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November 17

scenario - the best type of telescope?

Hi. Let's say a person wants to buy a telescope. I'll list the scenario. What type of scenario is best for said person? I'm asking for approximate diameter of mirror/lens, type of telescope (eg. Achromatic refractor, Apochromatic refractor, Newtonian reflector, Dobsonian reflector, Schmidt-Cassegrain, etc), approximate weight, approximate cost (CAD), approximate f/ratio, approximate limiting magnitude, etc. I'm not asking for brand name, location of where to buy, exact price, exact objective diametre, exact parameters, etc. Here's the scenario. Said person:

is not buying from a department store
lives in a light polluted reigon
will be observing from said person's residence
lives in an area of approximately 2000 people per sq km
lives in an area where most winter, and a few autumn and spring nights drop below freezing
is especially interested in planets, comets, and deep-sky objects

Said telescope:

should not cost over $300 (CAD, excl. tax)
should include eyepieces and accesories
should not weigh more than 20 pounds
should be able to be easily carried through a 1-m wide corridor
should not take up more than 1 sq m or total area space
should not break on the first use
should perform better than a 2 in department store refractor
should be easy to have finderscope accurately positioned
should not take more than 30 mins to locate a mag. 5 object
should not require constant maintenance
should not completely fog up under 0C in less than 5 mins
should not take more than 10 mins to set up or carry back to its storage area
should have a warranty lasting more than 3 months
should not have an owners' manual of more than 25 pg long
should not completely go out of focus due to a very slight accidental tap
should not have an extremely wobbly mount
should have eyepieces capable of at least 70x and not go over its maximum theoretical magnifacation
should have a limiting magnitude of better than mag. +8
should not capture too much light pollution when observing within 60 degrees of a streetlight
should have a maximum FOV of at least 10 arcmins
should actually work.

Sorry if my parameters wer a little too exact, but I only want vauge answers. If nessecary, please indicate some good brands. Thanks. ~A H 1^(TCU) 00:11, 17 November 2007 (UTC)[reply]

You've mixed in technical specifications with some rather arbitrary looking manufacturing specifications (length of manual? come on, is that really going to be your limiting factor between one and another? my washing machine has a longer manual than that and that doesn't mean very much—I only flip through it when something is not operating right). I really doubt anyone on here is going to have that much knowledge about specific telescopes to be helpful; you'd probably get a better response by just saying you're interested in a relatively cheap telescope that isn't too hard to use and will let you see some neat stuff in the sky. You also seem to have very exacting standards for a telescope that you don't want to cost very much at all—that's usually not how things work with precision instruments. Even if you came to me with questions like that about something I knew more about (e.g. buying a new laptop) it would take me AGES to check through the stats of any given option to see if it really lined up with your desire. Better to ask more generally and then check through the stats yourself on whatever people offer up. --24.147.86.187 (talk) 19:18, 17 November 2007 (UTC)[reply]

Here's my advice. Get the biggest brand name (not Walmart Christmas special) reflecting telescope you can afford and most of the other requirements on your list will be under control. I suggest you buy a Celestron AstroMaster 130EQ. Most of your requirements are within 10% or so. Also, download Celestia for free to provide some 3D context to the objects you're looking at. Good luck! DeepSkyFrontier (talk) 19:39, 17 November 2007 (UTC)[reply]

Hi. Um, the problem with a 130mm reflector is, does it actually cost less than $300 dollars, including mount and eyepieces? Also, the scenario takes place in a light-polluted reigon. It would also be rather heavy, and might be hard to fit in the back of a car (eg. from the store). What is the best aperture for this scenario? I'd think that approximately 100mm would be suitable, what do you think? How did you decide on Celestron, 130mm, and reflector? Is it because you own one yourself? Thanks. ~A H 1^(TCU) 22:57, 17 November 2007 (UTC)[reply]

My advice is just advice. It costs slightly more than $300, but you may be able to find one for closer to $300 if you shop creatively. It weighs 24 lbs. You can use the internet to find more information on it if you have any other questions. If you are in a light polluted area you should get the strongest, widest aperture telescope possible. It's a matter of contrast. Dim stars seen through an orange haze is worse than bright stars seen through the same haze. I personally wouldn't buy this telescope, but I certainly wouldn't buy anything smaller either. Celestron makes good equipment. I hope you find what you're looking for. DeepSkyFrontier (talk) 08:12, 18 November 2007 (UTC)[reply]

