Wikipedia:Reference desk/Archives/Science/2007 May 28

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May 28

What is the largest encyclopedia ever existed?

I have read from a webpage (from wikipedia) that Wikipedia is larger than Britannica. I would like to know if other (online, offline, like books) encyclopedias exists (or existed) who are larger than wikipedia. Thank you.

See Wikipedia:Size comparisons, Wikipedia is one of the biggest encyclopaedia known to human kind, depending on how you see it, maybe the biggest. --antilived^{T | C | G} 11:17, 28 May 2007 (UTC)

Though it will never be, nor aspire to be, as big as The Library of Babel. ;-) --140.247.240.18 14:10, 28 May 2007 (UTC)

Ironically, even with the larger article and word count, Wikipedia is readily available in electronic form and can thus be miniaturized to small volume (such as a few hard-hard-drives - I don't know how many terabytes it would take to save the whole encyclopedia, but this collection of Public Domain classical music just set my own system back a couple of gigabytes....) Nimur 15:03, 28 May 2007 (UTC)

The pages themselves are usually fairly small, it's the images that take a bunch of space. Still, even with all the images on commons and all the text on en., I imagine it would be very possible to store almost the entire encyclopedia on a single computer... actually, it sounds like a fun project :) go through the dumps and clean up articles and release them in a permenent "checked" version. Kinda like the Wikipedia CD, but online -- Phoeba Wright^OBJECTION! 16:07, 28 May 2007 (UTC)

Such projects have existed (in both commercial and non-commercial forms) but I think they have not had a lot of success. Nimur 00:34, 29 May 2007 (UTC)

The trouble is that a CD doesn't have enough space to store even a tiny subset of Wikipedia. If you download that version (I have), it's really disappointing compared to the real thing. With the online version, you can type almost anything into the search box and there will be an article about it. With the CD version, you're very unlikely to find what you want unless you ask about something fairly obvious (a 'big' topic) - or happen to get lucky and find that a featured article was written about that topic. Even when you find an interesting article, there are often hardly any links in them (because the articles that would have been linked to aren't there) and whilst all of the pictures seem to be there - they are only present in thumbnail form - you can't click on them to make them bigger. So quite honestly, the CD version isn't much use - except perhaps for schools where they need a 'sanitized' version of WP that doesn't have vandalism-induced obscenity or articles that are not suitable for children and where articles won't change between a kid referencing them and their paper being graded! But if you opened up the scope and decided on a 50 DVD boxed set of 'sanitized' Wikipedia - what then? That would truly be an endless task - nobody (not even a fairly large team) could possibly read all 1.8 million articles, grade them all and pick 'good' stable versions of them all. There is just too much stuff. Take a look at Special:Newpages - it's a list of all of the new articles that are being created. Notice that there are on average maybe three or four new articles created every minute of every day 24/7. You can't even read those new articles as fast as they are being are created - let alone have any hope to clean them up, grade them and decide whether they are worthy of your Mega-DVD boxed set. An enormous amount of effort has gone into making the CD versions - any hope of making something even twice that size is beyond what effort is available to do it. SteveBaker 05:05, 29 May 2007 (UTC)

Awhile back, I estimated that if I considered only content that currently exists (i.e., pick an arbitrary date, such as "now"), and freeze-framed the Wikipedia, it would take my entire life (80 more years, generously estimating...) of reading about 2000 articles each day to read everything on the encyclopedia. Without sleep, this would come to about 100 article per hour - almost reasonable! And of course, this would neglect everything written in the entire remaining duration of my life; that would be a lot of content which I would simply never see.

The philosophical implications are profound - if you are learning non-stop, every minute of every day, with no breaks for sleep, you could never possibly be able to claim to know the entirety of human knowledge before you die. This is almost a strong argument for religion - "take these facts on faith because it will take too long to explain them."

Furthermore - I can foresee a not-so-distant future when science or mathematics have evolved to the point where youngsters do not have time to learn elementary algebra and calculus, because they must be trained in far more advanced mathematics so that they are prepared to design the engineering and science projects of the next generations. As such, only a few esoteric historians will ever "know" how the underlying mathematics work. The education system will be designed to teach advanced concepts directly, without need for the "supporting foundation" which would take too long. Already, it takes nearly 10 to 20 years of mathematical training before someone can understand plasma fusion or stellar physics. What will happen when even more complicated concepts are needed to make our society work?

Thus enters Wikipedia, which provides quick summaries of complicated issues and topics. Individuals select which pieces of knowledge are necessary and plow forward... towards... progress. Nimur 02:53, 30 May 2007 (UTC)

There will always be need for some sort of "fundamentals" that will need to be taught, but the calculus of Newton and Leibniz may not always be fundamental. However, the idea of changing the fundamentals taught, maybe not as you have described, has been tried before, see New Math. To quote the article, "the New Math produced students who had 'heard of the commutative law, but did not know the multiplication table.'". I'm not a product of this, so I wouldn't know. But we always need to remember, whatever direction the teaching of mathematics turns, there will always be a practical aspect. Root⁴(one) 14:58, 30 May 2007 (UTC)

But what about using extreme learning techniques like subliminal messegeing coulnd't you learn more if you used one of those types of learning. so shouldn't the question be how much can the brain store not how fast can we get it stored in the brain?

Why is Ostrich meat Red?

If it is part of the bird family, why is Ostrich Meat Red and not the same as a Turkey or chicken, perhaps this is a trivial question but, I would like all the facts on this matter for peac of mind. —Preceding unsigned comment added by 41.241.11.179 (talk • contribs)

See White meat article. Dr_Dima.

