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April 16

Saving the Titanic

If water had been pumped into the rear compartments, to prevent the nose from diving to the point where water flowed over the bulkheads, could the Titanic have been saved ? If not, could the sinking have been delayed ? StuRat (talk) 02:37, 16 April 2012 (UTC)[reply]

That's a novel idea. Unfortunately, there would still be nothing to stop the water from continuing to come in through the gash in the side and the bulkheads would have been topped eventually. But the ship might have sunk more sideways (i.e. more like the Costa Concordia) rather than perpendicular. Maybe that would have bought them some time. However, if they had thought of that kind of thing ahead of time, they probably would have done some other things better too. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:21, 16 April 2012 (UTC)[reply]

But most other things would have increased cost, like more lifeboats, compartments which sealed at the top or down the center, a double hull, etc. Planning costs very little money. (Testing the rivets wouldn't have cost much either, so that they should have done.)

I had assumed that the top of the bulkheads was above the waterline, initially, or else those bulkheads would be rather useless. Are you saying that the top would still have dived below the waterline, even if the rear compartments were equally flooded ? Also, why would this action cause it to list to the side ? StuRat (talk) 14:46, 16 April 2012 (UTC)[reply]

If we speak about preventing it, then not braking at turning the same time would probably have saved her.--79.116.78.83 (talk) 16:19, 17 April 2012 (UTC)[reply]

StuRat has asked a very valid question. If the damaged area of the ship could have been prevented from sinking deeper the rate of flow of seawater into the ship would not have kept increasing. (The rate of flow of any fluid through an orifice or similar is proportional to the pressure difference on either side of the orifice.) A few years ago I saw an excellent TV documentary in which this idea was investigated. The producers built a scale-model of Titanic out of perspex and floated it in a water tank. They investigated the effect on trim, depth of the damaged area etc. of adding water to the various compartments in between the watertight bulkheads. The outcome was that flooding one or two of the rear compartments before the bow compartments became too full showed promise, but the damaged area was too long, and too many of the watertight compartments were filling with water, so that this strategy wouldn't have saved the ship. Dolphin (t) 03:30, 16 April 2012 (UTC)[reply]

I doubt that they would have had the machinery for doing such pumping anyway, even supposing they would have been game enough to try such an experiment. When, as a ship's command officer, you are confronted with such an emergency, you execute pre-planned action, not try impromtu experiments. What perhaps of more interest is this: It has been established that the hull plate rivets were substandard in material (too brittle), and were not installed correctly. If the rivets were good rivets and installed correctly would the ship have sunk? The answer seems to be "no, probably, well, maybe it might have". I haven't seen any definitive answer on this. And then there was the order given to go "full astern", which becasue the centre shaft could not go in reverse, was really an order to "go half astern and kill most of the rudder effectiveness". I haven't seen any definitive answer on whether another order such as "port full astern" would have resulted in a miss, or just a bigger bang. Remember the ship did not get a full trials test, where such emergency orders were normally tested. Ratbone60.230.222.160 (talk) 08:03, 16 April 2012 (UTC)[reply]

I seem to recall reading somewhere a claim that it would have been better to have rammed the iceberg head on. This would have destroyed the bow, but left most of the remaining watertight compartments undamaged, and the ship afloat. Hardly an instinctive reaction though. AndyTheGrump (talk) 14:18, 16 April 2012 (UTC)[reply]

I heard that, too, but the problem is that the captain lacked the info to make such a decision. (He would have to have known the damage would be so severe from a glancing blow that it would sink the ship.) StuRat (talk) 14:51, 16 April 2012 (UTC)[reply]

