Science desk
< October 31	<< Oct \| November \| Dec >>	November 2 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

November 1

Can a 40 feet transit bus fit into a wide-body aircraft such as the Airbus A330, Airbus A380, Boeing 747, or Boeing 777 as cargo?

Can a 40 feet transit bus fit into a wide-body aircraft such as the Airbus A330, Airbus A380, Boeing 747, or Boeing 777 as cargo? For example, could a 40 foot transit bus manufactured by Gillig or New Flyer fit into an Airbus A380, the world's largest passenger airliner if the aircraft was converted to a cargo only aircraft? WJetChao (talk) 08:15, 1 November 2015 (UTC)[reply]

Hello WJetChao. There are many websites with dimensions on bus widths and many others with cargo aircraft door widths. Large busses are 2.5 to 2.6 meters wide. Large cargo aircraft have loading doors that average 3.4 to 3.5 meters wide, though some are much wider. Accordingly, most busses can easily fit into large modern cargo aircraft. Cullen³²⁸ Let's discuss it 08:46, 1 November 2015 (UTC)[reply]

Not so fast, there, Grasshopper. Er, Dolphin. Some leaping creature, anyway.

I took as an example the New Flyer Excelsior bus and the Boeing 747-400 family freighter. As Dolphin says, the cargo door width is not a problem, and neither are the length or the weight. But the height is. The cargo door opening at the front of the plane is only 2.49 m high (98 inches); see page 18. Even the side door, which I suspect would not be workable with an object the length of a bus, is only 3.1 m high (123 inches); see page 15. Without doing something special you could not put this bus, which is at least 3.2 m high, onto this plane. The problem, of course, is partly that it's taller than a standard shipping container. I haven't looked at the other planes the OP mentioned; maybe one of them will take it. --70.49.170.168 (talk) 11:50, 1 November 2015 (UTC)[reply]

It would fit on top of a Boeing 747 Shuttle Carrier Aircraft, but that's probably cheating. Alansplodge (talk) 14:59, 1 November 2015 (UTC)[reply]

It would definitely fit into an Airbus A400M Atlas (17.71 m x 4.00 m x 3.85 m) or an Antonov An-225 (43.35 m x 6.4 m x 4.4 m). The Airbus Beluga and Boeing Dreamlifter appear to be larger still, but our articles don't actually have numbers for the maximum cargo dimensions. It wouldn't fit into an Airbus A330-200F, that has the same 96" height limit as the 747 (see Airbus' website). Tevildo (talk) 18:06, 1 November 2015 (UTC)[reply]

The Beluga's dimensions are 37.7 m x 7.1 m x 7.1 m (here). You could probably get two buses on top of each other in there. Tevildo (talk) 18:11, 1 November 2015 (UTC)[reply]

I think that the Boeing C-17 Globemaster III could handle a bus as well. Maybe two. Cullen³²⁸ Let's discuss it 18:15, 1 November 2015 (UTC)[reply]

According to this Airbus webpage, the Airbus Beluga can handle cargo 16 feet wide and 16 feet high. Cullen³²⁸ Let's discuss it 18:25, 1 November 2015 (UTC)[reply]

Indeed. I think the answer to the OP's question is "No" - a bus is too big for a converted passenger jet, it would have to go in a specialist cargo aircraft. Tevildo (talk) 21:04, 1 November 2015 (UTC)[reply]

The Beluga is a converted passenger jet, based on the Airbus A300. Cullen³²⁸ Let's discuss it 22:03, 1 November 2015 (UTC)[reply]

There's no requirement that the bus must be driven in. It would fit in on its side - though it's a rather tight fit, so you might want to take the wheels off. MChesterMC (talk) 10:05, 2 November 2015 (UTC)[reply]

