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January 26

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What animals have sense of overeating, and what animals don't?

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I heard that goldfish do not have sense of overeating, and in spring seasons cows are eager to fresh grass so that they sometimes overeat.--chaoxiandelunzi (talk) 01:22, 26 January 2014 (UTC)[reply]

I don't have a list for you, but the reason must be that those animals can't overeat. This would be primarily because the food they eat is so low in nutritional value, that it's difficult to get too much. Cows might fall into this category, as nearly constant eating of grass is required for them to get enough calories.
Also, some animals might lower their digestive efficiency when they eat more. The simplest way to do this is just to defecate whatever is in their digestive system when they overeat. This can all happen without any thought from them. Rodents, for example, seem to defecate constantly, with little control over it.
Incidentally, this also seems to be part of how human's control their weight. That is, if you eat 10 times your calorie requirement one day, you don't absorb anywhere near all of those calories. There are also some unfortunate people who never feel full, but they do tend to become obese.
Then there are animals that can grow indefinitely, or, in the case of single celled animals, some split into two. StuRat (talk) 01:54, 26 January 2014 (UTC)[reply]
Some dogs will overeat. Beagles and Labradors are classic examples. Other dog breeds won't. HiLo48 (talk) 02:11, 26 January 2014 (UTC)[reply]
  • Dogs tend to overeat, because like most carnivores, they encounter large prey on an irregular basis, rather than nibbling continually. (For the reason of continually eating, it seems unlikely most ruminants, which chew cud, will overeat under normal circumstances.) There are a huge number of articles on dogs overeating at google. My German Shepherd would do it when given a chance, especially with rawhide bones, then throw up. She had surgery once for "bloat" when she ate the string a turkey was wrapped in, and a lot of the trash that had absorbed the turkey fat. Bloat is a common cause of death for dogs. Unfortunately, looking for "not" a t google is difficult, so looking for animals that don't overeat is a difficult proposition. μηδείς (talk) 02:57, 26 January 2014 (UTC)[reply]
  • Yes, my brother had a dog that died from that. In his case the culprit was a grilled cheese sandwich, which apparently caused a blockage, which in turn caused the intestinal gas to build up until he couldn't breath. So, the dog died of flatulence. (They drove to the animal hospital as soon as they noticed his trouble breathing, but it was too late.)
  • Also note that humans overeat for much the same reason. That is, we evolved when big meals were few and far between, so it made sense to eat all you could in an attempt to avoid starvation during lean times. Fruits and veggies would have been available more often, but it's hard to get fat from those. StuRat (talk) 04:00, 26 January 2014 (UTC)[reply]
@ StuRat I don't have a list for you, but the reason must be that those animals can't overeat. This would be primarily because the food they eat is so low in nutritional value, What about fish being farmed where they are regularly fed a high protein diet to help them get as large a possible? - what about pigs or poultry that are fed high calorie food to increase their weight and cows are increasingly being kept in sheltered accommodation with artificial food that has to be limited to prevent over fattening. Cows might fall into this category, as nearly constant eating of grass is required for them to get enough calories. See above.
Also, some animals might lower their digestive efficiency when they eat more. The simplest way to do this is just to defecate whatever is in their digestive system when they overeat. This can all happen without any thought from them. Rodents, for example, seem to defecate constantly, with little control over it. So do sheep but sheep farmers go to lengths to fatten their animals, and successfully do so. So constant defacating doesn't control the weight of sheep. And dogs do not defecate more after a huge meal, they sit about for hours half asleep digesting it and turning it into body tissue. Rats and mice will get huge when fed excessively in captivity, so they don't use increased defecation to control their weights. Have you got a source that gives any support to your extra defecation theory.
Incidentally, this also seems to be part of how human's control their weight. That is, if you eat 10 times your calorie requirement one day, you don't absorb anywhere near all of those calories. Agreed but that is irrelevant to the question. The occasions when someone eats 10 times their recommended daily intake (lets say 20,000 cals) are vanishingly rare, Morgan Spurlock didn't get close to halfway that amount. There are also some unfortunate people who never feel full, but they do tend to become obese. This has nothing to do with what the OP asked. Stu, sometimes it is necessary to stop and think answers through, your answers often contain way too many weasel words that allow you to ease out when challenged. Caesar's Daddy (talk) 15:03, 26 January 2014 (UTC)[reply]
Perhaps I should have included the seemingly obvious: "This would be primarily because the food they eat is so low in nutritional value, in their natural environment." Obviously, we can supply high calorie foods to people or other animals, and they have not evolved a way to handle those, so they often get fat. StuRat (talk) 00:07, 27 January 2014 (UTC)[reply]
Then there are animals that can grow indefinitely, or, in the case of single celled animals, some split into two. StuRat (talk) 01:54, 26 January 2014 (UTC)[reply]
I can't speak for the entire species, but I've known Budgerigars that would stuff themselves until they made themselves sick if given unlimited access to something like a honey-coated seed bar or a long sprig of millet (i.e. leaving the thing in in the bird's cage and thinking it'll last a few days, but then discover that he's eaten it really quickly). Maybe it's something like eating a whole tube of Pringles and it being fine on the way down, but then paying for it afterwards... --Kurt Shaped Box (talk) 15:34, 26 January 2014 (UTC)[reply]

Which is brighter, the Moon or Venus?

