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December 19

Getting a Shine On

Why do the planets of the solar system, seen from earth, ahine like incandescent stars? Also, why does the earth's moon, seen from earth, shine rather like a fluorescent lamp? The astronauts who walked on the moon did not see the surface shining like that. Things are supposed to get dimmer as we get farther away. Instead, the planets and the moon get brighter. – GlowWorm —Preceding unsigned comment added by 98.17.46.132 (talk) 00:12, 19 December 2008 (UTC)[reply]

The light is simply reflected sunlight - the planets and moons (including Earth and our Moon) reflect sunlight - and against a dark sky - they look bright. I disagree that the astronauts walking on the moon didn't see the surface shining - they certainly did. That's how they could see the ground around them! When you stand in your back yard - looking at the grass - what you are seeing is light reflected off the surface of the earth (your lawn)...when you stand on the moon - it's the same deal. When you're in orbit around the moon? Same deal. When you are on the earth looking up at the moon? Same deal.

The business of things getting dimmer as they get further away is a bit more complicated. Light from a square meter of the surface of the moon reflects an amount of light into your eyes that's inversely proportional to the square of the distance you are from it. When you get twice as far away - you see four times less light from that square meter. BUT (and this is the important thing) as you get further away, each square millimeter of your retina is being lit up by more square meters of the moon - because perspective is making the moon look smaller. These two effects PRECISELY cancel out. That's why the grass on a distant hillside looks just as bright as the grass on the lawn you are standing on.

Now (and this is where it gets complicated) - this effect where the light from each square meter gets less and less - but more and more square meters pile up to push light into your eye - only lasts until the object starts to get very small. When it's small enough, all of the light from the object is reaching your eye already - now, as you get further away from it, the light from each square meter gets less - but you're already percieving all of the square meters already - so now the object starts to get dimmer as it gets further away.

The moon is so close that the 'cancelling out' effect is still working - so the full moon looks just as bright from here on Earth as it does when you're standing on it. (Well - except that our atmosphere blocks some of the light).

I'm not explaining this as well as I'd like to - but I hope you can follow what I'm saying. SteveBaker (talk) 00:36, 19 December 2008 (UTC)[reply]

(edit conflict, great explaination, SteveBaker)Hi. The reason is that the planets and moon reflect light from the sun. For example, the moon's surface is as dark as ashphalt. In fact, it only reflects about 8% of the sun's light. As you get closer to the Moon, its surface magnitude becomes higher, meaning the "density" of the brightness is less as it is more spread out. Looking down at an asphalt road on Earth on a clear day does not look bright, but if the entire Earth was covered in asphalt, then you may be able to see the reflection as ALL of the asphalt is now visible. Almost any surface will have some degree of reflection, and since the moon and planets are large objects, being able to see the entire object will make it appear brighter. However, if you were looking at the moon from say Mars, it would appear much dimmer than from Earth as it is farther away. See also albedo. I'm sure someone with a better explaination will be soon to come along, however. Hope this helps. Thanks. ~A H 1^(TCU) 00:41, 19 December 2008 (UTC)[reply]

(EC with SteveBaker and AH). Get in a really dark room, with no light at all. Hold a white ball infront of a black piece of paper. Do you see it? No? That's because the light is not on. Now, turn on the light. See the ball? Why? Cuz the light from the room is bouncing off of it. Now, the solar system is a REALLY BIG ROOM and the moon and the planets are still just balls, and the sun is a REALLY BIG LIGHTBULB. Same deal. --Jayron32.talk.contribs 00:43, 19 December 2008 (UTC)[reply]

Also, field of view plays into this as well. If you are standing on the moon looking straight down at the ground, you see only about two degrees wide clearly (a degree is a way of measuring how big something looks, a fist held at arm's length looks about 10 degrees wide). However, your periphial vision allows you to see about 160 degrees horizontally and about 100 degrees vertically. When you're standing on the moon, not only are you only able to see a small patch of the Moon if you're looking around, you can only see an even smaller patch at any one time. (If it's night on the moon, you would have a much harder time seeing, unless the Earth is lighting your sky [see Earthshine ]. You wouldn't be able to see the nighttime portion of the moon during the day and just barely at night.) Since from the Earth, you can see the entire moon with the naked eye without having to look around, you see the entire moon reflecting light into your eyes. ~A H 1^(TCU) 00:49, 19 December 2008 (UTC)[reply]

With regard to grass on a distant hillside looking just as bright as the grass at a person's feet, that is talking about two different things. The grass seen on a distant hillside has a far greater area than the grass seen at a person's feet when the angular span is the same for both views. A bigger object reflects more light. In contrast, an object of a given size seen from a distance will be less bright than the same object seen close up. When the same object is at a distance, less light energy is received by the eye. GlowWorm —Preceding unsigned comment added by 98.17.46.132 (talk) 06:05, 19 December 2008 (UTC)[reply]

