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May 12[edit]

Eye of a hurricane question[edit]

I was not able to find the answer to this query in the article on Eye (cyclone). Anyway, I remember a movie from many years ago where one of the characters (it may have been Larry Hagman)was in his yacht and got caught up in a hurricane. Although he tried to out run it, alas, the hurricane caught up to him. However, he was able to ride out the hurricane and survived to find himself in the eye. While in the eye, he was rescued by a navy sub!

OK. I'm quite certain that this could not happen but am wondering about the physics of why not. I am sure that there have been boats, ships, etc. who have had to weather the storm and found themselves in the eys. What would prevent a sub from surfacing within the eye (to either affect a necessary rescue or to perform some scientific experiment/observation)? 99.250.117.26 (talk) 00:55, 12 May 2011 (UTC)[reply]

Nothing. Though, it would be risky. Plasmic Physics (talk) 01:18, 12 May 2011 (UTC)[reply]

I don't even see why it would be risky, except that there would be a certain amount of wave activity. Looie496 (talk) 01:57, 12 May 2011 (UTC)[reply]

A large amount of wave activity. Not only can there be waves over 100 feet high, but in the eye the waves are chaotic and coming from all different directions, since the wind which causes these waves is convergent towards the eye. I remember seeing footage from a hurricane hunter (I believe it was Hurricane Gilbert showing whitecaps around 50 feet high breaking over each other in all directions: not the kind place you'd want to have two water craft anywhere near each other. That said, how strong was this hurricane? In a weak one it might be plausible.

The question concerns the surfacing of the sub, ignoring interactions with the other vessel. Even subs are a risk of capsizing in such turbulance. I also say it's risky, because you have track the eye - you don't want to be caught outside of the eye. Plasmic Physics (talk) 03:22, 12 May 2011 (UTC)[reply]

Also notice that the yacht would have had to pass through the eye wall, which is absolutely the worst part of a hurricane. Perhaps in the story he did that while it was only Category 1, then it grew to something bigger ? StuRat (talk) 06:54, 12 May 2011 (UTC)[reply]

From a meteorological standpoint, it depends on the strength, direction, speed, and physical size of the hurricane, among other things (intensity trends, etc.). The eye wall is most severe in the northeastern quadrant, and slightly weaker as you get further south and west. Also, some hurricanes are lopsided, especially as they begin to transition into extratropical storms. In such a morphing cyclone, the northern semicircle may possess hundreds of miles of convection while the lower-level center of circulation is exposed on the southern side. Obviously, as noted above, waves probably wouldn't be all that high, even in the eye, where waves from all directions converge. A marginal Cat 1 may only have 15 or 20 foot waves, especially in the formative stages, and the 100-foot trough to crest monsters we hear about are exceptionally rare, and there's no way most vessels would be able to ride that out in the first place. Juliancolton (talk) 20:18, 12 May 2011 (UTC)[reply]

Hospital planning[edit]

questions are........1eqment planning for 3oo bedded hospital? 2-role of varius people involved in hospital project? 3-planning of various department like .... opd , casualty, blood bank , cssd ,i.c.u. cath lab ,it department, how to planning for 3oo beed hospital ans n brief detail.... 4-norms for operation designing in hospital? —Preceding unsigned comment added by Twinkeptel87 (talk • contribs) 06:25, 12 May 2011 (UTC)[reply]

This looks like a homework question. The Wikipedia Reference Desks don't answer homework questions - if we did you wouldn't learn anything! Tell us how far you have reached and what aspects are causing you trouble and we will give you some hints to help you on your way. Dolphin (t) 07:50, 12 May 2011 (UTC)[reply]

Our article Hospital has a section about Hospital#Buildings that may interest. This article and these guidelines give detailed planning information. Cuddlyable3 (talk) 10:30, 12 May 2011 (UTC)[reply]

Tips on building an LCD[edit]

I'm considering building my own Tiger LCD game, but this would require building an LCD from scratch in order to etch the correct patterns onto the electrode layer. I could use an off-the-shelf graphic LCD, but I'd like to stay true to the actual Tiger games.

I found a post that pretty much describes the way I thought of doing it at http://www.avrfreaks.net/index.php?name=PNphpBB2&file=viewtopic&t=40355 (ignoramus, Jul 17, 2006 03:06 PM), but the post is more of an overview than a guide.

My questions are:

What is the recommended liquid crystal for this purpose?

How exactly do you coat the glass with the liquid crystal?

For the polarizers, would this work if it is cut and both pieces are in the correct orientation?

And where can you buy the liquid crystal and ITO glass? I see Sigma Aldrich has both, but they are quite expensive. They sell ITO coated plastic film for cheap, but it is also very thin, and I'm not sure if it would survive etching (and it also couldn't be heated, I assume).

