# Wikipedia:Reference desk/Archives/Science/2008 September 8

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# September 8

## Significant figures - fun stuff

If you're a professional scientist, you're probably laughing right now, but sadly I am but a poor high school student and sig figs are imprinted on us as the universal method of...well, messing up otherwise perfectly functional math. Anyway, I have a quick question regarding sig fig arithmetic: Suppose you have two numbers of equal orders of magnitude and the same number of sig figs, and you subtract one from the other. How many sig figs does the answer have? For example:

```0.125 g - 0.106 g = 0.019 g
```

Then you go off and use the 0.019 g in other calculations. Do you treat it as 3 sig figs (because the data used to obtain it had 3) or 2 (its face value)? In other words, are sig figs based on the numbers you put in, or the lowest point achieved during calculations? Another example:

```25.983 amu - 24.986 amu = 0.997 amu
```

Again, are there 5 or 3 sig figs? Thanks, FlamingSilmaril (talk) 00:03, 8 September 2008 (UTC)

What you wrote as an answer is correct. The rule that the answer should have the same amount of SigFigs as the number with the least amount of SigFigs only applies when you are multiplying or dividing. When adding or subtracting, the answer should have the same amount of decimal places as the number with the least amount of decimal places.
So for the first one, the answer is "0.019" and has 2 SigFigs. For the second one, the answer is "0.997" and has 3 SigFigs.
Acceptable (talk) 00:42, 8 September 2008 (UTC)
Unfortunatey, that rule is too simplistic and ends up with false precision. --Tango (talk) 00:45, 8 September 2008 (UTC)
It might be best to thing of it in terms of the size of the possible error. 0.125 means 0.125 +/- 0.0005 and 0.106 means 0.106 +/- 0.0005, so the sum becomes (0.125 +/- 0.0005) - (0.106 +/- 0.0005), the maximum value that could give is 0.1255-0.1055=0.200 and the minimum is 0.1245-0.1065=0.0180, so the answer is 0.019 +/- 0.001. That can't be written in terms of significant figures. If we say "0.019 (2sf)" then we mean 0.019 +/- 0.0005, which is more precise than what we actually have, so that would suggest we need less significant figures, so we try 1sf, which gives 0.02 +/ 0.005, which is less precise that what we actually have. Therefore, the best we can do (if you want to stick with significant figures) is to quote it as "0.02 (1sf)", but you've lost some precision by doing that. To quote it was "0.019 (2sf)" (or, even worse, to 3sf) would be false precision, so you certainly can't do that. If you need a final answer in terms of significant figures, your best option is to do all the calculations using figures as precise as you can and carrying through all the +/- terms as I did above and then round the final answer to the most significant figures you can justify (this will almost always be less than what you started with). --Tango (talk) 00:45, 8 September 2008 (UTC)
The +/- notation certainly seems like a more accurate way to depict uncertainty, but for now I have to use sig figs. What was confusing me was that 0 in the tenths place - the tenths place was significant in both the input data values, but I guess it sort of drops out when you subtract? FlamingSilmaril (talk) 01:11, 8 September 2008 (UTC)
If you don't want to go through the hassle of carrying through all the errors, you should at the very least knock one sig fig off the final answer, it's virtually never going to be accurate to the same as you started with (so in your first problem above, state the answer as "0.02 (1sf)" (you don't have to specify the "(1sf)" part, it's implied by the fact that you haven't written any trailing 0's, but it might be best to just to make it clear that it's an approximate answer). The error represented by a given number of significant figures depends on the magnitude of the number, when you subtract two numbers that are close together you get a number with a smaller magnitude, which means the same error is represented by fewer significant figures (however, the actual error is greater than the error you started with, hence the loss of an extra digit). --Tango (talk) 01:23, 8 September 2008 (UTC)
Just say no to sig.figs. It's useful as a very rough rule to try to avoid quoting more digits of precision in your answer than in your source data. But please - only use that for "back of envelope" calculations. For anything formal - or anytime the answer matters - you have to do as User:Tango says - you have to track the maximum and minimum possible value for every number in your source data and carry that error range through every step of your calculation.
For some bizarre reason, highschool teachers have got it into their heads that the sig.figs thing is a vitally important, cast-iron technique that all scientists use all the time - so it has to be drilled into the heads of their victimspupils. In truth it's a kinda scruffy rule of thumb that basically stops people from claiming far too much precision in their results. It allows you to glance at a calculation someone gives you and be immediately suspicious of it if there are more digits in the result than in the input.
Furthermore, you need to know what the nature of those errors are. Suppose you're taking the average of ten numbers. If those numbers came from (say) a human measuring something with a tape-measure - then the errors are likely to be random - and normally distributed. Calculating the average of 10 numbers increases your precision because those errors average out. But if the error came about because the person was using a computer program that truncates input data instead of rounding it - then all of the error is in the positive direction and the average of those ten numbers is no more accurate than the input data. It's a very subtle business - and not one that I trust to these rules put out by well-meaning math and science teachers to schoolkids.
Of course, most kids fall asleep during this stuff because it seems so divorced from reality. A practical anecdote might help here: I'm in computer graphics for video games - and having a really good feel for the size of the errors and the sensitivity of the math to those errors ends up being critical. We deal with colors on the screen of your computer that are integers in the range 0 to 255 - about two and a half "sig figs". But the math that determines how light reflects off of shiney objects entails taking the direction of the incoming light, the surface direction and the direction of the eye/camera, then you calculate the cosine of the difference between the angle of incidence and the angle of reflection and raising it to the power of the shininess factor. OK - but the shininess factor is often a big number - 20 to 200 might be typical. Well, when you subtract two numbers that are similar in size - then raise the result to the power of 200 - the number of sig figs you need in your input data skyrockets! We typically need numbers that are accurate to one part in a billion to get good results - so the input data needed something like 8 or 9 sig figs. Urgh! On the other hand, if the object isn't shiney - so the coefficient of that calculation is more like 1 - then the input data only needs about 3 sig figs. We can take practical advantage of that by storing and handling the data to rougher precision for dull objects than for shiney ones. That means we need to send less data between the computer and the graphics card for dull objects - that means we can draw a lot more dull objects than shiney ones. THAT means that we can tell our artists that if they have their monsters NOT be slimey - they can have more monsters.
Proper error understanding means more monsters!...Math in action where it matters!  :-)
SteveBaker (talk) 01:24, 8 September 2008 (UTC)
Unfortunately, he can't just say no, he has no choice but to do his homework in the way he teacher wants it done. Sig figs are useless for doing any kind of calculations, where they are useful is in writing down uncertain numbers. It's much quicker to write "52.50" and have everyone know it means "52.50 +/- 0.005" than write "52.50 +/ 0.003", so you do it even though you're losing some precision. It's a good rule of thumb for knowing you precise your starting values are and for stating how precise your final figures are, but for everything in between, they're useless. They're often useless even for that, as an example I just came up with shows: Consider the sum 135+120, both figures are to 3sf. If we work out the correct answer we get 155 +/- 1, if we want to save space by not explicitly writing the error term we need to find a number of sig figs that doesn't mean we end up with false precision. 3sf clearly isn't going to do it, but it turns out neither does 2st since rounding to 2sf would give us 160 +/- 5 (don't get me started on how to round 5's!), which means the answer is somewhere between 155 and 165, but the answer might actually be 154, so we have to go all the way to 1sf and say it's 200 (1sf), which is correct but 50 times less precise than it could have been, so the number is probably completely useless. --Tango (talk) 01:55, 8 September 2008 (UTC)