We have an excellent article on albedo - the extent to which an object diffusely reflects light from the Sun. It says that the average albedo of the Moon is around 7%. This is much lower than the albedos of fresh snow (80%-90%), desert sand (40%) or green grass (25%), so an astronaut walking on the Moon will perceive the surface as rather dull. Now if the Moon's albedo were much higher than this, say about 30%, then it would reflect 30% of the sunlight falling on it. As the Moon is almost exactly the same angular size as the Sun, this would make moonlight from a full Moon on a clear night 30% as bright as direct sunlight - so a moonlit night would be as bright as a dull overcast day (clouds reflect about 70% of incident sunlight). This is clearly not the case, so a low albedo figure for the Moon is reasonable. The only reason the Moon appears to be especially bright in the night sky is by contrast against a dark background. Gandalf61 (talk) 14:52, 19 December 2008 (UTC)[reply]

Here is my explanation of this phenomenon. Everyone has seen a diagram of how light emanates from a point source. The diagram shows lines radiating fanwise from the point. Actually, of course, the lines are imaginary. But four of the lines, adjacent in 3 dimensions, will strike an imaginary spherical shell at any given distance from the energy source. The shell is centered on the energy source. The 4 lines and shell portion form a pyramid with a rounded base. As distance from the source increases, the total light energy encompassed by those 4 lines is spread over a larger base. So the energy decreases on each unit of area on the base (such as one square millimeter). Also as distance increases, within each additional increment of distance the pyramid comes closer to being a tube with a square cross-section. Therefore at a great enough distance, as distance increases further, there is very little change in light intensity on the aforesaid square millimeter. This can also be thought of in terms of a cone becoming more like a round pipe. As distance increases, the area of the image decreases at a slower rate than total light intensity decreases. (The image is the base of the pyramid or cone). Therefore an image becomes brighter as well as smaller. (An image is made up of a great many points issuing or reflecting light. The light from each point follows the principle described above.) Eventually, at a great enough distance, a crossover point is reached at which the rate of decrease in light intensity is insufficient to increase the brightness. The image then becomes dimmer and dimmer as distance increases. — GlowWorm —Preceding unsigned comment added by 98.17.46.132 (talk) 16:04, 19 December 2008 (UTC)[reply]

P.S. I just made some changes near the end of the above explanation, then changed it back again. It's tricky to think about. I'm still not happy with the explanation. – GlowWorm —Preceding unsigned comment added by 98.17.46.132 (talk) 18:27, 19 December 2008 (UTC)[reply]

I'm certainly not happy with it. I'd be happier, i.e. less inclined to mutter "nonsense", if there were an algebraic expression of the distance at which "an image becomes brighter as well as smaller." —Tamfang (talk) 17:37, 28 December 2008 (UTC)[reply]

The sky is falling! Positive or negative feedback mechanism?

Hi. Apparently, as the troposphere warms as a result of global warming and greenhouse gases, the upper atmosphere, ie. the stratosphere to the ionosphere, is cooling rapidly. This is causing the upper atmosphere to shrink and move toward the ground. Since the extra air would be pressing on the troposphere, the part of the atmosphere that affects our weather, would this be a positive or negative feedback, or neither? Would it be a positive feedback since the extra atmosphere would increase the pressure in the troposphere and thus increase its heat content and also trap in the existing greenhouse gases, or would it be a negative feedback by absorbing the Sun's heat before it reaches the troposphere and by absorbing tropospheric heat as it attempts to radiate into space? In a comment on wunderground.com, it appears a 17C cooling has occured in the ionosphere, affecting satellites as the atmospheric pressure there has decreased by 2-3%. Could this be related to the gigantic chunks of ice falling from a clear sky reported by Discover magazine? Could this also be why the air temperature over parts of Antarctica have apparently cooled? Does the cooling of the stratosphere also decrease stratospheric ozone? Thanks. ~A H 1^(TCU) 01:26, 19 December 2008 (UTC)[reply]

Gas contracts when it is cooled, but a given volume does not become heavier. So a cooler stratosphere does not press more heavily on the troposphere. The stratosphere just becomes thinner. —Preceding unsigned comment added by 98.17.46.132 (talk) 02:47, 19 December 2008 (UTC)[reply]

Worse still - the troposphere is getting warmer - it's going to expand - surely that's going to push the upper atmosphere higher?

