Wikipedia:Reference desk/Archives/Science/2012 September 10

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September 10

Average age of first childbirth

In a "natural" human (as in before the Neolithic Revolution), what was the average age at which a woman had her first child? How about a man? --140.180.254.18 (talk) 05:24, 10 September 2012 (UTC)

The Wikipedia article titled Advanced maternal age has some information on recent trends, but I don't see anything from Neolithic times. Still, its a lead and there are references. You could start your research there. --Jayron32 05:55, 10 September 2012 (UTC)

Would that even be possible to determine? I guess women dying during pregnancy leave two skeletons, but apart from that... Ssscienccce (talk) 08:05, 10 September 2012 (UTC)

Actually, it would not be that difficult at all, provided there were enough skeletal remains from the population in question. The pelvic inlet in the skeleton of a woman who has given birth is significantly wider than in the skeleton of a woman who hasn't. Add to that demographic data about fertility, mortality and age structure, and you could get a very good idea about the age of first birth in a given population. All the more so as childbirth itself was one of the leading causes of death among women of that time, especially in young adolescents, so there would be plenty of early adolescent female skeletons that probably died giving birth or shortly thereafter. Dominus Vobisdu (talk) 08:40, 10 September 2012 (UTC)

Are you sure about the "leading cause"? Childbirth complications were a leading cause of death for women in early modern times, but the major risk factor there are infectious diseases like childbed fever. These would be a lot less widespread in hunter/gatherer societies than in more densely populated civilisations. Child birth would still be a significant risk, but possibly less than one would think from simple extrapolation. --Stephan Schulz (talk) 09:46, 10 September 2012 (UTC)

If the average age of first childbirth is hard to determine, how about the average age of childbirth in general? I'd look at the molecular clock to see how many generations had passed between times X and Y, and use radiocarbon dating to determine the absolute time difference. --140.180.247.208 (talk) 17:06, 10 September 2012 (UTC)

Couldn't one assume that the average age is not long after the earliest possible age for reproduction? What would be the factors tending to delay reproduction? The question as originally posed referred to a "natural" human and one before the Neolithic revolution. I can only imagine cultural factors as held by the group of people to which one belonged inhibiting the natural biological urge to reproduce, but I find myself doubting that such cultural constraints would exist as their efficacy in promoting the wellbeing of the group are not obvious to me. Bus stop (talk) 17:12, 10 September 2012 (UTC)

One doesn't 'assume' an average, one measures it. AndyTheGrump (talk) 17:33, 10 September 2012 (UTC)

AndyTheGrump—The original question wonders aloud about the "average age at which a woman had her first child".[1] Whether we "assume" or "measure", we are still addressing the question. I think in some instances assumptions can be worthwhile. The original question was not necessarily concerned with the average age of all births, but rather the average age at which a woman had her first child. Also the original question concerned the average age that a man sired his first child. Yes, measuring for this would be distinctly preferable. But how could one measure for this? Bus stop (talk) 17:49, 10 September 2012 (UTC)

List of youngest birth mothers may be relevant. Bus stop (talk) 18:25, 10 September 2012 (UTC)

Oh, don't be silly Bus stop. All the Grump has to do is go back in his time-machine and select a sufficiently large representative cohort and follow them for a few generations, (as our current-day archaeological evidence of the neolithic period is far to sparse top draw any firm affirmation). However, the earliest time that a female can conceive (whether they primate, meerkat or pussycat) is not a good datum (or posit) (or assumption) as females tend to mate when they are ready. I don't wont to get into the nature/nurture on this, because obviously, they can still be wearingly short-white-ankle-cotton-socks, when they meet some rich guy/pop star/etc., and the money pheromone brings them into heat. --Aspro (talk) 18:57, 10 September 2012 (UTC)

black sunglasses or brown sunglasses for maximum UV protection

I should use black or brown sunglasses for uv protection? how sunglasses work? — Preceding unsigned comment added by 101.63.161.197 (talk) 07:12, 10 September 2012 (UTC)

Blocking UV rays would have no connection with glasses looking brown or black in visible light. It would be an entirely separate attribute of the glasses. HiLo48 (talk) 07:46, 10 September 2012 (UTC)

