December 24

I ran across an image out there in the great wide Interweb, of a closeup of the index and middle fingers of a man who appears to be suffering what can only be a very uncomfortable condition (Warning: rather disturbing image: http://ninjapants.org/files/index.php?image=goodmorning.jpg) All of the detail of the image lead me to suspect that the image is authentic, but I'm at a complete loss at to the condition from which this poor person may be suffering. Does anybody have any ideas? – ClockworkSoul 05:32, 24 December 2007 (UTC)[reply]

I'd say without too much hesitation, and apologies for any disappointment, that this image has been altered. There are several clues that hint towards this. Firstly the bizarre image of an apparent serious medical condition is presented with a desktop and mouse in the background. If it had been photographed in some sort of 'medical' context it would look more realistic. Second off, the image shows some odd anomalies that indicate some sort of photographic manipulation. Look at the fingerprint whorl on the finger on the left, it appears completely misplaced. Whorls are normally, but not exclusively, on the pad of the digit. There are also some other signs of post-production manipulation around the edges of the 'what-ever-they-ares'. There is a complete absence of any inflammation around these peculiar 'wounds': it is conceivable that if they were long duration injuries they may not have visible signs of inflammation, but in my experience they would. Next, the inside of the finger pulp does not look like that (unless you are some sort of alien), the image shows the matrix of possibly some sort of fruit or vegetative specimen. Finally, the whole medical appearance looks entirely improbable, what are those white 'teeth' looking bits, where have they come from and what is keeping them there. As you say the rest of the images on this site seem to be authentic, some pretty odd, but not manipulated. I would be very pleased if someone could demonstrate with a reasonable degree of authority that this is some sort of traumatic condition, but for the moment I am not any bit convinced.Richard Avery (talk) 10:31, 24 December 2007 (UTC)[reply]

You're right, it's a fake, as covered by Snopes AndrewWTaylor (talk) 11:09, 24 December 2007 (UTC)[reply]

Lampreys. Of course. Many thanks! – ClockworkSoul 05:29, 25 December 2007 (UTC)[reply]

Just trying to find out the difference. Ive read alot of lorcets are with tylenol and lortabs are with aspirin but none of these sources are doctors or nurses or pharmacists that definitly know the answer. so please anyone that is knowledgeable in this feild answer only and let me know if what info i have dug up myself is correct. thanx.

(just unindented the question) --Ouro (blah blah) 06:39, 24 December 2007 (UTC)[reply]

http://www.druglib.com is a good place to go to look up the ingredients of common drugs. Our article Hydrocodone lists the active ingredients of Lorcet and Lortabs - but notice that there are many variations of both product with different dosages, etc. There appears to be only one variation ("Lortab ASA") that contains asperin - all of the other Lortab/Lorcet varients contain acetaminophen (also called "paracetamol" outside the USA). Acetaminophen is also the active ingredient of Tylanol but it would not be strictly true to say that either Lorcet or Lortab contains Tylanol - although the result may well be the same. For a definitive answer, you should read the ingredients on the actual medication bottle and for medical advice in general, you should see a doctor. SteveBaker (talk) 14:05, 24 December 2007 (UTC)[reply]

How many litres of air (not just oxygen) does the average person intake with each breath? -- Danilot (talk) 11:17, 24 December 2007 (UTC)[reply]

See Lung volumes - this suggests it's about half a litre. I also found this, which gives some stats on volume per minute under various conditions. AndrewWTaylor (talk) 11:27, 24 December 2007 (UTC)[reply]

Is it possible that we are all floating in a big black hole? Is there a way you can tell when you are in a black hole? Is it possible that whole observable universe is floating in very big black hole? If is possible, how big must it be?

Hevesli (talk) 11:22, 24 December 2007 (UTC)[reply]

No, it is not possible. The tidal forces would have torn us into atoms. See Black hole#What happens when something falls into a black hole? ›mysid (☎∆) 12:02, 24 December 2007 (UTC)[reply]

I disagree with Mysid. If the black hole is large enough, tidal forces are small at the event horizon. The Schwarzschild radius is proportional to the mass m, while tidal forces are proportional to 1/r³; therefore, tidal forces at the event horizon are proportional to 1/m³. Icek (talk) 13:02, 24 December 2007 (UTC)[reply]

Indeed - Mysid is way off the mark. With a sufficiently large black hole, there would be negligable tidal forces - even within the event horizon. SteveBaker (talk) 13:51, 24 December 2007 (UTC)[reply]