So what is this mechanism that's making the upper atmosphere cool? Is it because CO2 is trapping the sunlight in the lower atmosphere - reducing the amount of sunlight travelling back out through the upper atmosphere? SteveBaker (talk) 04:10, 19 December 2008 (UTC)[reply]

I was at the American Geophysical Union Fall Meeting 2008, and (being among other things, a researcher of ionospheric physics) I happened to spend a good amount of time in the Space Weather and Climate Modeling sessions. There are hundreds of thermodynamic models for global climate. Some only model ocean temperatures. Some model solar flares. Some model electric field interactions between low-altitude and high-altitude winds. Some generate statistical aggregates of empirical observations. A search for "troposphere climate feedback" in this fall's meeting abstracts generated over forty results - that is, forty new-as-of-this-Fall scientific approaches to that specific problem ([1], for example). The point I am trying to make is that it is the research scientist's choice to model the effect you describe as positive or negative feedback and there is not currently much scientific consensus (let alone, predictions or model-to-data matching). Nimur (talk) 05:10, 19 December 2008 (UTC)[reply]

:-), also at AGU. Dragons flight (talk) 06:06, 19 December 2008 (UTC)[reply]

If the climate had positive feedback it would quickly spiral out of control. Considering that it didn't, I think we can safely say the feedback is generally negative. Of course, since there is vastly more than one degree of freedom, positive and negative feedback is a gross oversimplification, but at least we can say that in the last five billion years, the climate never got so bad it wouldn't go back. — DanielLC 17:42, 19 December 2008 (UTC)[reply]

It is a common misunderstanding that a positive feedback implies a runaway feedback. A positive feedback chain can be understood iteratively as x1 = x0 + a*x0, x2 = x1 + a*x1 = x0 + a*x0 + a²*x0, ... If a is less than 1, the geometric series

\sum {x*a^{i}}={x \over 1-a}

, and one has a finite and controlled positive feedback. It is only when the response to a perturbation is larger than the perturbation itself that one has a runaway feedback, i.e. a > 1. The climate system has a great many positive and negative feedback, and in fact many of the most important are positive feedbacks, but obviously those feedbacks aren't large enough to cause the climate system to break down. Dragons flight (talk) 18:03, 19 December 2008 (UTC)[reply]

how to reduce alcohol or menthol smell but keep the cool function

I need to know how to reduce the alcohol/ menthol smell but keep the cool function.

By answer to my e-mail : email removed—Preceding unsigned comment added by 124.121.5.186 (talk) 04:18, 19 December 2008 (UTC)[reply]

Drink smoothies.--Shantavira|^{feed me} 08:58, 19 December 2008 (UTC)[reply]

You got an answer when you asked this a few days ago: Wikipedia:Reference_desk/Science#the_method_to_reduce_alcohol_or_menthol_scent. We aren't going to send it to your email. ~ mazca ^t|c 14:35, 19 December 2008 (UTC)[reply]

IMMUNOLOGY

Which is the most potent antigen presenting cell(APC) for naive T cells? Which is the best antigen presenting cell? —Preceding unsigned comment added by 117.196.65.28 (talk) 07:42, 19 December 2008 (UTC)[reply]

Our antigen-presenting cell article addresses this question directly (regarding the most potent APC for naive T cells). For your second question, "best" is not specific enough to answer directly because there is no single characteristic to define "best". --Scray (talk) 11:45, 19 December 2008 (UTC)[reply]

Conjoined twins longevity question

Why do conjoined twins have short lives? 124.180.116.201 (talk) 08:43, 19 December 2008 (UTC)[reply]

Did you read conjoined twins? Many of them lead long and fulfilling lives.--Shantavira|^{feed me} 09:00, 19 December 2008 (UTC)[reply]

Well, yes - but an awful lot of them die at birth or within a very short time afterwards. The average longevity has to be a lot worse than for normal twins. The "why" of that is likely to be different depending on how they are conjoined - but very often internal organs are shared or highly distorted or even compressed against bones that "shouldn't be there". SteveBaker (talk) 22:49, 19 December 2008 (UTC)[reply]

Demineralised water

Today I bought some demineralised water to top up my car battery. It said on the bottle, 'not for human consumption'. Now, I don't plan on drinking it, but if I did surely it would be fine wouldn't it? —Preceding unsigned comment added by 58.165.239.250 (talk) 11:39, 19 December 2008 (UTC)[reply]

Yes. What they most likely meant is that their production facility did not applied and not certified by food safety/quality control organs/bureaucrats. Vitall (talk) 11:50, 19 December 2008 (UTC)[reply]

Distilled water is safe to drink, but if you only drink distilled water, you can die of thirst. --131.188.3.20 (talk) 13:34, 19 December 2008 (UTC)[reply]

Do you have a cite for this? There are a lot of myths about distilled water. APL (talk) 14:02, 19 December 2008 (UTC)[reply]

Pure distilled water is rather horrible-tasting but I've never seen any evidence to suggest that you could die of thirst if you drank only that. Certainly you'd miss out on some of the minerals you normally get from water, but that's a minority of your daily mineral intake anyway. Our article on Purified water contains a long and waffly section about the benefits and drawbacks of drinking it (with a neutrality tag too!). To add to the original question, I agree that it's more likely to be an ass-covering safety precaution - there's possibly an increased risk of bacteria, etc, in the water if it hasn't been prepared and stored with drinking in mind. They'd rather not get sued if someone does drink it. ~ mazca ^t|c 14:33, 19 December 2008 (UTC)[reply]