Hi. I very much appreciate all this info, but there are a few things I'm still not clear about. I heard that very large telescopes are especially sensitive to heat waves, image shake, and can capture too much light pollution when viewing close to a streetlight. Now, one of my biggest concerns is cold air. When I bring a telescope outside on a cold day, the eyepiece fogs up, and the objective sometimes does too. When I bring it back inside, the whole instrument fogs up, and stays fogged for hours. When it's really cold I even have to worry about frost forming on my telescope. Is there a way to prevent exccesive fog from forming that isn't terribly expensive? At what aperture do atmospheric waves, light pollution, etc, really start to hamper the image quality? Finally, is a 130mm telescope fittable in the back of a car? By "car" I mean a small car, not a van or truck. Do Equatorial mounts allow you to locate a celestial object more easily? Do Celestron telescopes have the finderscope accurately aligned, so that there is no need for constant adjusting? Are the mounts steady enough so that one knock will not send the FOV far away from the intended target? Also, since this is new to me, do these telescopes require collimating? How do you collimate? Will adjusting the telescope to point from one area to another of sky cause the colimation to go off-focus? Do you collimate by inserting your hands into the tube, or by adjusting something from the back? Do EQ mounts need a source of electricity? Do telescopes in a store (excl. tax) usually sell for more or less than the price listed online? If a store has telescopes on display, where is the actual section where you find the telescopes you actually buy and take home? Will near-zenith objects need climbing on a chair, or can you lower the mount in a way that doesn't disrupt the EQ alignment? How accurate does the RA-Dec need to be to get the object in question into view withought constant adjusting? How long does a telescope like the one you mentioned usually takes to set up? Sorry if I'm asking a lot of questions, but this is mostly new to me. Thanks. ~A H 1^(TCU) 17:35, 18 November 2007 (UTC)[reply]

Sorry Dude, too many questions. I really don't know much about this telescope in particular. To prevent fogging, you just need to let the telescope reach the temperature you're going to use it at before hand. Put it outside in a safe place, for instance, for half an hour before use. It's the inside-outside trips that'll get you. Find some way to blow dry air (not your breath) onto the lens if fogging persists. Watch where you breathe. Fogging shouldn't be a serious issue on any night with good seeing anyway. Just don't use your telescope next to streetlights. Go to the backyard. If you can't see any stars with your naked eye, don't expect to see anything interesting with a telescope either. Best prices are usually found online, but the cost of shipping often cancels a lot of the advantage. I don't know what stores you have near where you live, so finding a good price is up to you. Camera shops have the best selections and the worst prices. You might get lucky because this is really an entry level scope and shops may sell it close to its suggested retail (or below, possibly) in order to earn a customer when you're ready to graduate to a $5k scope. Again, I have no idea what you're going to find. Do keep your eyes open on eBay though. I've used eBay for over ten years and I'll never understand why people don't trust it (if used carefully, that is). If the finderscope isn't aligned properly, you should be able to effect a durable fix in a matter of minutes. As for the rest of your questions, this is marketed as as simple, easy-to-use telescope. As long as you take care of it, it shouldn't ever require any dramatic adjustments. Some of your questions leave me feeling that you're testing my limits ;) Good luck. DeepSkyFrontier (talk) 19:20, 18 November 2007 (UTC)[reply]

CGPM news; possible redefinition of the kilogram

Does anyone know what's going on at the 23rd CGPM? It's supposed to be almost over, but I can't find any mention of news from it, or even something like a conference program. They could be redefining the kilogram as we speak! The BIPM web page is hardly helpful.

