Wikipedia:Reference desk/Archives/Science/2007 September 4

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September 4

Atkins diet and neuroglycopenia

In a previous question, it was suggested by one editor that the reason the human brain is so loathe to burning fats in the place of sugar is the absence of fatty deposits around the brain and skull. But it has since occured to me that I don't recall hearing of mass cases of neuroglycopenia in Atkin's dieters. Does the brain begin processing fat from other regions of the body only after some time passes, relative to when other organs start? Are neuroglycopenia symptoms present early in starting the Atkin's diet? Or is the diet itself just high enough in carbs to avoid this? Could someone please dispel my confusion? Someguy1221 00:15, 4 September 2007 (UTC)

Your body can convert ingested proteins into glucose: see gluconeogenesis. As well, the brain can draw a substantial fraction of its energy from the ketone bodies, which are also produced from amino acids (protein components). TenOfAllTrades(talk) 00:59, 4 September 2007 (UTC)

Yes, but if the brain apparently lags behind the rest of the body in this regard, it logically seems it should present a problem at some point while starting on an Atkin's diet. Someguy1221 01:40, 4 September 2007 (UTC)

You perhaps misunderstood the previous response. The question was primarily concerning people feeling dizzy after vigirious exercise or perhaps if they haven't eaten when their body had been expecting food. Starvation or abnormal food sources are a completely different matter. Any glucose produced by the body can be used by the brain. Indeed it is done so, selectively. The problem is that gluconeogenesis is a fairly slow process (hours) so in the interim you may have a low blood sugar level which will particularly affect the brain as it can't use other sources when you have a low blood sugar level. Nil Einne 19:28, 6 September 2007 (UTC)

Normal adults can maintain an adequate blood sugar supply to the brain for over 3 days of starvation, even without use of fats for gluconeogenesis. Ketones represent additional fuel beginning after about a day of starvation. Dietary protein can be used for gluconeogenesis. A fourth compensation occurs as glucose transport into the central nervous system is enhanced to increase extraction from even borderline amounts of blood sugar. It is however possible that for an occasional person an Atkins diet may reveal a defect of blood sugar defense that would have gone unnoticed on a normal carbohydrate diet. alteripse 02:01, 4 September 2007 (UTC)

Ice and Hydrogen Bonding

First, a disclaimer from a RD regular: this is somewhat of a homework question, but I'm not looking for answers, just making sure I understand the material. I know ice floats in liquid water, and I'm told that this is because the hydrogen bonds keep the different atoms at 'arms' length" from each other (that's how my textbook puts it anyway). What I don't understand is that hydrogen bonds attract things toward each other, so why would the increase in hydrogen bonds make the molecules spread out and make ice less dense? My current understanding is that the increase in hydrogen bonds means that each individual molecule is being pulled from more "sides." Sort of like how having two friends pull on each arm will stretch you apart, not condense you together. Is this understanding correct? I've looked at the articles on hydrogen bonding and ice, but they don't quite get into the level of detail I'm looking for. Can someone more knowledgeable than me confirm or deny my understanding? --Ybbor^Talk 00:31, 4 September 2007 (UTC)

The water molecules are pulled by hydrogen bonding to get to the stable configuration of ice. But, like overexcited schoolchildren, hot molecules bounce, spin and vibrate too much to stay in any orderly formation, and those thermal forces are greater than the hydrogen bonding.Polypipe Wrangler 01:07, 4 September 2007 (UTC)

Ummm okay. I'm not sure that answers my question. To put it more simply: are the molecules being pulled further apart because more hydrogen bonds are acting on them, and pulling them in more directions? --Ybbor^Talk 01:09, 4 September 2007 (UTC)

No. In order for what you said to work, it would require an infinately large piece of ice. Ignore Polypipe; it looks like he's describing why ice melts. — Daniel 01:21, 4 September 2007 (UTC)

The hydrogen bonds are merely the reason why ice crystallizes in in a certain crystal structure which has a lower density than the unordered state. Cacycle 01:36, 4 September 2007 (UTC)

One way of looking at it is that, on average, each molecule in pure water can only form at most four hydrogen bonds: one from each of its two hydrogens, and two from the oxygen to hydrogens of other molecules. This means that, to minimize their energy, the molecules prefer to arrange themselves so that each one has only four neighbors. But if you take, say, balls of play-doh and try to arrange them in a lattice so that each one has four symmetrically spaced neighbors, you'll note that the resulting structure has lots of gaps: in a densely packed lattice each ball/molecule would have up to twelve neighbors. As ice melts, some of the hydrogen bonds break and bend, allowing the molecules to collapse into a denser, less orderly arrangement. —Ilmari Karonen (talk) 02:21, 4 September 2007 (UTC)