Isn't the issue that it would explode your cells like distilled water on an amoeba will ? That is, the water will travel through the cell walls until the level of dissolved material is equal both inside and outside the cell, at which point there will be so much water inside that the cell walls will breach. The water would, however, mix with dissolved material in your saliva and stomach, so it wouldn't be pure distilled water once it contacts the stomach walls. However, if you drank enough, perhaps it would still be close enough to distilled water after this mixing to cause a problem. StuRat (talk) 16:07, 19 December 2008 (UTC)[reply]

No, an amoeba swells or bursts in distilled water because it evolved to live in slightly salty water. The fact that it's distilled water is irrelevant: fresh water would have the same effect, as our amoeba article correctly says. It's different with humans, as we evolved to drink fresh water. We don't care about the slight difference between fresh and distilled water either, except for the taste. Distilled water won't make our cells explode any more than fresh water will. Mine's a pint. --Heron (talk) 17:51, 19 December 2008 (UTC)[reply]

Humans have more salts in our cells than there are in fresh water, although less than ocean water. One result of this is how our skin wrinkles up when we take a bath in fresh water. StuRat (talk) 18:46, 20 December 2008 (UTC)[reply]

I do not see why the salt and other minerals and ph balance would not be supplied by other food. The International Space Station has a water recovery system which distills urine, and they will in the future use that for part of their drinking water. The article says nothing about possible danger from drinking the distilled water. We do not provide medical or nutritional advice, but a Google Book search [2] will show many books, some sounding quacky, advocating the drinking of distilled water. Men on ships have been drinking distilled water for 150 years [3] , [4]. "The Doctor's Heart Cure" (Sears, 2004) oppositely cautions against drinking distilled water, saying "Long term use of distilled water can lead to mineral deficiencies that can cause heart beat irregularities and hair loss." He cites "Day, C., 'Why I say no to distilled water only,' Health and Beyond Weekly Newsletter,' reprinted at mercola.com/article/Diet/water/distilled_water_2.htm ^{[unreliable fringe source?]} , but that site says the page is no longer available. Pros and cons of drinking distilled water are discussed at "Plain talk about drinking water," page 97 (2001) by Symons, published by the American Water Works Association. Any results from medical/nutritional textbooks from respected publishers such as university presses, or recent reliable peer reviewed scientific publications? Note: the article Purified water says that "demineralized" or "deionized" water may have had the mineral salts removed by ion exchange resins, which do not remove organic molecules, viruses, or bacteria, so it could be full of things which could make you sick. It is not steam distilled. Edison (talk) 18:03, 19 December 2008 (UTC)[reply]

But even if distilled water doesn't have minerals and stuff that you need - that doesn't mean that the water will harm you - it's the lack of minerals in your diet - which could be gotten in a bazillion other ways. That's like saying eating bananas will kill you because they don't have enough vitamin C. (Actually, I have no idea whether bananas have vitamin C or not - but I'm sure you understand my point). I'm pretty sure that true, pure distilled water is very safe to drink - but in all likelyhood, what was in that bottle of stuff you got to put into your car is no longer that pure. Aside from anything else, it's sure to have picked up impurities from the bottle it's sold in - and it's very possible that they used some kind of toxic substance as the mould release agent or something. The labelling criteria for stuff that goes into cars isn't as strict as for food items - so it may not actually be pure distilled water at all! After all - "High Octane Gasoline" doesn't have any more "Octane" in it than regular stuff - it's other additives that produce the EFFECT of higher octane levels. SteveBaker (talk) 21:44, 19 December 2008 (UTC)[reply]

Partially distilled water, ie. previously boiled water, is usually safe to drink. However, if you drink too much water that does not contain any electrolytes, you can die from water poisoning. ~A H 1^(TCU) 21:53, 21 December 2008 (UTC)[reply]

But you can die from that with mineral water, water from your kitchen tap - or distilled water. That's not something that's specific to distilled. SteveBaker (talk) 17:15, 23 December 2008 (UTC)[reply]

Water intoxication is remarkably rare. The human body has an excellent homeostatic mechanism involving antidiuretic hormone and aquaporins. As Vitall states, the disclaimer is because the company has not applied through the regulatory authority to confirm that the product is fit for human consumption. Axl ¤ [Talk] 22:49, 25 December 2008 (UTC)[reply]

If you only drink distilled water, you can die of thirst.
— 131.188.3.20

This is not true. Axl ¤ [Talk] 22:50, 25 December 2008 (UTC)[reply]

Isn't the issue that it would explode your cells like distilled water on an amoeba will ?
— StuRat

This is not true either. Serious symptoms of water intoxication appear once the serum sodium concentration drops below about 120 mmol/l (corresponding to osmolality of about 240 mOsm/kg). A sodium concentration below 110 mmol/l greatly increases risk of coma and death. [Normal serum sodium concentration is 135–145 mmol/l. Normal serum osmolality is 280–296 mOsm/kg. Pure water's osmolality is 0 mOsm/kg.] So a person would die long before lysis of cells became a significant factor. Axl ¤ [Talk] 23:02, 25 December 2008 (UTC)[reply]

beta sisteral

what is beta sisteral —Preceding unsigned comment added by 71.199.15.172 (talk) 12:35, 19 December 2008 (UTC)[reply]

I believe you are looking for beta-sitosterol. -- kainaw ™ 13:40, 19 December 2008 (UTC)[reply]

Symbol for Litre: uppercase or lowercase?

