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Why doesn't the common cold infect nonhuman mammals? Comet Tuttle (talk) 01:55, 19 February 2012 (UTC)[reply]

It does. Not exactly the same viruses as infect humans (they have evolved to use us as host and wouldn't be very effective in other hosts), but animals can certainly get colds. I just Googled "animals getting colds" and found loads of sites talking about it. --Tango (talk) 02:09, 19 February 2012 (UTC)[reply]

Note that outside the world of farming, you're very unlikely to get large numbers of strange animals packed together as closely as humans get in shops, offices, cinemas, trains, etc, which limits the potential for a disease like the cold to spread, which is why you might not see dogs or cats catching colds as often as humans do, and why cage-bound animals like hamsters may appear to never catch infections. Smurrayinchester 09:24, 20 February 2012 (UTC)[reply]

Although it's intuitively clear that females prefer to mate with the winner, I want to know how common is that in general among mammals? In humans, it's certainly not always the case. 88.8.66.12 (talk) 02:32, 19 February 2012 (UTC)[reply]

I think most animals that engage in mating contests fight until one contestant submits rather than having either contestant get significantly injured (although there are certainly exceptions). I those cases, I expect the contest would just restart if one of the contestants tried to mate with the female after submitting. --Tango (talk) 02:52, 19 February 2012 (UTC)[reply]

Is there any animal in which it's "always" the case? --140.180.6.154 (talk) 07:29, 19 February 2012 (UTC)[reply]

It would be fairly safe to say in species whose main reproductive groups are usually one male and a group of females, where the male holds his place through fending off intruding solo males from time to time, this would always be the case ("always" being a very strong word, as there can be sneaky exceptions). Things like lions, gorillas, and even kangaroos tend to hold to the 'mate only with winning male' rule. In humans this may have been more common in more historic societies where a powerful chief or similar could maintain a large group of females, such as the stereotypical depiction of a harem. --jjron (talk) 07:54, 19 February 2012 (UTC)[reply]

It "may have been" - but it almost certainly wasn't. In any case, biologically speaking 'historical societies' are a blip: we are the result of 'prehistorical' ones. And what what little evidence we have from contemporary hunter-gatherer societies suggests that something approximating to 'serial monogamy' was more likely to have been the norm. And if you are going to compare us with other mammals, it seems logical to compare us first with our nearest relatives, the apes. Not that this helps much, given that one can find more or less anything from indifferent promiscuity Bonobos to (apparent) monogamy Gibbons. As for Gorillas, I well recall a lecture on primate sociobiology, which demonstrated all to well that neither the 'dominant male' silverback, nor the 'submissive' females would seem to got much excitement out of mating - hardly surprising, given the silverback's equipment (think of the cap on a ballpoint pen), or the duration of the encounter, which rarely exceeded ten seconds. If you want to understand human sexuality, study people... AndyTheGrump (talk) 08:21, 19 February 2012 (UTC)[reply]

Gorrilas don't pair up - generally one male that can dominate other males will collect a harem. I recall an article in Scientific American a few years ago where the author had established that if a species pairs up to raise young to independence (eg humans, many species of birds), sex is highly pleasurable for both sexes, and sex is frequent, playing a bonding role, and rape is unusual - and where species do not pair up (domestic cats being an excellent example) to raise young, sex is only marginally pleasurable for the male, and probably not at all for the female. For such species, sex is only for reproduction and probably constitutes rape. The frequency and pleasurability of sex does not map at all to the closeness of one species to another, but extremely well to how the young are raised. Generally, species that pair up, the pairs are less likely to influenced by how dominant the males is with other males, and almost all males get partners. In species that do not pair up, weaker males may never get any sex. All this seems to hold, more or less, for reptiles, birds, and mammals. So, yes, you can learn something of human sexuallity by studying completely different species. Ratbone121.215.64.242 (talk) 10:12, 19 February 2012 (UTC)[reply]

Elephant seals, horses (actually all extant equids), springbuck (and many other antelope), and many other herding species have dominant teritorial males who fight to maintain control of a territory and the sole right to mate with females in the territory. Roger (talk) 09:29, 19 February 2012 (UTC)[reply]

