Wikipedia:Reference desk/Archives/Science/2009 May 9

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May 9

Cockatiel Blues

What do you guys think of this? Is this Cocktatiel really performing a spontaneous blues improvisation? --Kurt Shaped Box (talk) 00:29, 9 May 2009 (UTC)

No doubt the bird has heard that tune played 546 times and learned to imitate it. Along the same lines, have you seen Snowball (Cockatoo)? Looie496 (talk) 02:17, 9 May 2009 (UTC)

Yep, I'm certainly aware of Snowball and others like him. I actually started that article... ;) It seems that parrots really do have some appreciation of human music... --Kurt Shaped Box (talk) 16:20, 10 May 2009 (UTC)

Thought Experiment: Perpetual Energy

Of all the systems in a car, I believe the brakes are the most amazing. They will reliaby work just the same with no gas and a dead battery. With a minor press on a pedal, 2000 lbs of automobile will come careening to a dead stop from 100 mph in just a few dozen feet. I had no idea my foot was that strong!

I understand that liquids can't be compressed and that is the "magic" of hydraulics. My question is: what would happen if that same effortless press on a pedal was used to power up turbines instead of stopping a giant machine barrelling at ungodly speeds? Surely the energy produced would be able to overtake the effortless pedal press with more to spare! Sappysap (talk) 01:55, 9 May 2009 (UTC)

You seem to be confused about either the first or the second law of thermodynamics. The first says that energy is conserved. The kinetic energy of the car doesn't just disappear, neither does it go into your foot. The brakes convert the kinetic energy into heat (for example by pressing a disk against the wheel and letting friction slow it down). See File:Ceramic brakes.jpg for a striking visual manifestation of this heat.

The second law says that entropy always increases, or in less precise and more practical terms, energy can't be converted back from heat. This is why it's impossible to make a brake go in reverse and spin up a turbine rather than slowing something down. Braking a car is an inherently irreversible process, because the entropy increases so much. —Keenan Pepper 02:55, 9 May 2009 (UTC)

And just so we're clear, hydraulic brakes aren't fundamentally different from any other brake system. Ultimately, they also convert kinetic energy into heat by means of friction. The hydraulics simply transfer the kinetic energy before it is turned into heat. —Keenan Pepper 03:00, 9 May 2009 (UTC)

(Edit conflict) You seem to have confused two separate sets of energy. Your "effortless press" uses a small amount of energy generated in the cells of your leg muscles. This merely moves the brake pedal, which through the medium of the hydraulics (perhaps supplemented by servo-motors powered by your car's engine or battery) moves parts of the mechanisms of the Disk brake on each braked wheel, pressing the brake pads against the brake disks. The energy required is small, and is independent of whether your car is moving or not.

Your moving car, as you are aware, possesses a large quantity of kinetic energy which has been converted from the stored chemical energy of its fuel by its engine. By your lightly pressing the brake pads against the wheels, this kinetic energy is converted by friction into heat: no additional energy is generated by your operation of the brakes. This is not the same as somehow exerting a new large force opposite to the car's motion in order to stop it, which would indeed require an amount of new energy comparable to the car's kinetic energy (though somewhat less as road friction, internal mechanical friction and air drag are already helping to oppose the car's motion). Ask yourself, since energy can neither be created nor destroyed, only converted from one form to another, where would this new energy have come from?

This form of braking is called Dynamic braking, and the dissipated heat energy (which started as chemical energy in the fuel you paid for) is lost/wasted. In some recent vehicles, some of the energy converted by the brakes in order to slow the car is not dissipated as heat, but converted into a storable form, usually mechanical (for example by spinning up a flywheel) or electrical (in the existing or a supplementary battery), for re-use. This type of brake is called a Regenerative brake, but it too is not somehow creating additional energy, merely saving some of the energy that would otherwise have been lost.

My explanation has been broadly conceptual, but I expect another responder will set out the relevant physical equations in mathematical form, if that will help. 87.81.230.195 (talk) 03:12, 9 May 2009 (UTC)

The reason you can produce so much force in the brake pads with a fairly light touch of your foot is because your foot moves downwards several inches - but the pads only have to move inwards by a tiny fraction of an inch. That provides the 'mechanical advantage' you need to exert that kind of force...kinda like leverage...but with a liquid. Of course, many cars also have power-assisted brakes - and in that case, power (sometimes from the vacuum system of the engine, sometimes via a hydraulic pump attached driven from the serpentine belt - but increasingly, electrical) is applied to help your foot do the work.

As others have pointed out, the energy from the cars motion gets converted into heat and is dissipated by the disk (or drum) in the braking system. After a lot of heavy braking, the disks can get so hot they they'll actually glow - and if they get too hot, the hydraulic brake fluid might boil - this is called "brake fade" and it's exceedingly dangerous because the gasses produced by the boiling brake fluid is easily compressible (unlike the liquid which is impossible to compress) - so if the brakes get too hot, when you push down on the pedal, all that happens is that the gasses compress, the brake pads stop moving and quite suddenly you have no brakes!

That's why on long hills, you should slow down using the engine (down-shifting the gearbox) rather than using the brakes for large amounts of time. If you do have brake fade, then in an emergency, you can gently apply the parking brake - which is typically connected using a cable system that won't suffer from the heating problems. But you have to be pretty careful because you can easily lock up the wheels and put the car into a skid. SteveBaker (talk) 03:30, 9 May 2009 (UTC)

will ... what you are'nt understanding is that work is constant , so that brakes just applied the work you've done in avery

effective way on the wheels , its just like hydrolic machines , belozers , were the engine do more than 3000 rpm will its juut

make the vhicle to left the bocket up less than a meter . its just transfere like 40000 rpm to left the bocket up ward like

a meter. thats it .

you can explain it by imagining the multi speed bicycle , in some speeds you have to cycle you legs two times to revolve the

wheel once , which is easy ... in other cases you need to cycle one time to revolve the wheel 5 times which is too hard ,its

need mush bower than your legs can produce .--Mjaafreh2008 (talk) 10:52, 9 May 2009 (UTC)

MSUD in Adults??

