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April 4

Black hole gravity

A black hole has gravity so strong that light can't escape. How does its gravity escape? 71.215.74.243 (talk) 03:49, 4 April 2012 (UTC)[reply]

Well, only objects with mass is trapped by the gravity well; grivity does not have mass. Plasmic Physics (talk) 03:55, 4 April 2012 (UTC)[reply]

Agreed. If gravitons exist, they must be massless. StuRat (talk) 04:32, 4 April 2012 (UTC)[reply]

Photons are also massless.--Modocc (talk) 04:48, 4 April 2012 (UTC)[reply]

as are synagogues.--188.157.191.113 (talk) 20:31, 4 April 2012 (UTC)[reply]

...although weighty topics are often discussed. StuRat (talk) 04:44, 6 April 2012 (UTC) [reply]

See gravitational wave. In M theory gravitons are not stuck to the brane (like particles with mass are) and so are free to radiate into higher dimensional space. SkyMachine ⁽⁺⁺⁾ 04:45, 4 April 2012 (UTC)[reply]

(ec) That doesn't help — photons are also massless, but cannot escape a black hole. It's well known that there are lots of difficulties reconciling quantum mechanics with general relativity; I wouldn't be surprised if this were one of them (or at least one problem with describing gravity in terms of virtual-graviton exchange).

Photons have no rest mass, but that seems rather an esoteric thought, as they are never at rest, and do have mass when in motion. By comparison, I believe that gravitons never have any mass. Any physicists here to verify this ? StuRat (talk) 04:51, 4 April 2012 (UTC)[reply]

Photons and gravitons both have zero invariant mass, which is what physicists usually mean by the word "mass" these days when not otherwise specified. I am reasonably sure that gravitons, assuming they exist, do indeed have nonzero mass in the sense of mass-energy — if they didn't, how could they affect anything else? You also skipped over my second paragraph, which shows that the same problem exists with photons and electric forces. --Trovatore (talk) 04:57, 4 April 2012 (UTC)[reply]

Note also that black holes can definitely have charge, and their charge can be felt via electrostatic forces, which according to QED are mediated by the exchange of virtual photons, even though real photons cannot escape the black hole. I don't know what the resolution of this problem is, or even whether there is an agreed resolution, but it does appear to show that the problem is not unique to gravity. --Trovatore (talk) 04:47, 4 April 2012 (UTC)[reply]

See General relativity. Gravity is not what you think it is. --Jayron32 04:25, 4 April 2012 (UTC)[reply]

For black holes to gravitate, their gravitons, if these exist (they have not yet been measured (perhaps due to extinction) and are only theoretical) would need to "escape"... in which case, the gravitons could not be interacting with themselves very often. Analogously, photons that get absorbed/emitted are able to form waves with minimal interactions with each other. --Modocc (talk) 04:45, 4 April 2012 (UTC)[reply]

Try Google for "how does gravity escape a black hole?" You get loads of links, like [1] [2] [3][4] etc. Some points from these:

How does the electric field escape a charged black hole?

Gravitons don't have to escape; the information needed is all present in the original collapsing star.

Virtual particles can do all sorts of unreasonable things, like escape the event horizon. They can't carry information, but they effectively mediate the force...

Wnt (talk) 05:50, 4 April 2012 (UTC)[reply]

You say virtual particles can't carry information (and I've heard it said before), but what does that actually mean? It is those virtual particles that tell us the mass and charge of the black hole - isn't that information? --Tango (talk) 06:06, 4 April 2012 (UTC)[reply]

Any information released would be random and unrelated to the original information of particles that fell into the black hole, appart from only a few sum total values such as total mass (which you infer from size of the event horizon), sum charge. SkyMachine ⁽⁺⁺⁾ 07:19, 4 April 2012 (UTC)[reply]

Quantum entanglement is handy of course, especially whenever information is lacking..., --Modocc (talk) 08:44, 4 April 2012 (UTC)[reply]

I like the stringball/fuzzball theory of black holes: where everything is crushed into a quivering, tiny non-zero yarn ball of superpositioned superstring where entanglement means nothing. Plasmic Physics (talk) 08:58, 4 April 2012 (UTC)[reply]

Gravity can't escape a black hole. Gravity is the curvature of spacetime. Small fluctuations of that curvature, i.e., gravitational waves, travel at the speed of light. Gravitational waves originating from inside a black hole can't escape the black hole any more than can light. For example, suppose an AM CVn star gets sucked into a supermassive black hole. To an outside observer, as the binary star approaches the event horizon, the frequency of the gravitational radiation from the binary star will slow down to doing nothing more than just producing a static increase in the black hole's Schwarzschild radius, very similar to how the light from the binary star will be observed to redshift down to nothing as it approaches the event horizon. An AM CVn star's orbital period is so short, and a supermassive black hole is so huge, that after the binary star has crossed the event horizon, it will still have enough time to produce a few more periods of gravitational radiation before it hits the gravitational singularity at the black hole's center. But those last few periods of gravitational radiation will never make it outside of the event horizon, just like the last bit of light produced won't. Red Act (talk) 15:30, 4 April 2012 (UTC)[reply]

As in electromagnetism, the gravitational field depends on a retarded potential. The source of the retarded potential outside a black hole is the collapsing matter before it became a black hole. It isn't the mass inside the black hole (if it were, that would imply that light could escape the black hole). -- BenRG (talk) 17:21, 4 April 2012 (UTC)[reply]

Asking how a graviton could escape gravity seems somewhat similar to me as asking how an electron could avoid getting electrocuted... --Mr.98 (talk) 01:49, 5 April 2012 (UTC)[reply]

What's the right word? gravitate? - do gravitons gravitate? Plasmic Physics (talk) 03:32, 5 April 2012 (UTC)[reply]

Before deciding what they do, we should probably have some evidence they exist, and since a) there is yet no evidence that they do and b) there are perfectly workable theories of gravity that don't need them too, it doesn't really matter for this discussion. General relativity provides a very good, mathematically rigorous and very consistent with experimental evidence model which doesn't require us to invent properties for a particle no one has yet found and which no one can even agree on what it should behave like enough for us to know how to look for it. --Jayron32 03:40, 5 April 2012 (UTC)[reply]

Everyone agrees on what a graviton is: it's a quantized vibration of the gravitational field. It's hard to imagine a quantum theory of gravity that wouldn't have that. Part of the problem here is the idea that quantum forces are transmitted by particles, whereas classical forces are transmitted by fields, and therefore you need to throw away all of your classical knowledge and approach things in a completely different way when you go quantum. That's not true. Forces, whether quantum or classical, are transmitted by fields. There's also a virtual-particle picture of forces, both quantum and classical. It's often useful for calculation in the quantum case, and not so usefully classically, but if you're just trying to understand the nature of forces, then that shouldn't matter.

Anyway, gravity does gravitate. In the virtual-particle picture that shows up as interaction vertices involving only gravitons (whereas there are no interaction vertices involving only photons). It makes sense to ask whether a gravitational wave can escape a black hole, and in particular it makes sense to ask whether a real graviton (the weakest possible gravitational wave) can escape a black hole. I don't know if it makes sense to ask if a virtual graviton can escape a black hole. Probably it doesn't make sense. -- BenRG (talk) 07:15, 5 April 2012 (UTC)[reply]

A black hole at the distance farther than event horizon is any star,"the effect of supernova in our world is much much more than black hole"suppose that our sun comes to be any black hole all solar system family will remain in their placement ,the gravity fields are equal and wonderful for you is that orbitals will remain same.--Akbarmohammadzade (talk) 14:34, 5 April 2012 (UTC)

 the wonderful subject which have not been worked on is reverse
world creating in such planetary system. the observatories in thus field
of study might see world reverse.--Akbarmohammadzade (talk) 14:34, 5 April 2012 (UTC)[reply]

In fact I love supernovae ,for their light celebrating at start , their palpitating as our heart , for their expanding as the universe and their role in our existence with sending heavy elements here--Akbarmohammadzade (talk) 15:33, 5 April 2012 (UTC)Italic text[reply]

force carrier particles in a black hole

If the fundamental forces of nature are carried in particles/waves, as quantum physics maintains that they are, then how are gravitrons (for mass) and charge carriers able to exert influence on the outside world, seeing as they cannot escape the black hole? Thus, a black hole should not be able to exert the full amount of force due from its mass nor its charge. Magog the Ogre (t • c) 15:45, 2 March 2013 (UTC)

   A (theoretical) graviton is itself massless, so is not affected by gravity.

How can it "carry mass" yet itself be massless ? Think of it as carrying

information, like if you carry a bankbook with your bank balance written

on it, but that doesn't mean you have the actual cash on you. StuRat (talk) 16:38, 2 March 2013 (UTC)

       StuRat, sometimes the answers you give are very useful, but I

wish you would stop answering questions where you don't have the slightest

idea what you are talking about. Massless particles are indeed affected
by gravity, for example photons are redshifted when they pass through a

gravitational field. Regarding Magog's question, there is no accepted theory of quantum gravity currently, but if one is ever developed, it

will probably have the gravitons created in the region outside the

event horizon. The story is clearer for charge carriers. Quantum field

theory allows for the creation of an electron-positron pair out of

the vacuum, with one of them falling into the black hole and the

other escaping to infinity -- this mechanism allows electric field
propagation without anything actually escaping across the event
horizon. Looie496 (talk) 16:53, 2 March 2013 (UTC)

               To clarify, you mean particles with no rest mass can
be affected by gravity, due to the mass they gain due to relativity

effects at high speeds, right ? Yes, I neglected to include that (and.

our article on gravitons really should say it has 0 "rest mass", not

just 0 "mass", but I can't figure out how to change it). StuRat (talk) 17:26, 2 March 2013 (UTC)

                   They don't gain mass, but they are affected by
gravity because space time is curved. Light has a mass of exactly 0,
but it cannot escape a black hole due to the curving of space time.
Magog the Ogre (t • c) 17:38, 2 March 2013 (UTC)

                       But, again, photons have zero rest mass, but
do have relativistic mass. StuRat (talk) 20:55, 2 March 2013 (UTC)

 Gravity affects everything equally

(equivalence principle); the mass really doesn't matter, whether it's rest mass or relativistic mass. Also, relativistic mass has never been a very useful concept, and isn't usually taught any more.

It leads people to think that rapidly moving objects should collapse
into black holes, for example, which they certainly don't. -- BenRG

(talk) 23:30, 2 March 2013 (UTC)

           So, in theory then, a black hole could have a hugely
positive charge, but only the content which lay on the event

horizon would be able to affect the outside world in terms of charge? (I realize that incoming material does not fall into a

black hole but forever stays on the event horizon... but what

about material that was present when the black hole expanded, or material that was at the center of the original star, say a

magnetar, when the black hole formed) And what does this say

about the speed of gravity? Magog the Ogre (t • c) 17:08, 2 March 2013 (UTC)

               Matter does cross the event horizon. You can't see
it cross from outside, because no light from the crossing can ever

reach you, but it does cross. The electromagnetic field of a black hole has the same strength as the total field of the matter than went

into the black hole, but it doesn't come from inside; it is a "fossil
field" (retarded potential) from before the matter crossed the event
horizon. The same is true of the gravitational field. This doesn't

change (as far as anyone knows) in quantum mechanics; nothing that happens inside the event horizon affects anything outside. As for the virtual particle picture, I think this is just an example of a case where it isn't very helpful (as Count Iblis sort of says below). -- BenRG (talk) 23:30, 2 March 2013 (UTC)

Virtual particles are just mathematical tools to do computations,

they are unphysical and don't stick to the equations of motions

of real particles. E.g. in case of Compton scattering, two Feynman

diagrams are needed to compute the amplitude. In one of these

diagrams, the emitted photon leaves the electron before the photon

that is going to be absorbed has arrived. Count Iblis (talk) 18:45

, 2 March 2013 (UTC)

exact che soale mahi .... absolute favorite question!!

Why is pressure a scalar?

Consider this figure, which shows the forces due to pressure on an infinitesimal quantity of fluid. It's clear that p_y(y)=p_y(y+dy)+ (infinitesimal), or else there'd be an infinite acceleration in the y direction. Similarly, p_x(x)=p_x(x+dx) + (infinitesimal). But pressure is treated as a scalar, which implies/assumes that p_x=p_y. What's the justification for this assumption? It's not surprising, physically, that this would be the case if the fluid were incompressible, but is there a more rigorous justification for the claim that pressure is a scalar? 65.92.5.132 (talk) 04:21, 4 April 2012 (UTC)[reply]

In the general case, you have a stress tensor. Intuitively, I think it's the case that the stress tensor has to reduce to a single number, the pressure, in equilibrium conditions in a material that has zero tensile strength, but I can't say I'm sure about that. --Trovatore (talk) 04:26, 4 April 2012 (UTC)[reply]

Almost. In the most general case (nonzero viscosity and/or nonequilibrium system), you can have shear stresses, too. In a static system and the case of zero shear modulus, shear stresses must vanish and the stress tensor has to be diagonal in all reference frames. The only rank-2 tensor that is diagonal in all reference frames is a multiple of Kronecker delta (a tensor with all diagonal components equal to each other),

p\delta _{ij}

where p is pressure. This fact is known as Pascal's law (though he obviously discovered it experimentally long before people knew about tensors or shear moduli.)--Itinerant1 (talk) 05:35, 4 April 2012 (UTC)[reply]

And to put this into more intuitive terms, consider the figure from the original question, and draw a plane at a 45 degree angle through the box. If

p_{y}(y)\neq p_{x}(x)

, you can calculate that the force exerted by the fluid on one side of the plane on the fluid on the other side is not perpendicular to the plane. --Itinerant1 (talk) 05:43, 4 April 2012 (UTC)[reply]

This is why I visualise pressure as equivalent to energy density, which is a scalar quantity. I used to think that since force is a vector quatity, pressure must also be a vector quantity; that pressure is the average quantity of force applied at a right angle a surface. Plasmic Physics (talk) 06:19, 4 April 2012 (UTC)[reply]

Okay, so what I've understood so far is that 1) in general, the forces on a small fluid element won't be represented by a scalar, but by a tensor, so that the force on the top of a small cube of fluid will not in general be the same as the force on the side, and 2) in the case of either no shear forces or static fluid, it can be shown that the tensor reduces to $p\delta _{ij}$ , which means that we can talk about pressure as a scalar (the proof of this is clear when there are no shear forces, but why does this hold in a static fluid?).