[Links to Sigma Aldrich pages :

ITO: http://www.sigmaaldrich.com/materials-science/material-science-products.html?TablePage=9548901

Liquid crystals: http://www.sigmaaldrich.com/materials-science/material-science-products.html?TablePage=16378837

Links to other sources are appreciated.]

I was unsure if this should go in computing or science, but since I am asking about chemical sources and the use of the chemicals, I decided it was more apt to be here.

Thanks in advance! —Preceding unsigned comment added by EEPROM Eagle (talk • contribs) 14:14, 12 May 2011 (UTC)[reply]

May I ask why you don't just use a standard LCD with a grid of pixels ? StuRat (talk) 19:04, 12 May 2011 (UTC)[reply]

No reason that I can't. The choice is basically for the accuracy of how mass produced LCD games usually work, and for learning and fun. I'd certainly be taking a much more difficult (and expensive) route this way. :) EEPROM Eagle (talk) 19:56, 12 May 2011 (UTC)[reply]

This is a very specific question, I think you'd be lucky to find someone here who has done specifically what you are looking for.. Why don't you join the avrfreaks board and contact the user? I've worked on a few AVR projects and was an active member there a few years ago, i have very fond memories of that board, probably the best support forum I've ever participated in. Some truly smart, helpful and funny folks there. Vespine (talk) 23:36, 12 May 2011 (UTC)[reply]

My partial understanding of Liquid crystal displays is that the pattern is not etched but deposited as a transparent conducting film of Indium tin oxide after painting a mask on the non-conductive areas. The liquid crystal is not painted on the glass, it is just a drop of fluid that is squeezed between the plates which are pressed together with thin spacers. Cuddlyable3 (talk) 13:26, 13 May 2011 (UTC)[reply]

I contacted him and he said he used an eyedropper to add the liquid crystal, so that makes sense with what Cuddlyable3 said. In addition to that post, I had read this wikiHow article about etching the glass, so I think it may work either way (etching the glass or creating a mask before the ITO is deposited onto it.

I still don't know where to find the supplies or what liquid crystal to get. I've looked on Alibaba for ITO glass, but am not sure about minimum orders or payment. I also checked there for liquid crystal but couldn't find much.

Thanks for all the help you've provided so far! EEPROM Eagle (talk) 22:34, 13 May 2011 (UTC)[reply]

This supplier may be what you seek. Cuddlyable3 (talk) 23:01, 13 May 2011 (UTC)[reply]

Hmmm, looks interesting, but they seem to list more uses regarding temperature sensitivity, and the article here for cholesterol, under "cholesteric liquid crystals", says that it "changes colour when its temperature changes." I could see that being an issue. :) EEPROM Eagle (talk) 01:31, 14 May 2011 (UTC)[reply]

Spy plane[edit]

I would like to know the name of the spy plane that have a large radar dome on top. Note: its not a turbo prop plane. Link to article appreciated. --Tyw7 (☎ Contact me! • Contributions) → Gotta catch 'em all! 15:21, 12 May 2011 (UTC)[reply]

That would be AWACS (airborne warning and control) planes - they're not exactly "spy" planes, their role is monitoring air traffic for unauthorized flights or missile attacks -- Ferkelparade π 15:23, 12 May 2011 (UTC)[reply]

THanks. No wonder I couldn't find it. I was trying to search for spy planes. --Tyw7 (☎ Contact me! • Contributions) Wikipedia, the free encyclopedia. 15:42, 12 May 2011 (UTC)[reply]

The E-1, E-2 and E-3 had large external radomes. A few other "nonstandard" American military aircraft have also had radomes mounted during testing and special operations. More recent advances in directional antennas, synthetic aperture RADAR, and phased array antennas have mostly obsoleted the need for an external radome. Nimur (talk) 19:43, 12 May 2011 (UTC)[reply]

I was looking for the Boeing E-3 Sentry. --Tyw7 (☎ Contact me! • Contributions) → Do a good turn daily 21:39, 12 May 2011 (UTC)[reply]

mass of an atom in kinetic collisions[edit]

We note that electrons occupy most of the space in a molecule or an atom, but the nuclei have most of the mass. I am curious then, but the kinematics of a collision between two molecules or two atoms -- how exactly does an electronic collision lead to the movement of nuclei? We note that if a very small atom or molecule (like hydrogen) bounces off a very heavy one, the small atom cannot lose much of its kinetic energy, due to the law of elastic kinetic collisions base on the ratio of the two masses. How is this mass brought into play, if it is stored in the nucleus? Elle _{vécut heureuse} ^{à jamais} (be free) 19:48, 12 May 2011 (UTC)[reply]