Some strange thought that I just had: Could the expansion of the universe be described by the whole observable universe undergoing tidal forces? Viewed in the frame of a resting observer at different distances from 3/2 * r_S to r_S from the black hole's center, the circumference around the black hole gets larger the nearer you come to the event horizon (where it diverges). This provides for the expansion in 2 dimensions, while tidal forces explain the expansion in the remaining direction. But I've not done the math yet, and it's of course not the same for the falling observer. Icek (talk) 13:09, 24 December 2007 (UTC)[reply]

What do you mean by "floating in a big black hole"? We couldn't be inside a black hole—the pressures would be far too high to support anything like life. Could we be floating around a black hole, in orbit? Sure, but you'd really be asking "is our solar system rotating around a black hole at the center of the galaxy", which is, if I recall, what they think is at the center of the Milky Way (see Sagittarius A*). --24.147.86.187 (talk) 16:00, 24 December 2007 (UTC)[reply]

Wouldn't we be seeing some really weird effects, like everything that's between us and the center is completely still, and everything outward is accelerated? Of course, if the radial difference is in a different dimension then it might not be quite as obvious, but a few smart apples would be able to tell what's going on. Also, it would have to be a GIANT black hole for us to still be alive right now. --ffroth 20:14, 24 December 2007 (UTC)[reply]

There's no exact mathematical definition of what a black hole is, so there's no right answer to this question. The interior portion of the Schwarzschild solution makes a plausible cosmological model, and I'm sure it was considered for that role at some point in the history of modern cosmology. It's even homogeneous (the singularity is in the future, not in a particular spatial location). But it's a terrible fit to observation in other respects, and the other standard black hole solutions aren't any better. For example, they're all contracting in two spatial dimensions and expanding in one -- or vice versa if you use the time-reversed white hole solution -- so they predict redshift in some directions in the sky and blueshift in others (I think).

Contrary to what some people have said, the Schwarzschild interior is perfectly inhabitable in principle if m is large enough. How large? Er, I don't know, but it needs to be at least circa (c³/G)t, where t is the amount of time you want to survive before hitting the singularity. For t on the order of the age of the universe, that comes out to about the mass of the observable universe. But don't read anything into that -- it was bound to happen because of dimensional analysis alone.

In the

big crunch scenario the ultimate fate of all the matter in the universe is to hit a black hole singularity. Since a black hole event horizon is the boundary between stuff that necessarily hits the singularity and stuff that doesn't, you could argue that everything in the universe is already inside the event horizon of this final black hole. But this is a dubious use of the word "inside", since there's no outside (and no event horizon/boundary either). I'd tend to say that this pathological case shouldn't count as a black hole interior. The traditional big crunch scenario is ruled out in any case. it's possible that the quintessence

might cause a recollapse (depending on what the quintessence actually is), but I don't know how much this recollapse would resemble black hole solutions.

I'm also tempted to answer "humbug". Black holes get way more press than they deserve. The universe is pretty interesting whether or not it resembles a black hole. -- BenRG (talk) 07:33, 25 December 2007 (UTC)[reply]

I am reading a historical account of a battle and there is a claim of one, but the account is old and I'd like to verify it before putting in into the article; I'd assume we have means of calculating and verifying if a solar eclipse could have occurred in a specific region on a specific date? I couldn't find a comprehensive list or how to in our articles on Wikipedia, unfortunately. --_{Piotr Konieczny aka Prokonsul Piotrus| talk} 17:56, 24 December 2007 (UTC)[reply]

There is a

list of solar eclipses as well as a section on historical eclipses, neither of which are particularly helpful in your case. There's also this NASA image to the right, so they obviously have some record of 1000 years of solar eclipses.

-- MacAddct  1984 ^{(talk • contribs)} 18:07, 24 December 2007 (UTC)[reply

]

A quick search of NASA's site came up with this comprehensive list and this image, which is of a solar eclipse on August 12, 1654 and looks like it goes right over the Ukraine! -- MacAddct  1984 ^{(talk • contribs)} 18:14, 24 December 2007 (UTC)[reply]

Ha,

this will make a nice DYK then :D --_{Piotr Konieczny aka Prokonsul Piotrus| talk} 18:22, 24 December 2007 (UTC)[reply

]

This eclipse map may be easier to read. The path of the total eclipse of 1654 Aug 12 goes right through the middle of the Ukraine. - Nunh-huh 18:32, 24 December 2007 (UTC)[reply]

Ok, a related technical question than: can we use one of those images under

Yes, this one is marked as a NASA map, and for the other one here it says, "Permission is freely granted to reproduce this data when accompanied by an acknowledgment: 'Eclipse Predictions by Fred Espenak and Jean Meeus (NASA's GSFC).'" MilesAgain (talk) 08:13, 26 December 2007 (UTC)[reply]

