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March 4

iodine reacting with titanium?

I have a titanium mug/cup. After drinking some milk I forgot to rinse it out and it sat and became gross. I decided to add some tincture of iodine along with some water to help sanitize the cup. After adding this in letting it sit for about a minute it changed color and started a foul smell that made my nose sting. Is it possible it reacted with titanium to release some sort of harmful gas or substance? — Preceding unsigned comment added by 64.38.198.61 (talk) 00:08, 4 March 2012 (UTC)[reply]

The elemental iodine (I₂) is probably reacting with the titanium in the same way it reacts with aluminium. The product of that reaction would be titanium iodide, which is a very soluble salt, but it would also generate heat which volatilises some of the remaining elemental iodine. That stuff's not very nice to breath in. 203.27.72.5 (talk) 00:55, 4 March 2012 (UTC)[reply]

According to the titanium alloy article, titanium is usually alloyed with aluminium anyway. The reaction between iodine and aluminium is very spontaneous and can result in iodine vapor. 203.27.72.5 (talk) 04:28, 4 March 2012 (UTC)[reply]

The thing is I didn't see any actual cloud of vapor rise from the cup I just smelled it is that normal for this type of reaction? — Preceding unsigned comment added by 64.38.197.212 (talk) 12:04, 4 March 2012 (UTC)[reply]

If there had been enough vapor for you to see it in the air it would have been at a lethal concentration. Tincture of iodine only contains a few percent of iodine and not all of it is elemental either. I can imagine it would have been enough to smell but not enough to see by several orders of magnitude. 203.27.72.5 (talk) 20:14, 4 March 2012 (UTC)[reply]

Differences in sizes of thighs

why a person may have a diference of 2 inches from one thight to the other also from one arm to the other. what could be thje cause? — Preceding unsigned comment added by 98.87.197.110 (talk) 00:56, 4 March 2012 (UTC)[reply]

I suppose the question you really ought to ask yourself is; why should they be the same? You may be interested in symmetry in biology and laterality 203.27.72.5 (talk) 01:38, 4 March 2012 (UTC)[reply]

I'd expect the right arm to be a bit more muscular in the right-handed, and left in the left-handed. The thighs I'd expect to be more even, though, as walking is how they get the most exercise and we usually use both legs equally for that. If one leg is longer than the other, or they have a limp, then that may cause them to preferentially rely on one. I suppose certain types of work might do that too, like people who lay carpet and always use the same knee to push it along the floor. StuRat (talk) 01:45, 4 March 2012 (UTC)[reply]

Rod Laver is a prime example of arm disparity (the ATP World Tour site describes it thusly: "Laver's 5-foot-8-1/2, 145-pound body seemed to dangle from a massive left arm that belonged to a gorilla ...") Clarityfiend (talk) 05:50, 4 March 2012 (UTC)[reply]

Iron(III) electron configuration

Why does the Fe³⁺ ion have the electron configuration [Ar] 3d⁵ rather than [Ar] 4s²3d³? --108.225.112.237 (talk) 02:19, 4 March 2012 (UTC)[reply]

Because the pairing energy of the electrons is greater than the difference in the energy of the orbitals. If you draw out the orbitals in your two configurations above, you'll see that there are more unpaired electrons when you remove them from the 4s orbital first. See ionization of the transition metals 203.27.72.5 (talk) 02:57, 4 March 2012 (UTC)[reply]

Drawing it out looks like this:

Config	Drawing
Fe⁰	4s[↑↓] 3d[↑↓][↑][↑][↑][↑]
Fe³⁺ [Ar] 3d⁵	4s[] 3d[↑][↑][↑][↑][↑]
Fe³⁺ [Ar] 4s²3d³	4s[↑↓] 3d[↑][↑][↑][_][_]

203.27.72.5 (talk) 04:07, 4 March 2012 (UTC)[reply]

The interesting situation the "unpaired" idea raises and that OP doesn't mention as an option is why not 4s[↑] 3d[↑][↑][↑][↑][_] (all unpaired, but also populating the normal ordering 4s rather than just later-filled the 3d). As the article 203 linked mentions, stability also relates to "half-filled" (in this case d) not just "unpaired". DMacks (talk) 17:45, 5 March 2012 (UTC)[reply]

Unknown possible succulent from Southern California

Santa Barbara Mission succulent

This plant has grown from old cuttings years ago found at the Santa Barbara Mission, California. Does anyone know what the plant name is? Or is it possible a hybrid of some succulent?