The nuclei repel each other because they have similar electric charges. Dauto (talk) 20:25, 12 May 2011 (UTC)[reply]

This doesn't explain the Franck-Hertz experiment. I'm pretty sure the nuclei never come close to each other for nuclear-nuclear repulsion to be a significant force. Elle _{vécut heureuse} ^{à jamais} (be free) 20:50, 12 May 2011 (UTC)[reply]

(edit conflict)To answer the other question about mass distribution: for single atoms, a solid sphere with evenly distributed mass and a hollow, massless shell with a concentrated mass at the exact center will behave exactly the same. Atoms, however, only have one degree of freedom for its kinetic energy, that being translation, or movement through space. Molecules, depending on their shape, can have additional degrees of freedom, which include rotation and vibration. The organization of an atom certainly does have an effect on the results of collisions; however in the bulk substance, it makes MUCH more sense to think of these effects statisticly, rather than trying to treat them as a series of discrete events. Take something as simple as two gas-phase water molecules colliding, but from a classical mechanics sense. Depending on the angle, orientation, and incoming speed, such a collision could result in exhanges of kinetic energy between the water molecules which could result in either molecule changing its linear speed or trajectory (translation) or it could set one or the other molecule spinning (rotation) or it could introduce some bending and flexing within the molecule (vibration). It turns out that things like vibration and rotation are quantized at the molecular level, so treating them in a "classical, Newtonian" sense doesn't actually make much sense. Treating these motions in a statistical sense, as Ludwig Boltzmann's life work was about, makes more sense and quantum mechanical treatment of these modes of motion is how we actually make these calculations, see Rigid_rotor#The_quantum_mechanical_linear_rigid_rotor for a discussion of the mathematics behind calculating the rotational dynamics of a diatomic molecule. This is why temperature exists as a physical property in the first place; considering all of the multitude of motions involved in a substance on a individual molecular level isn't particularly helpful in understanding the behavior of the substance. --Jayron32 20:53, 12 May 2011 (UTC)[reply]

(edit conflict) Atoms and/or ions collide all the time; we have different names for various scenarios, because we use different physics and different equations to model them. If electrons are present on one atom, but not on the other, the phenomenon is called Rutherford scattering. If electrons are present on both atoms, the collision is described via chemical kinetics or "reaction kinematics" (a quantum chemistry treatment is needed for all but trivial examples). If electrons are not present around either atom, the result is nuclear scattering, whose effects can be dictated by either electrostatic or nuclear forces. Nimur (talk) 20:55, 12 May 2011 (UTC)[reply]

okay, it's just that I've been making these assumptions since forever. They tell me that hard spheres is a reasonable simplification -- I can believe that. But why should it behave like it has homogeneously distributed mass when most of it is in the centre? What about Franck-hertz -- which sounds more like electron-electron scattering, but instead the energy lost during elastic collisions is closer to a collision between a 0.000548 amu e- and a 200.59 amu mercury atom? Surely the electron is only encountering the electron cloud? Why does the electron cloud behave like it has so much mass? Elle _{vécut heureuse} ^{à jamais} (be free) 21:07, 12 May 2011 (UTC)[reply]

Because the electrons are so light compard to the nucleus, the electron cloud can adjust itself very fast when a charge approaches it. This is the rationale behind the Born-Oppenheimer approximation. The electron clouds at some given time are approximately the solution of the stationary Schrödinger equation, where you take the positions of the nuclei fixed. To a good approximation, the nuclei then interact with each other via an effective potential that you get when you add to the mutual Coulomb interaction the screening of the perturbed electron clouds.

You can exactly work out the case of two Hydrogen atoms in perturbation theory. If you take the two protons to be at fixed positions, then each electron is in the field of two protons and the other electron. If you treat the extra terms as a perturbation, you find that to second order in perturbation theory, the ground state energy is shifted. The ground state of this system drops in energy as the two atoms become closer to each other. Since this is the energy at rest of the atoms, this is the effective potential energy of the two atoms (this is the van der Waals interaction). So, atom-atom scattering at low energies can be described using this effective potential. But you see that this works precisely because the electrons are so light. Count Iblis (talk) 03:03, 13 May 2011 (UTC)[reply]

So is the hard sphere assumption (neglecting H-H bond formation -- just looking at a collision event) valid for this case? In the Franck-Hertz experiment, how does an energetic "free" electron feel the mass of the mercury nucleus? Elle _{vécut heureuse} ^{à jamais} (be free) 05:09, 13 May 2011 (UTC)[reply]

Well, what is de-Broglie wavelength of a 10 eV electron compared to the size of an atom? So, if such an electron interacts with the electron cloud, what you get is (with significant probability) a coherent interaction with the entire electron cloud that takes place on a time scale that is larger than the time scale on which electron cloud perturbations relax to new equilibrium states by transfering momentum to the nucleus.