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Subsequent discussion highlighted the transit of Venus

As seen from Earth.

I have to clarify to make it not seem a stupid question. Of course, we see more light from the Moon — its luminosity is higher. But that's at least partly because we see it as a (much) bigger disk, so that doesn't count. What I want to know is, which is brighter at a given spot inside the disk.

Venus is only about 70% as far from the Sun, so the intensity of light falling on it should be roughly twice what falls on the Moon, and its albedo is also higher, by a factor of more than 5. So I would expect its luminance to be about ten times that of the Moon.

But that's if we can actually see Venus as a disk, which I'm not so sure about. According to phases of Venus, a person with exceptional visual acuity can resolve Venus as a disk under ideal conditions. But ideal conditions for that would be when Venus is closest to the Earth, and of course it can't be full under those conditions.

So with that as preface, which is brighter? --Trovatore (talk) 03:35, 26 January 2014 (UTC)[reply]

Interesting Q. Of course, as you noted, the brightness of Venus as viewed from Earth will vary with it's distance from Earth, which also affects it's phase. The brightness of the Moon also varies, but presumably you want the lit section when it is less than a full Moon, and that part might not vary by much. So, then, the question becomes, is the Moon always brighter, is Venus always brighter, or is it sometimes one and sometimes the other ? StuRat (talk) 03:53, 26 January 2014 (UTC)[reply]
Well, actually, as long as you can resolve Venus as a disk, its distance from Earth doesn't matter (much). The intensity of Venusian light falling on the Earth falls off as the square of the distance, but so does the solid angle that Venus subtends, so that cancels out. The brightness diminishes a tiny bit because of scattering by the not-complete-vacuum of interplanetary space, but that's a minor correction.
On the other hand, if the solid angle is already below what you can resolve, then it doesn't cancel out anymore. --Trovatore (talk) 03:57, 26 January 2014 (UTC)[reply]
  • What matters is albedo times cross section--the apparent area of the face (visible disk) reflecting towards us. Venus is usually almost a point source compared to the Moon. Keep in mind also that this is proportional to the square of the apparent diameter. If the face of the Moon is ten times wider than the face of Venus, it has 100 times the area, and hence 20 times the brightness if Venus has five times the albedo. (I picked ten times out of the air as an example.) Obviously the distance determines the apparent size, but the relative apparent size is all you need to know for the comparison. μηδείς (talk) 03:58, 26 January 2014 (UTC)[reply]
    It doesn't matter whether it's a point source "compared to the Moon". It matters whether it's a point source as far as our eyes are concerned. If you can resolve it as a non-point-source, even though much smaller than the Moon, then the apparent area cancels out. --Trovatore (talk) 04:05, 26 January 2014 (UTC)[reply]
No, that's wrong. The brightness is independent of our ability to distinguish between whether it's actually a point source or not. (Indeed, there's no such thing as a point source--that's a subjective effect of our perceptual systems, not a real phenomenon.) The actual diameter (relative to one's viewpoint) squared (assuming the same phase and hence shape) times the albedo is all that matters. I won't be responding to this again, it's not a matter of argument. I suggest you ask the same question at the math desk if you want confirmation. μηδείς (talk) 04:18, 26 January 2014 (UTC)[reply]
Medeis, I'm not sure you read my whole question. If you care, you can re-read the bit starting with "I have to clarify". --Trovatore (talk) 05:17, 26 January 2014 (UTC)[reply]
It depends how you're measuring it and at what point in their phases. According to Apparent magnitude, the moon is brighter than Venus for their respective greatest possible brightness. Generally, at new moon phase, the moon is dark, while Venus is bright at most any phase, until it slips behind the sun or otherwise gets lost in its glare. ←Baseball Bugs What's up, Doc? carrots05:04, 26 January 2014 (UTC)[reply]
Hmm, so the question there is what they mean by "brightness", for objects we can resolve as a disk. Do they mean the total light you see, or the total light per, say, rod cell? I'm asking about the second one, and it's not clear which one they mean. --Trovatore (talk) 05:25, 26 January 2014 (UTC)[reply]
(ec)The apparent magnitude article doesn't mention disks or points, so that's not totally clear. Also, it's not clear to me why a couple of stars have greater apparent magnitude than the sun, when those objects are (presumably) safe to observe through a telescope, while the sun obviously is not. And even the full moon can be pretty freakin' bright through a telescope, though I don't think it would blind you like looking at the sun would. ←Baseball Bugs What's up, Doc? carrots05:41, 26 January 2014 (UTC)[reply]