Can an adult have MSUD. Can it be contracted in teen or adult years. What else might be causing this Maple Syrup smell in my sons room and bathroom?? —Preceding unsigned comment added by 68.195.147.98 (talk) 03:03, 9 May 2009 (UTC)

Sorry, the Reference Desk doesn't give out medical advice. Please consult a doctor. Tempshill (talk) 03:12, 9 May 2009 (UTC)

The second half of the question is not a request for medical advice, though. There are many things that could be causing maple syrup smell in the room - maybe a stash of hidden pancakes? Nimur (talk) 04:32, 9 May 2009 (UTC)

I'm with Nimur on this one. The substance that is responsible for maple syrup odor - both in pancakes and in MSUD - is sotolone, and, naturally, we have an article about it. --Dr Dima (talk) 07:48, 9 May 2009 (UTC)

Earthquakes

Is it possible to find the centre of an earthquake? —Preceding unsigned comment added by 174.6.144.211 (talk) 03:14, 9 May 2009 (UTC)

Yes - they can measure the arrival time of the seismic waves at remote seismology stations and by comparing those times, figure out where the earthquake was happening. Those shock waves move pretty fast but with accurate clocks, it can be done to a reasonable precision. SteveBaker (talk) 03:17, 9 May 2009 (UTC)

Well, to the extent that there is a well-defined "center". Earthquakes, especially big ones, don't happen at a single point. There's a section of the fault that ruptures. So it's more like it happens along a line (in the non-mathematician's use of the word line — for mathematicians that would be a one-dimensional manifold, possibly with boundary :-). --Trovatore (talk) 03:41, 9 May 2009 (UTC)

Actually, earthquakes happen (mostly) along a surface, (a two-dimensional manifold, possibly with boundaries.) And the motions vary with time, so we need to ass a third dimension. A classic example would be an earthquake that progresses along a fault. -Arch dude (talk) 07:34, 9 May 2009 (UTC)

Also see Epicenter. 71.236.24.129 (talk) 08:59, 9 May 2009 (UTC)

The epicentre, as mentioned above which is where the rupture begins. It then propagates within the fault surface until it runs out of stored elastic strain energy. In many cases the epicentre is distinctly offcentre for the rupture e.g. both the 2004 Indian Ocean earthquake and the 2008 Sichuan earthquake initiated right at one end of the fault segment that moved. We know this because of the distribution of aftershocks, which define pretty exactly the extent of the rupture. In terms of the felt intensity the area with greatest damage is often near the epicentre and for historic earthquakes, that happened before instrumental recordings, this is how the epicentre is estimated. In some earthquakes. however, such as the 2002 Denali earthquake the maximum intensity was felt at the other end of the fault surface that ruptured from the epicentre. That earthquake also illustrates another possible complexity, it began on a thrust fault jumped to the strike-slip Denali Fault before jumping again onto the Totschunda fault, another strike-slip structure. Finally, most earthquakes hypocentres (the actual point of initiation) occur within a small range of depths, normally 10-15 km, known as the seismogenic layer. This is the strongest part of the crust, as increased confining pressure makes fracturing progressively more difficult until the temperature rises sufficiently for ductile processes to become important. Most of the rupture propagation, therefore, occurs either laterally or upwards, so if you looked at the area that slipped, the epicentre is often nowhere near the centre of the rupture at all. Mikenorton (talk) 12:29, 9 May 2009 (UTC)

Are convergent, divergent, collisional, and transform boundaries fault lines? —Preceding unsigned comment added by 174.6.144.211 (talk) 04:20, 10 May 2009 (UTC)

All types of plate boundaries have faults associated with them, but only for some of them is there a single fault surface along the boundary itself. Subduction zones pretty much act as simple fault surfaces, as do transforms in oceanic crust. Some continental transforms such as the Alpine Fault in New Zealand, form single fault surfaces along the boundary but others, such as the San Andreas Fault form part of a deformed zone within which there are many active faults. Divergent boundaries in continental crust form rifts with many active normal faults, such as the East African Rift. Collisional zones such as the Himalayas have huge zones of deformation, sometimes 100s of km across, within which there are many seismically active fault zones. Mikenorton (talk) 11:51, 10 May 2009 (UTC)

I know that it is possible to calculate the depth of a hypocenter by using p waves, but is it possible to calculate the location of a hypocenter? —Preceding unsigned comment added by 174.6.144.211 (talk) 23:31, 10 May 2009 (UTC)

Electric circuit for highpower sawtooth wave output.

I need a sawtooth wave generator and i've searched the internet for it; but all the circuit i found have very low current output.The output i need is shown in fig.

File:Output Sawtooth.JPG

Where can i find my desired circuit diagram? —Preceding unsigned comment added by Shamiul (talk • contribs) 04:06, 9 May 2009 (UTC)

You're going to need a pretty hefty amplifier. Have you considered a PA system connected to a signal generator? You could take the output off the speaker-lines and connect it to whatever your load is. My back-of-the-envelope calculations are suggesting that this is going to need an average power output of around 600 watts, which is awfully big for an audio system. Are you really sure you need 25 amps at 40 volts? Nimur (talk) 04:30, 9 May 2009 (UTC)

Thank you for your answer, but i don't mean amplifier. The main circuit should deliver the output required that can be directly delivered to the load. I want to know the procedure that you used to calculate the power output, in case if i had made mistakes. —Preceding unsigned comment added by 123.49.45.67 (talk) 04:57, 9 May 2009 (UTC)

