Wikipedia:Reference desk/Science

May 27

Momentum of the world trade center collapse

The official theory for why the world trade center and WTC7 collapsed at nearly free fall speed is that the heat of the fires in those buildings, plus structural damage, weakened at least one floor; and then the floors above the damage crashed through the floors below the damage, all the way to the ground, under only the force of gravity. This theory proposes that the mass of the collapsing part increased with the addition of more dislocated floors, and that the velocity of the collapse increased against the resistance of the lower structure, and that the momentum of the collapse increased more rapidly with its vertical descent[1][2]. All of this is necessary to account for the nearly free fall speed of the collapse, i.e. nearly the same speed as an object falling in a vacuum, an observation which is agreed upon by the official account, and multiple video and seismic measurements. Additionally this theory proposes to account for material ejected sideways during the collapse by the pressure of condensed air pushed laterally by the collapse. But the official theory leaves many questions unanswered. First of all, how does the velocity of the collapse increase against resistance from the lower structure, considering that free fall speed is the maximum falling speed in a vacuum of any object no matter what the mass of the object? Secondly how does the momentum of the collapse increase more rapidly if the mass of the collapse is increasing, considering that more massive objects gain momentum more slowly than less massive objects? Thirdly wouldn't an increase in the rate of the momentum change mean that an outside force must be acting on the collapse, since the rate of change of momentum is the force of gravity, and the net force of gravity of a closed system is constant according to the conservation of energy? Fourthly how does the collapse compress air despite the open core which should allow the air to escape without resistance? Fifthly how does compressed air eject heavy perimeter columns as far as 450 feet horizontally, into the World Financial Center 3[3]? Sixthly if the fire is so widespread that all of the columns, interior and exterior of at least one floor failed then why is the fire not visible around the entire perimeter of the building? Seventhly if the steel was heated to the point that it weakened then why was the steel not glowing red? Lastly, how does this collapse hypothesis account for the pulverization of the buildings? Thanks for your assistance. Oneismany 00:05, 27 May 2007 (UTC)

That is a lot of questions. I will make a start:

First of all, how does the velocity of the collapse increase against resistance from the lower structure, considering that free fall speed is the maximum falling speed in a vacuum of any object no matter what the mass of the object? The collapse does not archive the free fall speed, it *is* slowed down by the resistance. However the resistance is not very large compared to the gravitational pull, so the slowing is a small effect, and the collapse *almost* archives the free fall speed.

If the resistance of the lower structure is such a small effect that it barely decreases the velocity of the collapse, then how did the structure hold the weight of the upper part to begin with? Presumably the building could stand up against gravity for 30 years since it didn't previously collapse. Oneismany 15:58, 29 May 2007 (UTC)

Secondly how does the momentum of the collapse increase more rapidly if the mass of the collapse is increasing, considering that more massive objects gain momentum more slowly than less massive objects? The gravitational force is proportional to the mass of the collapse. The momentum gain of the collapsing mass is proportional to the force pulling it. So the momentum gain increases with the collapsing mass. Whoever said that massive objects gain momentum more slowly than less massive objects is telling garbage.

Newton's law of universal gravitation $F=G{\frac {m_{1}m_{2}}{r^{2}}}$ is not garbage. Acceleration is proportional to altitude. Force is also defined as mass times acceleration. F = m₁ a₁, so a₁ = F / m₁, so $a_{1}=G{\frac {m_{2}}{r^{2}}}$. This means that, if force is constant, instantaneous acceleration of an object is inversely proportional to the mass of that object, i.e. the rate of change of velocity decreases the greater the mass of a body. As momentum is instantaneous velocity times mass, this means that more massive objects gain and lose momentum more slowly than less massive objects. This not a new result. It is well known that speeding up and slowing down in a heavy vehicle is more difficult than in a light vehicle. Yet the progressive collapse hypothesis contends that as the collapse gained mass the momentum increased faster which is impossible. Mass is intertia, more mass means more resistance to force and more self-gravitation. Oneismany 15:58, 29 May 2007 (UTC)

Thirdly wouldn't an increase in the rate of the momentum change mean that an outside force must be acting on the collapse, since the rate of change of momentum is the force of gravity, and the net force of gravity of a closed system is constant according to the conservation of energy? The momentum change is the momentum change. It is caused by the force of gravity, but it is not the force of gravity. The force of gravity on the collapsing top part is proportional to the mass of that collapsing part. As the mass of the collapsing part increases while it picks up more rubble, the force of the gravity on that part increases too. I cannot see how the conservation of energy would be related to any of this.

The rate of change i.e. the derivitave of momentum, is force, it is not caused by force. As discussed above, force is mass times acceleration, and is also proportional to altitude. If the collapse is a closed system and gravity is the only force acting on the collapse, the conservation of energy says that the net force cannot increase, i.e. the momentum cannot speed up. If the momentum is speeding up, then there must be an outside force acting on the collapse. Oneismany 15:58, 29 May 2007 (UTC)

• You may be interested in our articles on the various conspiracy theories surrounding the collapse of the World Trade Center. See Controlled demolition hypothesis for the collapse of the World Trade Center and 9/11 conspiracy theories. Our article (which you've linked already) on the collapse of the World Trade Center contains a great deal of information, as well as links to a wealth of scholarly reports. This Desk probably isn't a good place to rehash in detail all of the conspiracy theories around 9/11, nor to debunk them. TenOfAllTrades(talk) 01:25, 27 May 2007 (UTC)

Ugh, do we really need to bring up the conspiracy theories here? It's obvious none of them are correct. We're all hiding from the obvious truth. The martians want us to be afraid, so they filled the buildings with nano machines that eat at it like termites. 9/11 is just a cover up! -- Phoeba Wright^OBJECTION! 01:40, 27 May 2007 (UTC)

I don't think that calling conspiracy theory is appropriate here. A conspiracy theory usually makes arbitrary claims that cannot be falsified so there is no point in discussing these claims. All questions stated here are in a form where they allow for an answer or at least some sensible reply. Why should we not try to give the replies?

-- Phoeba Wright^OBJECTION! 02:16, 27 May 2007 (UTC)

First of all, how does the velocity of the collapse increase against resistance from the lower structure, considering that free fall speed is the maximum falling speed in a vacuum of any object no matter what the mass of the object?

The speed of collapse clearly cannot be the free fall speed in vacuum. There has got to be some slowing due to drag and the energy that has to be given over to collapsing the lower floors. However, that slowing is likely to be pretty negligable once there are enough floors of rubble pancaked together. So claiming free fall is likely to be an exaggeration - but a very slight one.

If the resistance of the lower structure is such a small effect that it barely decreases the velocity of the collapse, then how did the structure hold the weight of the upper part to begin with? Presumably the building could stand up against gravity for 30 years since it didn't previously collapse. Oneismany 15:58, 29 May 2007 (UTC)

Secondly how does the momentum of the collapse increase more rapidly if the mass of the collapse is increasing, considering that more massive objects gain momentum more slowly than less massive objects?

Momentum is mass times velocity. The mass increases as more and more floors join the falling rubble pile - the velocity increases because it's accellerating under gravity. Accelleration under gravity is independent of mass - remember Gallileo dropping the two balls off the leaning tower of Piza (actually, that never happened - but it's a good story!).

Momentum is mass times instantaneous velocity. Instantaneous velocity is independent of mass, that is the lesson of Galileo. But instantaneous acceleration is proportional to altitude, see above. If mass is increasing then it should slow the increase in momentum because mass (inertia) is resistance to momentum (motion). Oneismany 15:58, 29 May 2007 (UTC)

Thirdly wouldn't an increase in the rate of the momentum change mean that an outside force must be acting on the collapse,

Yeah - it's called "Gravity".

Gravty is not a net outside force on this system because the force of gravity has always been pulling the towers down and they were balanced against the force of gravity, as evidenced by the fact that they stood for 30 years without collapsing. If there is a new force acting on the system it is not gravity. Oneismany 15:58, 29 May 2007 (UTC)

since the rate of change of momentum is the force of gravity, and the net force of gravity of a closed system is constant according to the conservation of energy?

The rate of change of velocity is due to gravity...yes. The force due to gravity is indeed constant (for constant masses) - but force equals mass times ACCELLERATION - so under constant force with constant masses, the accelleration is constant - which means that the velocity is increasing - and since momentum is mass time VELOCITY, so is the momentum. Conservation of energy is not involved here - the kinetic energy gained by the falling rubble is equal to the gravitational potential energy it loses. If this were not true, nothing would ever fall at all!

But force is not constant, it depends on altitude, see above. The instantaneous acceleration depends on altitude and if mass is increasing then the acceleration (force per unit of mass) is decreasing. A greater mass is pulled by a greater force, only if acceleration is constant, which it is on the Earth's surface or at a certain altitude, but not if the altitude is not fixed. Conservation of energy stipulates that the total force of an isolated system cannot increase or decrease. The force of gravity is not an outside force because it is not new to this system. If the momentum is speeding up then the force is increasing, which means there must be an outside force. Or, the theory is wrong. Oneismany 15:58, 29 May 2007 (UTC)

Fourthly how does the collapse compress air despite the open core which should allow the air to escape without resistance?

Air has viscocity - it can't escape freely through all of the openings - as each floor collapses on the floor below, the air between the two has to go somewhere - some (no doubt) is pushed down the central core - some escapes through holes in the falling rubble pile - but the rest is compressed - it builds up pressure - and that pressure can blow out windows and even more substantial parts of the structure.

That is a plausible explanation except that there is no reason to suppose the air pressure would break through the glass or concrete before it escapes by another path. It would need to be proved. The observation that glass and concrete are blowing laterally does not prove it is done by air pressure, it is possible that the air pressure was not high enough to do that and it would have to ba accounted by some other hypothesis. Oneismany 15:58, 29 May 2007 (UTC)

Fifthly how does compressed air eject heavy perimeter columns as far as 450 feet horizontally, into the World Financial Center 3[4]?

If the pressure is high enough - that's perfectly possible. Suppose it built up a pressure similar to what's inside a car tire...that's not that much...right? Well, that's 35psi over atmosphere. 35 pounds of force pushing against every square inch of a concrete column. Imagine a foot-thick concrete column. Every foot of length weighs about 150lbs - and has a surface area exposed to the blast of 144 square inches. A 35 psi air blast pushes against the column with a force of 35x144=5000lbs! That's plenty enough to blow it a long distance. But more than that - if the column was on one of the floors close to where the planes hit - it would have to fall 1000 feet or so - that took many seconds - so even if it was moving laterally quite slowly - it had many seconds to move outwards - and would therefore land a long way from the foot of the building - but with all that high pressure air jetting out sideways from the building - I don't have any problem imagining it travelling 450 feet outwards. From 1000 feet - that's only about a 20 degree angle from the vertical,

Again it is possible that enough air pressure might be able to push material laterally, but it would have to be proved that this theory fits the data, which hasn't been done. How massive were those perimeter columns? I read that they were 30,000-60,000 tons. Also if they came from a high point on the towers presumabley the air pressure was less because the "collapse" was, er, slower at that point, according to the theory. Oneismany 15:58, 29 May 2007 (UTC)

Sixthly if the fire is so widespread that all of the columns, interior and exterior of at least one floor failed then why is the fire not visible around the entire perimeter of the building?

