think too much". ("Physics" is simply "how the world works" on some level.) --Mr.98 (talk) 16:32, 29 November 2012 (UTC)[reply]
I agree: physics is simply "how the world works." But, consciousness is not a question of physics. The methods of physics (and, by extension, the other natural sciences) can be used to explore mechanisms that are related to consciousness, but it's really in the domain of philosophers to define what consciousness is - because natural sciences don't even attempt to answer that type of question. If you undertake a well-rounded study of the entire body of knowledge on the topic of consciousness will, of course, learn all sorts of things that we do know through physics (or bio-physics, or biology). We've learned incredible things about the way neurons operate; how they grow and interconnect; the neurochemists can explain how neurons signal, and neurologists can describe all that we've measured using the electroencephalograph. But none of these explorations have ever answered the question, "what is consciousness." That is not the way experimental science works; and for all that we learn about the mechanisms of our world, the scientific method never address the motives of the universe. So, natural science will never put the philosophers out of a job; for all we conclude about "how" the world works, we still need them to address the issue "why." Nimur (talk) 17:08, 29 November 2012 (UTC)[reply]
My point in invoking physics being "how the world works" is that such isn't a very useful definition of it if you actually want to do something with physics — if people had stopped there we wouldn't know very much. As for consciousness in general, I don't think consciousness is only a question of philosophy; I disagree that it is a "why" question. There is a hugely significant "how" question at the center of human cognition. Once we understand that better — how you go from a lump of cells to a seemingly unified awareness, one capable of generalizing inquiry to such a degree that it can even sit around asking itself why it thinks about itself so much — I think we'll have a much better understanding of what we physically mean by the term "consciousness." I find both the assertions that "we basically understand it, just scale up what we know about neurons" to be inadequate; I find the "don't think about it, it's just a problem for squishy philosophers" to be even more so. Fortunately there are plenty of people who are actually doing the work to try and figure out how the mind works out there. --Mr.98 (talk) 18:30, 29 November 2012 (UTC)[reply]
Modern Maccs are aware of their existence, That's why they started to call themselves iMac. OsmanRF34 (talk) 17:03, 29 November 2012 (UTC)[reply]

SteveBaker has almost the correct answer. If you just take into account that I can't be aware of the exact data in my head and consider how I should (in general) consider the abstract space defined by the possible data sets consistent with it representing me, then that's pretty much the answer. Count Iblis (talk) 21:01, 29 November 2012 (UTC)[reply]