How do you know that? ←Baseball Bugs ^{What's up, Doc?} carrots→ 07:55, 10 September 2012 (UTC)

As well as the comment below, it's worth pointing out that UV rays are invisible to humans, so the visible colours of glasses are irrelevant to whether they block UV. HiLo48 (talk) 08:37, 10 September 2012 (UTC)

Because it's in our very article, Sunglasses. The color and protection factor can be varied independently. Now to answer the actual question, the original purpose of having a variety of colors for the lenses stems from the fact that sunglasses distort colors and contrast, and can impact your depth perception. The different colors have subtly different effects, and so you may prefer one over the other, although some colors are used purely for their cosmetic appearance. See Sunglasses#Lens for more. For the actual protection offered by the sunglasses, you'll have to look at the labeling on the sunglasses themselves. In the United States, European, and Australia, there are regulations on Sunglasses that require labeling to indicate the level of protection offered, and you can read about those at Sunglasses#Protection. I don't know if such regulations exist in India, to which your IP address belongs, but hopefully you will find some manner of label. Someguy1221 (talk) 08:23, 10 September 2012 (UTC)

That was the right answer. :) ←Baseball Bugs ^{What's up, Doc?} carrots→ 08:45, 10 September 2012 (UTC)

The color of the glass doesn't tell you if it blocks UV. Cheap dark plastic sunglasses may be worse than no sunglasses at all. And what Someguy says... Ssscienccce (talk) 08:30, 10 September 2012 (UTC)

Indeed. Actual real glass (as in silicon dioxide is opaque to UV light. It's why you can't get a sunburn through a window. Nearly all sunglasses have plastic lenses however, and many types of plastic do not offer any protection to UV light. --Jayron32 06:31, 11 September 2012 (UTC)

Not true, not true at all. Silicon dioxide is opaque to ultraviolet light below 200nm, where you'll find germicidal lights emitting, but is almost perfectly transparent to UVA and UVB [2]. Someguy1221 (talk) 07:52, 11 September 2012 (UTC)

I should clarify. The UV blocking ability of "glass" is highly dependent on the manufacturing process: some forms of silicon dioxide are opaque in a broad range of the UV, while other forms are basically transparent to many forms. The article you just cited seems to be dealing with fused silica, which is specifically manufactured for its UV transparency; fused silica cuvettes are used in UV-vis spectroscopy, for example. The introduction of that paper, where it discusses the four types of manufacturing processes used in producing the glass used for their experiemnts, are all used to produce this high purity fused silica. Most common glass isn't made the way they describeHowever, most people don't have fused silica windows, and common glass as is usually found in such uses is often Soda-lime glass, which is far more opaque in the UV. This article quotes a figure of 85% blocking of UVB rays for ordinary window glass, which makes it about SPF30. So, you can get sunburned through glass, but it takes a LOT longer to do so. --Jayron32 12:31, 11 September 2012 (UTC)

Actually, I have a pair of clear safety glasses that are apparently 99.9% protective against UVA and UVB radiation. The reflective tint on the outside surface coating of the glasses go from blue to indigo-violet near the centre, ranging toward orange and yellow near the edges. ~AH1 ^(discuss!) 18:27, 11 September 2012 (UTC)

I don't doubt that, but what makes them so? Does it have to do with the glass manufacture process, a dopant added to the glass, a coating on the glass, etc. etc. There's more than one way to skin a cat. --Jayron32 03:52, 12 September 2012 (UTC)

Allergies

This is not a request for medical advice. I am just wondering, what can cause allergies to spontaneously develop? When I was younger, I never had any allergies at all, but whilst in Japan, after about seven years I started to develop hay fever - or, in this case, specifically an allergy to the pollen of rice plants (which was hell, because I was living in the countryside surrounded by rice fields). Then, after a while I noticed that my stomach just below my navel developed a criminally insane itchy rash, which I later found out was an allergy to nickel (my belt buckle was causing it). My friend told me that in the past five years he has developed an allergy to something or other, but has no idea what. What causes allergies to develop? KägeTorä - (影虎) (TALK) 08:52, 10 September 2012 (UTC)