That doesn't really answer the question. Ostrich meat is not dark like chicken drumsticks; it looks like beef. I'd like to know why, also. I would guess it has something to do with the fact that ostriches do not fly. --Tugbug 20:17, 28 May 2007 (UTC)

Maybe ostriches are more muscular than other birds. Coolotter88 01:24, 29 May 2007 (UTC)

According to our article on red meat, "the main determinant of the color of meat is the concentration of myoglobin. The white meat of chicken has under 0.05%; chicken thigh has 0.18-0.20%; pork and veal have 0.1-0.3%; young beef has 0.4-1.0%; and old beef has 1.5-2.0%." So the question is, why does ostrich meat have more myoglobin? Because it is both large and flightless, like mammals such as cattle, ostrich muscles need to generate a lot of energy to carry the weight of the ostrich along the ground at speeds up to 50km per hour. Making energy requires oxygen, and myoglobin is the major oxygen-carrying pigment of muscle tissues. Another reason ostrich meat is so red is due to its relatively high pH (5.8 < lowest raw pH < 6.2), which results in a richer colour for a number of reasons. Rockpocket 01:50, 29 May 2007 (UTC)

Exactly. More red = more myoglobin (myoglobin contains iron, makes it red) = more exertion capacity of the muscle. Chickens and turkeys don't fly much, they run; therefore leg meat is dark, breast meat is white. Same for ostriches. Ducks fly a lot; breast meat is dark. More than that: tunas are fast long-distance swimmers; tuna meat is red. Gzuckier 19:02, 29 May 2007 (UTC)

Lagrange points with other bodies nearby

Lagrange points

Why don't the Lagrange points of, say, Sun–Earth get ruined by the other bodies in the solar system? Is it that the gravitational pull from the other bodies average out since they don't move along with the Sun–Earth system and therefore pull in different directions at different times? —Bromskloss 13:41, 28 May 2007 (UTC)

As explained at Lagrangian point#Stability, the effect of the rest of the solar system on Lagrange points is enough to complicate the orbits quite a bit, but (the rest of the solar system being quite light and a long way away) is not enough to de-stabilize the L₄ and L₅ points. It does destablize the orbits about the other three points, but this is small enough to just require slight correcting maneouvres every so often. Algebraist 17:05, 28 May 2007 (UTC)

OK, but the other objects aren't necessarily lighter and farther away than Earth, right? —Bromskloss 19:45, 28 May 2007 (UTC)

Well, they are all much further away - they may not be lighter though - the outer planets are all pretty huge. Since the gravitational forces due to those more distant bodies is inversely proportional to the SQUARE of the distance - their effect diminishes rapidly as they get further away. The distinctive factor about L4 and L5 (as opposed to the other Lagrange points) is that they form peculiar (unless you are a mathematician!) 'virtual' gravity sources - if you are close to one of those empty points in space, you'll get pulled towards them - you can even orbit them! This makes an object at one of those two points fairly insensitive to small disturbances - since this virtual gravity well will pull them back in again. But at the L1/2/3/6 points, the balance between the various forces is unstable - if you are exactly at the Lagrange point, you're theoretically in perfect balance between gravity of Earth and Moon and the "centrifugal" forces involved - but any small displacement from that point will result in you being pulled further and further away. So at (for example) L1, a teeny tiny occasional tug from (say) Jupiter - however small - will be enough to gradually move you away - and eventually you'll smack into the Moon or the Earth or get flung out of the Earth/Moon system altogether. SteveBaker 20:00, 28 May 2007 (UTC)

You say others are further away, but what exactly do you mean? There are surely instants when other planets are closer to one of the points than both Sun and Earth. —Bromskloss 20:26, 28 May 2007 (UTC)

Actually - I didn't read the question properly. I was thinking about the earth/moon lagrange points - not the Earth/sun. But still, the L1/2/3/6 points are unstable and the L4/5 points are stable - so in all likelyhood the same thing applies. SteveBaker 22:16, 28 May 2007 (UTC)

L6?? —Tamfang 07:11, 31 May 2007 (UTC)

Radioactivity

Knowing the half-life of a substance like Polonium-210 or Plutonium-239, how do I calculate how many alpha particles would be ejected, theoretically on average, per hour or so from a given sample size (say, 1 microgram of each)? --140.247.240.18 14:17, 28 May 2007 (UTC)

You start by computing the number of atoms in a microgram using the mass of a single polonium atom. Then you convert the half life to a decay rate by using the formula you can find in Radioactive_decay under the section decay timing. Now compute the number of atoms that are left after one hour using the exponential decay formula slightly above in the same article. The difference between both numbers is the number of atoms that decayed. The number of aplha particles is the same, provided that the decay sequence does not produce multiple alphas per atom.

That sounds reasonable. Thanks! --140.247.240.137 19:26, 28 May 2007 (UTC)

Distillation of ethanol, temperature

When a mixture of ethanol and water is heated and reaches the boiling point of ethanol, will the temperature stay the same (approx. 78.5°C) until all the ethanol has evaporated, or is it possible to heat such a mixture to, say 85°C, at ambient pressure? --62.16.173.45 14:34, 28 May 2007 (UTC)

The mixture will approximately stay at the same temperature until the ethanol has evaporated. It is not totally accurate, there are some effects when you go into detail. Destillation talks about the variation of the boiling point of ethanol in a solution. There is also the possibility of Superheating.

Thank you! --62.16.173.45 18:15, 28 May 2007 (UTC)

Actually, the boiling point of the mixture will change as its concentration changes, and approach the boiling point of water (100°C) as the ethanol concentration goes to zero. The boiling point of a mixture of ethanol and water has a curious dependence on concentration: Pure ethanol boils at 78.4°C; with decreasing concentration of ethanol the boiling point first decreases (!) toward a minimum of 78.15°C at 95.6% (by weight) ethanol, and thereafter increases towards 100°C at 0% ethanol. The 95.6% mixture is called an azeotrope. The evaporating gas will not be pure ethanol but a mixture of ethanol and water vapors, with the mixture depending on temperature in its own way. The vapor will in general have a higher concentration of ethanol than the liquid, which is why you can concentrate ethanol by distillation. At the azeotropic temperature, however, the vapor mixture will have the same relative concentrations of ethanol and water as the liquid. The azeotropic concentration of 95.6% ethanol is therefore the most concentrated ethanol that can be produced by direct distillation of an ethanol/water mixture (without addition of other compounds). --mglg_(talk) 16:17, 29 May 2007 (UTC)

Thanks a lot! Do you have any specific info on the boiling points at various ethanol concentrations, and on the corresponding relative concentrations of ethanol and water in the vapour phase? Is there a table somewhere on the net where I can look it up? And also, although I initially restricted the question to what would happen at ambient pressure, I am also interested in learning about the dependency on pressure, if a table that includes pressure is available. --62.16.173.45 17:03, 29 May 2007 (UTC)

• From the OP: A web search gave a partial answer to the last question. I will post a follow-up question shortly. --62.16.173.45 20:27, 29 May 2007 (UTC)

ARUN ARUN QUES ABOUT PENIS

to what extent I should retract my foreskin to wash smegma?