There are no shear forces in static fluid, because the shear modulus of a fluid is virtually zero, so the only possible source of shear forces is viscosity, and those are zero as well if there fluid is static.--Itinerant1 (talk) 18:14, 4 April 2012 (UTC)[reply]

But, I still have a problem. Even when the forces are treated as a tensor, pressure is still treated as a scalar. For example, in a Newtonian fluid, the diagonal components of the stress tensor $T_{ii}$ is equal to $-p+2\mu {\frac {\partial u_{i}}{\partial x_{i}}}$ , where μ is the kinematic viscosity and p is the pressure, clearly treated as a scalar. So, what gives? Is pressure being defined as a scalar? 65.92.5.132 (talk) 15:03, 4 April 2012 (UTC)[reply]

According to this, yes. It is defined as 1/3 the trace of the stress tensor.--Itinerant1 (talk) 18:26, 4 April 2012 (UTC)[reply]

Thanks. 65.92.5.132 (talk) 01:16, 5 April 2012 (UTC)[reply]

there are 4 kinds of topology in Set{1,2},and 29 in Set{1,2,3}, is exist a Formula in a Set {1,2,3,4,5,6...n}?

there are 4 kinds of topology in Set{1,2},and 29 in Set{1,2,3}, is exist a Formula in a Set {1,2,3,4,5,6...n}? — Preceding unsigned comment added by Cjsh716 (talk • contribs) 04:34, 4 April 2012 (UTC)[reply]

A topology τ on a set X is a subset of X which satisfies a number of rules. Thus any valid τ must be an element of the power set of X, and thus the number of topologies must be less than or equal than 2ⁿ where n = |X|. So how can there be 29 possible topologies of the set {1,2,3} as 29 > 2³ = 8? (Or am I misunderstanding something about your notation?) -- ToE 06:37, 4 April 2012 (UTC)[reply]

The 29 appears in our article Finite topological space#Number of topologies on a finite set, so I struck my comment above and will withdraw from this discussion until I better understand what is going on. -- ToE 06:42, 4 April 2012 (UTC)[reply]

A topology is a set of elements of the power set, not just an element of it, so the number of topologies is bounded by 2^(2^n), not 2^n. AndrewWTaylor (talk) 08:19, 4 April 2012 (UTC)[reply]

29 doesn't look much like it would be 2^(2^n) of anything. I guess 2^(2^3-3)-3....--188.157.46.184 (talk) 08:28, 4 April 2012 (UTC)[reply]

try http://oeis.org/A000798 --Digrpat (talk) 09:13, 4 April 2012 (UTC)[reply]

~~In most cases n~~Not every subset of pow(X) is a topology, so 2^(2^n) is an upper bound, not the actual number of topologies. AndrewWTaylor (talk) 09:32, 4 April 2012 (UTC)[reply]

To quote the article: "There is no known simple formula to compute T(n) for arbitrary n." You can write a program to enumerate all subsets and count the ones which are valid topologies. There are probably more efficient algorithms, but I don't know what they are. Dcoetzee 10:04, 4 April 2012 (UTC)[reply]

There's an efficient algorithm here: Posets on up to 16 points--Itinerant1 (talk) 10:11, 4 April 2012 (UTC)[reply]

How is heat energy converted to mass?

Mass is converted to energy in nuclear fusion and fission. Energy can be converted to mass, too. I see that Mass–energy equivalence#Practical examples says that each kilogram heated 1°C gains 1.5 picograms of mass. Where and what is that new mass from heat? 71.215.74.243 (talk) 08:09, 4 April 2012 (UTC)[reply]

It is in the motion of the particles, because heat energy is (mostly random) motion. A fast-moving particle simply has more relativistic mass than a slow-moving one. There are no additional particles making up the mass difference. 157.193.175.207 (talk) 09:09, 4 April 2012 (UTC)[reply]

Mass is the energy of an object at rest. If you add heat to an object, it will gain mass, simply because mass and rest energy are synonymous. The equation E = m c^2 has to be understood as a trivial statement where the factor c^2 does nothing more than convert between conventional units we use for mass and energy. Count Iblis (talk) 15:02, 4 April 2012 (UTC)[reply]

~~You means 'at rest' in your frame of ref. In someone elses frame moving relative to yours, the object would have energy. Yes?~~ Struck edit by banned user. Franamax (talk) 16:06, 4 April 2012 (UTC) — Preceding unsigned comment added by 92.28.95.250 (talk) 15:12, 4 April 2012 (UTC)[reply]

Yes, then the object would have more energy in addition to the rest energy. That additional energy is called "kinetic energy". Count Iblis (talk) 15:49, 4 April 2012 (UTC)[reply]

can some women not orgasm from intercourse, but only clitorally

or if they are unable to do the former, is it considered a dysfunction, what is the cure or can be done (other than obviously wearing socks). --188.157.46.184 (talk) 09:29, 4 April 2012 (UTC)[reply]

It amuses me that some men seem to think that the only acceptable orgasm is a vaginal one. Oh, and the reference to socks. --TammyMoet (talk) 09:31, 4 April 2012 (UTC)[reply]

It's an easy mistake to make - after all, reproduction is through intercourse and men orgasm that way. Having just read http://www.scarleteen.com/article/advice/the_great_no_orgasm_from_intercourse_conundrum however ,I must apologize Tammy, I didn't know this fact. Specifically, my partner thought of herself as having a 'dysfunction' in this way, causing emotional problems like she was 'broken' in some way. To me, I love getting her off orally, and after she has come intercourse is also good for her, but she is upset that she can't come from that. Actually I just learned this would put her in the majority. What do you recommend, if she has come orally and we are having intercourse? Should I touch her as well, or have her touch herself, or use a vibrator or what.. Or is intercourse just a chore in reality for her, despite what she might say about liking it (even if she doesn't get off from it). 188.157.46.184 (talk) 09:37, 4 April 2012 (UTC)[reply]

Cunnilingus aint so bad. Some women have their orgasm hang ups too. But you should probably address this question to Dan Savage if you want the expertize on this, or atleast subscribe to his Savage Love podcast. SkyMachine ⁽⁺⁺⁾ 09:44, 4 April 2012 (UTC)[reply]

Um, thanks for the recommendation, but I think this place can answer the question fine. I do love going down on her, she comes very hard like that, then asks me to put it in, and after a while she is upset that she isn't coming that way too. What do you recommend? One thing would be, we could just stop after a couple of minutes of intercourse... or, use a vibrator/finger her clit as well during it. I don't know, nothing works especially great, and more importantly she has hangups about it and will go as far as to start crying. But as I've just found out, this is perfectly normal (and doesn't bother me at all - the fact that she comes clitorally). Recommendations? 188.157.46.184 (talk) 09:48, 4 April 2012 (UTC)[reply]

In short: yes, there are women who cannot orgasm from vaginal penetration. This is not a dysfunction, but an individual variation. If she's not able to handle penetration for whatever reason - and it sounds like she's really not able to right now, if she can't bear it even with added clitoral stimulation - it's probably best to avoid it in favor of other things. If this frustrates you, just let her know and I'm sure she'll do her best to find a way to make sure you're enjoying yourself too. Most importantly make sure she doesn't feel broken or messed up, because she isn't, just different. Dcoetzee 09:59, 4 April 2012 (UTC)[reply]

You can digitally (with fingers) or with a vibrator stimulate the G Spot to arouse the underside of the female prostate. This is much harder to do in regular intercourse. SkyMachine ⁽⁺⁺⁾ 10:07, 4 April 2012 (UTC)[reply]

this is a good recommendation. are there any positions incidentlaly that are particularly good at stimulating the g-spot. I don't care if I put it in or not, I just want her to have fun. She's the one that asks me to. 188.156.25.146 (talk) 10:13, 4 April 2012 (UTC)[reply]

Well digital stimulation it doesn't matter, so long as she is comfortable. The roman way is your best bet for G Spot stimulation during intercourse. SkyMachine ⁽⁺⁺⁾ 10:24, 4 April 2012 (UTC)[reply]

Tammymoet, I don't know why you find my socks reference weird, just google "socks orgasm" -'When they gave the couples socks to wear, about 80% of the couples were able to achieve orgasm compared with 50% previously in this'. given that you can't come vaginally (not specifically at you, at anyone) is vaginal intercourse still 'fun'? or should we not be doing it at all. 188.156.115.31 (talk) 11:44, 4 April 2012 (UTC)[reply]

Well I also wondered what the socks referred to (did Tammymoet say it was weird - I don't think so) because in my 50 plus years of a pretty open life I had never heard about this unlikely use of socks. Placebos eh? don't you just luvvum! Caesar's Daddy (talk) 13:20, 4 April 2012 (UTC)[reply]

Just so long as it isn't some veiled message alluding to the perpetrating of sock puppetry, which it very easily could be read as. SkyMachine ⁽⁺⁺⁾ 13:41, 4 April 2012 (UTC)[reply]

This sounds cool (news link [5]) - but come on, it was a study of 13 people. That's 4 out of 13 people different, in a study which can't have a very good control ... I'm not even hunting for this one in the literature. The orgasm may blow your socks off, but the study? Not so much. Wnt (talk) 19:21, 4 April 2012 (UTC)[reply]

I'll just say that the best positions for stimulation for the woman are with her on top, or from behind on the side (according to Alex Comfort's Joy of Sex). Actually, why don't you get a copy of "The Joy of Sex" and work your way through it together? (by the way I'm still laughing at the sox...) --TammyMoet (talk) 18:29, 4 April 2012 (UTC)[reply]

Yep, they're a joke this season. A Quest For Knowledge (talk) 18:40, 4 April 2012 (UTC)[reply]

Live engineering status for UK TV and radio broadcast towers

I'm currently unable to receive digital radio; as two unrelated receivers experience the problem, I expect that the transmitter is currently off or operating on reduced power. Here in Yorkshire we've had some late snow and high winds and some places have experienced power outages and unplanned voltage excursions, so that's understandable. I think the DAB signal comes from Bilsdale transmitting station - I'm getting DVB-T okay, so it's possible that the radio transmitters alone are off (perhaps Crown Castle, in their infinite wisdom, have chosen to prioritise Cash in the Attic over The World at One (grrr)). I'd hoped I could find a website that shows the current status of the various transmitters on that (and in general other) towers. While it's mostly of academic interest to me, I'm sure it's vital for professional TV and radio installers (who don't want to waste time investigating a customer's reception when the problem is a failure at the broadcast tower). I can't find such a site, either for planned or exceptional outages - does such a resource exist? 46.208.221.86 (talk) 12:58, 4 April 2012 (UTC)[reply]

I've noticed that such info is rarely made publicly available. I suspect the reason is that they don't want advertisers (or customers, in the case of publicly supported stations) to get a true picture of how unreliable they really are, as this could adversely affect their income. As for the installers, I would expect them to carry a portable antenna with them, so they can verify that there is a signal, before working on the customer's equipment. This would also allow them to determine the optimal location and direction for the antenna, before they begin installation. StuRat (talk) 15:03, 4 April 2012 (UTC)[reply]

There used to be a weekly program on British TV aimed solely at installers. It gave details of new transmitters or relays, the frequency bands that the handful of analog stations were broadcast on, and the polarity of the antenna. This also listed planned outages and low power events. This was usually on between Carole Hersee and Pages from Ceefax; I've not seen it for 15 years at least. -- Finlay McWalterჷTalk 15:44, 4 April 2012 (UTC)[reply]

The BBC has a 'Reception Problems' tool at https://faq.external.bbc.co.uk/templates/bbcfaqs/emailstatic/interferencePage?entryID=&id=POKHPGEVQP0B1IPKD0SE2UO6DM&moduleID=&reset=true. You can enter your postcode and find out what's going on in your area. Without knowing your postcode I can't really tell what the problem might be, but just for the hell of it I made up a Bradford postcode, and found out that there's some -engineering work going on on Ilkley Moor (hopefully they've remembered to wear a hat) and the Keighley transmitter is operating at reduced power. - Cucumber Mike (talk) 15:36, 4 April 2012 (UTC)[reply]

Using a Helmsley postcode, I have established that the Bilsdale transmitter was off air until 13:41, so although you will have missed The World At One, hopefully you were back up and running in time to catch The Archers. Jill is overruled. Meanwhile Tony is keen to help. Sounds riveting... - Cucumber Mike (talk) 15:42, 4 April 2012 (UTC)[reply]

Thanks Mike, that's exactly what I was looking for - and it's entirely consistent with the pattern in which the radio stations came back to life. I doubt anyone would forget their hat 200m above Wharfedale, especially today. 46.208.224.194 (talk) 23:02, 4 April 2012 (UTC)[reply]

They might if they'd been "a-courting" Mary Jane. {The poster formerly known as 87.81l230.195} — Preceding unsigned comment added by 90.197.66.205 (talk) 17:27, 5 April 2012 (UTC)[reply]

empty space

if we were able to go out deep into empty space and encapsulate a "volume" of empty space. would that capsule be able to then either compress to a zero space or expand to a very large space without affecting/being affected by the interior empty space? — Preceding unsigned comment added by 165.212.189.187 (talk) 14:55, 4 April 2012 (UTC)[reply]

Pretty much, yes. However, since it's not a complete vacuum, there would be a limit to how far you could compress it. StuRat (talk) 14:59, 4 April 2012 (UTC)[reply]

How about expanding it? would you be "creating" additional space in the universe or "transferring" it from the outside of the capsule to the inside? — Preceding unsigned comment added by 165.212.189.187 (talk) 15:17, 4 April 2012 (UTC)[reply]

If you enlarge or shrink a container in a vacuum, you aren't enlarging, shrinking, creating or destroying the space within the container; all you're doing is moving the container's walls within space. Red Act (talk) 15:39, 4 April 2012 (UTC)[reply]

So, yes, you are "moving space from outside to inside" when you expand the vacuum, even if it's completely enclosed, and without affecting whatever empty space is inside. 188.36.162.23 (talk) 16:00, 4 April 2012 (UTC)[reply]

You are treading dangerously close to that line where you make so many impossible assumptions that science can't give you an answer. There is no such thing as an impermeable container that you can change the enclosed volume at will. I think you might be confused by the idea of a piston filled with fluid; when you pull it up, the pressure inside decreases. But this pressure is only caused by molecules bumping into each other. Nothing is being done to the actual space within the cylinder (whatever that means). If you had instead somehow had a perfect vacuum (which, I must reiterate, is impossible) and pulled up or pushed down on the piston, nothing special would happen. -RunningOnBrains^(talk) 16:28, 4 April 2012 (UTC)[reply]

You aren't "moving space from outside to inside". Space doesn't have a state of motion. When you imagine space going into your box, you are imagining the space as being at rest relative to some lab frame, through which the side of the box is moving. It doesn't work that way. -- BenRG (talk) 17:36, 4 April 2012 (UTC)[reply]

What doesn't work "that way"?; Then how does it work?165.212.189.187 (talk) 17:45, 4 April 2012 (UTC)[reply]

if the box has the mechanisms to expand while keeping what is out out and in in then what would happen?165.212.189.187 (talk) 17:52, 4 April 2012 (UTC)[reply]

You can't move quantities of space around. Space isn't a substance. When the box expands, the amount of volume inside the box is increasing and the amount of volume outside the box is decreasing, but that's exactly what it means for the box to expand. You can't divorce the two, because they're just two ways of describing the same concept. The only way for the box not to increase the volume inside is by not expanding. Rckrone (talk) 03:24, 5 April 2012 (UTC)[reply]

Hmmm, maybe Casimir_effect#Measurement applies? I don't know - is restricting virtual particles like "space" being compressed? - I never even really understood why we have to believe in vacuum fluctuations rather than just minute fluctuations of charge within the conducting plates, like London dispersion force. Wnt (talk) 17:22, 4 April 2012 (UTC)[reply]

You don't have to, and shouldn't. As explained in hep-th/0503158, the Casimir force is just a relativistic quantum correction to the electromagnetic force. -- BenRG (talk) 17:36, 4 April 2012 (UTC)[reply]

That plugs a hole ... thanks!!! But the Casimir effect article is written entirely in terms of virtual particles. I added quotes and a link to the reference, not wanting to make a major revision to a basic physics article I scarcely understand on my own, but somebody should increase the prominence of this saner-sounding interpretation. (I mean, it's like we were describing electric power transmission in terms of holes...). Wnt (talk) 19:13, 4 April 2012 (UTC)[reply]

I don't want to edit it because I think I would be opposed on the grounds that Jaffe's paper isn't mainstream. Most papers still get this wrong. Physicists who read Jaffe's paper change their minds, but nonphysicists have nothing to go on but numbers. I've been waiting for more physicists to come around, but it's taking a long time. -- BenRG (talk) 20:40, 4 April 2012 (UTC)[reply]

