Wikipedia:Reference desk/Archives/Science/2009 March 10

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March 10

blood alchohol content

Can you give blood while you are wasted, and if you do, does the guy getting the transfusion end up with the hangover? 12.216.168.198 (talk) 00:06, 10 March 2009 (UTC)

If you are obviously drunk, I doubt anyone would take your blood - it would be reckless. If you are less obviously drunk then they would probably discover it when they screen the blood (which they always do) and wouldn't give it to anyone. They might use it for other purposes, or they might just throw it away. --Tango (talk) 00:45, 10 March 2009 (UTC)

The giving of blood is always (well almost always) accompanied by screening to determine the donor's safety as well as the recipient's safety. The donated blood undergoes testing (the little test tubes on the side of the pouch) for Hepatitis B, Hepatitis C, HIV and syphilis. If you are wasted, the medical folks wouldn't want to go to all that trouble. You'd be given some orange juice and some crackers and be sent on your way - drunkards are the easiest to spot as they show up frequently. In any event, the usual transfusion of one pint of blood would become diluted with the existing blood (about ten pints) and thus would be its effect. -hydnjo (talk) 00:54, 10 March 2009 (UTC)

Hydnjo didn't say so explicitly, so I will: donated blood is not tested for its alcohol content. You're unlikely to be allowed to donate if you show up drunk at the blood center. However, if you are a donor that is contacted outside of ordinary blood center opening hours, because you have a special blood type or HLA type that a specific patient needs, you would probably be allowed to donate even if you had consumed a moderate amount of alcohol. The alcohol would be diluted even more than hydnjo suggests, as it would not just stay in the blood, but enter the extracellular fluid as well. --NorwegianBlue ^talk 23:08, 11 March 2009 (UTC)

You two seem to be forgetting three very important facets of this problem. 1) If you're receiving blood, chances are you don't have a solid ten pints in you. You've probably lost a significant amount of blood from a surgery or trauma, and that's why you need someone elses. 2) the recipient tne pints to begin with, such as a child, and 3) the alcohol in the donors blood could (and probably will) interact with drugs in the recipient. If you're receiving blood, chances are high that you're under anaesthesia. No matter how you cut it, it is EXTREMELY IRRESPONSIBLE to donate blood when drunk if you could otherwise do so sober. At the very least, inform the doctors so they can make the decision upon receipt of your blood. It would be a very poor surprise for the doctors when their multi-system trauma recipient crashes on the operating table because the alcohol in his donor blood interacted with the opiates the medics gave him. --Shaggorama (talk) 14:30, 15 March 2009 (UTC)

If Shaggoramas comment was directed at me, I'd like to make it clear that I did not intend to encourage anyone to appear for blood donation if drunk. My point was that a blood bank would consider accepting a donation from someone who had consumed a moderate amount of alcohol, if that someone was contacted outside of normal working hours because of an urgent need for their blood or platelet (HLA more rarely HPA) type, if the person informed about the situation. The blood bank would, of course, have to decide whether the alcohol in the donor's blood would could represent a danger to the patient in question. The only situation I can imagine where it might be a problem would be an exchange transfusion to a newborn with HDN. Interactions of the tiny amount of alcohol in the donated blood with drugs would be the least concern in a multi-trauma situation - trauma teams operate on drunk patients all the time. And the volume argument is also based on a misunderstanding. When a patient has lost a considerable part of his or her blood volume, the primary concern is volume replacement, which is done using saline solutions or colloids, not blood. The second concern is heamostasis (plasma and platelet concentrate), and the third concern is oxygen transport (erythrocyte concentrate = "blood"). The main reason why blood centers defer donors who are drunk, is that they might not answer reliably to the screening questionnaires used to decide whether they are eligible for donation. --NorwegianBlue ^talk 20:43, 16 March 2009 (UTC)

Rainbows

There are several things I have always wondered about rainbows. Can they ever be in the opposite color order (i.e. Violet, Indigo, Blue, Green, Yellow, Orange, Red instead of vice versa like normal), such as from an unusual angle? Is there really an "end" to the rainbow? Are Infrared and Ultraviolet actually on the rainbow, but invisible to humans? --76.212.103.145 (talk) 01:22, 10 March 2009 (UTC)

Yes about the opposite order, see our rainbow article to read all about that part of your question. It says that "More rarely, a secondary rainbow is seen, which is a second, fainter arc, outside the primary arc, with colours in the opposite order, that is, with violet on the outside and red on the inside." -hydnjo (talk) 01:40, 10 March 2009 (UTC)

...and yes (in principle) about the IR and UV - although it's possible that some of that light is absorbed by the rain droplets instead of being refracted...but certainly SOME UV and IR must be there. Rainbows are actually complete circles - but the ground kinda gets in the way of seeing all of them. From high altitude aircraft and in other special situations, you can actually see full circles. SteveBaker (talk) 03:12, 10 March 2009 (UTC)

If you're ever in Maui, you can see the complete circle of a Rainbow when standing on Haleakala (assuming the time of day and cloud cover are just right). Someguy1221 (talk) 08:53, 10 March 2009 (UTC)

Maui eh? Could I request some funding from the WikiMedia foundation to test that? I might need to be there for a couple of weeks so that the conditions will definitely be right. SteveBaker (talk) 13:04, 10 March 2009 (UTC)

But how do you expect to drive your car full of gold to Hawaii ? :-) StuRat (talk) 17:33, 10 March 2009 (UTC)

Well, now I'm not sure I should try. All of our 'experts' below warn me about how much heavier the gold will be in a warm climate - so in hauling it from Fort Knox (average March temperature: 40F) to Hawaii (somewhere in the high 70's), I could easily break an axle!  :-P SteveBaker (talk) 16:05, 12 March 2009 (UTC)

With regard to the 'end of the rainbow' part of the question... No, because a rainbow is not an object as such. It's more like a reflection. If you are looking right at a rainbow the sun will always be behind you. So if you are looking at a rainbow and so is someone else who is standing a mile away, you will both see the rainbow in different places.91.111.86.221 (talk) 21:39, 10 March 2009 (UTC)

Eye-testing telescopes?