MChesterMC, I do not think that your caveat applies to the Beluga, a converted passenger jet with a 16 foot wide by 16 foot high cargo door. Upright or on its side, a bus fits into it easily. However, it reminds me of an anecdote I heard many years ago about a truck one inch taller than a highway overpass getting wedged into place. Powerful tow trucks couldn't extract it, and smart men stood around scratching their chins. A young boy said, "Let the air out of the tires", and then it was done. Cullen³²⁸ Let's discuss it 03:43, 3 November 2015 (UTC)[reply]

Chemical notation

I'm surprised that we don't have an article (or a redirect) for chemical notation. Or where would we have definition of parentheses, as in "Al(l)"? BTW, the reason I needed to look this up was because of this edit, which looks like a good faith misunderstanding to me. — Sebastian 08:50, 1 November 2015 (UTC)[reply]

See Chemical equation and look under Common symbols. We also have Chemical formula and Chemical nomenclature. Dolphin (t) 09:06, 1 November 2015 (UTC)[reply]

Funny, I only now (thanks to your edit summary) realized that I had misread the letter, which I presume is what misguided the IP editor, too. I read "Al(l)" as "Al(I)", which can look the same depending on the font used. Since both of these occur in the article, I'm wondering if there's a better way to distinguish them. One thing I did notice is that the notation used in the article is inconsistent; often – as in "Al(L)_n(s)" (which, if I may add that parenthetically, has yet another use of a parenthesis) – the physical state is written as a subscript. Could that distinction help disambiguate "Al(l)" from "Al(I)"? — Sebastian 18:49, 1 November 2015 (UTC)[reply]

I agree it is potentially ambiguous to use Al(l) to mean both liquid aluminum and aluminum in oxidation state +1, and yet this ambiguity occurs in Aluminum! Perhaps a chemist would say it is always possible to resolve the ambiguity by considering the surrounding context. Where there is a risk of ambiguity I like the idea of using Al(liq) but I can't cite a reliable published source to support my preference. Dolphin (t) 06:44, 2 November 2015 (UTC)[reply]

I've sometimes seen the state-of-matter detail in italics and/or subscript, but ACS Style Guide says it goes in-line as plain text. What if the (l) for liquid is simply omitted? It's not key to the reaction itself and the surrounding text can emphasize that due to the high heat involved in the reaction, the metal is produced as a liquid. Or else include an explicit charge of zero as superscript (to emphasize that the charge is changing), which makes it clearer that this other detail is not also charge. DMacks (talk) 20:08, 2 November 2015 (UTC)[reply]

The whole Aluminium#Production and refinement is a mess. The "relevant reactions" for producing a certain chemical have that chemical as a reactant. And interleaves ideas about the chemistry with the physics/engineering but omits parallel details or relationships (no idea what the cathodic reaction is in particular, or that the C reactant at the anode is the anode itself). DMacks (talk) 20:16, 2 November 2015 (UTC)[reply]

Intelligence failure in the technological age.

In the modern era of the digital/technological age, why does intelligence fail? Surely digital capabilities should improve intelligence capabilities? --Copstwurst (talk) 09:52, 1 November 2015 (UTC)[reply]

This is a Humanities question, not science. "Intelligence failure" doesn't mean that the microphones broke down or the drones shut off their transponders and flew toward Antarctica. It means that spies (the people we hire to spend their lives lying and deceiving people) didn't tell us the truth. Was Nigerian yellowcake an intelligence failure, or an intelligence success? Depends on whether the spies' purpose is tell Americans the truth, or start the war the President's family wanted. Wnt (talk) 11:57, 1 November 2015 (UTC)[reply]

@Copswurst: intelligence and measures of are human constructs, facilitates Illusion of control, explored in Philosophical zombie, Simulacrum, technology reduce pressure on brain to process, memorize viaRed queen hypothesis Mahfuzur rahman shourov (talk) 15:18, 1 November 2015 (UTC) @Copstwurst:mention correctMahfuzur rahman shourov (talk) 15:19, 1 November 2015 (UTC)[reply]