In a recent response to Density (follow on from the above), one of the respondents used the symbol "ml" for millilitre, which I am always tempted to correct (I find "mL" to be much less ambiguous, and consider "L" to be the preferred symbol for liter). I decided to check my facts, at least superficially, and was somewhat surprised to see that the lead to our Litre article states that "l" (lowercase) is an accepted abbreviation for litre. Of course there are other sources to check, and I did find that in the U.S. the accepted abbreviation for litre is "L", at least according to NIST. To me, given the importance of legibility in many fields, using uppercase "L" is a no-brainer. I would be interested in any comments from the RD (sorry if this is a FAQ - I'm relatively new here). --Scray (talk) 12:36, 19 December 2008 (UTC)[reply]

My experience is that 'l' is the only acceptable abbreviation. Litre#Symbol indicates this is a national difference. Algebraist 12:41, 19 December 2008 (UTC)[reply]

I'm quite ashamed that I did not read down that far on the Litre page. Thank you - that settles this question for me. --Scray (talk) 12:54, 19 December 2008 (UTC)[reply]

Algebraist, I see you study at Oxford and am guessing you are English? Although l and ml are often used in shops, etc. At University we use L/mL and l/ml synonymously. --Mark PEA (talk) 13:54, 19 December 2008 (UTC)[reply]

Typically, units which aren't named after people have lowercase symbol, but for the litre an exception was made to allow L as well as l, as the latter looks like the digit 1 in some fonts. -- Army1987 – Deeds, not words. 19:48, 19 December 2008 (UTC)[reply]

According to International System of Units: "The American National Institute of Standards and Technology recommends that "L" be used instead, a usage which is common in the US, Canada, Australia (but not elsewhere). " - so everywhere else on the planet, stick with 'l'. But the odd thing is that the litre is neither an SI base unit nor (to my surprise) one of the 'approved' derived units (ie it's not in the standard list of 20 derived units in SI derived unit). So I think the "correct" thing is simply not to use it and to stick with either decimeter-cubed: dm³ or (more likely) 10^-3m³. That's backed up by Litre which says pretty much the same thing. Obviously, this only matters in strictly formal settings - everyone who uses SI will be very clear on what is intended by either 'l' or 'L'. SteveBaker (talk) 21:35, 19 December 2008 (UTC)[reply]

Shouldn't be any real surprise there. The liter is not and never will be a part of the SI, because the SI is a "coherent" system of units (each unit is a unitary combination of the base units; the unit of volume is the base unit multiplied by itself three times). That unit, the cubic meter, could have a name that could become part of SI, but the name sometimes used for it has not been made a part of SI.

The liter is officially recognized among the units acceptable for use with SI. That's the most it ever will be; it won't be a part of SI. But even after the introduction of SI, the CGPM has had three different resolutions dealing with the liter, IIRC.

You are obviously too young to have gone through the punishment some of us had to deal with, in the pre-1964 days when we had to learn the difference between a liter and a cubic decimeter, or between milliliter and a cubic centimeter, which then differed by about 28 parts per million. That was the result of an ill-advised redefinition of the liter in 1901, causing a couple of generations or so to waste a lot of time learning a distinction that usually didn't matter at all in any real-world measurements. Gene Nygaard (talk) 08:42, 22 December 2008 (UTC)[reply]

And, in fact, when the International System of Units was introduced in 1960, the liter/litre wasn't even among the units acceptable for use with SI. At that time it didn't even have a powers of 10 relationship to the base units. It was only after the 1964 restoration of the definition as a cubic decimeter that the liter was declared acceptable for use with SI. Gene Nygaard (talk) 08:44, 22 December 2008 (UTC)[reply]

Symbols for the litre (CR, 101 and Metrologia, 1980, 16, 56-57)

Resolution 6

The 16th Conférence Générale des Poids et Mesures (CGPM),

recognizing the general principles adopted for writing the unit symbols in Resolution 7 of the 9th CGPM (1948),

considering that the symbol l for the unit litre was adopted by the Comité International des Poids et Mesures (CIPM) in 1879 and confirmed in the same Resolution of 1948,

considering also that, in order to avoid the risk of confusion between the letter l and the number 1, several countries have adopted the symbol L instead of l for the unit litre,

considering that the name litre, although not included in the Système International d’Unités, must be admitted for general use with the System,

decides, as an exception, to adopt the two symbols l and L as symbols to be used for the unit litre,

considering further that in the future only one of these two symbols should be retained, invites the CIPM to follow the development of the use of these two symbols and to give the 18th CGPM its opinion as to the possibility of suppressing one of them.