BTW, shouldn't this guy be wearing a mask or something? What if he sneezes on that thing? —Keenan Pepper 00:51, 17 November 2007 (UTC)[reply]

P.S. Which is a more perfect sphere, that thing, or one of the Gravity Probe B gyros? —Keenan Pepper 00:52, 17 November 2007 (UTC)[reply]

The kilogram is overdue for change. Firstly, it's the only remaining SI unit that's defined in terms of a single physical object - and that's not a good thing. Secondly, by comparing the standard kilogram to various copies, it's becoming obvious that the standard kilogram is slowly changing mass (by 30 micrograms already - that's A LOT!). The ideal definition should be in terms of some exact number of some particular kind of atom. Defining the kilogram this way is definitely something that needs to be done because if the mass of the standard kilogram is changing - and since that is the very definition of the kilogram - what it means is that all of our scientific measurements are slowly becoming less and less accurate - and that's a disaster! However, I hope that what the Australians are doing is building a new kilogram that WILL NOT be "the" standard - but merely an example of a mass containing the right number of atoms of whatever atom is finally chosen (probably silicon - and that photo looks a lot like a ball of silicon and it's about the right size to be a kilogram of silicon). This can then be compared to the mass of the present standard kilogram - simply to verify that the new and old standards agree to within some reasonable precision. SteveBaker (talk) 16:31, 17 November 2007 (UTC)[reply]

The problem with defining the kilogram as a number of atoms is that it's very hard to accurately measure the number of atoms in a solid. According to Wikipedia, the uncertainty in Avogadro's number is around 50 parts per billion, which is larger than the total drift in the standard kilogram (30 parts per billion). Even if the kilogram does get redefined in this way, I think it will be quite some time before synthesis of a kilogram from the new definition will be easier or more accurate than copying a physical artifact. -- BenRG (talk) 18:05, 17 November 2007 (UTC)[reply]

Defining X number of Y atoms to be exactly 1 kg is equivalent to creating an exact redefinition of Avogadro's number. And the point is not that it should be easy. Surely defining a second as a number of vibrations of a specific atom, or defining a meter as the distance light travels in specific fraction of a second are not easy definitions. Nearly all clocks and meter sticks are still based on comparisons to physical objects that were created based on those definitions. The point is to have a fixed definition that can be reproduced by indepedent experts at different places and times in order to ensure that the standards of measurement don't accidently change over time. In other words it is a way to ensure that a standard kilogram produced in 2010 means the same mass as one produced in 3010. Of course this is dependent on someone showing they can produce a kilogram standard today from a number of atoms definition (or similar) with an accuracy comparable to the present uncertainty in the standard. Dragons flight (talk) 18:54, 17 November 2007 (UTC)[reply]

But the point is that right now, the very definition of the kilogram is a moving target - this year, a kilogram is a different number of silicon atoms than it was last year. A definition of a firm number for Avagadro's number (and by implication defining the mass of the kilogram) creates a standard that is absolutely firm and unchanging for all time. Our ability to make a mass exactly equal to one kilogram becomes difficult - but we'll know exactly what the standard. I'd also argue with the 30 migrograms error - in truth, we have no idea what the error is. The only way to see if the kilogram standard is changing mass is to compare it to replica standard masses. But what if all of the kilogram standards are in error? Maybe the error in the replica masses is 100ug - but the official standard is 130ug off? How would we know? SteveBaker (talk) 20:23, 17 November 2007 (UTC)[reply]

Thanks, but none of your responses really answer my question. I already knew all that stuff. I just want to know what actually happened at the recent CGPM. —Keenan Pepper 20:12, 17 November 2007 (UTC)[reply]

You may want to see our article on the Watt balance. The Photon (talk) 05:11, 19 November 2007 (UTC)[reply]

Some very old technology

Does anyone know anything about this device: the "Heiss patent amalgamator for saving fine gold from sand, gravel and pulp." I can find nothing on line. Click through on picture if you want more context. - Jmabel | Talk 04:37, 17 November 2007 (UTC)[reply]

Also, if someone is expert on technology from this era (circa 1900), there may well be other pictures on Commons:Category:Seattle and the Orient you could help describe better: there are several hardware store interiors, etc., with many objects I could not identify. - Jmabel | Talk 04:42, 17 November 2007 (UTC)[reply]

Not an expert, but know a little about mining. The picture appears to be a machine used in the amalgam process of gold recovery. Finely crushed ore and mercury would be mixed together. The mercury forms an amalgam with any gold present, and the amalgam can be readily seperated from the tailings (waste material). Gold is then recovered from the amalgam by heating. DuncanHill (talk) 04:50, 17 November 2007 (UTC)[reply]

There were dozens and dozens of different designs for things like this in the early 20th century. I'm not sure that this one is anything special—it looks pretty standard. If you search for "amagamator gold sand" on Google patents you can find dozens of similar devices. --24.147.86.187 (talk) 05:29, 17 November 2007 (UTC)[reply]

physics

What happens to its Kinetic Energy when a freely falling object stops on reaching the ground ?