I understand - and I agree. 100%. I just don't want students to graduate from high school imagining that the sig figs approach is any more than a rough rule of thumb. It's not taught that way - which is a major problem. When people start telling me that this is some kind of solid rule that "scientists" use - I get upset! In my job, having a solid feel for where precision get's lost - how to avoid losing it - where you need it and where you don't - is literally the difference between a great-looking video game and a terrible one. On the other hand, when someone on the WP:RD says that 220 million tons of coal is enough to last the UK for 400 years - and I do the math and wind up with just one year, it's nice to NOT say 0.845 years - and it's nice to know that it really couldn't really be 400 years because of roundoff errors - but it's a 'back of envelope' calculation and it's really not necessary to carry through anything more than a rough idea of the numbers of sig figs I'm using. So - as I said - the sig.figs approach is fine for 'back of envelope' calculations - but useless for serious work where the results actually matter. If an engineer designs a bridge that I have to drive over...I want accurate error tracking! SteveBaker (talk) 02:43, 8 September 2008 (UTC)
Ya gotta love WP:RD...in three hours I've gotten more helpful info on precision than in a year's worth of high school chem. Thanks a lot, everyone. This was very insightful! FlamingSilmaril (talk) 03:05, 8 September 2008 (UTC)
Thanks for your kind words - but PLEASE learn what your teachers teach - you need to get through exams, and if they require you to learn a pile of crap - go learn the pile of crap. Feel free to forget it later - but you need those grades! SteveBaker (talk) 17:29, 8 September 2008 (UTC)
Yeah, well said, Steve... I remember when I learned the real relationship between sig figs and careful error propagation, I thought, "I am way too old to be learning this for the first time." --Allen (talk) 07:26, 8 September 2008 (UTC)
Yep - me too. SteveBaker (talk) 17:29, 8 September 2008 (UTC)
Unfortunately, once you make it out of high school and into the realm of "professional science," there are all kinds of new silly specifications such as font-size and formatting minutia necessary for paper-publishing (which is in itself a dark and horrible cavern). It seems the same people who sit on school boards determining that SigFigs are relevant to high-schoolers are also sitting on paper-reviewing committees. I recommend obtaining and thoroughly learning a good word-processing suite which can handle formatting AND sig-figs so you can focus on the scientific task! As far as your data-processing, most of us do not have the constant need for real-time performance that Steve does, so we can perform over-kill and use more precision than necessary. But this is hardly a substitute for understanding the mathematical concept of error propagation. Even 32 or 64-bit calculations will only return what you asked them to return. Nimur (talk) 14:24, 8 September 2008 (UTC)
Or work in a discipline that uses software designed for scientific writing, rather than fighting with an office suite... -- Coneslayer (talk) 16:17, 8 September 2008 (UTC)
This stuff isn't only about writing research papers. Professional scientists do lots of actual practical work without ever writing a single paper in their entire lives. The discipline of maintaining correct precision throughout your calculations isn't just about getting your paper accepted - it's also about not having your bridge collapse - or (as in my case) making cooler graphics for your Xbox. That's an important thing - so many people do math and science in school, fondly imagining that they'll NEVER need to use any of this stuff in "real life" - when in fact, they really should. SteveBaker (talk) 17:27, 8 September 2008 (UTC)
If you are adding ten numbers, and each is +/- 0.1, why would you assume that the sum could be off by 1.0, which is the sum of the uncertainties? Wouldn't it make more sense to treat each one as a normally distributed uncertainty and take the square root of the sum of the squared errors? This would be the square root of 10 times 0.12 or .32? Edison (talk) 19:00, 8 September 2008 (UTC)
It depends on where your data comes from. For example, I once worked with some data-collection software with a strong bias for "+" in its data, so a total error of +1 was far more likely than a total error of -1 --Carnildo (talk) 23:25, 8 September 2008 (UTC)
Sometimes you do do things like that, it depends on the type of measurement. If the error is normally distributed, then you're right, but sometimes it isn't. One (rather bad) example would be where the error comes from the data having been previously rounded, then "+/- 0.5" doesn't means "there's a 68% chance it's between 0.5 less and 0.5 more" it means there's 100% chance, and the error isn't normally distributed, it's uniform. In that case, ten lots of +/- 0.1 does make +/- 1. (That's a bad example because you should never use data rounded off to a precision so low that it's a significant source of error, but there might be the odd occasion where you have no choice.) --Tango (talk) 09:56, 9 September 2008 (UTC)
Significant figures should be thought of as a way of introducing accuracy; it is not intended to be a rigorous mathematical tool. Perhaps sig figs are best thought of as another name for relative accuracy--that is, the error margin of a certain value compared to the value itself. To show how the relative accuracy is useful, consider a census. If a tiny community's population is reported to within 5 individuals, nobody would be impressed, but it would be shocking if the population of the entire U.S. is reported to within 5 individuals.
The example 0.125 g - 0.106 g = 0.019 can be thought of as: (1) There are two measurements that are reasonably accurate to within a certain absolute error margin. (2) When subtracting one number from another, the result should have double the absolute error margin. (3) Since the absolute error margin has been doubled and 0.019 is much smaller than 0.125 or 0.106, the error margin makes up a larger percentage of the value. (4) Intuitively, 0.019 should have less sig figs than the two original values.
I remember being taught sig figs in science class and calculating the actual error margins out of curiosity. The concept really is almost useless, and for that reason my teachers never required us to round answers appropriately on evaluations. It's useful to know that writing 2/3=0.6666666666666666666666666666666666666666666667 is not necessary, but that's likely as far as your chemistry teacher is going to go. --Bowlhover (talk) 05:00, 9 September 2008 (UTC)
This opens up another whole can of worms of course. If your data truly does have a normally distributed error pattern - then statistically, one sample can have an error that's a million times bigger than the "typical" value and even though you're averaging 10,000 samples together, that one outlier is enough to blow your error estimates off the chart. Worse still, if you multiply something with a normally distributed error by something else with a normally distributed error - you end up with a gaussian-squared error profile. Tracking the NATURE of the error through all of your math becomes a major mathematical nightmare. But at this point, you're more into the area of full-blown statistics than simple error tracking. However, even in relatively simple cases, this kind of thing becomes an issue. The round-off error in a floating point number in a computer depends on the magnitude of the number - the round-off error in an integer does not. SteveBaker (talk) 15:55, 9 September 2008 (UTC)
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## Dark skin and dominant/recessive traits

I know that darker eyes are a dominant trait contrasted with lighter eyes like blue eyes, which is a recessive trait. But what about darker skin? Is that a dominant trait as well contrasted with lighter skin? Is lighter skin a recessive trait compared with darker skin? ScienceApe (talk) 00:03, 8 September 2008 (UTC)