It might not have worked anyway. Upon head-on collision, a shock wave would have propagated back through the hull. Given the bad rivets, the shock wave could have caused plates to fall off well back from the point of impact & crush zone. It is speculated by some that not all plates came off because of direct impact with the iceberg as it is. Don't forget that the emergency engine & rudder changes were made, as they should have been, by a relatively junior officer - the captain was not on duty at the time. The officer in command may or may not have considered that a head-on collision would have resulted in the immediate death of possibly hundreds of 3rd class passengers, whereas steering around the iceberg had a chance of complete success. Ratbone121.221.27.78 (talk) 23:46, 16 April 2012 (UTC)[reply]

Ramming this low-ice- burg or “Growler" (as it was technically termed) head-on would have made sense. The damage would just have been in the bow (the Titanic was made of steel which flexes – other wise' wave action would brake a ridged structure up). Instead, a long gash traversed along several watertight bulk heads. However, if the bullhead doors had been left open, then the inward flow of sea water could have gone to the rear of the ship, where sufficient pumping capacity was situated to keep the vessel afloat until dawn. The benefit of hind-sight made all this all the more poignant. Lateral thinkers might even say, that if just the first class passengers alone, had only thought for themselves, then bed linen could have been dropped over the side to restrict the inflow of water. Ballard's research suggest that the gash was only about 2 to 4 inches wide in depth and close to the water line. Sheet material would easily get sucked against it. Had Jacques Cousteau with his lateral thinking mind had been on board, I'm sure that Titanic would have reached America still afloat (but perhaps with not a bottle of good Château left on board). The simplest explanation for such a monumental disaster may simply be that it represented such new technology that the designer and operators had not properly thought out the worst case scenarios and these oversights lead to a domino effect of bad decisions. --Aspro (talk) 22:11, 17 April 2012 (UTC)[reply]

Keeping a ship afloat after taking a hole below the waterline is not such a "new technology". Even wooden ships from the Age of Sail were saved after accidents/cannonballs pierced them under the waterline, with linen, cotton and pumps. --79.116.85.11 (talk) 16:41, 18 April 2012 (UTC)[reply]

I doubt that dropping a bedsheet over the side has a realistic chance of plugging up a 4-inch gash. Now maybe if the passengers had drunk all the wine, recorked the bottles and left them in the flooding chambers as a flotation device? :) Wnt (talk) 18:16, 20 April 2012 (UTC)[reply]

Spartan by Wavefunction Inc.

Does anybody here know how to interpret the calculation outputs generated by the Spartan molecular modelling software? I have to use this program for my university paper, and they do not provide any secondary sources. Plasmic Physics (talk) 08:03, 16 April 2012 (UTC)[reply]

I do, not at a professional level, but I've done it before many times. What do you want to know, orbital energies? --Ben (talk) 09:25, 16 April 2012 (UTC)[reply]

How to guess which MO coefficient set corresponds to a particular MO image. How to create a wave function from the MO coefficient sets, using |0.30| as the minimum allowed value. How to translate a wave function into a bond-type (sigma, pi, anti, non, etc.). How to explain how Spartan determines the number of basis functions and electrons to use in a semi-empirical calculation.

Plasmic Physics (talk) 12:46, 16 April 2012 (UTC)[reply]

There should be a list of MOs, with their energy in Hartrees and eV, the LCAO coefficient for each AO making up the MO, and some other stuff. There's probably a number for each orbital, starting with 0 or 1 for the lowest energy one (core orbitals, the 1s orbital of the heaviest atom, probably). The energies are with respect to the HOMO, if I remember correctly. Anything with a negative energy is bonding, zero energy is non-bonding and positive energy is antibonding.

Guessing AO coefficients from an MO image can be tricky. The smaller the coefficient on an AO, the smaller the part of the MO on that atom.

The wavefunction corresponding to an MO is the sum of (the wavefunction for each atomic orbital multiplied by its coefficient). The wavefunction for an entire molecule is the sum of its MOs, probably multiplied by some coefficient, but I'm less sure about this. Chemists normally deal with things at the MO level.

You can tell bonding/non-bonding/antibonding from the MO energy, as mentioned above.