You should never retract your foreskin in a way that feels in painful to you. If you are still growing up, your foreskin may not be fully retractable. That is completely normal.

Right. The condition of the foreskin not being fully rectractable is called phimosis, and our article on the subject states that it is common for boys and vanishes when growing up. For older teenagers and adults, it can, however, cause inconvenience or pain. So if you are still young and it is not painful, don't worry, otherwise, better go and ask a doctor for advice. And don't get worried by the phimosis article -- to my taste it is a bit technical and medical, so if you really want to get clear information, don't feel embarrassed to simply ask a doctor. Simon A. 17:09, 28 May 2007 (UTC)

Our article on phimosis is really a bit strange. I would suggest to read this instead: http://www.nocirc.org/publish/pamphlet4.html.

Red Rice - Health benefits

What are the health benefits of the Red Rice available in India? Is the nutrition value the same as Brown Rice? - Priya

The red rice article claims that it is a "pest" or weed, because less edible grains grow per plant. I don't know if it is actually unhealthy. Nimur 00:38, 29 May 2007 (UTC)

Regression analysis for tidal water levels

I'm doing some studying on the behavior of tides. I've made a big table of water levels at different times over two days, and I made a big graph of it. Of course, the tides go according to a sinusoidal function, but both the gravitational effects of the Moon and the Sun affect tidal water levels, so therefore, the function of water levels would be in this form:

(where L is water Level and t is Time)

L = A + B sin(C(t)) + D (sin E(t + F))

Where A would be the mean water level, B would be the amplitude of the lunar component of water levels, C would represent the period of the moon, D would be the amplitude of the solar component to water levels, E would represent the period of the sun (it would be equal to 2π*24), and F would be an offset between the start of the lunar cycle and start of the solar cycle.

I'm trying to figure out the numerical values of all those lettered coefficients. Now, I could figure all of those out with astronomical/oceanographic data (the period of the sun, the moon, the mean water level) besides B and D. For those two, I'd need to do a regression analysis from the data I tabulated earlier.

I don't know how to do regression analysis by hand (though I guess I could figure it out or try to learn it), so I looked all over the internet for a regression analysis program. I found a bunch that do regression analysis, but none of them do sinusoidal regression analysis. A polynomial regression analysis would work fairly well except that to find out the corresponding sinusoidal function, I'd have to do some super-complicated reverse Taylor series...

So my question is: Does anyone know a freeware/shareware/demo program that can do a sinusoidal regression analysis for something like this? If not, is there an easy way (without spending hours messing with taylor series) to figure out how to find the corresponding sinusoidal-sum equation from a polynomial equation? Jolb 16:53, 28 May 2007 (UTC)

First, you won't see the dependency on the Sun with only two days of data. Second, local conditions such as the shape of the coastline may strongly affect the data. Now to your question: Whenever you have function which you thin is composed of a sum of sines with different amplitudes and phases, you want to do a Fourier analysis. I am sure that there is free software to do this, but you can also do it easily with even only a little programming skills, or even with a spreadsheet program. Unfortunatly, our articles on Fourier transform and Fourier analysis do not breally strike me as very accessible for beginners, so a good textbook on engineering math might be a better choice to learn about it. But for a starting point, try the following (and then learn from your textbook, why this does what it does): Assuming that you have measured the tide at N times t_i and got the values h_i, calculate

$W_{c}(T)={\frac {1}{N}}\sum _{i=1}^{N}h_{i}\cos(2\pi t_{i}/T)$

and

$W_{s}(T)={\frac {1}{N}}\sum _{i=1}^{N}h_{i}\sin(2\pi t_{i}/T),$

where T is time length that you expect to be a period. Change T a bit, from a bit less than half a day to a bit more than it. When is $W_{c}^{2}+W_{s}^{2}$ maximal? If you have more data, try also with values for T around one month. Have fun. Simon A. 17:22, 28 May 2007 (UTC)

Put the data into CoolEdit, then use the digital filtering facilities to remove , in turn, the lunar frequency and then the solar frequencies. The amplitude of the remaining signals will then give you the coefficients. The 'dc' level of the complete Unfiltered) signal will give you A. F is more tricky as it represents a phase shift-- cant think how to do that yet.--Tugjob 15:13, 29 May 2007 (UTC)

Another suggestion. Fourier series analysis doesn't work very well if you only have a short time length of data and the data doesn't cover an exact integral number of periods, or if you are not exactly sure what the period is. A more general method which works for all sorts of problems is to do a least-squares fit to the data using the Solver facility in Microsoft Excel. Put your sample times in one column, and your data in the next. In the third column, put your trial formula for L, just as you wrote it above. To do this, you will need to make a separate table somewhere of the six unknowns A, B, C, D, E and F. Put reasonable guesses for these into the table, and make the formula use them, together with your column of t values, to generate your "fitted" values for L in the third column. In the fourth column, calculate the difference between your data (column 2) and your fitted curve (column 3). In the fifth column, calculate the square of the numbers in the fourth column. Then add up all the numbers in the fifth column. This will be the sum of the squares of the differences between your data and your fitted curve. You want to make this number as small as possible by proper choice of A through F. Plot your data, the fitted values and the differences (columns 2, 3, 4 versus 1) on the same graph, and play around with the unknowns to get something close to a good fit. Then use the Solver function in Excel (under Tools|Solver; if it doesn't appear there you may have to add it in first using Tools|Add-ins|Solver Add-In). Make the target cell (the sum of the squares of the differences) a minimum by changing the cells A through F. Watch what happens on your plot to see if it's working properly. When Solver finishes, the numbers in your A through F cells will be the best fit values for these parameters. The residual differences from column 4 will show you what's left over after you subtract your fitted curve from the data. It could be random noise, or it could show that there are other influences in your data. All that's needed for the technique to work is a reasonable starting guess, and more measurements than the number of parameters you are trying to fit. Good luck. --Prophys 11:51, 31 May 2007 (UTC)