I would imagine that it was all funneled upwards through the hollow center of the building - the aircraft tore their way fairly deeply into the center of the structure - and those were BIG buildings. But in any case, the glass, concrete and steel around the sides of the building furthest from the impact were still relatively intact. All of those columns only failed at the last moment when the whole thing pancaked down - there just wasn't time for the flames to move outwards at that time.

Okay, that sounds reasonable, but the calculations by Bazant et al propose a total collaps of at least one floor which presumably includes the perimiter and the core columns. Additionally steel conducts heat so presumably the whole floor or group of floors would have to be heated until all the steel reached the same temperature. And what about those people that were standing, alive and waving at cameras in the holes made by the planes (in the North Tower anyway), precisely where you would expect the temperature to be greatest if the jet fuel was concentrated there? Oneismany 15:58, 29 May 2007 (UTC)

Seventhly if the steel was heated to the point that it weakened then why was the steel not glowing red?

It must have been - deep inside the building. But one of the reasons for the collapse was that the steel they used was of a grade that softens significantly long before it's at red heat. Remember - those beams were under a lot of force before any of this happened - they were designed with some large safety factor - but as the heat increased - and because it was there for so long - the crystalline structure of the metal would have started to degrade long before it got red hot. I don't know for sure whether there was enough of a safety factor to allow them to get red hot - but I kinda doubt it. But in any case - it would have been the beams deep inside the structure that were undergoing the most heating - and you couldn't see them from the outside.

Again that sounds reasonable except when you take into account that steel conducts heat and a whole floor or group of floors failing would mean that presumably the perimeter columns would have to be heating up too. Now there was what looks like molten metal pouring down the side of the building during the collapse but that doesn't even fit the 'fire-weakened-steel' hypothesis, though it doesn't necessarily refute it either because it could be something else that only looks like molten metal .... Oneismany 15:58, 29 May 2007 (UTC)

Lastly, how does this collapse hypothesis account for the pulverization of the buildings?

I dislike your use of the term 'hypothesis' here - I think we're beyond that stage. But anyway - a bazillion tons of steel and concrete hitting the lower floors at a hundred miles an hour (or whatever it was) has an insane amount of crushing power. When it all finally hit the ground everything in the lower part of the rubble would obviously have been crushed to powder. No surprises there! The amount of kinetic energy in that falling structure was incredible - the energy had to go somewhere - and most of it would have gone into the upper levels pulverising the lower levels.

A hypothesis is an attempt to explain data, the term doesn't reflect the degree of success or failure, but OTOH it does imply that the attempt is somewhat provisional. The progressive collapse hypothesis is a provisional hypothesis, since it does not yet account for all the data. In particular it is at odds with Newtonian mechanics vis-a-vis the conservation of momentum and the conservation of energy, discussed earlier. It proposes that the structure beneath the crashing upper floors gave way easily despite being more massive than the upper floors. It proposes that the momentum increased because the mass of the collapse increased, yet instantaneous acceleration of an object decreases with increased mass. It proposes that the momentum speeded up yet it does not account for the increased force. The method by which the kinetic energy (momentum) increased so rapidly has not be adequately explained; the potential energy of the building is irrelevant if the kinetic energy cannot substantially increase. How does a building crush itself into powder? Oneismany 15:58, 29 May 2007 (UTC)

So - I don't think there is anything whatever surprising here. What happened to those buildings was precisely what good science predicts would happen. There is no mystery.

On the contrary, this is an astounding result unprecendented in scientific history! Jet fuel is a phenomenal form of propulsion if it can cause a steel framed building to pulverize itself! Or, the Twin Towers were phenomenally weak. Or, this theory is insufficient to account for the phenomena. Oneismany 15:58, 29 May 2007 (UTC)

SteveBaker 06:04, 27 May 2007 (UTC)

Thanks, Steve.

Speaking of "good science", it's also worth noting that those buildings actually survived the impacts of large jet aircraft traveling at speed. That's some pretty insane crushing power, too, and yet both structures had enough redundancy to withstand it -- as they were designed to. With the memory of the B-25 that had hit the Empire State Building in mind, it was a design goal that the Twin Towers be able to withstand aircraft impact -- and it's a testament to the engineers who designed them that they did. However, it was not a design goal that they be able to withstand intense, widespread fire fueled by a plane's full load of jet fuel -- and of course that's what did the buildings in. —Steve Summit (talk) 06:23, 27 May 2007 (UTC)

That is assuming that the conclusion is true, that the fire is what did the buildings in. But a theory cannot assume its conclusion, or it is circular reasoning and explains nothing. Oneismany 15:58, 29 May 2007 (UTC)

Yep - I agree. It was amazing they stayed up as long as they did - it was a testament to clever engineering. The fact that they did survive for almost an hour with that kind of damage gave something like 80,000 people the time they needed to escape the two buildings. The cost of designing and constructing buildings to withstand every single eventuality would be prohibitive - so you've got to go with the most probable events - and being hit by an airliner while it's completely full of fuel when you are such a long way from the routes these planes are supposed to fly was a long shot indeed. One could certainly argue that they were actually over-engineered. As terrible as the loss of 2700 lives was - that (and the fourteen who died when that bomber hit the Empire State Building) are the only people in all of history to have died as a result of large planes smacking into tall buildings. That's less than the number of people who die in US road accidents every single month. Risks that small are statistically hardly worth protecting against. SteveBaker 06:48, 27 May 2007 (UTC)

(edit conflict)

And yet, we had already learned in the years since 1972 (and before the WTC collapse) that steel-frame buildings are vulnerable to being weakened by prolonged fire. The beams in the World Trade Center had minimal (by modern standards) fire protection; modern building codes require quite a bit more. I know those codes were reexamined after 9/11, but I'm not sure whether changes were made.

You can't engineer against every foreseeable catastrophe, it's true, but you can learn from the ones that happen. One silver lining to be plucked from even the worst disaster is the empirical evidence it can yield that you could never afford to reproduce experimentally. We may not choose to require newly-built skyscrapers to withstand intense, widespread fire fueled by a plane's full load of jet fuel, but the knowledge gained from the collapse of the Twin Towers may yet make other buildings safer in the face of more realistic threats that they can be expected to encounter. —Steve Summit (talk) 07:22, 27 May 2007 (UTC)

One of the more interesting anecdotes is about asbestos. As we know, flaking asbestos causes asbestosis. Since it can only be causes by asbestos, lawyers had a field day with it and it became a huge liability so it had to be removed everywhere. As I understand it, replacement fireproofing is not nearly as effective or adhesive as asbestos. The planes would have damaged whatever fireproofing existed so I doubt the change would have saved the buildings but it would be interesting to see if asbestos would have given more time for evacuations. Incidentally, I am not minimizing the health consequences of asbestos, but if it didn't have a very specific disease it probably wouldn't have been removed. For example, if it just caused generic cancer that couldn't be attributable directly to asbestos, lawyers wouldn't have been so succesful. It wouldn't surprise me if the asbestos replacement was just as lethal but in a general non-attritubale way. --Tbeatty 07:00, 27 May 2007 (UTC)

If the buildings were so well designed then how did the upper floors crash through the lower structure so quickly (in the progressive collapse theory)? Oneismany 15:58, 29 May 2007 (UTC)

One other thing: recently, a span of the steel reinforced Bay Bridge in San Francisco collapsed after a fuel truck caught fire underneath it. I believe it has just been repaired. <gasp> a gasoline fire sufficiently weakened the steel so that it collapsed under it's own weight even though it was designed to withsand many times it's weight in traffic when the temperature was much lower. This is not a new phenomena and explosives and conspriacy theories are not required to explain that gasoline or jet fuel fires could cause structural collapse. --Tbeatty 07:36, 27 May 2007 (UTC)

Not the Bay Bridge itself but part of the "Maze" of elevated ramps on the Oakland side. I don't know exactly where it happened, but if the truck was at ground level it could have hit a support, which would have contributed some effect. —Tamfang 08:45, 28 May 2007 (UTC)

• The bridge that melted was at latitude 37.827463 N, longitude 122.29274 W. (In Google Maps, the map view has apparently been updated to remove the upper road, but the satellite view still shows it.) --Anonymous, May 29, 2007, 22:54, edited 22:56 (UTC).

Not all "steel" melts at the same temperature, it all depends om the composition of the alloy. Some steel alloys are tuned to handle heat, shear, impact, etc., and there is a vast range of possible trade-offs. Gasoline is not the same thing as jet fuel, and 'low test' gasoline burns hotter than jet fuel. You must examine all of the specifications of the relevant substances. Without more information, the Bay Bridge incident cannot be compared to the Twin Towers. Besides, just because part of the WTC steel may have melted does not mean that is what happened to cause the buildings to implode in about 10 seconds (which is the time according to the official theory). Additionally the official theory denies that the heat was great enough to melt steel. Who said anything about a conspiracy, anyway? Oneismany 15:58, 29 May 2007 (UTC)

I noticed the "requirement" that the steel glow red before it is structurally weakened. This is so outrageously false I can't even imagine how it's been propagating through the conspiracy theorist community. Anyone who has ever soldered will know that some metals can melt (turn entirely liquid) and never even remotely glow. Incandescence explains the process of glowing; steel mentions the melting point of several common alloys; all the facts are easily obtainable. Nimur 18:34, 27 May 2007 (UTC)

It is not a requirement, it is just a question. Steel is known to glow before it gets hot enough to melt; if you say that steel that is heated to weakening need not glow, then you must know more about it than me. Oneismany 15:58, 29 May 2007 (UTC)

Finally, I would like to comment on the scientific method. Whatever quantitative, theoretical explanation we have, experiment must always trump preconceived predictions. This is the tenet of modern science. Empirically, it has been observed that when an airplane hit the World Trade Center, the building collapsed. At the risk of sounding crass, this experiment has even been repeated. Two out of two trials have demonstrated the result of a large commercial airliner hitting the tower. If, hypothetically, somebody's structural model or simulation of the tower predicted that it should withstand that impact, then that simulation or model is wrong. Nimur 18:34, 27 May 2007 (UTC)

The building did not "collapse" when the airplane crashed into it. It did not "collapse." It exploded, and much later than the collision, despite evincing virtually no structural weakness in the intervening time. The experiment was not repeated because one building burned (smoldered) longer but collapsed faster than the other. OTOH WTC7 did actually collapse, six hours after it was damaged, but no airliner had crashed into it. The simulation or model in question proposes that fire caused each of these collapses, but this theory does not fit the data. Ergo the theory is insufficient. If our theory is wrong then we need to propose a new theory. Oneismany 15:58, 29 May 2007 (UTC)

SteveBaker wrote: "I would imagine that it was all funneled upwards through the hollow center of the building - the aircraft tore their way fairly deeply into the center of the structure - and those were BIG buildings." Just to note: the original poster was talking about World Trade Center 7, which was NOT touched by the airplanes. It's a different building than the two towers. Our article says that the NIST (as of 2006) can't account for the fall of building 7, with the lead investigator saying "truthfully, I don’t really know. We’ve had trouble getting a handle on Building No. 7."

I'm generally not one for conspiracy theories, but to my knowledge the government still does not know exactly how WTC7 fell, so it's still an open question. The questions of the original poster, in other words, are necessarily not the deluded ramblings of a conspiracy theorist. zafiroblue05 | Talk 18:41, 27 May 2007 (UTC)

"We don't have enough data" so (because we don't like Bush) lets assume it was the largest conspiracy the world has ever seen involving dozens of goverments (not only US citizens died in the attacks) and tens or hundreds of thousands of people "who know the truth". That would explain everything wouldn't it? Mieciu K 19:59, 28 May 2007 (UTC)

Well it is possible that the buildings exploded because of Muslim mind rays, on the other hand. Oneismany 15:58, 29 May 2007 (UTC)

Is the universe getting lighter, heavier or neither?