(And derived units are products of powers of SI base units; the category in which the litre, the minute, the hour, etc. are is "Non-SI units accepted for use with the International System of Units".) -- Army1987 – Deeds, not words. 00:15, 20 December 2008 (UTC)[reply]

Per Wikipedia:Manual of Style (dates and numbers): "For reasons of legibility, the preferred symbol for the unprefixed liter is upper-case L." --—— Gadget850 (Ed) ^talk - 01:03, 20 December 2008 (UTC)[reply]

Burt Rutan comments on V-22 Osprey and 1960's tiltrotor prototype

I recall reading an interview with Burt Rutan, designer of Spaceship One and other revolutionary aircraft. In this online article, he had some negative comments about the V-22 Osprey. He compared it unfavorably to a prototype tiltrotor aircraft developed in the 1960's or thereabouts. I believe it may have been developed by Bell. Anyway, Rutan argued that the old prototype was sigificantly superior to the modern Osprey in various flight characteristics, and especially in durability and ruggedness.

I have run extensive Google searches, trying to locate this article. Can anybody recall where this article might have appeared, or where I might otherwise find Rutan's comments and views on the Osprey? Thank you. —Preceding unsigned comment added by 69.255.102.84 (talk) 13:15, 19 December 2008 (UTC)[reply]

optics

newton propounded corpuscular theory of light and he is against the wave theory. but an experiment known as newtons rings is a popular one which is considered as proff of wave theory. to me this is a contradiction. newton gave corpuscular theory, but the experiment in his name is proving the wave theory. explain this paradox. —Preceding unsigned comment added by 118.95.105.71 (talk) 14:36, 19 December 2008 (UTC)[reply]

Just because it's called Newton's Rings doesn't mean Newton designed it, it was designed to test Newton's idea, or that even if it was designed to do so, that it didn't wind up proving him wrong. Names are just names:) DMacks (talk) 16:35, 19 December 2008 (UTC)[reply]

Meh. Neither the corpuscle theory or the wave theory are particularly correct; they both have some aspects that explain some behaviors of light. Light is light, however, and that it sometimes behaves like a wave and sometimes like a particle is not lights problem. Its ours... --Jayron32.talk.contribs 17:59, 19 December 2008 (UTC)[reply]

This is another good example of Stigler's law of eponymy in action. --Bowlhover (talk) 18:43, 19 December 2008 (UTC)[reply]

According to Newton's rings, Newton was the first to analyze the phenomena (but not the first to describe it). Dragons flight (talk) 18:51, 19 December 2008 (UTC)[reply]

"Light is light, however, and that it sometimes behaves like a wave and sometimes like a particle" is little misleading. To explain some of its behavior, we use wave theory and to explain some other behavior we use particle theory. It is not that when light is wave it is not particle. Light is both particle and wave, see Wave–particle duality. manya (talk) 06:41, 20 December 2008 (UTC)[reply]

How the heck does a car move?

Or to be more precise: given the power and torque curves, gear ratios, wheel and tyre sizes etc. of a standard Honda S2000, launched from a full-ball standing start and only gearing up at 9,000 rpm each time, what would the displacement-, velocity- and particularly the acceleration-time graphs look like?

Asked in a different way (what I'm REALLY asking): how does the power, torque and gearing on a car combine to translate into motive force? Say we compare three cars of equal power, a 177 kW 208 Nm high-revving Honda S2000 with its standard (short) gearing, a 177 kW 300 Nm Seat Leon Cupra with the Honda's gearing, and then the same SEAT with its standard gearing. Would be interesting to compare the 3 cars plotted against each other for displacement, velocity and acceleration.

For the purpose of the exercise ignore things like air resistance (and then add it back in once we're done with the simple case), as well as tyre slippage and drivetrain losses etc. and assume the cars are the same mass. And of course linearise the power curves piece-wise to make the calculus easier.

I got into a horrible tangle trying to determine the power as a function of time. It is a function of RPM clearly. The RPM is a function of velocity and gearing. Velocity v = ds/dt. However P = dW/dt where work W = F x s. Force F = ma = m d²s/dt². You can see it gets horribly complicated at this point. Help! Zunaid 14:56, 19 December 2008 (UTC)[reply]

It would have thought our article vehicle dynamics would provide an overview, but it doesn't. There is some information in our Wikibooks Subject Automotive Engineering, but I don't know if that's what you need. So, short of actually spending years studying automotive engineering, I would recommend buying a good old-fashioned paper book about it. — Sebastian 20:14, 19 December 2008 (UTC)) corrected 20:20, 19 December 2008 (UTC) [reply]

Our vehicle dynamics article is just a list it seems, and the Wikibook doesn't cover these topics unfortunately. Zunaid 13:23, 21 December 2008 (UTC)[reply]

Well, the thing to remember is that the engine, the transmission and the wheels are all connected - if we assume no slippage in the clutch and no wheel spin then the RPM and the Speed are linearly related for any given gear ratio - you can look up the exact ratio of the gears in your gearbox in most decent car spec sheets (or maybe in the owner's manual) - but be careful because there is also gearing in the differential and there may be another fixed gear in the engine someplace after the tachometer...you need to multiply all of those ratios together to get the relation between the RPM of the engine and the RPM of the drive axles...and from that and the circumpherence of the tyres you can go from engine rotations per minute to wheel rotations per minute and from that to distance travelled per minute and then miles-per-hour. So you can stop thinking about RPM and speed separately (except when shifting gears).