Blind sugar-eating dogs

A friend of mine once told me that sugar makes dogs go blind. Is that true? --Taraborn 18:23, 28 May 2007 (UTC)

It seems really unlikely - dogs are going to get sugars in their diet no matter how healthily you feed them. I guess the most likely issue is if the dog is diabetic. Blindness is a common symptom of untreated diabetes - and it's likely to be exacerbated by increased dietary sugar. But a healthy (or at least non-diabetic) dog shouldn't have any problems. None of the dog books I have mention sugar in their long lists of 'normal foods' that can be lethal to dogs (chocolate and grapes for example). SteveBaker 19:49, 28 May 2007 (UTC)

Is chocolate toxic to gulls, as a matter of interest? I've always erred on the side of caution and never given them any food that's remotely chocolatey. I know that it's toxic to parrots. Anyone know? --Kurt Shaped Box 21:04, 28 May 2007 (UTC)

Yes, theobromine, the stimulant component of chocolate (often mistaken for caffeine, which is only present in chocolate in trace ammounts), causes theobromine poisoning in many animals (dogs, horses, parrots) because they are unable to metabolise it quick enough. Laïka 22:05, 28 May 2007 (UTC)

I wonder what the toxic dose for a human would be? --Kurt Shaped Box 22:21, 28 May 2007 (UTC)

See Death by Chocolate? ;) --Krsont 22:40, 28 May 2007 (UTC)

I believe sugar can cause diabedes, which might in turn cause blindness. I could be wrong, though. Mdwyer 21:45, 29 May 2007 (UTC)

Good point. Our article on diabetes in cats and dogs says in fact that high blood sugar levels damage the eyes of dogs. Simon A. 07:54, 30 May 2007 (UTC)

Is it possible to get to proxima centauri in a practicle timescale?

I did some research about faster than light travel. It made me think that interstellar travel was possible, though it is overwelmingly expinsive. Is most of the second sentence variable? Fquantum ^talk 20:48, 28 May 2007 (UTC)

You don't need to travel faster than light to travel to nearby stars - it just takes a while. But yes, it'd still be prohibitively expensive, since you'd need a large ship for a voyage that takes a generation (at least). WilyD 20:51, 28 May 2007 (UTC)

Even at the speed of light it would take 4 years (though it would seem less to those actually travelling, per relativity). It's way beyond our current technology to get there. With our fastest spaceship it will take 17,000 years. See interstellar travel. --h2g2bob (talk) 21:18, 28 May 2007 (UTC)

Interstellar Travel is not a good place to look for sources, see one of the categories for details. Spacecraft propulsion would be a better place to look for sources. Thanks anyway.

Our fastest ship would arrive there a useless hunk of junk since the reliability of our engineering isn't good enough to last 17,000 years without maintenance. The solar sail approach (with massive lasers doing the work from orbit somewhere) is probably the way to go - but we don't have the technology or the funding or the will or enough knowledge of the destination to do that. Right now, we're better off building HUGE orbital telescopes - or perhaps a telescope array on the back side of the moon - we'll get more information about more stars more quickly and more cheaply than with robotic probes. SteveBaker 22:12, 28 May 2007 (UTC)

You can try it with an Ion drive powered by a Radioisotope thermoelectric generator. Both technologies are no longer science fiction, they already flew in a space craft (Deep Space 1 and Galileo (spacecraft)), however the performance of the Gallileo RTG would not be sufficient to power the Deep Space drive. You can use the most powerful planned rocket Ares V and exploit a Gravitational slingshot. Finally if you accept a flyby you can omit the break thrust (a huge contribution).

If you take this ideas to design a spacecraft, you will certainly get there a lot faster than 17,000 years. That number is quite old and based on some spacecraft velocity, that was the record at that time. It completely ignores the effect of longer firing times. I did some calculations (not the same, some similar idea) a few month ago and it resulted in something closer to 200 years. The project may be very very ambitious, but it is not completely science fiction. The worst problem would probably be reliability and the RTG power to weight ratio.

See Interstellar travel, Interstellar probe, Project Daedalus and Project Longshot. There was a heated discussion recently on the talk page of Gliese 581c as to whether human technology could send a probe to a star several lightyears away. My view on this is that sending a probe to a star several lightyears away, to be launched within say the next hundred years, and to have a travel time of a few hundred years, is an engineering problem rahter than a basic science problem. This is in the same sense that scaling up 1945 World War 2 German V2 rockets to the 1969 Saturn rocket which sent astronauts to the Moon was an engineering problem. It takes awesome amounts of energy to achieve an acceleration to an appreciable fraction of the speed of light (and deceleration at the other end if you want more than a snapshot as you pass. One technology with promise for a high speed flyby is to use near-earth lasers (perhaps on the moon) to accelerate the probe. This avoids having to carry the power source. Solar sails have some potential for accelerating a probe to a fairly high velocity by the time it leaves the solar system. Chemical rockets do not appear useful for interstellar travel if you want your probe to reach a star in less than tens of thousands of years. Ion engines have some promise. Edison 15:24, 29 May 2007 (UTC)

The solar sails are in fact under trial in experimental spacecrafts, but they lose thrust with distance from the sun, which makes them a silly choice for interstellar travel. Precisely hitting a spacecraft beyond neptune with a laser sounds plain impossible to me.