You set the number of electrons when you make your model in Spartan. Unless you tell it to make your molecular a dication or a trianion or whatever, it assumes the whole thing is neutral and gives each atom the same number of electrons as it has protons. You can tell sigma and pi bonds by looking at their images. Look for no nodal planes (sigma), one (pi), two (delta) and so on.

I attended a workshop on computational chemistry a few years and kept my report: http://www.benjamin-mills.com/bristol/theoretical-class-2of3-MO-calculations.pdf. It's not Spartan, it's Gaussian, but the output and interpretation are similar, I think. --Ben (talk) 14:03, 16 April 2012 (UTC)[reply]

I'm not sure that, that is the way it works. I had a look at model answers to an old semester test, and attempted to reverse engineer them to develop a method for solving the problems. I figured out that if all coefficients for an atom in a MO set are below that limiting factor, then that MO is non-bonding. Whether the MO eigenvalues/energies are positive or negative does not seem to factor into it. I still need to distinguish between the remaining MO types (sigma, *sigma, pi, *pi). Plasmic Physics (talk) 02:49, 18 April 2012 (UTC) Trifluoroborane: how can I tell that Ψ₇ = −0.46Ψ_2py(B1) − 0.51Ψ_2py(F1) − 0.51Ψ_2py(F2) − 0.51Ψ_2py(F3), is a π orbital, and not a π antibonding orbital? Plasmic Physics (talk) 04:47, 18 April 2012 (UTC)[reply]

Rather than reverse engineering model answers, why don't you just ask the lecturer who set the questions? I'm assuming in the case of BF₃ the y-axis is perpendicular to the FBF plane. If so, you've got a π bonding MO because the consitutent AOs are all the same phase (they're all multiplied by negative coefficients). Sketch a quick diagram to convince yourself. Or just use Spartan to visualise the MO and see for yourself. What is the energy of Ψ₇? --Ben (talk) 11:12, 18 April 2012 (UTC)[reply]

I tried asking the lecturer, but he said that if I don't understand his lectures, that there is nothing that he can do about it, except direct me to the software manual. From the provided coordinate reference, the molecule is alligned along the xz plane with one BF axis alligned with the z axis. Ψ₇ has an energy of -18.45932 eV and is a confirmed pi orbital, but what about other MO wavefunctions containing no py components, how do I determine them? Plasmic Physics (talk) 13:40, 18 April 2012 (UTC)[reply]

I'd be very annoyed if a lecturer dismissed my query like that. I would complain if I were you. Do his lectures give criteria for assessing whether an orbital is bonding, antibonding or non-bonding?

The article Antibonding summarises things nicely: if an orbital puts electron density mostly outside the region between two nuclei, it is likely to be antibonding. The clue's in the name: if electrons enter an orbital and that causes the bond between two atoms to weaken or break, then the orbital is antibonding.

A quick way of assessing whether an MO is antibonding is to look for nodal planes between nuclei. The more internuclear nodal planes, the more antibonding the MO. --Ben (talk) 14:34, 18 April 2012 (UTC)[reply]

I am annoyed with the lecturer, considering he does not even care to know the specifics of my query. Apparently, hs response does not merit a complaint, my educational advisor is aware of the conversation between the lecturer and myself.

His lectures gives criteria for assessing bond types, but only of electron density maps, not of MO wavefunctions. In the upcoming semester test, we will be issued with a printed semi-empirical output file, where we have to answer questions such as:

"Write down an expression for the wave function of MO 7. Draw a diagram depicting this molecular orbital and explain whether or not you think it is a π molecular orbital."