Is there some sort of system for qualifying the end-result image from a telescope/lens? I often hear the quality of lenses (such as the mirrors in HST, and newer telescopes) in terms of the accuracy of the actual atoms of the lens, or the "closeness to spherical perfection", but post-processing to increase accuracy/resolution is obviously going to factor in to the final quality of the image. Is there some sort of standard "eye test" for telescopes? 219.102.220.90 (talk) 01:34, 10 March 2009 (UTC)

I'm not sure if it's what you are looking for, but Adaptive optics (especially the external links) and Shack-Hartmann might be a start. A google search seems to yield more in depth information such as on this page. Much of the technology originally developed for measuring telescope optics has now been adapted for use in providing customized laser eye surgery, so you will find a mix of references to the two fields. -- Tcncv (talk) 05:32, 10 March 2009 (UTC)

The Foucault test (AKA knife edge test) is a simple way to test the quality of a telescope's mirror. I don't know what test can be used for a lens. Dauto (talk) 06:15, 10 March 2009 (UTC)

The Strehl ratio is often used to characterize the optical performance of telescopes. -- Coneslayer (talk) 14:06, 10 March 2009 (UTC)

Great, I found my answer (multiple actually) within the links. Thanks! 219.102.220.90 (talk) 23:51, 10 March 2009 (UTC)

Bananas

Do greener bananas have more vitamins/minerals/nutrients in them than a yellow banana? Basically, do they lose their nutritional value as they ripen? Dismas|^(talk) 03:00, 10 March 2009 (UTC)

Bananas are usually shipped green and ripened by exposing the fruit to ethylene gas so that consumers find fruit that meet the desired grade on the retailers "ripeness chart". The major changes during ripening involve starch being converted to sugar. The fact that the chlorophyll in the peel is changed can be ignored for bananas because we don't eat the peel. Minerals are pretty hardy so you can expect them to remain at the levels the fruit had when it was picked. What those levels are depends on the soil, the health of the plant and a couple of other factors. Vitamins are more sensitive. Plant vitamin A isn't metabolized very efficiently anyway, so I wouldn't worry too much about that. Some of the Vitamin B will be used up in the conversion of starch to sugar, but values you find like in our article are based on ripe bananas anyway. Ripening makes them easier to digest, so some of the nutrients will be more accessible to the body. Too tired to go through the list, looking at the label at banana may help. Nutrient content of natural fruit varies anyway. I doubt any changes during ripening would have a bigger influence than that. 76.97.245.5 (talk) 05:14, 10 March 2009 (UTC)

Besides, who wants to eat green bananas? Dauto (talk) 15:51, 10 March 2009 (UTC)

Thanks guys, now I can't get This Frigging Song out of my head. Beautiful... --Jayron32.talk.contribs 16:53, 10 March 2009 (UTC)

You could listen to Fried Green Tomatoes (soundtrack), instead. StuRat (talk) 17:21, 10 March 2009 (UTC)

So, to summarize, don't eat green bananas, as ripe bananas have plenty of nutritional value and won't be nearly as likely to make you sick. StuRat (talk) 17:29, 10 March 2009 (UTC)

Thanks, but I actually prefer green bananas. Dismas|^(talk) 18:07, 10 March 2009 (UTC)

Just don't confuse them with plantain, which isn't edible raw until really ripe! --TammyMoet (talk) 19:43, 10 March 2009 (UTC)

Green Eggs and Ham Anyone? cheers, 10draftsdeep (talk) 21:32, 10 March 2009 (UTC)

Encoder for remote application

please tell about the encoders that are used in remote application —Preceding unsigned comment added by Teja45 (talk • contribs) 05:29, 10 March 2009 (UTC)

You will have to be more specific regarding what you mean by "encoder". And you can expand on what "remote application" is while you're at it. -- Tcncv (talk) 05:41, 10 March 2009 (UTC)

I use gzip when I can, whether my application is remote or local. Nimur (talk) 13:53, 10 March 2009 (UTC)

It's possible the original question was about television remote control electronics? The encoding, or modulation, varies for each manufacturer and model series. Nimur (talk) 02:38, 11 March 2009 (UTC)

I was thinking it might have something to do with using an Enigma machine in a region where there was no electricity. -- Tcncv (talk) 03:25, 11 March 2009 (UTC)

Maybe it's about online applications using remote procedure calls? Microsoft suggests using one of Internet Information Services Security, SSL, or a proxy server controlling which machines can access RPC.[1] This is a fun guessing game! —Preceding unsigned comment added by Maltelauridsbrigge (talk • contribs) 16:42, 11 March 2009 (UTC)

Effect of heat on the weight of gold?