But, can females "move abroad" to another male? Is some kind of unpredictability in the process? 188.76.228.174 (talk) 14:39, 19 February 2012 (UTC)[reply]

Having watched Red Deer in London's royal parks often enough during the rutting season, I'd say that the hinds are well-motivated to 'move abroad', often taking advantage of confrontations between stags to break away from the herd - whether they'd behave in the same manner in a more natural environment, where there was a risk of predation involved with being alone, I'm not sure. It is also questionable as to whether the herds have 'territories' at all - instead they wander around as a group within an area shared with other herds. It would be interesting to learn just how cohesive a 'herd' is over time anyway, during the rut. AndyTheGrump (talk) 18:43, 19 February 2012 (UTC)[reply]

This is NOT a request for medical advice - I am just currious. A friend has been told she has 3 different types of breast cancer (in one breast!) - i.e. each tumour is of a different type. How often does this occur? Ratbone120.145.28.184 (talk) 03:32, 19 February 2012 (UTC)[reply]

The best data is for the US state of Connecticut, in which slightly under 2% of cancer patients were diagnosed with at least two primary tumors simultaneously [1]. I'm not sure what the probability is for three simultaneous cancers in the same breast, but presumably it's even smaller. Although if a person is genetically predisposed to developing breast cancer, then perhaps the chance is actually not small at all. Someguy1221 (talk) 10:34, 19 February 2012 (UTC)[reply]

Agreed. If all of the cancers were independent events, this would be extremely rare. However, whichever factors caused one of the cancers are likely to have also contributed to the others. A genetic predisposition is one factor, but there could also be exposure to radiation, carcinogenic chemicals (perhaps due to smoking), poor nutrition, age, obesity, a sedentary lifestyle, and a depressed immune system (due to stress, disease, etc.). StuRat (talk) 21:28, 19 February 2012 (UTC)[reply]

Many thanks, especially to Someguy. The papers hows that it is unusual to get 2 simulataneous tomours. Since this would include simultaneous tumours of the same type, tumours of diferent type must be a sub-set, so three diferent types must be quite rare. Ratbone121.221.28.225 (talk) 14:49, 20 February 2012 (UTC)[reply]

How does intake of food acids affect the human body? Thanks. 112.118.205.3 (talk) 09:18, 19 February 2012 (UTC)[reply]

Humans cannot survive without proper intake of essential amino acids --SupernovaExplosion ^Talk 12:03, 19 February 2012 (UTC)[reply]

Certain essential vitamins are also acids. Without Ascorbic acid, for example, you can get scurvy. The name "ascorbic acid" even means, broadly, "prevents scurvy". --Jayron32 14:02, 19 February 2012 (UTC)[reply]

Do carnitine have any real effect in reducing body fat? This says "studies do show that oral carnitine reduces fat mass". Then why do it says there is no scientific evidence for this fact? If a person start taking carnitine supplement, will he become dependent on supplement i.e. face problems with normal fat metabolism after supplement withdrawal? --SupernovaExplosion ^Talk 15:35, 19 February 2012 (UTC)[reply]

That link says "Although L-carnitine has been marketed as a weight loss supplement, there is no scientific evidence to show that it works. Some studies do show that oral carnitine reduces fat mass, increases muscle mass, and reduces fatigue, which may contribute to weight loss in some people." So, the increase in muscle mass may be more than the decrease in fat mass, so won't necessarily lead to weight loss. And "some studies" hardly sounds definitive. StuRat (talk) 21:24, 19 February 2012 (UTC)[reply]

Acetyl-L-carnitine and lipoic acid are recommended supplements for older people, the older you are the less your body produces of them, see e.g. here:

Finally, I take 400 mg of lipoic acid and 1,000 mg of acetyl-L-carnitine (ALCAR) daily. This is based on the research by LPI's Tory Hagen on the role of these "age-essential" micronutrients in improving mitochondrial function and energy metabolism with age, and the research in my own laboratory indicating that lipoic acid has anti-inflammatory properties and lowers body weight and serum triglycerides in experimental animals. In addition, lipoic acid is well known to stimulate the insulin receptor and improve glucose metabolism, and is used in Europe to treat diabetic complications.