However, we may not neccesarily be given such an easy MO to analyse, where their phases all align so nicely. Plasmic Physics (talk) 23:02, 18 April 2012 (UTC)[reply]

I can easily assess density maps, because it is just a matter of finding nodes. Guessing the density map from the wavefunction is difficult. Plasmic Physics (talk) 23:12, 18 April 2012 (UTC)[reply]

It'll be a fairly straightforward case, I imagine. Use the AO coefficients to sketch the MO. A negative coefficient inverts the relative phase. The bigger the coefficient, the bigger you draw the AO. Because it's only a sketch, just draw the AO that contributes the most for each atom. You could practise a few examples and check your answers here if you like. --Ben (talk) 23:21, 18 April 2012 (UTC)[reply]

OK, how about this: Ψ₄ = −0.56Ψ_2s(B1) + 0.37Ψ_2px(F1) − 0.37Ψ_2px(F2) − 0.43Ψ_2pz(F3), is it sigma with three nodes?

Or this: Ψ₅ = 0.40Ψ_2px(B1) − 0.45Ψ_2pz(F1) − 0.32Ψ_2px(F2) + 0.42Ψ_2pz(F2) + 0.47Ψ_2px(F3), what is it? There is head-on mixing between F1, F2 and B1, but side-on mixing Between B1 and F3.

+
z
−
| F2               F1
|          B1
|
|          F3
|_____________− x +

Plasmic Physics (talk) 00:22, 19 April 2012 (UTC)[reply]

http://www.brynmawr.edu/chemistry/Chem/sburgmay/chem231/MOpics/BF3mos.jpg and http://www.dartmouth.edu/~chem64/64%20pdf%20files/PS3A.pdf might help you visualise the expected results. From the numbers you quote, I drew the following sketches: http://www.benjamin-mills.com/Wikipedia/PlasmicPhysics-BF3-MO-question.png. Ψ₄ looks like we have sign convention problems, so just flip the phase on the boron 2s and you have a sensible sigma bonding MO. Ψ₅ looks fine, but it's weird that F2 has a 2p_x component whereas F1 does not. Compare with your output. And visualise the MOs in Spartan. --Ben (talk) 11:00, 19 April 2012 (UTC)[reply]

p.s. Ψ₅ looks like a combination of sigma and pi bonding. The interactions between F1 and B1, and between F2 and B1 are sigma-ish. But the F3-B1 interaction is definitely pi, as it's side-on overlap of p orbitals with a nodal plane right down the F3-B1 internuclear axis. --Ben (talk) 11:04, 19 April 2012 (UTC)[reply]

The coordinates given are (in Angstroms):

Atom X Y Z

1 B B1 0.0000000 0.0000000 0.0000000

2 F F1 1.1313727 0.0000000 0.6531983

3 F F2 -1.1313727 0.0000000 0.6531983

4 F F3 0.0000000 0.0000000 -1.3063966

Does this not indicate the orientation I provided? The reason that psi-5 looks that way is because the x component coefficient falls below the limiting factor of |0.30| and is considered negligible and so is excluded from psi-5. Plasmic Physics (talk) 11:48, 19 April 2012 (UTC)[reply]

If the positive z-direction is downward then it would ake a lot more sense. Psi-4 would then be anti-sigma, and psi-5 would be still not be anti-bonding. But it must be an either/or situation, I am required to choose sigma or pi for psi-5, how then do I choose which one predominates? Plasmic Physics (talk) 12:00, 19 April 2012 (UTC)[reply]

Consequences of not chewing food properly.

My mom always used to tell me to chew every mouthful of food 32 times. What are the consequences if you don't chew your food enough? Naively, it seems like larger lumps of food would be harder to digest - so could less chewing be an effective diet trick? — Preceding unsigned comment added by 216.136.51.242 (talk) 14:54, 16 April 2012 (UTC)[reply]

The idea that each bite of food should be chewed 32 times is a diet fad invented in Victorian times by a businessman named Horace Fletcher. He came up with elaborate justifications for why this would be true, but there really isn't a good scientific basis for the practice. At best, perhaps you'll be less likely to overeat if you force yourself to eat really slowly, although according to this article by a nurse, anyway, there haven't actually been any clinical trials to scientifically examine "Fletcherism", as the practice is called.[1] Red Act (talk) 15:47, 16 April 2012 (UTC)[reply]