What is the effect of heat on the weight of gold? Does the weight of gold decrease with heat or it remains the same.Sathyass (talk) 10:50, 10 March 2009 (UTC)

Weight is the force on an object due to gravity, and based upon the mass of an object. The mass of gold will not change if you heat it up or cool it down, so the weight should stay the same. Then again, I haven't got a degree in anything like this yet, I'm just applying stuff I know. —Cyclonenim (talk · contribs · email) 11:06, 10 March 2009 (UTC)

Actually, any closed system will increase in mass when heated, due to mass-energy equivalence (specifically, Mass-energy equivalence#Practical examples), but the actual change in mass will be pretty pathetic, right up to temperatures that will bring the gold to near vaporizing. I specifically recall that very delicate attempts to measure the change in mass of a heating crystal to have failed due to the thermodynamic noise exceeding the expected change in mass. I'm not sure it's ever actually been measured. Someguy1221 (talk) 11:14, 10 March 2009 (UTC)

So what are you saying it happens even though it can't be experimentally observed? What sort of mass does this heat become? As I understand it mass energy transformation phenomenon applies only to nuclear transformations. Heat to mass sounds fully theoretical based on the experimental results you described.--OMCV (talk) 11:37, 10 March 2009 (UTC)

Mass/energy applies to everything and heat exists in a solid as energy particles called phonons, the mechanical vibration equivalent of a photon of light. Putting in some figures, the specific heat of gold is 0.2291 kJ/Kg-K and its meting point is 3129K. The heat contained in 1kg of gold at its melting point is 0.2291x3129=716.9kJ. The mass equivalent of that is 716.9x10³/C² = 7.97x10^-12kg, about eight billionths of a gram. You won't measure that on your bathroom scales, or even your weightwatchers scales. SpinningSpark 12:05, 10 March 2009 (UTC)

However, if someone has ten tons of gold they could lend me, I think I could get this experiment to succeed as the mass increase would then be in the μg range. SpinningSpark 12:22, 10 March 2009 (UTC)

We should consider putting it in the trunk of a very small car and driving very fast to see if kinetic energy will also make it heavier! Actually, I have the right car(s) for doing that in - so just send me half of the gold and we can do both tests at the same time. SteveBaker (talk) 13:01, 10 March 2009 (UTC)

No, gold doesn't become heavier when it gets hotter. What you are probably thinking of is that (like most materials), it EXPANDS when it gets hotter. But as it gets bigger - so it also gets less dense - and because weight is dependent on size multiplied by the density - the two things exactly cancel out and your gold stubbornly DOES NOT increase in weight. The effect the previous posters have been babbling on about it quite utterly negligable and someone needs to kick them in the shins and tell them to shut up and not confuse the questioners. SteveBaker (talk) 13:01, 10 March 2009 (UTC)

Steve, you're not suggesting we deny scientific fact just because it has subtle complexities? The above effects are put in context (the word "negligible" was used; and quantitatively, the numbers came out to 1 part in a trillion). The mass-energy equivalence does exist, even though it is fairly negligible, and to deny it for the sake of simplicity is disingenuous. Nimur (talk) 13:58, 10 March 2009 (UTC)

It's a matter of APPROPRIATE detail. We could whitter on all night about how the planet loses a little mass due to the transfer of energy from everything else into the gold so the earth loses mass and that reduces it's gravitational pull....yeah, yeah, very clever, I guess it allows you to strut your Physics 101 stuff - but it's a completely, totally UTTERLY useless answer to give to someone who came here to ask a simple question and get a simple, straightforward answer. Which is "NO!!" Everything beyond that is pseudo-intellectual bullshit. At the level of heat that you can get gold up to without it boiling away, the gains from mass/energy equivelence is VASTLY less than the amount of additional gold evaporating off the surface - or rubbing off on the surface it's resting on or...any number of other irrelevent things. An OP who has to actually ask this question clearly doesn't know enough physics to understand your ridiculous answers. You AREN'T telling the WHOLE story because it's essentially impossible to do so - you arbitarily cut off the approximation at some point...and that point was WRONG for the degree of complexity demanded by the questioner. All you'll do is end up convincing our OP that gold does indeed get heavier when you heat it up - which is BULLSHIT at any conceivable useful level of description. If you guys can't answer questions appropriately - then don't answer them. Were here to help people not show off our arcane knowledge of the utterly irrelevant. SteveBaker (talk) 23:16, 10 March 2009 (UTC)

Steve you are completely wrong; you are drawing an awful distinction between people. Saying that some are to foolish to be able to take a straight answer, and that we have to dumb it down for them. The fact is the original poster as a question to which the answer "yes, but it is negligible", since that is the answer we should not say no. It is extremely arrogant of you to dismiss the questioner as too foolish to understand the full answer. And even if that is the case, it is better for them to know there is a complex answer that they dont understand; rather than believing in a false simplification that is not true. The fact is the question gives no guidance as to the appropriate level of detail to answer at, and so we should not assume that they are stupid. Your condescending nature does grate on some users of this desk. —Preceding unsigned comment added by 129.67.37.225 (talk) 12:16, 11 March 2009 (UTC)

(ec) It depends on on precisely how you want to define weight. If we take it as, the magnitude of the gravitational force on an object, then the discussion above has reached the correct result. The increase in thermal energy corresponds to a (vanishingly small) increase in mass, leading to a corresponding increase in the gravitational attraction between Earth and the gold.