Unfortunately, it's not that simple, since there isn't just "dark skin" and "light skin", there is a whole range of colours. If someone with very dark skin and someone with very light skin has child, it will generally have skin somewhere inbetween. Talking about dominant and recessive traits only really works when the trait is governed by a single gene, which isn't the case with skin colour. --Tango (talk) 00:30, 8 September 2008 (UTC)
It's not even actually the same with eye color, which is controlled by a number of genes but can usually be simplified. --98.217.8.46 (talk) 13:24, 8 September 2008 (UTC)
I haven't read it, but we have Skin_color#Genetics_of_Skin_Color_Variation. --Allen (talk) 12:01, 8 September 2008 (UTC)
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## Robustness or lack thereof in "B-rex" (dinosaur question)

I know the validity of the robust/gracile morphotypes is uncertain, but does "B-rex" (MOR 1125)--the Tyrannosaurus specimen that was discovered to have medullary bone--fall into the robust or gracile morphotype? I know the robust morphotype is generally considered as more likely to be female, but am curious as to whether or not MOR 1125 supports this or not. 75.211.143.162 (talk) 01:47, 8 September 2008 (UTC)

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## Is this radiant energy exploitation possible?

1) Let there be a plane where the temperature is the same on both sides.

2) Let the plane have a hole in it. lets call one side the sending side and the other the receiving side.

3) Assume that there is a device on the sending side that will redirect and focus all ambient radiant energy that arrives in its collector and that its collector is the same size as the hole. Therefore the amount of radiant energy passing through the focus of the device would be the same as that passing in one direction through the hole in the plane.

4) Assume that this device is so arranged as to focus its output on the hole so that its focused energy is radiated through the hole in addition to what normally would pass through the hole.

5) In this case the receiving side should increase its energy (temperature) levels until its emissions through the hole matched those of the sending side. The energy (temperature) levels of the receiving side would be increased until its emissions matched those of the IMPLIED energy (temperature) levels of the sending side.

This is of course completely contrary to the 2nd law of thermodynamics as we would be moving energy from a colder to a hotter body and we would be decreasing entropy at the same time.

Some calculations -

Assume temperature of both sides starts at 300K (27 C) and a hole size 1 cm.

Therefore the amount of radiant energy passing normally through the hole is -

300^4 /10000 = 810,000 energy units (units of measure are not important because we work back to temperature)

We double the amount of energy passing through the hole by using the device.

810,000 * 2 = 1,620,000 energy units

The implied temperature of the energy passing through the hole is therefore

SQRT(SQRT(1,620,000*10000)) = 356.76K (83C)

Therefore the temperature of the receiving side will have to increase over time to 356K to balance the amount of energy passing through the hole.

Can you help and point out what I am missing? Or perhaps this actually works and for some reason that I cannot see, I don’t actually violate any laws of science.

Thank you.

John —Preceding unsigned comment added by Habanabasa (talkcontribs) 06:49, 8 September 2008 (UTC)

You cannot have your assumption 3 device, which in itself violates the second law (the directed radiation has less entropy than the original undirected radiation). --Stephan Schulz (talk) 06:57, 8 September 2008 (UTC)

Sounds a bit like Maxwell's Demon. DMacks (talk) 16:57, 8 September 2008 (UTC)

I agree with Stephan - step 3 is impossible - this "focussing" device breaks at least the second law of thermodynamics - since that's bogus, the whole concept fails. Maxwell's Demon is a more subtle problem - and the reasons why it fails are really rather subtle. SteveBaker (talk) 18:25, 8 September 2008 (UTC)
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## will the world ever end?

i would like to know if the worldwill ever end because people have been saying we are going to go through the black hole. is it true? —Preceding unsigned comment added by 124.179.167.13 (talk) 08:35, 8 September 2008 (UTC)

Probably not, no. Unless you hear about it everywhere - all newspapers, all TV channels etc - then it's most likely a hoax. The life on Earth will probably end in a few billion years, though, when the sun burns out and expands. -- Aeluwas (talk) 09:12, 8 September 2008 (UTC)
You're probably referring to the (fairly tiresome) controversy surrounding the Large Hadron Collider. No, it's not true; you can read more about the LHC and black holes here. In short, though: there's no reason to believe that there's any threat to Earth. Well, not from the LHC, anyway. (Also, you might want to think this through a little bit: if the world was known to end real soon now, don't you think there'd be some kind of serious discussion about it on a level that you couldn't help but notice instead of vague rumors and whatnot?) -- Captain Disdain (talk) 09:17, 8 September 2008 (UTC)
Certainly the world will eventually end. It's not likely to be a black hole that does it - but our own sun. As the sun starts to run out of fuel, it will gradually expand outwards, getting cooler and redder - swallowing up Mercury, then Venus and possibly also the Earth itself. But even if we don't get swallowed by the sun, the heat will boil away our oceans and probably blow away our atmosphere - eventually melting the rocks and everything mankind has ever worked for. Fortunately, the sun is barely halfway through it's fuel supply so this horrible event is many billions of years away. So mankind has plenty of time to think about building gigantic spacecraft and to start finding a younger star to live nearby. SteveBaker (talk) 17:20, 8 September 2008 (UTC)
There is an article on the end of the world which talks about even more ideas. Adding humans to the planet greatly increases the chances that something diffferent will end the world. Even if the LHC cannot destroy the earth, a hypothetical man made black hole that is big enough hypothetically could. Graeme Bartlett (talk) 21:21, 8 September 2008 (UTC)
Sure, there are all kinds of hypothetical ways we could destroy the world. A black hole is one of the more unlikely ones - nuclear war is several orders of magnitude more likely, for example. --Tango (talk) 09:59, 9 September 2008 (UTC)
It depends on what you mean by "end of the world". End of human civilisation? Probably in the next few million years. End of life on Earth? Likely to take millions, if not a few billion, years. End of the Earth as a discernable entity in space? Wait for the Sun to go nova, as described above. End of the universe? See what the article says. But destroyed by a microscopic black hole, or by an amazing planetary alignment, or just because we hit a date in some calendar that happens to coincide with the end of a cycle? Probably not in my lifetime. Confusing Manifestation(Say hi!) 23:09, 8 September 2008 (UTC)
Our sun will not "go nova". It will expand into a red giant and then collapse into a white dwarf. You may also want to read up on the difference between a nova and a supernova. --Tango (talk) 09:59, 9 September 2008 (UTC)
No, it won't produce a nova - for that you need a binary star. But at the end of the Red Giant phase, there would be a "helium flash" - which would be just as exciting as a nova if you happened to be nearby! The hydrogen in stars the size of the sun gradually runs out - forming helium as it does so - and the core of the star collapses while the outer layers are expanding to form the red giant which is what will kill the earth. When the core has collapsed enough, (long after the Earth is toasted to a cinder) the sun will reach a temperature that is enough to start a helium fusion reaction (producing carbon). That's not a gradual thing - it happens for just a few seconds - and during that time the star pumps out a hundred billion times it's normal amount of energy! It's not a nova - but it's pretty damned impressive! This blows off the outer "red giant" stuff to form a "planetary nebula" while the core collapses into a white dwarf which eventually starts fusion reactions with carbon and onwards to higher and higher atomic weight byproducts. However, that exciting "helium flash" won't happen until the Earth is long-gone. SteveBaker (talk) 15:34, 9 September 2008 (UTC)
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## sufuric acid

how can i prepare a solution of 0.13 M of sulfuric acid —Preceding unsigned comment added by 213.207.34.76 (talk) 08:38, 8 September 2008 (UTC)