My son thinks the universe is getting lighter because stars are constantly converting mass to energy (fusion), where energy (photons) has no mass. I am under the opinion that in a closed system (albeit a very big one), these is no net change in mass. Could someone help us amateurs settle this arguement? 66.66.169.162 02:41, 27 May 2007 (UTC)

I don't know too much, but I don't think it's possible to lose mass, just move it around. -- Phoeba Wright^OBJECTION! 02:58, 27 May 2007 (UTC)

Errr... so I guess this equation means nothing to you: E=m*c^2. --Taraborn 19:24, 30 May 2007 (UTC)

We currently have an unexpected edit war on this topic. It seems nobody can ever settle this issue but my personal opinion is: No, there is no net change in the mass of the universe. Actually I believe that it is not only constant, but precisely zero, but that second part is just speculation.

The total mass and energy of the universe are not well-defined quantities. Actually, there's no global conservation of energy in General relativity. The density of mass in the universe is decreasing as it expands, and the density of energy is also decreasing. If you want to restrict the problem a bit, forget about general relativity, and just ask about whether or not mass is destroyed in nuclear fusion, then it's a different question. Unfortunately, it's also a question without a simple answer. The mass of a system is in general not the same as the sum of the masses of its constituent parts, so it depends on what you want to look at to find a net change. A photon has no mass, but a system of several photons (say, a bunch of photons in a box) can have mass. --Reuben 04:15, 27 May 2007 (UTC)

tell your son he asked a great question, though - a very entertaining thought Adambrowne666 04:54, 27 May 2007 (UTC)

Yes, it's definitely a great question. The problem is that cosmologists are still working on understanding the question, let alone knowing the answer! --Reuben 06:48, 27 May 2007 (UTC)

When you are talking about things on a cosmological scale - with relativity and all of these other weird effects getting involved, it's better to talk about conservation of mass+energy - not just mass alone. The two things are interchangeable. Imean, sure, your star is emitting massless photons - but elsewhere in the universe, photons are being absorbed and producing new mass. As far as we know, mass+energy is conserved perfectly. In that sense, the amount of 'stuff' in the universe isn't changing at all - although it does change form. SteveBaker 06:53, 27 May 2007 (UTC)

oooh, but brings up another question. The universe is expanding. That seems to imply that the general theory of relativity warpage of space is decreasing as the mass-energy is spread out more. That would seem to imply that the gravitational influence of the universe on any specific atom is decreasing over time. Does that mean it's getting lighter? --Tbeatty 07:43, 27 May 2007 (UTC)

But there's not any global law of conservation of energy in GR. And mass+energy doesn't quite work either, as mass is already included as a pat of energy. There are local conservation laws, but the total energy and mass are in general undefined. --Reuben 14:43, 27 May 2007 (UTC)

Lead acid batteries

When charging lead acid batteries does it mean that when the charge cycle gets shorter and shorter that instead of having used the battery less and less that the battery is actually loosing capacity until it finally appears to charge almost immediately but actually have no capacity left? Is this a charactteristic of lead acid batteries? 71.100.5.249 03:36, 27 May 2007 (UTC)

Yes, this commonly occurs in old batteries. See lead-acid battery for some background. Nimur 04:36, 27 May 2007 (UTC)

It's exacerbated by the fact that you should avoid totally discharging lead-acid batteries - they are designed to be kept fully charged pretty much all of the time (as in a car for example). When the capacity of the battery gets worse, it's more and more likely that you'll somehow run it completely down - so the battery gets damaged more and more because it's capacity gets less and less and you run it down more and more frequently. It's interesting that this is the complete opposite of Nickel-Cadmium (NiCd) batteries which should always be run completely down before recharging them - and failure to do so will cause them to have shorter and shorter discharge times. SteveBaker 05:21, 27 May 2007 (UTC)

How about Nickle-Metal Halide batteries? Should they be completely discharge or kept fully charged by trickle charge? 71.100.5.249 07:55, 27 May 2007 (UTC)

Trickle is fine according to our memory effect article, q.v.--Shantavira|^{feed me} 09:30, 27 May 2007 (UTC)

more:em waves

hi thanks for previous answers... it looks my project idea was strong enough that someone has edited it. kindly lemme kno about the proceedings. anyway one more query whether electromagnetic waves are deflected by magnetic fields ..as strong as in an atom.(yes/no)?.why Sameerdubey.sbp

I think you are looking for your previous question? Here. As for whether electromagneic waves are deflected by magnetic fields, your best bet for an answer is Electromagnetic radiation (which does answer that very question). Best of luck with the project. --TeaDrinker 08:28, 27 May 2007 (UTC)

Year 12 Chemistry

While studying for exams i came across several things which I was unsure of:

1. CH3CHCOOHCH3 - is this Propan-2-oic acid or 2-methylpropanoic acid (I think it is the former, though not sure.)
2. Chloroethane can react with NaOH to produce Ethanol. Can it react with H2O in the same way? Can H2O be thought of as HOH (similar to NaOH)?
3. Concerning Gas Chromatography, as the components move through the column they separate due to their differing degrees of absorption to the mobile and stationary phases. When a component is attracted to the stationary phase, can it be said that it has condensed, and then the attraction to the mobile phase has made it evaporate again? Is this a correct way of looking at the process?
4. Also in Gas Chromatography, why is the coil heated? Is it to vaporise the sample, or to ensure that it stays vaporised as it moves along?
5. When titrating ammonium hydroxide (volumetric analysis), why is a back titration used rather than a direct titration with HCl. The answers so because ammonium hydroxide is volatile and may evaporate. It also says because it handling should be limited in its undiluted form due to safety reasons. Could another answer be: because ammonium hydroxide is a weak base and wouldn't provide a clear end-point when titrated directly with HCl?
6. During the production of sulfuric acid (Contact process) if the number of passes over the catalyst beds is decreased, what effect would this have on the reaction rate? Why?
7. In the equilibrium reaction 4NH3 + 3O2 <--> 6H2O + 2N2 , if water vapour is added to the system, what is the effect on the forward reaction rate? (The answers say unchanged, but i thought decreased, b/c according to Le Chatelier's principle, the back reaction will increase to get rid of some H2O.
8. During the contact process, what type of chemical reaction is 2SO2 + O2 <---> 2SO3? (Is catalytic oxidation correct?)

Thank-you very much D3av 09:17, 27 May 2007 (UTC)

I'm in year 12 doing chemistry aswell, I will try to answer your questions.

1. CH₃CHCOOCH₃ is an ester, you cannot have a carboxylic acid group in the center of a carbon chain, as it uses 3 of the carbons bonds
2. NaOH is ionic H₂O is stable covalent bonding, and thus cannot be thought of as a hydrogen hydroxide.
3. .
4. .
5. .
6. .
7. The forward reaction should be unchanged, but the back reaction will accelerate giving a net shift towards products
8. .

Mnay of your questions are clearly from a different course to mine, or I cant remember from GCSE. So sorry I cant answer them all.

Ok let me give these a shot

1. It looks like 2-methylpropanoic acid. Also known as isobutyric acid. Although your notation seems a little off as the CH₃ is hanging off in the middle of nowhere.
2. H₂O can sometimes be thought of that way in the liquid/aqueous form. It looks like you are probably working in aqueous conditions, so yes, the reaction would occur (at least according to my organic chemistry book). See Haloalkane#Reactions_of_haloalkanes. In other forms, I agree with the previous post.
3. Kind of? The analyte is attracted to the stationary phase, but it's not quite condensation. See adsorption. Note the "D". Not aBsorbtion.
4. By the time the sample gets to the coil, it should already be vaporized (by the injector).
5. I pretty much agree with all of those. Ammonium hydroxide appears to be weak base, but I believe the main reason is that by the time you titrate to the endpoint, a considerable amount may have evaporated. (Ammonium hydroxide being aqueous ammonia). As an analogy, imagine doing a titration of carbon dioxide in pop (soda).
6. I'm actually not sure. Maybe someone more familiar with catalysis can answer.
7. I think this one might be tricky. Adding water vapor will shift the equilibrium, meaning the rates are equal to each other. So I imagine that adding water vapor will increase the rate of the backward reaction, also increasing the rate of the forward reaction. I'll have to get back to you on this.
8. Correct. See Contact process.

Let me know if you have any follow-ups. --Bennybp 15:06, 27 May 2007 (UTC)

Thanks very much for your help. After asking some other people similar questions, I agree with all your answers. The answer to Q3 was especially helpful.

If anybody can help further with questions 6 and 7 , that would be fantastic.

[Further to Q7, some things to think about: say you have 2SO2 + O2 <---> 2SO3 at equilibrium. If you increased pressure, you would expect both forward and back reactions to speed up (more pressure always means faster rate). Yet increased pressure would make the equilibrium shift to the right (less particles). Does "shifting to the right" mean that the forward reaction speeds up? Does this mean the back reaction slows down? Seems like there are some contradictory points here. Can someone please clear up which parts are true?] D3av 08:35, 28 May 2007 (UTC)

Ok I've seem to have found a page with a better explanation for question #7 here. Basically, I think you (and me a little bit) are trying to convert Le Chatelier's principle from "equilibrium position" to "reaction rate". At equilibrium, the rates are always equal (that's the definition I think). When that is disturbed, the rate for one direction increases, which kind of disturbs the other side, etc, until it returns to equilibrium. At the new equilibrium, the reaction rates may be different from their initial rates, but still equal to each other. Also, the equilibrium is constant for any change except for temperature, I believe. Here is another good page. It looks like maybe wherever you got your answers from may have over-simplified, or was maybe not clear enough (i.e. at the moment you add the water vapor, the forward rate is unchanged, while the backward reaction rate increases.) Hope this helps :) --Bennybp 16:12, 28 May 2007 (UTC)

Thanks for those websites D3av 09:58, 30 May 2007 (UTC)

The Pill

I've been often reminded that the pill is not a 100% reliable form of contraception, so what are the chances of getting pregnant on it. I know someone whos on it, and theire period is several days overdue, is the most likely explanation really that she is preganant, or can these things vary? Although she said its not prone just to be late like this.