It's pretty clear (to me at least) that 32 times is too many - and it surely would have to depend on the type of food. But what does happen if you chew less times than you naturally would if you weren't thinking about it? 216.136.51.242 (talk) 15:57, 16 April 2012 (UTC)[reply]

There could be a choking hazard if the pieces of food are larger than your esophagus can handle. Also, eating too much, too fast can cause an upset stomach (not sure why, but it's happened to me). StuRat (talk) 16:23, 16 April 2012 (UTC)[reply]

Chewing that much would slow down the rate of food intake and possibly result in the feeling of satiety before all of the food has been consumed, and prevent over eating. SkyMachine ⁽⁺⁺⁾ 22:26, 16 April 2012 (UTC)[reply]

If I recall from high school health class, chewing starchy foods helps to mix them with saliva which somehow aids metabolism. Also swallowing big unchewed chunks of meat could make choking more likely. Edison (talk) 23:43, 16 April 2012 (UTC)[reply]

Found a ref that chewing is an important part of the digestion of starches, since the amylase in saliva converts the starch to sugar during and after the chewing process: [2]. It does not specifically prescribe 30 chews per bite. Edison (talk) 23:48, 16 April 2012 (UTC)[reply]

IMO it might just be a way of preventing you from gulping your food at the dinner table. If you take a long time to eat your food, it's less likely that you will overeat. That aside, I've read a study where they feed rats pre-ground food and they put on a lot more weight than the controls. So maybe better chewing helps you pick up more nutrients. The exact final effect though is hard to speculate about. Staticd (talk) 05:56, 17 April 2012 (UTC)[reply]

If you think of the human body like an industrial process, it takes raw material into the comminution circuit where it is ground up to increase the surface area and mixed with water to make a suspension. Then it enters a leaching tank where it is agitated in an acidic medium to extract the valuable constituents. The better the comminution, the more efficient the leaching process is. So theoretically, better chewing will lead to better absorption of nutrients, but our bodies seem to be fairly well adapted to getting the most out of what we eat, so I'd say the normal amount of chewing is instinctively the optimum (but I'm no biologist, so don't take that for gospel). 120.155.205.57 (talk) 10:09, 17 April 2012 (UTC)[reply]

So it's reasonable to assume that chewing ones food less would be an effective dieting trick without adverse consequences? 216.136.51.242 (talk) 15:01, 17 April 2012 (UTC)[reply]

That sounds like a request for medical advice. 203.27.72.5 (talk) 21:22, 17 April 2012 (UTC)[reply]

Agreed. Sorry, 216, that's one we can't answer. Talk to a doctor or dietician. — The Hand That Feeds You:^Bite 21:27, 17 April 2012 (UTC)[reply]

I'm trying to visualize chewing a spoonful of applesauce 32 times. Beef jerky, I could see. But applesauce? Yuch. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:54, 19 April 2012 (UTC)[reply]

Red leaves

I know that Chlorophyll can make plants green and that's how they feed. But many plants have red leaves, especially in the spring, plants like Roses, Rhodendrons, some Maples, and Photinia. What is the chemical which turns leaves red, and is it as effective as Chlorophyll in feeding plants? 69.62.243.48 (talk) 17:06, 16 April 2012 (UTC)[reply]

There is some information in Autumn leaf color that you might find helpful. — Preceding unsigned comment added by 148.177.1.210 (talk) 17:11, 16 April 2012 (UTC)[reply]

You're conflating several different things here:

1) Flowers, such as roses, are not for the photosynthesis, their primary purpose is to attract pollinators, such as bees and hummingbirds.

2) Some leaves are red when healthy. This is what you seem to be asking about.