A small percentage is converted to sound energy, smaller percentage converted into heat energy, the rest is transffered into the earth.220.237.156.78 (talk) 09:45, 17 November 2007 (UTC)[reply]

Way wrong.

This is a completely inelastic collision, so the speed imparted to the Earth is given by the formula at Inelastic collision#Equations of Motion. To avoid subscripts say v is the object's initial speed and m is its mass; M is the Earth's mass and V is the speed imparted to the Earth by the collision, its initial speed being 0. Then we have mv+0 = (M+m)V, but since the object's mass is negligible compared to the Earth's, we can replace M+m by just M, and we have mv = MV. Therefore (mv)² = (MV)² and therefore MV² = (MV)²/M = (mv)²/M = (mv²)(m/M); you'll see in a moment why I did this.

Now, the initial kinetic energy of the object is mv²/2 and that of the Earth and object afterwards is (M+m)V²/2, which again we can simplify to MV²/2... but by the above calculation, this equals (mv²/2)(m/M), or m/M times the initial kinetic energy. In other words, only about m/M of the object's initial kinetic energy -- a tiny, tiny fraction -- becomes kinetic energy of the Earth.

In fact almost all of it goes into heat. --Anon, 10:10 UTC, November 17, 2007.

In addition, if the object breaks, then some of the kinetic energy is converted into surface energy, as a result of breaking chemical bonds within the object. Gandalf61 (talk) 10:58, 17 November 2007 (UTC)[reply]

Treating the Earth a single rigid body is not appropriate. Assuming nothing breaks at impact (i.e. you don't break the object, or leave a hole in the dirt), then the largest fraction will be acoustic energy in the Earth (e.g. pressure waves eminating from the point of impact). This eventually becomes heat, but not immediately. As the first anon said, most of the energy goes into the Earth. Dragons flight (talk) 17:41, 17 November 2007 (UTC)[reply]

The first anon contrasted "converted to sound energy" and "converted into heat energy" on the one hand with "transffered into the earth" on the other. Since the original question was about kinetic energy, it seems clear to me that "transferred into the Earth" meant that it becomes kinetic energy of the Earth. On that basis the original answer was wrong and that's how I was responding to it.

Also, since we're talking about an object that falls to the ground and stops, we are talking about either the ground or the object "breaking", or more generally deforming, on impact. I think most of the energy will go into that, which means it then goes directly into heat; but I must admit to not knowing how to calculate the amount that goes into sound waves within the Earth.

--Anon, 06:28 UTC, November 18, 2007.

Virtually all of the recoverable deformation of the ground goes into pressure waves. Normally you would not refer to something as "breaking" unless the deformation is permanent. The fact that something is briefly compressed is not the same as it being broken. Dragons flight (talk) 07:35, 18 November 2007 (UTC)[reply]

Agreed, but if the deformation/compression is temporary, the object won't stop dead on impact: it'll bounce. That's not the case we were asked about. --Anon, 22:40 UTC, Nov. 18.

Bear in mind that all energy is conserved, thus the landing object, say a ball, will undergo all the opposite effects of a falling ball. For example, the falling ball will tend to

A)Speed Up, B)Cool down, C)Lose Pressure and D) Expand.

We know this because the ball, when it lands, will tend to

A)Slow Down and Stop, B)Warm Up, C) Compress and D) Compact - opposite of 'Expand'.