Nimur appears to have arrived at the power requirement by a standard RMS integration of the voltage waveform multiplied by 25 amps. However, in his audio amp solution the 5 volt offset would have to be added back after the amp because audio amps do not usually pass dc, the required power rating of the amp is therefore a little less. You do not say why this is not an acceptable solution, and without details of your application it is difficult to give you precise help. Possibly bandwidth is a concern, 10kHz would certainly go through a high quality audio amp but the harmonics are likely to get mangled, badly distorting the waveform from linear. There are certainly plenty of audio amps out there in the right power range at affordable prices. In any case, any solution is liable to be a low power waveform generator stage plus a power amplifier stage as Nimur says, although the power stage might be just a power transistor working off a 40 volt rail if your waveform generator has an open collector output. I think you will struggle to find a ready made circuit diagram to do this - high voltage ramp circuits were common in the days of crt televisions, but I can't think offhand of many high current applications. So to give you some specific suggestions: for the waveform generator there is the 555 timer which has been around since the stone age and the internet is littered with circuit diagrams and application notes. Here is a page describing some alternative waveform generator ICs. If you need great precision of the waveform then you might want to consider a synthesised arbitrary waveform generator which are available as modules/pcbs from some manufactures. For the output transistor, there are many available with the required current and voltage rating. SpinningSpark 10:37, 9 May 2009 (UTC)

Yes, that's how I got it... I think the original questioner should really provide context. When you say you don't want an amplifier, I'm a little worried - because any real circuit which will generate such high voltages and currents IS an amplifier. If you're a novice electronics enthusiast, you should not be playing with kilowatt-scale, high-amperage systems, because they can kill you. (Let's be clear here - 40 volts and 25 amps is deadly). What are you trying to do? With some context we can give you a better and more realistic answer. Nimur (talk) 15:05, 9 May 2009 (UTC)

By the way, when you asked the same question last year your image was deleted for the same reason this one's going to be if you don't put a licence tag on it soon. SpinningSpark 20:35, 9 May 2009 (UTC)

As a retro alternative circuit, how about a 5 volt DC generator in series with a DC generator, the field current of which is regulated by a motor driven rheostat such that it sweeps through the required output voltage? A 600 watt generator is not all that large or expensive. Edison (talk) 01:40, 10 May 2009 (UTC)

That wouldn't be very energy-efficient! Half the power would be burned over the rheostat! You really shouldn't be encouraging resistive-loss as an effective method for generating a desired voltage. This is especially bad if you're really going to pump the output signal at 25 amps through it! That sucker's going to need a heat sink! Nimur (talk) 16:15, 10 May 2009 (UTC)

Is your first name Heath? SpinningSpark 02:15, 10 May 2009 (UTC)

Downshifting vs. braking

Inspired by a SteveBaker answer above: Is it really better to downshift than brake when going down a hill? Won't this cause excess wear on the transmission or gearbox or some other part of the car that, unlike brakes, was not designed to be worn so much? Tempshill (talk) 04:10, 9 May 2009 (UTC)

Hmm? I don't see where there's any extra stress on the drivetrain- it's the same as what's going on most of the time while you're driving. Brakes slow you down by friction- the pads wear out as you use them. And they're emphatically not meant to be used constantly for long periods of time, because they'll heat up and fail (see brake fade). The drivetrain is meant to be turning pretty much all the time. Friday (talk) 04:15, 9 May 2009 (UTC)

When I lived in Colorado and Montana I did a lot of mountain driving. I could always tell who the flat-landers were by the excessive amount of brake lights they displayed on the down grade. --jwalling (talk) 04:41, 9 May 2009 (UTC)

In Colorado and Montana you'd also likely have encountered quite a few people driving cars with automatic transmissions. They are a lot more common here than in Europe. 71.236.24.129 (talk) 08:51, 9 May 2009 (UTC)

Not exactly the same, but maybe you're interested in jake brake. Shadowjams (talk) 09:02, 9 May 2009 (UTC)

i think gears were'nt designed to slow the vhicle , because at high speeds its not right to shift back from the fourth to the

first , and in some cases it could break the sustem . i think its better to use brakes instead except when its dangrous to do

so at high speeds or at mountain roads were such action could cause the car to go off the road .--Mjaafreh2008 (talk) 09:49, 9 May 2009 (UTC)

I live in a mountainous country, where we are taught to always downshift when driving downhill. The article Måbødalen disaster may be of interest. It was in part caused by the bus driver's lack of experience in mountain driving. --NorwegianBlue ^talk 11:40, 9 May 2009 (UTC)

But you aren't trying to slow the vehicle, you're trying to prevent the vehicle speeding up. That's a very different thing. If you want to slow down, use the brakes, if you want to maintain a constant speed, put the car in the right gear for that speed (which will be a lower gear when going downhill than when on the level). --Tango (talk) 13:27, 9 May 2009 (UTC)

Let's be PERFECTLY clear about this. For routine slowing down - brief dabs on the brakes separated by minutes of time, typical in-town driving - then you most certainly should use the brakes. That's what they are there for. However, if you are going to have your foot on and off the brake - or lightly pressing the brake for (let's say) a minute or more - then you're driving extremely dangerously and sooner or later you'll boil your brake fluid and you could very easily DIE as a result of your failure to understand the principles of controlling your vehicle. This most often happens on long downhill sections (but it can happen if you're driving agressively on the freeway at high speeds). In these circumstances, you MUST use the engine to slow yourself down. On a stick-shift, you need to shift down progressively (ignoring the horrific racing noises from the engine!) until your speed is under control - where you want it to be. The engine and transmission most certainly are designed to be able to do this - you aren't damaging it any more than when you stamp on the gas pedal to accelerate away from a stop. On an automatic, taking your foot right off of the gas pedal may be enough - but if you're still accelerating then you need to use the brakes one time to get your speed down - and then put the transmission into '2' or even '1' for the remainder of the hill. People who worry that they are ruining their transmission or engine need to understand that they are ruining their brakes by NOT doing this - heating their brake disks/drums until they are literally glowing red hot is going to cause them to warp and to wear the brake pads prematurely. If you do experience brake fade, the seals on brake slave cylinder can be ruined and you can get all sorts of crud in your brake lines. I can't overstress the importance of this. It's one of the commonest mistakes people make when they aren't used to driving in hilly terrain. Failure to understand how brake fade comes about kills a good number of people every year - you should have paid more attention during driver's ed. classes! SteveBaker (talk) 14:13, 9 May 2009 (UTC)