Put 0.13 mol of sulfuric acid in a 1l flask and fill up with water. (molar mass of sulfuric acid is in the table at the artikle)--Stone (talk) 09:41, 8 September 2008 (UTC)
Don't do that! For safety reasons, it is very important to add sulfuric acid to water and not water to concentrated sulfuric acid. As I understand it, putting water into concentrated acid can cause the acid to spatter. Wanderer57 (talk) 13:06, 8 September 2008 (UTC)
... as shown in this sign, ALWAYS add acid to water. Nimur (talk) 14:08, 8 September 2008 (UTC)
Always add acid? I often prefer my water acid-free -- MacAddct  1984 (talkcontribs) 14:58, 8 September 2008 (UTC)
The 'add acid to water - not water to acid' rule is important. The reason is that the resulting liquid can indeed splash, 'fizz' and splatter. If you screw up and add water to acid then the liquid that's splattering is concentrated acid...not good. If you do it the right way around and pour the acid into water - then the splattering is of water - or at least, very dilute acid. But even when you do it right - you have to be really careful. SteveBaker (talk) 17:12, 8 September 2008 (UTC)
Also the heat of solution seems less sudden. Acid-to-water, a small amount of acid dissolves, so a small amount of heat in a large container of water. Water-to-acid, lots more energy released by the solvation and for ultimately dilute solutions there is usually a very small solution volume with all that energy, so lots more heat. I've cracked several flasks by adding water to acid and by not using sufficient cooling. DMacks (talk) 17:35, 8 September 2008 (UTC)
Sulfuric acid is usually provided as a concentrated solution in water. (Usually 18 M, although your stock solution may be different - examine the container to be certain.) After determining the volume of dilute solution you need, calculate the number of moles of sulfuric acid you need for that volume (for 1 L, it's 0.13 mol; for 100 mL, 0.013 mol). Then determine the amount of concentrated solution you would need to give you that molar amount of sulfuric acid. (Once you get the hang of it, you can save time on the calculation by using the c1*v1 = c2*v2 trick.) Measure that out, and add it to almost-but-not-quite the amount of water you would need to bring the total volume to the desired volume, and once mixed bring the solution to the final volume with a small amount of water. (The convoluted procedure is needed because volumes aren't always additive.) This proceedure should have been explained in your beginning laboratory course - if you are unsure about your technique, I would recommend seeking skilled supervision before working with concentrated acids (... and then ask them to show you how to do it). -- 128.104.112.147 (talk) 15:48, 8 September 2008 (UTC)
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## TV detector vans

As many will be aware, everyone in the UK who watches television is legally obliged to own a TV licence. The money from these funds the BBC, since it is a non-commercial broadcaster. Anyway, enforcement of TV licence ownership is a big issue; see Television licensing in the United Kingdom#Licence fee enforcement. As that article states, TV detector vans are used in an attempt to trace homes that are watching TV without a licence. But there have long been suspicions that these vans are fake and just employed as a deterrent. This article states that the technology used by the vans is top secret. So, are TV detector vans really able to detect when a TV is on in a house? And what is this supposedly top secret technology? --Richardrj talk email 08:59, 8 September 2008 (UTC)

The "top secret" technology is most likely nonsense perpetuated from a simpler and more credible credulous age because:
1. I don't see how any "detector van" can possibly detect people who are watching non-broadcast television e.g. via cable or the internet (but, yes, they do still need a television licence).
2. This information leaflet from the BBC Licensing Authority says "TV Licensing’s policy is to visit all addresses where people inform us that no television is in use at the property" and "TV Licensing’s policy is to visit all addresses where people inform us that a black and white television is in use at the property" (and it makes no mention at all of detector vans !). So if the Licensing Authority makes follow-up visits at all properties which claim they do not need a licence or claim they only need the cheaper black and white licence, why invest in complex and expensive technology as well ? Gandalf61 (talk) 10:39, 8 September 2008 (UTC)
Do you mean a more credulous age? Algebraist 10:43, 8 September 2008 (UTC)
Yes, indeed. Gandalf61 (talk) 10:51, 8 September 2008 (UTC)
You can't generally do it sitting in a van, but it is fairly easy to send an electrical pulse down a cable line and estimate the number of splits and connected devices by measuring the reflections that come back. It only works if the number is small, so you need to be close to the end user, but it is a fairly standard technique for detecting illegal cable taps. Broadcast would be far harder, I'd imagine. (Unless the secret tech is "binoculars" and you are peeping in people's windows.) Dragons flight (talk) 10:46, 8 September 2008 (UTC)
Re. 1, investigation of the TV licensing authority website FAQ reveals that you don't need a license if you don't watch television as it's broadcast. Currently, this even means that you can watch the BBC iplayer without a license (as it only has programmes on-demand). [1] AlmostReadytoFly (talk) 10:55, 8 September 2008 (UTC)
I'm fairly sure it does streaming radio but if the BBC ever gets it into their head that they can charge me a licence fee just because I own a PC capable of internet access and could potentially go to their iPlayer website they can sod off. Its one thing when there are 3 out of 5 channels so you don't really have an excuse of "I don't watch the BBC channels" because you're likely to, but when its several billion webpages it gets silly TheGreatZorko (talk) 10:59, 8 September 2008 (UTC)
It does indeed do streaming radio, but the radio license has been abolished. I'm not sure how they would plan to cope with all the people telling them to sod off if they introduced live internet TV broadcast. I wouldn't be surprised if that's one of the reasons it hasn't happened. AlmostReadytoFly (talk) 11:13, 8 September 2008 (UTC)
Um cable? Why would they have to worry about cable? All they need is for the cable TV network to give them a list of subscribing households for them to check against their list of TV licensees. If someone has cable but not a TV license, well just pay them a visit... Same with paid satellite. I'm not sure if you need a license if you only use FTA satellite but even if you do, it's not that easy to hide a satellite dish so even if they can't detect you receiving satellite TV, they can probably detect your satellite. Nil Einne (talk) 11:37, 9 September 2008 (UTC)
Back to the question... It is possible to detect a flyback transformer in use. Until the current flat-panel televisions became popular, the flyback transformer was a very important part of all televisions. Detecting it is nearly identical to how (good) radar detectors, laser detectors, and anti-radar detectors work. They detect the emission of radio frequencies from the electronics. In the U.S., all electronics that I see have an FCC sticker on them that states that the radio emissions are within FCC guidelines - which basically states that it is emitting some sort of radio frequency. Usually, it isn't strong enough to be detected more than a few inches from the device. With the older televisions, I tested the range and was able to detect a television turning on and off from across the street. It was almost as powerful a signal as blenders and vacuum cleaners. Then, my whole "lets have fun with radio signals" ended when I was caught driving through neighborhoods and opening all the automatic garage door openers. -- kainaw 12:19, 8 September 2008 (UTC)
Could this principal be used to verify which channel is being watched? The claim was that the detection vans could tell which channel was being watched, since during the days of terrestrial television only two channels were BBC and hence liable for a licence. Jdrewitt (talk) 12:29, 8 September 2008 (UTC)
No - the law has always been that you have to have a licence regardless of what channels are being watched. If you never watch BBC, you still have to have a licence. --Richardrj talk email 12:41, 8 September 2008 (UTC)
The old analogue method of detecting the local oscillator in a turned on set could determine the channel being watched by the frequency detected. As the oscillator would emit a frequency that was always 39.5Mhz above the tuned in channel a simple bit of arithmetic would pinpoint what they were watching, eg - if a frequency of 793.75Mhz was received, subtracting 39.5Mhz would leave a detected value of 754.25Mhz which lies between the 751.25Mhz - 757.25Mhz band which is analogue channel 56 or BBC1 to you and me. I seem to recall something about the European Court of Human Rights either issuing a ruling or threatening action about this particular data capture as it was deemed outside of the requirements of enforcement to take note of someone's viewing habits. Nanonic (talk) 12:44, 8 September 2008 (UTC)
In this century of the fruit bat, when we now have plasma screens, satellite dishes, cable, digital TV, TiVo and TV tuner cards, I still don't see how it is feasible to remotely detect each and every instance of live TV viewing or recording. Far cheaper and easier to visit the relatively small number of properties that do not have a TV licence - which is what the Licensing Authority actually says it does. Gandalf61 (talk) 13:06, 8 September 2008 (UTC)
Actually, what they say is that they visit all properties whose owners claim not to have a TV (or to have a black and white one only). I can't find anything in that pdf that says what they do if you just ignore their rude and threatening letters; my own very limited experience is that they send more letters and do nothing. Algebraist 13:14, 8 September 2008 (UTC)
My experience is that they send the threatening and rude letters regardless of whether or not you have a TV license, so I'm not sure there's any point in getting on at all. --Tango (talk) 13:22, 8 September 2008 (UTC)
I think the main way they catch people these days is by requiring an address whenever you buy a TV and then checking to see if you buy a TV license (or already have one) within a month of purchase, or whatever. It's not hard to get around that, of course. --Tango (talk) 13:22, 8 September 2008 (UTC)