In this way, all energy is conserved. Interestingly, the Cosmos seems to be falling, judging solely from the effects it's undergoing. Peter Lamont (November 17)

This makes no sense to me. Why do you think a "landing" object will "undergo all the opposite effects" to a "falling" object? "Falling" and "landing" are not time-reversals of each other. Also, it's obvious that falling objects speed up because of gravity, but the other three effects are news to me. What makes falling objects cool down? The answer is, they don't usually cool down, and if they do it's caused by some other effect like evaporation. This response is incorrect and I urge the original asker to ignore it. —Keenan Pepper 20:25, 17 November 2007 (UTC)[reply]

enerhiya

ano poh ba ang enerhiya? —Preceding unsigned comment added by 124.106.204.177 (talk • contribs) 10:15, 17 November 2007

ano ba yun. hydnjo talk 15:34, 17 November 2007 (UTC)[reply]

What language is this? —Keenan Pepper 20:31, 17 November 2007 (UTC)[reply]

I'm pretty sure it's Tagalog. MrRedact (talk) 20:55, 17 November 2007 (UTC)[reply]

GAA triplet repeats

Is it correct to wikilink "GAA triplet repeats" (found in Friedreich's ataxia) to Trinucleotide repeat disorders? Lova Falk (talk) 15:03, 17 November 2007 (UTC)[reply]

Yes, see this. --JWSchmidt (talk) 19:59, 17 November 2007 (UTC)[reply]

Thank you! Lova Falk (talk) 08:47, 18 November 2007 (UTC)[reply]

Waterboarding

Could waterboarding be used as a medical procedure against alcoholism and drug addiction? If not, why not? 71.100.6.233 (talk) 15:04, 17 November 2007 (UTC)[reply]

How about - "Because it's a really horrible, painful and terrifying torture technique" ? SteveBaker (talk) 15:58, 17 November 2007 (UTC)[reply]

Also, because it is probably illegal, at least under US law. In the past we have prosecuted water boarding as a war crime, and as a "Violation of the Laws and Customs of War". -- dcole (talk) 21:17, 17 November 2007 (UTC)[reply]

Why hasn't the US government then required that this be acknowledged by persons in authority, namely the new Attorney General? Isn't the failure to require such acknowledgment evidence of a double standard or is waterboarding an acceptable and legitimate means of persuasion when people do not agree or see things exactly your way? 71.100.6.233 (talk) 22:26, 17 November 2007 (UTC)[reply]

Probably because until now no one told suicide bombers how much more effective kidnapping, rape and waterboarding can be on the psyche. Taxa (talk) 10:41, 19 November 2007 (UTC)[reply]

I think I should point out that if somebody actually is a suicide bomber, then any attempt at waterboarding as a "means of persuasion" will almost certainly fail... 130.88.79.77 (talk) 14:52, 22 November 2007 (UTC)[reply]

Shards of flesh from a self-detonated explosion have psyche too. 199.76.152.229 (talk) 03:15, 24 November 2007 (UTC)[reply]

Friedreich's ataxia

In Friedreich's ataxia it says: "Median age of death is 35 years, while females have better prognosis with a 20-year survival of 100% as compared to 63% in men." I'm not a physician, but isn't that preposterous? Even healthy women don't have a 20-year survival of 100%. (I have googled, but I can't find information on the 20-year survival of women.) Lova Falk (talk) 15:50, 17 November 2007 (UTC)[reply]

I would assume that the intended meaning is that the disease doesn't kill any of the women in their first twenty years, not that none of them die. I agree that the wording is misleading. I'll see if I can find a source. TenOfAllTrades(talk) 15:55, 17 November 2007 (UTC)[reply]

Hmm. This article (PDF) (Klockgether et al. (1998) "The natural history of degenerative ataxia: a retrospective study in 466 patients" Brain 121(4):589-600) indicates that "...survival was not influenced by gender", though women tended to become reliant on walking aids or wheelchair-bound sooner than men. TenOfAllTrades(talk) 16:16, 17 November 2007 (UTC)[reply]

Nazi freeze/atomic bombs

I found an interesting old New York Times article the other day, dated January 14, 1945 (headline is Nazis talk less of new V weapons, for those of you with ProQuest access). An interesting excerpt:

Swedish scientists, who have a reputation of being extremely well informed and just about as wide awake as their German colleagues, are most skeptical of German claims and rumors.