Well, this is a bit exaggerated. It's pretty hard to lose your brakes in a modern, ordinary-sized passenger car, unless they are in bad shape to start with. The danger is only serious for heavy vehicles like trucks or buses, or if you do something really stupid like keeping your foot on the brake while pressing the accelerator. It's definitely better to use engine-braking as much as possible, but you don't have to be paranoid about it. I've numerous times descended steep switchbacky mountain roads where I had to brake the whole way down for thousands of feet -- being in low gear helped of course but only to a limited degree. Looie496 (talk) 17:21, 9 May 2009 (UTC)

? Please quote sources, and re-read Steve's response and the article I linked to. --NorwegianBlue ^talk 19:11, 9 May 2009 (UTC)

If you were in first and still going too fast then, of course, you have to use the brake, you have no choice, but you were still risking your brakes overheating. --Tango (talk) 20:19, 9 May 2009 (UTC)

You sure have a choice. Try shifting to reverse. --NorwegianBlue ^talk 20:36, 9 May 2009 (UTC)

Ok, you don't have a good choice! --Tango (talk) 21:50, 9 May 2009 (UTC)

I think we are confusing two different issues: The use of either brakes or the transmission to slow down the car versus the use of either brakes or the transmission to maintain speed on a hill. The brakes should always be used to slow the car down, but the transmission should be used to maintain the proper speed when going down hill. That's because brakes are not designed to be applied for long periods of time, but in short bursts. --Jayron32.talk.contribs 04:36, 10 May 2009 (UTC)

Why do you advise against taking advantage of the engine's ability of slowing down the vehicle, say, when leaving a motorway? Used with proper technique, you can drive very smoothly that way. Combined, of course, with using the brakes. --NorwegianBlue ^talk 10:44, 10 May 2009 (UTC)

Generally - because brake pads are cheaper to replace than clutches, you should use the brake for short decelerations. But if you aren't concerned about wear and tear, then either approach works and is safe for brief periods. I confess that I use engine braking more than I probably ought. The big issue is only with prolonged use of the brakes - which is just plain dangerous. SteveBaker (talk) 14:20, 10 May 2009 (UTC)

Steve, do you have a reliable source for this advice? Thanks. Axl ¤ [Talk] 06:28, 13 May 2009 (UTC)

Climate change vs. Global warming

Are there simple guidelines to determine which term should be used in a discussion. In my mind, Global warming is a major cause for the effects of Climate change on regional scales. It's my impression that 'climate change' is used often when 'global warming' would be more accurate. --jwalling (talk) 04:27, 9 May 2009 (UTC)

Global warming is one form of climate change (i.e a subset). It's also possible to have global cooling and other forms of climate change. We are currently experiencing massive human induced climate change in the form of global warming which is expect to get worse but obviously when you refer to global warming it doesn't have to refer to this particular instance nor does climate change have to mean global warming. Nil Einne (talk) 04:37, 9 May 2009 (UTC)

The ambiguity of 'climate change' (warming or cooling) supports my view that 'global warming' is more accurate when discussing the current global climate. 'Climate change' is appropriate for an unspecified region or uncertain period of time, but if you do specify a region and a time period which is showing warming, such as the Arctic for the next 10 years, then 'Climate warming' is more accurate. -- jwalling (talk) 05:35, 9 May 2009 (UTC)

Well firstly I'm presuming you've already specified on the planet earth and over the past 100 or so years and the next 100 or so years. Otherwise it's still confusing (are we referring to global warming on Venus? 1 billion years ago? 1 billion years from now). But even then arguably anthropogenic global warming (a favourite of denialists) or human-induced global warming is more accurate (or more correctly more precise since global warming is by definition a form of climate change so it's very rare you can be wrong if you specify climate change instead of global warming, it's just that you are not being precise as to what form of climate change you are talking about) since it is what we are observing (not say global warming due to sunspots). If you want to go further, perhaps global warming due primarily to an increase in global greenhouse gas levels as a result of human activities. Even further global warming due primarily to an increase in global greenhouse gas levels as a result of human activities including extensive use of fossil fuels, large scale changes in land use...... I think to some extent both are precise enough. There is only one form of massive global climate change we are currently observing and expect as far as I'm aware and that is global warming. So if you want to talk about climate change from a global sense in the next 100 years then of course that will primarily be about global warming not global cooling or anything else. It also depends on what you're talking about. For example the scientific opinion on climate change is about all climate change we expect in coming years and as I've mentioned this is global warming. Similarly the Intergovernmental Panel on Climate Change is supposed to study all climate change and of course global warming is of primary interest to them. In other words, IMHO it's more a matter of semantics and trying to argue which one is more accurate or perhaps IMHO since as I've already stated more accurate is a bit of a misnomer, which term is better applied to a given situation is mostly a pointless waste of time outside of times where it matters e.g. the naming of wikipedia articles, it's better to concentrate on more important things liking convincing those who are fooled by the denialists that there isn't one big conspiracy and climate change in the form of global warming is happening and is going to cause major problems and yes it is caused primarily by human activity. Nil Einne (talk) 08:46, 9 May 2009 (UTC)

The problem with the terminology is that the increase in greenhouse gasses is causing global warming - but also other kinds of climate change (increased hurricane strength and/or frequency, rising sea levels, changes in drought patterns, etc). So the term "Climate Change" better suits the whole range of problems that are being caused. However, some of these effects are secondary effects of global warming - without global warming, most (if not all) of the other effects wouldn't be happening. However, there is at least ONE other effect that we can describe as "Climate Change" which is totally unrelated to global warming or the greenhouse effect - and that is the destruction of the ozone layer. That was not caused by greenhouse gasses like CO2 and Methane - but instead by Chloroflourocarbons released into the atmosphere. Destruction of the ozone layer causes increases in ultraviolet radiation - which is harmful in all sorts of ways. Worrying about the ozone layer has become less trendy than concern over the greenhouse effect - but it's also a major problem. I think it's reasonable to say then that the term "Climate change" is more encompassing than "Global warming" which is in turn more encompassing than "The Greenhouse effect". So, although I confess to tending to use the terms interchangeably - if we are being super-careful:

• If you are talking about CO2, Methane and other gasses that we're dumping into the upper atmosphere - then you need to talk about "The Greenhouse Effect".
• If you are talking about organohalons, chloroflourocarbons, nitric oxide and related gasses causing destruction of ozone - then "Ozone layer depletion" is the correct term.
• If you are talking about gasses dumped into the lower atmosphere that cause plant destruction and soil pollution as they are washed out of the atmosphere by rain - then you need to talk about "Acid Rain".
• Any or all of the previous three terms could be described as "Atmospheric pollution".
• If you are talking just about the rise in globally and seasonally averaged air temperatures - then you should say "Global Warming" - because there are other causes (and mitigating factors) for that beyond the greenhouse effect - things such as the decrease in the planets' albedo due to melting ice caps and glaciers and the increase in albedo due to the contrails from high flying jet aircraft.
• If you are talking about all of the consequences of all of the things we're doing to the planet's atmosphere and hydrosphere - then "Climate Change" is a more appropriate term - because there are other causes of that - such as changes in water usage patterns and evaporation rates, destruction of the Ozone layer, the effect of large cities creating local hot-spots, etc.
• If your concerns range beyond that, then "Oh Shit!" may be the term of last resort!

SteveBaker (talk) 13:45, 9 May 2009 (UTC)

marine diesel engine

working of air distributar Bose09 (talk) 07:10, 9 May 2009 (UTC)

Diesel engine has some information under medium-speed engines and low-speed engines but it is a bit scanty I'm afraid. SpinningSpark 11:08, 9 May 2009 (UTC)

Kc calculation

"The equilibrium constant, Kc, for the reaction to form ethyl ethanoate from ethanol and ethanoic acid, C2H5OH + CH3CO2H=CH3CO2C2H5 + H2O, at 60 degree Celsius is 4.00. When 1.00 mol each of ethanol and ethanoic acid are allowed to reach equilibrium at 60 degree Celsius, what is the number of moles of ethyl ethanoate formed?" This is the question I have to solve. I know you guys dont do homework questions and I dont want the answer to the question either. I just want to know how I am supposed to solve this question, i.e the working. Thanks. —Preceding unsigned comment added by 116.71.59.203 (talk) 08:38, 9 May 2009 (UTC)

Use the first formula given here. yandman 11:50, 9 May 2009 (UTC)

But the question doesn't give the amount of water produced. I cant use that formula. —Preceding unsigned comment added by 116.71.59.203 (talk) 11:57, 9 May 2009 (UTC)

You don't need to be given that...you need to figure it out (at least algebraically) from the info that is given. You're not told "how much ethanol is present at equilibrium" but "how much ethanol is starting before equilibrium has occurred". The amount of reactants are decreased by the identical (considering a stoichiometry of a balanced reaction) that the amount of products increases. I am sure you did examples like this in class or in your textbook. DMacks (talk) 17:37, 9 May 2009 (UTC)

So by "figuring it out", Id say that the amount of water is also 1 mole? I haven't really understood this part. —Preceding unsigned comment added by 116.71.33.115 (talk) 05:10, 10 May 2009 (UTC)

Maybe and maybe not. You start with 1 mole of starting materials and 0 mole of products (per question as written). As you can see in the reaction, every bit of water that is formed comes from ethanol, so if the full 1 mole of starting material is consumed, you get 1 mole of product, and that would leave leave you with 0 mole of starting material. The whole idea of equlibrium and Kc is that the reaction doesn't run all the way (which would consume all of the starting material) but only runs partway. "Some" starting material gets converted to product, "the rest" remains as starting material. And Kc defines exactly the ratio of those two quantities. You know the total (perhaps SM+P?) and the ratio (formula for Kc and its given value), now solve for the variables. DMacks (talk) 08:07, 10 May 2009 (UTC)

teeth

I have a very weak teeth , i visit alot of dentist by they were'nt alot of help

now after all that advance in evry field , i find it wierd that they had'nt find

asolution for cavity , its brrety simple .

some kind of amouth wash that will produce a thin film at the surface of the tooth

so it wont be in contact with food left , so no cavity will form .

realy i dont think its that hard , we built a space ship , rockets , submarines , airplanes

i think its aprrety easy task .

i said that because i visit the dentist so much , and its frightning each ti.

--Mjaafreh2008 (talk) 09:37, 9 May 2009 (UTC)

Okay. Do you have a question? -- Captain Disdain (talk) 10:26, 9 May 2009 (UTC)

Answers removed, after discussion on the talk page.

Mjaafreh2008, as Captain Disdain pointed out, you aren't really asking a question. Please rephrase your question, keeping in mind that we cannot give medical advice. --NorwegianBlue ^talk 16:55, 11 May 2009 (UTC)

Class of drug

Is Clopidogrel an Anticoagulant? If not what's the difference? (No medical advice.)71.236.24.129 (talk) 11:02, 9 May 2009 (UTC)

In the ATC classification system, clopidogrel has the ATC number B01A C04, and is classified as an antithrombotic agent. This ATC group has its own Wikipedia page: ATC code B01. If you check it out, you'll find that a traditional anticoagulant such as Warfarin is in the same group. The difference is that clopidogrel inibits platelet aggregation, whereas anticoagulants inhibit the proteins necessary for coagulation. --NorwegianBlue ^talk 11:25, 9 May 2009 (UTC)

(After EC) The usual definition of an anticoagulant is a substance that stops the blood from clotting. Anticoagulants like heparin or warfarin directly affect the clotting factors and prevent the entire process from getting started. Clopidogrel is an antiplatelet drug, which prevents platelet activation, the process by which platelets form a kind of plug to stop bleeding. The drug therefore prevents aggregation of platelets but does not interfere with the rest of the clotting cascade. It may seem a subtle distinction since the end result of both classes of drugs is to interfere with the formation of blood clots, but there are differences in mechanism and what the various drugs are used for, so it's a meaningful separation. --- Medical geneticist (talk) 11:38, 9 May 2009 (UTC)

A difference that also may be of interest is in their indications: clopidogrel is typically used to prevent arterial thrombosis after medical procedures such as the insertion of a coronary stent, anticoagulants are typically used to prevent venous thrombosis. --NorwegianBlue ^talk 12:00, 9 May 2009 (UTC)

Thanks. It's a bit clearer now. 71.236.24.129 (talk) 14:01, 9 May 2009 (UTC)

How a factory works

Is there any book/package/work/whatever that will teach you exactly how a factory works? Or if you want to create a factory you must pick up the pieces from here and there? --Mr.K. (talk) 11:53, 9 May 2009 (UTC)

Please clarify: are you asking about a factory or a Factory pattern? --NorwegianBlue ^talk 12:04, 9 May 2009 (UTC)

If it's the latter: A piece of software that "constructs class instances" (all three of those words having very specific meanings) according to some set of parameters - then you REALLY need to ask this question on the computing section of the ref desk.