Not mainstream? He's writing from the theoretical physics department at MIT. Can you get more mainstream than that? Wnt (talk) 22:39, 4 April 2012 (UTC)[reply]

Sure you can. I don't feel like digging up references, but it's easy to find distinguished particle physicists effectively contradicting Jaffe in published papers. I think there are examples in Jaffe's paper. On the other hand, I'm pretty sure that none of them cite Jaffe. I could try to make a case for Jaffe on the grounds that, seven years later, no one who has read the paper has said that it's wrong. I'll think about it. -- BenRG (talk) 00:25, 5 April 2012 (UTC)[reply]

the OP is basically asking, if you have a collapsible structure with webbing, for example, in space, and you open the structure so that instead of containing 10 cubic centimenters it contains 100 times as much volume. Anything weird or strange happen? No. Where does the extra 90 cubic centimeters of volume come from? Outside of the original confines of the box. This is weird to the OP because if it were on Earth, and an enclosed container contained 10 cc of air, and then 100 cc of air when opened, then it must have a leak. In space no leak is required. (The OP is almost thinking in terms of ether, I think, and my answers were along those lines). In this way of thinking: when you open the webbed box ether magically moves from outside to inside, penetrating the wall. In fact since space is empty, nothing needs to move. I guess if it were a theoretically really impermeable membrane then the few particles per million that space has since it's not a perfect vacuum would be reduced to a lower ratio. (this would especially apply if you're in a part of space with a low density cloud of particles) But the OP is asking about the vacuum part, not the paricles part, and isn't asking or thinking about space with a cloud of particles in it, but mostly empty deep space. 188.157.140.201 (talk) 18:15, 4 April 2012 (UTC)[reply]

What's wierd to me is that you presumed that that was wierd to me. — Preceding unsigned comment added by 165.212.189.187 (talk) 18:57, 4 April 2012 (UTC)[reply]

I don't get it, why even post the question? Something had to be weird to you, as it's a completely straightforward question... what made you ask it? 188.157.140.201 (talk) 19:15, 4 April 2012 (UTC)[reply]

I don't know, I guess I'm arguing for the ether theory. I thought that there would be an upper limit to the expansion of the container besides its obvious physical limitations of having enough matter. Doesn't the energy increase as the container expands? Isn't the space inside sequestered from the space outside? If there was even one atom inside the container would it eventually rip apart or will the atom stop the expansion? — Preceding unsigned comment added by 165.212.189.187 (talk) 19:42, 4 April 2012 (UTC)[reply]

Well, the point of empty space is stuff can move through it freely. If that weren't so... it wouldn't be empty space. But if there are atoms in it (and there are - outer space (the interstellar medium) is not literally empty space - then you could eventually concentrate them to something substantial. See Bussard ramjet, for example. Wnt (talk) 20:03, 4 April 2012 (UTC)[reply]

(edit conflict)Ah here we go; I think I get what you don't get now. You're asking about some ideal situation where you somehow got a "true vacuum" (aka nothing), which really is physically impossible, but I'll try to go on. You should re-read my explanation above. It seems you have a misunderstanding about how pressure works. You are not doing anything to "space" when you, say, pull up on a piston. When you put something, like say a marshmallow, in a vacuum, it expands. But this is not due to some negative force that pulls it apart: this is due to the air bubbles trapped inside (since it was made at atmospheric pressure) exerting an outward force on the marshmallow. Likewise, at sea level, if you pull up on a piston to lower the pressure inside, the resisting force you feel is not that the air or space or anything within the piston is resisting you. The force that is resisting is the atmospheric pressure pushing down on the piston. We don't notice air pressure in our every day experiences because it's essentially the same all around us; there are no pressure differences, or gradients. But atmospheric pressure is quite strong: about 10 kPa, or 10,000 Newtons per square meter. To put this in common units, on every 1 square centimetre (0.16 in²) of surface, the air is pushing down with 1 newton (0.22 lbf) of force, the weight of 0.1 kilograms (0.22 lb) (these units seem small, but even for a decent-sized surface they add up quickly!). If the pressure is the same in the piston as out, there is no net force. But when you pull up, the pressure inside the cylinder decreases, so you must apply the same force as was previously applied by the inner pressure.

Sorry, I've gotten on a bit of a tangent. The point that I think you need to understand is this: pressure is caused by the force of collisions of fluid molecules against a surface. Say you have a box that has a certain pressure of gas inside. If you expand that box, so now the gas has a greater volume and a lower pressure, you have not applied some magical negative force on the molecules, you have simply allowed them more room to move, which means fewer collisions with the side of the box on average, so a lower pressure. "Space" is not a thing in the classical sense, it is the absence of things, and so your container is not interacting with it. Saying that after expansion, there is "more space" in the box, so "where did it come from" is not a valid question, as space is nothing. I hope I have answered your question. -RunningOnBrains^(talk) 20:17, 4 April 2012 (UTC)[reply]

Yes, I was misinterpreting the "negative pressure" idea. can you have infinite negative pressure? — Preceding unsigned comment added by 165.212.189.187 (talk) 20:25, 4 April 2012 (UTC)[reply]

Well, there's the Big Rip idea. If I remember correctly, dark energy = a cosmological constant. The more dark energy, the higher the negative pressure on the universe, which somehow causes space itself to expand. With infinite negative pressure, space expands infinitely. Unless I got the sign wrong. I never really understood this stuff ... it seems as bizarre as some of the talk above to me. How can pressure cause space itself to expand/contract? Wnt (talk) 20:49, 4 April 2012 (UTC)[reply]

"Pressure" itself has no effect on space. With regard to dark energy, "negative pressure" is shorthand for the effect dark energy is having on spacetime. It's not actual pressure in the barometric sense, but a repulsive force. — The Hand That Feeds You:^Bite 21:37, 4 April 2012 (UTC)[reply]

No, it is honest to goodness negative pressure. A more familiar name for "negative pressure" is "tension". I'm not sure where the "negative pressure" terminology came from—it seems pointlessly obscure.

Pressure and tension do gravitate, and tension gravitates negatively. This is a gravitomagnetic effect—pressure is like a current of mass (more precisely, of momentum). -- BenRG (talk) 23:23, 4 April 2012 (UTC)[reply]

Didn't someone just ask why pressure is scalar (has no direction)? Thus isn't negative pressure quite different from tension, as tension most definitely has a direction!! (just pull a boat from far away and it will move toward you - quite different from trying to pull up on it from a helicopter. So negative pressure != tension! 134.255.48.97 (talk) 23:44, 4 April 2012 (UTC)[reply]

Tension can be omnidirectional, though it's harder to think of realistic examples (a gas can't be under tension). One example is a sphere of metal that shrinks when cooled surrounded by a spherical shell of metal that doesn't shrink (or shrinks less). When it's cooled, the sphere should be under roughly constant tension. Another example is a boat with a bunch of ropes attached, pulling in all directions. Pressure can also be unidirectional (e.g., water pressure in a pipe, or pushing a boat). -- BenRG (talk) 00:25, 5 April 2012 (UTC)[reply]

Water pressure in a pipe isn't unidirectional, it's just being resisted in some directions and not in others. The pressure is still there in the directions towards the walls of the pipe. --Tango (talk) 05:03, 5 April 2012 (UTC)[reply]

hydrocortisone

If someone uses a ear drop that has 1% hydrocortisone in it, how much hydrocortisone is absorbed into the bloodstream? — Preceding unsigned comment added by 64.38.226.84 (talk) 21:08, 4 April 2012 (UTC)[reply]

It's very difficult to say without a physical examination. It depends on things like earwax, age, and liver metabolism. 71.215.74.243 (talk) 21:15, 4 April 2012 (UTC)[reply]

Yeah, there's no direct answer for this. It's going to depend on the individual's body chemistry, as well as how much of the drops are used. — The Hand That Feeds You:^Bite 21:40, 4 April 2012 (UTC)[reply]

And whether they subsequently lie on their side, and if so, which side. --Colapeninsula (talk) 09:14, 5 April 2012 (UTC)[reply]

A clinically insignificant amount. Do the math. An ml of 1% hydrocortisone will have 10 mg of hydrocortisone. It is roughly 20 drops. Typical dosing might be 4 drops tid, or about 6 mg a day. When given by mouth, 5 mg a day may be full replacement/full suppression for a 2-5 year old child, but only a small amount of the otic hydrocortisone gets absorbed, much less than 6 mg. refalteripse

Frank Sherwood Rowland

You have put in the wrong portrait. (4 April 2012 at 3:13 PDT) — Preceding unsigned comment added by 66.241.92.137 (talk) 22:13, 4 April 2012 (UTC)[reply]

You should post this to the talk page for that article, along with any proof you have (preferably the correct pic). StuRat (talk) 22:16, 4 April 2012 (UTC)[reply]

Looks an awful lot like the guy in the Telegraph obituary photo. Clarityfiend (talk) 02:53, 5 April 2012 (UTC)[reply]

Header linked for convenience. Richard Avery (talk) 07:21, 5 April 2012 (UTC)[reply]

April 5

Types of copper

What are the types of copper? — Preceding unsigned comment added by Anto Christopher (talk • contribs) 07:51, 5 April 2012 (UTC)[reply]

Do you mean phases? Plasmic Physics (talk) 08:02, 5 April 2012 (UTC)[reply]

Our article copper refers to isotopes, alloys, oxidation states, ores, and many other topics that might be relevant depending on what you mean by "type". --Colapeninsula (talk) 09:16, 5 April 2012 (UTC)[reply]

Well, there are those who cop a dollar and those who cop a feel, so that's two right there. :-) StuRat (talk) 17:20, 5 April 2012 (UTC) [reply]

An "expert-witness" in this morning's New York Times mentions 14 million alloys of copper, though I'm still not sure if he's speaking hyperbolically or literally. In any case, ASME and ASTM publish numerical and nomenclature standards for alloys of copper. There are catalogs full of this stuff. I only use two types of copper, "pure enough for electrical wiring" copper, which is some 99.9% copper, and "machine copper," McMaster calls it brass "Alloy 330" which is alloyed with tin, iron, nickel, and a bunch of other gunk, to make it hard enough to hold its shape when milled or worked. Nimur (talk) 15:36, 5 April 2012 (UTC)[reply]

The exact quote is

He said it would be nearly impossible to prove that Rubedo is a unique mixture of metallic components, adding, "It may be one of the 14 million alloys that people have cooked up over the decades."

14 million sounds about right for the total number of metallic alloys that have been cooked up and characterized over the past few centuries. --Carnildo (talk) 00:17, 6 April 2012 (UTC)[reply]

volume measurer

everyone knows the story of Archimedes' "Eureka! Eureka!" (I have it!) as he ran dripping naked through the streets of Athens, when Archimedes figured out how to determine for the King whether his "pure gold" crown was adulterated. Gold was the heaviest of the metals used, so if he put the supposedly pure Gold irregular crown in a full tub (as he just realized getting into one himself) the amount of water displaced would show the volume of the crown in terms of an exactly equivalent volume of water: the weight of displaced water is easy to measure for them, and if this was heavier (more volume, i.e. less dense crown at same weight) than that displaced by an amount of known pure gold WEIGHING the same as the crown (a simple pan balance lets you assemble such nuggets), in whatever shape or size -- then, since gold was the heaviest of the used metals, it must have been adulterated by something lighter, and the crown would be adulterated. The story continues that the crown was submerged, then pure gold weighing the same as the 'pure gold crown' was submerged, and when the former volume of water weighed more than the latter, the king's pilfering goldsmith was executed.

now my question today is really about how we measure volume today.

How do we measure the true volume of irregular objects? Is it by submersion? What about this idea: simply put it into an open container, close it, pump in x cc of air from outside, and see how much pressure goes up. Would this tell you how empty the (known volume) container was, i.e. how much volume was in it? (I think this would be true because if it's nearly full, all but 100 cc of empty space, the pressure would double. but if the 100 cc represented 1/10th of the available free space in there, the pressure would increase by far less. what do you guys think of this idea? (it depends on being able to pump an exact amount of air and to measure pressure accurately, and to have an enclosed container with known volume and no leaks. No water needed and it just takes a second! 188.6.92.6 (talk) 14:45, 5 April 2012 (UTC)[reply]

in practice what is the usual way of measuring the true volume of something today? submersion still or another technology? 188.6.92.6 (talk) 14:47, 5 April 2012 (UTC)[reply]

The displaced liquid method is today still a good, easy, and accurate way of measuring volume of an odd-shaped object. In industry, 3D scanning, either articlulated probe type, laser type, or camera type, is gaining ground. In 3D scanning, probing the surface is used to build up the shape in a computer, which can then calculate the volume.

Your air pressure method is not particularly good, because a) it is easier to measure liquid volume accurately than it is for pressure, b) you have to know the air volume pumped in accurately as well, as source of error, and c) the pressure of air varies in proportion to temperature, introducing yet another source of error. Raising the pressure of air increases its temperature, which will then slowly come back down, dropping the pressure, complicating the measurement still further, as you would have to wait until it stabilises.

More often than not though, the material in the object is known, so all you have to do is weight it. If not, many objects of complex shape can be visuallised as several simple shapes joined together, so you can easily calculate the total volume.

Ratbone124.178.37.185 (talk) 15:28, 5 April 2012 (UTC)[reply]

isn't the temperature of air in a room pretty homogenous? I would think you could just have a simple piston-pump say, containing the exact volume of the whole container (like a giant syringe) with the containing part of the syringe/piston open to the air to freely circulate until the device is used. Meanwhile the temperature is measured. The object is placed inside, the syringe/piston is plunged, and if the container is empty, the pressure doubles. The less empty it is, the more the pressure increases over double pressure. The piston doesn't need to go all the way down, if the container is nearly full you don't have to subject the object to great pressure... I would think far harder than accurately pumping air (easy peasy with a piston) maybe you are right that measuring pressure is hard. But isn't that also easy - I mean can't you just have a membrane on the side that flexes out with pressure, or a kind of barometer attached to the whole thing (filled with water if you want) that is calibrated to 0 at room pressure and temperature and just goes up like a normal weather barometer? It seems this is also a solved problem.... how hard can measuring pressure be? 188.6.92.6 (talk) 15:35, 5 April 2012 (UTC)[reply]

Trouble is, while you are perfoming your measurement, the room aircon cuts in or someone opens the door. So right away you've got a temperature change of 1 or 2 degrees, perhaps more. That alone will make your measurement eeror larger than even a sloppy measurement of displaced lequid volume. When you raise the prerssure of air, you inherently increase its temperature (gas laws), and the temperature will then slowly decay back to room temperature, lowwering the pressure. So, you would have to wait until it stabilises. Not a good method. Ratbone124.178.37.185 (talk) 15:42, 5 April 2012 (UTC)[reply]

First of all, I believe you. But could you help me understand the physics more deeply. I can imagine for example a vessel (the whole container) with slots that are all open, everything is circulating freely. Let's say we can only measure the air temperature to +/- 5 degrees. (huge variation). You can have temperature measurements at various locations inside and outside. Now after placing the object inside (it might not be room temperature!) you press a button, and within 1/10th of a second the slots blast closed, a cylinder of compressed air is fully released, and you measure the new pressure. You know that the cylinder had been at x temperature roughly (you can measure it) and you know that it will cool off during decompression, but can't you calculate for this effect? Could you walk me through an actual example. I have a hen I want to see the volume of that can withstand 3x air temperature without ill affects. I put it into the machine which has enough room for the average sized hen to displace 20% of the air. I press the button, the slots close, the tank inside decompresses so that the typical 80% of the remaining space will be at 3x pressure, and we measure the difference between 3 times normal pressure at the temperature (+/-5 degrees) and what we actually get, the slots open to neatralize pressure, the doors slide open, and the hen has been volumed. (like 'weighted'). Now could you walk me through the physics of where we go wrong here? (meanwhile, air is being pumped into the pressure container again for the next hen).