Imagine trying to pack strong bar magnets in a box. The magnetism keeps wanting to pull them into a single long string, which is not helping your trying to pack them into a nice tight rectangle. Same with H2O molecules; the peculiar shape with the hydrogens as little + bumps 90 degrees apart and therefore the - charge acting at the opposite end doesn't pack as well when the - and + line up nose to tail as when you can squoosh them together at whatever angle you like. Gzuckier 15:31, 4 September 2007 (UTC)

Roche limit

If a body be in an eliptic orbit that passes through the Roche limit of what it's orbiting, will it be torn apart while it's inside, but reform when it leaves? — Daniel 01:17, 4 September 2007 (UTC)

That depends on the fluidity of the body, the time it spends and the distance inside the Roche limit. For the vast majority of the cases, the answer is "no", but it is possible. The melting ice comet 73P/Schwassmann-Wachmann disintegrated without any tidal forces being responsible. ←Ben^B4 02:42, 4 September 2007 (UTC)

The friction during breakup will release heat, and thus energy, from the desintegrating body. And so my natural conclusion would be that it might not orbit out again. I'm not sure precisely what would happen, however. Someguy1221 02:50, 4 September 2007 (UTC)

As far as I can tell, that comet didn't reform, so I don't think it has much to do with it. Any estimate on how long an orbit like the one I'm suggesting could last? Do planets or satalites ever have that eccentric of orbits? — Daniel 03:04, 4 September 2007 (UTC)

Maybe this issue is precisely why planets and satelites don't have such eccentric orbits? Someguy1221 03:15, 4 September 2007 (UTC)

No, I think that's because planets formed about where they are, in roughly circular orbits, from the protoplanetary disk around the early Sun. They are also massive, so it requires a close pass by another major object to force them into highly eccentric orbits. This also has to do with why Pluto was excluded from the list of planets, as it seems to have a different origin. StuRat 12:09, 4 September 2007 (UTC)

Biking in the Rain

I commute on a bike to school every day, a few miles over rough sidewalk and across some heavily-traveled streets. If it rains, what should I do to keep myself and my books dry, and keep from getting hurt? (If this counts as "medical advice" or something similar, expert references would be just as good.) Black Carrot 02:33, 4 September 2007 (UTC)

mud guards on the wheels are especially helpful. It stops a muddy line being sparayed onto your back. Put your books into a waterproof bag. Wear a raincoat witha hood - bright yellow is good for visibility. But be aware that sound is blocked a bit from your ears, so it is harder to hear, and tell which direction the sound is coming from. Try to avoid riding in the dark as it's much harder to see in wet dark than dry dark. A bike rack on the back is good. Put something waterproof on the bottom and clip your bag on the top. How do you carry your books normally? Graeme Bartlett 02:55, 4 September 2007 (UTC)

It's best if the raincoat is a breathable fabric, (like GoreTex). I've seen some people using kayaking bags if it's particularly wet (they're waterproof and the opening folds over a couple of times to keep the rain out). Even if you use a waterproof bag, an extra plastic bag can't hurt. Use a flashing red rear light to make you more visible. And bring dry shoes and socks. Nothing you can do will stop your feet from getting soaked! Flyguy649 ^talk _contribs 04:20, 4 September 2007 (UTC)

Please check with your local law-enforcement agencies before attaching a flashing light to your bicycle, as it may be illegal. DuncanHill 18:19, 4 September 2007 (UTC)

One prob you might have is that it doesn't look like rain, so you don't wear rain gear, but it rains anyway. I suggest one of those compact plastic ponchos stuffed in your bag (which should be waterproof), for days when you guess wrong. StuRat 11:52, 4 September 2007 (UTC)

If you wear a helmet, pack a small towel at the bottom of your bag; doesn't take up much room and always nice to have dry hair / face afterwards. Nothing worse than trying to dry your face with those damn green paper towels. Lanfear's Bane —Preceding unsigned comment added by Lanfear's Bane (talk • contribs) 14:51, 4 September 2007 (UTC)

I modified my rainy day bike inspired by pictures of European commuting bikes. Besides mudguard/fenders, I stretched plastic sheet over the fender stays (the sort of coathanger wire things that hold the fenders on one end and attach to the bike frame or fork on the other) and the frame and fork to cover most of the sides of the wheel. It helped. Gzuckier 15:35, 4 September 2007 (UTC)

I did this for a few years, commuting several miles along heavily trafficked roads. I came up with a bunch of things that help. You can place your books in a plastic bag inside your backpack - surprisingly, the worst rain wetness always affected the bottoms of my books, not the top. Presumably this was from road splash. Avoid the obvious large puddles, they make you more wet. Wear a poncho, if you want, but it doesn't help a whole lot in heavy rain. I found that despite all of my efforts, I was always soaked through by the end of the bike ride, if the rain was heavy. In this case, I carried a pair of socks (several, actually) - because if only one part of me is dry, I prefer it to be my socks. Some people go all out and carry an entire change of clothes, but I never found that to be particularly helpful. Finally, you can use this opportunity to develop a philosophical outlook which anticipates catastrophic soaking, and adapts accordingly. This philosophy can be applied in all aspects of life, and creates a disaster-tolerant, resilient world-view. Nimur 15:54, 4 September 2007 (UTC)

never heard of a biking cape?--88.110.3.68 03:16, 5 September 2007 (UTC)