Count Iblis (talk) 00:43, 20 February 2012 (UTC)[reply]

In that case, it's quite possible that the supplement will have no effect on those who already have enough. StuRat (talk) 05:47, 20 February 2012 (UTC)[reply]

Yes, but people over the age of 35 are biogically "old", so they may already need it at that age. Someone at age 20 doesn't biologically age much over the course of a year, but someone at age 40 will biologically age much more significantly in a period of one year. If you could keep the rate of aging the same as it is for 20 years old regardless of age, then you would live for many thousands of years. It has been suggested that taking cheap over the counter supplements like vitamin D, acetyl-L-carnitine, fish oils etc. could significantly slow the aging process. In a recent article some persons were were interviewed who have taken such supplements for many years, one of them is 85 years old, but he is way above average fitness for someone of even age 60 let alone 85. He still competes in athletics, he participates in the 110 metres hurdles. Count Iblis (talk) 15:02, 20 February 2012 (UTC)[reply]

Number 26 is what i'm trying to solve. I have calculated Cluster II has diameter twice as long as Cluster I. The answer for number 26 is square root of 2 or approximately 1.4142 times smaller. I was trying to solve it by approaching the period orbit formula ways. I know that it has to do something with how fast the star rotate around the clusters. It doesn't seem to me that the formula for period orbit can be applied to this. Can anyone show me how to get the answer and what formula to use? Thanks!Pendragon5 (talk) 20:06, 19 February 2012 (UTC)[reply]

Watch your "rotations" and "revolutions"! While a galaxy rotates, its individual stars revolve about the galactic center. Globular clusters can't even be said to rotate as their star motion is pretty random (which is why they are not discs), though an individual star may revolve (or orbit) about the cluster. I've also changed your section title to reflect the nature of your question; hope you don't mind. -- ToE 00:52, 20 February 2012 (UTC)[reply]

This strikes me as a poorly worded question as its setup discusses the angular diameter (as observed from earth) of the globular clusters, leading to the conclusion that Cluster II has twice the diameter of Cluster I, though we are told they are of the same mass. It then asks about the ratio of the angular velocities of stars at the outer edge of the two clusters -- presumably with respect to the center of their respective cluster. Perhaps that should go without saying, but having just discussed the observed angular diameter, I would prefer if they explicitly indicated that they were not asking about the observed angular velocity.

Motion of stars within globular clusters is described at Globular cluster#N-body simulations, but that is certainly not what they are expecting to be solved here. I assume, as does the OP, that they are discussing stars which remain on the edge of the clusters in fairly circular orbits. (In highly elliptical orbits, their orbital angular velocities would be greatly reduced in the vicinity of apoapsis.) If Orbital period#Small body orbiting a central body applies (should it?), then the period is given by ${\displaystyle 2\pi {\sqrt {a^{3}/GM}}}$ and the ratio of their periods would be 2^3/2, not the 2^1/2 suggested by the answer key.

But wait! Perhaps they have been asking about the observed angular velocity of stars moving across the face of the clusters all along -- would that throw in the factor of two we need? Alas, the faster star is in the closer, denser cluster. Its orbital angular velocity is 2^3/2 times that of the star in the more distant, less dense cluster, so with an orbit half the size, its orbital velocity is 2^1/2 times greater, and being half the distance from sol, its proper motion would be 2^3/2 times greater. (The coincidence coming from the setup of the problem.)

Were I the OP, I'd be rather perturbed that the two apparent errors encountered in the answer key (see #Formula doesn't work why?) both involve calculations of orbital periods. -- ToE 00:23, 20 February 2012 (UTC)[reply]