Especially is this so in the cases of the so-called freeze and atomic bombs, the former numbered V-3 in Reich propaganda releases. About the freeze bomb Swedish scientists say its principle is well known.

The German method of freezing has been the subject of much detailed study by Swedish chemists. The Swedes have come to the conclusion that at present it is completely impossible to manufacture an effective freeze bomb weighing less than eighty tons. They admit the possibility, however, that the Germans have been able to combine extreme cold and explosive effects in a somewhat lighter projectile.

Regarding the atomic bomb, Profs. Theodor Svedberg and Georg von Hevesy, both Nobel prize winners in physics and atom experts, agree that such a bomb is very far from realization and that there are scientific reasons to believe that the engine will prove a complete "flop" if one day in the distant future it is produce.

No other discussion of these particular weapons is in the rest of the article. All jokes about their great estimate on atomic weapons aside (the US crash program to make a bomb was pretty incredible—it is not what you would necessarily expect a country to have invested in, and did require scaling up brand-new physics work to previously unprecedented industrial levels, so I give them quite a break in being wrong), my questions were:

Any guesses as to what the "freeze bomb" would actually be and how it would actually work? What were the "well known" principles? Googling "freeze bomb" and many variants turns up very little actually reality and a whole lot of bad sci fi.
Would such a weapon really have much combat value, or would it simply be a bizarre Nazi science terror weapon?

Pure speculation is invited and encouraged. My understanding, by the by, is that the actual V-3 weapon was a long-range cannon, so it's possible/likely that any other information leaked was just propaganda and/or misunderstanding by reporters, etc. --24.147.86.187 (talk) 19:40, 17 November 2007 (UTC)[reply]

Sounds like the science fiction effect at work (like the CSI effect, only describes how people think all things are possible back in the '40s and '50s because of all the pretty paintings of flying cars on the cover of "Popular Science" ;). All I can reason is that the bomb is actually a delivery vehicle for a large quantity of liquid nitrogen or other cryogenic fluid (which, if used in normal bomb-sized amounts, evaporates before it can cause anything but superficial damage to anyone but a few hatless and hapless bystanders). 80 tons all at once, however, would be enough to overcome the rate of evaporation and possibly overwhelm ventilation systems in underground bunkers and kill the inhabitants through severe frostbite and/or asphyxiation. I'm trying to imagine how you could manipulate the dew point using a pressure wave, but the worse that could happen would be, well, frost. Only problem is that when pressure rises, so does the dew point (which is the same as the frost point, assuming its cold enough). Making things colder is a lot harder than making them hotter in an unconfined space like a battlefield. So, all my meager reasoning takes me farther away from the feasibility of capitalizing on such a route. For creating a low pressure zone, you could use something like a fuel-air bomb, but that doesn't last very long. And it raises the temperature of the air. I think that a freeze bomb is a manifestly dumb idea, worthy of the best traditions of Nazi psuedoscience. DeepSkyFrontier (talk) 20:04, 17 November 2007 (UTC)[reply]

I can easily believe that Hitler would have directed his scientists to develop such a weapon. Some think (see Horrifying Utopias) that Hitler believed in a theory of fire and ice, attributable to the fiction writer Hanns Hoerbinger—that things could be expained in terms of a conflict between the two forces. Fire played a major role in Nazi ritual; torchlit ceremonies were standard fare. It is even supposed by some that this belief in the destiny of Aryan fire contributed to the Nazi defeat at Stalingrad where German soldiers froze to death for lack of winter clothing, unneeded by Hitler's Aryan fire-beings. Given all this, it would have been natural for him to want to counter the fire of his enemies with Nazi ice. Speculative enough fer ya? --Milkbreath (talk) 22:06, 17 November 2007 (UTC)[reply]

A couple of (very) sci-fi thoughts: 1) Magnetic refrigeration bombs. 2) Create a field capable of dampening all molecular motion (although I think the uncertainty principle would nix this one). 152.16.59.190 (talk) 06:18, 18 November 2007 (UTC)[reply]