Most people resolve planets as disks even if they are unaware of it. Disks don't twinkle like point sources. Find the objects at night that don't twinkle and they are planets. There is no star (or planet or moon) that has an apparent magnitude greater than the sun (-26 or so) from the surface of the earth. There are stars that would be brighter if the earth was located the same distance from them but we have an average star so that is expected. Similarly, the sun has a larger apparent magnitude on mercury than earth. the answer to the question is whether you want to compare the same arc angle or the same distance. If venus were the same distance to the earth it would appear significantly larger than the moon. If you compare the brightness whether they consume the same amount of area (i.e. venus is further from the earth than the moon but they appear the same size), it's a matter of scaling it's current arc to the moons along with the brightness. --DHeyward (talk) 20:32, 26 January 2014 (UTC)[reply]
This is a tricky thing to get your head around. Consider this - the physics of light transport says that the brightness of a light source diminishes as the square of the range. When one light source is twice as far away as another, it's four times dimmer. HOWEVER, if you look at a tree that's (say) 10 meters away, it doesn't look 100 times brighter than a tree that's 100 meters away. That's because although there is indeed ten times as much light arriving at your eye from the close-up tree - that light is distributed over more rods and cones in your eye than the light from the more distant tree. When you double the distance to an object, all of the light from it is focused onto a quarter the number of rods and cones. The net result is that the diminution of light with distance exactly cancels out the fact that that light is focused onto a smaller area of the retina. Hence objects tend to maintain the same perceptual brightness, no matter how far away they are.
BUT (and this is the key here) once the light source is so far away that it's entire image is focused onto just one rod/cone - the effect of it's light begin shared over more or less cells stops happening - and a true range-squared diminution effect sets in. That's what we truly mean by a "point source" - a source that focuses to a point that's smaller in size to a single rod or cone cell.
So from a pure optical perspective, you'd expect objects of the same brightness to look equally bright no matter how small or how far they are from you UNTIL they get small enough to be focused on just one rod/cone - and then to start getting rapidly dimmer with size/range.
What greatly complicates this is that our brains know all about this effect - and they work to counteract it. This results in a translation of true brightness against perceived brightness that's hard-to-measure, and possibly different between individuals.
SteveBaker (talk) 05:37, 26 January 2014 (UTC)[reply]
Ah, so your last two sentences are interesting (the rest of it is just a restatement of what I already said, but that's fine because I'm not sure everyone got it). Do you have any further information on this "brain" effect? --Trovatore (talk) 05:44, 26 January 2014 (UTC)[reply]
To make a long story short, Venus is brighter because it's closer to the Sun, with nice white fluffy clouds. Venus is at 0.7 AU, so it gets about double the light, plus it has an albedo of 70% or so versus something like 12% for the Moon (according to quick web searches). I wonder if every once in a while, geologically speaking, the Moon gets hit with a comet and a coating of hoarfrost makes it look five times brighter for a few generations? So Venus should be roughly 12 times brighter per unit area than the Moon. Of course, your optics may be blurry, in which case it will look dimmer, and how much depends on how blurry. Wnt (talk) 05:51, 26 January 2014 (UTC)[reply]
I think I already said all that. Question is, what's the bottom line? For human eyes. I understand that it may depend on distance, phase, and the particular human. --Trovatore (talk) 05:53, 26 January 2014 (UTC)[reply]
Well, according to angular diameter, Venus can sometimes be larger than an arcminute, sometimes much smaller. Though there's the wrinkle that when it is largest, it is 'new', so not bright at all, and close to the Sun. Practically, Venus at a good viewing angle should be less than half of that arcminute, and you'd still be seeing it less than half lit. According to discussions at naked eye there is some disagreement among sources about the practical resolution of the human eye, but it may go down to an arcminute in some people. I see no reason to suppose that some human doesn't have focus which is more or less perfect. Looking for some of the numbers online I found an interesting blog [1] which estimates diffraction- and cone photoreceptor based resolution at around 0.2 arcseconds, or maybe a bit more. The math looks believable, so... a curious coincidence seems to be that Venus makes up about as much of the sky as a photoreceptor makes up of the retina, give or take a factor of 4 or 5. But... the wheels come off the wagon at the next step. The thing is, retinas don't see light, people see light. Photoreceptors are averaged, contrasted, what came as a simple measurement becomes an aesthetic sensibility. It is the result of elaborate retinal processing. And just because one cone cell is shouting that Venus is brighter than the ground in broad daylight doesn't mean that that's how we perceive it. That final, subjective process seems to defy ready examination. Wnt (talk) 16:18, 26 January 2014 (UTC)[reply]

Just by the way, in case anyone is interested, there is a related discussion at talk:Moon; the article currently claims that the Moon is the brightest object in the sky after the Sun, which is true as most people probably interpret the word "brightest", but possibly not true in a more technical sense. Anyone who has thoughts on how to resolve the issue at that article is invited to comment on that talk page. --Trovatore (talk) 05:56, 26 January 2014 (UTC)[reply]