So, you have a significant probability of an elastic interaction of the electron with the entire atom. You can compare this to how you have a significant probability that a neutron colliding with a nucleus in a crystal lattice will not produce a phonon, and instead the whole lattice will recoil as a whole. So, how can the neutron then feel the mass of the entire lattice instead of the single nucleus? Of course, this is all due to basic quantum mechanics, just compute the Debye Waller factor at zero temperature. Count Iblis (talk) 16:01, 13 May 2011 (UTC)[reply]

Identifying distant sources of light[edit]

A recent article (BBC News) noted that a gamma-ray emission (lasting six days), emanating from a small region of the Crab Nebula, had been detected in April. The article caused me to wonder about two questions:

Can scientists determine definitively whether the flare originated in the Crab Nebula or from a source behind (from our point of view) the nebula?
What process do scientists use to calculate, based upon observation of a distant light source, the distance from Earth of that distant object? Perhaps we can, if it will simplify the explanation, ignore considerations introduced by the metric expansion of space?

Thank you, -- Black Falcon ^(talk) 20:37, 12 May 2011 (UTC)[reply]

Nuclear processes do not emit light evenly over a broad spectral area, but emit light with the characteristic emission spectra of the involved elements. Astronomers can examine the spectral distribution of a light source, determine which elements are involved, and so measure the red shift of the received light. This allows them, by Hubble's law, to determine how distant the source is. -- Finlay McWalter ☻ Talk 20:46, 12 May 2011 (UTC)[reply]

Hubble's law would only apply in general, there are exceptions. So, you want to look at the red shift of the Crab Nebula and see if it matches the red shift of the gamma-ray source. If they match, chances are it's in the Crab Nebula. StuRat (talk) 20:49, 12 May 2011 (UTC)[reply]

Yes, that's true. Not only for cosmic oddballs that don't move away as one would generally expect, but particularly within gravitationally bound structures like galaxies. The Andromeda Galaxy article briefly discusses measurement of its rotation by observing the relative redshifts of the approaching and receding sectors of its disk, over and above the overall redshift of the whole thing. -- Finlay McWalter ☻ Talk 21:27, 12 May 2011 (UTC)[reply]

As a simple example consider the characteristic double-line emission spectrum of sodium (img). If you see a double-line, but at a different frequency, it's likely that you're seeing sodium, but red-shifted (or occasionally blue-shifted) due to its relative motion; you measure the distance between where you see the double line and where it should be (from your perspective) and that tells you the redshift. In practice things are more complicated, as there are many elements emitting, and some elements have much more complicated spectra (like iron). And some "sources" are really multiple things moving with different relative speeds, that happen to occupy the same piece of the sky (so, for example, you might see the sodium lines redshifted to two different places, because they're coming from two unrelated sources). But figuring out what's what is nevertheless tractable. -- Finlay McWalter ☻ Talk 20:58, 12 May 2011 (UTC)[reply]

Thank you both! Your comments and the articles to which you pointed—Hubble's law and Redshift, and from them Doppler effect—have clarified the issue quite a bit. Again, thanks, -- Black Falcon ^(talk) 21:49, 12 May 2011 (UTC)[reply]

This is not a new gamma ray source, but fluctuation in the brightness of a source that is always visible (and is, in fact, the brightest gamma ray source in the sky). Crab Nebula#Distance talks about how the distance to the Crab Nebula is measured. Cosmic distance ladder talks about the various methods of judging distances in astronomy. Hubble's law can't be used to find the distance to the Crab Nebula; it only applies at the scale of galactic superclusters and larger. -- BenRG (talk) 22:01, 12 May 2011 (UTC)[reply]

2011 Mississippi River floods[edit]

How many coal slurry ponds and how much coal slurry has been washed into the Mississippi River Valley so far in the 2011 Mississippi River floods? Can the Old River Control Structure keep most of Baton Rouge and New Orleans from flooding? 99.39.5.103 (talk) 22:35, 12 May 2011 (UTC)[reply]

For the latter, the ORCS is not primarily a flood control structure -- the Bonnet Carré Spillway and the Morganza Spillway are the two primary flood relief systems in Louisiana. The Bonnet Carré is already 2/3rds open and the US Army Corps of Engineers is readying the Morganza if needed. Current projections put the current floods below the Project Design Flood, the theoretical flood which these structures are intended to counteract. That said, even ideal operation of all structures along the Mississippi will not prevent Baton Rouge and New Orleans from flooding; indeed, the river gauge at Baton Rouge is already 3.5 feet into "major flood" level with New Orleans hovering at flood level. — Lomn 03:50, 13 May 2011 (UTC)[reply]