On a more practical level, the apparent weight of the material – the net downward force experienced by the gold while it sits at rest on the Earth's surface, and what you would measure if the gold were sitting on a very sensitive balance – would actually decline slightly. As the temperature of the gold increases, so too does its volume. By Archimedes' principle, the gold will displace a bit more air, reducing very slightly its apparent weight. TenOfAllTrades(talk) 13:18, 10 March 2009 (UTC)

This is probably the most relevant effect. Hot materials could create interesting convection patterns, and those will interfere with whatever sensitive balance or scale you use to measure the object. These will probably dwarf any of the effects due to mass-energy equivalence. But, if the scale were in a vacuum, those effects would be negligible. All told, though, these are instrumentation problems, not actual mass or weight increases. Nimur (talk) 14:04, 10 March 2009 (UTC)

Actually, if you heat the object inside an oven and then wait for all the convection stop, the aparent weight would increase slightly because the density of the air inside the oven decreases more markedly then the density of the gold object itself. Dauto (talk) 15:42, 10 March 2009 (UTC)

Whether we consider the relativistic increase in mass due to mass-energy equivalence of the added heat, or we consider the decrease in apparent weight due to displacement effects WRT the air the gold is in, this all belies the fact that these calculations are largely intellectual curiosities, and have little bearing on actual measurable effects. They may be calcuably real, but are realisticly imaginary. All of these effects are on the order of micrograms per ton; at those ranges, any device sufficiently resiliant enough to not be crushed by that sized lump of gold would not be sensitive enough to measure the mass or weight difference at the different temperatures. And any scale which may be sensitive enough to measure microgram amounts would be crushed by a one ton sample of gold. So yes, one can calculate a meaninglessly small number that would represent the change in mass or weight of a hot vs. cold object, but such a calculation, while rigourously true WRT the theory, would be in practice impossible to actually measure. Its an interesting thought experiment, but has no practical purpose. --Jayron32.talk.contribs 16:50, 10 March 2009 (UTC)

Never tell a physicist than a change is 'in practice impossible to actually measure'; you'll never hear the end of it. (On the plus side, you'll get a brand-new high-precision balance out of it.) The (volume) coefficient of thermal expansion for gold is given as 42ppm/K, so about 250 degrees' (celsius or kelvin) temperature change will result in a 1% change in volume. 25 degrees will suffice for a 0.1% change.

One thousand grams of gold will occupy a volume of about 50 cubic centimeters (milliliters, mL), 1% and 0.1% changes in volume are respectively 0.5 and 0.05 mL. The density of room-temperature air is roughly 1.2 kilograms per cubic meter, equal to 1.2 g/L or 1.2 mg/mL. Assuming constant temperature air, the buoyancy change would therefore be .6 milligrams or 0.06 milligrams — that's a bouyancy effect of between (roughly) 0.1 and 1 part per million. Is that tiny? Absolutely. But it's well within the capabilities of ordinary analytical chemistry instruments. A very quick Googling finds this balance – a steal at just under ten thousand dollars! – with a 0.1 ppm repeatability. Do some more detailed shopping and I guarantee that you'll be able to find commercially-available balances that are even more precise.

Of course, those high-precision balances are all going to be rigorously temperature and atmosphere controlled, because changes in air temperature and humidity would otherwise have a significant effect on measurements. Both the bouyancy of the object being weighed and the components of the balance have to be carefully accounted for. As Dauto noted, it's virtually impossible to precisely weigh a sample that is at a different temperature than the surrounding air, and a warming the air reduces its density more than a similar temperature change would reduce the density of the gold. The reduced buoyancy of both the gold and the balance's weighing components will alter the apparent reading of balance. The precise effect on the balance's reading will depend on the particular weighing mechanism the balance employs, but typically the displayed measurement will be higher with increasing temperature. TenOfAllTrades(talk) 17:41, 10 March 2009 (UTC)

Jayron, eventhough the relativistic dependence of an objects mass on its temperature is indeed very small and as far as I know has never been directly measured, it's still a real effect and there is one very important practical reason for us to be talking about it, namely it enhances the reader's understanding about modern physics. One common misconception I've seen many times before is that Einstein's mass-energy equivalence applies only for nuclear reactions. Dauto (talk) 18:04, 10 March 2009 (UTC)

If an atom or ion of gold were accelerated in vacuum in a particle accelerator to a significant fraction of the spped of light, would any increase in mass be detectable? I've been told that since early cyclotrons, relativistic mass increase of electrons had to be taken into account or the thing would not work. Granted an electron has tiny mass compared to a gold atom, but Particle accelerator seems to say that gold atoms are accelerated to several GeV per nucleon. Is a relativistic mass increase taken into account in the generation of the accelerating fields? Is the high energy gold particle "hot" or just "fast?" Or is "heat" just a property of numerous nuclei? Edison (talk) 19:29, 10 March 2009 (UTC)