If you're talking about an actual physical factory...then it's still not got much to do with science and you might get a better answer on the miscellaneous desk where there are more people to respond to the question.

SteveBaker (talk) 13:18, 9 May 2009 (UTC)

Every factory is different. Some are highly mechanized, and some are just big empty rooms where people work, sewing or bending sheet metal or some other process. If you have a question about a specific type of factory, we can answer that better. Maybe you would find the machining, assembly line, and manufacturing articles helpful. Nimur (talk) 13:39, 9 May 2009 (UTC)

Most factories are based on a Workflow analysis. So if you are thinking of a brick and mortar version that would be a good starting point. Apart from that we'd really have to know what type of factory you're interested in. There are significant differences between building a meat processing plant, a saw-mill or a coffee machine manufacturing plant (OR). Even restrictions on site selection (where you could/should build it) would be quite different. 71.236.24.129 (talk) 13:58, 9 May 2009 (UTC)

Yes, every factory is different. I just wanted to know if any of you have ever seen a book/work/or whatever that explained exactly how you could build your own factory of whatever. I don´t want you to describe how a factory works, I just need the bibliographic reference. --Mr.K. (talk) 17:49, 10 May 2009 (UTC)

When you say 'build', do you mean like construction of the building? Or like design of the inner mechanical workings? Or like making all the machine parts? Or building the workforce and management structure? Or what? Are you thinking start with a plot of greenfield land and end up with a working factory?

A warehouse with workers stood at tables could be a factory, and all you'd need to build that would be the construction skills to make a one-room building (and the workers and tables and supplies and know-how and...) Something to make pharmaceuticals would be very different, which in turn would be completely different from a factory for smallscale cake production which would, in turn, require quite different skills to building a facility to process volatile petrochemicals.

So, are you just looking for a book that describes how to build a simple building? Like, bricklaying skills and suchlike? O_o 80.41.104.79 (talk) 21:12, 10 May 2009 (UTC)

Net charge of earth

Is the net charge of the earth is +ve ? Why?? —Preceding unsigned comment added by 117.98.99.182 (talk) 13:24, 9 May 2009 (UTC)

I don't think so. I'm pretty sure that the Earth as a whole entity has neutral charge. There are locally charged regions (a lot of times, charge builds up in the atmosphere); and while we have a radiation belt full of charged particles, there are both electrons and positive ions to balance out. Take a look at this Discussion on the Earth's Net Electric Charge. In this published paper, a scientist from the Navy Research Lab states "As a consequence of this discussion, any statement made up to now on a net electric charge of the earth remains in doubt. It may be that the net electric charge of the earth (as globe or as planet) is zero or near zero - always or in the average - but we do not know a way to prove or disprove it." This paper is a pretty good introduction - it defines the problem well, mentions several experimental and theoretical approaches, and describes the various difficulties in all these methods. Nimur (talk) 14:04, 9 May 2009 (UTC)

Hi Nimur, I saw that paper but cannot read all of it. Does it debunk the common statement that the surface of the earth has a high negative charge (balanced by positive charge in the upper atmosphere/ionosphere)?. Or is it merely saying that past attempts to enumerate this charge precisely are misguided? SpinningSpark 14:25, 9 May 2009 (UTC)

The author refines the question by asking whether the question is "net charge on the earth, volumetrically", "net charge on the earth's surface, integrated over the entire surface," and "net charge on the surface in a local sense". Without doubt there is a net charge on the earth's surface in a local and transient sense, and this is generally believed to be nulled out by atmospheric electric charge, sometimes re-equalizing with lightning. The author discusses the "spherical capacitor" model, with a charged surface (lithosphere) and a charged ionosphere boundary, but investigates whether that can really apply to a surface that is 70% oceanic. Further, there are discussions about net current, vs. net charge, and whether one can exist without the other. There is discussion about Earth's space environment (being a conductive plasma, any net charge on Earth would be able to flow off into space and equalize to zero). In each of these scenarios, the author provides the essential basic-physics approximation, and attempts to derive some consequences, before presenting conflicting experimental evidence to suggest that the basic-approximations do not hold well. His concluding remarks are that there are no practical methods to measure a net-charge on Earth, because of the various confounding factors; and that the theoretical explanations give various answers ranging from "large positive charge", "neutral," "large negative charge", "dynamic and fluctuating with zero-mean," yet none of these theories are practical to verify experimentally. Nimur (talk) 14:43, 9 May 2009 (UTC)

In specific answer to your question, SpinningSpark, he does not "debunk" the common statement about charge-splitting between lithosphere and troposphere/ionosphere. (As I mentioned above, this explanation is widely understood as the basis for a lot of tropospheric weather, and has been clearly observed in a variety of experimental setups). However, he does make two counterpoints regarding the extension of this model to the entire globe. First, characterizing the magnitude is much more complicated than it might seem (my experience in this area strongly supports such a stance). Second, the capacitor-model of charge-splitting may not be applicable over the entire earth's surface, (certainly not "instantaneously", and probably not even in the time-average-sense). Nimur (talk) 14:46, 9 May 2009 (UTC)

A 'Science Advisor' on physicsforums.com says that Earth is "neutral to about one part in 10^26", but I can't see where he gets that figure from. As mentioned above, the Earth is bathed in a plasma that would conduct away any excess charge. Perhaps the next question should be "What is the net charge of the Solar System?" --Heron (talk) 17:55, 9 May 2009 (UTC)