Some allergies can develop after a certain amount of exposure. For example, this paper starts out with "Beekeepers are strongly exposed to honey bee stings and therefore at an increased risk to develop IgE-mediated allergy to bee venom." [3]. Increased exposure may or may not play a role in the allergies that you describe. My understanding is that allergy development is not well understood in general. (start WP:OR): I was "immune" (not allergic) to poison ivy for years, until I had a very large exposure. Now I am normally susceptible (/end OR). This is not medical advice, but reference information, etc. etc. SemanticMantis (talk) 14:36, 10 September 2012 (UTC)

Very true SemanticMantis. Shortly after leaving school I had a very mild exposer to 'work'. Even now, 50 years on, I'm still trying to recover. The only respite I have found, is to lay down in a darked room and stare at the back of my eyelids. Fortunately, during those episodes (which are frequent) I can often relieve the boredom by turning my monitor's screen brightness level way-down-low and edit Wikipedia. Note: My method of staring at the back of the eyelids is in no way to be taken as medical advice for this very serious condition – consult a specialist.--Aspro (talk) 18:19, 10 September 2012 (UTC)

More on "not well understood in general". This [4] and similar studies have been in the popular science news over the past few years. The authors suggest that increased exposure during childhood may be linked to lower incidence of allergies in rural-raised children, compared to their urban counterparts. So timing and amount of exposure seems to regulate whether the exposure will increase or decrease the likelihood of (some) allergies. In short, there is no simple answer. SemanticMantis (talk) 19:23, 10 September 2012 (UTC)

• All allergies require a period of sensitization to the allergen -- sometimes it is rapid, sometimes slow. Our allergy article describes the process, in the "Pathophysiology" section, but unfortunately in terms that probably only a biologist can make sense of. Looie496 (talk) 19:24, 10 September 2012 (UTC)

• • Also of note is that there is a significant genetic component to allergies, see Allergy#Genetic_basis. This is again one of those nature-nurture unseperatable things: a person may have a genetic predisposition for an allergy, but without preliminary exposure to sensitize them to the allergan, the allergy itself may never fully develop. A person without a genetic predisposition could happily bathe in the same allergan all day and never develop any problems, a person with that predisposition may be fine on a few exposures, but then get "sensitized" and go all anaphalactic without any warning. My father developed a severe bee-venom allergy in his 40s after a being stung by a half-dozen bees; he'd been stung before and never had a bad reaction. Other people can be stung regularly and hundreds of times, and never develop that allergy. It is likely that it was in his genetic makeup to be allergic, but needed a trigger to bring the allergy on. If it wasn't in his genetic make up to have the sensitivity, it may have never developed into a full allergy. --Jayron32 19:31, 10 September 2012 (UTC)

Another component might other changes in the environment. For example, I developed hay fever (to ragweed) about eight years ago (verified by scratch test, etc.). I've lived in the same general area of SW Ontario all my life and never had a problem, but now spend every Aug-Oct period sick as a dog. So, what changed? Did my body require thirty years to gradually build up a sensitivity to ragweed? It seems unlikely, but it's obviously possible. Now, three years ago, we went on a camping vacation to an area about 200 km away. It was the end of August and Murphy's Law required that I forget all the allergy medication at home. However, my whole family (who are also all horribly allergic to ragweed) and I spent the three or four days there with nary a sniffle or itchy eye. So, what caused that? Different species of ragweed? Less irritation from general city air pollution? There are so many different factors that it's hard to say. Maybe your rice allergy developed due to a new strain being planted that changed the pollen just enough to trigger your outbreak? Or perhaps you'd moved during the off season and got exposed to a slightly different kind of pollen. Or maybe some other pollutant acts in concert with the allergen to create the hay fever. It's obviously a complicated bit of biology. Matt Deres (talk) 16:38, 11 September 2012 (UTC)

Anecdotal evidence: it would seem that allergies such as ragweed are more common among those born in an area where the plant pollen is native, though I could be wrong. Also, food "allergies" such as lactose intolerance are a bit more complex: most people who are Han Chinese are not lactose intolerant (citation needed), but the rate of intolerance reaches ~90% among second-to-third-generation populations of Chinese descent in North America. ~AH1 ^(discuss!) 18:23, 11 September 2012 (UTC)