If anything is accelerated to a significant fraction of the speed of light, it's relativistic mass will change significantly. It's taken into account, although it doesn't amount for anything dramatically different than the thing accelerating more slowly. It's just 'fast', not 'hot'. Heat is a statistical thing, and as such, you need a lot more than just one particle. (Or as one professor put it: "Temperature is something you measure with a thermometer") Speaking of gold and relativity; The yellow color of gold is also an effect of special relativity. The high nuclear charge causes the innermost electrons to have high kinetic energy; so high that they acquire relativistic mass. This, in turn, leads to a change in energy. Specifically a p-band level which is in the ultraviolet in silver, is shifted down into the blue. Hence, gold absorbs blue light and appears yellow. So you could say that if Einstein was wrong, gold would be silver-colored! --Pykk (talk) 20:02, 10 March 2009 (UTC)

Yep, the increase in relativistic mass abosolutely has to be taken into account in accelerator design. Early cyclotrons could only accelerate particles up to a few percent of the speed of light; as the particles gained mass, they fell out of sync with the alternating, accelerating radio frequency field. Later evolutions of the cyclotron led to the synchrocyclotron (which varied the RF frequency to compensate for the more massive particles) and the isochronous cyclotron, which uses a nonuniform magnetic field. Modern synchrotron technology, meanwhile, carefully tunes both the accelerating electric field and the bending magnetic field to follow the increasing mass and energy of a particle beam. TenOfAllTrades(talk) 21:31, 10 March 2009 (UTC)

All of this is all fine and wonderful - but you're doing the OP no favors at all. The first answer from Cyclonenim was 100% clear and correct - no elaboration was required. Almost all of the answers since then have SUBTRACTED clarity - SUBTRACTED truth and piled confusion onto confusion until the sum total of all of the replies is a big steaming heap of CRAP. All you've done is collectively either turned off the poster or confused the heck out of someone who needed a simple answer to a really simple question. The simple word "NO" would have been a million times better than the 20 paragraphs of irrelevent twaddle. I'm angry and disgusted. We are hear to answer peoples question - that's the objective here. This is a low point in the annals of WP:RD/S. SteveBaker (talk) 23:16, 10 March 2009 (UTC)

Amen brother. Everything he said. X 2 from me. --Jayron32.talk.contribs 23:20, 10 March 2009 (UTC)

You may be right but I prefer not to patronise the reader and assume they can handle complexity. Dauto (talk) 00:02, 11 March 2009 (UTC)

It's not patronising to make an assumption as to the level they're ready for based on the question. If someone asked "I know that babies are made when the sperm and the egg meet, but how does the sperm get to the egg?" you could be pretty sure that, assuming they weren't pulling your leg, an answer involving hormones and the mechanisms of sperm motion would be unhelpful. Clearly they need to understand some more basic principle first. You might find, after answering that, that they ask for further information, in which case you can go into more detail. But not before they understand the larger point. 79.66.56.21 (talk) 08:08, 11 March 2009 (UTC)

No but if they ask "does the egg get heavier when its hot" you don't say no if the answer is yes. —Preceding unsigned comment added by 129.67.37.225 (talk) 12:19, 11 March 2009 (UTC)

I'll bet good money that a hot egg is no heavier than a cold one. APL (talk) 15:12, 11 March 2009 (UTC)

The proper place for a metadiscussion about how to answer questions at the Desk is on a talk page (either that of the user involved, or on the Ref Desk talk page). Can we please take the bickering elsewhere? TenOfAllTrades(talk) 13:05, 11 March 2009 (UTC)

What's wrong with just giving the poor guy a simple answer and, if you must, adding a note that the universe is always more complex if you need it to be? Let me try :

"The gold will stay the same weight. Assuming that it's not getting so hot that it boils away, but even then it would stay the same weight if you count the weight of the vapor. See Conservation of Mass. (By the way, I'm ignoring Einstein's E=MC², because that effect won't even be noticeable unless you've got about a million pounds of gold.) "

There. That was easy. APL (talk) 15:12, 11 March 2009 (UTC)

@SteveBaker: No. (I'd explain why but I'm assuming you wouldn't understand.) – 74  18:44, 11 March 2009 (UTC)

That's unfair. Steve's assumption is reasonable. Zain Ebrahim (talk) 08:37, 12 March 2009 (UTC)

Running dog

My dog Jerka runs after her anus, turning very fast, always counter-clockwise. I was told by a friend, a physicist, that this is due to us living in the northern emisphere. But how can it be? Should it mean that she would turn the other way around in the southern emisphere?? Thanks, it's not a veterinary request.--131.114.73.84 (talk) 12:21, 10 March 2009 (UTC)

No - that's not the reason and if your friend is truly a physicist then (s)he is making a joke at your expense. Your friend is referring to the coriolis effect - but that only applies to very large scale things like hurricanes and ocean currents - not to small dogs! The reason your dog is going this is because she is bored or because she has infected Anal glands. If she also 'scoots' along the floor - sitting down and pulling herself along with her front feet - then it is the latter and you need to go see a Vet ASAP. If she's not scooting then it's probably just boredom - but you might want to get her checked out anyway. Dogs who engage in this 'tail chasing' behavior do get into the habit of doing it in just one direction - but sometimes you can get them to do it in the opposite direction too by gently pointing the tail down the other side of the body...but boredom isn't good. Play with your dog more - get toys - walk her more frequently, and if you can find someplace safe like a dog park where she can run with other dogs off-leash, that would be a very good thing too. SteveBaker (talk) 12:52, 10 March 2009 (UTC)

Your friend is pulling your leg. Dauto (talk) 15:48, 10 March 2009 (UTC)

You may also want to read our article on Stereotypy, a serious psychological condition in animals, usually brought on by extreme isolation or lack of interaction. If you are concerned that your dog has this problem, please seek medical advice from a veterenarian. --Jayron32.talk.contribs 16:42, 10 March 2009 (UTC)

Sparrow behavior

Hi. I feed birds in my back yard, so I often have a crowd of avians around, on the feeders, on the ground, etc. This morning, I noticed the Passer domesticus doing something I haven't seen before. Near the dish of water is some dry ground with nothing growing. At least two male House sparrows were sort of... bathing in the dust? They would lower their breast to the ground, and then shake, kind of like a dog drying itself. It seems that dirt would get in their feathers. What are they doing?