Technology can improve intelligence gathering - reduce the rate of disasterous failures - but you're asking why there are any failures at all! There will always be failure so long as intelligence isn't 100% perfect. But 100% perfection requires governments to know 100% of everything that's going on absolutely everywhere. Even if 100% perfection were technologically possible - the issues of privacy and data security would be crippling. We don't actually want 100% perfection - we want an appropriate balance between privacy rights and security from evil-doers. It's tough to put together a scientific answer to that question because it's all about cultural, political, personal and emotional responses to invasions of privacy and such like. SteveBaker (talk) 16:25, 1 November 2015 (UTC)[reply]

In the case of battlefield intel, we aren't so worried about privacy there, so 100% correct intel would be a good goal. (And I'd have to think that civilians there would be willing to give up their privacy if it meant an end to us bombing the wrong targets.) StuRat (talk) 03:00, 2 November 2015 (UTC)[reply]

Human intel is still a critical ingredient. The US military has recently relied on "signature" targeting, which means killing people such as men wearing all black and carrying guns in the target area. Of course, in the Middle East, that's no guarantee they are the bad guys. You need people on the ground to tell the good guys from the bad (and, of course, who is good or bad is a matter of opinion, as Osama Bin Laden was considered a good guy by the US while battling the Soviet Union in Afghanistan). StuRat (talk) 03:04, 2 November 2015 (UTC)[reply]

A matter of opinion or a matter of how much cheese a rat gets for pointing out the "bad guys". Zero cheddar points for negative results. It's a science question. InedibleHulk (talk) 03:30, 2 November 2015 (UTC)[reply]

Neutron photography

I was making up a chart at neutron reflectometry recently, noticing the variation between elements in coherent scattering, incoherent scattering, and absorption. The neutrons can penetrate tens of millimeters into many materials. It is possible to use specular reflection for crystallography, but often at high angles there is simply random scattering. Still...

What would common landscapes - people, forests, houses - look like as illuminated by a neutron source? (Assuming either that you can detect at very low 'illumination', or can see the result of an instantaneous flash, like a Hiroshima bomb) At first I would think that everything would be just kind of white, from random backscatter, except that some materials absorb more than others, and I'm not sure how the coherent reflections would affect an ordinary photograph.

But there is also a matter of wavelength. The numbers I put in that paper were for neutrons with 1 angstrom or perhaps 1.8 angstrom wavelengths and seemed quite consistent, but in theory neutrons can have all sorts of wavelengths. And scattered back, their wavelengths should change from the recoil. So I'm thinking there ought to be such a thing as color neutron photography.

Is there a sense of this out in the literature somewhere? I wonder if the world seen by neutrons is as beautiful as by photons' light. Wnt (talk) 12:20, 1 November 2015 (UTC)[reply]

From a quick glance at the neutron reflectometry article, I note that nitrogen has quite a high scattering length - so probably very foggy, since the air is no longer transparent. MChesterMC (talk) 09:58, 2 November 2015 (UTC)[reply]

Engineering professional registration

Does registration for professional status after getting an engineering major take a significant amount of time outside working hours? — Preceding unsigned comment added by Clover345 (talk • contribs) 15:08, 1 November 2015 (UTC)[reply]

Regulation and licensure in engineering is not simply a matter of "getting an engineering major [degree]" (in some jurisdictions that may not even be necessary). Whether gaining PE status takes much time outside of working hours will depend significantly on whether one's workplace permits doing such work on company time. — Lomn 17:10, 1 November 2015 (UTC)[reply]

framerate human brain

OP curious — Preceding unsigned comment added by Mahfuzur rahman shourov (talk • contribs) 15:25, 1 November 2015 (UTC)[reply]

No - the brain is a massively parallel and asynchronous system. There is no specific "frame rate". That said, there are various regular waves that can be detected - and each individual neuron has "firing rate". But as a whole, there is no particular number that could be assigned to major processing cycles. SteveBaker (talk) 16:00, 1 November 2015 (UTC)[reply]