See Combined oral contraceptive pill#Effectiveness. --Allen 12:12, 27 May 2007 (UTC)

Also, morning-after pill. --TotoBaggins 13:36, 27 May 2007 (UTC)

Which is illegal in many places...Mr.K. (talk) 20:01, 27 May 2007 (UTC)

Keep in mind we can't give medical advice- but many of the newer birth control pills can also shorten or even eliminate periods. -- Phoeba Wright^OBJECTION! 15:03, 27 May 2007 (UTC)

Get her to check the label/advice sheet which comes with the pill. If you're in the UK, you can phone up NHS Direct or ask a pharmacist. --h2g2bob (talk) 19:25, 27 May 2007 (UTC)

Handycam

I have a handycam with a swivel LCD screen.Any idea how the display information on this gadget and other cell phones are sent to the small LCD screen? Also it senses whether the screen is tilted and automatically inverts the picture.Any idea how that works??~~

The data is sent over a ribbon cable or some similar connector. There is probably a video processing integrated circuit or microprocessor, and the text/data is probably composited into the image before it is displayed on the screen. As far as the tilt-sensor, there are many possibilities. A small encoder or even an analog potentiometer is the most likely sensor, or perhaps something as simple as a rotary switch similar to a motor switcher. Because the screen is rigidly attached, it would not make sense to use some other direction-sensor such as an accelerometer or mercury switch, though a conceivable Rube Goldberg camera might want to use those sensors for something fun and trivial. Nimur 18:39, 27 May 2007 (UTC)

The data is usually converted to serial form so that it can be sent through the hinge on a thin, flexible serial cable, rather than a wide, fragile ribbon cable. They use electronics like the Smart Mobile Hinge Link (rubbish name, but they probably stopped trying to make up catchy names after they came up with "TOSLINK") from Toshiba, which is based on LVDS. --Heron 22:07, 27 May 2007 (UTC)

Any idea what can cause the tilt mechanism to fail overtime?~~

Most likely, mechanical stress on the hinge connector. Nimur 17:19, 28 May 2007 (UTC)

I would have thought it would be the repeated flexing and unflexing of the cable rather than the mechanical components in the hinge. SteveBaker 13:47, 29 May 2007 (UTC)

By "hinge connector," of course, I mean the electrical connector in the mechanical hinge. Steve Baker's clarification is more valid than my wording... Nimur 02:38, 30 May 2007 (UTC)

Question about a disease

I am not asking for medical advice. Recently, my friend said that he had a "disease" that made him immune to bacterial infections and caused him to have a high white blood cell count. My question is, does this "immuno-proficiency" (I just made that up) disease exist? or is my friend just making up a bunch of crap? Coolotter88 15:00, 27 May 2007 (UTC)

Uh, wouldn't that be leukemia? -- Phoeba Wright^OBJECTION! 15:04, 27 May 2007 (UTC)

I doubt the bacterial immunity though, since leukemia causes high white blood cell production that is basically useless to immunity. Splintercellguy 16:14, 27 May 2007 (UTC)

Your friend is pulling your leg. alteripse 17:56, 27 May 2007 (UTC)

Uh, I've read that in a short story in a sci-fi magazine. Let me see if I can find it. – b_jonas 09:53, 28 May 2007 (UTC)

Are you sure he just doesn't have mold? --Tbeatty 04:44, 29 May 2007 (UTC)

What body parts are seen as interesting or valuable to keep?

Being an anatomist, or perhaps even a taxonimist, it would make sense that they have an admiration for the body (anything alive having a body, not just humans).

Let's say with this admiration, they have a desire to collect fragments or wholes of bodily tissues and/or bones.

For the bones, the skull is definately something that would be kept. For tissues, perhaps the brain and the heart, but what other things (anything that can be retrieved from the body) can be collected as "interesting 'data'"? PitchBlack 16:29, 27 May 2007 (UTC)

Penis? Having a collection of stuffed/mounted (NO PUN INTENDED AT ALL) animal penises on the wall would certainly be a talking point at parties... --Kurt Shaped Box 16:37, 27 May 2007 (UTC)

And likely the last party you'll ever be at... --Wirbelwindヴィルヴェルヴィント (talk) 19:26, 27 May 2007 (UTC)

It's my party and I'll get my penises out if I wanna. --Kurt Shaped Box 20:18, 27 May 2007 (UTC)

I did read once about a luxury ship or house that the owner had decorated with many lavish and strange things... the leather on the seats of the bar were made out of the foreskin of a whale's penis. Anyone know what that was? I'd like to remember -- Phoeba Wright^OBJECTION! 20:56, 27 May 2007 (UTC)

see Icelandic Phallological Museum Bendž|Ť 09:47, 28 May 2007 (UTC)

Human horn? Assuming we are talking about artistic value rather than trafficking in human organs, any highly unusual case would be interesting for at least a specialist. A brain tumor the size of baseball-sized hail, a twinned spleen, that sort of thing. As far as animal parts, the most bizarre thing which was once commonplace that comes to mind would have to be the elephant-foot ashtrays. Personally, I would rather display an old trepan and leave the squishy stuff alone. Eldereft 19:58, 27 May 2007 (UTC)

Sounds like the whole Bodyworlds exhibit. Root⁴(one) 23:08, 27 May 2007 (UTC)

I was thinking more of just average body parts, not bizzare or unusual things. And now, lets talk about humans. There's skulls, blood samples, maybe semen (but that would be hard to get right?), heart, brain, maybe all the bones crushed and powdered... what interesting (but average) things can be collected from the human body? PitchBlack 00:35, 28 May 2007 (UTC)

Hard to get? Get a 5 cent plastic cup from wal-mart, go up to a guy on the street, give him the cup and a 20$ bill, show him to a public restroom, and wait. -- Phoeba Wright^OBJECTION! 03:21, 28 May 2007 (UTC)

And maybe a magazine or two... It's much easier for a girl to find a guy to reproduce with than reverse. --antilived^{T | C | G} 05:21, 28 May 2007 (UTC)

A German fellow of my acquaintance, who is a caretaker for a property up in Desolation Sound, has quite the collection of mounted bear penis bones. He is apt to introduce them as effective conversation starters when given the chance, though I have observed that with certain company the subject has rather a chilling effect on the immediate discourse. As to Pitch's particular question, human skin comes readily to mind. I've heard of some libraries Special Collections containing Books bound in human skin, and I vaguely recall recently reading something about tattooed individuals making provisions for their "artwork" to be saved/displayed subsequent to their deaths. -- Azi Like a Fox 11:50, 28 May 2007 (UTC)

Penis *bones*? --Kurt Shaped Box 14:35, 29 May 2007 (UTC)

Yes. See baculum. --Carnildo 23:13, 29 May 2007 (UTC)

I wish I had one of those. --Kurt Shaped Box 22:29, 30 May 2007 (UTC)

Do we have an article on electrical burns?

I've created and populated Category:Burns, but can't find an existing article on electrical burns. Am I missing a different names or something, or does it just not exist? Carcharoth 16:33, 27 May 2007 (UTC)

I'm not sure if there's significant additional information on electrical burns; as far as I know, the burning is caused by resistive heating from the current. However, death by electrocution is often caused by ion imbalance in the heart. Though an electrical burn may accompany death by electrocution, I don't believe they are the same. You can create an article anyway, so long as you document your sources! Nimur 18:43, 27 May 2007 (UTC)

Interesting. Thanks. Unfortunately, my only source so far is Wikipedia:Reference Desk... :-) Carcharoth 18:49, 27 May 2007 (UTC)

Sorry, don't have an exact citation, but I have seen electrical safety films which showed electrical burn due to high voltage contact (7200 volts AC) which said that the burning was deeper in the tissue than typical from burns due to fire. They showed the worker's arm opened up like a beef roast to remove internally burned muscle tissue. (He survuved and was an object lessson to use all protective gear). In other words, electrical burns are different from thermal burns. Edison 14:47, 29 May 2007 (UTC)

Contraception

You know how they say condoms are only like.. 98% effective and the pill is about 97% effective.. so what would the percentage effectiveness be for using both? It obviously wouldnt be 195% as there is still a risk, and I doubt it would be an average of the 2 as using the condom with the pill would increase effectiveness more than 97 percent, not decrease the effectiveness of the condom from 98% etc. Christopher

It is more a question of mathematics. The answer is most easily written as that the condom removes 98% - the pill then removes 97% of the remaining 2%. This means there is a 0.06% chance because 2-(2*0.97)=0.06 81.93.102.185 17:06, 27 May 2007 (UTC)

1 - (1 - 0.98) * (1 - 0.97) = 1 - 0.02 * 0.03 = 1 - 0.0006 = 0.9994 = 99.94%

Ohanian 18:40, 27 May 2007 (UTC)

Ohanian's math is correct assuming that there is no interaction between the two random processes; I don't know if that claim can be made for something as complex as an epidemiological study of contraception effectiveness. There are other problems with this statistic - is the device "ineffective" if it fails due to improper use? As such, you should consider all these statistics rough estimators. Nimur 18:46, 27 May 2007 (UTC)

It's easier to think of it like this: the failure rate of the pill is 3% of the 2% it is needed. ${\frac {2}{100}}\times {\frac {3}{100}}={\frac {6}{10000}}=0.06\%$ As Ohanian said, this is 99.94%. --h2g2bob (talk) 19:14, 27 May 2007 (UTC)

Extra caution is required for the effectivity figures, as they may be averaged over users with different compliance (medicine) or carefulness. Especially the effectiveness figures for condoms varies widely: If you count only people who know how to propely use it, you get something well above 99%. If you take into account those poeple, who mix is with latex-dissolving lubricant, don't squeeze the reservois, put it on too late, loose it, or do any of the many other little mistakes possible, the figure is much lower. Simon A. 06:10, 28 May 2007 (UTC)

To simplify the above a bit, the same people who are likely to misuse a condom (say put it on wrong) are also likely to misuse the pill (miss taking it frequently), so the two are not independent events. Thus, the chances that both will fail is much higher than the 0.06% calculated for independent events. StuRat 05:21, 29 May 2007 (UTC)

I have seen much lower effectiveness rates stated for condoms than the 98% rate stated. For one thing, they break sometimes. The Columbia University Health Center site, GoAskAlice [4] says that out of 100 women whose partner uses a condom "typically,". 14% get pregnant in a year. For those who use condoms "perfectly, 3% become pregnant.The "failure rate" is misleading in that it implies 100% of couples having frequent sex without contraception would achieve pregnenacy each year. That only happens in soap operas. That highly relevant number does not often appear. Many couples trying to have children fail to achieve pregnancy without medical intervention. Edison 14:49, 29 May 2007 (UTC)

Washing hands and static electricity

I just washed my hands, and upon walking past my TV found that only the hairs on the parts of my arms that were washed, were standing on end. The rest of the hair wasn't. How come? 81.93.102.185 17:01, 27 May 2007 (UTC)

That seems opposite to the normal behavior, where static charge builds up in dry environments. Maybe all your hairs would have stood up, but some form of crusty buildup on the unwashed parts overcame the coulomb force? Nimur 18:53, 27 May 2007 (UTC)

Or drying your hands and arms rubbed the hairs so they were already standing a little, and thus loose and closer to the telly. Eldereft 20:07, 27 May 2007 (UTC)

If you want your hair to stand up on a Van de Graaff generator it has to be very washed and unconditioned. Same with a CRT. --Zeizmic 20:13, 27 May 2007 (UTC)

I suspect you washed the oil off the hairs that stood up. Oily hair doesn't stand up. This is why "moisturizer" (oil) is used on hair to prevent "fly-away hair". StuRat 05:00, 29 May 2007 (UTC)

Shock

From Shock:

Opinion varies on the type of fluid used in shock. The most common are:

• Crystalloids - Such as sodium chloride (0.9%), dextrose (5%) or Hartmann's solution.
• Colloids - For example, synthetic albumin (Dextran™), polygeline (Haemaccel™), succunylated gelatin (Gelofusine™) and hetastarch (Hepsan™).
• Combination - Some clinicians argue that individually, colloids and crystalloids can further exacerbate the problem and suggest the combination of crystalloid and colloid solutions.
• Blood - Essential in severe haemorrhagic shock, often pre-warmed and rapidly infused.