3) Other leaves only turn red when they die. This is the autumn leaf color referred to above. StuRat (talk) 17:16, 16 April 2012 (UTC)[reply]

Specifically, springtime red leaf colors in those plants are due to anthocyanins, which play a variety of roles for the plant. Primarily, they defend against cold, as well as some insect herbivory. These compounds do not "feed the plants". If anything, they prevent too much light from burning leaves (more on that at photoinhibition). (Post WP:EC, the OP is clearly talking about red leaves on roses. Young spring leaves of many species are quite red, then the anthocyanins fade out as they are not needed, and the color of the chlorophyll comes through.) SemanticMantis (talk) 17:22, 16 April 2012 (UTC)[reply]

I am not conflating anything. I'm not talking about rose petals, I'm talking about rose leaves, which are red in the spring prior to the roses blossoming. And I am not talking about autumn leaves. 69.62.243.48 (talk) 18:32, 16 April 2012 (UTC)[reply]

OK, my mistake. Petals are a type of modified leaf, BTW. StuRat (talk) 18:41, 16 April 2012 (UTC)[reply]

In short, it's complicated and no one is really sure! Red leaves is on my to-do-someday list. There are some interesting papers found here. Possible reasons are to protect the young leaves from sunlight, fungi and insect herbivores. In reality, it is probably a mix of all three and possibly others. SmartSE (talk) 13:30, 19 April 2012 (UTC)[reply]

Running at 2000 feet altitude

I weigh 230 pounds. When I exercise at the gym treadmill which is at 2000 feet, I can run a mile in about 10 and a half minutes. At gym treadmills down at sea level I can go faster, like 9:30 for a mile and then i can keep going. Also I can run over 7mph down there for several minutes, which i can't do for more than 45 seconds at 2000 feet. But i'm wondering if it's just a different brand of treadmill that's fooling me or is it the altitude? Thank you.Cyrus Burbank (talk) 19:12, 16 April 2012 (UTC)[reply]

The main limiting factor in how much exercise you can do is how quickly you can get oxygen to your muscles. At lower altitude, there is more air, so more oxygen, so it is easier to get it to your muscles. See altitude training for one application of this (well, an application of the reverse, really). --Tango (talk) 19:17, 16 April 2012 (UTC)[reply]

Still, offhand, it sounds to me like a fairly dramatic difference for only 2000', which is not that high. It's possible, I guess. But if there's a possibility that the treadmills are just different, it's hard to know which is the controlling factor. --Trovatore (talk) 19:28, 16 April 2012 (UTC)[reply]

You could try using a piece of running equipment called "a road" instead of a treadmill. It has several advatages: 1) The government pays for it (maybe that means you in the end, but you pay whether you run on it or not). 2) Better ventilation. 3) As you run along, the view changes. 4) You get to meet other people (maybe even other runners) along the way, unless you live in the back of beyond. You can measure distances to set points using your car's tachometer. Running a mile on a road at sea level will then be directly comparable to a mile on a road at 2,000 feet. Apologies for the sarcasm, but it was an easy target. Alansplodge (talk) 20:00, 16 April 2012 (UTC)[reply]

You measure distances with a tachometer? I suppose you could, if you take it out of the car, count how many times you have to turn it end-over-end to cover the road, and multiply by the diameter of the tachometer. --Trovatore (talk) 20:35, 16 April 2012 (UTC) [reply]

D'oh! I meant Odometer - I knew that mileometer wasn't the correct name and I conflated it with a Tachograph. Alansplodge (talk) 21:42, 16 April 2012 (UTC)[reply]

More seriously, you're unlikely to find comparable roads at sea level to those at 2000'. Also, treadmills are padded, and some people who find they get injured running on roads may have better luck on treadmills, though I don't offer any warranty on that. --Trovatore (talk) 20:37, 16 April 2012 (UTC)[reply]

No shoulders on roads, winding roads, sometimes snow (still). Mountain lions.20:14, 16 April 2012 (UTC) — Preceding unsigned comment added by Cyrus Burbank (talk • contribs)

I fully retract my suggestion - being eaten by a lion (mountain or otherwise) is not an aid to fitness. Alansplodge (talk) 21:46, 16 April 2012 (UTC)[reply]