I looked at the same problem when I saw ChrisJBenson's first edit to the Moon article, but decided not to get involved because of the two different meanings of "brightness". Perhaps we need links to luminance and luminosity, as well as the existing note link to apparent magnitude. Dbfirs 09:08, 26 January 2014 (UTC)[reply]
The term the OP is looking for is surface brightness. I'll calculate the surface brightnesses of both Venus and the Moon tomorrow, but I'd be very surprised if Venus doesn't have a higher surface brightness. --140.180.249.191 (talk) 09:34, 26 January 2014 (UTC)[reply]
The full or near-full moon is so bright that it wreaks a degree of havoc with astronomy observations in general. Venus looks very bright when it's near a crescent moon. The question would be, at what point (if any) does Venus get close enough to the moon that the reflected light of the moon "drowns it out?" Or does Venus stay visible (to the naked eye) even when very close to the edge of the moon's disk? ←Baseball Bugs What's up, Doc? carrots13:20, 26 January 2014 (UTC)[reply]
Thanks, 140.180 for the link to surface brightness. I think we should mention in the article that Venus has a higher surface brightness (subject to your checking of the calculation, and BB's observation of no "drowning out" to confirm it). Dbfirs 16:56, 26 January 2014 (UTC)[reply]
My concern, as I mentioned at that talk page, is that (whatever its technical definition), the term "surface brightness" sounds like something you would observe near the surface of the object, whereas I'm talking explicitly about what you observe from Earth. --Trovatore (talk) 19:24, 26 January 2014 (UTC)[reply]
Bugs's test is very nice, by the way. I don't ever recall seeing Venus just on the limb of the full Moon, but it would seem, based on abstract considerations, that you ought to be able to see it, no matter how close it is. If you can't, I'd like to know why. --Trovatore (talk) 19:29, 26 January 2014 (UTC)[reply]
Oh, I just realized — that can't happen; the geometry doesn't work. When the Moon is full, that means it's on the opposite side of the Earth from the Sun. You can't look that direction and see Venus. Still, it ought to be possible to see Venus right on the limb of a crescent Moon, so close that it actually looks like a bump on it. I have never seen that. --Trovatore (talk) 19:37, 26 January 2014 (UTC)[reply]
Oops, you're right, it wouldn't work. The question remains as to how much (if any) the full moon blinds the naked eye to any objects near it, such as bright stars or oft-seen outer planets. I know from anecdotal evidence that fainter objects such as comets show up much better on a moonless night. ←Baseball Bugs What's up, Doc? carrots21:14, 26 January 2014 (UTC)[reply]
That's certainly true. I still think it's a good test, conceptually, even if it can't actually happen that way. Rephrase: Imagine a helicopter carrying a searchlight, pointed at you, in such a way that it reproduces the full Moon in brightness and apparent size, and have it fly almost between you and Venus in such a way that Venus would appear just on the limb. Would you be able to see Venus, and would it look brighter than the searchlight? From the discussion so far, I think yes, but I'm not quite sure. --Trovatore (talk) 21:55, 26 January 2014 (UTC)[reply]
Venus is always visible, even when right beside the bright side of the Moon. For proof, see this video of an occultation of Venus by the Moon. --Bowlhover (talk) 22:04, 26 January 2014 (UTC)[reply]
That's lovely, thanks! However — that's for a camera's optics. It doesn't quite answer the question for a human viewer. Still, great find, thanks again! --Trovatore (talk) 22:08, 26 January 2014 (UTC)[reply]
(140.180 here, didn't realize I got logged out)
According to this source, the surface brightness of Venus is 1.5 magnitudes per square arcsecond. The Moon's brightness when full is magnitude -12.7 (according to apparent magnitude), and its radius is 0.25 degrees, so it has a surface brightness of 3.3 magnitude per square arcsecond. So Venus is brighter per square arcsecond, by a factor of 5.
Surface brightness is independent of distance. It doesn't matter whether you measure it from Earth, from the Moon, or from Venus--Venus will always have a surface brightness of 1.5 mag/sq arcsec, and the Moon will always have a surface brightness of 3.3 mag/sq arcsec. --Bowlhover (talk) 22:04, 26 January 2014 (UTC)[reply]
That's actually one of my concerns about reducing the question to "surface brightness". The surface brightness of Sirius is much greater than the surface brightness of the Moon, but you wouldn't be able to see Sirius next to the Moon. The solid angle subtended by Sirius is so much smaller than what your eye can resolve that the surface brightness is pretty much irrelevant. For Venus, though, it may not be. --Trovatore (talk) 22:17, 26 January 2014 (UTC)[reply]
In astronomy, "brightness" almost always refers to the total amount of light emitted from an object. So does apparent magnitude. So does the layman's usage of the word "brightness", since nobody would say that Venus is brighter than the Moon. --Bowlhover (talk) 22:09, 26 January 2014 (UTC)[reply]
Really? If I saw live what the video you posted showed, I would certainly say Venus was brighter than the Moon in that shot. --Trovatore (talk) 22:12, 26 January 2014 (UTC)[reply]
Another consideration might be, which is more likely to cast a shadow? ←Baseball Bugs What's up, Doc? carrots22:53, 26 January 2014 (UTC)[reply]
The weather we've been having? I think they're both sure to cast a shadow. :) Wnt (talk) 16:57, 27 January 2014 (UTC)[reply]

Surround sound for two ears

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I was thinking about real surround-sound vs. virtual surround-sound (where timings are altered to make sound reach ears ever so slightly out of sync). Given that I only have two ears anyway, shouldn't virtual surround-sound be every bit as good as actual surround-sound? I think I've noticed real-life situations where I cannot tell whether a sound originates in front or behind me but I think that's an inherent flaw in a two-ear system. --78.148.110.69 (talk) 15:09, 26 January 2014 (UTC)[reply]