That seems to lead to the conclusion that there are now, and always, Platonic abstractions that are not only conscious, but conscious of every possible quale. I find that — let's say unlikely. --Trovatore (talk) 10:24, 30 November 2012 (UTC)[reply]
  • Consciousness is usually used to mean awareness of one's surroundings, like "regaining consciousness". What's being talked about above is usually called self-awareness, although Julian Jaynes, whom everyone should read (albeit with a shaker of salt), used consciousness in this latter sense. μηδείς (talk) 21:27, 29 November 2012 (UTC)[reply]
    Actually, what I find is that people often say self-awareness when they just mean awareness (in the sense of qualia — mechanical devices are "aware" of their surroundings in the sense that they have sensors capable of responding to them, but we rarely posit that they experience actual phenomena). I am not sure how the self- part snuck in there, because it's really not about a sense of self; it's just about a sense of anything, period. Phenomenal consciousness is another term that's better than self-awareness for the concept under discusson, and also illustrates the distinction with awareness of surroundings (for example, I am phenomenally conscious when dreaming, but not conscious of my physical surroundings). --Trovatore (talk) 19:08, 30 November 2012 (UTC)[reply]
    • "Awareness" is the common element there: of one's own existence and of one's surroundings. In short, "alive". Does a mechanical device such as a toaster, a car or a computer have an "awareness" of its surroundings? Yes, to the extent that it's been programmed to. But does it have awareness of its own existence? A sense of "self"? Maybe. But where's the evidence? Matrix is an interesting (while highly derivative) story. Have we been "programmed"? If so, by what? A computer? Natural evolution? God? Or all the above? ←Baseball Bugs What's up, Doc? carrots03:36, 30 November 2012 (UTC)[reply]
The latest news on AI seems to be this, from just up the road from my house. "Spaun can recognize numbers, remember lists and write them down. It even passes some basic aspects of an IQ test, the team reports in the journal Science." Matt Deres (talk) 16:48, 30 November 2012 (UTC)[reply]
The difficulty with consciousness is that we are not independent witnesses. If we were only allowed to use the evidence measurable with scientific instruments of the people around us - ignoring our own built-in biases completely - I think we could conclude that there is nothing special that human brains can do that couldn't be programmed into a highly complex and (presumably) non-conscious computer. Occams' razor would tell us that there is no such thing as consciousness. The HUGE problem with that conclusion is that each of us as individuals is seeing their own personal consciousness "from the inside". That (IMHO) is the only evidence that there is anything at all special going on here. We have no definition for what this "experience" is. We can only infer that people other than ourselves have this "thing" because they seem to say the same kinds of things about the experience as we do - and since we're all members of the same species, it seems likely that we're all more or less the same in this regard. So how could we possibly tell if some other animal or computer were to be conscious in the sense that we feel ourselves to be? Without a definition, or a way to quantify and measure it - it's kinda difficult to come up with any kind of science-based response. SteveBaker (talk) 19:01, 30 November 2012 (UTC)[reply]
Very true, which is exactly why the whole area of phenomena/qualia is so problematic for your "scientistic" world view, Steve. There are folks like Daniel Dennett who more or less explicitly reject the whole field for exactly this reason, and they can come up with very clever ways of framing this rejection, but it's still (IMO) a willful refusal to deal what's in front of their noses (or behind their noses, or in their noses, or you get the idea). We clearly are phenomenally conscious, and science's attempts at "explaining" this are all category errors. In my view they always will be, and this will remain a fundamental limitation on science. Don't get me wrong; I love science. But I don't think it can explain phenomenal consciousness. --Trovatore (talk) 19:39, 30 November 2012 (UTC)[reply]
I would be very suspicious of someone who did not believe that they themselves were conscious. It raises questions of their self-worth, self-respect, regard for the well-being of those around them, and ultimately trustworthiness and believability.165.212.189.187 (talk) 20:10, 30 November 2012 (UTC)[reply]

Excipients and generics

When substituting a brand name for a generic drug, could it happen that the latter is less efficient due to a wrongly chose excipient? Are any real cases known? For what I understood, some substances are not easily absorbed unless they are mixed with another appropriate substance, so the excipient is also somehow part of the medicine (not just there to bulk up the real active medicine). OsmanRF34 (talk) 12:44, 29 November 2012 (UTC)[reply]

Our article on generic drugs notes that most regulatory regimes call for some test of measured bioequivalence to the brand name drug that they imitate. That article's footnotes will guide you to more specific criteria (which in any case will depend on the particular drug's route of administration, intended use, and so forth) but will almost always include things like measurement of the amount of active drug in the bloodstream at various times after administration. In the U.S., the FDA requires generics to pursue an Abbreviated New Drug Application prior to approval. TenOfAllTrades(talk) 13:22, 29 November 2012 (UTC)[reply]

electronic multivibrator repetion rate

i wish to build (for a modelling project) a DC circuit thst will alternately flash 2 lights. An integrated circuit called a 555 timer is capapble of doing this. The input for such device is an RC network, and I wish to have a repetition rate of about 0.3 to 0.5 Hz. at the output.

The MAIN question is how to calculate the proper input resistor/capacitor values in order to control (change) the repetition rate. A corollary question involves the output voltages. (I can utilize either 1.5, 9 or 12 voltlamp bulbs) Perhps I should ask an IC manufacturer/else consult an (unknown) application book. Thought that Wikipedia would be a good place to start. If I had an address, I could send a prliminary schematic of what I will be doing.

Thank you in advance for your assistance.