But lactose intolerance isn't an allergy, it's the discomfort of having gut flora digest the lactose, causing gas, rather than enzymatic digestion. If your point is correct, could it not simply be because Han in China don't even carry the gut flora that cause the gas? μηδείς (talk) 02:42, 12 September 2012 (UTC)

Relativity theory: a body moving while emitting an energetic radiation

According to Einstein's formula, when one calculates the current mass of a resting body that has just emitted an energetic radiation $E$ , one should subtract $E/c^{2}$ from the original mass the body had had before it emitted the radiation. Note that Einstein did not have to explain whether the original mass he referred to - is the original total mass - or the original rest mass, because he referred to a resting body, i.e. to a body whose rest mass is equal to its total mass. However, how about calculating the current mass of a moving body that has just emitted an energetic radiation $E$? Should one subtract $E/c^{2}$ from the original total mass - or from the original rest mass? Note that this matters very much, because if one subtracts $E/c^{2}$ from the original rest mass, then one subtracts $\gamma E/c^{2}$ from the original total mass. If you think you know the answer (which I don't), then please explain also the full considerations which could have enabled one to get the answer by oneself. 77.127.218.31 (talk) 09:28, 10 September 2012 (UTC)

I'll give it a try. There are three quantities involved here: energy, momentum and mass. These are related through the four-momentum of a particle. This is a vector with four components (E, p_x c, p_yc, p_zc). The length of this vector is computed as ${\sqrt {E^{2}-p_{x}^{2}c^{2}-p_{y}^{2}c^{2}-p_{z}^{2}c^{2}}}=mc^{2}$, where m is the mass of the particle, this is invariant (what you call "rest mass"). For a particle at rest, the four-momentum is (E_p, 0, 0, 0) = (mc², 0, 0, 0). Let that particle emit a photon of energy E in the positive x direction; the four-momentum of the particle photon is (E, p_xc, 0, 0) = (E, E, 0, 0) (the momentum is equal to the energy because the photon has mass zero. Now apply conservation of energy (for the first - actually zeroth -- component) and momentum (for the other components), to determine the four-momentum of the particle after emission (just the difference between the four-momentum of the particle before emission and the four-momentum of the photon), it is (E_p − E, −E, 0, 0). With E_p = m, you find that the mass of the particle after emission is ${\sqrt {m^{2}-2mE/c^{2}}}$. This shows that the particle after the emission cannot be the same as the one before. Hence, if the particle is an elementary particle it decays to another particle type (examples?). If it is an atom, it changes its internal structure (an electron drops to a lower energy level). If you insist on using "total mass" (but this concept is redundant because it is nothing but the energy), then you can simply say that the photon's energy is subtracted from the total mass, although most physicists today prefer saying that the photon's energy is subtracted from the particle's energy. In terms of rest mass, it is somewhat more complicated, as I've shown. --Wrongfilter (talk) 15:03, 10 September 2012 (UTC)

.

Thank you for your detailed response. I appreciate it. I have two comments, the second one being more important:

.

1. You write: "the four-momentum of the particle is (E, p_xc, 0, 0)", but you probably meant: "the four-momentum of the photon is (E, p_xc, 0, 0)". Later you write: "the momentum is equal to the energy", but you probably meant "the momentum multiplied by the velocity is equal to the energy". Later you write: "With E_p = m", but you probably meant: "With E_p = mc²". Later you write "it is nothing but the energy", but you probably meant "it is nothing but the energy divided by c²". Later you write: "you can simply say that the photon's energy is subtracted from the total mass", but you probably meant: "you can simply say that the photon's energy divided by c² is subtracted from the total mass".

.

2. Let me take your sentence that contains (in my opinion) the most direct answer to my question: "If you insist on using 'total mass'...then you can simply say that [in order to get the mass of the particle after the emission] the photon's energy [divided by c²] is subtracted from the total mass [the particle had before the emission]". Please notice that the total mass of a particle at rest is simply its mass - i.e. its "rest mass" (as I call it), so your sentence mentioned above lets me conclude that: "[in order to get the mass of the resting particle after the emission], the photon's energy [divided by c²] is subtracted from the [rest] mass [the resting particle had before the emission]". However, this conclusion contradicts your (well-reasoned) conclusion that "the mass of the particle after emission is ${\sqrt {m^{2}-2mE/c^{2}}}$ ", doesn't it?