For an analysis of the energy requirement and gaps between present and required technology for an interstellar probe, see a paper presented at the 46th International Astronautical Congress, October 1995, Oslo, Norway by Geoffrey A. Landis (physicist and part-time science fiction writer), Ohio Aerospace Institute, NASA Lewis Research Center [1]. This paper does some calculations for a 30kg probe with a 760 m sail, to be powered by laser pulses for a flyby of a solar system, to send back observations by optical signal. Still requires 54GW of power with an energy cost of billions of dollars. If we were discussing in the 1830's the possibilities of electric motors powering industrial machines and trains, the costs would have looked equally prohibitive because of inefficient motors and batteries as the only sources of electric current, but these things were extremely practical within a lifetime. Edison 21:22, 29 May 2007 (UTC)

Precisely hitting a spacecraft beyond neptune with a laser sounds plain impossible to me. - but it's not a matter of hitting the spaceship with the laser so much as it is steering the sail to keep it over the center of the beam. If the mirror were maintained in a roughly parabolic shape, that tendancy to steer so as to stay aligned with the beam would be entirely automatic. That means that the challenge for the crew of the laser here on earth is just of keeping it as still as possible. Situating the laser in space - preferably far out from the influence of planetary gravity. But we can keep the Hubble telescope lined up sufficiently accuracy to take sharp photos of very distant objects with exposures measured in days - this sounds like a similar level of difficulty. The laser could also be somewhat defocussed as the mission continues - it'll deliver less power that way - but it would greatly reduce the steering problem. The bad thing would be if the spacecraft lost the beam completely at some point - that would be hard to fix with communications delays measured in years. SteveBaker 23:29, 29 May 2007 (UTC)

Also, remember that getting there is only half of the challenge. Once you've reached the other stellar system, you'll have to stop by, and it takes a massive amount of energy to decelerate a ship to a reasonable speed later on (just as much as the ship's kinetic energy, actuallt.) If you don't do this, your probe will simply zip around the star and vanish in the void of space. — Kieff | Talk 22:28, 29 May 2007 (UTC)

So you'd have to start reversing thrust at roughly the halfway point of the journey? --Kurt Shaped Box 22:33, 29 May 2007 (UTC)

With a solar sail, you'd want to use the light from the destination star to slow you down. Then various slingshot and aerobraking manouvers would get you where you needed to be. SteveBaker 23:35, 29 May 2007 (UTC)

Glasses effect

Is there any corrective effect of glasses directly on eyes? For instance if one wore glasses for 1 hour, would there be a slight sight improvement without them? --Brand спойт 21:52, 28 May 2007 (UTC)

I believe that some lenses can train the eye to focus more clearly, although it will depend on why you need glasses and how they are prescribed. I personally wore glasses until around when I was 9, and my vision is actually better than 20/20 now. -- Phoeba Wright^OBJECTION! 22:32, 28 May 2007 (UTC)

i'm not sure about the training aspect but i doubt it. the basic principal behind glasses does NOT affect how ur eye works--rather it changes the incoming image so that it can be suitable for your eyes. take them away and you're back to the image your eyes can't properly focus.

There might be effects on the lense of the eye. The lense will be in a different state of compression with or without the glasses.

This is OR but my personal experience is that I see poorer if I wore my glasses all day and then took them off. My eyes are simply not adjusted to the far away focal point and needs some time to return to my normal eye sight without glasses. --antilived^{T | C | G} 10:27, 29 May 2007 (UTC)

With children (whose eyes are still developing), glasses can be used for actual corrective effects (by forcing the kids' eyes to adapt to a certain focal condition). With us old codgers, that doesn't work.

Atlant 13:02, 29 May 2007 (UTC)

You might also find our article on orthokeratology interesting. Special contact lenses are worn by the patient at night. These lenses (with time) reshape the lens of the eye, correcting vision. TenOfAllTrades(talk) 19:11, 29 May 2007 (UTC)

What would happen if I were to eat lugduname?

What would happen if I were to eat Lugduname? would it burn a hole through my tongue? --Krsont 22:29, 28 May 2007 (UTC)

You mean if you ate crystals of pure lugduname? I can't find anything that says it's toxic, so it probably wouldn't kill you, and I don't see how it could "burn a hole through" your tongue. (Don't take my word for it, though.) However, it probably wouldn't actually taste 200,000 times sweeter than sugar. That figure means you can dilute lugduname 200,000 times as much as a sugar solution and it will taste just as sweet, but if you tasted pure lugduname, the sweetness receptors on your tongue would become saturated (i.e. almost all of them are bound to a lugduname molecule all the time), and they would send the maximum amount of nerve signals, which is still a lot less than 200,000 times as many. —Keenan Pepper 15:42, 29 May 2007 (UTC)

Net Ionic equation

Is there any website where you do a net ionic equation?

Ionic equation has information. If you learn how, you can do a net ionic equation anywhere! On the bus, in a treehouse... Nimur 00:47, 29 May 2007 (UTC)

Simple Static Mechanics Problem

I have 3 coins lying motionless on a frictionless surface. The 3 coins are all the same mass and size. The coins are arranged such that each coin is touching the other 2 coins. (the lines connecting their centers form an equilateral triangle) I apply a force F to a coin (#1) directly towards another coin (#2). In that instant, what are the forces on coins 1, 2 and 3 (the remaining coin)? Aepryus 00:00, 29 May 2007 (UTC)

The point of contact will be a theoretical "single point" on each circular edge. The force will act along the normal to the mutual tangent. This will be the same as the line connecting the centers of each coin (if they are all the same size). Since these lines form an equilateral triangle, you will have 60-degree angles. Now you just need to assume that the force you apply is totally delivered to the opposite edge of the coin (no compression); a bit of trigonometry will tell you the numerical relations of the force. It will also depend on where you push the first coin (i.e. which point you are touching); in the ideal case, your force will always act normal to the tangent line of the circle. Nimur 00:51, 29 May 2007 (UTC)

Assuming static equilibrium, the total force must sum to zero in each orthogonal direction.