The box looks like this:

T

 __S__S__S___
S T        T S
S     HH     |  T
S    _HH_    S
S   |    |   S
|txx|____|___|  T

T
Thusly:
S   = slots normally open until door is closed and button is pressed
T   = temperature measurement device, of which there are several both inside and outside the box
xxx = cylinder pumped full of air (to the extent you would like for best results/best equation)
t   = a thermometer attached to the cylinder of air.
      over time if the box isn't used for long, cylinder,
      which would be hotter when you first pumped it full of air,
      would cool to room temperature.

      It's important to know when this happens,
      as after this point you will cool the box by releasing it
H   = hen.

now with the proper equations, couldn't this work quite well? Also, if you know how conductive the air container is you can calculate just based on the time it was last discharged and the temperature outside (i.e. the gradient) and the materials it contains, how slowly it cools off after being pumped full. If it's like a thermos it could retain the heat quite well, and you would get the same temperature of air out as you put in (instead of colder air after the warmer container dissipated it under extra pressure). This makes a lot of assumptions, of course, but could you walk me through why the equation fails to be meaningful? 188.6.92.6 (talk) 16:09, 5 April 2012 (UTC)[reply]

Archimedes may have used his principle of buoyancy to determine whether the golden crown was less dense than solid gold.

Not an answer to the question, just a comment on the story that introduced it. In the version of the story I've heard (in a podcast somewhere on this website), Archimedes did not measure the density by measuring the water displacement. Instead, he balanced the crown on a scale with a gold reference sample, and then immersed the apparatus in water.This is suggested by our article Archimedes#Archimedes' principle, see illustration reproduced here. --NorwegianBlue^talk 00:03, 6 April 2012 (UTC)[reply]

Sounds like an excessively complicated way to frighten the life out of poor chickens, should they manage to survive the overpressure. Wickwack124.178.55.81 (talk) 05:48, 6 April 2012 (UTC)[reply]

Volume measurer name ?

Is there a standard device for measuring volume, and, if so, what is it's name and do we have an article on it ? StuRat (talk) 02:12, 6 April 2012 (UTC)[reply]

Lots of things measure volume. Graduated cylinders, volumetric flasks, syringes, beakers, pipettes. It all depends on for what purpose one is measuring volume. --Jayron32 04:49, 6 April 2012 (UTC)[reply]

I mean something automated. You drop the object in, and it spits out a number on it's display. Doesn't this exist ? StuRat (talk) 06:34, 6 April 2012 (UTC)[reply]

It does now, Stu! Silence denotes consent. You want to go into business with me? I have a killer business plan. 188.156.29.4 (talk) 07:30, 6 April 2012 (UTC)[reply]

How about this: a 3D, raster scanning, laser that gives an accurate approximation? Plasmic Physics (talk) 14:32, 6 April 2012 (UTC)[reply]

Like this? DMacks (talk) 15:11, 6 April 2012 (UTC)[reply]

Stu, these guys are trying to get us to abandon our business plan because an insanely expensive device technically already fills this space, with lasers and precision software. We can manufacture ours in China for four-fifty if we don't mind the compressed air tank exploding and the air ducts not closing hermetically after 200 cycles, but for twenty or thirty dollars we can cut this whole "3D raster scanning laser approximation thing" which no farmer would EVER use to actually volumize a Hen, if you'll patent this with me (I think we have 90 days since I spilled the beans here) we are going to strike it rich. They say build a better mouse-trap, but you know what's better than a surface-plotting, mouse-modelling, laser-guided mousetrap? A cheap one!! Are you in or out? 188.6.92.6 (talk) 16:24, 6 April 2012 (UTC)[reply]

Just send me a check each month, please. :-) I imagine the laser scanner would have a problem with hollow objects, while submerging in a liquid would work better, as long as there are holes to let the liquid in and air out. I wonder if water is the best option, though. It certainly is cheap, but it's tendency to soak into things like fabric might be a problem. It might also be important to rotate the object and vibrate it, to get all the air bubbles out. Then again, considering all these issues, perhaps using a gas instead of a liquid would be better. StuRat (talk) 18:18, 6 April 2012 (UTC)[reply]

I read your user page with your philosophy. ("I believe that the movement of wealth from the poor and middle-class to the rich is a serious problem, eventually destablizing a nation, and should be countered with a heavily progressive tax system (up to 90%)") I assume, since this is quite profitable, that you would like me to just send you 10% of the money, with 90% going to the government. Of course, the government is free to spend it on whatever the current president wants, a war we disagree with for example - but doesn't actually have to take any risk on my invention. Do I have this right? So let me take on all the risk, for the chance to give civil servants who didn't do anything for it control over how 90% of it is spent. Do I have your philosophy right? If so, I look forward to sending you 10% of your money soon... 134.255.105.197 (talk) 21:45, 6 April 2012 (UTC)[reply]

That 90% is for rich people, earning a million dollars a year or more. If you want to send me $100K a year and the gov the rest, I'm fine with that. BTW, I also believe in direct democracy, meaning we all get to vote on how the money is spent, as opposed to back room deals with corrupt politicians giving it to whoever donated the most to their campaign funds. StuRat (talk) 21:59, 6 April 2012 (UTC) [reply]

retrograde rotation of venus

Venus opposite rotation is remained mystery for duration of its discovering time till now ,and it has been refereed to some hypothetic events such as external object impact with planet which supposed it changed its rotational direction . HOw does this occure?--Akbarmohammadzade (talk) 14:54, 5 April 2012 (UTC) In fact Iwant to have any discussion ,in addition by let of Wiki site I ask you to see (http://www.gsjournal.net/Science-Journals/Research%20Papers/View/4029 )to know how i am thinking about the subject--Akbarmohammadzade (talk) 15:01, 5 April 2012 (UTC)[reply]

If you've already written a published paper on this, then why would you ask us about it ? StuRat (talk) 17:06, 5 April 2012 (UTC)[reply]

All the planets have a tilt to them, some more than others, and even the sun has over a 7 degree tilt. This last point was discussed recently here: Wikipedia:Reference_desk/Archives/Science/2012_March_12#Why_does_the_Sun_have_a_7.25_degree_tilt_.3F. As for the mechanisms, I suppose a moon-sized object could strike at a shallow angle, counter to the old rotation direction. However, I think it more likely it would strike at close to a right angle to the direction of rotation. Thus, Venus continued to rotate, rather than reversing direction, but the axis of this rotation was changed. This isn't necessarily an impact from a single object, either, there may have been many. This leads to a model of the early solar system with far more large objects in it than it does at present, since every surviving planet appears to have had it's rotation disturbed. StuRat (talk) 17:10, 5 April 2012 (UTC)[reply]

It is not a peer-reviewed journal, so I suspect that the OP might not a specialist in the subject. As far as "external impact events", I don't understand why these exotic theories are needed. The explanation that makes the most sense to me is that its rotation is retrograde due to its thick atmosphere. Although the exact evolution to its current state is up for debate, with an atmosphere 92 times denser than Earth's at the surface, Venus should have had more than enough time to spin down from a conventional orbit due to tidal effects. After it had reached a near-zero rotation, it's not inconceivable that drag from the atmosphere could set it spinning in either direction, and since it seems that the dominant winds in the atmosphere are retrograde, it should be of no surprise that its rotation is retrograde as well. -RunningOnBrains^(talk) 06:19, 6 April 2012 (UTC)[reply]

Isn't the retrograde motion of the atmosphere due to the retrograde motion of the planet, not the other way around ? What do you think causes the motion of the atmosphere ? And I see how a planet would eventually stop revolving, down until it became tidally locked, but I see no way for it to start spinning again without an impact. StuRat (talk) 06:39, 6 April 2012 (UTC)[reply]

I don't see why you would imagine that Venus's atmosphere would make a significant difference. Internal redistributions of angular momentum do nothing to affect the total angular momentum of the planet. A planet needs to interact with an external objects (e.g. other planets or the sun) in other to accomplish a net change in angular momentum. Besides which, the Earth is covered with a fluid (the oceans) significantly more massive than Venus's atmosphere. Dragons flight (talk) 19:00, 6 April 2012 (UTC)[reply]

Yes, but the oceans are contained do basins, and so rotate with the earth and don't have a general direction of flow, while the atmosphere has a net retrograde motion in comparison to the surface. The earth's atmosphere is the number one reason for changes in the length of day. However, I see your point about external torques, my false assumption was that the jet stream and Hadley cell structure would be similar for venus in the past, but this is simply not true, because these structures only form on a rotating planet. Still, the atmosphere could have been responsible for slowing its rotation, then it would take a much less significant impact event to turn the planet retrograde.-RunningOnBrains^(talk) 21:48, 7 April 2012 (UTC)[reply]

Escaping from a black hole

It's standard to point out that even light cannot escape from a black hole. Likewise, if a cannonball is fired very fast outward toward the event horizon, the gravity will decelerate it to zero and then negative velocity. But a cannonball has no countervailing force accelerating it to offset the gravity. What if, instead of a cannonball, we have a rocket that is continuously firing its thrusters? Could a time profile of generated thrust be found that would let the rocket escape? Or would the required thrust rise to infinity as the event horizon is reached? Duoduoduo (talk) 15:36, 5 April 2012 (UTC)[reply]

The escape velocity at any point within the event horizon of a black hole is at least the speed of light. No object with mass cannot be accelerated to (much less beyond) the speed of light, and so no rocket operating within the bounds of known physics can escape. It would fall more slowly, but still it would fall inexorably. As I understand it, though, your cannonball example is fallacious. Within the event horizon, gravity is sufficiently strong that you cannot move away from the singularity, even temporarily. So there's no launching a cannonball "up but not quite far enough"; it can only fall. — Lomn 15:54, 5 April 2012 (UTC)[reply]

the scape velocity for outer distance from event horizon is same as first star ,the sun radius is 1/200 AU , the escape velocity from its surface is 617 Km/s .if it came to be for example any black hole ,the necessary velocity will be equal .Akbar mohammadzade IRAN — Preceding unsigned comment added by 78.38.28.3 (talk) 16:11, 5 April 2012 (UTC)[reply]

This is not true. What will remain the same is the escape velocity from the Sun's former radius. Escape velocity is dependent on the radius from the center of gravity. So the escape velocity of a black-hole sun would be greater than the speed of light within its event horizon. -RunningOnBrains^(talk) 06:46, 6 April 2012 (UTC)[reply]

The article escape velocity says: It is the speed needed to "break free" from a gravitational field without further propulsion. But my question involves further propulsion.

By analogy with escape from Earth: An unthrusted cannonball, propelled only by its initial explosion to just less than the escape velocity, would not quite make it out; but if you add some more thrust continuously, couldn't you make it out at a slow speed? Duoduoduo (talk) 16:11, 5 April 2012 (UTC)[reply]

No. Within the event horizon, gravity is constantly drawing you in at a rate greater than the speed of light. Your hypothetical rocket, therefore, would have to have, at some point, a velocity also greater than the speed of light in order to escape. Per known physics, this requires infinite energy and so cannot be done. Alternately, consider light: it is always traveling at the speed of light (the black hole does not slow it down), and yet is unable to escape. How, then, can a rocket that can never reach the speed of light ever escape? Rather, the geometry of spacetime is sufficiently altered that no escape is ever possible. The images at Black_hole#Event_horizon illustrate this. This is also my point above about the problem of the cannonball analogy: no paths within a black hole's event horizon exist that move closer to (or even remain the same distance from) the event horizon; all possible futures move strictly towards the singularity. — Lomn 16:23, 5 April 2012 (UTC)[reply]

Thanks, Lomn -- very clear and helpful. Duoduoduo (talk) 16:52, 5 April 2012 (UTC)[reply]

It's not very accurate. Gravity doesn't draw you in at a rate. It accelerates you. As you said, ordinarily you don't need to accelerate to the escape velocity to escape from something, so the argument that "you would have to accelerate to the speed of light to escape, therefore you can't escape" makes no sense and people shouldn't use it. However, it is true that you can't escape from a black hole by any means. -- BenRG (talk) 16:58, 5 April 2012 (UTC)[reply]

My understanding is that accelerating (in any direction) while within an event horizon actually causes an object to fall faster. --Tardis (talk) 14:03, 6 April 2012 (UTC)[reply]

Inside the event horizon, "inward" and "outward" are no longer spatial directions, but they are timelike, i.e. like future and past. That's what the Schwarzschild metric is saying. Icek (talk) 20:07, 5 April 2012 (UTC)[reply]

I can see where the OP's confusion comes from. One thing that is often glossed over in a casual discussion of escape velocity is that it is only the velocity at which you will escape the planet's gravity. It is not a velocity you must achieve to escape the planet's gravity. Indeed, one could envision an imaginary spacecraft with near-infinite fuel that could escape Earth's gravity at walking speed; all they would have to do is adjust their thrust slowly downward as they moved away.

Another thing that is glossed over is that escape velocity is dependent on the distance from the center of gravity of the object you are escaping from (see the equation at the top of the escape velocity page), but the number that is most often cited is the escape velocity at the surface. The further you get from, say, Earth, the lower your escape velocity will be.

In regards to your original point though, you can't think about these cases classically, because black holes are, by definition, non-classical objects. I'm not really knowledgeable on the in-depth mathematical details of black holes, so I presume Icek knows what he's talking about, but I know enough to know that even in a thought exercise where your spacecraft somehow wasn't annihilated by the ridiculous tidal forces that the concepts of acceleration, velocity, direction and distance will become non-intuitive inside the event horizon (even though, technically, physics can't say anything about that which lies past the event horizon).-RunningOnBrains^(talk) 06:46, 6 April 2012 (UTC)[reply]

Body fat monitor

How do body fat monitors (example) work? How accurate are they? --SupernovaExplosion ^Talk 15:42, 5 April 2012 (UTC)[reply]

I don't know much about this, only enough to tell that the relevant article is Bioelectrical impedance analysis. --NorwegianBlue^talk 23:38, 5 April 2012 (UTC)[reply]

Thanks for the link! --SupernovaExplosion ^Talk 04:19, 6 April 2012 (UTC)[reply]

Caramel Catastrophe!

I had some delicious leftover caramel that I'd made--I put it in a plastic tupperware container (one I've regularly used in a microwave) and put it in the fridge.

Later, when I was microwaving it to drip onto ice-cream, the caramel melted through the bottom of the tupperware and burned the heck out of my hand!