LizardLis (talk) 03:53, 4 March 2012 (UTC)[reply]

Looks a lot like Kalanchoe fedtschenkoi. The link has almost no information but a couple of images that look a lot like your plant. Here is a link to more information. (why they refer to it as an 'air plant' seems a bit odd) Richard Avery (talk) 08:08, 4 March 2012 (UTC)[reply]

Noise-cancelling headphones and microphones

Am I right at assuming that the active noise control that these devices incorporate works better for continuous sounds (like a turbine running) than for unexpected sounds (like music or someone speaking)? XPPaul (talk) 12:53, 4 March 2012 (UTC)[reply]

Yes, there will inevitably be a delay between the microphone picking up the sound and the speaker producing the cancelling sound, due to the signal processing that needs to be done. That means varying sounds are less well cancelled than constant sounds. --Tango (talk) 16:51, 4 March 2012 (UTC)[reply]

In fact, there is a relationship between frequency response and settling time. Part of a noise-cancelling headset is a controller that attempts to drive the audible sound in the headset chamber to exactly equal the electronic signal input. A significant amount of the mathematics of control theory is the study of how rapidly such a controller can respond to changes, without introducing instability. Adding phase lag has a stabilizing effect, but also slows the response time. In the formal study of linear systems, an arbitrary sound signal can be decomposed into a sum of sinusoids - using the Fourier transform - so the engineers design parameters for a "continuous" tonal wave model, and then apply those results to an arbitrary sound signal that is a sum of sinusoids. I believe the Bose headsets even come with a bode plot in the documentation. Here's a very nice undergraduate research report, Noise Cancelling Headphones, detailing the simulations and showing actual measured results. Nimur (talk) 18:25, 4 March 2012 (UTC)[reply]

Heavier Objects falling faster due to air resistance

In a vacuum, I understand that under most scenarios, same-sized objects with different masses fall towards the earth with the same acceleration and arrive at the same time (ignoring the fact that the heavier objects attracts the earth towards itself with a tiny bit more acceleration than the lighter object).

But with air resistance, the heavier object will have a higher terminal velocity because the force of gravity on it is higher and hence, a higher velocity is require to create the higher air resistance required to balance the downward force. So in a situation where two balls of same size, but different masses are dropped from a very high location such that they have enough time to reach terminal velocity, the heavier ball will land on earth first, correct? But what if they are dropped not high enough, such that they do not have enough time to reach terminal velocity? Will they land on the earth at the same time? In other words, does the air resistance difference change the acceleration of the two balls or does it only affect the final maximum velocity? (Please treat this at the college-level) Thanks! Acceptable (talk) 17:43, 4 March 2012 (UTC)[reply]

What does "college level" mean? College-level physics students should be able to solve this analytically for an arbitrary linear or nonlinear air resistance model, so the question is irrelevant; "solve analytically in terms of the air resistance modeling operator." I know at least a few college physics professors who would kick you out of their classroom for asking such a question (because it belies that you don't know basic physics)! Physicists can be pretty stodgy; not everybody is as jovial Feynman. Anyway, we had to take these sorts of things in stride. For the college-level non-physicist, you might find it helpful to review some common, simple models of air resistance. The most common simple model is viscous drag, where the force due to air-resistance is proportional to velocity (not to mass); consequently the acceleration component due to air resistance has a mass dependency. Our Wikipedia article calls this "Stoke's drag," but I don't recall ever using that terminology. We called it "a first-order model of drag," where I went to college. If you believe this model is a good approximation for your scenario, then one of your questions is already answered: yes, the air resistance will cause a net effective acceleration that depends on mass. Anyway, let's try to turn you into a college-level physicist: Step 1. Write the equation of motion for the object, including the air-resistance term. Step 2. Solve this equation for time; so now we know how long it takes to fall from a certain height. (This answers your other question - each object will land at a different time. Solve the equation to determine exactly when this occurs). Step 3. Consider every parameter in that equation as a potential variable. (Hint: you will probably see mass, gravitational acceleration constant, distance the object fell, and air resistance model parameter. Step 4. Is there any combination of these parameters that you can vary, within reasonable physical limits, that would cause object 1 to land before object 2? I think you will see that the answer is obviously "yes." We can construct some height, some ratio of masses, and some model of air resistance, such that either object may land first. Using simple first-order drag, this is not possible; the lighter object is always falling more slowly at all times, and therefore always lands later. Nimur (talk) 18:40, 4 March 2012 (UTC)[reply]