Do you mean "orbital period" when you said OP? I'm kind of lost on the wording you used, sorry if i'm slow at understanding, when you said "Perhaps that should go without saying,". What should go without saying what? And you said "I would prefer if they explicitly indicated that they were not asking about the observed angular velocity." then at the bottom you said "Perhaps they have been asking about the observed angular velocity of stars moving across the face of the clusters all along". So are they asking about the observed angular velocity or not? And what do you mean by moving across the face of the cluster? You mean it's moving in the straight line across the diameter of the cluster? And some more info i forgot to add, perhaps it would help you explain to me better. Since the angle of Cluster II is half of the angle of Cluster I and they both have the same angular diameter so that's mean Cluster II and twice as further away as Cluster I. Therefore Cluster II has twice diameter as Cluster I. And i'm also confused on this statement "the faster star is in the closer, denser cluster". We don't know which one Cluster is denser than which. And sorry that i don't understand the second to the last of the paragraph of yours. Alright based from our calculation we know that Cluster I is rotating 2^3/2 faster than Cluster II then the next few lines... I have been trying to understand this but eventually i still don't get it. It's ok if you don't satisfied my confusing, i know you have tried you best to explain to me. If there is something you can explain further, please do.Pendragon5 (talk) 01:42, 20 February 2012 (UTC)[reply]

"OP" means "original poster", ie. the person that asked the question (so that's you!). --Tango (talk) 01:49, 20 February 2012 (UTC)[reply]

Angular velocity is always with respect to some point, and I assume that the question is asking about the orbital angular velocity of the star and not the observed angular velocity from earth, and "perhaps that should go without saying" because if they were interested in the latter, they would probably have referred to it as proper motion. It is a bit confusing because I think they must be making a lot of unspoken assumptions. From my understanding of globular cluster, their stars typically do not follow elliptical orbits, but instead move about in complicated (chaotic?) patterns as they interact with the other bodies in the cluster until they are eventually ejected, and that this ejection process leads to a very slow evaporation of the cluster. For any calculation to be possible, I figure that they must be speaking of stars that don't just happen to be at the outer edge of the clusters at this moment, but which maintain a fairly circular orbit about the cluster at its outer edge. I don't know if this is the common behavior of stars at the periphery of globular clusters, but what else could they expect you to calculate?

Next, I'm assuming that if the two stars doesn't pass too close their neighbors, a "raisin embedded" shell theorem should allow us to calculate the orbital period via the Orbital period#Small body orbiting a central body formula, and this yields the 2^3/2 ratio. I understand that this is the value you got as well. (This is the point where I'm hoping that an astronomer will butt in and point out what we are doing wrong.)

Since the answer key says 2^1/2, I then tried to figure out how to make that work, possible by using the fact that one cluster was twice as far away as the other, and hoping that this 2 might cancel out a 2 in the 2^3/2 yielding 2^1/2. This could only come into play if we were supposed to be comparing the observed angular velocities -- the proper motion -- all along, and since no other configuration was specified, I considered that in which the stars' motions were entirely transverse (at right angles to the line of sight). This happens as the stars are moving across the face of the clusters as they are presented to earth. There is no real reason to assume any of this; I am grasping at straws, trying to make the answer key's 2^1/2 work.

So we know that the star at the outer edge of Cluster I (the smaller, denser cluster -- denser simply because it is of the same mass but half the diameter) is moving at an orbital angular velocity of 2^3/2 times that of the star at the outer edge of Cluster II, right? But the orbit size around the outer edge of Cluster I is half the size of that for Cluster II, so to maintain that 2^3/2 orbital angular velocity advantage, it only needs to be moving 2^1/2 times as fast as the other star in terms of actual orbital velocity (in km/s or furlongs per fortnight or whatever) . But any motion in the closer cluster is twice as apparent as the same motion in the one twice as far away, so the faster star has 2^3/2 times the proper motion as the slower star, not the 2^1/2 I was grasping at. (So no, I don't think they were asking about observed angular velocity; it was a wild goose chase for me, but a red herring for you.) This arriving at the same 2^3/2 value is just a coincidence setup by the problem in which the two clusters have the same observed angular diameter, and thus the one which is twice as far away has twice the physical diameter as well.