How do you, electrically, take an analog input and isolate only the signal that's on a certain frequency? This seems like a fundamental component of radio recievers, but I can't decipher our extremely-technical radio articles to figure out how it works. Just what is happening when I turn the dial on an old radio? --ffroth 20:01, 24 December 2007 (UTC)[reply]

An

resonance frequency, and so it captures the signal specific to that frequency while excluding others. By adjusting the inductor (the L in LC), you can tune the circuit to match the radio's sensitivity to the frequency you want to listen to. Dragons flight (talk) 20:12, 24 December 2007 (UTC)[reply

]

You use a filter. The LC circuit is an example of this. How it works is that energy is alternatively exchanged between a current with magnetic field in the inductor, and a charge with electric field in the capacitor. This happens at a particular frequency. When you add a loss this does not oscillate for ever, and instead you get a filter that roughly lets through a range of frequencies. There are also crystals that ring like a bell and also allow only a narror range of frequencies. Graeme Bartlett (talk) 20:28, 24 December 2007 (UTC)[reply]

Makes sense. But where in this do you connect the actual signal wire? And how do you set the frequency? Are the capacitors/inductors somehow powered separately from the LC circuit? --ffroth 20:35, 24 December 2007 (UTC)[reply]

Hm, well I found Variable capacitor, but I don't understand it. How does the capacitance have anything to do with frequency? --ffroth 20:38, 24 December 2007 (UTC)[reply]

The resonant frequency is a function of the magnitudes of both the inductor and the capacitor. Make either variable and you can vary the resonant frequency (and so, pick your station). Most radios use a variable capacitor; older automobile radios often used variable inductors. Nowadays with electronically-tuned radios, the variable capacitor is more often a varicap than a physically-adjustable capacitor.

Atlant (talk) 00:26, 25 December 2007 (UTC)[reply]

(edit conflict) Resonance is the key. Here's an over-simplified rendition:

There are basically three sections in an old radio that deal with higher-than-audio frequencies, in this order: the RF (radio frequency) amplifier, the mixer, and the IF (intermediate frequency) amp(s). The RF amp does nothing more than amplify (add power to) the signal from the antenna, amplifying the whole mess of jumbled frequencies across the band it's designed to receive. We'll come back to the mixer. The IF section is tuned to pass only one frequency, the intermediate frequency, which becomes the new carrier for the actual audio signal to be detected later.

The real magic happens in the mixer. Two signals are fed into the mixer: the output of the aforementioned RF amp, and the output of the local oscillator. These beat together, producing harmonics all over the place. The local oscillator sits there the whole time ringing at a single frequency determined by where you have the knob set. It is designed to oscillate at a frequency different from the frequency of the station you want to tune by a difference equal to the intermediate frequency. The mixer and IF effectively perform an analogue subtraction, and the difference is the IF. The IF stage can only pass the intermediate frequency, so it acts like a filter. A side benefit is that the IF is a lot lower than the RF and is therefore easier to control and contain. Darned clever, eh? Some inventions are obvious once somebody's done it, but this ain't one of them.

To answer the other question, when you turn the knob you are rotating a row of semi-circular aluminum discs such that they interleave to a greater or lesser extent with similar fixed semi-circular discs, thereby changing the capacitance between them. This capacitance dictates the resonant frequency of the local oscillator. --Milkbreath (talk) 20:43, 24 December 2007 (UTC)[reply]

Grr, that doesn't make any sense. How can you use analog interference to change frequencies? That would be utterly destructive! --ffroth 03:28, 25 December 2007 (UTC)[reply]

The "magic" of heterodyning is that this mixing in the frequency domain does not destroy the signal in question. Instead of destroying the signal, it demodulates it. Nimur (talk) 08:03, 25 December 2007 (UTC)[reply]

Nothing gets demodulated in the mixer stage. When you beat two frequencies together, you get interference. It's just like the way you get a "beat" note when two musical notes clash. If you and another person whistle the same note, and one of you lets his note go slightly sharp or flat, you will hear the beat, which is the difference frequency. The difference frequency is explained here. Only the difference frequency that is equal to the IF will pass. The original modulation is obviously retained. --Milkbreath (talk) 13:38, 25 December 2007 (UTC)[reply]

Your explanation is incorrect. Beating happens when two signals (of similar frequencies) are combined additively. A mixer works by combining two signals multiplicatively. --72.94.50.57 (talk) 16:05, 25 December 2007 (UTC)[reply]