Now hold on a second, wouldn't clinicians and EMTs and whoever know if crystalloids or colloids further exacerbate the problem, or not? [Mac Δαvιs] ❖ 17:24, 27 May 2007 (UTC)

This is a debate that dates back at least 50 years. There are advantages and disadvantages of both fluids. No two patients in shock are exactly alike, and it is rather difficult to recruit people arriving at the ER in shock into a truly randomized trial. Animal research can be found to support either position. Everyone does agree that rapid restoration of circulating volume is essential for survival. Search medline for "shock resuscitation fluid" and you find lots of more detailed discussions of the issue. alteripse 17:54, 27 May 2007 (UTC)

Lactated Ringer's solution

Why is Lactated Ringer's solution call "lactated?" [Mac Δαvιs] ❖ 18:45, 27 May 2007 (UTC)

Our article, which you linked, states that there is NaC3H5O3 (Sodium lactate) added. Nimur 18:50, 27 May 2007 (UTC)

Precisely. Though I was going to quote the "28 mEq of lactate = 28 mmol/L." bit. Still, I'd be curious to know if you (Mac Davis) got to that article via my burns question above, with stops at Category:Burns and burn (injury) along the way? :-) Carcharoth 18:52, 27 May 2007 (UTC)

I did :) [Mac Δαvιs] ❖ 21:03, 27 May 2007 (UTC)

And which also actually led to my shock question above, in which I went from burn (injury) to Hypovolaemic shock to Shock (medical). [Mac Δαvιs] ❖ 21:07, 27 May 2007 (UTC)

Experiments which can be performed with a TV

I have a 6-year old TV which is performing rather well, but which soon will have to be phased out. Are there any scientific experiments I can conduct with a TV (emittance of photons, electromagnetic fields, etc), and/or parts of it, even experiments which may be fatal to the TV? I am hoping to get a good list of experiments worth doing, so as to conduct a bit of original research. Thanks a lot for your help. I chose this category instead of Miscellanous because of the answers I am hoping for... :) 81.93.102.185 20:45, 27 May 2007 (UTC)

This will depend on the type of TV. A CRT experiment may be fatal to YOU. -- Phoeba Wright^OBJECTION! 20:59, 27 May 2007 (UTC)

I think he meant it was a CRT-TV, and that he wasn't going to take it apart (as in, "not fatal to the TV")

Definitely. The Frankin Bells is a nice little project that you can do pretty fast, and is complicated enough to be a science project.[5]. Can't find the article on it, though. [Mac Δαvιs] ❖ 21:03, 27 May 2007 (UTC)

You can see the magnetic field of a strong magnet if you put it close to the screen. It distorts the colors of the picture. Try it with some magnets. See if you can predict the distortion patterns caused by sets of magnets in different arrangements. This experiment may cause permanent damage to the TV, but it is safe for you. Please don't play with the electrics inside, ok?

You could connect and face a video camera to the TV and put the system into a feedback loop. It might be interesting to see what kind of visual a digital camera could make when the camera is rotated, zoomed, up real close, off to the side (but with the TV still in view. Then add the magnet and make things even more fun. ;-) (but not to the camera!) Root⁴(one) 23:23, 27 May 2007 (UTC)

Be sure to play with the lighting if you do that as well. Make things Pitch black and occassionally turn on a flashlight or light a match. Root⁴(one) 23:26, 27 May 2007 (UTC)

Thanks, these are all good ones that I will look into. Keep them coming if you want to, the whackier the better (and I wouldnt mind suggestions that means the end of my TV :))81.93.102.185 23:46, 27 May 2007 (UTC)

You could take the entire electronics out, put a new glass in front and make it an aquarium. Maybe not very scientific, but certainly cool and it destroys the TV. ^^

I wouldn't. The flyback transformer remains charged (~30,000 volts or 30 kV) even when the TV is unplugged. While the amperage is low (sub-20 mA, if memory serves), if you don't know exactly what you're doing, your curiosity can be fatal. Even those who are qualified sometimes make mistakes - a good friend of mine was thrown ~2 meters and got a concussion and a dozen stitches for being in a hurry while fixing one. -- MarcoTolo 01:27, 28 May 2007 (UTC)

Contrary to popular belief, not everybody who opens a TV set is instantly electrocuted. However if you think that pulling the plug is sufficient precaution, you should probably not try this one: http://www.aquahobby.com/tanks/e_tank0307d.php

The magnet (which may indeed destroy the TV) and the feed-back camera thing are without doubt the best experiments to play with - do the magnet thing last in case you do damage the TV. If the TV does get some permenant-seeming problems, turning it off an on again several times over - with at least a 30 second wait between each time - will sometimes restore it to normal operations. Failing that, if you can get a hold of a 'degaussing coil' you will almost certainly be able to fix it. If you are doing the feedback camera experiment - you might want to bring in a large mirror too - that produces some interesting effects). While you have the camera and TV handy - you might ask students why they look different when they see themselves on a TV compared to seeing themselves in a mirror. There are two interesting differences - one is that when you look at yourself in a mirror you can't avoid staring right at your own eyes - no matter how you turn your head. The camera can avoid that. Also, the mirror appears to swap your left and right - but the camera does not. When you see yourself on TV, you truly see yourself as others do - in the mirror, you always see a mirror image of yourself. This is why nobody ever thinks that their driver's license photo looks like them. In fact, it's probably a very good likeness - but you're only used to seeing yourself in the mirror so the "you" that you recognise isn't the real you - it's a mirror image! Anyway - it's fun to have kids try to use a mirror to comb their hair or something - then switch them over to using the TV+camera setup to do the same task! It's quite disorienting when you switch back and forth between them. SteveBaker 03:48, 28 May 2007 (UTC)

Don't believe the all hyp that suggests you will need a new tv when analogue signals are phased out. All you need is a set-top box and the right leads. Your tv doesn't even need a scart socket. Oh, and a decent aerial. That's always assuming you can receive the digital signal. I can't and I live in a big city.--Shantavira|^{feed me} 08:13, 28 May 2007 (UTC)

Bjork opened up her tv to see how it operates. You can be like Bjork. (yes, see warnings above) (subtitled version for deaf/HoH) —Pengo 03:24, 30 May 2007 (UTC)

Faeces in urine

My cousin has been talking about how he has been having bowel problems etc, and has had blood present in his urine, and now tells me that he has found fecal matter in his urine, as confirmed by doctors. He has been getting tests for it, but I am just wondering if anyone has any ideas as to what it could be. Christopher

What's the medical term for coughing up faeces, anyway? I used to drink with this old drunk who told me that he had that happen to him when he had cancer. --Kurt Shaped Box 21:20, 27 May 2007 (UTC)

The article on fistula might be helpful to you. Actually, looking at it, it's a shitty article (so to speak), but several of the external links at the bottom are more accessible and informative. And specifically, it could be this form of fistula. Anchoress 21:49, 27 May 2007 (UTC)

The Effects of Music on Plants

Yes, this is for school. But, my science partner and I have found nothing. We are testing how much music is good for a plant. We were just wondering if anyone else has done anything on that and can tell us the result. Or found any articles, thanx --Ninjawolf 23:10, 27 May 2007 (UTC)

Try some of the articles in these google searches: GOOGLE SCHOLAR and plants "music therapy" study. Anchoress 23:29, 27 May 2007 (UTC)

Mythbusters have done it, and it proves that death metal is the way to go. :p --124.197.9.207 antilived^{T | C | G}(forgot to log in) 00:23, 28 May 2007 (UTC)

Although at that volume, good luck with your neighbors -- Phoeba Wright^OBJECTION! 04:43, 28 May 2007 (UTC)

You might want to include a discussion about causality, correlation, and the scientific method when you discuss this experiment or any results for your final project. Sometimes, experiments can show a correlation (a "link") between things (such as music volume and plant growth). Sometimes, repeating the experiment many times can confirm this "link" (strong correlation). If this link was strong, you should investigate a causative link ("music causes plant growth") - but first you have to demonstrate the correlation by a repeatable controlled experiment. Once this is established, you can try to find a good reason why this link exists. The best way to start is examining various physical laws and how they apply. I have not performed this experiment, but I doubt you will find a correlation between ambient music level and plant growth; my understanding of plant tropism leads me to believe that there is no effect. Nimur 14:58, 28 May 2007 (UTC)

Actually - I don't think it's impossible that loud rock music has a beneficial effect on plant growth. This is just a hypothesis - but it's not entirely unreasonable: Think about this - the plants are being kept in an enclosed greenhouse without the natural airflow caused by the wind. As they absorb sunlight and CO2 and emit oxygen and water vapor through their leaves, the lack of natural airflow might result in a localized depletion of CO2 and an excess of O2 and H2O in a layer close to the leaves - resulting in the plant being unable to photosynthesize as efficiently as it would in the wild. OK - so add some really loud rock music - full of low frequencies and abrupt changes in frequency - might this not shake the leaves and enable a natural mixing of gasses from near the surface of the leaf? A gentle Brahms lullaby with very little bass content - and probably played much more quietly than the rock music might not be enough to cause that agitation of the leaves. Similar arguments might apply to talking to your plants (oh jeez - I just realised that I might have to apologize to my mother for what I said when I caught her talking to her plants...ack...the sacrifices we make for science!). Sure, it's a long shot - but it definitely sounds kinda plausible given the megre experimental results we have to go on and I don't think it's a good idea to dismiss this idea without more experiments. However, I most certainly don't expect that the result would be "Plants have auditory organs and can 'think' and as a result can hear and appreciate music and they prefer Motörhead to Brahms"...that's just crazy talk. But if my wild (and utterly untested) hypothesis turned out to be correct - then merely providing a slow speed fan close to the plants might prove even more effective than Spinal Tap (even if turned up to 11). SteveBaker 05:24, 29 May 2007 (UTC)

Heck, even if you were just scaring off whiteflies it would give you a healthier plant. Gzuckier 18:59, 29 May 2007 (UTC)

Agreed; hence the need for a carefully controlled experiment. If the airflow is causing the growth-rate change, and the music is causing the airflow, then this result is repeatable. It will boil down to a matter of word-mincing: does music make the plant grow, or does air-flow make the plant grow? The answer to that question depends on whether the music and the airflow are separable effects. Nimur 02:44, 30 May 2007 (UTC)

Eggs and matches

There is a question on the Miscellaneous desk that has to do with chemistry: Wikipedia:Reference_desk/Miscellaneous#Eggs. Also, can you answer me how hard it is to reverse the combustion of a safety match? A.Z. 23:40, 27 May 2007 (UTC)

It depends on what you mean by "hard". In theory it would be possible to reverse combustion. In reality...not so much. Combustion is a chemical reaction - in order to reverse the process one would have to be able to add ridiculous quantities of energy and figure out how to re-align all the atomic/molecular structures of the match pre-combustion. In all practical terms, reversal is not possible. -- MarcoTolo 01:18, 28 May 2007 (UTC)

It's not so much to do with chemistry as it is information theory and nanotechnology. The word "reverse" in the question is just too vague in this context. Forget about whether the chemistry can be driven in reverse. Think instead of these kinds of questions:

• Since almost all of the combustion products went up in smoke - drifted off into the atmosphere and dissipated widely, how would we retrieve all of them in order to use them as inputs to the 'reversing' process?
• What about reconstructing the insanely complex DNA molecules in the wood of the matchstick that was burned away?
• We don't have a way to record the precise structure of the wood fibres in enough detail that we could reconstruct them even if we knew how!