I didn't leave it in for an excessive amount of time, and my friend says it has something to do with the 'heat capacity' of sugar---can any of you science-types shed some light on what that may or may not mean? Does that mean that sugar can hold a very high amount of heat without becoming a sugar vapor or some such? (please note, I'm not referring to the brand: Tupperware--I'm actively promoting the genericide of that brand)66.30.10.71 (talk) 15:51, 5 April 2012 (UTC)[reply]

a large heat capacity means that it can hold a lot of heat energy without it's temperature increasing (feeling hot). OR, If it's hot it can lose a lot of energy (and maybe burn more stuff) before it's temperature reduces. So heat capacity is not directly responsible for melting through the bottom. Sugar does not normally vaporize. It melts and then begins to char to form caramel. If you keep heating it it will go all the way to charcoal. Most foods will never heat up enough to melt the plastic as once they reach the boiling point of water, any more energy supplied to them from the microwave will just be carried away by the steam that is formed. In the case of your caramel, no such luck. the water formed by charring reaction was too little to carry away the heat in the steam. All that energy instead raised the temperature of the caramel enough to melt the plastic. Staticd (talk) 16:26, 5 April 2012 (UTC)[reply]

I don't know, don't you make caramel in a very hot pan before the sugar 'melts' - obviously too hot for tupperware (you wouldn't put sugar in a tupperware container and put it in a microwave and heat it until it became caramel. so maybe getting it to melting temperature is equally hot, even though it was 'already caramel'. just a theory. 188.6.92.6 (talk) 16:39, 5 April 2012 (UTC)[reply]

It's never a good idea to microwave plastic containers. Long before they actually melt, they leach toxic chemicals into the food. I suggest glass or ceramic for your microwaving needs. StuRat (talk) 17:00, 5 April 2012 (UTC)[reply]

What about the containers in which prepared ("ready") meals or dishes are sold explicitly to be microwaved within them (some of which I save and re-use), and the plastic kitchenware specifically sold for repeated use in microwaves (which I also use)? I agree, of course, that any plastic container not clearly designated for microwave use should be avoided. {The poster formerly known as 87.81.230.195} 90.197.66.205 (talk) 17:37, 5 April 2012 (UTC)[reply]

They might be better than foam take-home containers, which seems to melt immediately, but still I bet pretty much any plastic they sell as "microwaveable" is likely to melt and/or give off liquid chemicals and fumes before glass and ceramics designed for the microwave. One worrying sign is if tomato sauce heated in a plastic bowl turns the bowl red, and the color won't come out. This show a level of interaction between food and container which should not be allowed. StuRat (talk) 19:31, 5 April 2012 (UTC)[reply]

I'm not sure, but if the caramelization involves enough charring, the substance would become electrically conductive and absorb a lot of energy from the microwaves. (There's a somewhat related demonstration where a beer bottle is microwaves on its own - nothing happens - then a torch flame is applied to it for a moment, and remicrowaved ... and the glass melts) Wnt (talk) 18:57, 5 April 2012 (UTC)[reply]

The caramel could simply be heating very unevenly (it's viscous, so a local high-energy spot might not redistribute very well), leading to "that one spot" getting hot enough to melt through. I would assume that there was enough caramel total and that it is polar or ionic enough to absorb sufficient microwave energy, but if not (or it can't conduct the heat and MW energy away effectively), wherever the energy concentration is greatest just keeps hitting the plastic itself. DMacks (talk) 21:46, 5 April 2012 (UTC)[reply]

AFAICT, the microwave energy will be absorbed by the rotational energy levels of the -OH groups in caramel, same as water. Need'nt be conductive. From there equipartition theorem calls for it's redistribution. Staticd (talk) 11:37, 6 April 2012 (UTC)[reply]

My point is that the redistribution is slow...microwaving a frozen block of meat gives some areas that get hot enough nearly to start to cook while others are still frosty. It's unlike a mug of water, where the molecules are free to move around and allow physical redistribution of the "hotter" ones so the whole thing winds up heating more evenly. I assume that's why "defrost" is a long time of cycling of a lower energy level instead of the 100% "boil water" mode. DMacks (talk) 11:59, 6 April 2012 (UTC)[reply]

Agreed. I've found that just about everything turns out better when microwaved longer on a lower setting. Pizza, for example, doesn't turn out burnt at the edges and frozen in the center (although it's still soggy, so I prefer a convection oven). StuRat (talk) 18:13, 6 April 2012 (UTC)[reply]

When you add sugar (or salt) to water it elevates the boiling point. This can be above the melting point of the plastic which can be as low as 105°C. Oil also boils at high temperatures, and so microwave containers have warnings about these kinds of foods. Graeme Bartlett (talk) 22:56, 6 April 2012 (UTC)[reply]

And thermoplastics (soft plastics) don't so much have a melting point as a wide range of temperatures over which they slowly change from solid to tacky to gel-like to runny. Anything beyond completely solid is problematic if you intend to eat the contents, as toxic chemicals will leach out at much higher rates then. StuRat (talk) 23:31, 6 April 2012 (UTC)[reply]

Instantaneous communication at a distance

Suppose I have a stick of fixed length. I assume it is absolutely incompressible. I press one end of the stick, and the other end moves (and so does an incompressible object that the far end is touching). Have I communicated instantaneously at a distance? How does this relate to the relativistic notion that (is this right?) information travels no faster than light speed? Duoduoduo (talk) 16:52, 5 April 2012 (UTC)[reply]

Yes, that would violate the laws of physics. Thus, a perfectly incompressible solid is not possible. StuRat (talk) 16:57, 5 April 2012 (UTC)[reply]

This is a common question on the reference desk. See WP:Reference desk/Archives/Science/2012 January 16#Information transfer Faster Than Light - DIY and WP:Reference desk/Archives/Science/2008 January 10#Communicating Faster than Light Using a Rotating Rod for starters. -- BenRG (talk) 17:16, 5 April 2012 (UTC)[reply]

Perfect rigidity is excluded by relativity. In an object composed of matter, the atoms are held together by electrostatic forces. When every position along the object is not accelerating, the atoms are found with an equilibrium seperation, similiar to a spring. When a position along the object is accelerating, the seperation between atoms at that point compresses to a minimum relative to their acceleration. As with any process, it takes time to be compressed (however short), before the minimum is reached. Since all the atoms inside the object are free to move and compress, the whole object can compress as a cumalative effect.

If you have a long pole, and pull very fast on it, it will stretch momentarily before returning to its normal length. If you push the pole, it will shorten. Even the motion of atoms relative to each other, is constrained by the speed of light; the distant end of the stick will only respond to a push on the near end at a rate slower than the speed of light.

This is the behaviour of matter: no matter how rigid something is, nothing is pefectly rigid, and every thing can compress and stretch. Plasmic Physics (talk) 23:10, 5 April 2012 (UTC)[reply]

Thanks, Plasmic Physics. You say "Perfect rigidity is excluded by relativity". I thought it was excluded by the physics of the small, not by relativity. Is it not possible to conceptualize an (empirically non-existent) physics of the small involving perfect rigidity? If so, It seemed to me that this kind of instantaneous communication at a distance would be possible, and would not violate relativity because nothing physical is moving at the speed of light. Is it really relativity that precludes perfect rigidity? Duoduoduo (talk) 20:27, 7 April 2012 (UTC)[reply]

Follow-up

These two questions may be identical, but I had actually been thinking of something similar for a while -- say a hypothetical "see-saw" was constructed that spanned several light years through deep space. The object is a single piece and rotates along a hinge (like a see-saw). When one end is raised, the other is lowered. Assuming the far end is raised with great speed, when would those at the near end begin to perceive a lowering? My initial feeling is that, since the atoms and molecules of the "see-saw" cannot be accelerated beyond the speed of light, the other end would not be affected until such a time as the displacement of the atoms "rippled" through to the other side. Probably an objective observer (viewing the entire object from the side) would see a slow warp in the object until the "ripple" reached the terminal point and "straightened out" the object. Is this an accurate description? Evanh2008 (talk) (contribs) 02:23, 8 April 2012 (UTC)[reply]

Sure, but it wouldn't be any less odd or strange to the observer than watching a flexible bar bend and then straighten out. Just on a really huge scale. --Jayron32 02:29, 8 April 2012 (UTC)[reply]

Have they ever tried multi-stage balloons?

They launched a 1.35 million m³ balloon before, for the world altitude record (~50 km). Why not tie a barely inflated small balloon to it, so that by the time the carrier balloon bursts (or stops rising?), it's payload has expanded enough to be buoyant? Sagittarian Milky Way (talk) 18:10, 5 April 2012 (UTC)[reply]

The expansion won't change the buoyancy. It is displacing a greater volume of air, but that air is reduced in density by the same amount (the pressure inside and outside the balloon need to be equal). The buoyancy is determined by the number of molecules of lifting gas and the difference in molecular weight between the lifting gas and air. The number of molecules of gas in a given volume at a given temperature and pressure will be the same, regardless of what the gas is, so one molecule of helium will always displace one molecule of air. --Tango (talk) 18:50, 5 April 2012 (UTC)[reply]

(ec)I don't understand. Why have a second balloon rather than allowing the carrier balloon to vent lift gas so it doesn't burst? Wnt (talk) 18:50, 5 April 2012 (UTC)[reply]

If you vent your lifting gas, you won't have the lift any more. Balloons don't usually have much more lift than is needed to counter their weight, so there isn't much you can vent before you start falling again (and you might as well have just not had that bit to start with). --Tango (talk) 22:17, 5 April 2012 (UTC)[reply]

how about this one guys?

why not just shoot a giant artillery round up that is itself a piece of artilley, timed so that when it reaches its apex, right before it starts falling again, it shoots its payload? Oh, and guess what the payload is... right, another smaller round of artillery and so on, right down to a miniature little gun that fires a little pobe into space? All the heavy artillery can parachute down for another fun go just the same. Why is this better than just one giant piece of artillery firing straight into space? Because whatever it fires doesn't have to undergo such a huge explosion, just enough to make a nice arc, and the next one likewise, and so on and on. See http://en.wikipedia.org/wiki/Schwerer_Gustav - it weighs 1300 tonnes - let's trim it down to 700 tonnes without a lot of track and equipment and guiding stuff, etc, and fires a 7-ton shell 23 miles. I suppose at a 45 degree angle. If instead it shot straight up, surely it could reach 10 miles straight up, no? Then we repeat: the 7-ton (7000 kg) projectile is itself a piece of artilly that, at the same ratio, can fire a 70 kg piece another 10 miles up in the air. We are at 20 miles. The 70 kg piece itself is able to fire 0.7 kg i.e. 700 grams, which ought to be enough to launch a 7-gram projectile another 10 miles (especially due to much lower air resistance), getting close to reaching low-earth orbit http://en.wikipedia.org/wiki/Low_Earth_orbit . Admittedly we need to trim bit more to really reach 10 stages of matruschka artillery, i.e. the 100 miles that's required for low-earth orbit, but I think we can actually do better than being able to launch only a projnectile that is 1/100th of the piece that is firing it, since all we need is an explosive chamber (no wory about reloading it on the spot), a mechanism to keep it upright right before firing (for example it could deploy its chute and fire up through a whole in the center whenever it is hoizontal (parallel to the ground) according to simple level device that doesn't weigh much. Sure, it would lose a bit of altitude after deploying chute, but you don't have to worry about timing and trajectory and all that stuff...very simple. Basically like those handheld artillery pipe things from vietnam era warcraft (don't remember what they'e called). Just a pipe you put explosive in and a shell, aim and ignite. So, what's wrong with this way of getting to space repeatedly, always retrieving the larger and larger pieces of 'artillery' as they fall with a gps tracker? 79.122.54.224 (talk) 18:57, 5 April 2012 (UTC)[reply]

Sounds like a multi-stage rocket, only less efficient and more complex. I always thought a rocket launched from a balloon or balloons would be the best combo. StuRat (talk) 19:26, 5 April 2012 (UTC)[reply]

A multistage rocket would be much gentler on the payload. You've specified that the payload is a space probe, and delicate electronic devices like a space probe tend to not fare so well when subjected to explosive forces. Red Act (talk) 19:53, 5 April 2012 (UTC)[reply]

See Gerald Bull, the HARP Project, and the supergun. All the parts are out there somewhere, dunno about the drawings though. I think the missile was designed to fire while still in the gun barrel, but not positive on that. Franamax (talk) 19:33, 6 April 2012 (UTC)[reply]

how about THIS one guys

I don't get multistage rockets, if you're not jettisoning a LARGE amount of your weight at each stage (as with my artillery) then isn't the following easiest: fill a giant balloon in ana aerodynamic shape with jet fuel. Put a small hole on the bottom with the ignition, the pressure from the balloon and gravity keeps the ignition fed and the balloon (being rubbery) contracts by itself. So you go from a huge balloon filled with jet fuel to a tiny one. Why waste all this metal on 'stages' and so on? It would be better if we could suspend the belloon, of course, so it doesn't have to stay rigid but that's tough. I guess you could have a long pole that you hang the aerodynamic balloon on, that siphons from the top and has ignition at the bottom (past the bottom of the balloon, that way the rocket-shaped balloon is suspended the whole time, but this adds an unnecessary pole to the whole thing that weighs exta. illustration 1:

Diagrams of proposed space launch vehicles

ILLUSTRATION 1 (NOT IMPORTANT)

  P      P = PAYLOAD
  H     H (for hook) = TOP OF POLE, SUSPENDING IT ALL
 B+B    , B = BALLOON, + = SIPHON  | = POLE, HERE UNDER COMPRESSION, NOT TENSION X = FIRE
BB|BB
BB|BB
BB|BB
BB|BB
BB|BB
BB|BB
BB|BB
BB|BB
BB|BB
BB|BB
BB|BB
  |
  |
  |
  |
  X
 XXXX
XXXXXXX

basically, why have multiple stages instead of just jet fuel and an elastic balloon filled with it and in the shape of a rocket?

Another option (bette I think) for aerodynamics is if you can find some very light metal that can withstand continuous jet-fuel burning, then you put the ignition a good distance ABOVE the aerodynmaic balloon, so that it's igniting straight down but doesn't reach the balloon of jet fuel, which feeds up through the a tiny line that is way off to the side out of the way of the flames, like this

       
           [^]
          /[ ]   <--- [payload]   <<-- FAR LEFT: FUEL LINE, HELD OUT AWAY FROM IGNITION FLAME
fuel-->  / [_]
line    /  [x]  <--- ignition point and can be diected by payload for correct guidance
       /   x|x
      /   xx|xx  <-- flames licking the metal (x = flame)
     /      |     
    /       |  <- far left - fuel line
   /        |  <--- metal to suspend the aerodynamic balloon, now under tension, not compression  
  /         | 
 |          |
 |          |  
 |          + <-- good insulation on the line so it doesn't heat up too much fom the above flames
 |          +
 |          +  
 |__________+
 |          +
 |          +  
 |         (F)
  \        (F)
   \      (FFF)
    \     (FFF)
     \    (FFF)
      \   (FFF)  <--- balloon with jet fuel (F) in it
       \  (FFF)       could just be canvas with rubber bands around it
        \ {FFF}
         \{FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          {_F_}


it could also have two fuel lines for balance, thusly:




           [^]
          /[ ]\
         / [_] \
        /  [x]  \
       /   x|x   \
      /   xx|xx   \
     /      |      \
    /       |       \ 
   /        |        \
  /         |         \
 |-+-+-+-+--|-+-+-+-+--|
  \         |         / 
   \        +        /  
    \       +       /   
     \      +      /    
      \     +     /     
       \    +    /      
        \   +   /       
         \ (F) /
          \(F)/
          (FFF)
          (FFF)
          (FFF)
          (FFF)
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          (FFF)
          {FFF}
          {FFF}
          {_F_}

Hang it on a long pole that siphons from the top (due to the pressure from the balloon) and near the top of pole put the ignition. Now you go all the way to space with a shrinking balloon that weighs less and less. 79.122.54.224 (talk) 21:32, 5 April 2012 (UTC)[reply]

I haven't read your whole idea because your premise is wrong. Multistage rockets *do* jettison a large portion of their mass. Take the Saturn V rocket that took man to the moon, for instance. The first stage (excluding fuel) was 2,300 tonnes. The rest of the rocket (including the fuel for the other 2 stages) was 621 tonnes. By jettisoning the first stage and using the much smaller second stage engine, they were able to lose about 80% of their mass. --Tango (talk) 22:17, 5 April 2012 (UTC)[reply]

The idea of keeping the fuel in a "balloon" is not that much different from keeping it in the casing of a rocket, except that a rocket will actually work flying up through the air. And when the balloon shrinks, what you're really saying is, keep the extra bits of outer shell even though much less fuel is being held by it; whereas the multistage rocket dispenses with the bits of extra shell when the fuel they enclose is burned.