I didn't see anything in the house sparrow article under "Behavior", but I suspect it's a more general bird behavior. Anyone? -GTBacchus^(talk) 16:47, 10 March 2009 (UTC)

I would guess that this is a way to get rid of parasites. Another possibility is that it masks their scent. StuRat (talk) 17:10, 10 March 2009 (UTC)

http://www.stanford.edu/group/stanfordbirds/text/essays/Bathing_and_Dusting.html --Digrpat (talk) 18:03, 10 March 2009 (UTC)

Just sounds like a dust bath, which many animals enjoy. Weirder: anting (bird activity). --Sean 19:45, 10 March 2009 (UTC)

Very interesting. Thank you all. -GTBacchus^(talk) 03:04, 11 March 2009 (UTC)

I started a page, but s.o. has already ripped out the structure and lots of the contents. I'm throwing up my hands in despair and am waiting for the delete to hit. Anyone else willing to waste some effort is welcome to try (Lisa4edit) 76.97.245.5 (talk) 03:43, 11 March 2009 (UTC)

It seems to be a good stub; I don't see why it would be deleted. Thanks for your work. -GTBacchus^(talk) 15:10, 11 March 2009 (UTC)

Energy loss in Reflection (intensity loss) [formula wanted please]

Moved from Maths desk --Tango (talk) 17:01, 10 March 2009 (UTC)

I have been working on a solution to Multi-path dispersal in optical fibers But i need to know how much energy is lost every time the light hits the fiber wall. As a percentage or a decimal if that is how it works. i.e.

Intensity_Out = 0.95 * Intensity_In

and also, the refractive index of optical fiber.

BenC303 Benc303 (talk) 16:57, 10 March 2009 (UTC)

The fresnel equations will tell you how to compute the reflectability of a interface given the refractive idexes of both materials as well as the incidence angle. Dauto (talk) 19:00, 10 March 2009 (UTC)

I don't think the Fresnel equations are going to be helpful here — optical fibers rely on total internal reflection to keep the light in the fiber core. Do we have any optical engineers in the house? TenOfAllTrades(talk) 23:21, 10 March 2009 (UTC)

Yes, you are right, and yet the Fresnel equations are the right answer to the question. We can only assume that either Benc has a different thing in his mind than the normal use (or kind) of optical fiber or that s/he doesn't know what s/he is doing. Dauto (talk) 23:45, 10 March 2009 (UTC)

I'm an A level student, that has lots of random theories, and ive been testing this one out using interactive physics, modeling light as photons, unfortunately, my knowledge of how light acts in wave form inside an optical fiber is very limited. Even in TIR there is a very small percentage of light lost on every reflection..(well, thats what we've been taught).. and if there isn't, then it only makes my idea better. But as i say, my knowledge is limited and my ideas will most definitely need refining. is there an email i can send my work to someone who's interested? Benc303 (talk) 23:20, 11 March 2009 (UTC)

benefits of wanking

Hi wikipedia,

Something that i've always wanted to know.. I keep finding articles about the benefits of sex (like this one[2] but as someone who hasn't got laid in a really really really long time, can i get similar benefits from flying solo? Or, should i whore myself out for the benefit of my immune system/ prostate gland/self-esteem? Thanks81.140.37.58 (talk) 17:32, 10 March 2009 (UTC)

If you want a definite answer, you'll have to ask a doctor, we can't provide medical advice. You may find Masturbation#Benefits useful, though. --Tango (talk) 18:01, 10 March 2009 (UTC)

Here's some ideas: it increases self-knowledge of your own sexuality; it helps you sleep; relieves pain and stress; it may help prostate health in men (it has this in common with sex) and reduce menstrual pain and yeast infections in women; as well as providing hours of cheap amusement in these straitened financial times.[3][4] --Maltelauridsbrigge (talk) 16:49, 11 March 2009 (UTC)

"it has this in common with sex" - I have a bulletin for you: it is sex. -- JackofOz (talk) 19:58, 12 March 2009 (UTC)

Stuff like this is generally contested, but here's an example of one study done on the topic. ~AH1^(TCU) 22:58, 13 March 2009 (UTC)

We need to settle a debate: Athletic vs Non-Athletic People

Classmate states that the more athletic a person is, the more they will have to urinate and defecate, frequency and amounts. If a person is overweight or obese, they do not need to urinate and defecate as much (frequency). So, he's saying that fatter people hold in "poop" longer than they should, healthwise. I told him that this was bull poop and that it depends on your diet and genetics (maybe?). Is there really a difference? --Emyn ned (talk) 17:50, 10 March 2009 (UTC)