You should use the word "denser", not "heavier". Yes, the denser object will accelerate faster, since the air resistance force is a smaller portion compared with the downward force (weight). F=ma applies, with F being the weight minus the air resistance. (You didn't mention the surface of the balls, so I will assume they are identical, as some surfaces will increase air resistance more than others.) StuRat (talk) 18:41, 4 March 2012 (UTC)[reply]

Apart from air resistance there are two more effects that makes the lighter body fall slower through the air. The first one is the bouyancy which decreases the downward force without changing the inertia of the body, and thus decreases the downward acceleration of the body. The second one is the induced mass which increases the inertia of the body without changing the downward force, thus also decreasing the downward acceleration. (There does not seem to be an article on induced mass in wikipedia!) The kinetic energy of the air flowing around a moving body represents a work done by accelerating the moving body, and this has the effect of an induced mass. For a sphere the induced mass is half the mass of the air displaced by the body. A sphere having mass M and volume V has the acceleration

a=g{\frac {M-\rho V}{M+\rho V/2}}

where ρ is the mass density of the air, and g is the gravitational acceleration near the surface of the earth. In vacuum we have ρ=0 and so a=g. These two effects makes the lighter body fall slower even when the air resistance can be neglected! Bo Jacoby (talk) 19:08, 4 March 2012 (UTC).[reply]

See Added mass for the article! :-) --Modocc (talk) 20:05, 4 March 2012 (UTC)[reply]

Pascal's principle

1) Is this a good description of Pascal's principle: if we construct a small cube with sides of area δA around a point in a fluid with pressure P, then fluid will generate a force pointing outwards from each side with magnitude PδA?

2) Does Pascal's law hold for compressible and/or viscous fluids?

Thanks 74.15.139.132 (talk) 17:58, 4 March 2012 (UTC)[reply]

You could probably derive (1) from the principle, but in itself does not sound like a good description. Did you take a look at Pascal's law? If the fluid is compressible then you will get a sound wave from the point that you apply pressure. This means that things will change with time. Also with this wave or viscosity you could well get loss of energy in your fluid, which will heat it up, and possibly change the pressure, confusing the simple formulation of the law. Once things have settled down though and stopped moving you should still be able to apply the law. (Some fluids like mud or tomato sauce or cornflour in water will not obey the law though, as they can have some solid like properties and maintain a pressure differential). Graeme Bartlett (talk) 22:05, 4 March 2012 (UTC)[reply]

Dimension

Is there a hierarchy of dimensions? I read about flatland and reading about string theory i think it implies about dimensions having a hierarchy. I think i can see how cubes are next to squares, but how are we sure that time is the 4th dimension? 203.112.82.2 (talk) 18:51, 4 March 2012 (UTC)[reply]

OP here, for clarification, i read 'about' flatland and string theory, but i dont understand them at all. im really no expert. 203.112.82.2 (talk) 18:53, 4 March 2012 (UTC)[reply]

I don't know what you read about flatland, but it's a novel about the life and times of personified geometric shapes. I don't really see what meaningful conjecture it could contribute with respect to the academic study of physical dimensions.203.27.72.5 (talk) 20:40, 4 March 2012 (UTC)[reply]

Its always mentioned when i try to find something about dimensions here in wp, i know its a novel, but we ca n use it as analogy for picturing other dimension. 203.112.82.1 (talk) 23:01, 4 March 2012 (UTC)[reply]

The suggestion that "time is the fourth dimension" is not really meaningful, as your choice of axioms can render it true, false or neither. Flatland and similar jeux d'ésprit remove a spatial dimension, but assume that time is still there - it is simply irrelevant to the question of dimensions. Most theories of space-time (such as Minkowski's) continue to treat time as fundamentally different from the spatial dimensions. Theories that require more dimensions may regard them as all interchangeable, or as grouped into fundamentally different kinds, and may have a hierarchy within a particular theory; but there is no theory-independent hierarchy of dimensions.--ColinFine (talk) 20:58, 4 March 2012 (UTC)[reply]

I'm copying here, a short discussion I've had one month ago about Flatland:

I would recommend reading Flatland and its "related works" for more info about dimensions. Time is not a dimension AFAIK. Disclaimer: I am not Stephen Hawking Von Restorff (talk) 11:52, 2 February 2012 (UTC)[reply]

You are completely wrong in your suggestion of this work, Von Restorff. Flatland is not about dimensions. It is about social classes. It's just a boring book with mere historical value. 88.14.194.205 (talk) 14:45, 2 February 2012 (UTC)[reply]

Lol. It is also about social classes, but also about dimensions. Did you read it? I wouldn't describe it as boring, but I do not know what kind of books you like. Most of the related works have little to do with social classes btw. Von Restorff (talk) 00:24, 3 February 2012 (UTC)[reply]

(I'm 88.14). I didn't say anything about the related books, which appear to be much better. And no, Flatland is no good work for someone trying to understand something about dimensions. Flatland is so much about dimensions as Terminator is about time travel. Neither has made any contribution to the topic or have anything to say about the physical side of it. Both have a different leitmotiv. WKB52 (talk) 13:29, 3 February 2012 (UTC)[reply]

Think what you want, but for those interested in physics, I still think Flatland is a rather flat book (which has little value as literature either. WKB52 (talk) 20:59, 4 March 2012 (UTC)[reply]

I'm not sure what you mean about a hierarchy of dimensions; my best guess is compactified dimensions? Wnt (talk) 02:25, 5 March 2012 (UTC)[reply]

Another relevant concept is symmetry. Nearly all known physical laws exhibit symmetry of one form or another. Chirality is a special case of asymmetry in many physical laws; it is, for example, why we must use a "right hand rule" and calculate cross-products. In this respect, the "x," "y", and "z" axes must be defined according to a certain sign convention; this is not a degree of freedom. Other than this sign convention, the axes are essentially interchangeable, as a result of symmetry of almost all relevant physical laws; for example, gravitation is spherically symmetrical; electrostatic force is spherically symmetrical; most of electrodynamics is almost spherically symmetric but exhibits chirality. The remaining fundamental physical forces are usually modeled with explicit discussion about their symmetry relations. Unfortunately for the "layman," by the time you get to study electroweak interaction and the strong interaction, symmetry relations are expressed in generalized coordinates, so you'll have to do a bit of fairly intellectually demanding work to relate this back to your conventional "x,y,z" cartesian coordinate space. Nimur (talk) 19:10, 5 March 2012 (UTC)[reply]

Fuel synthesis cost

How much would it cost, per retail gallon of gasoline, to synthesize fuel from waste carbon or oceanic carbonic acid? I know that it depends primarily on the cost of hydrogen production which fell very substantially because of some catalytic anode platings discovered around 2007, but I'm having trouble getting the numbers. Doty Energy's economic overview (not peer reviewed) suggests it would cost the equivalent of $55 per barrel of oil using nighttime wind and special patent-pending processes -- see question 13 on Doty's FAQ. But the peer reviewed Musadi et al (2011) says 109 UK pence per liter wholesale using traditional, non-patent processes and the North Sea market 24-hour contract price for existing offshore wind power plants. That was just 10 pence/liter above the price from petroleum when that article was written, and the price of oil has gone up substantially since. The Navy and Naval Research Lab do all kinds of calculations but never get in to the dollar cost. For example these 2012 Naval Research Lab slides say, "A 100 MW power plant could produce 41,000 gal. of fuel / day" (page 28.) If power is 5 cents per kilowatt hour, that works out to $2.93 per gallon on a shipboard plant. How much savings could you get on a land-based plant where presumably there would be more opportunities for waste heat recovery? How about if you were only synthesizing during off-peak electricity hours? 71.212.231.71 (talk) 20:51, 4 March 2012 (UTC)[reply]

One comment on comparisons: There are substantial taxes on gasoline/petrol. Whether these taxes will also be added to synthesized gasoline/petrol will therefore make a major difference in determining the net price. Subsidies might also come into play. StuRat (talk) 04:05, 5 March 2012 (UTC)[reply]

More importantly, what would be the point of turning electricity into liquid fuel? There will obviously be inefficiencies along the way, so you would be better off just using the electricity in the first place. If you want to produce liquid fuel using waste carbon, you'd be better off getting an algae to do the hard work for you. The costs of this are coming down rapidly. SmartSE (talk) 10:56, 5 March 2012 (UTC)[reply]