Edmund 71.200.89.43 (talk) 17:38, 29 November 2012 (UTC)[reply]

The 555 timer IC article has some equations regarding RC values and oscillator frequency, and also some specs about source/sink power. DMacks (talk) 19:11, 29 November 2012 (UTC)[reply]
You could try using LEDs as they draw way less power than a bulb.Staticd (talk) 18:31, 30 November 2012 (UTC)[reply]

Unidentified Bird in Neighbours

About a minute into today's Neighbours episode, there was a brief shot of a black and white bird on a chimney. Usually I don't have too much of a problem identifying the birds featured in the show, but this one has me stumped. One of the writers suggested that it might be a Magpie, but the shape of the head and the markings don't seem quite right to me. Did anyone catch the episode or can ID the bird from this blurry screenshot [8]? - JuneGloom Talk 21:02, 29 November 2012 (UTC)[reply]

I don't know - but it would be a good idea to ask over at Wikipedia talk:WikiProject Birds. The guys on there are really good at this - in fact, they spend much of their time identifying birds in photos. --Kurt Shaped Box (talk) 21:11, 29 November 2012 (UTC)[reply]
I live in the same state in Australia that Neighbours is filmed and other then the magpie, the other very common black and white bird here is the Magpie-lark, was it that? Vespine (talk) 21:24, 29 November 2012 (UTC)[reply]
I couldn't see the photo on my work computer, but I looked at your screen shot on my phone and I'm pretty sure it's the 3rd bird I would have guessed: a Grey Butcherbird, (not to be mistaken for Butcher Bird). The photos here looks more like your screenshot then the photos in our article. Vespine (talk) 21:43, 29 November 2012 (UTC)[reply]
I don't think it's a magpie, the beak is quite short, and looks hooked to me. That suggested a raptor, so I browsed List of birds of Australia... after also looking here [9], my best guess based on coloration is the Australian Hobby, perhaps a juvenile. It also seems to be about the right size (hard to get scale from the screenshot), and our article says the Hobby is common in urban parks. That being said, I have no specific expertise in this area, and might be way off :) SemanticMantis (talk) 21:51, 29 November 2012 (UTC)[reply]
After the Magpie suggestion, I did think Grey Butcherbird or maybe Pied Butcherbird. But their beaks appear to be too long. - JuneGloom Talk 22:28, 29 November 2012 (UTC)[reply]
Yes, my first impression was raptor and Semantic Mantis seems to have it. μηδείς (talk) 01:09, 30 November 2012 (UTC)[reply]
Ok, it's NOT a raptor. It's sitting on a house brick which has a length of 230mm, even a juvenile hobby is longer then that. My 4th guess: Willie Wagtail, fits the size, but i'm just not 100% sure about the white marking, it looks like it's coming just slightly up the shoulders, the only photos I can find of a wagtail it seems more of a distinctly chest patch...Vespine (talk) 03:07, 30 November 2012 (UTC)[reply]
It doesn't look like the wagtail, whose throat is black, and beak is different, it looks just like the hobby. And you can't judge the size given the tail is obscured and the bird's angle is foreshortened. μηδείς (talk) 04:30, 30 November 2012 (UTC)[reply]
In my estimation, this is not identifiable without further information. It is not Grallina, not Cracticus, not Gymnorhina. It is also not any sort of Wagtail. It is also not an immature Pycnonotus jocosus which has a visible white malar slash not evident here. The undertail is white, and this eliminates a whole lot of possibilities. The tail is also relatively long, and this eliminates the possibility of it being one of the black-headed races of Daphoenositta. It is, however, obviously a Passeriformes, and certainly not a Raptor. What can I not eliminate therefore? I can't eliminate, at least not on this poor photo, some sort of immature Myiagra though the head jizz seems wrong to me for that bird group. The bird appears to possess, however, an evident eye-ring even as bad as this photo is. I am also not sure that it might not be, as somebody mentioned above, one of the hooded races of Pycnonotus barbatus, perhaps an immature, perhaps an escape. Because of the eye-ring there might be another group that should be looked at, and that is somethng from Melithreptus though I can not discern here the peculiar whitish nucal half-collar usually associated with the genus even when juvenile. So, take it from one that has analysed bird photos for ID for almost thirty years, this is not identifiable with no further information, and from just this one bad photo. Anybody telling you that it is certainly this, or that, is just talking through his hat. Only possible ID's can come from this and based on pure conjecture.Steve Pryor (talk) 06:56, 30 November 2012 (UTC)[reply]
Well! We got told. :) Great reply, thank you.. Not a raptor:PVespine (talk) 07:15, 30 November 2012 (UTC)[reply]
Links to those species would be useful, this is the reference desk. Without that it's just alot of handwaving. μηδείς (talk) 17:36, 30 November 2012 (UTC)[reply]
Nobody here has claimed to provide a definitive answer, and I agree that a blurry photo is not enough to get a rigorous ID. In the spirit of edification, can you tell us why you rule out Raptors, and are so sure of Passiformes? Is it an overall "feel", or are there certain characteristics that you are using to key? I think many of us are genuinely interested in getting better at this sort of thing, and I for one would appreciate more detail from an expert. SemanticMantis (talk) 22:18, 30 November 2012 (UTC)[reply]