While I can see that muscle tone will have an effect on elimination (it's often given by nurses as the reason elderly people suffer from constipation, for example), I can't see that it will have an effect on retention. Water retention will depend on the size of the bladder, the efficiency of the kidneys, the amount of salt consumed, and so on. However, fecal retention may well have something to do with size because you've got to put all that poo somewhere! There is an urban myth that John Wayne had retained half his bodyweight in poo, according to the post mortem. I would have also thought that the amount consumed will affect the amount eliminated too. --TammyMoet (talk) 19:38, 10 March 2009 (UTC)

There's a scam out there that says something along the lines of "people retain too much fecal matter, and this causes toxins to enter the body, and you can stop all this and improve your health by giving us money for X cure". Except for cases of constipation, where laxatives might be warranted, this is pretty much all a scam. As for the quantity of feces, that's determined by the quantity of food eaten. Athletes and the obese both tend to eat more food, so I'd say a thin, but sedentary, person would likely produce less. However, this is no indication of general health. Indeed, the fewest feces might well be produced by an alcoholic who gets 100% of his calories from booze and never eats anything solid. Also note that healthy foods tend to be low on calories and high on nutrients. The "low on calories" part requires that more food be eaten to get enough calories, and hence more feces will be produced. The quantity of urine produced isn't so closely linked to the quantity of water consumed, because water is also lost in sweat, respirations, etc. StuRat (talk) 20:24, 10 March 2009 (UTC)

Does that mean I win? --Emyn ned (talk) 17:47, 11 March 2009 (UTC)

Yes, it does. StuRat (talk) 01:17, 12 March 2009 (UTC)

Looking for a biochem comic

Hi, I wonder if anyone can find this:

It's a short comic about biochemistry and genetic manipulation in particular, but in an educational way. There's a kid and a teacher, or something, and the teacher is explaining what is and isn't possible with genetic manipulation. Eventually, they make a microbe that generates a filmy layer and also free hydrogen, which it gathers up to make a balloon that rises into the air.

I think it was a few years back, and it was online. Possibly it used flash or something, and might be associated with a scientific thing?--163.1.210.162 (talk) 20:33, 10 March 2009 (UTC)

Just so you don't think we're ignoring you - I had a good dig around - but didn't turn up anything. If there are ANY more things you can remember about it - it would help. SteveBaker (talk) 01:55, 11 March 2009 (UTC)

Loop-the-loop on a skateboard

I saw one of those video clip shows featuring various nasty accidents. One clip showed a skateboarder attempting a loop-the-loop, failing, and landing on his head. Luckily he survived with a light concussion. The reason for his fall was presumably insufficient centrifugal force to hold the rider to the roof of the full pipe he was in. I feel that if he had been travelling faster, he might well have completed the loop successfully. I was thinking though, was there anything else he could have tried to improve his chances of success (eg. crouching down on his board at the start of the loop)? Astronaut (talk) 22:29, 10 March 2009 (UTC)

Crouching would have made things worse by reducing his speed at the top of the loop and increasing the effective radius of his trajectory. Dauto (talk) 23:27, 10 March 2009 (UTC)

But...if you go around standing up - your moment of inertia is higher than if you were crouched - so that suggests that more energy goes into producing that rotation. Since the only source of energy here is your initial kinetic energy, it follows that you'll have less speed at the top of the loop if you're standing up. It's not entirely clear to me which is best...but my 'gut feel' says that crouching is better. SteveBaker (talk) 00:33, 11 March 2009 (UTC)

Not quite, in this case. Your moment of inertia is smaller in a crouch if you're rotating about your center of mass — but that's not happening here. The skateboarder's rotation will be about the center of the loop, and moving mass closer to the center of the loop reduces the moment.

Alternatively, you can convince yourself using a conservation of energy argument. The kinetic energy of the moving skater entering the bottom of the loop will be converted to (gravitational) potential energy as the skater comes through the top of the loop. If the skater is in a crouch, his center of mass ends up higher at the top of the loop (closer to the track) than if the skater is standing erect; the crouching skater has to give up more of his kinetic energy. TenOfAllTrades(talk) 02:06, 11 March 2009 (UTC)

This is the easiest explanation to understand that I've seen posted so far. You can play with rotating frames of reference and moments of inertia all you like, but Conservation of Energy will not be violated. Being crouched while at the top of the loop will require more gravitational potential energy, so the skater will be moving slower. At some critical minimum speed ($v_{tangential}={\sqrt {gR_{loop}}}$), the skater will fall out of the loop. Interestingly, one can also calculate the speed at the top from conservation of energy, since ${\frac {mv_{i}^{2}}{2}}=mg\delta h+{\frac {mv_{top}^{2}}{2}}$. This will fully constrain the minimum initial velocity for a given loop radius if the skater does not want to fall out of the loop. Equivalent to a stable circular orbit, half the energy should be kinetic and half should be potential (at the point of zero contact-force against the loop - e.g., instantaneous separation from the "floor"/"ceiling"); this is the stability boundary. Nimur (talk) 02:19, 11 March 2009 (UTC)

Two points. First: energy is conserved only when there is no external work and the contact force with the ground can do work here when you crouch or stand. Second: What was that babbling about equal amounts of kinetic and potential energy? Dauto (talk) 02:52, 11 March 2009 (UTC)