Polynomial in several variables

Somehow I stumbled on the polynomial $R(x,y)=\prod _{i=1}^{n}\prod _{j=1}^{n}(x_{i}-y_{j})$. Now I wonder what that is, that's to say, does this polynomial occur naturally in some context of math or other science? 93.132.150.3 (talk) 13:50, 9 May 2009 (UTC)

It doesn't look particularly useful to me, and I don't think it's a polynomial it's represented in a way that doesn't look very polynomial-ish. It appears to be some kind of norm function (though I don't think it satisfies some of the properties). It looks like it takes two vectors of length n, and combines them to give a single scalar value. Did you notice, though, that if any value of x is equal to any value of y, that the result is zero? Maybe that's a feature, or maybe it's a bug... Nimur (talk) 14:00, 9 May 2009 (UTC)

You don't just stumble on something like that as if you found it lying in the street, tell us what page it is on or otherwise give context. SpinningSpark 14:36, 9 May 2009 (UTC)

(After ec) I first mistook for some kind of volume, which it is clearly not, at least not the common kind. In the context I found it, the variables are free symbolic variables, so I didn't think of it evaluating to zero ... but yes, it's a feature, not a bug. I deliberately did not mention any details on the domain of the variables so I wouldn't miss any suggestions for other kind of variables. 93.132.150.3 (talk) 14:38, 9 May 2009 (UTC)

Take a look at Cauchy determinant SpinningSpark 14:48, 9 May 2009 (UTC)

Great, that's the polynomial! The way I came on it was that I had a look at signature of permutations where the second proof uses a similar polynomial and I was wondering if the somehow arbitrary restriction to i<j could be avoided. Unfortunately, if you use the above polynomial for that you get the square of the signature, which makes it not very useful for that purpose. 93.132.150.3 (talk) 15:07, 9 May 2009 (UTC)

It is also similar to the Resultant. 81.98.38.48 (talk) 11:11, 13 May 2009 (UTC)

cloud formation

I have searched google and wikipedia and could not find the answer to this question. What attracts the water vapor/ice particles together to form clouds? Ionic or static charges? The wikipedia page for cloud formation simply says "When surrounded by billions of other droplets or crystals they become visible as clouds".

Is the attractive force between these billions of particles simply a natural phenomenon and no one knows why, or is there a commonly accepted answer?

thanks for your help,

HarveyHarveyalton (talk) 15:08, 9 May 2009 (UTC)

For clouds to form, two things must happen, the air must be below the dew point and there must be seeds (eg dust) around which water vapour can nucleate. This can happen with rising warm air, as the air rises in the atmosphere the pressure falls which causes the dew point to fall. If the pressure falls far enough, the air becomes saturated (ie relative humidity is 100%) and water droplets start to form (if there are cloud condensation nuclei present). Other processes can cause air to rise and trigger cloud formation. Amongst these are the collision of warm and cold regions of air in frontal systems and orographic lift. SpinningSpark 15:24, 9 May 2009 (UTC)

Just a few corrections: the air need not be below the dew point, but only equal to it, for a cloud to form. This is because the dew point, by definition, is the temperature at which the air is saturated with water vapor. Also, it is not the pressure fall which causes rising air to become saturated, but the temperature change due to adiabatic cooling. But these are minor quibbles.

I think the OP misunderstands the concept of a cloud: The cloud particles are not physically bound to each other, they merely reside in the same area. Cloud droplets appear together (as a cloud) because air temperature can not change rapidly in a short area without equalizing itself quickly. Therefore, you won't have just small areas here and there where the temperature is at the dew point, but rather a (relatively) large area on the order of a few hundred or thousand feet. And since the atmosphere has no shortage of cloud condensation nuclei, almost any area at the dew point will also have cloud droplets, and thus a cloud.-RunningOnBrains 16:14, 9 May 2009 (UTC)

Pur water filter.

We have one of those Pur water filter gadgets that bolts onto the kitchen tap. It has an exchangeable charcoal filter cannister and a little battery-operated gizmo with a red/green LED.

The LED lights green when you run water through it - and the light turns red when the cannister needs changing.

Does anyone know what the sensor is actually measuring? I kinda suspect it's just measuring flow rate and integrating that to figure out the amount of water that's run through the filter...but I'd like to know for sure.

SteveBaker (talk) 16:07, 9 May 2009 (UTC)

I used to work in a laboratory that used de-ionised water in industrial quantities. We checked for when the filter chemicals needed changing by measuring the resistivity of the water. This can only be done when the water is actually running - as in your device apparently. So the facts fit but its still only a guess. SpinningSpark 16:27, 9 May 2009 (UTC)

If you have one of these the promo video on that page (click the cinemachine) says the LED indication is based on amount of water used. SpinningSpark 16:36, 9 May 2009 (UTC)

The one I'm thinking of is this one (actually, the one on the left of that picture). SteveBaker (talk) 17:38, 9 May 2009 (UTC)

I'mm not sure resistivity is a usable measure for simple charcoal-filtration. I would expect that lots of ions aren't filterable that way, so the detector would have to be calibrated for the expected presence and amount (fluoridation, hardness, etc.) for your particular supply. Ain't gonna get 10MΏ without reverse-osmosis or at least some well-designed ion-exchange resins. DMacks (talk) 18:55, 9 May 2009 (UTC)

I agree, there is a big difference between filtered drinking water and laboratory quality de-ionised water. I expect it is simply measuring throughput and knows the life expectancy of the filter cartridge. --Tango (talk) 20:13, 9 May 2009 (UTC)

That's my guess too - but I have a deeper concern. What cheap flow rate sensor might they be using? These gizmo's are pretty cheap. Is it possible that they just measure the amount of time for which water is present? (I'm kinda skeptical that these things are doing a good job). SteveBaker (talk) 22:39, 9 May 2009 (UTC)

I'm curious - why do have one of these in the first place? Do you live somewhere with water that doesn't taste very nice? As far as I know, that's the only genuine use to these things. Have you done double blind taste test to check if it is really making a difference? --Tango (talk) 00:31, 10 May 2009 (UTC)