Oil is a very dense way to store energy. You need lots of very heavy batteries to run an electric car, or a fairly small tank of petrol. You can also make use of the existing cars and petrol stations, rather than needing to replace all the cars with electric cars and the petrol stations with plug sockets. --Tango (talk) 19:39, 5 March 2012 (UTC)[reply]

I'd like to see a serious projection of algal oil which suggests it will ever produce fuel under $50/gallon. The survey at [1] discusses several reviews and recent projects, none of which even come close to that figure in practice. On the other hand, wind power is about 1.5 cents per kilowatt hour during off-peak hours, which given the Naval Research Laboratory figure above would mean less than a dollar per gallon synthetic fuel. 138.86.194.162 (talk) 19:44, 5 March 2012 (UTC)[reply]

But the low off-peak price is because nobody has figured out a way to use all the electricity then. Once you put in place a way to use that electricity on a large scale, then there's no longer a reason for the utility to offer a reduced price in order to increase off-peak sales. Something similar happened with ethanol production from corn/maize in the Unites States. At first it looked like a good idea, because corn was dirt cheap. However, once large-scale ethanol production was started, corn prices shot up, not only increasing the price of ethanol, but also meat from animals which are fed corn, alternative crops which were scrapped in favor of growing corn for ethanol, etc. This type of diseconomy of scale must be considered. StuRat (talk) 23:12, 5 March 2012 (UTC)[reply]

The price will only be driven up to the point where it ceases to be worth doing - the profits basically get shared between the electricity companies and the oil-synthesising company (with the electricity companies taking as much as they can while still leaving it profitable enough for the oil-synthesising company for them to still consider it worth doing). --Tango (talk) 13:21, 6 March 2012 (UTC)[reply]

Right, and the price of regular petroleum would be expected to drop off, if the synthetic oil was produced at the same price, due to the increased total supply. The same thing happened with ethanol, where the price in the US is lower than gasoline by exactly the amount the fuel value is reduced. However, since about the same amount of petroleum is used to produce ethanol as if it were used directly to make gasoline, no increase in supply occurred, and the overall prices did not go down. StuRat (talk) 21:38, 6 March 2012 (UTC)[reply]

Tunnel through the center of the earth

Gravity as a function of depth

I really thought I saw an animated GIF here on the reference desk a year ago or so illustrating a tunnel pendulum of a person jumping into a tunnel that's been bored through the center of the earth to the other side; but I can't find the article! Do we have an article on this hypothetical tunnel? Comet Tuttle (talk) 19:13, 4 March 2012 (UTC)[reply]

Gravity train, perhaps? Tonywalton ^Talk 19:17, 4 March 2012 (UTC)[reply]

That was it; thank you! Comet Tuttle (talk) 18:27, 6 March 2012 (UTC)[reply]

If you wanted to avoid the magma, could you do the same thing with an Chord_(geometry) instead of a diameter, at least if it were close enough to the diameter? I guess you would have some side friction (toward the center but if it were a maglev type setup would it still work? In the illustration I imagine x at, say, 1 o'clock and b at, say, 5 o'clock -- would that avoid the magma? Would it work? --80.99.254.208 (talk) 19:45, 4 March 2012 (UTC)[reply]

Short answer: "yes". It's mentioned in passing in the first bullet point under Gravity_train#Mathematical_considerations (no matter where the endpoints are situated) and nicely illustrated here, which is a link from the article. Tonywalton ^Talk 19:54, 4 March 2012 (UTC)[reply]

"In the illustration I imagine x at, say, 1 o'clock and b at, say, 5 o'clock -- would that avoid the magma?" See the article on Earth's crust. The answer is no. 203.27.72.5 (talk) 20:55, 4 March 2012 (UTC)[reply]

If one really means magma (i.e. liquid rock), then the answer is probably yes unless you happen to intersect a localized pocket of magma near the surface. The mantle is a solid! Many people have the image that the crust floats on a sea of magma, but that image is simply wrong. Actual magma is rare, localized, and usually located near the surface. Most of the bulk of the Earth is a solid (specifically a rheid) that deforms and flows under stress in a manner similar to the way that glaciers flow, or how cheese deforms when pressure is applied. But it is a solid, nonetheless. Beneath the mantle, the core contains a very large volume of liquid iron (with nickel and some other things), but that is also not the same as magma. Dragons flight (talk) 22:26, 4 March 2012 (UTC)[reply]