Circular motion is stable if the potential energy equals the kinetic energy. At the apex of the circle, the instantaneous representation of the kinetics are identical to that of an orbit (the force is applied towards the center of the motion, and the orbit is only circular if the kinetic energy equals the potential energy. After this instant, the analogy breaks down, because gravity acts downwards, which is not pointing towards the center of the loop, at all other points. Nimur (talk) 03:07, 11 March 2009 (UTC)

Nimur, you are seriously confused. That principle of equal kinetic and potential energy for circular orbits that you are using simply doesn't exist. For instance, a planet in a circular orbit has negative potential energy and positive kinetic energy and the sum of the two is negative. Dauto (talk) 03:19, 11 March 2009 (UTC)

Potential energy is only defined up to an additive constant (you can choose the 0 point arbitrarily), so it only makes sense to talk about changes in potential energy. The change is potential energy will be equal to the change in kinetic energy (with the opposite sign), but that's true regardless of the path of motion. So I have no idea what Nimur was thinking of. --Tango (talk) 12:46, 11 March 2009 (UTC)

It's a bad analogy. And you are right, it is only valid up to an additive constant; but, if you accept that the initial height is "zero" and solve the conservation of energy equation I posted above, you will see that half of the initial kinetic energy is transferred to potential energy when the skater reaches the top. Of course, the change in potential is always equal the the change in kinetic energy; but I'm talking about the ratio of $KE_{final}/KE_{initial}=1/2$. If any more kinetic energy is lost, then the loop is too high for the initial speed, and the skater falls down. This is useful if you want to solve for the minimum initial speed to make a full loop; you can write the equations however else you find convenient. As for the planetary orbits, again, this is valid only to an additive constant; it's a functional analogy only for a particular reference potential. I'll see if I can dig up this in my mechanics book which thoroughly works out the orbital equation (I believe it's in Chapter 7 or 8). My description of my application of this technique was sloppy, sorry if it confused anyone. To summarize, it is just a special-case solution of the conservation of energy for a zero-contact-force at the top of the loop. Nimur (talk) 14:29, 11 March 2009 (UTC)

I'm gettig $KE_{final}/KE_{initial}=1/5\,$. I think you should read that chapter again. Dauto (talk) 15:41, 11 March 2009 (UTC)

Well, I get 1/3. (Assuming by "final" and "initial" you mean "top" and "bottom" - the final will be equal to the initial if you do the full loop). $v_{\mathrm {top} }={\sqrt {gr}}$ so $KE_{\mathrm {top} }={\frac {1}{2}}mgr$. The change in potential energy between the top and bottom is $\Delta PE=-mgr$, so the kinetic energy at the bottom needs to be $KE_{\mathrm {bottom} }=KE_{\mathrm {top} }-\Delta PE={\frac {3}{2}}mgr$ so ${\frac {KE_{\mathrm {top} }}{KE_{\mathrm {bottom} }}}={\frac {1}{3}}$. Is there is mistake in there somewhere? --Tango (talk) 16:47, 11 March 2009 (UTC)

Yes, there's a mistake in there somewhere! The height is twice the radius, so $\Delta PE=-2mgr$, so $KE_{\mathrm {bottom} }={\frac {5}{2}}mgr$, so ${\frac {KE_{\mathrm {top} }}{KE_{\mathrm {bottom} }}}={\frac {1}{5}}$. Dauto is correct. --Tango (talk) 16:50, 11 March 2009 (UTC)

Woops. I guess I was mistaken; I should have done the equations out on paper. The important thing is that there is an analytic solution to the problem. Nimur (talk) 19:40, 11 March 2009 (UTC)

Just from experience with this sort of thing, it makes a big difference if you start crouching and stand up or "push" when entering the transition. Since a full loop is a continuous transition, I would put my money on the skater starting crouched and continuously transitioning to a standing position at the top of the loop for best results. You can actually see most people do this when they do make the loop, and then bend their legs again to absorb the impact of the exit transition. —Preceding unsigned comment added by 99.255.228.5 (talk) 00:47, 11 March 2009 (UTC)

As a matter of fact, your moment of inertia with respect to rotations around the center of the trajectory will be smaller if you are standing. Gut feel is a terrible way to do science. Dauto (talk) 00:49, 11 March 2009 (UTC)

Quite. The one key advantage to remaining in a crouched position is it is easier to maintain your balance. (The contact point of skateboard with track will be closer to your center of mass when you're crouched than if you're standing erect.) TenOfAllTrades(talk) 01:46, 12 March 2009 (UTC)

Try that: get a longish object that has one end havier than the other (a baseball bat) try balancing it on your hand with the heavy side up. Try it with the heavy side down (Should be easier according with your explanation). Let us know what happened. Dauto (talk) 16:54, 12 March 2009 (UTC)

Conservation of angular momentum suggests that if you go into the loop while crouching, then straighten up while already beginning to loop, you should pick up speed by straightening (the extra kinetic energy comes from your having to exert force using your muscles to straighten up). Ballet dancers use this effect when they begin a pirouette with their arms outstretched, then pull their arms inward as they spin. 207.241.239.70 (talk) 01:06, 11 March 2009 (UTC)

Yes, crouching and standing again will help. The sooner you stand back up the better (as long as you are already in the loop). The worst thing you could do would be to crouch and forget stand up again. Dauto (talk) 02:04, 11 March 2009 (UTC)