The comment above is correct. You might want to look at flicker fusion for one possible interpretation of an effective frame rate for the human visual system, but it's actually far more capable than that suggests in special cases: hence the need for things like high frame rate video in VR systems. See this link for Palmer Luckey's opinion on this. -- 16:14, 1 November 2015 (UTC)

The simple way of stating it is that the human visual system doesn't have "frames". Video systems work by displaying static images in rapid succession. These are the "frames" that "frame rate" refers to. Originally it referred to the individual frames on a film strip, but the term generalizes well to electronic displays like CRTs and LCDs, which work similarly, by drawing an image and then overwriting it with a new one. Our brains interpret these images as a single sequence due to persistence of vision. --71.119.131.184 (talk) 22:23, 1 November 2015 (UTC)[reply]

@Mahfuzur rahman shourov, SteveBaker, and 71.119.131.184: I happened upon a possible answer in [1]. I did not make the considerable effort required to properly understand the paper, but the gist was that patterns of firing of place cells in the hippocampus tends to change at a particular point (I forget now if it was high or low) in the cycling of slow gamma waves. As a result, a rat's internal conception of its own position tends to change as if it had a frame rate of 25 to 50 Hz, I think it was. Wnt (talk) 20:37, 12 November 2015 (UTC)[reply]

The trouble is that gamma waves are only one of at least seven identifiable 'waves' that occur in the brain: vis Delta wave – (0.1 – 3 Hz), Theta wave – (4 – 7 Hz), Alpha wave – (8 – 15 Hz), Mu wave – (7.5 – 12.5 Hz), SMR wave – (12.5 – 15.5 Hz), Beta wave – (16 – 31 Hz) and Gamma wave – (32 – 100 Hz). These are not nice multiples of each other - they constantly vary their frequencies - often independently of each other. So this is in no way an identifiable "frame rate"...at least not in the sense of a movie - or even of the "main loop" of a piece of software. It's possible (I suppose) that each of these waves represents the 'frame rate' of a different subsystem - but it really doesn't seem that way. SteveBaker (talk) 20:56, 12 November 2015 (UTC)[reply]

@SteveBaker: Well, the key concept in that paper is of a "static attractor" - a particular pattern of firing that is maintained stably when the rat is in a particular place, and changes as it moves to a different one. The static attractor changes in sync with the slow-gamma. What makes this a frame-rate in my mind, therefore, is that the static attractor is the frame - the particular fixed image which then becomes labile between frames. Now I'm not saying that this is universal throughout the brain (though the gamma wave article seems to make some implications in that direction); it is surprising enough to me to find frames exist at all, that I won't be picky about them being consistent throughout all activities! Wnt (talk) 22:44, 12 November 2015 (UTC)[reply]

Certainly it's interesting that one part of the brain operates at one rate - but we're not being asked about one specific part of the brain and I'm certainly not happy with telling our OP that the conclusion is that there is a frame rate for the brain of 32 to 100Hz. That's flat out not true - the brain as an entirety is an asynchronous thing - hence all of these seven so-far detected frequencies controlling different kinds of subsystems. They aren't even exact multiples or submultiples of each other - and the relationship between those seven frequencies isn't even constant over time. Many of those wave patterns are poorly understood in humans. The study to which you refer was done in rats - and some of the waves that they exhibit are hardly even measurable in humans - so it's extremely dangerous to pin that down as an answer here. To say that there is any kind of frame rate for the human brain as a whole is exceedingly misleading...regardless of some parts of conscious behavior being synced specifically to the gamma wave rate in rats. SteveBaker (talk) 23:10, 12 November 2015 (UTC)[reply]

How To Calculate Inductance?