But, if you dug a tunnel through the mantle, the tunnel's low internal pressure would allow the mantle to become magma. It would then of course fill the tunnel. 203.27.72.5 (talk) 01:48, 5 March 2012 (UTC)[reply]

No, because any meaningful tunnel would have to be reinforced to prevent its own collapse under the ambient pressure. If it has enough integrity to resist being crushed by the pressure, then it will equally well prevent magma formation. Of course, actually building such a tunnel is a wildly impractical proposition. Dragons flight (talk) 03:56, 5 March 2012 (UTC)[reply]

Note that as shown in the diagram I've inserted to the right, the gravity within the Earth is not directly proportional to the distance from the center, as you might expect from the combination of inverse-square gravity and mass proportional to the cube, given that the gravity of a sphere is equivalent to that of a point at its center of the same mass. Because the core is so much denser than the mantle, you would see simple harmonic motion only for objects in a tunnel somewhere within the core. Also, the tunnel needs to run from pole to pole or else they will of course veer out of it as the world turns. Wnt (talk) 02:22, 5 March 2012 (UTC)[reply]

Wouldn't it actually be your own radial velocity that causes you to hit the side (and not the world turning as such)? Being from the surface, you have a higher velocity than the parts of the earth further in, so as you go down, you will catch up with the wall which is moving slower. 203.27.72.5 (talk) 04:14, 5 March 2012 (UTC)[reply]

It's easiest to just model it in the Earth's own non-inertial reference frame. In that frame, you and the tunnel are motionless with respect to one another, and this choice of reference frame introduces no lateral fictitious forces, so you will fall straight down the tunnel, provided it goes straight through the center of the Earth. Switching back to an inertial reference frame in which the Earth, rather than the frame itself, is rotating, what you would see is that the tunnel is turning as you fall through it, at a sufficient rate to keep you the same distance from the walls at all times. Someguy1221 (talk) 04:22, 5 March 2012 (UTC)[reply]

I don't quite get what you're saying there. I see it as working like a spinning dancer pulling their arms in towards themselves and thereby rotating faster. As you go in closer to the Earth's center, won't you complete revolutions faster? 203.27.72.5 (talk) 05:11, 5 March 2012 (UTC)[reply]

If you stay in that mindset that right is right, left is left, even as you rotate, you will see that you are not falling straight down. Think about it. The moment you started falling, you were also moving to the side. Initially, the tunnel was moving to the side with you, as it is dug right into the ground you were standing on. And if you keep the same right/left you were going by when you started falling, you will find that the center if the Earth is no longer directly below you (seeing that requires you to rotate your frame of reference). And why is that important? Because the Earth is now pulling you sideways! Robbing you of that lateral momentum. It's difficult to imagine, and that's why it's best to put yourself in the most convenient reference frame, even if it is not inertial. Someguy1221 (talk) 05:54, 5 March 2012 (UTC)[reply]

In a non-inertial frame you'll be thrown off course by Coriolis force; in an inertial frame one can plainly see that the non-zero angular momentum at the surface will not be present when someone passes directly through the center. Wnt (talk) 17:03, 5 March 2012 (UTC)[reply]

Side question

Let's suppose someone comes up with a good reason to create some kind of tunnel or at least some kind of hollow tube going all the way through the earth. Presumably you would have to reinforce it somehow to keep it from closing up (with or without earthquakes). Presumably also the interior of the earth is pretty hot in at least some places. So my questions are: (1) What is the hottest temperature likely to be encountered; and (2) given that temperature, is there any substance that could stand up to it without melting and/or dissolving? ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:22, 6 March 2012 (UTC)[reply]

The hottest temperature that would be encountered is ~6000 K. There is no material that is solid at that temperature at standard pressure, but then again you're at the center of the Earth, where at over 3 million atmospheres, simple iron is solid. Now, normally when people talk about these tunnels it's a vacuum inside so you don't have to send the person or vehicle through what would get to be very dense air at the center. I don't know what that would mean for the material lining the inside of the tunnel. Someguy1221 (talk) 06:34, 6 March 2012 (UTC)[reply]