Furthermore, torque-at-the-wheels and force are also linearly related for any given wheel diameter (again, assuming no wheel-spin). So you can take the standard torque-at-the-wheels-versus-RPM graph that you get out of a dynamometer and use that to directly calculate linear force versus speed in each gear. Dyno curves for pretty much any car can be found from an enthusiast somewhere - or you can take your car to a dyno center and for some outrageous amount of money (less than $100 - but more than $50) you can get that graph. OK! Now where getting somewhere!

Now we need to turn force into acceleration - and it's tempting to grab our old friend F=Ma and say that the accelleration depends on linearly on the force for a given mass - which is more or less constant except for plus fuel and passengers. That would be OK if there was no friction and no drag and the car was on level ground...and I know you're happy to ignore those things. But sadly, you can't...not if you want to get anywhere close to the right answers. Air resistance particularly is an enormous force. At 100mph it's by FAR the biggest force...and (critically) it's what limits the top speed of most cars. In my MINI Cooper'S - the rev limiter cuts in in 5th gear at about the top speed of the car...if it were not for air resistance, I could shift into 6th gear and go even faster...but when I do that, the car actually slows down because there isn't enough torque in that gear to overcome air resistance at the top speed of the car.

Air resistance is simply not a reasonable thing to ignore because at anything much over about 30mph it BY FAR dominates the results you're going to get.

OK so now what? Well, you need to know the drag force on the car and the amount of rolling resistance. The coefficient of drag for most cars is available in the spec sheet you get from the car dealership or the manufacturer's web site. You also need to know the viscosity of the air (Wikipedia is your friend!), the frontal cross-sectional area of the car (you can estimate that fairly well as the height times the width of the car) and a few other things. This can be plugged into the equation in Drag (physics) to get you an equation relating speed to the drag force.

So - at any given speed - the forward acceleration is:

The force we got from converting the torque/RPM curve into a force/speed curve

...minus...

The drag force comes from the equation in Drag (physics)
The rolling resistance (friction) in the transmission, tyres, etc. (We'll talk some more about this in a moment).

This gives us a NET force at any given speed, for a particular gear.

OK - so NOW we can grab a hold of F=Ma and figure what happens if your foot is on the floor at some given speed. It would be nice to toss some math in here and get a graph - but sadly the torque versus RPM curve isn't a nice mathematical function - it's a weird shaped mess that's different for every car on the planet. (And it's different in winter and summer - on one brand of gas versus another, etc).

So using math to plot the position or speed of the car versus time requires some ikky small-time-increment numerical integration.

For that we can use a computer. We can say what the speed is now - plug that into the math and get out the acceleration now. Then we can say that over the next (say) millisecond, how much will the car accellerate - we can use that to get the new speed one millisecond into the future - and plug THAT into the same pile of math to get the speed 2 milliseconds from now - and so forth.

This computer program is something I've written for my MINI and it works great - it fits reality very accurately. However, there are a couple of gotcha's. One is that you have to shift gears - that takes time - during which the torque from the engine drops to zero - the force from the engine goes away and drag and friction start to kill your speed until you hit the next gear. Assuming you're driving smoothely, it's reasonably save to say that you matched the RPM's correctly - so all you need to do is to switch the gear ratio in the bowels of all that math - and take away the engine torque for as long as it take you to shift.

So - you do all that and you get a 0-60 time that looks quite a bit better than the manufacturer's spec sheet (if they are being honest) - and when you try to make the car do that - it won't. That's mostly because you need practice at actually driving the car like that - but with a lot of highly disciplined effort - you can get the car to do EXACTLY what the computer says it will. Hooray! Simulation works!

There is one more 'gotcha' I missed. What is the rolling resistance and the friction in the car's drive-train? This is never written down on any car spec sheet and there isn't a reasonable way to calculate it...so now what?

Well, here is what you do. You take your car - get it up to a reasonable speed then stand on the clutch - the car will slow down...and down...and down...and eventually stop. The amount of time that takes is dependent on the drag and the friction in the drive train/tyres/etc. We can plug the numbers for the free-wheeling car into our car simulation software and get an answer for how long it would take for the car to slow down from (say) 60mph to (say) 30mph if there was ZERO rolling resistance (ie just air resistance)...and we can plug in different values for the rolling resistance until we get an answer out that matches what we see in the real car. Sadly, this number is not linear with speed (at least it isn't in my car) - so you have to measure how long it takes the car to go from (say) 100mph to 90mph, from 90mph to 80mph...and so on all the way to zero mph. This is tedious (and, on open roads, downright dangerous!) - but if you are determined - you can do it and plot a graph of friction versus speed.