...(insert a million other similarly cogent arguments here!)...

which taken together means that it's utterly impossible in all but the most theoretical sense. So you have to start relaxing some of the requirements...well, maybe we don't have to restore every single molecule to it's original position - maybe we're allowed to use new materials - but if that's allowed then can we just toss out the original burned match and bring in a nice new one and claim that we "reversed" the effects? Either end of that spectrum of possibilities is laughable. But where in the middle of that range of possibilities the questioner stands determines the answer to the question. But if forced to come up with a one word answer - it would have to be 'no'. It is truly a dumb question.SteveBaker 03:30, 28 May 2007 (UTC)

As I said on the other reference desk, isn't the real answer to the question to do with entropy? Once you light a match or cook an egg, you are disassembling a well-structured object and producing a disordered system. The entropy of the system has increased. Reversing entropy is very, very difficult to do. We only have localised increases in entropy on this planet because the Sun is providing the energy needed, but with a large decrease in entropy for the Sun (ball of hydrogen converts to photons scattered throughout the universe). See also Arrow of time, Irreversibility, and Entropy (arrow of time). Carcharoth 06:23, 28 May 2007 (UTC)

I don't think it's a dumb question. And in an extremely loose sense, photosynthesis in trees is a reverse reaction of burning wood. Burning a match turns wood and oxygen into carbon dioxide, water, and energy; a growing tree does the opposite. --Allen 15:51, 28 May 2007 (UTC)

As I said - it depends on how loosely you are prepared to interpret the question. Tossing out the old match and taking a new one out of the box could be said to have reversed the reaction if you are prepared to interpret the question sufficiently loosely - and if you interpret it as a requirement for a perfect reversal in every possible sense then it's clearly impossible because of Heisenbuerg's uncertainty principle. So - pick wherever along that range you feel happy - but you still aren't answering the question. SteveBaker 19:44, 28 May 2007 (UTC)

You're right; photosynthesis is no better an answer than the others. But I think even questions that don't have a clear answer can be useful. Not everyone has thought about these issues enough to see the spectrum of interpretations you laid out, and having some sense of the endpoints and some points along the path can increase their knowledge and understanding. --Allen 20:39, 28 May 2007 (UTC)

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 [6]. 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: [7]. 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:[8](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)

May 29

Black Body Radiation

At what rate do things like man-made satellites, space ships, and planets radiate electromagnetic energy?

I know that's a very general question, so I'll give some background. I was thinking about man-made satellites and space probes. Satellites have electronics in them, and in those electronics are batteries and resistors. So since batteries have a resistance and current runs through them, there will be a certain amount of heat energy coming from those electronics. So it seems to me that since there's no air for the satellites to radiate their heat energy into, they'd either overheat or they'd have to radiate electromagnetic energy to cool down. So for something like a satellite, how much power can they radiate in the form of electromagnetic energy? Is this referred to as albedo? And do satellites have to run on very low current or can they run with current similar to personal computers? Jolb 00:53, 29 May 2007 (UTC)

Satellite electronics are similar to ground electronics with a few major differences. They must be radiation tolerant; they must be low-power; and they typically have failsafe mechanisms built in to "reboot" in case of failure. What this typically means is that the upper boundary of available components for satellite applications are a few months or years behind the "cutting edge" ground systems (slower processors, less memory, etc). What they lack in "benchmark" performance specs, they make up in reliability, low-power consumption, and ability to function in space. As far as cooling: some amount of heat is clearly radiated away (see thermal radiation, which can occur even in total vacuum). Some satellites also have cooling systems or heat exchangers. Although it varies for each spacecraft, it is probably safe to say that there is more heat gained due to solar radiation than due to the internal electronics. Nimur 00:58, 29 May 2007 (UTC)

Mmm... I think heat exchanger doesn't describe what goes on in vacuum. Satellites, manned or otherwise, can have some very elaborate radiators, however. Gzuckier 19:09, 29 May 2007 (UTC)

You're right that that's a very general question, and the answer, as you might expect, is also very general.

Satellites usually have a heat rejection system, not only because their internal circuits etc. generate heat but also because the side that's facing the sun tends to get very hot. Heat is generally lost using a space radiator (redlink! that needs fixing!) which rejects heat to deep space even if it's subject to an environmental heat input on its surface. Electrical equipment is mounted on cold plates and heat transferred from these to the radiators using heat pipes. The heat rejection of a space radiator can be estimated using:

$G_{s}\alpha \cos \theta +{\frac {Q_{w}}{A_{r}}}-\sigma \epsilon T^{4}=0$

where

$G_{s}=$ solar flux

$\alpha =$ surface absorptivity

$\theta =$ angle between surface normal and solar vector

$Q_{w}=$ waste heat rejected by radiator

$\sigma =$ Stefan-Boltzmann constant

$\epsilon =$ infrared emissivity

$T=$ operating temperature of radiator (Kelvin)

Satellites tend to use a little power as possible, because more power usage means more weight of power generation and control systems, which means more cost (both development cost and launch cost are dependent on weight). Satellite powers range from very low (a couple of hundred watts or less) to very high (multiple kilowatts). An average medium-sized spacecraft consumes about 1kW in total, which accounts for the requirements of the instruments plus propulsion, stabilisation/control, communications, thermal control etc. Hope that helps a bit. --YFB ¿ 01:30, 29 May 2007 (UTC)

There is abundant energy available to a satellite from the sun via photovoltaic panels which receive over a kilowatt per square meter, and which weigh from .5 kg to 10 kg per kilowatt per Solar power satellite. 12% efficiency would yield over 100 watts per square meter. Our article on the International Space Station says nothing about the amount of power available, but [[9]] says it has about an acre of solar panels and [[10]] says the solar panels plus rechargeable batteries for when they are in the shade make 110 kilowatts available. The ISS uses liquid ammonia radiators to eliminate the waste heat from the systems. Edison 15:43, 29 May 2007 (UTC)

The whole topic of cooling in a vacuum is a nontrivial subject. I recall an SF story from my youth with the somewhat inept hero on the moon cheerfully mixing up an entire barrel of epoxy at one time. All the experienced hands run for cover because a few minutes later... kablooie. Heat of the chemical reaction in that volume built up too fast for radiation to keep it down to a non-kablooie level. Gzuckier 19:09, 29 May 2007 (UTC)

My cooler pump???

hey friends i,ve bought a water cooler pump but it isint working well.at times when required supply is given then also it dosent works and produces some humming sounds or moves with low speed.after giving it a break of about 15mins it again starts working.can any1 give me wats exactly wrong with it.. as a solution should i fix some timer circuits in it(i do have timer of 1 minute cycle) 0663Sameerdubey.sbp

Could it be overheating, which causes it to shut down to protect itself from damage ? StuRat 04:29, 29 May 2007 (UTC)

That's a reasonable suggestion - but I don't think it works because the pump is humming and/or moving slowly. If it was a thermal cutout or an overvolt protection circuit or some kind of stall protection, it would just shut off. I think we need more information. What is this water cooler pump for? Can you give us an idea of how big it is - what current it draws, etc?

Does it need to be primed? --Tbeatty 05:15, 29 May 2007 (UTC)

If it did, why would waiting 15 minutes help? SteveBaker 13:43, 29 May 2007 (UTC)

exactly it is a 240V water pump with no overheating cutoff type circuits in it.actually it is a simple ac motor with a long shaft and veins fitted on the other end to lift water.such motors are sometimes used in small ventilator exhausts. Sameerdubey.sbp0663

It's probably an "induction" electric motor. Such motors can fail for several reasons:

• Failed bearings; they may overheat and thermal expansion causes them to "sieze up". Later, when they cool, there may be enough clearance to allow the motor to turn again for a while.
• Failed starting circuitry. Single-speed induction motors often have a centrifugal switch that controls the starting. If thaat fails, the motor won't start turning, will usually hum loudly, and will rapidly overheat, tripping the termal protection device.
• Too much load (but this seems unlikely in a pump)
• Too little voltage (leading to a higher-than normal current draw and no starting or overheating). This may also be caused by too long a mains lead causing too much voltage drop in the wiring.

In an existing application that used to work, I'd guess that the fault is bad bearings or a lack of lubrication of the bearings.

Atlant 16:27, 29 May 2007 (UTC)

Space Elevator

I was reading the Space Elevator article and stumbled upon the fact that as the cable increases in hieght it mus increase in diameter, then it goes on to explain this could be avoided many various ways, but one the fact which intrigues that if you had a tower 100km tall then this would reduce Diameter increase. I was curious if an airship station was 100km in the air could the cable be attached to that?67.127.167.131 04:53, 29 May 2007 (UTC)

The problem with cables is that attach point/top of the cable has to support the weight of the cable. Think of the top layer of molecules of the cable being pulled on by the entire weight of the cable below it. there is a material science problem with creating a cable that could support 100 km of weight. How much does a 100 km of cable weigh and what is the tensile strength of the material? --Tbeatty 05:10, 29 May 2007 (UTC)

100km is far too high for any type of airborne object to support a large weight. The atmospheric density in 100km is so low, that you can do orbital flight in that height. I really dislike the space elevator article, it gives a lot of "back of an envelope" calculations concerning the physics, but it has little consideration of engineering. For example 100km is more than two orders of magnitude larger than the highest tower ever built, and that tower had to support some radio dishes, not a giant cable.

The odd thing about the space elevator (at least in it's most common incarnation) is that the cable is self-supporting. The idea is that centrifugal force one one end of the cable (which is higher than geostationary orbit) exactly counterbalances the gravitational forces on the section that's below geostationary orbit. There is an equation (see Space_elevator#Cable taper) that relates the tension in the cable to the distance above the ground - and in order to make the thing work at all, the cable has to have a fairly complex shape such that it tapers just enough to take the strain at each point - this is called the "taper". A cable supported entirely under gravity also has a variable tension - but it's not the same as in the elevator. Certainly you can dangle a cable under an airship - but the size of the airship needed in order to support 100km of cable would be utterly immense (not to mention that there isn't any atmosphere at 100km - so you can't possibly fly an airship that high). With a tower, the entire structure is in compression - so the structure has the most weight to bear at the bottom - so tall towers are commonly thicker at the base than at the top. But a cable dangling under an airship is different - it's in tension and the most tension is at the top - so ideally you'd want a cable that thick at the top and very thin at the bottom. The space elevator is a completely different beast - in an idealised form, it just floats there - it doesn't have to be anchored to anything at the top, middle or base! In practice, they'll probably shorten the segment that's above geostationary orbit and replace most of it with a counterweight - but the counterweight is a waste of resources - it might be better to have the elevator simply stretch out further and be it's own counterweight. Consider how the elevator would be constructed: The entire cable would be spooled up into a large satellite and launched into geosynchronous orbit using a conventional rocket - then the satellite would proceed to unspool the cable symmetrically - both up and down at the same time(!) - allowing the outer end's increasing centrifugal force to counteract the lower end's increasing gravitational tug in order to keep the satellite balanced in geosynchronous orbit. As the cable unspools, either a ground station (or more likely, a ship) would see the cable end literally drop out of the sky! In principle - there would be no tension whatever on the lower end of the cable. In practice, the effect of the wind on the cable would be considerable - so a large mass is needed on the bottom end in order to stop it flapping about and causing nasty instabilities. SteveBaker 13:23, 29 May 2007 (UTC)

Wouldn't the point at the middle of the cable (from where this satellite would unspool it) be under a massive amount of tension (with the half above pulling upward and outward, and the half below pulling downward)? Or is that fixed by the taper? Neil (►) 17:38, 29 May 2007 (UTC)

No, you're correct. The tension would be highest in the middle, and lowest at the two ends. The cable must taper to its widest point there. TenOfAllTrades(talk) 19:15, 29 May 2007 (UTC)

Yep - exactly - the cable (which is actually generally imagined to be a flat ribbon so that the elevator cab can get a grip on it) gets widest at geosynchronous orbit - and thinner at either end. The snag is that there are no existing materials that can do the job. We need carbon nano-tubes of sufficient length to be woven into some kind of a composite material - but we can't make them long enough or cheaply and in sufficient quantities yet. There are numerous other problems with the Space Elevator concept - not the least being that it's unlikely that humans will ever be able to ride in it because it moves too slowly and we'd be slowly cooked as we passed through the radiation belt. The necessary amount of shielding would weigh too much for the cable to support. Incidentally - if you think the space elevator is a daring and exciting piece of technology - check out the space fountain and the space rotor (for which we don't appear to have an article?!) SteveBaker 22:43, 29 May 2007 (UTC)

Allergies

Ok, so I'm not supposed to ask medical questions. But answer me with something, resources, or whatever you please.