.

By the way, what did you mean by :"I'll give it a try"? Did you mean a try to give the real answer - although you're not sure whether it's the real answer, or a try to explain the answer to me - although you're not sure whether you will succeed to explain it to me?

.

Again, I appreciate very much your detailed response. 77.127.218.31 (talk) 19:15, 10 September 2012 (UTC)

ad 1): "four-momentum of the photon" is what I meant, I've corrected it above. I usually do these things in units where c=1, hence I usually don't write c in any of of these equations. I tried to pay attention to this, but apparently I failed repeatedly... You're right on all accounts. ad 2): I understood "total mass" as being "relativistic mass", i.e. energy divided by c², i.e. the first component of the four-momentum. That's a frame-dependent and redundant quantity, and I very much dislike it. Before emission, the particle's "relativistic mass" is equal to its "rest mass", because we chose to work in a frame where the particle is at rest initially. Incidentally, the equation for the (rest) mass still has the frame-dependent E; in order to make it valid in any frame, E should be divided (right?) by $\gamma$. By "I'll give it a try", I mean "I'll try to remember/figure out how to do this" and "I'll try to explain it in a way that people can understand it". --Wrongfilter (talk) 20:44, 10 September 2012 (UTC)

Unfortunately, I suspect I didn't understand your reply to my second comment: Please note that (in my previous response) I was talking about a particle at rest (in our reference frame), i.e. about a particle whose $\gamma$ equals 1. How can this remove the contradiction I've pointed at? Namely: on one hand you claim that "If you insist on using 'total mass'...then you can simply say that [in order to get the mass of the particle after the emission] the photon's energy [divided by c²] is subtracted from the total mass [the particle had before the emission]". So, in our reference frame - in which the particle's $\gamma$ equals 1, the mass of the particle after the emission must be $m-E/c^{2}$ , where $m$ is the (rest) mass the particle had before the emission; However, on the other hand you claim that "the mass of the particle after emission is ${\sqrt {m^{2}-2mE/c^{2}}}$ ". Can't you see the contradiction (at least in our reference frame in which the particle's $\gamma$ equals 1)? 77.127.218.31 (talk) 21:38, 10 September 2012 (UTC)

The emitting body isn't at rest after the emission—there's a recoil. So $m-E/c^{2}$ is not its final rest mass, though it is its final relativistic mass in the frame where it's initially at rest. -- BenRG (talk) 22:58, 10 September 2012 (UTC)

Yes, thank you. I'd thought $\gamma =1$ because I'd ignored the recoil...

Btw, I think that, when considering the possibility of a mass emitting photons, there is no longer any justification for regarding the amount ${\sqrt {E^{2}-p_{x}^{2}c^{2}-p_{y}^{2}c^{2}-p_{z}^{2}c^{2}}}$ as "invariant" (as Wrongfilter and almost all phisicists regard it).

I also think that, when ignoring the concept of a "relativistic mass" (which Wrongfilter "dislikes" as he explicitly wrote), there's no longer any justification for the equation: ${\sqrt {E^{2}-p_{x}^{2}c^{2}-p_{y}^{2}c^{2}-p_{z}^{2}c^{2}}}=mc^{2}$ when we have no pre-assumptions, except for: logic, mathematics, the newtonian definitions for momentum (namely: the product of mass and velocity) and for energy (namely: the integral of velocity over momentum), and the three invariants: conservation of momentum, conservation of energy, and light speed. 77.127.218.31 (talk) 11:35, 11 September 2012 (UTC)

You're wrong about that. ${\sqrt {E^{2}-p_{x}^{2}c^{2}-p_{y}^{2}c^{2}-p_{z}^{2}c^{2}}}=mc^{2}$ is always an invariant. Note that invariant doesn't mean conserved. An invariant is a quantity that has the same value for all observers. A particle's rest mass is an invariant - that is all observers measure the same value for its rest mass. But when the particle emits a photon, its rest mass changes, so it is not conserved (Though the total rest mass for the particle + photon system is conserved). The new rest mass (after the emission of the photon) is a new invariant - that is all observers measure the same value for it. Dauto (talk) 14:22, 11 September 2012 (UTC)