It's more fun to think of ways of having really powerful fuel, like red oxygen, variants on inverse sodium hydride and other such bizarrities. Wnt (talk) 23:11, 5 April 2012 (UTC)[reply]

All this stuff about altitudes is beside the point. Space is an arbitrary altitude, and getting there briefly isn't very useful (but for silly people with too much money and astronomers with too little). What's useful is orbit, which allows you to stay up for a long time. Orbit is so much more than "getting up really high"; even if you build a 100km tall tower, when you stepped off it you'd drop like a stone. To be in orbit you need speed; absolutely huge speeds, in fact. The lowest elliptical orbital speed that article quotes is 6.5 km/s, which is 14500 mph. That's a massive speed, and to get to it you need to expend a massive amount of energy - that's why astronauts strap themselves to gigantic exploding bomb things and blast violently into orbit on a huge column of flames. Air launched systems (be it from balloons or aircraft like Pegasus) have some plusses from range safety and bad-weather avoidance, and some modest gain from avoiding some of the atmosphere, but you don't get to orbit without that massive speed, and you won't get a balloon going at 14500 mph. -- Finlay McWalterჷTalk 23:36, 5 April 2012 (UTC)[reply]

It's understandably hard to get your head around the forces involved, but if you did you'd see why the balloon idea is just silly. Have you ever had a water balloon explode in your hand because you tried to throw it too hard? Now imagine that you're not throwing something, but applying almost 10,000,000 pounds (4,500,000 kg) of force to it (forgive my abuse of units, fellow scientists). There is no "elastic material" in existence that could survive such a journey.

Your entire thought experiment is unnecessary, however. You seem to assume that we could somehow save the amount of fuel needed to get into space by cutting down on the weight of materials used to build the rocket. Using the space shuttle as an example, the fuel needed to reach orbit weighs 20 times as much as the object it is lifting! So you see, our efforts would be much better spent trying to reduce the weight of the fuel, as Wnt points out above. -RunningOnBrains^(talk) 07:13, 6 April 2012 (UTC)[reply]

Also, a balloon is going to be far less aerodynamic; it'll tend to bulge out rather than be long and thin. Rockets have fins etc for stabilisation in the atmosphere; without a rigid structure these aerodynamic features would be less useful (imagine a floppy airplane). --Colapeninsula (talk) 09:18, 6 April 2012 (UTC)[reply]

= be-all end-all

I have the be-all end-all of all fuel-containing solutions!!!!

This eliminates 100% of the enclosing container.

This is what you do.

You FREEZE the rocket fuel in a long rocket-shaped cone, and melt off the bottom as you use it!!! You can space the distance between the burning unit and the bottom of the frozen rocket fuel such that just enough heat goes up to keep melting it. Granted, you still need loads of rocket fuel, but you now have ONLY a payload at the top of the chunk of frozen rocket fuel and a drip container and burning unit at a variable distance at the bottom:

 P      P = payload
FFF
FFF
FFF
FFF     F = frozen rocket fuel
FFF
FFF
FFF
\D/     D = drip container, dripping down to burning unit
 |
 |      | = variable length somewhat insulated pneumatic aparatus for keeping the burning 
 |           just far enough to melt what you want, with a tube inside to drip from D to xx
/xx\
xxxx    x = blastoff fire
xxxx

For a mini rocket, how can you get ANY more streamlined than having NOTHING containing the rocket fuel? Depending on how fast you get into orbit, you can keep the whole thing in a cryogenic container until just the right moment, then blast off and hopefully get into orbit before it all melts. Does this math work, guys? — Preceding unsigned comment added by 188.6.92.6 (talk) 09:34, 6 April 2012 (UTC)[reply]

Your problem isn't maths, it is engineering. I can't see how that design could possibly work - what's holding the engine onto the frozen fuel? You are also mistaken in thinking that the container is a significant issue. Let's take the Saturn V first stage as an example. Unfueled, it weighed 131 tonnes. The fuel tank (empty) weighed only 11 tonnes. The tank is only a fairly small amount of the weight. Conventional, multi-stage rockets are very well designed and are very efficient - you're not likely to improve on them without a lot of training and experience (you would be better off trying to invent whole new ways of getting into orbit - so much work has gone into rockets that they are pretty much as good as they'll get). --Tango (talk) 12:17, 6 April 2012 (UTC)[reply]

I only read thru your second sentence, will be back later to finish. "what's holding the engine onto the frozen fuel?" - the engine is pushing up on the frozen fuel! even if it were not the frozen fuel would rest on it of its own pendulous weight, but the thrust makes this even greater than 1g (a lot greater than one g). so the same thing holds onto it as a server (waiter/waittress) who swings a plate from below to weave past something). more later. 188.6.92.6 (talk) 14:03, 6 April 2012 (UTC)[reply]

Tango. I've read the rest of your comment. First of all, I welcome any suggestions as to 'new ways' to get into orbit other than rocketry. Secondly, let us say that the fuel tank weighed only 11 tonnes. Now let me ask you another question: how much do you think those 11 tonnes cost? How about designing them?

How much is a safely and properly designed conventional rocket in MONEY TERMS, as opposed to a device that is meant to just push some frozen fuel up into the air by burning it off, just like an inverse candle: what could be simpler than a candle?

I am talking, amateur rocketry for example. If all you have to do is freeze the rocket fuel (or maybe even gasoline!!) in a special shape, place it onto this aparatus, and have it lift off without having to jettison anything or have any moving parts or do anything other than control how fast the fuel melts into its recepticle - which can be done with a simple feedback loop (it can be purely mechanical if you want) moving the rocket exhaust out whenever the recepticle is more than half full and retracting it again if it is less than half empty? One thing nobody has done, however, is done the calculations: does frozen fuel stay frozen long enough to make it to space? Rentry is seriously hot - how about getting to space to begin with?

This would be like a home kit you buy for very cheap, plus your fuel costs, a form to freeze your gas into and a trajectory, put it outside with your payload on top and blastoff into low earth orbit. I can't imagine anything simpler than this, if the physics allow it. 188.6.92.6 (talk) 16:17, 6 April 2012 (UTC)[reply]

If the fuel isn't securely attached it will just fall off. Rockets don't fly smoothly through the air, there is a lot of turbulance. You have also neglected to consider the oxidiser. There is no oxygen is space, so you have to take it with you. Rockets generally have two fuel tanks, one containing the fuel itself and one containing liquid oxygen. They have to be mixed in just the right ratio just before being fed into the engine or the rocket doesn't work. --Tango (talk) 16:24, 6 April 2012 (UTC)[reply]

Well we can't take frozen 02. i'll have to think about this... do you want to take up my other bright idea with me if stu doesn't? 188.6.92.6 (talk) 16:47, 6 April 2012 (UTC)[reply]

Freezing rocket fuel sounds like a poor way of creating a solid fuel rocket. Now I suppose if you can have caseless ammunition you ought to be able to design some kind of rocket made entirely out of fuel. Modeling different rates of burning, creating fuel with good tensile strength, controlling the rate at which it burns so that it somehow remains firmly pressed against a nozzle (or otherwise forms its own?) ... Some kid at Caltech should come up with such a thing sometime for a class project. Somehow I doubt it would replace sanely made rockets though. Wnt (talk) 04:31, 7 April 2012 (UTC)[reply]

I'm sorry, but all solid rocket fuel is frozen. If you mold it first, then it was liquid before it was frozen. Your point doesn't change my proposal at all. I am "that caltec kid." 134.255.3.231 (talk) 09:00, 7 April 2012 (UTC)[reply]

Perhaps you should read combustion chamber. In some toy rockets and fireworks, the casing is made of paper, or there is no casing at all. In powerful rockets, the operational chamber pressure is large (because it is a rocket engine). Chamber walls are necessary for structural integrity. High pressure hot gas impinging on explosive fuel or on load-bearing structure can cause structural damage, even if only a tiny gas leak impinges on a tiny portion of the structural material. As was discovered during the investigation into STS 51-L, a very small amount of exposure can cause total structural failure. You are suggesting to expose the entirety of the structural, load-bearing portion of the rocket to the internal environment of the combustion chamber. This will cause dangerous explosions. If you are indeed building small rockets, consider taking ample safety precautions. Even small toy rockets can contain large amounts of energy. Before you ever fire up your rocket, use water to test structural integrity at high pressures - well above your design pressure - because when your design fails, and the water rapidly depressurizes, it does not expand volumetrically, unlike hot gas. Remember that at high temperatures and pressures, and in the presence of oxidizer, solid steel can be more flammable than gasoline. Rocket science is difficult because the inside of a rocket motor is a very unusual environment. Many of your intuitions about material strengths, time-constants, and general behaviors of material are dangerously invalid in this chemical and physical environment. It is for this reason that rocket engineering relies so heavily on rigorous theoretical analysis and very careful experimental validation of subcomponents. Nimur (talk) 16:37, 7 April 2012 (UTC)[reply]

Dandruff and eyes

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.

April 6

Virtual work

Here's something I don't understand about virtual work.

By definition, $\delta W={\vec {F}}\cdot \delta {\vec {r}}$ . Normally, we write ${\vec {F}}\cdot \delta {\vec {r}}={\dot {\vec {p}}}\cdot \delta {\vec {r}}={\frac {\mathrm {d} ({\vec {p}}\cdot \delta {\vec {r}})}{\mathrm {d} t}}-{\vec {p}}\cdot \delta {\dot {\vec {r}}}$ and solve from there.

But couldn't we just say that $\delta W=\delta T=\sum _{k}{\frac {\partial T}{\partial q_{k}}}\delta q_{k},\ \mathrm {where} \ q_{k}=\mathrm {k^{th}} \ \mathrm {degree} \ \mathrm {of} \ \mathrm {freedom}$ ? It doesn't give the right answers, but physically I don't see why this shouldn't work. 65.92.5.132 (talk) 02:19, 6 April 2012 (UTC)[reply]

Hello, if T is kinetic energy then surely

{\frac {\partial T}{\partial q_{k}}}

is zero? For example, a point particle has

T={\frac {1}{2}}m{\dot {q}}^{2}

which isn't a function of

q_{k}

.

Perhaps we need:

$\delta W=\delta T=\sum _{k}{\frac {\partial T}{\partial {\dot {q_{k}}}}}\delta {\dot {q_{k}}}$ and then connect this to $\delta {\vec {r}}$ ? Sorry I get confused with generatised coords. 94.72.209.210 (talk) 13:07, 6 April 2012 (UTC)[reply]

There are situations where T can depend on

q

as well as

{\dot {q}}

. For example, if a point mass is attached to a rigid rod which is pivoted at the origin and rotating in the xy plane w/ angular speed

\omega

(the point mass is allowed to slide up and down the rod), then

{\vec {r}}=q\cos {\omega t}{\hat {x}}+q\sin {\omega t}{\hat {y}}

(q is the distance from the point mass to the origin) and so

T={\frac {1}{2}}m({\dot {q}}^{2}+\omega ^{2}q^{2})

. And besides,

\delta {\dot {q}}_{k}=0

by the definition of a virtual displacement. 65.92.5.132 (talk) 15:09, 6 April 2012 (UTC)[reply]

You aren't defining your terms, but it looks like you're using standard notation, where T is kinetic energy. If so, you forgot U (potential energy). Even when working in generalized coordinates, you must still use total energy. And, as pointed out above, kinetic and potential energy can both vary as functions of q, dq/dt, or any other generalized coordinate. Nimur (talk) 00:41, 7 April 2012 (UTC)[reply]

Sorry for not stating it, but I'm defining

\delta W

to be the total work done, including work done by any conservative force. Also, why would dependence on

{\dot {q}}

matter? Isn't

\delta {\dot {q}}=0

? 65.92.5.132 (talk) 02:34, 7 April 2012 (UTC)[reply]

Dreams: maker versus watcher

I am often struck by the disconnection between what things in dreams appear as and what they are interpreted as. In a dream, I can know that what I see as a tiger is 'really' a lion. Sometimes I've thought that the concept was real and its graphic rendition was fouled up, but this time it appears otherwise...

A dream I had last night involved the story of a little girl living out on the streets who was taken in by a caring family. After she told a social worker how much she loved it there and hoped she'd be adopted, her would-be parents took her up to an even nicer bedroom to stay. There's a sudden heavy thump on the ceiling that makes the hanging light in the room sway, and they tell her that's "their daughter Grace". Then they hurriedly lock the door leaving the girl inside, and as Grace comes down into the room she screams and runs over to a tiny window which she manages to get open and slide out through. As a fast-tempo musical theme from The Nutcracker (I think) starts up I see Grace, who is a giant snake with glowing eyes that follows the girl out the window and chases her around the manor's rooftop.

Now what was odd, though, is that all the while I'm watching this giant snake slither out the window and around the dormer windows this little girl was so desperately skirting, (eventually devouring a Garfield-like "pet cat" the girl had with her the whole time after she slammed a far door on it) I never once believed that what I saw as a giant snake was anything other than an older girl. I was still thinking, this is the cannibal freak daughter that was locked up in their attic chasing her. But the set-up - the thump on the ceiling, even the family adopting street urchins so that when they disappear nobody notices so that their "daughter" can have its preferred human meat - it all points to the person creating the dream knowing it was a giant snake the whole time.

Is there any term for this? Who makes the dream? Who watches the dream? Where are these people located in the brain? What mechanism notices incongruities between what we see and what we think we see, and how is it deactivated?