One of my friends as a strange type of allergy—doctors said they were normal, seasonal allergies, like to pollen. But she has them year-round, and they involve total body itching, as well as inside the ears and throat. That doesn't seem like any type of normal allergy that I know about that isn't from food, or more exposure. Perhaps she is just extra-sensitive? [Mac Δαvιs] ❖ 06:37, 29 May 2007 (UTC)

Total body reactions are not so uncommon, see Atopy. If you are exposed to the allergen all around the year, you will have the symptoms all around the year. This can happen for example if you react to the excrements of the House dust mite. If your friend is in doubt about the source of her allergy she can do a RAST test. This test is more reliable than skin tests. Oh, and please don't ask me why I know all this (*sneeeeeze*).

If the RAST test does not work, try the homeopathic remedy: the QX machine. This is an amazing machine based on NASA technology that can pick up allergens in your body (amongst a whole host of other problems) and can help you to treat yourself. Of course if your environment contains something like toxic mould, just be glad that you identified it and you could run away from it in time. Sandman30s 13:53, 29 May 2007 (UTC)

Amazingly there is an article (not a very good one) on that here: Qxci - could an admin please rename this to QXCI and I will edit the article? Sandman30s 14:06, 29 May 2007 (UTC)

"Based on the principles acupuncture, homeopathy, quantum physics, and electronics". Are these people for real, or is it a genuine hoax?

It is a very real, and very good machine. My family and many others have made use of consultations with homeopaths who use this machine. Sandman30s 14:39, 29 May 2007 (UTC)

Wonderful. That text has everything. Health, latin terms, experts, statistics, contradictions, spirituality, quantum mechanics, brain waves. The list is endless. Nobody will *ever* believe me that I didn't made that up myself.

I sent it to Quantum Xrroid Consciousness Interface per style. Nothing links there and the machine is obvious quackery, but it does get 10,000 G-hits, so I can give it the benefit of a doubt on being encyclopedic. Incidentally, pages may be moved by any logged in user - be bold! Eldereft 17:11, 29 May 2007 (UTC)

Well, thanks to the first guy. Any other comments? [Mac Δαvιs] ❖ 20:10, 29 May 2007 (UTC)

Heat and Waves ??any relation

hey friends ...in waves the energy disappears at trough and appears at crest ... if sound waves are made to travel in a heat transfer pipe then can we expect any change in heat transfer rates by the pipe. we may consider equal energy appears and disappears at crest and trough but crests might be thought to transfer energy to boundary of tube.and troughs interact with fluid in the vessel. orr if i perform the experiment in a open vessel ment for transferring heat. a double chambered vessel in series

                                              _ _ 
                            sound waves->----1 1 1
                                             1-1-1

0663Sameerdubey.sbp

(Fixed formatting so we can read the question!) SteveBaker 12:55, 29 May 2007 (UTC)

I disagree with your very first premise - energy doesn't "appear and disappear", it just changes form. Consider a water wave. At the peak of the wave the water is almost stationary - and it's higher than the average level of the water. It has gravitational potential energy. As the wave moves on, the water rushes downwards and forwards into the trough - at which point it's moving quickly - but now below the average level of the water. It's lost gravitational potential energy and gained kinetic energy. No energy is lost in the process - it's merely transformed from one form to another. The effect is much like a pendulum swinging - gravitational potential energy at the ends of the swing when the pendulum bob is moving slowly - kinetic energy in the middle of the swing when it's moving quickly. In sound waves, the trade-off is between kinetic energy and pressure. It's always just energy changing form - not being gained and lost. In the case of sound waves moving along your heat pipe, when the air (or whatever fluid you are using) is compressed, it'll heat up - when it decompresses again, it'll cool down. If you take away heat from the high pressure regions then I guess you'll dampen down the sound wave because as you cool the fluid the pressure will decrease making it less able to gain kinetic energy. So I suppose you could (in principle) deaden the sound and extract a little energy from it in the form of heat. But it's going to be tricky (or perhaps impossible) in practice because as the sound travels along the pipe it's alternately heating and cooling your heat pipe. Thermal inertia being what it is, I don't think your apparatus would work as a practical device. SteveBaker 13:04, 29 May 2007 (UTC)

I think it might-- see [11]

Yup. I used to read about these kinds of things in Popular Science 50 years ago. Gzuckier 19:13, 29 May 2007 (UTC)

A psychological experiment where half of the subjects took on the role of a guard and the other half the role of prisioners

I believe, if I am not mistaken, that several psychological experiments took place at major universities where 1/2 of the test subjects were given the role of guards and the other test subjects that of prisoners. To pinpoint exactly what I am looking for the experiment I am seeking was held in the confines of a prison-like atmosphere. 1/2 of the test subjects acting as "guards" (I believe, they were taking on the role of a brown shirt within the now defunct Nazi Regime). The other 1/2 of the test subjects were their prisoners. I recall, that the controls of the experiment were very loose allowing the test subjects acting as "guards" many freedoms over their prisoners.

The experiment showed how easily one can be manipulated in whatever direction the researchers wished them to be.

I am sorry as I cannot supply any more information than the aforementioned, nor the universities where these experiments took place. The experiments took place a number of years ago perhaps in the 90's.

The experiment would be similar to Golding's study of human behavior in his novel: "Lord of The Flies."

I thank you for your time and effort.

Antony baekeland 12:02, 29 May 2007 (UTC)

See Stanford prison experiment and Milgram experiment. There may be others. Anchoress 12:08, 29 May 2007 (UTC)

You might want to read around Jane Elliott who did a similar exercise (although a lot less violently) showing how discrimination can so easily come about based on almost ANY criteria you could come up with. Her experiment has been repeated dozens if not hundreds of times. Also, see The Third Wave for another kind of experiment. What's chilling about these things is not just that they work - but that they work so strongly that they tend to get wildly out of control. SteveBaker 12:24, 29 May 2007 (UTC)

Also Strip Search Prank Call Scam - very weird. SteveBaker 13:41, 29 May 2007 (UTC)

I remember seeing a film (Das Experiment) that fitted the criteria, and lo' and behold, it led me to the Stanford prison experiment. Icthyos 20:30, 29 May 2007 (UTC)

Seagulls

Recently I saw 2 seagulls. One was standing on the others back and sqwaking loudly, the one under neath was sqwaing more quitly in higher pitch. The on on the top was balancin using its wings. What were they doing?

This question should be moved to the Seagull Reference Desk, but that seems to be missing. Also, they were making more Seagulls. --Cody.Pope 14:00, 29 May 2007 (UTC)

They were 'making fuck', as an old friend of mine from India used to say... --Kurt Shaped Box 14:37, 29 May 2007 (UTC)

If the original poster is correct in saying the one gull was only standing on the other, and no mating was occurring, could this be some type of dominance display ? StuRat 16:33, 29 May 2007 (UTC)

Do you realize just how much blood would be spilled if a gull tried to climb on the back of another gull without permission? ;) --Kurt Shaped Box 16:36, 29 May 2007 (UTC)

No, it's los mating. 81.93.102.185 16:38, 29 May 2007 (UTC)

The gull on top didnt seem to be accessing the rear of the the lower gull and I watched them for about 2 minutes (perhaps he was having erection problems?). How long does it take for a gull mating?

Opposite of helium?

I was at a "mad science"- -type presentation, and the man demonstrated the effect of helium on his voice (making it higher). After that, he ingested another gas that had something of the opposite effect. However, I forgot the name of that gas. Would anyone have a clue what that may be? Abeg92contribs 14:05, 29 May 2007 (UTC)

See xenon. --Kainaw ^(talk) 14:26, 29 May 2007 (UTC)

Sulfur hexafluoride? Video here - cool, huh? --Kurt Shaped Box 14:40, 29 May 2007 (UTC)

(double ec!) Anything heavier than air would work: xenon, argon, sulfur hexafluoride. Safety first! Even helium has killed people (although Darwin was an accomplice). --TotoBaggins 14:45, 29 May 2007 (UTC)

It's probably a bad idea to try it with radon... --Kurt Shaped Box 14:48, 29 May 2007 (UTC)

Or xenon. It's an anesthetic. Argon should be cheaper anyway: it makes up almost 1% of the air. --Anonymous, May 29, 2007, 23:03 (UTC).

Argon wouldn't work, as it is only marginally heavier than air (Molecular weights of Ar: 18, O₂: 16, N₂: 14). You wouldn't hear the difference. Sulfur hexafluoride on the other hand makes a dramatic difference (molecular weight 70). --mglg_(talk) 23:39, 29 May 2007 (UTC)

Is there a heavier gas that's relatively safe to inhale, as a matter of interest? --Kurt Shaped Box 23:42, 29 May 2007 (UTC)

There is no gas heavier than SF6 apart from radon. CO2 is relatively safe to inhale. —The preceding unsigned comment was added by 88.109.29.179 (talk • contribs) 00:39, 30 May 2007 (UTC)

Well, "relatively" compared to, say, cyanide, sure. But you won't enjoy it. Concentrated CO2 is pretty nasty. --Trovatore 00:48, 30 May 2007 (UTC)

Yes as our article says:

Carbon dioxide content in fresh air varies and is between 0.03% (300 ppm) and 0.06% (600 ppm), depending on location (see graphical map of CO2 in real-time) and in exhaled air approximately 4.5%. When inhaled in high concentrations (greater than 5% by volume), it is immediately dangerous to the life and health of humans and other animals. The current threshold limit value (TLV) or maximum level that is considered safe for healthy adults for an 8-hour work day is 0.5% (5000 ppm). The maximum safe level for infants, children, the elderly and individuals with cardio-pulmonary health issues would be significantly less.

The Earths Atmosphere

Dear Sir or Ma'am,

IN your article on the Earth's Atmoshpere you wrote...

"For this reason, the Earth's current environment is oxidizing, rather than reducing, with consequences for the chemical nature of life which developed on the planet."

If hydrogen is able to escape the earths gravity and leek into outer space, thus oxidizing the Earth's current environment, what would those consequences be, specifically, for the chemical nature of life on earth as we know it.?