You probably mean I was wrong about the invariant. Yes, you're right. However, I still think I wasn't wrong regarding my last comment - about the relativistic mass. 77.127.218.31 (talk) 14:51, 11 September 2012 (UTC)

I don't understand what you mean by that last comment. In the particle's rest frame (where its momentum is zero), its energy is E₀ = mc² - Einstein's famous equation, with m the particles (rest) mass. The length of four-momentum is invariant with respect to coordinate transformations (in particular Lorentz transformations), its length is still mc² when viewed in a frame where its momentum is not zero (in fact the vector as a geometrical object is invariant, it's just its coordinate representation that changes). This is a very fundamental equation in special relativity. At this point, you may want to read a book on special relativity — I learned a lot from the first chapters of Bernard Schutz's "A first course in general relativity". --Wrongfilter (talk) 15:23, 11 September 2012 (UTC)

I mean that if you ignore the concept of relativistic mass then you can't deduce (what you call) "Einstein's famous equation" with m being the particle's (rest) mass. However, this is not the case when you do consider the concept of relativistic mass, because then you can mathematically prove Einstein's formula if E refers to the kinetic energy granted to a particle at rest, and m refers to the relativistic mass added to a mass at rest, and then you can generelize Einstein's Formula - also for the energy contained in a rest mass - by a simple induction, which takes into account the conceptual similarity between relativistic mass and rest mass (similarity in the sense of their not being distinguished in terms of conservation of momentum). To sum up, what I'm actually claiming - is that you couldn't have reached Einstein's formula (when referring to a rest mass) if you had ignored the concept of relativistic mass. 77.127.218.31 (talk) 15:45, 11 September 2012 (UTC)

Does anyone know what this plant is?

Hi all,

I have a small bush growing in my yard in Cambridge, MA. Its leaves are smooth and almond-shaped, and they have a marking on the top like someone covered them with a smaller spikier leaf and spray painted over them. The branches end in long tendrils with dozens of tiny tiny red buds on them. It grows out of multiple thin stems, and the whole thing is about two feet tall.

http://img.skitch.com/20120910-jk4qjt9rt6sqkg54casg23mh8a.jpg

(Is there a way to embed non-Wikipedia images?)

Thanks! — Sam 24.128.48.26 (talk) 11:32, 10 September 2012 (UTC)

Looks like this Persicaria. --TammyMoet (talk) 11:47, 10 September 2012 (UTC)

Thanks! The picture linked looked quite different, but looking it up online finds that you're completely right. Persicaria virginiana — Lance Corporal Knotweed. Thank you so much! — Sam 24.128.48.26 (talk) 12:30, 10 September 2012 (UTC)

It's often easier to identify plants once they are in flower. Yours appears to have finished flowering so we can't go much on the stalk that's left. But I've seen that plant quite a bit over the years. Some varieties of Persicaria are edible. --TammyMoet (talk) 18:16, 10 September 2012 (UTC)

Maldives in the Ice Age

Hi. I'm not sure how coral islands work. But I've read in the last ice age, sea levels were about 120 metres lower than now. So were the Maldives then bigger islands? Would they have had land as much as 100 metres above sea level? How big would they have been and where could I find a map. Thank you. — Preceding unsigned comment added by 142.150.38.84 (talk) 13:30, 10 September 2012 (UTC)

Probably not. The way that coral reefs form (see Coral_reef#Formation), they generally stay just ahead of sea-level rise. Coral atolls like the Maldives aren't that old, geologically. Most are likely less than 10,000 years old, according to our article, and thus if the Maldives existed during the lower sea levels of the last ice age, they would have also been comparitively lower in elevation, so they may not have been considerably larger. Atoll#Formation also has some information, but also notes the dynamic nature of atolls. --Jayron32 16:55, 10 September 2012 (UTC)

That makes sense logically, but this study reports finding the remains of a 120-thousand-year old reef only 14 meters below current sea level -- so it seems likely that there would have been about 100 meters of exposure at the LGM. Looie496 (talk) 19:12, 10 September 2012 (UTC)

When another ice age rolls around and the sea level drops again, the reefs will die, but they won't disintegrate, and the Maldives will grow until the sea level starts rising again. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 23:30, 10 September 2012 (UTC)