I understand the basics of this problem, but it strikes me as a being perhaps a little more subtle than you describe. Whatever, the force is that is being applied from coin 1 to coin 2, it's my sense that it should be twice as much as the force being applied from coin 1 to coin 3. (since the cos(60) = 0.5) But, how to determine what the force is, I have no idea. Perhaps the vector sum of the force on coin 2 plus the coin 3 should be a vector of length F (the original force). If F2 is twice that of F3, I can solve for the forces and get F2 = 2F/sqrt(7) and F3 = F/sqrt(7). But, that would seemingly modify the direction of the force entirely, which I'm not terribly comfortable with. (BTW, this isn't a homework problem; I'm trying to implement a numerical way to handle objects in contact with one another and it seems to be a non-trivial problem) Aepryus 01:41, 29 May 2007 (UTC)

Are coin-coin contacts also frictionless? If so, all contact forces will be perpendicular to their contact edge, that is, directed from the point of contact towards the center of each coin. Some things to note:

A. Coins 2 and 3 will be pushed away from each other, so there will be no force between them.

B. In the initial motion, coin 1 will remain in contact with coins 2 and 3.

C. The reaction force on coin 1 by coin 2 will of course be opposite and equal to the force on coin 2 by coin 1, and the same goes for coin 3.

D. The acceleration of each coin will equal (1/m) times the total force on that coin.

E. The total force on coin 1 equals the external force plus the reaction forces from coins 2 and 3.

The key is to consider the arrangement of coins after an infinitesimal time dt. Each coin has moved by an amount equal to its acceleration vector times (dt)²/2, and that acceleration vector equals the total force vector on the coin divided by the mass; thus displacement is proportional to force. You know the direction of initial motion of coins 2 and 3 (away from the center of coin 1). Knowing that, the only tricky part is to work out the initial direction of motion of coin 1 that allows it to stay in contact with coins 2 and 3 while satisfying conditions C–E above. This is just geometry. --mglg_(talk) 02:48, 29 May 2007 (UTC)

Note that as described (with no other forces keeping the coins stationary) this is not a "Static Mechanics Problem": the coins will move. --mglg_(talk) 02:58, 29 May 2007 (UTC)

This problem seems to be to me spectacularly more involved than I would have thought. I think I have a solution, which is the force on the coin 2, is 5/6F; the force on coin 3, is 5/12F and the force on coin 1 is sqrt(19/144)F.

I obtained this result by first noting that the force on coin 3 is going to be half the force on coin 2 (F₂cos(theta)). And by noting that the energy will be conserved. The force applied, F for a time dt, will cause a velocity such that: F dt = m dv. dv = F dt / m. Energy will be m/2 (dv)²; which means basically that the squares of the forces before and after should be the same. So, by using some somewhat messy geometry one gets 2 equations and 2 unknowns. (F₃ = F₂/2 and F² = F₁²+F₂²+F₃². F₁ can be obtained through geometry (calculating length of diaganol of a parallelogram).

So 2 questions: is this result correct? and is there an easier away of obtaining it? Aepryus 18:26, 29 May 2007 (UTC)

It is not true that the force on coin 3 is going to be half the force on coin 2. Coin 1 will not move straight towards coin 2, because the reaction force from coin 3 will push it towards the side (of the line between coins 1 and 2) opposite coin 3. Therefore the displacement of coin 3 will be less than half the displacement of coin 2, and the same goes for the ratio of the forces on those two coins. --mglg_(talk) 19:20, 29 May 2007 (UTC)

OK, I broke down and did the math. Consider, as I did above, the infinitesimal displacements that take place during an infinitesimal time dt right after the force is applied. The displacements of coins 1, 2, and 3 are {d_1x,d_1y}, d₂{0,1}, and d₃{-sqrt(3)/2, 1/2} respectively. To maintain contact between coins 1 and 2, these will have to satisfy:

d₂ = d_1y .

To maintain contact between coins 1 and 3, they will have to satisfy:

d₃ = d₁ • {-sqrt(3)/2, 1/2} = (-sqrt(3)/2)d_1x +(1/2)d_1y .

Because forces are proportional to displacements (during this infinitesimal time interval, see my earlier posting above)), the same relations apply to forces:

F₂ = F_1y

F₃ = (-sqrt(3)/2)F_1x + (1/2)F_1y .

The total force on coin 1 equals the external force plus the reaction forces from coins 2 and 3:

F₁ = F_ext - F₂ - F₃ .

Breaking the last equation into x and y components, we get:

{F_1x,F_1y} = F_ext{0,1} - F₂{0,1} - F₃{-sqrt(3)/2, 1/2}.

Equating the left and right sides separately for x and y components, and plugging in F₂ = F_1y, we get the equations:

F_1x = (sqrt(3)/2)F₃

2F_1y = F_ext -(1/2)F₃

Together with F₃ = (-sqrt(3)/2)F_1x + (1/2)F_1y (from above), these yield three independent linear equations in three unknowns. Solving this system, I get:

F₂ = (7/15)F_ext

F₃ = (2/15)F_ext

F_1x = (1/(5 sqrt(3))F_ext

F_1y = (7/15)F_ext .

I give no guarantees at all that I did this correctly, let alone transcribed it correctly here, but even if I didn't, I think you get the general idea of how to solve your problem. --mglg_(talk) 20:31, 29 May 2007 (UTC)

Hmm, your solution doesn't seem to conserve energy. Also, it seems to me that the coins would not be in contact with one another the instant after the force was applied, so I'm a little hesitant to use such a constraint. Im guessing if I where to remove the 3rd coin and use your techinque, I would calculate a force of F/2 on coin 2 and F/2 on coin 1. Which, I'm pretty sure shouldn't be the case. I would assume in that instance the force on coin 2 would be F and on coin 1 would be 0. Aepryus 22:18, 29 May 2007 (UTC)

Huh? Energy is conserved. The energy expended by the external force is F_extd_1y = F_extF_1y((dt)²/(2m)) = (7/30) F_ext²(dt)²/m. The kinetic energy of coin n is (1/2)m v _n² = F_n²(dt)²/(2m). Summed over all three coins for the above values yields a total kinetic energy of (7/30) F_ext²(dt)²/m, which is indeed equal to the energy input by the external force.