I suppose what perplexes me the most is, if one part of me had the ability to cobble together such an elaborate plot, why is a different and seemingly unconnected part able to watch it? Why isn't there just one place in the brain capable of conceptualizing the idea of "snake" and "little girl"? Wnt (talk) 13:24, 6 April 2012 (UTC)[reply]

(...crickets...) I don't think you'll get any refs for this one :). Dreams are so subjective, and you can easily appreciate how hard this would be to study scientifically. Anecdotally, I think I know what you mean, but I don't think it implies a "dream maker" vs. "dream watcher" dichotomy. Rather, I think it points toward the differences between what you "see" in a dream v.s what you "know" in a dream. Example: I've had dreams about people I know where (to borrow acting terms) the "part" of person X is "played" by person Y. Often, person Y is indeed a famous actor/actress, maybe because my brain is more inundated with images of the famous person, so their face fills in, even when the dream is about someone I know personally. This, to me, sounds similar to what you describe. You "knew" it was a cannibal human, but it was "played" by a giant snake. Does that make any sense? SemanticMantis (talk) 22:00, 6 April 2012 (UTC)[reply]

It sounds like we're observing very close to the same thing - which means there's something reproducible to it. But it sounds like in the case you describe the known version was indeed more fundamental than the seen version, which is interesting. I can't believe there isn't data on it though, at least, psychological anecdotal data, given how many shrinks tried to infer Freudian things from dreams - surely some must have taken a general interest in how they work, and come up with some such observations. (Actually, as I was fairly consciously aware while having it, this dream was rooted in a certain Commons thread where for a time I played the part of the social worker, but the less said there the better) Wnt (talk) 04:24, 7 April 2012 (UTC)[reply]

Just to pick up on SemanticMantis's point, the meaning of your dream will be apparent to you if you think hard enough about it. Dreams, as they are currently perceived by the psychological community, are a way of making sense of memories, and as memories are derived from your own experience, nobody else can tell you what a dream means. So the maker of the dream, the watcher of the dream, are all you. --TammyMoet (talk) 08:11, 7 April 2012 (UTC)[reply]

Middle East climate

Why is the Middle East so dry even though it directly borders parts of Europe and Asia that aren't? --204.184.214.187 (talk) 14:43, 6 April 2012 (UTC)[reply]

Not all of the Middle East is very dry; Israel, Lebanon and parts of Jordan and Syria have a more moderate climate. Also, the Middle East is not a unique dry part, but part of a much larger dry area spanning from northwest Africa to parts of west India. At Köppen climate classification, you can view a world climate map illustrating this. What exactly the cause of the drought is, may be complicated and is probably related to wind patterns, but I don't know much about that. -- Lindert (talk) 15:19, 6 April 2012 (UTC)[reply]

In a nutshell, it seems you don't get much rain around the Persian Gulf and the Red Sea because the atmospheric circulation known as the Hadley cell carries water away from the Middle East, and the mountains around the Iranian plateau (mostly, I think, the Zagros Mountains) trap what rain does blow in. This paper has more details, though it's rather dry and technical. Smurrayinchester 15:51, 6 April 2012 (UTC)[reply]

It's part of a dry subtropical lattitude band that stretches from the Sahara to the Gobi and also includes the US Southwest and adjoining northen part of Mexico. It's southern hemisphere counterpart includes the Namib and Kalahari in Africa, the Atacama in South America and the Central dessert in Australia. Roger (talk) 20:42, 7 April 2012 (UTC)[reply]

The answer is the Hadley cell, but not the reason given above. The reason the subtropical latitudes are so dry is that air rises in thunderstorms near the equator, it then must sink at some point. These desert regions are areas where there is a net subsidence, or sinking of the air, which in general suppresses the rising motions that contribute to rain. This is a very general, averaged effect, so local effects such as the Indian monsoon cause areas at similar latitudes to be quite moist.-RunningOnBrains^(talk) 21:57, 7 April 2012 (UTC)[reply]

Vitamin B12 absorption

I've read that vitamin B12 absorption from low dose supplements (less than about 50 micrograms) is limited to a maximum of about 1.5 micrograms, presumably because the limiting factor is the availability of enzymes needed for absorption.

How long does it take after taking one supplement before the body is able to absorp 1.5 micrograms from the next supplement? Count Iblis (talk) 15:58, 6 April 2012 (UTC)[reply]

Uniqueness of the attractor for Earth's climate

Earth's climate tendencies can be viewed as an attractor within which the actual climate evolves. For example, in the above discussion of the Middle East's climate, the Hadley cell carries water away from the Middle East, giving it a mostly dry climate; the Jet Stream over the US follows a time path that, while not precisely repeating itself annually, stays within certain limits; etc.

But is it known whether the observed global climate attractor is unique? In other words, if we held constant the locations of the continents and oceans and the orbit of the Earth, but started from initial conditions of say zero wind velocity everywhere (or started with new initial conditions after some massive external disruption), would we eventually return to the attractor that we are familiar with? Or are there multiple attractors, each with a different basin of attraction in the space of initial conditions? Duoduoduo (talk) 17:38, 6 April 2012 (UTC)[reply]

It partly depends on the timescale you look at. There is, I believe, a pretty strong case that Snowball Earth is also an attractor -- if the entire Earth were covered with ice, it would reflect so much sunlight that it would never warm enough to melt the ice. It is widely believed that the Earth has been in such a state several times, but that it eventually escaped because over the course of a few hundred million years so much CO2 accumulated in the atmosphere that it eventually produced enough global warming to melt the ice. That account isn't universally accepted, though. Looie496 (talk) 17:46, 6 April 2012 (UTC)[reply]

One big effect seems to be the motion of the continents. When they move they change ocean currents, and, to a lesser extent, air currents, causing massive changes in the climate. So, you would have to stop plate tectonics, among other things, to get a constant climate. The "snowball Earth", for example, might be caused by plate motions. So, I would argue there would be a single equilibrium point if none of the inputs changed, but, since they do, we end up with multiple equilibrium points, depending on the current inputs to the system. StuRat (talk) 17:58, 6 April 2012 (UTC)[reply]

My gut feeling is that the Ice Ages approximate such an attractor (I know the orbit is supposed to be involved, but have you looked at the comparison? It's not very convincing looking). In any case, the transition back and forth from Ice Age to not is so extreme, I think it would have a good chance to kick us out of any smaller attractors if they exist. Of course, climate has changed on that time scale in many places, e.g. North Africa which in Carthaginian times was better suited for horses than camels. Wnt (talk) 20:32, 6 April 2012 (UTC)[reply]

As Looie said, it depends on the timescale you are looking at. We will never reach a true attractor in climate (even if one might exist locally), because the solar and geophysical inputs are constantly changing: on short scales we have seasons, on intermediate scales we have the Milankovitch cycles, on longer scales we have drifting continents, and on even longer scales we have ever-increasing solar luminosity, and decreasing tides and increasing length of day due to the tidal acceleration of the earth-moon system. This is not to mention the occasional high-impact events such as supervolcanoes and impact events, and other bizarre occurrences (Messinian salinity crisis, megafloods) which can affect the earth for thousands of years, or human activities such as deforestation and anthropogenic emissions causing photochemical smog, ozone depletion, and global warming.-RunningOnBrains^(talk) 20:56, 7 April 2012 (UTC)[reply]

Explaining light

Some years ago I received from an aquaintance by e-mail a curious physics explanation. I need some help to provide a serious point-by-point answer. The following is the exact text as I received it, with no attempt to correct typos e.g. Cerb to Cerne , paricles to particles, degenartion to degeneration.

(text starts)That collider at Cerb, Switzerland is another non-starter. They are using something ( electrons/magnetism throigh a material conductor) to find out what conditions were like when there was nothing. It's a dead end. All you ever get when dividing matter into ever smaller paricles is another surface. No one has yet considered that there may be a dimension where light can not go less than 186k/sec., and that this dimension ( the big bang) ocurred when an anomaly, or time warp, retarded light and as it descended from super light, time came with it and the release of supercharged energy created heat, which when it expanded created matter. Desension of superlight from hyperspace couldn't dispell itself fast enough resulting in a dense coagulation, or degenartion of energy, into more solid form.

I've never understood why if the speed of light is a constant, its speed can become squared; as in E= Mc2. But if the speed of light can be squared ( in fact) there should be a universe, or dimension, where this is the standard. (text ends) 84.209.89.214 (talk) 23:52, 6 April 2012 (UTC)[reply]

Technobabble like this example can sound quite serious, but it is pure gibberish. A point-by-point explanation would be useless, because this text excerpt is absolutely meaningless. It name-drops several interesting topics in physics, and it misspells CERN, among other errors; but it doesn't actually say anything. Nimur (talk) 00:44, 7 April 2012 (UTC)[reply]

Perhaps you should reply in technobabble too ? "The light-force emanations vibrate into crystalline astral projection planes in non-isotropic Euclidean parallel multi-verses, thus ensuring the isotope strings of an invariant cosmology." StuRat (talk) 00:54, 7 April 2012 (UTC)[reply]

One fairly big problem is that there was not "nothing" at the Big Bang, there was an incredible amount of energy contained in an infinitely small space. The idea of the Large Hadron Collider is not to create "nothing from something", but to create as large an amount of energy in as small a space as possible. The rest is basically gibberish. Speed squared is not a speed, any more than distance squared is a distance (for instance, 5 inches squared is not 25 inches, but 25 square inches, which is a measurement of area). Smurrayinchester 09:05, 7 April 2012 (UTC)[reply]

just one easy to contradict point. Let's say a typical rat is 5-6 inches. If a person is 5-6 feet, that means that the length of a person in terms of rats is 12 * ratlength. But if 12 * ratlength is meaningful, then there must be a rat the size of a person. We just have to find it and put it on television. (this is the same argument used about the fact that you can 'square the speed of light' in e=mc^2, exactly the same as multiplying the length of a rat by 12. it doesn't have any bearing on the speed of light!) 188.156.106.2 (talk) 08:39, 7 April 2012 (UTC)[reply]

There are rats the size of people (infants, at least). See nutria. StuRat (talk) 16:43, 7 April 2012 (UTC)[reply]

Nutrias are nothing! Check out these guys. Evanh2008 (talk) (contribs) 21:23, 7 April 2012 (UTC)[reply]

Yep. Tecnobabble, and not even good technobabble. In addition, heat could not have existed before matter since heat is a property of matter. Evanh2008 (talk) (contribs) 09:03, 7 April 2012 (UTC)[reply]

Escape velocity

Is the escape velocity of the Earth's gravitational field in m.p.h. is equal to the length of the circumference in miles? 84.209.89.214 (talk) 23:53, 6 April 2012 (UTC)[reply]

No. Why would it be? That the numerical values, expressed in different units, are similar is merely a coincidence. Dragons flight (talk) 00:01, 7 April 2012 (UTC)[reply]

Earth's circumference in cubits is 8.8×10⁷ while Earth's escape velocity is 8.8×10⁷ cubits per hour. Lord! it's a miracle!

For any spherical body, the ratio between its circumference and its escape velocity is given by the following expression:

{\frac {{\sqrt {2}}\pi r^{\frac {3}{2}}}{\sqrt {GM}}}

Subbing in Earth's dimensions gives us 3575 seconds, which is coincidently very close to the 3600 seconds in each hour. Thus it's pure coincidence that an object travelling at escape velocity around Earth would take approximately one hour. Anonymous.translator (talk) 02:16, 7 April 2012 (UTC)[reply]

As a side note, "Eight Eights" (88888888) is probably the shortest mnemonic possible for memorizing both Earth's equatorial circumference and Earth's escape velocity. For someone living in ancient Egypt at least.Anonymous.translator (talk) 02:24, 7 April 2012 (UTC)[reply]

Escape Velocity = the speed at which a prisoner is pardoned after making a generous contribution to a retiring US President, or governor of a US state. StuRat (talk) 02:36, 7 April 2012 (UTC) [reply]

April 7

How much extreme weather damage can the world economy take?

The international reinsurance company Munich Re says that the amount of damage from extreme weather tripled from 1980 to 2010, and they expect that trend to continue.[6] Assuming that remains true, (1) how long until the damage from extreme weather causes the world economy to collapse? And, (2) assuming global wind power growth continues along its current trend, will it prevent such a collapse in time? 71.215.74.243 (talk) 02:52, 7 April 2012 (UTC)[reply]

At some point neither private companies not governments will be able to insure people living in high risk areas, like below sea level in a place regularly hit by hurricanes. This will result in people no longer being able to get mortgages to build there, and will thus limit damages from future storms, restoring the balance. StuRat (talk) 03:09, 7 April 2012 (UTC)[reply]

So do you think the world economy can absorb an infinite amount of storm damage? 71.215.74.243 (talk) 11:07, 7 April 2012 (UTC)[reply]

No, but an infinite amount of storm damage isn't possible, so that's pointless to worry about. You seem to be making a common extrapolation error. Whenever anything is increasing, all you have to do is assume it will forever increase at the same rate to conclude that it will eventually become infinite. The assumption that anything which is currently increasing must therefore increase forever is fundamentally flawed. StuRat (talk) 17:10, 7 April 2012 (UTC)[reply]

I agree. Let me rephrase my question: If the amount of economic damage from extreme weather continued to tripple every 30 years, do you think the world economy would still remain functioning indefinitely? Or are you saying that the damage must eventually level off before it causes economic collapse? 71.215.74.243 (talk) 22:20, 7 April 2012 (UTC)[reply]

Yes, it will level off or even reduce. For example, now that global warming is fairly well established, people will eventually stop building on eroding coastlines. StuRat (talk) 22:33, 7 April 2012 (UTC)[reply]

The premise is false. Data from impartial sources [7] show that there's no such increasing trend.

Also the global wind power cumulative capacity can't grow forever either, since the total amount of available wind power on Earth is limited. Anonymous.translator (talk) 03:18, 7 April 2012 (UTC)[reply]

You don't think that graph you linked to shows a trend? Wind power can't grow forever but there is enough for dozens of times expected human needs even assuming everyone started using US per-capita power. 71.215.74.243 (talk) 03:31, 7 April 2012 (UTC)[reply]

Looking at the dotted line, no, I don't see a clear increase. For most years the damage is roughly $10 billion while in the occasional "disastrous" years the damages spike up.Anonymous.translator (talk) 03:47, 7 April 2012 (UTC)[reply]

Do you think the average of the dotted line in, for instance, the first, middle, and most recent third of the graph comprises a trend? 71.215.74.243 (talk) 11:07, 7 April 2012 (UTC)[reply]

I'm pretty sure that as a publicly traded company whose core business is insurance, Munich Re has a legally mandated duty to be honest about the past claims they have honored and the future claims they expect. As a global company, I would expect they actually have a better picture of climate related losses than data that solely comes from the US. That said, I seem to recall that Munich Re also blames most of the trend on increased human development in vulnerable areas, and relatively little of it on actual changes in weather conditions. Dragons flight (talk) 05:11, 7 April 2012 (UTC)[reply]

One thing I left unstated is that most of the damage is due to people building in unsafe areas. Once we stop that, the problem is largely solved. StuRat (talk) 04:27, 7 April 2012 (UTC)[reply]

Yes. People need to remember such things as the fact that flood plains are for floods. HiLo48 (talk) 05:31, 7 April 2012 (UTC)[reply]

Indeed, Munich Re used to say that sort of thing, but not in the past several years. Their 2010 report linked above, for example, says, “In Germany, extreme precipitation resulting in floods is becoming increasingly common. This affects not only people living on rivers: there are more and more cases of heavy rain and flash floods. Anyone may be affected.” 71.215.74.243 (talk) 11:07, 7 April 2012 (UTC)[reply]

That's a rather meaningless statement, rather like saying that anybody can be struck by lightning, so you might as well play golf on top of a hill in a thunderstorm. While anyone can be affected by some type of bad weather, by no means is the risk equal in all locations, far from it. StuRat (talk) 16:16, 7 April 2012 (UTC)[reply]

One line on that graph is normalized for inflation but not for population growth or overall real estate value. Wnt (talk) 15:44, 7 April 2012 (UTC)[reply]

Not much, and the weather doesn't have to get a lot more extreme overall. According to a recent BBC Horizon documentary, we're now seeing significantly more fluctuations in the weather, while the average has only shifted slightly due to global warming. The latter is driving the former. A prediction of climate models is that cyclones can hit places where they have sitorically never occurred, e.g. Dubai seems to be at risk of being washed into the ocean by a category 5 hurricane.

A small perturbation is enough to let the World economy crash, because the economy has been mismanaged anyway. We were about to hit a global depression, simply due to mismanagement in 2008. So, it's a bit like flying a plane that could theoretically fly to the thunderstorm, but with incompetent pilots flying it, it's already at risk of crashing due to pilot error when the weather is fine. Obviously, you then don't want to fly a plane with these pilots through thunderstorm.