Ah, but uncertainty principle is part of Jewish physics, so Nazis do not have to worry about that one! ;-) --24.147.86.187 (talk) 16:55, 18 November 2007 (UTC)[reply]

Construction of a tower

Say I wanted to construct a 65 m tall hollow tower of steel that would be able to support its own weight plus a further 15,000 kilos, how would I calculate how thick the walls would theoretically need to be, assuming a density for the steel of 8g/cm³? Would 10cm be enough? --80.229.152.246 (talk) 21:19, 17 November 2007 (UTC)[reply]

What's the diameter of the tower, for one thing? If it's much less than 65m, then your first mode of failure will be buckling, for which the article lists the relevant equations. Otherwise, check out compressive strength. Basically, you need to know, in addition to density, the Young's modulus for steel. I might try working out the equations tonight, but try it yourself and see what you get. SamuelRiv (talk) 03:02, 18 November 2007 (UTC)[reply]

In addition to the above articles, the area moment of inertia article will give you the equation you’ll need for the area moment of inertia of a hollow cylindrical cross section, which you’ll need in the equation for the maximum axial load without buckling. And the structural steel article has a bunch of links to articles on various standard steel alloys, such as the popular A36 steel. Those articles will list the yield strength of the alloy, so you can ensure that there won’t be more compressive stress on the tower than it can handle. Finally, actually designing a real tower would also involve using a factor of safety. What value you’d use for the factor of safety would depend to some extent on what the tower’s being used for. MrRedact (talk) 03:39, 18 November 2007 (UTC)[reply]

The nature of the load really makes a difference. If that 15,000 kg is distributed vertically, like a bunch of platforms going up the side of the tower, then it won’t require steel as thick as if the load were entirely at the top. If the load has a lot of wind resistance, like if it’s a windmill or huge billboard or something, then it’s going to require thicker steel due to the force of the wind. Or if it’s a load that might be lopsided, like a room containing a bunch of people who might all stand on one side, then compressive strength at the top of the tower and the required resistance to buckling are both going to need to be greater than if the load is guaranteed to be well-balanced. MrRedact (talk) 04:35, 18 November 2007 (UTC)[reply]

Assuming this is not homework, then it sounds like you need some professional advice from a structural engineer. Gandalf61 (talk) 13:13, 18 November 2007 (UTC)[reply]

You must consider more than one failure mode. Assume that your tower is strong enough to support its own weight plus the specified load. Now apply wind, both sustained and gusting, unless you can guarantee there will never be any wind. Now consider oscillation. I once saw a structure designed by a new engineer to support a heavy object of several hundred pounds. It was strong enough, but by momentarily and repeatedly pressing with the fingertips, the whole structure could be made to oscillate in increasing swings. Does the structure have to withstand a load of people moving in synchrony? Edison (talk) 14:38, 18 November 2007 (UTC)[reply]

Sorry about the missing values. I meant to add that the diameter of the tower is 3 m. The 15,000 kg load is directly on top, and I will assume no air resistance (even though I know that that is probably as far away from the truth as possible). Oscillation and a factor of safety are unimportant. I'll have a go at some calculations with the links you have provided, but if you think I need more information, please don't hesitate to ask. Thanks. --80.229.152.246 (talk) 16:24, 18 November 2007 (UTC)[reply]

Am I right in saying that for my tower of outer diameter 3 m and inner diameter 2.8 m, the second moment of area would be 0.958893 m⁴? --80.229.152.246 (talk) 16:35, 18 November 2007 (UTC)[reply]

Furthermore, assuming that my value for the second moment of area is correct, would I be correct in saying that the maximum force that can be beared without buckling (assuming a Young's modulus of 210 GPa) is 1.17599 x 10⁸ N? It seems an awful lot. --80.229.152.246 (talk) 16:46, 18 November 2007 (UTC)[reply]

Actually, with a 3 m diameter tower, its slenderness ratio is only about 43, so it counts as a short steel column, and you don't even need to worry about buckling. You just need to worry about compressive stress. MrRedact (talk) 17:00, 18 November 2007 (UTC)[reply]