In theory, it could be. It is a lot easier to achieve with headphones than it is with loudspeakers. Headphones only require a channel for each ear (of which there are only two as you astutely observe). With loadspeakers, the entire sound field of the original room must be replicated. This can be done approximately only with a number of discrete loudspeakers, but is increasingly accurate the more sound channels and speakers that are used. See Surround sound#Creating surround sound. SpinningSpark 15:40, 26 January 2014 (UTC)[reply]
One thing traditional headphones haven't done is change the balance between the ears as you move your head. This is an important factor in making the sound seem "real". I would expect that this could be done fairly easily now, though. Does anyone know if such a product is on the market ? StuRat (talk) 20:37, 26 January 2014 (UTC)[reply]
The answer is no. There is a common misconception that sound only enters through the two holes in the sides of our head - when in fact, the skull itself vibrates, reverberates and has certain resonances, sounds travel through the brain matter three times faster than through the air and the external parts of our ears perform subtle changes to the sound to help with directional awareness. If our auditory system was literally like two microphones, then you wouldn't be able to tell whether a sound was coming from in front or behind you - or above or below. In truth, our brains are able to detect subtle changes to the sound due to it passing through bone and brain and use these clues to spatialize sound more effectively than the "obvious" model.
The first efforts to improve stereo recordings for people wearing headphones were to put microphones into a plastic head, designed to reflect and refract sound just like a person's real head and record live performances that way. This is incredibly inconvenient for most recording studios. Modern "virtual surround sound" attempts to do the same thing with software post-processing of the raw audio. However, both of these approaches are only somewhat effective because each person's ears and skull shapes are different.
Another problem with this is that you need to make the recording differently depending on whether you expect the end-user to listen through speakers or headphones - and whether the headphones are "cups" or "earbuds" since the former allow your external ear parts to change the sound - where the earbuds do not. This means that commercial recordings with this kind of processing are not really feasible - so the calculations have to be performed as the music is replayed.
All of this means that true surround-sound is always going to be better than the virtual kind - UNLESS the real-to-virtual processing is highly tuned to your particular headphones and even to your personal physiology. Since nobody has yet managed the latter, it's clear that "real" surround-sound is going to be more convincing...*if* it's done right...which is another story entirely.
SteveBaker (talk) 15:39, 27 January 2014 (UTC)[reply]

Fish oil's half life

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Hi there. I want to emphasize it is a biological, not medical question. I wonder if anybody has this information. What is the half-life of the fish oil taken as a vitamin/food supplement? Thanks, --AboutFace 22 (talk) 18:00, 26 January 2014 (UTC)[reply]

Are you sure you mean half life, as opposed, to, say, best by date? The former mostly applies to nuclear physics. The latter is printed on the bottle, and varies by product (e.g. liquid vs capsule) and method of storage (e.g. unrefrigerated or not). Check the label on the bottle. 88.112.50.121 (talk) 18:29, 26 January 2014 (UTC)[reply]
I would guess they are asking when it would deliver half of the vitamins as when it was new. Yes, it will vary widely, but perhaps we can come up with some estimates ? StuRat (talk) 20:39, 26 January 2014 (UTC)[reply]
I would guess the actual meaning is biological half-life — that is, how long the oil stays in the body. I'm not sure that's a meaningful question, for fish oil. It's not a drug. It's supposed to be metabolized, and the hope is that the outcome of that process is better than for other fats. --Trovatore (talk) 21:02, 26 January 2014 (UTC)[reply]
You are somewhat randomly focusing on the group of compounds called vitamins. Fish oil has little to do with that. "Half life" of those compounds is of even less practical use. You are not being helpful. 88.112.50.121 (talk) 21:13, 26 January 2014 (UTC)[reply]
It was actually the original poster who called it a "vitamin", which of course it isn't, but that's a little bit arbitrary (which nutrients are called "vitamins" and which ones are not is AFAICT largely a historical question rather than a clearly-defined meaningful categorization). --Trovatore (talk) 21:18, 26 January 2014 (UTC)[reply]

I did not expect so much confusion. When a person swallows a medication a curve called bioavailability may be measured. Let's say a peak blood concentration after one hour upon ingestion is so many mcg/dl in plasma. The question is how long does it take for that concentration to decrease by 50%. It is a standard parameter for many drugs. I called the fish oil a vitamin to avoid a suggestion that this is a medical question. There is no need to go into a long discussion on it. We can talk about omega-3 fatty acids, etc. I want to know the half-life of the fish oil, period. Thanks --AboutFace 22 (talk) 21:37, 26 January 2014 (UTC)[reply]

That seems like a fair question, though I don't know the answer. However, in case anyone does, you might clarify whether you really mean the oil molecules themselves in unmodified form, or if you accept the constituent fatty acids, either free or in some other combination. --Trovatore (talk) 21:43, 26 January 2014 (UTC)[reply]