On line degree of medical science

I would like to learn and earn degree: Bachelor of medical science (with curses of biochemistry of course), but it's important for me that it will be from base (mainly in chemistry), and that all the degree I will learn OnLine. Where is it a website like this? 95.35.67.215 (talk) 23:22, 29 November 2012 (UTC)[reply]

Lots of on-line degree programs are not widely respected. If you think about taking one of them, I would strongly advise that you check with a variety of medical schools that you might subsequently apply to, to find out if they accept that degree from that particular on-line institution as a qualification for admission. Duoduoduo (talk) 16:53, 30 November 2012 (UTC)[reply]
I doubt a complete online accredited medical degree exits at all. You can make a descent bacherol in other science at some serious online college, but medicine no. Comploose (talk) 17:45, 30 November 2012 (UTC)[reply]

November 30

Can someone familiar with nuclear physics comment on whether polywell fusion reactors are actually feasible or not? To a laymen it sounds awfully lot like a free-energy device scam. There's been very little peer review on this since the research team is under a publishing embargo.Dncsky (talk) 00:56, 30 November 2012 (UTC)[reply]

It's not a free-energy device scam — fusion is real physics — but with all fusion programs there are three major hurdles: 1. Can it reach ignition? That is, can it generate more energy through the reactions than it takes to start the reactions? So far none of the methods pursued have done this, yet, with the exception of thermonuclear weapons. 2. Can it produce useful net energy? That is, can you get electricity out of it? This requires both considerable efficiency and also in some cases elaborate means of extracting the energy from the fusion reaction. 3. Can it produce economically viable net energy? That is, can a fusion plant be economically competitive with other fuels out there? Another big unknown. The rub is that as of yet, we're still trying to find a method that satisfies #1. NIF was supposed to do it; it hasn't so far and it's not clear that it ever will. ITER is supposed to do it, we'll see. Once someone has found a way to do #1, it should be relatively easy to figure out a way to scale it up for #2. As for #3, it really depends on technical constraints we don't know yet, as well as issues unrelated to fusion technology (e.g. the price of natural gas).
I'm not a fusion scientist but my read of the Polywell article suggests that the people working on it really think it will work, and that they have been able to convince other experts at funding agencies that it might work. I don't think it's a scam, but that doesn't mean it will actually work. The fusion quest is nothing if not a field of failed dreams. So far. --Mr.98 (talk) 03:25, 30 November 2012 (UTC)[reply]
I am familiar with the current state of fusion research and I support every cent behind the 10 billion spent on ITER. Regardless of whether it will eventually work that money still needs to be spent. I am asking because the only publication I can find on the polywell reactor [10] seems to have been discredited more than a decade ago[11][12]. Yet they are still receiving funding from the Navy. Dncsky (talk) 04:27, 30 November 2012 (UTC)[reply]
The "References" list in the article is quite long and includes lots of scientific publications. I guess I'm not understanding why you think there is not much information published on it. The article shows it going through many funding reviews, many publications, and so on. The Navy has given them a few million — not chump change, but not a huge amount by research standards. Again, I can't evaluate the technical merits, but I don't see anything here that's against the laws of physics. The question is just whether it'll work or not, and that's not an easy thing to answer usually without spending some money on it. --Mr.98 (talk) 14:08, 30 November 2012 (UTC)[reply]

What are the global\general\basic ingredients of sand "regular" earth?