all this chat about detection is a little bit hypothetical surely. I've happily watched tv without a license for years (shhhhh, it'll be our little secret). I've had a few threatening letters but ultimately Capita (who are the faceless corporation who run the enforcement side) aren't agents of the law and so cannot come into my house without a warrant. Surely, when you get the knock on the door, you just insist they get a warrant then sneak out and hide your TV at your neighbours house? Now that increasing numbers of us live in apartments blocks, you don't even have to put your socks and shoes on..82.22.4.63 (talk) 20:04, 8 September 2008 (UTC)

I believe that the TV licence enforcement people don't need a search warrant. Remember - the UK doesn't have a constitutional right for individuals to not be searched as there is in the USA - in fact, we don't have a constitution. So I believe, the guy will simply enter your home anyway. Plus, if they do detect you are using a TV, that evidence can probably be recorded and used in court without them actually finding the TV set. My father always swore he was going to buy two VCR's - use one of them to record TV shows and the other to play them back into a TV that he'd remove the UHV demodulator from. The result would be that he would not own a device capable of viewing broadcast TV - and would therefore be off the hook legally speaking. Of course, like the rest of us with clever schemes, he just paid up for the license fee.
Of course they need a search warrant. You don't need to have a written constitution to have laws against breaking and entering and trespass. --Tango (talk) 23:50, 8 September 2008 (UTC)
Frankly, in the more confusing maze of modern TV watching technology, and with virtually 100% of the population watching TV anyway - the government should just turn it into a regular tax. This idea has been fought for a long time because the original idea was to have the BBC funded by the TV license fee in order to give them independance from government control - so they could be truly neutral in reporting government matters...but I think that could be achieved in other ways - such as requiring the government to fully justify any reduction in BBC funding relative to current inflation...but the idea of 100% independence is a good one.
SteveBaker (talk) 21:58, 8 September 2008 (UTC)

Just out of curiousity, how much is a television licence? I don't notice a price in the brochure... - Nunh-huh 20:11, 8 September 2008 (UTC)

According to the BBC:
"The annual cost of a colour TV licence (set by the Government) is currently £139.50. That works out at less than £12 per month - about 38p per day for each household. A black and white TV licence is £47. The licence (whether colour or black and white) is free if you are 75 or over, and half-price if you are registered blind, although you still need to apply."
So the full thing is roughly \$260 US per year. (And it's TOTALLY worth it - I'd rather pay that much for a couple of BBC channels without adverts than the roughly 2x that amount I pay for 200 channels of advert-laden crap.) And it's half price if you're blind? Wow - so the audio part of the signal is half the cost but the color part of the video signal is more than two thirds of the cost? Whatever happened to "A picture is worth a thousand words"?! There must be a huge market in black and white TV's for blind people!  :-) SteveBaker (talk) 21:45, 8 September 2008 (UTC)
Say rather that it's half price if you have a blind person in your household. Algebraist 23:32, 8 September 2008 (UTC)
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## Anti Gravity in Salalah Muscat ?

How does antigravity point work? In Salalah (Oman) there is a place where cars roll uphill from this point. Any scientific explanations? ref:- http://www.salalahport.com/salalah.asp

```            "...the anti-gravity thingy kicks in your vehicle instead of rolling down the
slope begins to roll back UP hill."
```

http://mybay.wordpress.com/2008/01/20/salalah-the-oasis-of-oman/

```             "...Instead of the car moving downwards, you will see your car moves upwards..."
```

Rehanrazak (talk) 11:39, 8 September 2008 (UTC)