I usually carry my books in my backpack. And I wear sandals, so I won't need socks unless it gets a lot colder. So, I think the main points have been:

• Put my books in a waterproof bag attached to the bike
• Include emergency equipment and rain gear
• Get lights, and avoid the dark
• Wear a hooded jacket (preferably of an athletic material)
• Get things to cover my wheels so they don't splash

Looking good. What do I do about my pants? How will attaching a bag of books and stuff to my bike affect its handling? What should I do about my chain and gears? Does water effect my bike's braking ability, since the entire mechanism is rubber pushing on rubber really hard? Black Carrot 05:37, 5 September 2007 (UTC)

They do make rain gear that includes overpants. Alternatively you can just carry dry pants and underwear to change into. Overshoes or boots are also a good idea. The book bag shouldn't have much affect on handling as long as it's somewhat near the center of gravity of the bike, and a rack above the rear wheel is usually pretty good for this. A basket in front of the handlebars, on the other hand, is a really bad idea for heavy items. The chain and gears work fine while wet, but should be covered to limit spray. You will need to oil the chain and gears more often if the bike is used, or even left out, in the rain. The rubber brakes should work reasonably well in rain, and bike tires aren't particularly prone to sliding on wet pavement or hydroplaning, but caution (driving conservatively) is still a good idea. Puddles can be a problem in many ways, though:

1) Hitting a deep puddle at high speed will make a huge splash of muddy water that will drench you.

2) It's also like hitting the brakes, and you may fall forward if unprepared.

3) There can be hidden obstacles in the puddle, like nails or potholes.

Therefore, try to avoid going through puddles, or slow way down if you must go through them. StuRat 16:10, 5 September 2007 (UTC)

Also, big knobby mountain bike tires are likely to handle mud and water better than thin racing tires. StuRat 16:18, 5 September 2007 (UTC)

In addition, make sure the part of your water bottle that touches your mouth when you drink can't get muddy, or you could get a disease from drinking it. You might need to keep it in a plastic bag. StuRat 16:22, 5 September 2007 (UTC)

I commuted by bike for many years in Seattle, a city known for its rain. For the books, I suggest waterproof rear Panniers attached a back rack on your bike. A pair like these is what I use--look for a roll-down top (rather than a zipper or buckle), and all waterproof material (rather than waterproof fabric which fits over the bag for rain). As far as handling, for light-weights I hardly noticed the bags were there, and even the heaviest (sometimes 40-50 lbs, 18-23kg) was manageable. I would also recommend finding a rack which attaches to your frame rather than your seat post.

For clothing, I found it was best to shower and change cloths completely when arriving at work. That way I could ride hard through anything without worry. This may not be an option, but it is great if you can do it. Regardless, it is a good idea to waterproof (as water gets really cold when biking). What I found worked best was a thin waterproof (bright yellow) jacket over any insulating layers. If it was warm-ish, I wore ever popular spandex (Lycra) shorts; cold weather I had waterproof warm-up pants, again over something to keep warmer. If it were very cold (below freezing) I might wear a bandanna to cover my ears, but otherwise made no effort to cover my head with anything other than a helmet (with or without helmet cover).

Lights are a good idea, and may be legally required where you live. Check your brakes as well; They may take more work to stop in the rain (but tightening them and/or getting new brake pads goes a long way). Hope this helps. --TeaDrinker 20:34, 5 September 2007 (UTC)

As a car driver I can offer the following advice to minimise your risks (based on UK conditions)

• LIGHTS - I cannot stress this enough. Bright clean lights make it MUCH easier for us to see you in poor visibility.
• Reflective/Flourescent Clothing - Reflective clothing reflects our headlight light back at as, and flourescent glows in the dark - you can get combined garments.
• Wear light clothing - Don't wear black - we can't see it!
• Wear a helmet - this can save your life
• Keep your bike properly mantained - bald tyres or worn brake blocks will not help!
• Obey the rules of the road - Jumping red lights, cycling on the wrong side of the road or the wrong way up one-way streets are good ways to get into accidents.
• Be aware you are more difficult to see than a car - never assume a driver has seen you.
• Don't cycle up the inside of vehicles - they may change lanes or turn into you.