Möller’s Cod Liver Oil is a pure natural product without preservative. An opened bottle should be kept for a maximum of three months in the refridgerator before it should be replaced. This is due to exposure to oxygen from light and air every time the bottle is opened. Hence, it is important to put the screw-cap on directly after consumption to prevent exposing the oil to oxidation (becoming rancid). 84.209.89.214 (talk) 00:07, 27 January 2014 (UTC)[reply]

Trovatore, sure I mean active ingradients, that is omega-3. --AboutFace 22 (talk) 00:19, 27 January 2014 (UTC)[reply]

You see, the first pass through the liver is important as we know. It may kill some portion, however that should not affect the bioavailability curve. Then usually chemicals are divided into free floating and plazma protein bound fractions. How easily the proteins part with the chemical may determine the half-life I am looking for. It is all a big question mark for me but is important to know. --AboutFace 22 (talk) 00:25, 27 January 2014 (UTC)[reply]

Here's a study measuring the bioavailability speficially (for both oil and capsules). My math isn't good but if yours is can you extract the number you need from the values and charts given? 184.147.128.82 (talk) 01:44, 27 January 2014 (UTC)[reply]
There are insufficient data points to be able to accurately calculate half-life from those graphs, but assuming an exponential decay from the peak of the graph we can obtain time constants of 2.104 and 2.187 days for the orange and blue curves respectively. The relationship of exponential time constant to half life is ln(2) giving half-lives of 1.46 and 1.52 respectively. If you wish to measure from the time of ingesting, the 24 hours to the peak of the curve must be added making 2.46 and 2.52 days respectively. However, this is likely to be an underestimate. If more datapoints were shown we would likely see the rounded top of the curve initially falling slower than an exponential. So the answer is likely to be 3 to 4 days based on the scanty information in those curves. SpinningSpark 11:13, 27 January 2014 (UTC)[reply]
According to this book it is 18 days, although that seems to be a rather promotional source and I would not entirely trust it to be accurate. SpinningSpark 10:14, 27 January 2014 (UTC)[reply]
The reason I haven't touched this one is that it is difficult to figure. I mean, exogenous fatty acids can end up going straight into myelin,[2] where depending on how and what you measure they can have very long or short half-lives.[3] So depending on the precise question you're really interested in, the interesting answer could be all over the place, minutes to months. Wnt (talk) 17:05, 27 January 2014 (UTC)[reply]
No, I think OP DOES mean half-life. Fish have phospherous in them, and that is mildly radioactive. So my answer would be have as many pills as you like, but stop when you start to glow in the dark. Myles325a (talk) 03:41, 28 January 2014 (UTC)[reply]

Myles325a - LOL. Thanks everyone who contributed. It is very helpful. --AboutFace 22 (talk) 15:03, 30 January 2014 (UTC)[reply]

Washing away a flu

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Does taking a shower or bath actually help get rid of a cold or flu? I've heard it said that doing so washes away toxins or waste that is sweated out and therefore helps the healing and recovery. Is there anything that you can cite that says that this helps? Dismas|(talk) 20:54, 26 January 2014 (UTC)[reply]

I can't see it helping by that mechanism, but the heat, humidity, and motion might help one to cough up phlegm. Strong smelling shampoos and body washes might also help this process.
And just feeling clean might make the person more willing to go out and face the world. So, they may not have a lower viral count, but might feel better at least. StuRat (talk) 21:23, 26 January 2014 (UTC)[reply]
During cold and flu season especially, the experts urge frequent hand-washing. Presumably that could work in both directions - cleanliness might reduce the chance of infecting someone else. ←Baseball Bugs What's up, Doc? carrots00:52, 27 January 2014 (UTC)[reply]
  • Basically, no. Any "toxins" that make you feel sick are immune factors in the blood like interferon or dead cell products in the blood stream. Viruses only live inside living cells such as the lining of the respiratory and gastrointestinal tract, and can travel through the blood as well. As mentioned, washing will help prevent transmission and the heat and humidity may be soothing, but it won't remove toxins or otherwise affect a flu or rhinovirus infection. Only small molecules can really leave the body passively through the sweat, like alcohol, and this only at a very-low rate, which is why you die if you don't have functioning kidneys, bowels and a liver. μηδείς (talk) 01:23, 27 January 2014 (UTC)[reply]
A quick rule of thumb I've developed is to ignore any folk advice that has to do with "removing toxins from the body" as it's almost 100% bullshit. Matt Deres (talk) 18:25, 27 January 2014 (UTC)[reply]
A favorite example of that was those foot pads that claimed to pull the "toxins" out of your body through the soles of your feet. They said that it worked 'naturally' just as trees flush waste products out through their roots...it had a very pretty graphic of the tree absorbing raindrops though it's leaves and passing the resulting 'toxins' out of the roots...so it must be true! SteveBaker (talk) 21:02, 28 January 2014 (UTC)[reply]

Vinyl record pre and post echo

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It is well known that pre echo exists in some vinyl records owing to the cutting process. Is it possible that post echo is also likely to be present if pre echo exists on a vinyl disc?--86.184.57.126 (talk) 21:06, 26 January 2014 (UTC)[reply]