What I forsee happening within a few decades, is droughts and floodings causing India and China to have to import huge amounts of rice and grain for a few years in a row. They have enough financial reserves to do so, but that would drive food prices up by so much that other countries will close their markets to prevent all their rice and grain from from being exported to India and China. That will then lead to collapse of free trade agreements and ultimately to the collapse of the World economy. Count Iblis (talk) 17:42, 7 April 2012 (UTC)[reply]

In that case, what they need is system to collect flood water and store it until the next drought. StuRat (talk) 18:25, 7 April 2012 (UTC)[reply]

So simple, yet so difficult! How should this be accomplished on the massive space and time scales necessary to help? Shall we start a lobby to subsidize rain barrels? SemanticMantis (talk) 19:21, 7 April 2012 (UTC)[reply]

In the case of underground aquifers, it may be as simple as redirecting rivers to refill them rather than draining to the sea. StuRat (talk) 19:47, 7 April 2012 (UTC)[reply]

Going back to the original question, I don't see what wind power has to do with any of this. Additionally, the correlation between severe weather events and global warming is unknown; indeed, we should see a reduction in Atlantic hurricanes in the next decade or two due to a downturn in the West African rainfall cycle. -RunningOnBrains^(talk) 20:33, 7 April 2012 (UTC)[reply]

I've read many times that it's not simply the number of Atlantic hurricanes that matters; supposedly global warming is predicted to increase their average intensity -- fewer less intense ones combined with more very intense ones = more damage. Is that not right?

Also, the problem is not solved once we stop building in unsafe areas (although obviously that is important too). An amazing percentage of the world's population has already built in unsafe areas -- there are lots of hugely populated coastal cities (and for good reason since coastal cities can have ports).

I think the questioner's point about wind power was simply that, to the extent that wind power supplants CO₂-releasing power generation, we can slow down or stop the growth of CO₂ levels in the atmosphere. S/he is asking whether present trends in the growth of wind power usage are fast enough to do that before we reach a tipping point. I'm pessimistic about it, but it's not all-or-nothing -- increased use of wind power might delay or minimize the extreme-weather effects of CO₂ even if not stopping them entirely. Duoduoduo (talk) 21:06, 7 April 2012 (UTC)[reply]

It could be true: certainly increasing the sea-surface temperatures in the absence of all external factors will increase the amount of latent heat available for the hurricanes to tap into, but I've seen studies that suggest that wind shear will increase dramatically over the tropical Atlantic with a warming planet, which would lead to less intense hurricanes. We simply don't have enough information to know for sure. One thing that is known, however, is that increasing sea levels will make residents increasingly vulnerable to the hurricanes that do occur. (I'm not talking Al Gore's nonsense about a 30m increase in sea levels: even a conservative estimate of 20 centimetres (7.9 in) in the next 100 years could have devastating long-term impact.-RunningOnBrains^(talk) 21:22, 7 April 2012 (UTC)[reply]

As far as wind power costs, even if the world managed to stop emitting fossil fuels in the next 20 years (an impossible goal, as China is clearly not going to try to stop burning coal any time soon), CO2 in the atmosphere is far from its steady state. The residence time of CO2 in the atmosphere is hundreds of years, so it would likely take thousands of years (without active intervention) to return atmospheric CO2 to it's current levels, never mind pre-industrial levels.-RunningOnBrains^(talk) 21:30, 7 April 2012 (UTC)[reply]

We not only need to stop building in unsafe areas, but move away from them. Ports can remain (and be hardened against foul weather), but the rest of the city should be moved. This will happen slowly, as storms destroy buildings in unsafe areas and the foolish people who built there are unable to get insurance and mortgages to rebuild there again.

Also, there's a natural life of a building, after which it would be rebuilt (except for a few historic sites). When the buildings in unsafe areas reach that age, nobody will be willing to risk their money to rebuild there. StuRat (talk) 21:19, 7 April 2012 (UTC)[reply]

The idea of moving away from the unsafe areas is interesting -- the Maldives, I've read, are shopping around in Australia for land they can eventually move their entire country to; and I read very recently that one of the South Pacific island nations is also shopping around for land. But these are tiny populations. It would take an extremely long time, with an extreme amount of social disruption and a massive amount of highly expensive infrastructure investment, to move everyone in say Mumbai farther inland. And then transportation costs to the port would be higher than now. So it's not going to happen to any significant extent, unless gradually rising sea levels concentrate people's minds to organize it. Duoduoduo (talk) 21:37, 7 April 2012 (UTC)[reply]

There's no need to move the entire city at once. Start with the coastal area, then gradually move inland. This could all happen over a century. For a case where it is happening rapidly, look at Christchurch, New Zealand, which has the misfortune of being on an active fault which destroyed the downtown area recently. StuRat (talk) 21:41, 7 April 2012 (UTC)[reply]

Male sexual activity and cognitive performance

How close have any studies come to establishing or disestablishing an effect of sexual activity on cognitive performance in men? Have there at least been correlational studies that controlled for the confounding fetal-androgen effect, or lab studies using the same combination of hormones that are released in orgasm? Neon Merlin 10:18, 7 April 2012 (UTC)[reply]

What kind of cognitive performance? PMID 19105078 is a recent review, and PMID 16490297 is an example study controlling for fetal androgen exposure. PMID 19250266 includes the study of prolactin, which is the only hormone conclusively shown to be released during male orgasm. 71.215.74.243 (talk) 11:38, 7 April 2012 (UTC)[reply]

Physics LC circuit

The natural frequency of LC circuit is 1 kHZ. How do we reduce it to 0,5 kHZ, using the equation: natural frequency = 1/(2 π sqrt{LC}) I think duplicating the T (2 π sqrt{LC}) is not the correct answer? Thank you in advance!--Atacamadesert12 (talk) 17:40, 7 April 2012 (UTC) — Preceding unsigned comment added by Atacamadesert12 (talk • contribs) 17:39, 7 April 2012 (UTC)[reply]

if you want the frequency to be halved then 1/(2 π sqrt{LC}) needs to be halved, i.e. LC needs to be four times as large. Dmcq (talk) 19:41, 7 April 2012 (UTC)[reply]

Or you could increase the value of 2π -RunningOnBrains^(talk) 20:21, 7 April 2012 (UTC)[reply]

Double it, to be precise, so π ≈ 6.28318530717958647652528677. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 20:26, 7 April 2012 (UTC)[reply]

Electronics on board Apollo 11 spacecraft

What were the specifications of the Apollo 11 spacecrafts' electronics? By this, I mean the motherboard, memory, processor, etc. Also, how differently would the spacecraft have been designed if it were designed today with modern electronics?--99.179.20.157 (talk) 18:40, 7 April 2012 (UTC)[reply]

Apollo Guidance Computer is a good start. Almost everything would be different if implemented today; not least you wouldn't have a room full of ladies hand-knitting the program. -- Finlay McWalterჷTalk 18:43, 7 April 2012 (UTC)[reply]

It was the memory core that got hand-knitted, not the program.--Aspro (talk) 19:02, 7 April 2012 (UTC)[reply]

Maybe your right Core rope memory--Aspro (talk) 19:10, 7 April 2012 (UTC)[reply]

Great link, thanks! SemanticMantis (talk) 19:22, 7 April 2012 (UTC)[reply]

(ec)I'm not at all sure that the motherboard concept even existed at the time. Please clarify whether you are interested only in the on-board computers or all of the electronics in general. Most "electronics" don't contain motherboards, processors or memory, those components are specific to digital computers. Roger (talk) 18:49, 7 April 2012 (UTC)[reply]

A good resource for finding your way around more recent spaceflight computer systems is Category:Radiation-hardened microprocessors, which lists a bunch of devices that are, or have been, used in spacecraft. -- Finlay McWalterჷTalk 18:55, 7 April 2012 (UTC)[reply]

You should also remember that the onboard computer systems were never intended to be general purpose computers; they were calculators and autopilot controllers designed specifically for the single purpose of automating certain timing operations for the manned flights to the moon. These computers have more in common with a cockpit instrument than with modern general purpose personal computers. For example, there is no "screen" on the infamous Apollo Guidance Computer; only a few cryptic numerical status indicators and some warning lights. However, if you compare this to, say, a RADAR altimeter instrument in terrestrial aircraft of that era, you see that AGC actually provides much more information and control, including automatic fly by wire control of some spacecraft and propulsion systems. Nimur (talk) 19:48, 7 April 2012 (UTC)[reply]

Using tachyons to blast something out of a black hole

Would it be possible to "rescue" people or objects from beyond the event horizon of a black hole by hitting them with a super-high energy beam of tachyons at just the right angle? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 20:37, 7 April 2012 (UTC)[reply]

I take you mean hypothetically. Well, hypothetically one might have better luck using a transporter. How do you think these questions up?--Aspro (talk) 20:46, 7 April 2012 (UTC)[reply]

AAAAAAAAAAAARGH!!!!! Whoop whoop pull up ^{Bitching Betty | Averted crashes} 20:55, 7 April 2012 (UTC)[reply]

So...is that the sound of you trying the transporter idea? -RunningOnBrains^(talk) 21:13, 7 April 2012 (UTC)[reply]

Maybe the transporter reassembled his molecules with those of a fly The Fly (1986 film)? Are hybrid flies allowed to hold a WP account?--Aspro (talk) 21:25, 7 April 2012 (UTC)[reply]

BE SERIOUS!!!!Whoop whoop pull up ^{Bitching Betty | Averted crashes} 21:28, 7 April 2012 (UTC)[reply]

Verily - but you first. --Aspro (talk) 21:36, 7 April 2012 (UTC)[reply]

I couldn't even begin to approach the highly theoretical math involved, but my understanding is that tachyons can't interact with normal matter, as this would create paradoxes. -RunningOnBrains^(talk) 21:33, 7 April 2012 (UTC)[reply]

The serious answer is "no, because tachyons do not exist as far as we know, and even if they do exist, we can't know anything about them, and even if we did know all about them, we couldn't manipulate them, and even if we could manipulate them, we couldn't use them to manipulate ordinary matter." 71.215.74.243 (talk) 22:42, 7 April 2012 (UTC)[reply]

Can you pull something out of a black hole? E.g., what if it was a microscopic black hole, or maybe a one-pound black hole -- could you stick part of a sturdy object partway in it and then pull it out? Duoduoduo (talk) 22:56, 7 April 2012 (UTC)[reply]

No. As summarized in the previous black hole thread, once within the event horizon, conventional "spacewise" directions become "timewise" -- all future paths move to the singularity; all paths back out require you to move backwards in time. Additionally, it only confuses matters to think about "an object" that stretches into a black hole. That whatever-it-is is in fact a large number of independent subatomic particles bound together by various forces, but those forces are overcome by the black hole once the particles pass within. — Lomn 23:07, 7 April 2012 (UTC)[reply]

How about creating a wormhole inside the event horizon? Plasmic Physics (talk) 23:58, 7 April 2012 (UTC)[reply]

Radicals and VSEPR

What geometries does VSEPR theory predict for radicals? For instance, does the methyl radical have a trigonal planar geometry or a trigonal pyramidal geometry? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 21:26, 7 April 2012 (UTC)[reply]

Lone electrons are treated just like non-bonding electron pairs. The methyl radical is triagonal pyramidal. Plasmic Physics (talk) 22:55, 7 April 2012 (UTC)[reply]

Thank you. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 02:17, 8 April 2012 (UTC)[reply]

Resolved

April 8

Illumination in lumen additive or parallel?

If various LED light emitters is fitted within the same "bulb". Will their light intensity add up as in total_lumen = lumen_1 + lumen_n + lumen_n+1.. Or will a light source made up of many parallel light emitters never be equal to a single (incandescent) light source of x lumen? That is parallel photons will not add up into less photons with the same energy but larger amplitude? because phase synchronization will not happen? Electron9 (talk) 00:58, 8 April 2012 (UTC)[reply]

I believe it's additive. However, our perception is not. That is, twice as many lumens doesn't seem "twice as bright". When lighting a room, it's helpful to have lights in each corner, to avoid any dark corners (this is especially true with dark walls, like wood paneling, where little light reflects into the corners). StuRat (talk) 01:15, 8 April 2012 (UTC)[reply]

Right. Photons do not join together or combine into fewer photons of different amplitude or frequency. 71.215.74.243 (talk) 01:23, 8 April 2012 (UTC)[reply]

Thus the whole idea about replacing incandescent light bulbs with multiple emitters is fundamentally flawed because it just won't be the same ? Electron9 (talk) 01:49, 8 April 2012 (UTC)[reply]

No, that's not at all what's being said. For the same total number of lumens, and same quality of light, the room will generally be perceived as being better lit if it has multiple, lower-lumen lights in multiple locations, instead of one bright light in one location. But if you wanted the more poorly lit look of a single light for some reason, you could just put the multiple lights in essentially the same location. Red Act (talk) 03:29, 8 April 2012 (UTC)[reply]

Agreed. Also, having the light sources spread out will be perceived as more light. StuRat (talk) 03:41, 8 April 2012 (UTC)[reply]

Strange zoo bird

Any idea what this bird is? I've gone through all the birds on its zoo's list and none seem to be it. 71.215.74.243 (talk) 01:05, 8 April 2012 (UTC)[reply]

I think it may be the Chestnut-breasted Malkoha - which is on the list. AndyTheGrump (talk) 01:23, 8 April 2012 (UTC)[reply]

More on it here - with another video. Surprisingly, it is one of the Cuculidae - the Cuckoo family. Common in Southeast Asia. AndyTheGrump (talk) 01:32, 8 April 2012 (UTC)[reply]

No doubt! Thanks, I don't know how I missed it from the accurate description linked from the list. 71.215.74.243 (talk) 01:36, 8 April 2012 (UTC)[reply]

Cool bird! "Unlike many cuckoos, it builds its nest and raises its own young." Marking as resolved. SemanticMantis (talk) 01:57, 8 April 2012 (UTC)[reply]

Resolved

Octane rating of jet fuel

What is the octane rating of jet fuel? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 02:30, 8 April 2012 (UTC)[reply]

Jet fuel is essentially kerosene (C8 to C16 molecules), not gasoline (C4 to C12 molecules). Octane rating is irrelevant to jet fuel, since predetonation is not a concern in a gas turbine. However, if it was measured it would be considered to be very high, since octane rating is a measure of fuel's resistance to predetonation, not its volatility. Acroterion (talk) 02:38, 8 April 2012 (UTC)[reply]

Do different chicken breeds have different tastes?

I'm not talking about their eggs, whose taste apparently depend not on their shell color, but the chicken feed. 66.108.223.179 (talk) 02:32, 8 April 2012 (UTC)[reply]

If so, it must not be very noticeable, or I'd expect various brands to claim they have better tasting breeds. StuRat (talk) 03:35, 8 April 2012 (UTC)[reply]

mercury spectrum

what elements may have a spectrum of wavelengths very similar to the spectrum of mercury — Preceding unsigned comment added by 197.255.118.206 (talk) 06:43, 8 April 2012 (UTC)[reply]

"Similar"? You can eyeball the visible spectra at http://www.umop.net/spctelem.htm Hg is in the lower left. 71.215.74.243 (talk) 10:16, 8 April 2012 (UTC)[reply]

sodium spectrum

what are the wavelengths of the red, yellow, yellowish-green, green, greenish-bue and violet colours of sodium . — Preceding unsigned comment added by 197.255.118.206 (talk) 08:15, 8 April 2012 (UTC)[reply]

physics

what is the difference between spectroscopy and spectrometry — Preceding unsigned comment added by 197.255.118.206 (talk) 08:31, 8 April 2012 (UTC)[reply]

Spectroscopy is the science of determining characteristics from a spectrum. Spectrometry is the measurement of spectra. In many cases they can be used interchangeably without too much ambiguity, but it's better to be precise. 71.215.74.243 (talk) 10:06, 8 April 2012 (UTC)[reply]