Exxolon 00:01, 6 September 2007 (UTC)

I've gotten a plastic jacket-and-pants set (for some reason, Wal-Mart had a wide selection and the local bike shop had none), and I'll get the rest as soon as I can. Thank you very much for your help. Black Carrot 17:46, 7 September 2007 (UTC)

needing to know what kind of spider i have found.????

i live in n.e. texas and was hunting the other day when i came across several spiders within about 2 acres. they all are black with white spots on them and around3/4 inch round and all had the same (crab shell back) on them. i have never seen these before and i cannot seem to find them on the internet. can someone please tell me what kind of a spider it was and also if it is dangerous or not.

thanks..masonstorm41102 —Preceding unsigned comment added by 4.90.32.54 (talk) 05:01, 4 September 2007 (UTC)

Can you post a pic ? StuRat 11:45, 4 September 2007 (UTC)

My first guess is a type of orb-weaver spider though. They're fairly common, and I know some have patterns like you mentioned, and fit the size you mentioned. Did it look something like this or this? --Wirbelwindヴィルヴェルヴィント (talk) 00:59, 5 September 2007 (UTC)

By "crab shell back", what do you mean? Do you mean it has spiny bits thrusting up along the edges? Does it look a bit like this [1]? If not, search on google for spiders of texas and see if you can find a picture of it. Black Carrot 05:41, 5 September 2007 (UTC)

I think you will find your answer here: http://entomology.uark.edu/museum/steatoda.html Looks pretty similiar to me. —Preceding unsigned comment added by 70.246.85.49 (talk) 20:43, 7 September 2007 (UTC)

Kinematics formula

How did the formula $v_{f}^{2}=v_{i}^{2}+2a(x_{f}-x_{i})$ come to be? How can you explain it in terms of integrals or visually using graphs? --antilived^{T | C | G} 05:17, 4 September 2007 (UTC)

To arrive at $\,x_{f}-x_{i}=v_{i}t+{\frac {1}{2}}at^{2}$ integrate acceleration with respect to time twice. x_i and v_i arise as the constants of integration. Further, knowing that uniform acceleration is defined as $a={\frac {v_{f}-v_{i}}{t}}$, and thus $t={\frac {v_{f}-v_{i}}{a}}$, this formula can be put in place of time in $\,x_{f}-x_{i}=v_{i}t+{\frac {1}{2}}at^{2}$. Simple algebra thereafter yields $v_{f}^{2}=v_{i}^{2}+2a(x_{f}-x_{i})$. Someguy1221 06:18, 4 September 2007 (UTC)

I understand $\,x_{f}-x_{i}=v_{i}t+{\frac {1}{2}}at^{2}$, but it's the steps after that to get that formula that I'm not quite sure of... --antilived^{T | C | G} 06:26, 4 September 2007 (UTC)

Well, once you substitute the new equation in for t, you get $x_{f}-x_{i}={\frac {v_{i}(v_{f}-v_{i})}{a}}+{\frac {1}{2}}a{\frac {(v_{f}-v_{i})^{2}}{a^{2}}}$, which (after multiplying each side by a) expands to $a(x_{f}-x_{i})=v_{i}v_{f}-v_{i}^{2}+{\frac {1}{2}}v_{f}^{2}-v_{i}v_{f}+{\frac {1}{2}}v_{i}^{2}$. This simplifies to $a(x_{f}-x_{i})={\frac {1}{2}}v_{f}^{2}-{\frac {1}{2}}v_{i}^{2}$. Hope that helps. Someguy1221 06:41, 4 September 2007 (UTC)

You could also note that the distance traveled is equal to the average velocity times the time taken: $x_{f}-x_{i}=\left({\frac {v_{i}+v_{f}}{2}}\right)\left({\frac {v_{f}-v_{i}}{a}}\right)$. Then multiply both sides by $2a$. -- BenRG 09:58, 4 September 2007 (UTC)

softwood shipping markings

can anyone help me I need a full list of wood shipping markings, the ones that are printed on each piece of wood with country , and the grades of wood.

Thank you

jasimps —Preceding unsigned comment added by 91.105.62.225 (talk) 06:11, 4 September 2007 (UTC)

Didn't you already ask this elsewhere ? StuRat 11:44, 4 September 2007 (UTC)

I think I saw it recently too, but I don't think it got much of an answer, which may be why it's been asked again (with seemingly the same response again. --jjron 07:09, 5 September 2007 (UTC)

Sutherland's constant

I need to know the parameters in the Sutherland equation to calculate the viscosity of methane. I've trawled the internet, and the textbooks I have, and can't find them anywhere. Does anyone know where I can get them? 84.12.252.210 08:15, 4 September 2007 (UTC)

• The book "Flow of fluids through valves, fittings, and pipe" should have the parameters, and a decent university library should have that book. --Sean 16:50, 4 September 2007 (UTC)

That's great, thanks. Hopefully I'll manage to get to a library fairly soon... 84.12.252.210 08:03, 5 September 2007 (UTC)

Charon ?

See [2].