Yes it is possible. The cutting stylus for a Gramophone record unavoidably transfers some of the subsequent groove wall's impulse signal into the previous groove wall. A quiet passage followed by a loud sound will allow one to hear a faint pre-echo of the loud sound occurring 1.8 seconds ( = 60/33⅓ ) ahead of time. This problem can also appear as "post"-echo, with a tinny ghost of the sound arriving 1.8 seconds after its main impulse. This link explains groove echo and shows photos of a recording lathe that is able to spread grooves i.e. "add land" to reduce the problem. Echos with shorter delay also occur due to analog tape print-through. 84.209.89.214 (talk) 23:55, 26 January 2014 (UTC)[reply]
Gd ans. Tx.31.55.102.122 (talk) 01:11, 27 January 2014 (UTC)[reply]

[Banned user's contribution deleted]

Thx but post echo on vinyl 33rpm discs occurs 1.8 sec after the event doesnt it? Ears should recover in that time.31.55.102.122 (talk) 01:11, 27 January 2014 (UTC)[reply]
How does post echo form? I assume the problem for pre echo is that when the stylus cuts the next groove, it pushes some of the laquer of the (now) wall between there and the previous groove back into the previous (the stylus pushing some of the material sideways rather than simply scraping it out). But the whole laquer from the location of the stylus in is not yet cut. And when the stylus does get to the next one and cuts there, why is it not able to cut out any deformation that may have occurred? DMacks (talk) 04:31, 27 January 2014 (UTC)[reply]

[Banned user's contribution deleted]

I agree that as the cutter deforms the master disk in the N'th revolution of the disk, it's imposing this new audio on top of the N-1th revolution. That should result in a 1.8 second pre-echo ("HELLO WORLD" becomes "helloHELLOworldWORLD") where you hear sound coming from 1.8 seconds into the future. It seems unlikely that the N-1th groove is going to have much impact on the path that the head takes when cutting the N'th groove so post-echo seems unlikely. What DOES seem likely is that the act of replaying the disk causes the problem. As the stylus is accelerated left and right and up and down, it's going to impose forces on both sides of the groove. That must somewhat distort the vinyl and result in both pre-and-post echo ("helloHELLOhello"). This seems like a more likely mechanism anyway because I presume that the master disk is cut into some kind of metal substrate that would be resistant to deforming the previous grooves.
Tape print-through can also produce both pre-and-post echoes - and those would be present even in the master disk if it was mixed down from tape. However, tape print-through is usually only a problem when tapes are stored for long periods without being periodically re-wound onto new spools. It seems unlikely that this would happen to any significant extend with most vinyl disks being made within days of the performance being recorded.
One particularly fun record is a Monty Python disk where they recorded two parallel grooves on one side of the disk - so you hear a completely different recording, randomly, half of the time! It would be interesting to listen for 'echoes' of something you're NOT going to hear - because that would clearly distinguish groove distortion from tape print-through. SteveBaker (talk) 20:30, 28 January 2014 (UTC)[reply]
You mean a multisided record? DMacks (talk) 04:06, 29 January 2014 (UTC)[reply]
I didn't know there was a name for it! I know they used to use them for gambling on the outcome of horseraces - where they'd make a disk with six parallel tracks - each narrating a horserace with a different outcome. The disk would start playing on one of the tracks at random - and people would then bet on the outcome. SteveBaker (talk) 21:36, 29 January 2014 (UTC)[reply]
"They're Off" was eight parallel tracks according to its entry in Unusual types of gramophone records#Parallel grooves. DMacks (talk) 22:34, 29 January 2014 (UTC)[reply]

Abandonment of wild rabbit burrows

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Do wild rabbit colonies abandon their burrows and dig new ones after a rabbit dies in there? I'm assuming they would hide underground if in pain, like dying cats will try to find a quiet "safe" place. Or is there a different reason for their movement? I live near a hillside which is full of rabbit holes, but actual rabbits seem to be mostly at one end. 86.144.47.4 (talk) 21:38, 26 January 2014 (UTC)[reply]

Assuming you're correct, and they do that, the obvious reason would be to avoid the diseases they might catch from the decomposing rabbit corpse in there. The smell might also attract scavengers. Of course, rabbits wouldn't reason any of this out, they would just have the instinct to dig a new burrow in that case (because those with that instinct would be more likely to survive and pass on their genes). StuRat (talk) 23:57, 26 January 2014 (UTC)[reply]

Staples on trees

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Is it more harmful to trees to remove staples on the bark or to leave them on the bark?--The Emperor's New Spy (talk) 23:34, 26 January 2014 (UTC)[reply]