Thus I fail to find any reasonable (or only moderately unreasonable) interpretation of the question yielding the answer in the key and must assume that either it is wrong or we are both making some mistake. -- ToE 20:10, 20 February 2012 (UTC)[reply]

Oh, I see how my "moving across the face of the cluster" would be confusing as their "at the outer edge" probable meant the observed circular (2D) edge of the cluster, not the spherical (3D) edge. In that case, my proper motion wild goose chase would apply only to those stars whose orbital plane was perpendicular to the line of observation. -- ToE 22:24, 20 February 2012 (UTC)[reply]

An alternative to encryption (one immune to quantum computing) would be a storage device that can be freely written to but is physically impossible to read without a password. Furthermore it would be good that at the time of writing it could be verified that (1) the device is set to use the (unknown to the verifier) password of the intended recipient and (2) the device will function as intended, i.e. the manufacturer has not rigged the device. Does a device of this kind exist, at least in theory? --145.94.77.43 (talk) 20:27, 19 February 2012 (UTC)[reply]

I wouldn't count a password as physically securing data. A physical switch you must toggle to put the device into read mode and another switch for write mode would be better. And they have to be real hardwired switches, not just ones that set some flag the software is supposed to use, because software can't be trusted. Being able to write to a device without reading from it does pose some challenges, though, like knowing if there's room for the data, verifying that it was written correctly, etc. The device would need to have the intelligence to check for such things itself. StuRat (talk) 21:19, 19 February 2012 (UTC)[reply]

"Being able to write to a device without reading from it does pose some challenges..." Well, if you can find a way to force users to write in Perl. . .--Atemperman (talk) 13:16, 20 February 2012 (UTC)[reply]

You seem to be describing something else. From their mention of encryption, it seems clear the OP wants some method to stop unwanted parties reading the data (e.g. after finding, stealing or confisicating the device) but doesn't want to use encryption out of fears quantum computing will break it. I don't see how what they're describing is going to work they're probably worrying too much about the encryption being broken rather then simply accidentally or being forced to reveal the password, but that's more of an aside. As for what your describing there seems little point. A hardwired write switch may have some use (although I don't think it's as foolproof as you believe). However there's not going to be much reason to use a switch to allow the device to be written too but not read. There may be some limited use in strange scenarios like where your copying data to the device but it also has info you don't want read. But in most cases it would make far more sense to have 2 or more devices and don't attach the device with sensitive data to something you don't trust. It's definitely not going to help in the OPs case unless your purported adversary is incredibly stupid (and if they are, they're not going to be able to break encryption even with quantum computers). Nil Einne (talk) 21:33, 19 February 2012 (UTC)[reply]

Yes, they used the word "physical", but what they were asking really had nothing to do with physical security. StuRat (talk) 00:36, 20 February 2012 (UTC)[reply]

Honestly, it looks fairly similar to public key encryption: the author of the file can't actually decrypt what he's writing; only the recipient can. I don't know of a device that bakes that sort of thing in, but there's no reason in practice that you couldn't rig up a software solution to apply that sort of thing. Note, though, that "quantum computing" won't enter into this at all. All encryption is, at some level, password-based, and so replacing one password (symmetric encryption) with another (public key encryption) isn't fundamentally altering that idea. — Lomn 01:32, 20 February 2012 (UTC)[reply]

What I was after was making something that is impossible to crack (even when the pass-phrase is much shorter than the data), as opposed to overwhelmingly improbable. --145.94.77.43 (talk) 04:05, 20 February 2012 (UTC)[reply]

I don't understand how you expect this to work at all. Imagine if you can only read it with the password, it doesn't make it impossible to bruteforce. Just keep trying with a different password. (That's bruteforcing!) In fact, if you really in some way can't 'read it' without a password, that makes it even easier since you know it's a failure so can quickly move on to the next candidate. (Depending on how you use encryption, you may only know if you failed to decrypt by the data not making sense.) Any sort of imposed limitation on number of tries, whether in software or hardware is of limited purpose since it can be overiden by using hardware or software which doesn't impose such a limitation. Perhaps if it physically takes a while to read, bruteforcing would be made more difficult but still not impossible. If reading is physically destructive and too many tries you kills the device it may seem this would work. However we get back to the first flaw in the concept. How do you design a device which can only be read with a password? However you store the data (magnetic, holographic whatever), ultimately there must be some way to read the data which the device contains and store it without bothering with any 'password'. Then you can just in software simulate whatever happens when the 'password' is used to read the device. Nil Einne (talk) 05:26, 20 February 2012 (UTC)[reply]

A good point about having a chance to guess the password anyway, but you are forgetting that the device is hardware in itself and cannot be overridden. Simulation is not a weakness either since the state of the device is unknown and by definition the device must somehow fend off any unwanted attempt to probe or tamper it. --145.94.77.43 (talk) 06:05, 20 February 2012 (UTC)[reply]