Is Charon, the largest moon of Pluto, also visible in this discovery photo of Pluto ? I see a dim dot at approximately the 2:00 position from the left photo of Pluto (shown by the arrow). I don't see that dim dot on the right pic (either near Pluto's original or new location), possibly because Charon is in front or behind Pluto in that pic. I'd like to label the caption accordingly, if it is Charon. StuRat 12:29, 4 September 2007 (UTC)

I seriously doubt that is Charon. However, it is 19,570 km from Pluto. Pluto is 2,390 km wide. So, if you consider the "dot" of Pluto to be a "pixel". The "dot" of Charon would be 8 pixels away. That is about how far away that faint dot is. The reason I doubt it is Charon is because it took so long to discover Charon - so wouldn't they have noticed it right away if it was on the discovery photos? Of course, they may have simply overlooked it. -- kainaw™ 13:26, 4 September 2007 (UTC)

How does the brightness relative to Pluto fit with the pic ? Also, do we know if there is a star or other object in that location ? StuRat 13:39, 4 September 2007 (UTC)

I'm not sure how you can calculate how many pixels that Charon should be away from Pluto on the image without a scale being there. You can't just assume that Pluto is 1 pixel wide. After all, many stars in the image appear much wider than 1 pixel wide, but in terms of angular diameter, even the close-by stars are going to be around a few milliarcseconds across. That's approximately the same as, say, a person's thumb width viewed from something like 300 miles away - and that's for the close-by stars. The maximum distance from Pluto to Charon is less than an arcsecond. I can't identify which particular stars are in the image at the moment (but they're in Gemini, near to Wasat) - so can't tell the scale of the picture - but I highly doubt if the resolution is good enough to show two faint bodies as separate objects. Most ground-based telescopes need adaptive optics to split Pluto and Charon. Richard B 15:15, 4 September 2007 (UTC)

In a plate like that, you can't easily tell the relative brightnesses. Also, the bloom on the plate means that you can't tell size due to the size of the planetary disk from the bloom caused by the brightness of the object. Pluto would have been a LOT smaller than "one pixel" on that plate. So I don't buy [[User:Kainaw's "8 pixel" argument. Charon is also a lot dimmer than Pluto - it has half the diameter (ie 1/4th the area) and a 25% lower albedo - so it should be about ten times dimmer than Pluto - those two dots look to be about the same brightness to me. But when Clyde Tombaugh discovered Pluto in 1930, he was using a 'blink comparator'. If anything visible besides Pluto was moving within the frame, he'd have seen it immediately and Charon would have been discovered on the same day as Pluto because they'd both 'blink' together. When the moon was discovered in 1978, the did a search back through old photos of Pluto to see if a moon had been visible in earlier photos - and the oldest picture they could find that showed any sign of Charon was from 1965. That photo is described as showing a 'bulge' in the image of Pluto that comes and goes over time - the bulge was therefore determined to be Charon and the effect of it's orbit causing the bluge to come and go. This means that with the telescopic magnifications available even as late as 1965, Charon wasn't even visible as a separate dot. With the technology of the 1930's there is no way that a 10x dimmer object could have been visible on the plate! So, no - the image in that plate most certainly isn't Charon - and User:StuRat would be wrong in changing the caption. SteveBaker 15:27, 4 September 2007 (UTC)

I don't think you're looking at the right object if you say they are about the same brightness. If you continue past the dim object in the 2:00 direction you will see a brighter object that is constant in both pics. That's probably the one you are looking at. Try to make out the dimmer object between Pluto and that object in the left frame. Also, were huge improvements in optical telescopes made between 1930 and 1965 ? StuRat 17:12, 4 September 2007 (UTC)

That's definitely not Charon. Here are some of the discovery images for Charon; it was only visible as an elongation of Pluto in 1978 [3]. The January 23 plate is more exposed than the January 29 plate, so the dim star you're seeing probably just didn't come through in the second image. Those plates were taken with a 13-inch telescope on the ground. Here's what Pluto + 3 moons looked like from the 94-inch Hubble Space Telescope in 2006: Pluto system 2006. The 1978 images were from a 61-inch telescope. So there's no way that Tombaugh could have resolved Charon so well with a 13-inch telescope in 1930. But let's do the math. Pluto doesn't move very fast, so Tombaugh was observing the motion of Pluto because of parallax as the Earth moved. The parallax over six days, at opposition, would be about (2*pi) * (1 au / 30 au) * (6 days / 1 year) = 0.0034 radians = 0.2 degrees. The max elongation of Charon from Pluto is (20,000 km / 30 au) = 4.5×10^-6 radians = 0.00026 degrees, or 1/800 the distance that Pluto traveled from one plate to the other. From the image, I get that Pluto moved about 177 pixels, and your small dot is about 12 pixels from Pluto, a ratio of 1:15 instead of 1:800. So the real Charon is about 50× closer to Pluto than that small dot is. --Reuben 19:10, 4 September 2007 (UTC)