Hey, I'm no physicist, I'm an ex–First Phone and repair tech. I figure the guy who wrote the Shrader knows more than anybody here, and it says: "...mixing, beating, or heterodyning one frequency with another in a non-linear circuit... The result is always at least four output frequencies: (1) one of the original frequencies, (2) the other original frequency, (3) the sum of the two frequencies, (4) the difference between the two frequencies." (author's italics) (Electronic Communication. Robert L. Shrader. McGraw-Hill:New York. 1975. p.396) Suffice it to say that interference happens, and a difference frequency is produced. --Milkbreath (talk) 19:08, 25 December 2007 (UTC)[reply]

See the Wikipedia articles on frequency mixer and superheterodyne receiver. --71.175.22.107 (talk) 07:11, 25 December 2007 (UTC)[reply]

One quick comment: An RLC circuit is the electronic equivalent of a mechanical damped harmonic oscillator. The inductor provides the inertia (it resists changes in current just like inertia resists changes in momentum), the capacitor provides the restoring force (it stores energy when displaced from equilibrium, like a spring), and the resistor provides the damping. So if you can see why a mechanical harmonic oscillator has a resonant frequency, it's not that hard to see why an RLC circuit has a resonant frequency. Instead of a mass moving back and forth periodically, it's charge carriers flowing back and forth. —Keenan Pepper 02:46, 25 December 2007 (UTC)[reply]

I just want to make sure.. in event of apocalypse or zombie invasion, does anyone actually know how this stuff works? I thought radios were super simple, this is insanely deep stuff --ffroth 07:22, 26 December 2007 (UTC)[reply]

Well, no, we don't know. Does anybody know what's going on when a magnet makes a paper clip jump up and stick to it, or what's keeping my ass in this chair right now? To paraphrase Heinlein, have you ever put salt on the tail of a wavicle? Sometimes the lemmings stay put, and their holes migrate. There is math that accurately describes a lot of this, and there are models that predict things pretty well, but it's all a mystery at the base level. --Milkbreath (talk) 13:01, 26 December 2007 (UTC)[reply]

We certainly can't explain why things like fundamental forces exist (and hence, at the deepest level, we don't know "why" a magnet attracts a paperclip) - but we certainly understand what's going on when you demodulate an AM radio signal. That's just simple math. Ptolomy (and later, Euler) figured out that:

 sin(x+y) = sin(x)cos(y) + cos(x)sin(y)  ...and...
 sin(x-y) = sin(x)cos(y) - cos(x)sin(y)

Add those two equations and the cos(x)sin(y) terms drop out and you get:

 sin(x+y)+sin(x-y) = 2sin(x)cos(y) = 2sin(x)sin(pi/2-y)

cos(y) is just a phase-shifted version of sin(y) - so you can demodulate a signal (x) by multiplying it by a 90 degree phase-shifted version of the carrier frequency (y). You'll end up with sin(x-y) - which is your recovered audio signal and sin(x+y) which is too high in frequency for the subsequent amplifier stages to deal with, so it's rejected. Crappy analog multipliers (aka transistors or vacuum tubes) allow some of the original signal through too - hence Schrader's assertion that there are FOUR signals present in the output rather than the two you wanted - but those are also way too high in frequency and are effectively rejected by the audio stage amplifiers or by the inertia of the electromechanical loudspeaker. QED

SteveBaker (talk) 16:05, 26 December 2007 (UTC)[reply]

• Basic AM radios are extremely simple. When you tune an old radio, you are changing the capacitance. The LC circuit is connected in parallel. In this configuration, it's a band stop circuit. It stops the frequency the LC circuit is tuned to and allows all other frequencies to pass thru. Basically it shorts out all the other frequencies and you are left with only the frequency it's tuned too. You are left with an amplitude modulated signal at the stage. A simple diode and capicitor is needed to demodulate this into audio signal. You now have a crystal radio. It is a fully functional AM radio requiring NO power source. Get a crystal radio kit to learn more. FM works similar but demodulation is more complex. NYCDA (talk) 16:55, 26 December 2007 (UTC)[reply]

• As for whether or not this stuff is "simple" or not, it depends what you are trying to do with it and how deep a knowledge of it you require. Remember that it took a gigantic laboratory (the
  Rad Lab) of top scientists to adapt radio waves for military use during World War II. Aspects of it are classical-physics simple, while other parts delve very deep into the quantum mechanical. --24.147.86.187 (talk) 19:30, 27 December 2007 (UTC)[reply
  ]