We usually assume a vacuum because the maths is really hard without one. In reality (to the extent that such a proposal could really happen, which isn't much) you would need to have roughly equal pressure inside and outside the tunnel or it would just get crushed. I can't see any way you could possibly have a tunnel strong enough to support the entire weight of the Earth above it (which is basically what the pressure is). The same issue then applies to the capsule you are sending through - the interior pressure of that would need to be 3 million atm or it would get crushed, so I think we can rule out sending humans that way. There might be some cargo you could send if you stretch the limits of imagination to breaking point. --Tango (talk) 13:26, 6 March 2012 (UTC)[reply]

You can always support the ambient pressure by reinforcing the tunnel with a rigid cylindrical shell that is "thick enough". On the back of my envelope, I get that a 8m diameter tube passing through the center of the Earth built from a hypothetical material with the same rigidity as room-temperature steel would "only" need to be about 5 km in diameter to resist being crushed by the weight of the whole Earth. Of course that's still wildly impractical, but personally, I'm a bit surprised that the tunnel walls would only need to be a few kilometers thick to resist the whole weight of the Earth. Of course, you'd have to cool the structure to maintain the rigidity and keep the low pressure hole in the center from melting, but that's a whole other problem. Dragons flight (talk) 13:59, 6 March 2012 (UTC)[reply]

Being 5km thick might resist the pressure, but there's other considerations with a structure of that size. Won't the magnetosphere induce a current in a conductive structure that long (I know next to nothing about anything electrical so I really don't know)? And you couldn't build a 5km thick steel pole that's one Earth-thickness in height due to the static load, so I can't imagine it working in what is essentially the world's biggest retaining wall. 203.27.72.5 (talk) 22:09, 6 March 2012 (UTC)[reply]

See [2] for a very funny analysis of a movie that depicted a "ship" that could travel through the earth, and why it won't work (at least not the way they showed it). 203.27.72.5 (talk) 22:02, 6 March 2012 (UTC)[reply]

Does conservation of energy for unsteady ideal fluid hold?

My text was able to show that for an ideal (incompressible and inviscid) and steady fluid in a gravitational field, the energy density $E={\frac {1}{2}}\rho u^{2}+\rho \chi +p$ is constant for any fluid element (ie ${\frac {DE}{Dt}}={\frac {\partial E}{\partial t}}+\mathbf {u} \cdot \nabla E=0$ ). Does this hold for an unsteady ideal fluid? If not, what causes the change in the mechanical energy of the fluid element? 74.15.139.132 (talk) 22:01, 4 March 2012 (UTC)[reply]

Edit: $\chi$ is the gravitational potential. 74.15.139.132 (talk) 22:01, 4 March 2012 (UTC)[reply]

Newton's cradle with 2 or more balls lifted

Newtons cradle 3 ball swing in a 5 ball system, notice the central ball swinging without any apparent interruption

The article Newton's cradle says "Intrigue is provided by starting more than one ball in motion". It sure does! In the case that 2 balls are lifted, how come the other side "knows" there were two? And not, for instance, one that was lifted twice as high? I can't find it in the article.

If there were 1,000 balls, and one would lift 76 balls on the left, how could the 923th ball possibly tell his neighbour to stay put, but "tell" the 76 to his right to start moving? If it's only done by transfering kinetic energy, how would one make a machine that can kick 10 balls to get 3 balls to stay put and 7 to go up? I you'd have extremely rigid balls, wouldn't you be able to transfer binary information (1 ball=0, 2 balls=1) faster than light? It's an amazing thing that I only knew as a gadget, but I completely missed it in high school curriculum. Joepnl (talk) 22:38, 4 March 2012 (UTC)[reply]

It is because of the conservation of energy and conservation of momentum. Yes, the balls do have to be very rigid. Bubba73 ^{You talkin' to me?} 22:49, 4 March 2012 (UTC)[reply]

The most incredible thing happens when you lift 2 balls but with a gap between them (use 2 hands) and let them go at not precisely the same time. The 1st ball you let go (2nd from the end) will act like the "end ball" and make the other end ball shoot up, then the 2nd ball you let go, which is the "real" end ball will make the 2nd ball on the other end shoot up, by then, the real end ball will be falling back down and contact the 2nd ball on it's way up, they'll rebound and all sorts of amazing seemingly chaotic, yet perfectly symmetrical patterns of clicks will transfer back and forth through the cradle. It doesn't last long because each collision takes energy out of the system, but for the few seconds it lasts, it's incredible to watch. Vespine (talk) 23:21, 4 March 2012 (UTC)[reply]

To expand on Bubba73's answer, the moving balls have both energy and momentum, both of which need to be transferred to the other balls. The balls are made of steel, so virtually no energy or momentum is lost by the balls deforming, and the way the balls are hung prevents any sideways motion confusing the results. Kinetic energy is related to mass and speed squared, while momentum is just related to mass and speed. This means that one ball dropped from 4 cm may have twice the energy of two balls dropped from 2 cm, but it has different momentum, and this combination of mass and energy is how the balls at the other end "know" how many balls were moved. If you try and solve the maths for Newton's cradle, you should find there is only one way to conserve both momentum and energy - have exactly the same number of balls move.