File:Analysis of Pluto discovery plates.png

This image is a crop of the result of overlaying the two images (red==righthand side, green=lefthandside). As you can see, they don't match up exactly. The dot pointed at by the green arrow is Pluto when the left photo was taken, the dot pointed at by the red arrow is where it was when the righthand photo was taken. Yellow areas are where nothing moved between the two photos - which (theoretically) would be everything except Pluto (and maybe, Charon)...except that they don't line up very well because the plates simply aren't that exactly registered - so everything on the left of the image has the red dot slightly to the left of the corresponding green dots - and everywhere on the right of the image, it's the other way around. In addition to Pluto, you can see (as StuRat claims) a faint green dot with no matching red dot off to the left of it - which I marked with a blue arrow. So - is this Charon? Well, there is no matching red-dot up by the red arrow - but I guess StuRat is going to claim that Charon moved either in front or behind Pluto. Sadly, that argument doesn't hold water. Charon's orbit takes 6 days, 9 hours - and those two photos were taken 6 days apart...so Charon ought to be in the same place (relative to Pluto) in both photos...and it's not. Sadly, the red image has a lot fewer stars in it than the green one you can see lots of green stars up above the green arrow for example so this putative 'Charon' is not the only example of a green dot that vanished in the red image. These could be due to clearer 'seeing' in the green image than in the red one - it's hard to tell. If you ramp up the gain on this image, you can actually see very faint red images of those stars - but you can't see even a trace of "Charon" to the top-left of the red arrow, no matter how much you boost the image. It's definitely not Charon. SteveBaker 19:35, 4 September 2007 (UTC)

Looks like some good evidence there, guys. Steve, could you show us that pic with the red cranked up ? I think I saw traces of most of the left plate's dim spots on the right plate, although right at the limit of perception. For the one I thought might be Charon, however, I saw no trace at all. As for it not being in front or behind on the right plate, it seems we only need to vary the orbital speed of Charon, the starting location, or the time between the plates (was it exactly 6 days or a few hours off ?) slightly to get that result (say 3/4 of an orbit). Charon being 1/50th as far away as it appears is more convincing, unless there is a major mistake in the figures. StuRat 20:33, 4 September 2007 (UTC)

Tombaugh's 1980 book (with Patrick Moore) Out of the Darkness includes a photo identified as "the best image" of Pluto then available, taken from a 155 cm plate. As enlarged for the book, the planet shows as a 6 cm wide disk -- but that's not its real diameter, just the telescope's circle of confusion. Charon is visible as a bulge causing one side of the disk to protrude an extra 1 cm.

The book doesn't mention what time of day the two discovery plates were taken. It does mention that the two images of Pluto were 3.5 mm apart, the size of the plates being 14x17 inches (356x432 mm). Each plate showed hundreds of thousands of stars. When blinking, Tombaugh systematically viewed one 10x20 mm area of the combined plates at a time until he had covered the entire plates. --Anonymous, 23:18 UTC, September 4, 2007.

Look: Firstly, Reuben's argument is pretty convincing there simply isn't enough angular resolution in those photos. Secondly, the historical record shows that even with 1965 technology, Charon didn't show a separate disk - and FOR SURE, the guys who first found Charon would have looked back at the original 1930's plates to see if it were there. If they didn't see Charon in them then we won't either. Thirdly (and, for me, most telling) we know that Tombaugh used a blink comparator on those two photos - which means that if Charon were really that visible it would have 'blinked' along with Pluto (because of the coincidence of there being 6 days between the photos and Charon's orbit being close to 6 days). So if Charon were visible in those plates, he would have discovered both moon and planet at the same time - and it took another 48 years of advances in telescope design for that to happen. Remember too that he was looking at the original plates so it would have been blindingly obvious if we could really see it in our scans of copies of those plates. So give it up. For sure that's not Charon - let's just end this pointless debate. SteveBaker 23:31, 4 September 2007 (UTC)

I've just managed to find the area in question. It's just over half a degree away from Wasat, and the spiral fuzzy thing in the top of the image is the galaxy NGC 2365. The brightest star in the image is around magnitude 8.5. Pluto moves just over 7 arcminutes during the 6 days. If the 177 pixels is accurate, then the scale of the image would mean that Charon would lie around 1/3 of a pixel away from Pluto at maximum separation, rather than the 12 pixels that the suspect image does. If you're still not convinced, then wikisky.org shows a star at that precise location: USNOA2 1050-05082251. It's at around 16th magnitude. See here Richard B 00:37, 5 September 2007 (UTC)