That doesn't make sense. "Hardware" vs "software" is irrelevant for this (it's just implementation detail). If you have a password, that password can be guessed. No information is more secure than the password. "Hardware" vs "software" is functionally meaningless here; cryptographically, all that's relevant is that you have password-protected data, and any means you have to legitimately enter a password constitutes a means for an attacker to enter a password. As for "the state of the device is unknown", obscurity is not a substitute for security. The world is rife with systems that were "secure" by virtue of a supposedly hidden secret (see: most industry DRM over the last decade). They're typically cracked within a year -- and those are industries with billions of dollars at stake! Now, tamper-resistant is its own thing, and it's very real, but you trade that against accessibility. You could make a box that self-destructs if someone tries to open it, or enters the wrong password, but you have to be willing to accept that you might destroy your own data if you type the password wrong. Of course, if they can bypass the self-destruct, or if you store the data in a second place for "backup", it's as if you didn't have any of that security at all. — Lomn 14:43, 20 February 2012 (UTC)[reply]

Indeed a key point 145 seems to be missing is there's arguably no way you can make your device tamper-resistance perfect. Sure you may make things more difficult for your adversary, but ultimately there's always the possibility they will break whatever safeguards you put inplace and can therefore overide any restrictions you attempt to enforce. Nil Einne (talk) 15:06, 20 February 2012 (UTC)[reply]

Physically impossible--no not in any known way. Difficult in practice--sure, see security token. 67.117.145.9 (talk) 22:59, 19 February 2012 (UTC)[reply]

I have the impression that most people worrying about the security of encryption are just being paranoid. — Preceding unsigned comment added by Ib30 (talk • contribs) 00:16, 20 February 2012 (UTC)[reply]

Nonsense. There are many examples in practice of nonsecure encryption and the negative consequences thereof. Modern secure algorithms are such only because people are constantly and professionally paranoid about them. — Lomn 01:32, 20 February 2012 (UTC)[reply]

Indeed. The pure mathemetical underpinnings of most modern widely-used cryptographic schemes have been pretty thoroughly scrutinized, and if implemented correctly are secure against any current brute-force computational attack. However, it's actually quite difficult to translate that mathematics into a useful software implementation without inadvertently introducing serious weaknesses. Bruce Schneier's blog is a great way to get daily bite-sized snippets of news about security and risk management in the modern era, including fairly frequent stories related to (sometimes good, but usually poor) uses of crytographic algorithms and technology. Three days ago, he covered this story, which involved a discovery by cryptography researchers that something like 1 in 250 public keys in the wild share a common factor, making them inherently insecure. (This was probably due to the use of flaky, not-quite-random random number generators in the creation of the keys.) TenOfAllTrades(talk) 05:24, 20 February 2012 (UTC)[reply]

"Most people" don't have almost any information of value, that deserves any effort to be protected against "quantum computing" or that deserves any effort to be cracked. And most people cannot protect their most valuable information like credit cards and bank accounts themselves, since their respective banks are at charge. So, in this sense it's nonsense to worry about common algorithms or about common programs. Just use them correctly and you are save. On the other hand, if you are into an industry which heavily depends on security (to protect your digital TV transmission, credit cards processing information, etc) then you'll have to be up-to-date to know what others are doing to crack your data. 88.8.66.12 (talk) 14:24, 20 February 2012 (UTC)[reply]

Again, nonsense. "Most people" (and why the scare quotes?) rely on secure encryption for sensitive data on a daily basis, whether they realize it or not. — Lomn 14:44, 20 February 2012 (UTC)[reply]

What? If they do not realize the use, why being paranoid? Most people do not have any need or possibility to manage the encryption they use. I doubt you understood the previous comments. Take a look at: that's Lomn. 88.8.66.12 (talk) 16:10, 20 February 2012 (UTC)[reply]

Save data to flash drive. Physically remove flash drive when done. Store in vault. How's that sound? (Yeah, I know, flash memory degrades eventually. But it does survive a trip through the wash). Guettarda (talk) 21:00, 20 February 2012 (UTC)[reply]