But since you're doing that - you can actually just use the graph of speed versus rate of slowing down with the clutch pedal pushed down (the deceleration) to directly measure the sum of rolling resistance, drive train friction AND wind resistance at a whole range of speeds. Knowing that deceleration curve (and the mass of the car) - you can produce an empirical measurement of the net forces on the unpowered car as a function of speed - so you can forget about all of those ugly drag calculations (which is good because some of the things like the coefficient of drag and the cross-sectional area are hard to calculate directly).

Now you can factor that into your math.

Yes - this isn't easy. Sorry about that! But I did it - and it works.

SteveBaker (talk) 20:18, 19 December 2008 (UTC)[reply]

Thanks Steve (I somehow suspected you'd turn up with a detailed answer. Thanks!). The S2000's torque curve is fairly flat (i.e. constant) in two sections (before and after VTEC at 6,100 RPM) so should lend itself quite nicely to being defined piece-wise as two constant functions (of which the first half only applies in 1st gear anyway). Otherwise numerical integration would be the more accurate way to go. I was just hoping to do some juicy differential equations but alas the problem is more practical than that as you've explained. Mmmm okay so I can't leave the drag out, that's a drag ;) There are a few at-the-wheels dyno charts of an S2000 online which I shall grab. I also know the speed vs RPM for each gear which I've already determined emperically, which would take account of all the "internal" gearing between the engine and the wheels. Would you be willing to give out your code? I'd love to play around with a few numbers and see what comes out. Have you by any chance produced graphs of the displacement, velocity and acceleration vs time using your program? Would be interesting to see them. Also, I think vehicle dynamics could do with your input. At the moment it is just a list article. Zunaid 13:23, 21 December 2008 (UTC)[reply]

Mysterious Music Box Malfunction

A music box in my house, which has not been wound-up in at least a year, went off on its own a few days ago. It played one chorus and then stopped. I don't know anything about the inner mechanisms of music boxes, but I'm curious as to what can cause this sort of malfunction. Any ideas?

It was by itself in a corner of a room and no one had been near it all day, so it was undisturbed by exterior forces. So there's probably something wrong with the mechanisms inside it. It's about a decade old. "Music box" may be a misnomer since it's not shaped like a box. It's actually a large snow globe on a stand that I assume houses the musical component, and sticking out of the stand is the key-shaped metal thing that you turn to wind it up and make it play. Maybe something (liquid? confetti?) from the snow globe leaked down into the music box part and set it off?

After it went off on its own we tested it and it still plays just fine when wound up. 96.233.7.70 (talk) 20:29, 19 December 2008 (UTC)Perplexed[reply]

Most music boxes are powered by a mainspring similar to some mechanical watches and clocks. My guess is that temperature changes caused either expansion or contraction of the Mainspring, which caused the music box to fire off. No need to alert Craig T. Nelson or anything... --Jayron32.talk.contribs 21:06, 19 December 2008 (UTC)[reply]

There has to be some kind of ratchet inside that holds against the tension of the mainspring - when that fails, all the energy in the spring is released at once. If it was already close to being wound down - this event might release enough energy to play a chorus or so before it utterly wound down. Why it chose that moment to come loose is anyone's guess - temperature changes - vibration - who knows? I have a large "30 day" clock at home (you only have to wind it every 30 days) - it has two mainsprings - one for the chimes and one for the movement. A few months ago, I'd wound up one spring and was in the process of winding up the other when the first mainspring's ratchet failed in some way - the key rotated amazingly quickly - and the 'wings' of the key whacked me across the knuckles several times - it was enough to draw blood and left me with a finger that was so badly bruised that I thought it was broken...fortunately it wasn't but I had to keep it in a splint for 10 days anyway. Never underestimate the power of a clockwork motor! SteveBaker (talk) 22:55, 19 December 2008 (UTC)[reply]

Another culprit could be a drop in humidity alone, or in conjunction with a temperature change. Humidity makes things "sticky", including increased friction that might have caused the motion to stop near the end of the winding, and drying out could decrease the friction enough to allow it to play a bit more. Humidity is especially important with wood and lacquer components. StuRat (talk) 01:44, 20 December 2008 (UTC)[reply]

Thanks so much. :-) 96.233.7.70 (talk) 20:45, 21 December 2008 (UTC)NoLongerAsPerplexed[reply]

Time delayed bombs in World War 2 had clockwork mechanisms to delay the detonation for some period such as 24 hours, to increase the terroristic value of the weapon. Many of these stopped running before detonation, often when they were nearing the end of the period and the clockwork was just starting to press on the detonator trigger. Decades later, it was common for one to resume its ticking and detonate, either because someone fund it while excavating for a building, or just because a truck hit a bump on the nearby road, or because of a spring thaw. Similar minor jarring or temperature changes could cause the music box mechanism to run a bit longer, to more melodious effect. Edison (talk) 21:02, 21 December 2008 (UTC)[reply]