Nice work, Richard. I found Charon's absolute magnitude listed as 0.9, that can't be the same type of magnitude, can it ? I also found another mystery. The fairly bright object in the upper, left corner of that cropped image provided by Steve above only shows in the left plate, and doesn't show up at Wikisky, either, even though far dimmer stars do show up there. StuRat 05:56, 5 September 2007 (UTC)

I don't know what it is, but I would have thought (but can't check until later this evening at least) that it's probably just an asteroid - particularly as it's close to the ecliptic. If it's a main belt asteroid, then its movement could usually be detected on photographs after only a few minutes, and the angular distance that Pluto appeared to move over those 6 days could be completed in a few hours - so you would expect that the asteroid wouldn't appear on the 2nd plate 6 days later.Richard B 11:28, 5 September 2007 (UTC)

<edit>Commenting on the brightness of the object, the USNO catalogue magnitude is 16.2 for that star, the absolute magnitude of Pluto is around -1.0, and of Charon 0.9, but at around 42 or so AU from the Sun, the apparent magnitude of Pluto was about 15.1, and Charon around 17.0. Mag 17.0 is around twice as faint as mag 16.2. Richard B 13:22, 5 September 2007 (UTC)

Thanks (so I wasn't all that far off as far as the brightness goes). StuRat 15:24, 5 September 2007 (UTC)

Hi Steve, I don't think our anonymous friend was arguing that the dot was Charon. He's just providing some supplementary information of interest. --Reuben 01:59, 5 September 2007 (UTC)

Yes - I know - I was replying to the previous message from StuRat. There was something of an edit conflict. I fixed my indent level to try to make that clearer. SteveBaker 02:47, 5 September 2007 (UTC)

Thanks, Reuben. I should perhaps have said explicitly that if the best image available in 1980 didn't come close to showing Charon as a separate dot, then obviously that a 1930 image from a smaller telescope would not. The other information was indeed intended just for interest. --Anon, 03:23 UTC, September 5.

I'm not arguing either, I just wanted to see if I was right in saying that there is no trace of that dot in the right plate, while there are traces of other dots, which appear just as dim in the left plate. I accept most of the arguments given and no longer suspect that dot of being Charon. Thanks for your help, everyone ! StuRat 05:40, 5 September 2007 (UTC)

Psychologist >>> Physician

Can a Psychologist become a Physician? (in the UK or US) Or should he/she study the whole career of medicine from the beginning? —Preceding unsigned comment added by 85.57.21.17 (talk) 17:25, 4 September 2007 (UTC)

Our article Medical school in the United Kingdom may help, however it would be advisable to contact medical shools directly to ask, as their response would depend on the precise nature of the existing qualifications. Generally, however, there are very few "graduate entry" places for medical training in the UK, and the great majority of doctors (we rarely call them physicians) will have undertaken the complete undergraduate route. I have no knowledge of the system in the USA. DuncanHill 18:07, 4 September 2007 (UTC)

I would guess anyone would have to go to the beginning of medical school to become a doctor. It's different from the UK as the above poster puts it about the "undergraduate route". In the US anyone can major in anything as an undergrad to get into med school, provided that you took the classes that the schools look for.128.163.224.222 20:13, 4 September 2007 (UTC)

There are doubtless a few graduate psychologists (in the U.S. you should have at least a Masters degree to call yourself a psychologist) who took scientific courses in their undergrad careers rather than liberal arts or psychology. Someone, for instance, who passed a premed program as an undergrad, could have then, for whatever reason, pursued a grad degree in psychology, then had second thoughts and applied to med school, and been able to go on in his education and finish the medical degree. If he had not taken the rigorous science courses in premed, he might have to take two years or more of refresher or catchup courses before being able to gain admission to med school. Some med schools have a more lax admission policy than others. Hardly any undergrad or grad courses in psychology would substitute directly for med courses. All that said, yes, a psychologist can become a physician, as can anyone else, if they are smart, willing to work hard, and have many years and hundreds of thousands of dollars to devote to the effort. Edison 03:28, 5 September 2007 (UTC)

You probably would want to read up on the differences between Psychologists versus Psychiatrists. You can get a PhD in psychology or you can go to med school and do a Psychiatry residency and fellowship. If you are already a psychologist, you would have to apply to med school to become an MD.
Mrdeath5493 03:55, 6 September 2007 (UTC)

Frequency

Does two-hourly mean every 2 hours or every half an hour? (Not talking about ound or anything like that). 20:59, 4 September 2007 (UTC)

I would actually say that it is ambiguous. The prefix "bi" (which most literally means "two") for instance, can mean "every two" or "twice every." Someguy1221 21:05, 4 September 2007 (UTC)

TO mean it would be every two hours, as opposed to half hourly. But what did the author mean? Graeme Bartlett 21:40, 4 September 2007 (UTC)

I've rarely if ever seen e.g. "bimonthly" to mean anything other than every two months. —Tamfang 01:51, 5 September 2007 (UTC)