Thank You,

Eugene Rosenquest

H₂ does in fact rise to the top of the atmosphere and escape permanently, so the consequences would be... that things are the way they are now. Did you mean to ask what would happen if H₂ could not escape? —Keenan Pepper 15:28, 29 May 2007 (UTC)

ARUUN ques. about clitoris

what is the maximum size of human clitoris?

Read Clitoromegaly. Anchoress 14:39, 29 May 2007 (UTC)

The normal volume of it expanded from arousal is a little bit less than what it can be. [Mac Δαvιs] ❖ 20:13, 29 May 2007 (UTC)

Heavy gases ans speker enclosures

Is there any record of the use of 'heavier than air' gases being used in loudspeaker cabs to make them appear acoustically bigger?--Tugjob 14:55, 29 May 2007 (UTC)

Well, short answer: I never heard of such. Longer answer: would be difficult; obviously, cabinet would have to be completely gastight, and that causes problems as the outside atmospheric pressure rises and falls. Even sealed enclosure cabinets need a tiny hole to allow them to equilibrate. Otherwise, nobody could use them in Denver, for instance, unless they assembled them there. Gzuckier 19:17, 29 May 2007 (UTC)

If you enclosed the gas within flexible gas tight bags what then?--Tugjob 20:36, 29 May 2007 (UTC)

It seems to have been done: [12]. Search within the page for "gas". --Reuben 20:55, 29 May 2007 (UTC)

Puking up turds?

What's the medical term for the condition that causes the digestive tract to work in reverse, leading to the patient vomiting faeces from the mouth? I used to know someone that had this happen to him as a result of alcoholism and cancer. --Kurt Shaped Box 15:18, 29 May 2007 (UTC)

Copremesis. There's an article at fecal vomiting. --Joelmills 15:29, 29 May 2007 (UTC)

Antiperistalis [13]

Thanks guys. I've created a redirect for that redlink... --Kurt Shaped Box 20:52, 29 May 2007 (UTC)

This happened on an episode of [House]Llamabr 20:59, 29 May 2007 (UTC)

I don't remember watching that episode. I know that fecal vomiting was mentioned on South Park... --Kurt Shaped Box 21:11, 29 May 2007 (UTC)

Vented cabinet.

In a bass reflex cabinet, what importance does the cross sectional area of the port have? I thought it was just the volume of air in the duct that was important.--Tugjob 20:41, 29 May 2007 (UTC)

See diffraction for some details on wave interaction with an aperature. Nimur 22:11, 29 May 2007 (UTC)

I cant see how that applies to my question. Can you explain?--Tugjob 23:03, 29 May 2007 (UTC)

Phase diagrams for mixtures of ethanol and water.

I recently asked a question about the boiling point of a mixture of ethanol and water, and its dependency on pressure. After some googling, I realised that what I actually was asking for is a binary phase diagram for a mixture of ethanol and water. The best I could find on an image search was this one (middle panel), which is what I'm looking for, but with very low resolution in my region of interest. This exellent tutorial explains how to read the diagram.

• Is anyone able to provide pointers to a similar diagram, with better resolution in the range 78-100°C?
• Is all the information I need for calculating boiling points and vapour composition at various pressures in the phase diagram, or does the phase diagram itself depend on pressure? The fact that the diagram ends at 100°C for pure water suggests the latter, since water would have boiled at a higher temperature if the pressure were higher than 1 bar. If indeed several phase diagrams are needed, does anyone have information about where to find them? My region of interest for pressure is from atmospheric to approx. 1.5 bar. --62.16.173.45 20:48, 29 May 2007 (UTC)

Sorry, I don't have a table to point you to. But the following may be conceptually helpful for the pressure question: At any given temperature, a given mixture of ethanol and water has a well-defined vapor pressure (or partial pressure) of ethanol, and a different vapor pressure of water vapor. The vapor pressures are defined as the gas concentrations that the fluid would be in evaporation/condensation equilibrium with. The vapor pressures do not depend on the actual external pressure. The fluid(s) will evaporate or condense until the actual concentrations of each gas is equal to the fluid's vapor pressure for that gas (which may itself change during the process because evaporation/condensation can affect the temperature and fluid mixing ratio). At a certain temperature, the total vapor pressure will equal the ambient pressure; this temperature is the boiling point, at (or above) which evaporation can take the form of bubbling. Increases in pressure by addition of a third, more-or-less inert gas (such as air) does not in principle change the vapor pressures, just the boiling point. --mglg_(talk) 21:37, 29 May 2007 (UTC)

I found a random old reference (Phys. Rev. 57, 1040–1041 (1940)) that contains partial pressure graphs for low temperatures (20-40°C). They got the data from the International Critical Tables, which might contain data for higher temperatures (more relevant for you) as well. You can find out in the online edition, but you may have to pay to use it. --mglg_(talk)

Which do you think is more likely to happen first?

A) Developing a spacecraft capable of carrying humans to Proxima Centauri and back within a reasonable time period (decades, as opposed to centuries).

B) Developing technology that allows humans to be preserved (cryogenically or otherwise) for a journey to Proxima Centauri and back that may take several centuries.

--Kurt Shaped Box 21:30, 29 May 2007 (UTC)

Probably annihilation of our species will occur before either of these is feasible enough for even an eccentric, self-funded enthusiast-investor to implement. Nimur 22:10, 29 May 2007 (UTC)

I would say (A) - but my personal belief (which I've gone into in some detail here before) is that the correct answer is more likely to be (C) That we'll first figure out how to scan the 3D structure of the brain in enough detail to place our psyche's into computers and dispense with these dangerous organic bodies. Once that happens, we can say several things about interstellar travel:

1. Since (as a block of digital data) we would be able to be transmitted at lightspeed - we could send a very slow ship anywhere we really wanted - and transmit our minds via radio or laser to the onboard computer when it eventually arrived. No time would seem to pass for us during the trip - and it would only take four years to travel four lightyears. In effect we could teleport anywhere where there was a sufficiently powerful computer at lightspeed. Very 'Star Trek'!
2. We could slow down the clock rate on the computer simulating our minds and time could be sped up to any degree desirable to make a fairly conventional trip happen with as much or as little time as we wanted it to. You might want to speed up your clock to normal speed (or perhaps even faster) for a minute or soe every day - just to check up on your craft - then fast-forward through the next 23 hours 59 minutes. Forty years would zip by in about ten days of perceived time.
3. With effective immortality - and a more or less unlimited ability to 'fast-forward' through time by slowing down our system clocks, our 'long view' of events would make use pay much more attention to these kinds of long-range goals than we do at present. Stuck here on earth would be a dangerous thing indeed for an immortal being - the odds of being squished by an asteroid or something nasty would certainly have me demanding that a backup copy of my mind be kept a nice safe distance away and updated by laser transmission at least annually. Proxima Centauri would do nicely.
4. Without the demands of a physical body - we wouldn't need oxygen, food, water, recycling or anything else like that - just a nuclear power cell good for the duration of the trip. So maybe this craft could be extremely small - perhaps the size of a coffee can - yet it could easily contain a virtual reality world of whatever size we needed to keep ourselves amused during the trip. A tiny craft like that could be accellerated to high speed much more cheaply than a honking great life-support system!

I believe we'll have the computing and scanning technology to do this for a few (rich!) people in 40 years and for everyone else in 50 years. It might be dangerous to be scanned and limiting to be 'stuck' in cyberspace - but if you were near death anyway, it would be worth it. SteveBaker 22:18, 29 May 2007 (UTC)

(B) could be a real downer for the human involved. Imagine setting off as THE FIRST HUMAN TO LAND ON AN EXTRASOLAR PLANET!, being frozen, getting there 500 years later and being unfrozen only to discover that spacecraft technology improved leaps and bounds in the time you were in stasis, the ships built after yours got there much faster than yours and the system was already teeming with humans. Not only are you now 500 years out of your time, instead of being a pioneer, you are now just a historical curiosity. --84.68.104.115 22:30, 29 May 2007 (UTC)

A.E. Van Vogt wrote a story like that ("Far Centaurus" I think). Clarityfiend 22:46, 29 May 2007 (UTC)

This question asks for speculation and is therefore disallowed under the present guidelines.--Tugjob 22:59, 29 May 2007 (UTC)

• When we achieve that, most likely both technology will be used. A spacecraft that reaches there, say about 150-200 years; and some bio-chemical ways to slow down the human metabolism perhaps in a hibernation or dormancy state so that he can last the journey, to conserve energy, and also to prevent him from being bored to death. I don't think Cryogenics will work, but increasing lifespan to 200 years is feasible. Afterall, we have doubled our life expectancy in the last century. --Vsion 00:19, 30 May 2007 (UTC)

B, I think. I saw on this show that a researcher is developing a blood-substitute (using fluorocarbons, I think) that will enable medical teams to immediately shut down an injured patient (e.g., after a 911 call or on a battlefield) and them revive him/her at the hospital. The point, I think, was that many patients die in between the arrival of medical care and entering a hospital. They called it "temporary death." They're not far away from developing this right now, and I imagine that it could be extended to space travel if needed. Getting spacecraft up to an appreciable fraction of the speed of light, however, is still far, far away. zafiroblue05 | Talk 01:21, 30 May 2007 (UTC)

There's also an episode of Naked Science where they talk about space travel. Scientists know of genes that allow some animals to survive being frozen (Wood Frog, for example), as well as animals that can withstand higher temperatures than other animals (some reptile, IIRC). They speculated that if those genes were put into humans, that humans might be able to survive the trip as well as the conditions at other planets. However, they did raise the question if those humans are still humans. --Wirbelwindヴィルヴェルヴィント (talk) 01:31, 30 May 2007 (UTC)

What would happen

What would happen if I ate a cactus. Not like a cooked cactus, I mean if I just went to some desert and knifed a cactus into bitesize pieces and shoved it down my throat taking painkillers if needed cause of the barbs.

Seemingly some cactus varieties are edible (http://www.foodreference.com/html/f-cactus.html, http://www.rawguru.com/cactus.html). I suspect that some are not edible/could cause harm so (as with any 'eating wild food') you should always be 100% sure before eating produce in the wild. ny156uk 23:18, 29 May 2007 (UTC)

You can eat raw nopales, assuming you remove the spines. If you ate the spines too, I suspect you would tear up your mouth and esophagus, but as I don't know anyone who's tried it, I can only speculate. You don't have to go to a desert, either - prickly pears grow in a lot of places. They're even common in Florida. --Reuben 00:48, 30 May 2007 (UTC)

Haven eaten both nopales and prickly pears, you obviously take the spines off. Nopales are eaten raw, and it's rather slimey, so a lot of people don't like it from what I gather. Prickly pears are eaten raw like a fruit, where the seeds aren't chewed, and are quite tasty. --Wirbelwindヴィルヴェルヴィント (talk) 01:23, 30 May 2007 (UTC)

mmm, nopales. Amazes me that some people might not like 'em. It's not like we're talking natto or something. --jpgordon^∇∆∇∆ 01:28, 30 May 2007 (UTC)

A cell's life?

Do you agree that a cell (any cell) leads only one kind of life - either a mitotic or a meotic life? Or are there cells known to exhibit both?Arun T Jayapal 23:25, 29 May 2007 (UTC)

I would check mitotic to start with.

I would start with your textbook.... --Wirbelwindヴィルヴェルヴィント (talk) 01:24, 30 May 2007 (UTC)

Biology

     what chemical the white blood cells produce that makes it functional?