I doubt if they would do much damage to the tree either way. However, they could cause injuries to people. Spikes in trees can be dangerous when some day they are cut down with a chain saw. I'm not sure if a staple would be able to damage a chain saw, but maybe large ones might. StuRat (talk) 23:54, 26 January 2014 (UTC)[reply]
There's a guy on one of the educational channels [Animal Planet] who builds elaborate tree houses.[Treehouse Masters] His process involves drilling pretty good sized holes in the sides of the trees and embedding something like a large metal "molly" as a basis for supporting the corners of the structure. If he thinks something like that is harmless to the tree, I wouldn't think staples would do much. ←Baseball Bugs What's up, Doc? carrots00:46, 27 January 2014 (UTC)[reply]
  • There are two potential problems from the tree's point of view. The first is that the tissues that carry the sugar and water (phloem and xylem) in a hardwood tree lie peripherally around the trunk, in the outermost living ring, just under the bark. A single puncture will hardly be problematic. But if you ring the bark it will kill the tree (slash and burn agriculture) by cutting off the flow to and from the top of the plant. There is sideways flow, but it is limited. So you can also kill the tree by cutting off significant flow of fluids at various heights. A large number of gashes can do this. And a disease like Dutch elm disease does this as well, by killing off spots of conductive tissue at various places, eventually ringing the trees.
That brings up the second point, that puncturing the bark allows the entry of insects and funguses like Dutch Elm and the Asian long-horned beetle. The solution is not to damage a tree too much, and to use a sealant you can get at a garden center either when you puncture it or if you remove objects puncturing it. μηδείς (talk) 01:08, 27 January 2014 (UTC)[reply]
Also more susceptible to viruses when the "pipes" are exposed. Better to leave things in (and not put them there to start), as long as they're not toxic (or dangerous to non-trees). I don't think rust has any harmful effects, but I'll check. InedibleHulk (talk) 02:18, 27 January 2014 (UTC)[reply]
Apparently there are "rust diseases", which make plants look rusty. Not sure I'm a capable enough Googler for this one. InedibleHulk (talk) 02:21, 27 January 2014 (UTC)[reply]
Oh right. "Iron oxide". That found me this shockingly sourced Yahoo Answer. Apparently rust is fine. InedibleHulk (talk) 02:26, 27 January 2014 (UTC)[reply]
I concur, there's no reason for the tree's benefit to remove any staples. See this. μηδείς (talk) 03:13, 27 January 2014 (UTC)[reply]
Damn. Now that will ruin a chainsaw. I couldn't help but check Snopes. InedibleHulk (talk) 03:36, 27 January 2014 (UTC)[reply]
Nice link. The narrative doesn't mention that although the bark and cambium will envelop things attached to them, they won't lift them off the ground, because the only parts of the tree that gets longer are the new shoots on the end of the twigs, the branches and trunk only grow wider, not longer. In London, timber merchants looking to acquire fallen trees for their wood often run a metal detector over the trunk to find out if it's full of shrapnel from The Blitz. Alansplodge (talk) 13:58, 27 January 2014 (UTC)[reply]
I believe that depends on the type of tree. Bamboo, for example, grows in length at each joint, allowing it to grow faster and kill off the competition by stealing it's sunlight. (Some might argue that bamboo is a grass, not a tree, but most people would call it a tree.) StuRat (talk) 21:22, 27 January 2014 (UTC)[reply]
Always been grass in my eyes. But no, won't argue. InedibleHulk (talk) 23:30, 27 January 2014 (UTC)[reply]
Also, I think some other tree species do grow upwards slightly at all points. The tree in the pic has a bicycle which has clearly been lifted off the ground, for example. I doubt if it was suspended in air when the tree grew through it. And if the ground washed away, I'd expect to see exposed roots. StuRat (talk) 00:34, 28 January 2014 (UTC)[reply]
The branch of a woody tree (including the crown stems) will only grow lengthwise in its first season, while it is green. (That even applies to grasses which become woody like bamboo.) Given the height of the bike, it must have been placed in the crook of a branch at the height where it is still found--and a new twig in its first year of growth would not have held its weight. μηδείς (talk) 01:23, 28 January 2014 (UTC)[reply]
Medeis is right and StuRat is wrong. See Will initials carved on the side of a tree always remain at the same height?. The bicycle must have been thrown over a branch which has since fallen off (as Medeis says above), or chained high up the tree as a practical joke. Alansplodge (talk) 16:26, 28 January 2014 (UTC)[reply]
  • The question that hasn’t been asked yet is : Why use staples? There are many good contact adhesives available now that will keep a notice stuck to a tree long after another Katrina has come and gone. Because bark doesn’t grow, the glue eventually falls off.--Aspro (talk) 16:51, 28 January 2014 (UTC)[reply]
Good question. I suppose staples are much quicker and easier to use. There are many tree barks that contact adhesives don't stick to. If you want to be ecologically friendly, don't use trees for your fly-posting! Dbfirs 17:05, 28 January 2014 (UTC)[reply]
True, but in many neighbourhoods, street lights out number trees these days. There is also now a plethora of suitable contact adhesives available (Try smearing yourself, first with supper glue, then do bit of tree-hugging. It will take the fire department -hours and hours - to peel you off). Also, try cutting out a Stencil and spray painting your flier on the sidewalk, walls, fences etc. Don't forget to add a QR code to link the reader to a website that gives further information.--Aspro (talk) 22:27, 28 January 2014 (UTC)[reply]
Cyanoacrylate isn't really a contact adhesive, it is just fast drying if used sparingly (and infuriatingly slow drying if you use too much). I like your suggestions for alternatives. Dbfirs 14:23, 30 January 2014 (UTC)[reply]