If you remove coin 3, forces F₂ = F₁ = F_ext/2 do make perfect sense. Coin 1 pushes on coin 2, so they will remain in contact and thus undergo the same acceleration; therefore they must feel the same total force. Because they together have twice the mass of a single coin, their acceleration will be half that of a single coin subjected to the same external force; thus the force on each coin must be F_ext/2. Why is this counterintuitive?

If the total force on coin 1 were zero, as you suggest, that coin would remain stationary (right?). Why would then coin 2 feel any force (and accelerate) at all?

As for your concern about coin-coin contact: Coins 2 and 3 will indeed move apart, but coins 1 and 2 must remain in contact, because coin 1 is being pushed (accelerated) against coin 2 by the external force, and must push (accelerate) coin 2 out of the way. Similarly, coins 1 and 3 must remain in contact (to start with). --mglg_(talk) 23:28, 29 May 2007 (UTC)

I think I was ambigiuous in the description of the problem. The force I'm looking at is an impulse force, not a constant force; and I'm really only looking at the instant the force is applied before any motion occurs (which is why I perhaps erroneously called it static). As far the 2 coin instance, imagine billiard balls. If you hit one billiard ball of equal mass directly into another billiard ball the energy and momentum is entirely transfered to the other ball; the first ball coming to a complete stop at the point of contact. (Assuming no angular momentum) A force is applied to coin 1, coin 1 then applies that same force to coin 2, every force has an equal and opposite force, so coin 2 applies an equal force in the opposite direction on coin 1. The net force on coin 1 is zero. Perhaps an even better example is newton's cradle, where the steel balls in the center don't move at all but transfer the force to the ball adjucent to them. As far as energy is concerned it should be proportinal to the force applied. (Fdt = mdv --> dv = Fdt/m) K = m/2(Fdt/m)^2 = (Fdt)^2/(2m). which becomes 1 = (7/15)^2 + (2/15)^2 + (sqrt(52)/15)^2 = (49 + 4 + 52)/225 = 105/225 which is not equal 1. Aepryus 00:37, 30 May 2007 (UTC)

I see – I suspected that maybe you were really thinking about impulses... In an impulse scenario, you shouldn't think about forces at all (since they are all infinite), but rather formulate the question in terms of impulses (integral(F dt)) and momenta (m v). In this case, I think the situation is somewhat ill defined. Consider a situation where there is a minute gap between coins 1 and 3. Then coin 1 impacts coin 2 first, transfers all its momentum to coin 2, and stops (as you explained). Thus coin 3 is never impacted, and remains stationary! If instead there is a minute gap between coins 1 and 2, coin 1 will impact coin 3 first, send it off at 60°, and then impact coin 2 at an angle, sending coin 2 moving upward and itself continuing straight to the right (if coin 3 was at the left side). Since initial situations that differ only infinitesimally yield qualitatively different results, you may want to reconsider the problem definition. --mglg_(talk) 01:11, 30 May 2007 (UTC)

(By the way, if you include the factor of two from (2m) in the denominator of your own energy expression, it will become 105/(2*225) = 7/30 which you will recognize from my kinetic energy result above. Why you think this should equal one I don't see; it should instead equal the energy expended by the external force, which I calculated above and has the same value. But that was all for the continuous force case, which is not what you are interested in.) --mglg_(talk) 01:18, 30 May 2007 (UTC)

Apologies to mglg for branching off the root instead of helping, but I'd like to propose a different approach. Let us assume, as can be proven, that coins 2 and 3 exert no force on each other. Then what we know, at any given instant where the coins form the 2-1-3 chain, is

1. $m({\ddot {\vec {x}}}_{1}+{\ddot {\vec {x}}}_{2}+{\ddot {\vec {x}}}_{3})={\vec {F}}$ (conservation of momentum for all coins)
2. ${\ddot {\vec {x}}}_{2}\|{\hat {\alpha }},{\ddot {\vec {x}}}_{3}\|{\hat {\beta }}\;\;({\hat {\alpha }}:={\frac {{\vec {x}}_{2}-{\vec {x}}_{1}}{|{\vec {x}}_{2}-{\vec {x}}_{1}|}},{\hat {\beta }}:={\frac {{\vec {x}}_{3}-{\vec {x}}_{1}}{|{\vec {x}}_{3}-{\vec {x}}_{1}|}})$ (conservation of momentum for coins 2 and 3 individually; forces are radial)
3. ${\ddot {\vec {x}}}_{2}-{\ddot {\vec {x}}}_{1}\perp {\hat {\alpha }},{\ddot {\vec {x}}}_{3}-{\ddot {\vec {x}}}_{1}\perp {\hat {\beta }}$ (coins are rigid and non-overlapping, and remain in contact)