Lets assume I have a an inductor which is wound on an iron core that forms a closed loop and another identical inductor wound on a similar iron core but with the core not forming a closed loop as the first but open at both ends.
How do I calculate the inductance for these two scenarios? — Preceding unsigned comment added by Adenola87 (talk • contribs) 18:33, 1 November 2015 (UTC)[reply]

Useful articles are Inductor, Magnetic core and Toroid. The A_L value of an iron core is frequently specified by manufacturers. The relationship between inductance and A_L number in the linear portion of the magnetisation curve is:

L=n^{2}A_{L}

where n is the number of turns, L is the inductance (e.g. in nH) and A_L is expressed in inductance per turn squared (e.g. in nH/n²). Bestfaith (talk) 22:51, 1 November 2015 (UTC)[reply]

The inductance L of an iron-cored coil may be calculated using: L = 3.2 × N^2 × μ × a/10^8 × l Henrys, where:

N = number of turns

a = effective cross-sectional area of coil in square inches

l = length of magnetic circuit in inches

μ = effective permeability

The length of the magnetic circuit is the length measured around the core at the center of cross-section of each magnetic path. For a rectangle-shaped core with the winding wound around one or more edges, this effectively means the mean perimeter of the core. For the more typical interleaved-E shaped core with the winding wound around the central element (think of it as two rectangles butted together), the magnetic path length consists of the mean perimeter around one rectangle plus the mean perimeter around the other.

The effective permeability depends on the type of steel used, on the AC and DC flux densities in the core, and whether the core laminations are interleaved or whether there is an air gap in the magnetic circuit. In order to calculate the inductance, you must know the properties of the iron or steel used, and whether the inductor is or will be subjected to a DC current flow. There may be no DC current flowing in the winding or windings, or the net effect of DC currents may be zero because the currents are of such magnitude and winding sense that DC flux is cancelled out. The calculation of μ is very complex and use is usually made of charts which show variation for multiple variable factors. I am unsure how to address the part of your question that relates to the open-ended inductor. Does 'open-ended' imply that the length of the magnetic circuit is very large (but not infinite)? Others here may know.

There is no simple calculation that will answer your question. Too many unknown factors. This response has been based on information in the "Iron-cored Inductors" chapter of The Radiotron Designer's Handbook, Fourth Edition, 1952, by F. Langford-Smith, with some of the above being direct quotes. There is far more information in that chapter than I can present here, and permeability graphs and charts are presented. Akld guy (talk) 00:07, 2 November 2015 (UTC)[reply]

Calculating the inductance for the closed core-loop core is somewhat straight forward but the second part of the problem with the open-ended inductor is where I am stuck. Just to clarify what is meant by "open-ended core", I have made a simple diagram to illustrate what I mean. Have a look below:

--Adenola87 (talk) 04:12, 2 November 2015 (UTC)[reply]

Yes that is exactly what I thought you meant. I'm having trouble with the open-ended one too. By the way, the dashed line in the closed-core image is what I described as the '~~median~~ mean perimeter', and is used to calculate the length of the magnetic circuit. That image shows it very well. I'm curious how you are determining the permeability of a core of unknown characteristics. Usually when designing an inductor or transformer, a manufacturer will start with iron or steel of known permeability. You don't have that luxury. Or have you stumbled on a stock of unused cores with manufacturer's data attached? Akld guy (talk) 04:30, 2 November 2015 (UTC) Akld guy (talk) 06:58, 2 November 2015 (UTC)[reply]

I am currently working on something and I am trying to establish a sort of relationship in the size of the inductance of two similar inductors with one having a closed loop core while in another instance the same inductor is just wound on an open loop core. I am in the preliminary stages of my design and I have not started to zero in on the specifics of the actual permeability of the core. I do know that most commercially available cores can easily exceed 1000. It will be nice if I can work some sort of ratio between two coils of similar parameters with one wound on a closed iron core and the order on an open core like in the diagram. — Preceding unsigned comment added by Adenola87 (talk • contribs) 04:55, 2 November 2015 (UTC) --Adenola87 (talk) 04:56, 2 November 2015 (UTC)[reply]

It's called an autotransformer. Pretty straightforward calculations in the article, I believe. --DHeyward (talk) 07:06, 2 November 2015 (UTC)[reply]