Eustatic sea level is a tricky subject these days. There is much proxy data but if coral growth is factored in, then paleo-environmental sea level readings are perhaps unreliable in those locations. ~AH1 ^(discuss!) 18:14, 11 September 2012 (UTC)

Cat

Can someone identify the brownish cat in the 21st (or third from bottom) picture in this article for me - [5]? - JuneGloom Talk 19:44, 10 September 2012 (UTC)

It's a caracal. μηδείς (talk) 20:01, 10 September 2012 (UTC)

I just ECed with Medeis, but was about to give the same answer. It looks like a caracal, or perhaps a cross including the caracal, such as the Caraval. --Jayron32 20:06, 10 September 2012 (UTC)

Our article suggests the caraval is spotted. Not to be confused with the also quite handsome Caracalla. μηδείς (talk) 20:43, 10 September 2012 (UTC)

Thank you guys. I couldn't remember what it was, but knew I'd seen one before. I assume the one in the article is domesticated and lives alongside the Cheetah. - JuneGloom Talk 23:07, 10 September 2012 (UTC)

Plate tectonics censored?

(Not sure if this is the right section.) From the agnatology article: "For example, knowledge about plate tectonics was censored and delayed for at least a decade because key evidence was classified military information related to underseas warfare." I can't find any reference to this online, nor am I sure what is meant by 'classified military information related to underseas warfare' in this case. Ratzd'mishukribo (talk) 21:08, 10 September 2012 (UTC)

Well, knowledge of the location of undersea trenches and ridges is important if you want to hide a sub from enemy detection, and those are caused by plate movements. And, when you plot all of those trenches and ridges, it's clear that they are like the seams on a baseball, not just the result of isolated events. StuRat (talk) 21:17, 10 September 2012 (UTC)

Much of the knowledge of the sea floor was gathered through military surveys, and the military's default state is "classify everything we know". It wasn't until the 1950s that a significant civilian effort to map the sea floor was undertaken. However, I don't know that the suppression was all that significant or that it was as widespread and obfuscatory as that statement makes it out to be. Harry Hammond Hess, a U.S. naval sonar operator and later geologist, was key in developing the modern theory of Plate Tectonics. I don't know that he was deliberately censored. --Jayron32 21:35, 10 September 2012 (UTC)

So if that is what was meant, is there indeed any source to the allegation that it was ever censored, or perhaps someone needed an example of subject matter for the the topic "agnatology"? Ratzd'mishukribo (talk) 02:25, 11 September 2012 (UTC)

'Censored' is the wrong word, to quote from a book by Naomi Oreskes "Another group at Lamont had focused on bathymetric data - measurements of the depth of the sea floor - primarily in the Atlantic. These data were highly classified, but Bruce Heezen (1924-1977) and Marie Tharp had found a creative means around security restrictions: a physiographic map, essentially an artist's rendition of the what the sea floor would look like drained of water, based on quantitative measurements, but without actually revealing them."[6]. So even if the data were classified, researchers still had access to them and found ways around the restrictions. Mikenorton (talk) 04:24, 11 September 2012 (UTC)

Is the concept of "hiding submarines among ridges" real or happens only in The Hunt for Red October (film)? I would have thought that mid-ocean ridges are way deeper than where submarines can go (a couple of deep sea research vessels excepted, and assuming they want to come back up.) 88.112.47.131 (talk) 05:50, 11 September 2012 (UTC)

They aren't all deep. Iceland is on top of one. See Iceland#Geology. StuRat (talk) 06:03, 11 September 2012 (UTC)

And do submarines hide there among rocks in real life? 88.112.47.131 (talk) 13:12, 11 September 2012 (UTC)

This article may be of significant yet peripheral interest: Timeline of the development of tectonophysics (after 1952). ~AH1 ^(discuss!) 18:12, 11 September 2012 (UTC)

I wouldn't expect so in peacetime, unless for training purposes, as it's risky. However, in war, when the enemy is out to destroy your sub, that changes the risk/benefit equation. StuRat (talk) 20:45, 11 September 2012 (UTC)

This page mentions wartime secrecy about plate tectonics evidence - essentially as StuRat and Jayron wrote - but also without a source. Ratzd'mishukribo (talk) 02:17, 12 September 2012 (UTC)