There's no anti-gravity; it's an optical illusion. See gravity hill and list of magnetic hills. -- Finlay McWalter | Talk 11:44, 8 September 2008 (UTC)
We have a "magic hill" here locally. Your car rolls uphill instead of down. Finlay is right, it's an optical illusion, which I proved first hand with a spirit level. Fribbler (talk) 11:47, 8 September 2008 (UTC)
(Sadly, your spirit level test is meaningless - if there really was antigravity - then it would make your spirit level work incorrectly too. The same thing is true for plumb-lines. All of those devices measure the direction of prevailing gravity - so if there WAS antigravity, they'd produce the same result as cars rolling, etc). SteveBaker (talk) 17:04, 8 September 2008 (UTC)
I suppose the spirit level would be valid if it were a magnetic (rather than anti-gravity) hill, but thanks for pointing this out, Steve -- I've noticed the problem myself before and wondered if I was just thinking too hard. — Lomn 00:17, 9 September 2008 (UTC)
A compass would handily rule out the possibility of a "magnetic hill". If it points "up" the hill - then yeah - it's a weird (VERY WEIRD) magnetic effect. If it carries on pointing North...then it's busted. Of course if the hill runs "up" to the North, that's not gonna prove anything...but with a reasonably sensitive compass, you can still figure it out. But the truth is that a freak magnetic effect would either:
• Be VASTLY stronger at one end of the hill than the other (magnetism is an inverse-square kinda thing)...to the point where if it could start a car rolling uphill at a distance of (say) 1 km from the magnetic source - then the force would be 1,000,000 times stronger at the "top" of the hill where the magnet would have to be - and nobody would be able to move the car once it got there.
...OR...
• Be more or less even in force all the way along the hill - in which case the magnet would have to be situated hundreds of miles away and be a field would have to be VASTLY stronger than the Earth's. If that kind of gigantic magnet existed, it would be detectable everywhere on the planet.
Hence, we know it's not magnetic without even having to do an experiment. SteveBaker (talk) 05:04, 9 September 2008 (UTC)
This is more fun by far than the original question. I propose we use two spirit levels connected by a long pole. We calibrate them to agree with each other in an area of known-good gravity and then take measurements every few feet in the area of the magic hill. If the two levels suddenly disagree we'll know there is a problem with the local gravity. APL (talk) 20:45, 9 September 2008 (UTC)
Well, if you intend to throw down the "weird side-effects" gauntlet: Antigravity would be easily detectable because the density of the air above the region of reversed (or reduced) gravity would be reduced. This 'lighter' air would rise like warm air does - but the winds blowing towards the antigravity source would instantly reduce in density - rising to be displaced with more heavier air...pretty soon you'd have a really impressive toroidal tornado thing. SteveBaker (talk) 03:15, 10 September 2008 (UTC)
There is certainly no such thing as anti-gravity. There are plenty of these "magic hills" around - and what's really going on is that the hill slopes in the direction you'd expect for things to be rolling downhill - but something about the surrounding topography fools your eyes into thinking that the hill slopes in the opposite direction. Sometimes it's because the surrounding terrain gives you a false idea of where the horizon is - sometimes it's because of a strong prevailing wind running up a valley that makes trees grow with a slight slant...whatever it is, it's never "antigravity". SteveBaker (talk) 17:02, 8 September 2008 (UTC)
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## Encyclopedic?

This is a section from the article about the herbicide Roundup.

"==Genetically modified crops==

In 1996, genetically modified Roundup Ready soybeans resistant to Roundup became commercially available, followed by Roundup Ready corn in 1998.[1] Current Roundup Ready crops include soy, maize (corn), sorghum, canola, alfalfa, and cotton, with wheat still under development. These cultivars greatly improved conventional farmers' ability to control weeds since glyphosate could be sprayed on fields without hurting the crop. As of 2005, 87% of U.S. soybean fields were planted to glyphosate resistant varieties.[2][3] The use of roundup ready crops has changed the herbicide use profile away from atrazine, metribuzin and alachlor. This has the benefit of reducing the dangers of herbicide run off into drinking water.[4] The use of roundup-ready crops has resulted in greater use of roundup, which has created a problem with weeds that are resistant to the herbicide. With greater use, it has become more likely that weeds that are not effected by the herbicide, survive and reproduce and proliferate.[5]"

PS I wonder if it says "conventional farmers" to distinguish them from eccentric ones. Wanderer57 (talk) 13:21, 8 September 2008 (UTC)
If that were the entire article I'd think it was a little unbalanced but the rest of the article contains extensive sections devoted to the many criticisms of Roundup. A section listing the products and the benefits to the farmer is not out of place, even if it is only part of the story. I suspect "conventional" here is to distinguish from "organic" but I don't know. --98.217.8.46 (talk) 13:47, 8 September 2008 (UTC)
While the opening sentences appear to come from the product brochure, the final two seem to balance the others? Saintrain (talk) 17:33, 8 September 2008 (UTC)
The reference desk is not an editing board for the rest of the wikipedia project; we're just here to help answer factual questions that people can't find answers to themselves. Issues concerning the content of an article are best dealt with on the discussion page of the article in question. --Shaggorama (talk) 08:27, 9 September 2008 (UTC)
And yes, "conventional" is often used in grocery stores that also stock organic produce (e.g., the Whole Foods chain). "Organic" means free from pesticides, bovine growth hormone, what have you; "convention" means "no such guarantee." So in this sense a conventional farmer is one who doesn't warrant that his crops are organic. --- OtherDave (talk) 21:40, 9 September 2008 (UTC)
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## Wellbrutrin and Libido

I have read somewhere that the drug Wellbutrin has caused some increased libido in women. Do you know why this happens? Is they a similar drug that has the same effect? Can anyone find this article? --Anilmanohar (talk) 13:52, 8 September 2008 (UTC)

Wellbutrin#Sexual_dysfunction is quite heavily cited. You might find more information in one of those references. AlmostReadytoFly (talk) 14:26, 8 September 2008 (UTC)
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## Cat external parasite

I just noticed a parasite under the fur of my kitten and I'm trying to identify it (New York). My first thought were fleas, however: (1) they seemed to long and thin to be a flea, they were approximately 3-4mm long, and (2) they scurried/ran rather quickly, which I didn't think was the way fleas moved. (Does this fall under the auspices of medical advice? I will be making a call to the vet later today, nonetheless) -- MacAddct  1984 (talkcontribs) 14:53, 8 September 2008 (UTC)

We can't give veterinary advice either - so for "what should I do to help my cat?" you certainly should do what you have said, and go to your vet. If you post a photo of the critters and ask "what are these? They live on a domestic cat in [insert country name here]", then I am sure we will do our best to help identify them. DuncanHill (talk) 15:24, 8 September 2008 (UTC)
Yup, just looking for identification for curiosity sake. -- MacAddct  1984 (talkcontribs) 18:01, 8 September 2008 (UTC)
Sounds like fleas to me. They can look like you describe and they can scurry like the devil. They're horrific little things. When my cat had them they were like little hard dark grains of rice. Gross to the max, as they say. Fortunately there's stuff you can put on them that takes care of them pretty well. We used Advantage or Frontline or one of those things. But check with the vet, esp. if it is a kitten. --98.217.8.46 (talk) 16:22, 8 September 2008 (UTC)
Even more creepy, I discovered what looked like an excellent example of the complete larval lifecycle of the flea (if that's what it is) on a cushion. -- MacAddct  1984 (talkcontribs) 18:01, 8 September 2008 (UTC)
Fleas will certainly scurry when they're in among fur. It's only on hard surfaces that they resort to hopping. A flea comb is very satisfying to use if you don't want to use the medications .46 suggested. --Sean 17:07, 8 September 2008 (UTC)
Thanks, I only ever knew fleas moving through they're famous (but apparently not the most impressive) jumping skills. -- MacAddct  1984 (talkcontribs) 18:01, 8 September 2008 (UTC)
Cats and Dogs WILL pick up these kinds of thing - they need protection - and something like Advantage is a good choice. However, with very young animals, you have to be super-careful. Lots of those chemical treatments can't be used on them until they get older. Kittens have to be more than 8 weeks old before you can use either Frontline or Advantage (for example). If your cat is a younger than that - then talk to a vet, tiny kittens can become aneamic from repeated flea bites. If older than 8 weeks - nuke the little brute with chemicals and keep doing it every month for the rest of it's life! You may also want to attack your carpet with something like this. SteveBaker (talk) 18:18, 8 September 2008 (UTC)
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## The newly-discovered Vernulles effect

I suspect it is meant to be Venturi or Bernoulli. Perhaps someone else will take a look at the description of the Vernulles effect.