Even in the absence of predatory cats, like the OP, I prefer myself to run on a treadmill as well. I'm a life-long runner (well, intermittently, but still over 21 years, more or less) and as I have gotten older, road running has gotten considerably more hazardous. For the first, the road is hard, and I am more suceptible to shin splints which are no fun. Secondly, running on grass or dirt shoulders is quite uneven. I used to run cross country running, but wouldn't dream of it today: my ankles and knees can't take the uneven ground and I'm quite likely to roll an ankle or something like that, which I could shake off in a week or so in my youth, but at this age is more of a pain. Treadmills remove all of those variables, they are softer on the legs, you aren't likely to step on a rock and roll your ankle. A ten dollar gym membership lets me run in front of a TV as long as I want, and I get full use of the full range of weight machines and other equipment on days when I want to do something other than run. But even so, running on a treadmill has its advantages, especially for geezers like me (if you consider 35 to be geezer, which I imagine most whippersnappers here do). --Jayron32 00:31, 17 April 2012 (UTC)[reply]

Just as a note, at 2000 feet you have, on average, 92-95% of the oxygen pressure than at sea level; seems hardly enough to explain the difference you describe, though it could be partially responsible. -RunningOnBrains^(talk) 21:42, 16 April 2012 (UTC)[reply]

I was also a pretty keen runner for most of my life until about a year ago. My job involves too much running up stairs and the older employees have frequent knee problems that I would rather avoid, so I switched to cycling for their sake. I wouldn't want to give up the fresh air and scenery by going to a gym treadmill. And mountain lions? Really? They're just there to provide motivation. 203.27.72.5 (talk) 02:39, 17 April 2012 (UTC)[reply]

This running calculator works out finish times at different altitudes. If you enter your 00:09:30 at sea level, it calculates your time should be 00:09:35.5 at 2,000 feet. It is apparently based on your VDOT. Alansplodge (talk) 21:57, 16 April 2012 (UTC)[reply]

Owl rotation

How and why does this and this (to an extent) work? Warning - both vids contain a song that will get stuck in your head. --Kurt Shaped Box (talk) 23:08, 16 April 2012 (UTC)[reply]

My understanding is that it is based on the bird's desire to keep its frame of vision relatively constant, independent of body orientation. It doesn't take too much imagination to see how this could be of adaptive value in a visual predator. Humans do this too, but to a lesser extent. Recently this [3] made the rounds, comparing a chicken's head stabilization to a human head, and human hands. I think somewhere in the chicken video documentation, I found links to a serious scientific talk or paper on the subject, but I can't find it at the moment. SemanticMantis (talk) 23:24, 16 April 2012 (UTC)[reply]

In fact, the video you linked links its video source on Youtube, which in turn gives its source as "K.E. Money 1962. THE VESTIBULAR SYSTEM OF THE OWL", you can probably track that down, but let us know if you don't have the proper access. SemanticMantis (talk) 23:27, 16 April 2012 (UTC)[reply]

See Vestibulo-ocular reflex. Red Act (talk) 02:34, 17 April 2012 (UTC)[reply]

Here is a much more technical article about the vestibulo-ocular reflex,[4] which among other things points out that an owl is capable of little or no eye movement within the head, so an owl's reflex involves more tilting of the head instead of movement of the eyes within the head. The opposite extreme is fish, which are capable of virtually no head movement relative to their body, so the reflex for them most purely involves movement of the eyes within the head. Other species are in between. Red Act (talk) 03:02, 17 April 2012 (UTC)[reply]

This topic was discussed on the science desk nearly four years ago, November 2008 - chicken head stabilization. There are some days I wish I did not have a very good memory. Nimur (talk) 19:27, 18 April 2012 (UTC)[reply]

Thanks for all the answers guys. I'll look into all the links you provided (sorry, I've had my head up my ass). --Kurt Shaped Box (talk) 21:14, 18 April 2012 (UTC)[reply]