(sorry for possible misterminology, i don't have basic knowledge in geology). thanks 79.182.153.70 (talk) 04:34, 30 November 2012 (UTC)[reply]

Sand regular earth? Are you talking about plain old Sand? As our article discusses, the chemical makeup of "sand" is highly variable, as it is defined based on its gross properties rather than its chemical composition. Our article gives some of the more common types of sand and what they are made of. Since you mentioned "earth", you might also be interested in soil. Someguy1221 (talk) 04:42, 30 November 2012 (UTC)[reply]
okey, it's becoming interesting. what do you mean by "it is defined based on its gross properties rather than its chemical composition". this sentence is very abstract to a layman like me. i ask what are the most basic and general and typical ingredients of sand (and soil) just like a little boy, a curious boy would ask about them, i rally am a total ignorant in this matter. you are doing an holy work by lifting me out of this mud of ignorance (nice example eh?). — Preceding unsigned comment added by 79.182.153.70 (talk) 06:59, 30 November 2012 (UTC)[reply]
In laymans terms, sand is pretty much any finely ground rock or mineral. If it looks like sand and it feels like sand, it's sand, no matter what it's really made of. In the same way wood is wood no matter what tree you cut it from. Someguy1221 (talk) 11:17, 30 November 2012 (UTC)[reply]
The most common material is quartz which would make a white sand. Also common are rock forming minerals such as feldspar and pyroxene. Other sand may be made from fragments of rock like basalt, or pieces of shell or coral. Black sand may contain ilmenite or particles or wood or charcoal or discoloured by iron sulfide. Graeme Bartlett (talk) 10:27, 30 November 2012 (UTC)[reply]
There is a nice table of the chemical composition of the Earth crust at Composition_of_the_Earth#Chemical_composition. The most abundant substances are silica, alumina, and lime, as far as I know all rock-forming materials (and hence possible sources for sand). --Stephan Schulz (talk) 12:27, 30 November 2012 (UTC)[reply]
Our article mentions quartz is common for sand in 'inland continental settings and non-tropical coastal settings'. It also mentions 'The bright white sands found in tropical and subtropical coastal settings are eroded limestone and may contain coral and shell fragments in addition to other organic or organically derived fragmental material'. There is a bit more useful info, I strongly suggest the OP read it if they haven't already. Nil Einne (talk) 13:47, 30 November 2012 (UTC)[reply]
As noted, 'sand' can be from just about any mineral that is ground by erosion and accumulates in one spot. I strongly recommend a visit to Hawaii to examine their...sand. In addition to 'white' sand (predominantly quartz, most common) you can find black sand (Punalu'u Beach, principally basalt), green sand (Papakolea Beach, colored by olivine), and red sand (Kaihalulu Bay, rich in iron compounds). TenOfAllTrades(talk) 13:50, 30 November 2012 (UTC)[reply]
Are you sure of the first statement? Haiwaii would generally be consider a tropical region and most sand there is likely to be coastal, and as I mentioned above (perhaps with an EC) our article suggests silica or quartz is actually often not the predominate material in white sand in such settings. [13] [14] [15] seem to agree that silica or quartz sand is not common in Haiwaii. Nil Einne (talk) 14:15, 30 November 2012 (UTC)[reply]
How about extremely small pebbles between X and Y mm diameter? Might it also help to identify what sand is not? EG: clay & soil, both with significantly more organic and fine powder-like material?165.212.189.187 (talk) 14:54, 30 November 2012 (UTC)[reply]
If the questioner goes to his neighborhood building center or hardware store and asks for "sand" for his child's sandbox, he is likely to get "play sand." This product has a Material Safety data Sheet which says it is made of "crystalline silica," which seems to be another name for silicon dioxide. It has been washed to reduce the dust and dirt. The other type of sand he would find in the store is "all purpose sand," which is darker and contains more dust. It is used for making concrete. Edison (talk) 15:50, 30 November 2012 (UTC)[reply]

Without NASA we wouldn't have computers

Someone told me this, is it correct? ScienceApe (talk) 18:08, 30 November 2012 (UTC)[reply]