Wanderer57 (talk) 16:47, 8 September 2008 (UTC)

I'm pretty sure they mean Bernoulli. But that section of the article is really badly written. Wikipedia is not a "HOW TO" guide. SteveBaker (talk) 16:58, 8 September 2008 (UTC)
Good eye wanderer. Don't be afraid to be bold and make edits yourself. I edited the article to remove the bad physics, but I'm no physics pro myself so someone here might want to check my edits. --Shaggorama (talk) 08:20, 9 September 2008 (UTC)
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## Hubble Space Telescope, and Image:NonFreeImageRemoved.svg

I wonder, after STS-125, will Hubble be able to get a better picture of Pluto than this? --Itwilltakeoff (talk) 18:58, 8 September 2008 (UTC)

No.
They are upgrading the wide-angle camera (no use for looking at teeny-tiny planets) and adding a new spectrograph (no pretty pictures there!).
The main thing they are there to do is to replace the gyroscopes, add new batteries and replace the heat blanket. They are also adding a docking ring so they can send up an unmanned craft to safely de-orbit the telescope when it finally does die. The mission is especially important because the gyros on the telescope are dying one by one and when the last one dies, the telescope will start tumbling - which will have three bad effects: Firstly, it won't be any use as a telescope, Secondly, a shuttle mission would be unable to fix it because the shuttles robot arm wouldn't be able to grab it without getting ripped to shreds, and Thirdly because de-orbiting the telescope safely would be impossible and it's a REALLY big spacecraft to come smacking into the middle of a city. Replacing the batteries is important for the same reasons - they power the gyros when the craft is in the earth's shadow and the solar panels don't work. The Hubble has always had problems with overheating - so the new blanket will help it stay cool.
The only other thing they are doing that MIGHT improve that picture is an upgrade to the Hubble's pointing mechanism. If you can point the telescope accurately then in principle you can get more precision when photographing planets by taking a bunch of photos at different times and with the camera aimed a fraction of the size of a pixel to one side or the other - then averaging the photos all together at higher resolution. But I doubt the new pointing mechanism is going to help that enough to make a better image.
This is why it's called "The Hubble Rescue Mission" and not "The Hubble Upgrade Mission". SteveBaker (talk) 22:13, 8 September 2008 (UTC)
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## Phase of the moon

What phase of the moon was seen in San Francisco, CA on August 14, 1972? --Anilmanohar (talk) 19:26, 8 September 2008 (UTC)

According to the navy it was a waxing (~4 day old) crescent. I hope this is about establishing an alibi and not astrology! Saintrain (talk) 21:02, 8 September 2008 (UTC)
And amazingly enough, the SAME moon phase was seen from everywhere else on earth on that date. -Arch dude (talk) 22:12, 8 September 2008 (UTC)
As the earth rotated while time passed, is there the possibility that folks at some other location saw a different phase of the moon? Does this ever occur, that there is a "phase transition" during one 24 hour day? Edison (talk) 05:00, 9 September 2008 (UTC)
Well sure, if you're seeing it at different times. Technically - you'll see a SLIGHTLY different phase when you are in a part of the world where the moon has just risen than at the exact same moment in time when your on the opposite side of the world and the moon is setting. That's because you're seeing the moon from a slightly different angle. So strictly speaking, one person could see it (say) fractionally before "full" and someplace else could see it fractionally past "full"...but it's a very tiny effect. SteveBaker (talk) 06:35, 9 September 2008 (UTC)
Yoursky can plot the sky for any location and probably any date you'd ever be interested in. Here is what the sky looked like above San Francisco, CA on August 15, 1972 at 02:00 UTC (19:00 Pacific Daylight Time on August 14). --Bowlhover (talk) 03:56, 11 September 2008 (UTC)
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## Speed of engulfment by LHC black hole

I know, I know, the LHC isn't going to create a black hole that will swallow us all up. But if it did... If a tiny black hole were created that was stable enough to continue existing, and strong enough to suck in the LHC, Switzerland and the rest planet, how long would that all take? Would the boundary expand at the speed of light, or would we have a few hours to get drunk and make a few phone calls?

— Sam 19:38, 8 September 2008 (UTC)