No, it is NOT called an autotransformer, which is an entirely different device than what we are talking about. The autotransformer is a special class of transformer that has only one winding. It has a tap connection (or more than one tap) on the winding to transform one voltage to another, and does not in any way isolate the input terminals from the output terminals as a normal two-winding transformer does. Here, we are not even talking about a transformer. We are talking about an inductor, whose principal application is in smoothing DC that has been obtained by rectifying AC, thus reducing hum and ripple. Akld guy (talk) 11:21, 2 November 2015 (UTC)[reply]

From a physics perspective, what's the difference? It's magnetic flux through a number of surfaces. Why wouldn't the 1:1 autotransformer case degenerate to an inductor with an open core? --DHeyward (talk) 02:42, 3 November 2015 (UTC)[reply]

The autotransformer is used only where AC is involved, since its function is to transform one AC voltage to another (or to a range of AC voltages that the user can select by moving the load to a different tap on the winding). The AC-only stipulation simplifies the calculations since the iron or steel core will not take on a permanent magnetization with only AC present. Now, the OP's question relates to an iron-cored inductor, and the only real use for an iron or steel cored inductor is as a smoothing choke in AC power supplies where the AC is first rectified to produce unfiltered, raw DC and is then passed through the inductor. The nature of an inductor is that it resists any increase in current flowing through it, yet tries to maintain the same current when the current flowing through it is decreased. The inductor therefore smoothes out the irregularities (the roughness) of the DC that resulted from rectifying the AC. Such metal-cored inductors are rarely encountered today, but in decades past they were used extensively in tube radios, TV's, amplifiers, speech equipment. Very large ones were used in telephone exchanges. The OP has not stated whether DC will be present in his inductor, but I see little practical use for an iron-cored inductor except as a smoothing choke as just described. To answer your question, 1. the presumption is that DC will be flowing, and 2. DC flow makes the calculation more complicated than for the case of the autotransformer by modifying the permeability of the core and thus introducing another variable. Akld guy (talk) 03:47, 3 November 2015 (UTC)[reply]

Permeability is not a factor for DC. It's straight IR drop. Reactance has a frequency component. It seems odd to me to use a frequency varying permeability, though, as it would be non-linear which would create intermodulation products. Dynamic access arrangements used in telco phones and modems were very linear and the permeability was frequency independent (at least in the bands of interest). Midcom was the last brand I used but that was many moons ago. The inductor you are describing would be useful as a choke but is also used in relays and also ground-fault interrupting equipment. --DHeyward (talk) 04:38, 3 November 2015 (UTC)[reply]

The effective permeability depends on the type of steel used, on the AC and DC flux densities in the core, and whether the core laminations are interleaved or whether there is an air gap in the magnetic circuit. That is a direct quote [my emphasis added] from page 243 of the Radiotron Designer's Handbook, Fourth Edition, 1952, which is freely available as a PDF online. I suggest that you read that chapter. The pages from 243 onwards show how to calculate inductance when AC and AC plus DC are present. Yes, inductors (that is, coils wound on a metal core) are used in relays, but it's not usual to refer to them as inductors because the properties of inductance are not exploited in relays. Relay windings act as simple electromagnets which move the armature when DC is applied to the winding. That does not exploit the reactive property of an inductor, X_L = 2πfl ohms, which is relevant only when a non-DC waveform is applied, such as the unfiltered DC that I mentioned above. If f = 0 (as in the case of DC), the reactance is zero, so essentially the relay winding is not being used as an inductor. Please try to answer the OP's question, instead of drawing me into irrelevant side issues. Akld guy (talk) 08:24, 3 November 2015 (UTC)[reply]