No, computers were already around before NASA was even founded. See e.g. the Atanasoff–Berry Computer and ENIAC. - Lindert (talk) 18:18, 30 November 2012 (UTC)[reply]
On the general topic, I have read that entire industries have come out of the space program, because sending rocket ships into orbit requires engineering designs that can vary by at most one part in 10 000, or something like that. I don't know what exactly we owe to the space program, although I know it isn't computers, and it isn't teflon. IBE (talk) 18:27, 30 November 2012 (UTC)[reply]
Orange Tang. SteveBaker (talk) 18:43, 30 November 2012 (UTC)[reply]
Certainly NASA didn't produce the first computer...who precisely did depends crucially on your definition of the word "computer"...Atanasoff–Berry Computer (circa 1942) is the most likely candidate...but there are reasons to say that it doesn't count. Arguably, NASA owned the first small "microcomputer". NASA paid IBM to build a 19" long computer (weighing in at 60lbs!) in May 1963. It was used by NASA on the Gemini program and had a fairly respectable 16k bytes of memory. However, there is always a problem with "Without A, we wouldn't have B" arguments. Clearly, IBM had the technology to build this thing - so if NASA hadn't paid them to do it, what is to say that six months or a year later, someone else wouldn't have? Since computers were already in fairly widespread use in 1963 - it was only a matter of time before someone else built a tiny one. There is no evidence whatever that NASA's machine was widely copied - or that it is somehow the progenitor of all computers that followed it - to the contrary, it was not much more than a dumb calculator with far fewer features than it's contemporaries. SteveBaker (talk) 18:43, 30 November 2012 (UTC)[reply]
It seems according to the Computer page the Z3 was first computer (obviously depending on your definition of computer) in 1941. Dja1979 (talk) 20:20, 30 November 2012 (UTC)[reply]
Here in the UK, everybody knows that Colossus was the first computer. I think we should have a List of computers claimed to be the first computer. Alansplodge (talk) 20:38, 30 November 2012 (UTC)[reply]
To which we could add Charles Babbage's Analytical Engine of 1837, "the first design for a general-purpose computer that could be described in modern terms as Turing-complete." Alansplodge (talk) 20:45, 30 November 2012 (UTC)[reply]
At the Computer History Museum in Mountain View, California, the exhibition hall is set up so that the first items you see are historical computation contraptions dating to pre-history; such artifacts as abacuses and cuneiform tabulations. The exhibition hall progresses forward through more advanced mathematical machines; Babbage engines; punch-cards and time-clocks (other intricate mechanical devices that could perform domain-specific computation); Curta peppermills; and finally, after you've gone through a whole row of historic inventions, you finally see the first of the electronic and electronic-digital-machines that start to resemble what we call a computer. Like any question of history, there is much room for debate and different perspective. You can navigate an online version of the "Revolution" - the first 2000 years of computing, and the Computer History Timeline. The museum used to be free and open to the public; but now charges an admission fee. Nimur (talk) 22:02, 30 November 2012 (UTC)[reply]

blocking in experimental design

I understand (more or less) what a block is in experimental design, but I don't get what the authors are talking about here when they say "In addition, although blocking subjects on initial interest in the target activity of course eliminated any between-groups differences in this variable,..." What is this "blocking subjects" referring to? In a block design, I thought the blocks were supposed to be arranged in advance, and furthermore, our article states "A nuisance factor is used as a blocking factor if every level of the primary factor occurs the same number of times with each level of the nuisance factor". This doesn't look like something you could arrange after the experiment. What's going on? IBE (talk) 18:23, 30 November 2012 (UTC)[reply]

I'm not sure what is confusing you. The article describes the blocking and the assignment of subjects to groups as arranged before the experiment, as far as I can see (p. 131, upper left). Looie496 (talk) 19:30, 30 November 2012 (UTC)[reply]

Electrical Properties of Tubes

What if instead of wires we were to use tubes (of copper, say for indoor wiring)? Would there be any advantages/disadvantages and are there any special electrical properties of such a configuration? Seems as if the current would travel on the outer surface mostly, but I'm just guessing, honestly. 66.87.126.32 (talk) 20:51, 30 November 2012 (UTC)[reply]