I think this post belongs on the language ref desk - it's hardly a matter of science. SteveBaker 22:07, 4 September 2007 (UTC)

It could be a matter of life-and-death if we're talking about dosages!! Wikipedia cannot give medical advise! If this is indeed a dose question, please contact a doctor! --Mdwyer 00:53, 5 September 2007 (UTC)

Actually although this will depend on where you live a pharmacist may be a better choice although both should either be able to answer your question or direct you to the person who can Nil Einne 15:03, 7 September 2007 (UTC)

Since hourly means every hour, I would take two-hourly to mean every two-hour, er, two hours. It's not a common usage. The other would be twice hourly. —Tamfang 01:51, 5 September 2007 (UTC)

I'd suggest the answer is in the hyphen. "Two-hourly" would suggest every two hours; "two hourly" would suggest two per hour. It may be nice to have some more context though. --jjron 07:06, 5 September 2007 (UTC)

Train service and [4] 22:37, 5 September 2007 (UTC)

I'm pretty sure it means once every two hours, but it'd be easiest to look at a schedule if you can get one. Black Carrot 17:53, 7 September 2007 (UTC)

meaning of convected in vortex dynamics

If an experiment claims that a vortex convects a pressure difference of 0.4.05 atmospheres downstream, is this the amount of pressure it looses along the way or how much it delivers upon impact. Thanks for your time Robin —Preceding unsigned comment added by 80.6.35.136 (talk) 21:51, 4 September 2007 (UTC)

I'm not sure I really understand the situation (0.4.05 atm?), but given the use of the term 'pressure difference' I would take an educated guess that you're looking at differences, and therefore how much pressure it looses. --jjron 07:01, 5 September 2007 (UTC)

Weather Systems?

I realize that this question in general, is incredibly broad, but where do weather systems originate?

What causes the changes in atmospheric pressure? Where do cold and hot fronts start? Is there a factor that causes these weather factors to change?

Thanks in advance,

Perfect Proposal ^{Speak out loud!} 23:10, 4 September 2007 (UTC)

Well, the sun shines on ocean water, ice, grasslands, desert, forest, etc. Each of those reflects sunlight back to a different degree and absorbs the rest. Hence (for example) desert sand gets really hot in the sunshine - but the nearby grasslands (say) do not. The air above the desert gets some of that heat and expands - the air over the grassland stays cool and does not expand. Now, the density of the air over the desert is less than that over the grass - so "hot air rises" - so the hot air goes up and is replaced by cool air from the nearby grasslands. Now you have a wind blowing from the cooler area to the hotter area. If the heat is over water, then some of the water will evaporate - cooling the ocean (and hence the air) and putting water vapor up there. The water vapor forms clouds - which being white, reflect the suns heat back out into space - further cooling the oceans beneath. Multiply all of these increasingly complicated effects by the size of an entire planet - and you get winds and clouds and all sorts of other complicated effects. Air doesn't move simply - it make swirls and vortices as it passes over mountains or as a cool 'wall' of air hits a warmer wall coming the other way.

Air pressure varies because (as I said before) warm air is lighter than cool air - so when warm air from a nearby desert (or whatever) blows past your home - the air pressure drops. When cool air from the snowy north blows your way, it's denser - heavier - and therefore (in general) exerts a higher pressure. 'Fronts' are places where the air temperature changes suddenly - as at the boundary between grassland and desert - or ocean and forest - or whatever. They move because of the 'hot air rises and cold air flows beneath it' - but because this is an insanely complicated system of whorls and vortices on a planet-wide scale, it's never that simple. Fronts are important because if a mass of warm, wet air is pushed upwards and cools off abruptly, it cannot hold as much water vapor - so it comes down as rain or snow. If that upwelling of air causes sufficiently violent turbulance then clouds can get charges of static electricity and you'll have lightning - or little vortices can be spun off causing tornadoes. It's all to do with the motion of the air - the transport of heat and water vapor.

The factors that cause the weather to change are the more or less random ones caused by all of this complex 'chaotic' (in the mathematical sense) motion - and of course things like the day/night cycle (everything cools off at night and heats up during the day - so again, air changes density and winds blow) - the seasons...everything affects everything else. So the weather changes - and on a small scale, it's very random indeed. Nobody can predict whether there will be a gust of wind where you are right now at 2:43pm tomorrow. On a medium scale, it can be predicted (Will it rain tomorrow? We can make a pretty good guess.) - and on a very large scale, it's completely predictable. (Will it be hotter in summer in Texas than in winter? Will it snow in Alaska this year?)

SteveBaker 00:35, 5 September 2007 (UTC)

Just a nitpick, but air doesn't "hold" a different amount of water vapour at different temperatures; gasses operate independently from one another. As the temperature drops in a given area, there will be a net increase in condensation, resulting in cloud formation. Matt Deres 16:25, 10 September 2007 (UTC)