So we can reduce the number of variables (currently 6) to 4 by using equations 2 and parameterizing ${\ddot {\vec {x}}}_{2}:=a_{2}{\hat {\alpha }},{\ddot {\vec {x}}}_{3}:=a_{3}{\hat {\beta }}$. Then equations 3 give ${\ddot {\vec {x}}}_{1}\cdot {\hat {\alpha }}=a_{2},{\ddot {\vec {x}}}_{1}\cdot {\hat {\beta }}=a_{3}$. Suppose (as is the case at $t=0$ in the given problem) that ${\hat {\alpha }}\!\!\not \!\|{\hat {\beta }}$, so that this information determines ${\ddot {\vec {x}}}_{1}$: writing M for the 2x2 matrix whose rows are ${\hat {\alpha }},{\hat {\beta }}$, and writing ${\vec {a}}:={\begin{bmatrix}a_{2}\\a_{3}\end{bmatrix}}$, we have $M{\ddot {\vec {x}}}_{1}={\vec {a}}\Rightarrow {\ddot {\vec {x}}}_{1}=M^{-1}{\vec {a}}$. Going back to equation 1, and being very clever with ${\ddot {\vec {x}}}_{2,3}$, we then have $m(M^{-1}{\vec {a}}+M^{T}{\vec {a}})={\vec {F}}\Rightarrow {\vec {a}}={\frac {1}{m}}\left(M^{-1}+M^{T}\right)^{-1}{\vec {F}}$. Let us set coordinates such that the line between coins 1 and 2 is the x axis, with coin 3 below: then ${\hat {\alpha }}={\begin{bmatrix}1\\0\end{bmatrix}},{\hat {\beta }}={\frac {1}{2}}{\begin{bmatrix}1\\-{\sqrt {3}}\end{bmatrix}}$ and ${\vec {a}}={\frac {F}{15m}}{\begin{bmatrix}7\\2\end{bmatrix}}$, so ${\ddot {\vec {x}}}_{1}={\frac {F}{15m}}{\begin{bmatrix}7\\{\sqrt {3}}\end{bmatrix}}$. Thanks to mglg for spotting my arithmetic errors. --Tardis 23:13, 31 May 2007 (UTC)

Sorry, I have just noticed this post. I haven't spent too much time analyzing it yet. But, I wanted to make sure that we are on the same page before I attempt that. In equation 3, you state that "coins are rigid and non-overlapping, and remain in contact". I think for my case it is not the case that coins need to remain in contact and actually I believe for t>0 they are not in contact.

My true motivation in trying to understand this problem better is so that I might be able to better understand the following paper: [2]. I am trying to write a physics engine that is able to handle the motion of objects resting upon one another. Here is my current application:[3](requires java 1.5 or higher) Before I try to delve into the mathematics of that paper I wanted to understand a concrete simple example.

On the issue of conservation of energy, my thinking is as follows: If I apply a Force F for a time interval dt on a mass m. I can calculate the change in velocity as follows:$Fdt=mdv\rightarrow dv={\frac {Fdt}{m}}$. There will be no potential energy in the system, so the total energy will be $E={\frac {1}{2}}mv^{2}\rightarrow E={\frac {1}{2}}m({\frac {Fdt}{m}})^{2}\rightarrow E=F^{2}{\frac {dt^{2}}{2m}}$ In a similar progression the energy of the system of the 3 coins should be: $E=E_{1}+E_{2}+E_{3}=F_{1}^{2}{\frac {dt^{2}}{2m}}+F_{2}^{2}{\frac {dt^{2}}{2m}}+F_{3}^{2}{\frac {dt^{2}}{2m}}$ Conservation of energy lets us assume the energy before and after will be the same. So by equating the 2 equations and dividing by ${\frac {dt^{2}}{2m}}$, one gets: $F^{2}=F_{1}^{2}+F_{2}^{2}+F_{3}^{2}$.

I greatly appreciate both of your help and any further help on this will be greatly appreciated. Aepryus 02:49, 5 June 2007 (UTC)

The crucial distinction you must understand is between a "continuous force" problem and an "impact" problem. The continuous force problem is what Tardis and I solved above (only for t=0 in my case); the impact problem is what you are thinking about when you mention Newton's cradle or colliding billiard balls completely transferring momentum. The difference between the two types of problem lies in the time scale of force application. For a moment, consider the coins not as idealized rigid bodies but as real elastic objects. If two such coins collide with one another, the regions of the coins near the impact point will start to deform, and will start to apply elastic forces to the two coins. After some time t0, the elastic forces will have accelerated the two coins enough that they have moved apart and are no longer in contact. For real metal coins, the impact contact time t0 is typically pretty short, on the order of a few microseconds. If the time of application of your external force is long compared to t0, then you are in the continuous force limit. If on the other hand the application time of your external force is short compared to t0, you are in the impact limit. In the impact limit, you can often use thinking involving sequential elastic collisions between pairs of coins: If for example we assume a minimal but non-zero gap between coins 1 and 3, then the impact problem becomes trivial: an external force F_ext imposed on coin 1 for a short time t₁ imparts a certain momentum F_extt₁ to coin 1; coin 1 then collides with coin 2 and transfers all its momentum in the way you described, and stops; coin 3 is never touched and remains stationary. (Note: this is not the solution to the three-coin impact problem, since a very different result is reached if one instead assumes a small gap between coins 1 and 2.)

The paper you refer to is explicitly considering the continuous force problem only ("We will not consider the question of impact in this paper; thus, we assume that the relative normal velocity of bodies at each contact is zero."). In your problem formulation, you ask about well-defined forces existing simultaneously between all the coins; by assuming that such forces exist simultaneously, you are in fact considering the continuous force problem. The characteristic feature of an impact problem is that everything changes very rapidly with time, and forces develop and disappear between different coin pairs at different times, so at no time is there a self-consistent quasi-static set of forces between all coins. On the other hand, when you calculated the energy being input by the external force, you assumed that only the mass of a single coin was being accelerated by the external force, which only makes sense in an impact problem: in the continuous force problem, the external force accelerates all the coins. Since the paper you seek to understand is treating the continuous force problem, that is what you should think in terms of. In that case, the classic Newton's cradle demonstration is the wrong model to have in mind; rather think of a finger slowly pushing on the first ball in the cradle, and making all the balls swing away as one mass. The continuous force case was solved by Tardis (generally) and me (for t=0 only) above. Please ask if any of that reasoning is unclear. --mglg_(talk) 16:41, 5 June 2007 (UTC)

Doh -- ok sorry about being so dense. Now, I just need to figure out both of the solutions... Aepryus 18:39, 5 June 2007 (UTC)