I believe this was done, in some places, specifically with the tube carrying either positive or negative, with a regular insulated wire inside carrying the opposite charge. There are several disadvantages, though:
1) Not flexible, so much harder to install and maintain, especially where bends are needed.
2) Takes up more space.
3) Requires more electrical insulation.
4) Since it's uncommon, people might not realize it's carrying current, and be electrocuted.
5) Access to the interior wires is more difficult.
Using the tube as ground/earth with both positive and negative insulated wires inside makes more sense, especially out-of-doors, where the tube provides additional protection from the environment, for the wires. The tube might also function as a structural support, say when using a flagpole as the ground/earth for a light placed on top, with wires running inside. StuRat (talk) 21:51, 30 November 2012 (UTC)[reply]
The one-conductor-inside-another layout is termed coaxial cable. As the article indicates, it's used mostly for RF signal transmission (where it has beneficial properties), although the configuration has been used for power in certain situations, though attempts to search for a good reference are swamped by mentions of coaxial power connectors. - By the way, the article Skin effect shows a three-wire-bundle high-voltage power line, mentioning that because of the skin effect, they're effectively one conductor, which is taking the tube-as-conductor idea one step further. -- 205.175.124.30 (talk) 22:20, 30 November 2012 (UTC)[reply]

When did the last visible (with naked eye) star appear in the sky?

Did Jesus looked at the same sky as us? Comploose (talk) 21:08, 30 November 2012 (UTC)[reply]

Pretty much, with a few exceptions. Some stars only become visible as a result of a nova or supernova. A new star being ignited might only very slowly become visible to us, as the dust clouds around it clear. The Earth's precession also makes different stars into the pole stars every so often (Polaris isn't always the North Star). There are also periodic comets which are only visible certain years, like Halley's Comet. See List of periodic comets. StuRat (talk) 21:18, 30 November 2012 (UTC)[reply]
Timeline_of_white_dwarfs,_neutron_stars,_and_supernovae is your friend :) Dr Dima (talk) 21:26, 30 November 2012 (UTC)[reply]
Proper motion also has an effect; a star might be nearer or further away, or in a slightly different place relative to others. But 2000 years is a pretty short timescale for such things; while there were differences, most of them were subtle. I don't think any significant naked-eye stars have appeared or disappeared in that time. AlexTiefling (talk) 21:28, 30 November 2012 (UTC)[reply]
They haven't appeared of disappeared, but some of them have moved noticeably. According to our article on Alpha Centauri:

"Edmond Halley in 1718 found that some stars had significantly moved from their ancient astrometric positions.[63] For example, the bright star Arcturus (α Boo) in the constellation of Boötes showed an almost 0.5° difference in 1800 years,[64] as did the brightest star, Sirius, in Canis Major (α CMa).[65] Halley's positional comparison was Ptolemy's catalogue of stars contained in the Almagest[66] whose original data included portions from an earlier catalog by Hipparchos during the 1st century BCE"

Alpha Centauri itself has a much larger proper motion than Arcturus or Sirius, and moves by 1 degree per millenium. For reference, that's about the width of your thumb at arm's length, or twice the angular diameter of the Sun or Moon.
So in conclusion, Jesus' sky would have looked almost identical to ours, with the exception that the north pole would have been in between Polaris and Kochab instead of very close to Polaris (see http://en.wikipedia.org/wiki/File:Precession_N.gif). --140.180.249.151 (talk) 23:22, 30 November 2012 (UTC)[reply]

Small angle formula

D = X · d / 206,265

What is the difference between D and d? They are both distances so can someone explain the difference to me in an easy way to understand?Pendragon5 (talk) 21:52, 30 November 2012 (UTC)[reply]

It appears to me that your formula is a special case of arc length of a circle, commonly denoted as ; but you've got it in a form where the radius angle is presented in normalized units (your constant 206,265). I didn't recognize that constant off the top of my head, but it wouldn't surprise me if it's related to, e.g., a special case of angular resolution. And, lo and behold, it's a conversion constant to arc-seconds, when d is the radial distance to the target, and D is the linear size of the object, and X is measured in radians. 206265, a unit conversion factor I don't have any common use for.
So, in plain english: d is the distance to the object, and D is the size of the object. This case-sensitive notation is a little ugly. Perhaps it derives from an era when ink was much more expensive. Nimur (talk) 22:27, 30 November 2012 (UTC)[reply]