I see. Thanks. Would I be correct in saying that I just need to work out the area of the wall at the top and then use the yield strength of the steel to find out what force the column could take, giving me about 2.25943 x 10⁸ N? Surely this should be less than the buckling force? --80.229.152.246 (talk) 17:08, 18 November 2007 (UTC)[reply]

No, you should only use the cross-sectional area of the steel in the tower. I.e., just use the area that's steel on a horizontal plane intersecting the tower. MrRedact (talk) 18:04, 18 November 2007 (UTC)[reply]

Sorry, I should have stated that doing as you suggest would be the same as doing what I meant, i.e. the tube is hollow all the way along. I'm still not convinced that I've got the right figure though. --80.229.152.246 (talk) 19:48, 18 November 2007 (UTC)[reply]

I get the same answer as you did (2.25943 x 10⁸ N). Clearly using 20cm thick steel is major overkill. MrRedact (talk) 20:59, 18 November 2007 (UTC)[reply]

Furthermore, could someone please point out any errors in the following calculations, because it does not seem right to me that the tower will fail first due to buckling than it will to compressive stress. Here goes:
Buckling: $F={\frac {\pi ^{2}EI}{(Kl)^{2}}}={\frac {\pi ^{2}\cdot 210,000,000,000\ \mathrm {Pa} \cdot {\frac {\pi }{64}}((3\ \mathrm {m} )^{4}-(2.8\ \mathrm {m} )^{4})}{(2\cdot 65\ \mathrm {m} )^{2}}}=117,599,000\ \mathrm {N}$
Compressive stress: $F=\mathrm {Cross\ sectional\ area} \cdot \mathrm {Yield\ strength} =(\pi \cdot (1.5\ \mathrm {m} )^{2}-\pi \cdot (1.4\ \mathrm {m} )^{2})\cdot 248,000,000\ \mathrm {Pa} =225,943,000\ \mathrm {N}$

I get the same answers as you do for both calculations, which surprises me too, since it goes against the rule of thumb about the slenderness ratio. Maybe we’re both doing something wrong, or maybe 210 GPa isn’t a very accurate estimate of the Young’s modulus of A36 steel. MrRedact (talk) 22:54, 18 November 2007 (UTC)[reply]

I think I may have an answer. The column is rather hollow, which I presume affects the slenderness rule of thumb quite a bit. That's the only thing I can think of. Thanks for all your help by the way; my engineering knowledge has improved enormously. --80.229.152.246 (talk) 17:39, 19 November 2007 (UTC)[reply]

Is there a difference between a sunrise and a sunset?

If one were given a stack of photos of sunrises and sunsets here on earth,would it be possible without any other information to tell the difference between the two?216.10.164.215 (talk) talking zero —Preceding comment was added at 22:54, 17 November 2007 (UTC)[reply]

If there are clouds like these in the pictures it might be possible. --JWSchmidt (talk) 00:35, 18 November 2007 (UTC)[reply]

If the date is known, a bright star may help, if available. Pallida Mors 01:22, 18 November 2007 (UTC)[reply]

See the previous discussion at Wikipedia:Reference desk/Archives/Science/2007 September 8#Sunrise vs. Sunset. hydnjo talk 02:56, 18 November 2007 (UTC)[reply]

If the land features are recognizable, you can tell east from west; but perhaps that counts as "other information". —Tamfang (talk) 10:49, 18 November 2007 (UTC)[reply]

There are more flying insects in the evening than in the morning (probably because evenings are warmer, and insects are quite sensitive to temperature). These, and other animals, or possibly even humans, might help. – b_jonas 11:52, 18 November 2007 (UTC)[reply]

Here's my guess. In the morning, the air is colder, right? So that means the air is denser. And that means the air's refraction index would be higher, right? So in that case the sunrise has more red-dish color. Correct me if I'm wrong. I do believe that a sunrise tends to look more beautiful than a sunset for some reason, generally speaking. Maybe its because you actually took the time to get up early that day. On a side note, I've looked for pictures of earth from space that show some kind of coloring in the day / night line divide. It's a subtle effect, and most earth shots don't show any noticable reddening, except when taken at relatively low altitudes (60 to 100 miles). --InverseSubstance (talk) 05:47, 20 November 2007 (UTC)[reply]