Even ignoring Hawking radiation, you would comfortably die of old age. Gravity is a very weak force, and black holes are only potent because they have huge masses. A micro blackhole is much smaller than atoms and thus has trouble eating anything. It could well take longer than the age of the universe to grow a TeV black hole to a size you would notice. Dragons flight (talk) 19:50, 8 September 2008 (UTC)
A tiny black hole would weigh very little - so it would have VERY little gravity. The only reason it could grow at all is because it's not only very lightweight but also very, VERY tiny. Since gravity gets much stronger the closer you get to something (and you can get very close indeed to a micro-black hole because it's so small), eventually it's gravity is as strong as a star-sized black hole - more than enough to suck things in. But by "close" I mean "much less than the size of an atom" (MUCH less). So these things would be attracted to the center of the earth - but they are so small and need to get so close to something that it might take them years before they'd actually happen to get close enough to suck in even one atom. At that rate of growth, the sun would have died and swallowed the earth long before the gravitation of the micro-black hole could get big enough to eat the planet. Hence, they really aren't of much concern. But if we believe in Hawking radiation (and that Hawking fellow is no slouch when it comes to the physics of black holes!) - then these micro black holes will "evaporate" in such an amazingly short amount of time that they essentially don't exist at all. So don't sweat it - you'll have plenty of time for your great-great-great-(...)-grandchildren to get virtual-drunk, send a few telepathic emails and ask the same question over on the Hyper-pandimensional-wiki-mega-pedia Ref Desk. SteveBaker (talk) 20:59, 8 September 2008 (UTC)
That's simply not true: a black hole exerts the exact same gravity as would any object of the same mass, no matter how close you come to it. Gravity does *not* become asymptotically strong is you come close to the event horizon. --baszoetekouw (talk) 06:50, 9 September 2008 (UTC)
Oh - but it most certainly is true. Ask yourself this: How come it's a black hole? To be black, the gravity field at it's teeny-tiny event horizon has to be so strong that light cannot escape from it...that's some seriously strong gravity - even though the darned thing weighs about as much a couple of atoms. Besides, the math is very clear: Per Sir Isaac Newton (the guy with the apple)... F = G x ( m1 x m2 ) / r2 when 'r' (the distance between two objects) gets small enough - 'F' (the force between them) can reach any arbitarily high value you'd care to choose. Since a black hole is a singularity (it has literally zero size), you can get to within zero distance of it - and F becomes infinite! With normal atoms, you can't get close enough to them without the strong nuclear force stopping you...but when one of the objects has collapsed into a singularity, you can get as close to it as you like...and sooner or later, it's gravity is as strong as any galaxy-eating monster. Of course, a billionth of a meter further away from it and the gravity is so close to zero you can't even tell it's there. But that's the entire point! SteveBaker (talk) 23:58, 9 September 2008 (UTC)
It doesn't become arbitrarily strong as you approach the event horizon, it becomes arbitrarily strong as you approach the singularity. However, the event horizon of such a small black hole would be very close to the singularity (closer than you can get to an atom) so, even though the mass is tiny, the gravity would still be very large (an escape velocity equal to the speed of light, by definition - give or take some GR effects). --Tango (talk) 00:03, 10 September 2008 (UTC)
That's true - it doesn't become arbitrarily strong at the event horizon - but it does become as strong as any star-eating black hole. The gravity at the event horizon of absolutely any black hole is (by definition) such that the escape velocity is equal to the speed of light. That has to be true of absolutely any black hole...no debate. And if you do cross the event horizon, the gravity will steeply rise to infinity at the singularity. But when you are at a distance comparable to the radius of an atom, the gravity has dropped to completely negligable amounts. It's asymptotic alright!
Guys, it doesn't become "arbitrarily strong" anywhere, it tends towards infinite strength the closer you get (in current theory anyway, God knows with Quantum Gravity). It would be some pretty bad physicists that were announcing that the results of their experiments were "arbitrary"! Deamon138 (talk) 03:04, 10 September 2008 (UTC)
Yowza. Okay, there are at least two different notions of "gravitational field" that could be relevant here. The coordinate-independent (i.e. generally covariant, i.e. measurable without outside reference) gravitational field goes to infinity at the singularity. It does not go to infinity at the event horizon. But the coordinate-independent field is the tidal force, which is not what people normally mean by "gravitational field". The acceleration of gravity—i.e. the amount you have to accelerate upward to stay at the same distance from the hole—goes to infinity at the event horizon. Also, the strength of the generally covariant field (tidal forces) at the event horizon is larger for less massive black holes, but the strength of the generally covariant field or the acceleration at a given distance from the center is roughly proportional to the mass. And "becomes arbitrarily strong" is just another way of saying "tends to infinity", at least to physicists. -- BenRG (talk) 11:49, 10 September 2008 (UTC)
To put it really simply - such a small black hole would have next to no gravity so things would only "fall" into it if they happened to crash directly into it by random chance. Remember, this black hole is smaller than an atom and atoms are mostly empty space, so it's very unlikely for anything to crash into it, it will just go through all the atoms. --Tango (talk) 21:24, 8 September 2008 (UTC)
I looked this up for a similar question a few weeks ago. I don't remember the exact figure, but suffice it to say that it is many many times longer than the expected lifetime of our sun. Not only will we be dead an gone long before that becomes an issue, but so will our entire solar system. Plasticup T/C 01:21, 9 September 2008 (UTC)
In point of fact, I am already drunk, so, checkmate, miniature black hole! --Sean 02:36, 9 September 2008 (UTC)
Is there a different mode of Wikipedia editing? Edison (talk) 05:01, 9 September 2008 (UTC)
See also: Wikipedia:Editing Under the Influence SteveBaker (talk) 05:31, 9 September 2008 (UTC)

Ridicule me if you will, but I have a couple of decent questions. First off, we know that the experiments at the LHC have been performed in nature numerous times. Is there a difference between the collision of cosmic rays and stationary matter as opposed to matter moving about as fast in opposite directions? Second thing: if such collisions occur in the entire universe ten millions times per second, what would happen if they occur sixty times more often in a very, very close space? Proven or theoretical answers would be appreciated. —Preceding unsigned comment added by 66.183.135.96 (talk) 01:01, 10 September 2008 (UTC) —Preceding unsigned comment added by Hey Lao (talkcontribs)

Well, I can't really answer specifically your questions, but as far as I'm aware, there isn't much difference between collisions in the atmosphere and that will take place at the LHC. The only real difference (and effectively the reason for the project) is that the LHC collisions are controlled by the physicists, and there are cameras and computers and god-knows what else, recording the results. Obviously this can't be done with the cosmic rays, so it's controlled nature is what sets it apart. Deamon138 (talk) 03:12, 10 September 2008 (UTC)
Actually, this is one of the more rational objections to the LHC. With cosmic ray collisions, it's overwhelmingly likely that an incoming particle moving at close to lightspeed will smack into something more or less stationary (a rock perhaps). Conservation of momentum means that the resulting mini-black-hole/strangelet/whatever will shoot off at a substantial fraction of the speed of light. But in the LHC, the two particles are approaching each other with equal and opposite speeds - precisely so that their momentum cancels out and the collision products will be more or less stationary, so they can be observed. So a 'problem' particle caused by a cosmic ray would be halfway to the moon before you could blink - but if a strangelet were to form inside the LHC, it would truly be the end of the world within not many seconds.
Mini-black holes are really no concern whatever...it's strangelets that are deeply scarey.
SteveBaker (talk) 03:31, 10 September 2008 (UTC)
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## Carbolic acid - uses

I was wondering if anyone can tell me if carbolic acid is effective in killing roaches. A neighbor told me it was but I cannot find the stuff in any stores. Any assistance would be greatly appreciated. Thanks! —Preceding unsigned comment added by Leathergurl (talkcontribs) 20:30, 8 September 2008 (UTC)

One reason you might not be able to find it is that it is rather toxic. Around the turn of the 20th century it was a very popular means of suicide (many stories from that time refer to people taking their lives with 10¢ of carbolic acid). The stuff you ought to be looking for is boric acid, said to be very effective against roaches, and much less toxic to humans [2]. - Nunh-huh 20:57, 8 September 2008 (UTC)

BORIC acid, okay, that explains why I couldn't find the stuff! Thank you so much.  :-) —Preceding unsigned comment added by Leathergurl (talkcontribs) 21:16, 8 September 2008 (UTC)

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## Control systems companies

Hi, I don't know which desk this goes under but does anyone know any companies (preferably well known ones) that in whole or in part are involved in control systems engineering? I've been looking for some but I can only find relatively small, unheard-of companies. Thanks. 202.37.62.221 (talk) 23:38, 8 September 2008 (UTC)

I found a list of universities involved in it... if you don't get any other suggestions you might be able to ask one of those schools for the names of some companies. Plasticup T/C 00:46, 9 September 2008 (UTC)
Lots of big companies use or support research in control systems engineering, you may for instance google "model predictive control" and StatoilHydro or the name of other big oil companies. But if you'r looking for a company that has control system engineering as their core business, many of these are companies in niche markets and thus smaller and less well known. But anyway, a quick glance at a list [3] of SCADA and automation companies quickly reveal ABB Group and Honeywell which I guess are fairly well known names. EverGreg (talk) 10:31, 9 September 2008 (UTC)

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