The reason it doesn't violate relativty, incidentally, is that the energy still travels as a wave from one to the other - the balls are held together by electromagnetic forces between the atoms, and electromagnetic forces are limited by the speed of light, so the compression wave that ripples through the balls must be slower than light too. Smurrayinchester 00:02, 5 March 2012 (UTC)[reply]

Thanks a lot for your answer. Actually, the relativity part is the easy one, the part that sounds logical. The math is way beyond my knowledge, I'm hoping this can be explained in non-mathematical terms? Let's say I'm hitting a golfball with a heavy club with 10 joules of kinetic energy, and than with a lighter (but faster) club with again 10 joules of kinetic energy. Wouldn't that "feel" exactly the same to the ball? If all energy and momentum was transfered into the ball without any losses and than you look at the ball you can tell by its speed that it has x kinetic energy but you can't tell if it was hit by a heavy or a light club, could you? (I mean that the ball can't tell another ball what happened to him) Joepnl (talk) 00:57, 5 March 2012 (UTC)[reply]

A golf club can't transfer all of its energy and momentum to the stationary golf ball unless it has the same mass as the ball. In an elastic collision the golf club will always keep some of the forward momentum and kinetic energy assuming it's heavier (and how much it keeps depends on exactly how the masses compare). For both momentum and kinetic energy to be conserved, this has to be so. The ball will "feel" the difference between the clubs of different masses in that the lighter club will give the ball more of its energy. Rckrone (talk) 03:22, 5 March 2012 (UTC)[reply]

"A golf club can't transfer all of its energy and momentum to the stationary golf ball unless it has the same mass as the ball" How can you reconcile that with the image shown above of the 3 balls transfering all of their kinetic energy to 2 balls? 203.27.72.5 (talk) 04:17, 5 March 2012 (UTC)[reply]

They don't. Two balls transfer their energy to another two balls, and the third ball never stops moving. Someguy1221 (talk) 04:24, 5 March 2012 (UTC)[reply]

Oh yeah, I see that now. 203.27.72.5 (talk) 05:15, 5 March 2012 (UTC)[reply]

I've seen the light, thanks all! Joepnl (talk) 20:23, 5 March 2012 (UTC)[reply]

Why don't I bruise?

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. TenOfAllTrades(talk) 23:09, 4 March 2012 (UTC)[reply]

Oh my... I don't recall asking for any advice specifically, but let me rephrase the question: Why do some people bruise less than others? Happymulletuk (talk) 23:16, 4 March 2012 (UTC)[reply]

Did you look at the bruise article? There is a fairly thorough looking section that appears to answer your question.Vespine (talk) 23:28, 4 March 2012 (UTC)[reply]

I did yes, but my question relates to those amongst us that don't appear to bruise at all (or at least very little). Specifically, I'm interested in the mechanism(s) or lack thereof that causes some folks to swell, but not bruise. My original question stated this quite well I thought, but alas, it was edited out of existence. Happymulletuk (talk) 23:38, 4 March 2012 (UTC)[reply]

And so that section explains that "Condition and type of tissue, age, gender, color of skin, disease, location, force and genetics" all play a part in how you will "appear" to bruise. I don't see how you could get a more specific answer.. Vespine (talk) 00:43, 5 March 2012 (UTC)[reply]

Since bruising is caused by blood leaking out of ruptured capillaries, here are some factors affecting the actual size of bruises, not just their appearance:

1) How "thick" the blood is. People given blood thinners often bruise far more easily than normal, even without any apparent trauma. This also can happen in the elderly.

2) Platelet count (since platelets help form blood clots) and other disorders which prevent rapid clot formation. Hemophiliacs may get large bruises easily.

3) Blood pressure, with higher BP causing more bruising. StuRat (talk) 03:33, 5 March 2012 (UTC)[reply]

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