I pretty much know what an auto-transformer is and how it works but what I am talking about here is is an inductor. Sure there are similarities in how most of these devices that have some sort of coil windings (like transformers, inductors, etc) operate since all them operate by means of electromagnetic induction but every one of them have their on peculiarities and functions. What I am interested in here is something which is just basic and what I am trying to do is to establish how much the inductance of the inductor with the open core will be different from the one having a closed core. Most of the calculations you need for simple designs do not need to be over complicated and extremely precise. It depends on the nature of the project you are working on. The inductance of the inductor with the closed core is easy to calculate. The problem is with the other inductor that has an open core. I found this article [[2]]. I found this formula there: $\mu _{eff}={\frac {\mu _{r}}{1+k(\mu _{r}-1)}}$ I have not been able to verify the formula yet. I must say that I never heard anything like demonetization factor mentioned in the formula before. — Preceding unsigned comment added by Adenola87 (talk • contribs) 08:20, 3 November 2015 (UTC)[reply]

Which thing can resemble to atlas and axis?

I tried to think on door and its pivot but it's not so close. Maybe you have other idea for that? 78.111.186.40 (talk) 19:13, 1 November 2015 (UTC)[reply]

Please expand the question. How do atlas and axis go together? Are you looking for a (better) metaphor for something else? —Tamfang (talk) 22:57, 1 November 2015 (UTC)[reply]

He means the atlas (anatomy) and axis (anatomy). Sagittarian Milky Way (talk) 23:43, 1 November 2015 (UTC)[reply]

What actually rotate around what?

What actually rotate around what? Does the atlas rotate around the axis or axis rotate around the atlas? (I've pictures on google that show both of them and that's why I'm confused. 78.111.186.40 (talk) 20:02, 1 November 2015 (UTC)[reply]

See Atlas (anatomy) and Axis (anatomy) for the relevant articles. Neither of them rotate "around" the other - the skull rotates against the atlas at the atlanto-occipital joint; the atlas rotates against the axis. All the vertebrae rotate around the spinal cord. Tevildo (talk) 20:50, 1 November 2015 (UTC)[reply]

Hmmm, it's a complicated design... Not all of the rotation is in the atlas-axis. This site describes the atlas-axis relationship as so: A pivot joint is made by the end of one articulating bone rotating in a ring formed by another bone and its ligaments. Think of a metal washer twisting around a bolt. The dens articulates with the facet on the atlas, as well as the transverse ligament, and this articulation provides the head with approximately 50% of its movement. Wnt (talk) 01:12, 3 November 2015 (UTC)[reply]

Carbon surface chemistry

Hi I'm trying to visualise the surface of a diamond - there wouldn't be enough carbon atoms for the tetrahedral structure to be maintained would there be? I'm probably getting it wrong, but are there C=C double bonds there - like the surface of a water drop? Le Sanglier des Ardennes (talk) 23:16, 1 November 2015 (UTC)[reply]

It's a crystal and the face will be along crystal boundaries and axis. There doesn't need to be anything but the crystal structure but the face and angles will be deterministic. The type of diamond cuts and shapes are determined this way. The axis and face are very important to know prior to cutting. See Cubic crystal system and Diamond cubic. File:Diamond Cubic-F lattice animation.gif is an interesting animation as it shows certain face alignments as they rotate into view. Alignments (at least for similar structure in silicon) are often noted as <010> or <111> and can determine many properties. None of it changes the basic tetrahedral alignment though. A double bond would be a crystal defect. --DHeyward (talk) 07:38, 2 November 2015 (UTC)[reply]

The surface of a diamond has each carbon atom terminated by hydrogen atoms. Compared to the overall number of carbon atoms in any reasonably-sized diamond, this number of hydrogen atoms (a single atomic layer of them) is too minuscule to be counted in any reasonable way in comparison to the number of carbon atoms, and yet, they still exist. this is a nice image of it. There are synthetic diamonds which have surfaces terminated with other functional groups (this article provides a review of some of them), for a variety of chemical reasons, but standard diamond is usually hydrogen terminated. --Jayron 32 02:59, 3 November 2015 (UTC)[reply]

@Jayron32:@DHeyward: Thank you kindly; an excellent clarification. Le Sanglier des Ardennes (talk) 04:13, 3 November 2015 (UTC)[reply]