The water quality here in Round Rock, Texas is generally pretty good - but there are certain times of year when the turbidity of the water goes off the charts and despite assurances that it's safe, it looks disgusting - and the filter takes care of that. I have not done double-blind taste tests - I'm more concerned about the actual, published data from the water utility. But overall, I'd rather we had this than have my wife attempt to destroy the planet by buying bottle water - which would be the sure and certain result of not using it. SteveBaker (talk) 14:03, 10 May 2009 (UTC)

Steve, I can do the standardized test for drinking water turbidity. Can you collect some water on a day when it is very turbid and send it to me? It would be interesting to see if it exceeds what you water utility is reporting (or it exceeds the established limits for these things. ike9898 (talk) 15:58, 12 May 2009 (UTC)

My Miele vacuum cleaner shows when its dirt bag is full by a mechanical indicator of the air pressure difference across the bag. I suspect the charcoal water filter is monitored the same way, by the water pressure difference. If so, a red light could change to green when the tap is turned partly off. Cuddlyable3 (talk) 00:13, 10 May 2009 (UTC)

Oohh, Miele keeps my clothes clean and my whites white! Also the best vacuum cleaner in the world. I'm always suspicious of automatic indicators for replaceable items though - they can be equally easy to calibrate to meet corporate revenue targets. Steve, I'd suggest a call to the customer support desk (and persistence) or appropriate use of a screwdriver. I'd agree though with the comments above that it uses pressure differential, which really has nothing to do with whether or not the charcoal is active anymore - it just tells you when the filter is clogged. If they work on the water mains and send down a blob of black goo, you might still be getting good water, just not very much of it Franamax (talk) 00:56, 10 May 2009 (UTC)

Since they sell a similar gizmo with an all mechanical indicator I'd be rather surprised if they didn't just use the battery operated LED for the indicator. The rest of the setup is probably the same as for the mechanical thing. I don't want to take ours apart. So, I can't tell you what exactly the mechanism is. My guess would be nothing more fancy than an impeller like thing moved by water flowing through the unit. (No pressure increase like in our article.) It likely turns some wheel and lever setup till it hits the battery contacts. Here's an experiment you can do when your filter is full. Just remove it and put it back in. I guess your LED will return to green. (Our mechanical indicator gets reset when I do this.) 71.236.24.129 (talk) 04:23, 10 May 2009 (UTC)

This patent describes a monitoring system for such a filter (not necessarily your filter though). It consists of a switch activated by the flow of water and a timing circuit that counts how long water has been flowing across the total lifetime of the filter. Replacement is indicated when the timer reaches a preset value. The switch is a simple on-off, and so doesn't measure actual flow but rather assumes some typical flow rate. Dragons flight (talk) 04:25, 10 May 2009 (UTC)

Possibly also of interest this patent describes a system for measuring actual flow rate based on pressure difference (and not just on-off). Dragons flight (talk) 04:36, 10 May 2009 (UTC)

Light bulb

What does it mean to have a 60W, 120V light bulb? I would assume it means that the light bulb transfers energy at a rate of 60W and that the potential lost across the light bulb is 120V. The reason I ask this is because I had the following homework question: Two 60W, 120V light bulbs are connected in series with each other and are in parallel with a 1600W, 120V air conditioning unit. The battery has an emf of 120V. What's the current intensity?

To me, there seems to be a contradiction. Both light bulbs have to lose 120V in total, so they can't be both 120V light bulbs, right? Normally I would have said that the total power of the system = 1720W, which equals E*I. Solving for I, we get 14.3A. The correct solution, however, is 13.59A, which I was able to get by the applying the formula P=V^2/R to each light bulb and the AC unit. But working backwards from this, we get that the potential across the light bulbs is 60V (to be expected, but it's still not 120V) and the power is 15W. Can someone help? Thanks. —Preceding unsigned comment added by 65.92.6.148 (talk) 18:43, 9 May 2009 (UTC)

You're on the right track and see an interesting problem! The voltage-rating of the bulb is just a maximum allowable, not the definite drop. A passive component such as a resistor would have a voltage-drop of "whatever potential is applied across it" by definition:) But if you put "too much" potential across it, you'll cause more current to flow than it can handle or more power dissipation (i.e., heat) to it can withstand or something like that. DMacks (talk) 18:48, 9 May 2009 (UTC)

So if I understand you correctly, 60W is also the maximum allowable power, right? But since neither of those (voltage or power) describe this situation, how can we use them to answer the problem? —Preceding unsigned comment added by 65.92.6.148 (talk) 19:15, 9 May 2009 (UTC)

For your problem think of a bulb (and even the AC!) as just a resistor, and the watt and voltage ratings as an indirect way of defining its resistance. Does that help ? Abecedare (talk) 19:27, 9 May 2009 (UTC)

(ec)Because knowing the maximum power and voltage allows you to calculate the resistance. Once you have the resistance you can calculate the voltage and power in a new situation. But beware, this only works in homework questions, in real life the resistance changes quite dramatically at different voltages because the temperature of the filament is different. SpinningSpark 19:30, 9 May 2009 (UTC)

The OP sees a contradiction where there is only this error: A light bulb that is called a "120V light bulb" never has to lose 120V. It just "needs" that voltage to shine at its full strength which is 60W. When two such bulbs are connected in series the applied 120V is split between them so each bulb gets 60V. They shine dimly but are still "120V light bulbs", that's just their maximum rating.

Clearly the homework question was designed to be a simplistic exercise using Ohm's Law for DC circuits and ignores some practical realities Cuddlyable3 (talk) 00:01, 10 May 2009 (UTC)

An air conditioning unit draws a lot of power. I never heard of one that works from a battery. Even the cooling unit on a refrigerated truck works from a small gasoline en;gine. For a homework question maybe we can accept the imaginary situation, But I think it would be better to stay with real-life situations unless there is a definite reason for going into the reaalm of imagination. Otherwise, the imaginary aspect may confuse the student. —Preceding unsigned comment added by 174.130.249.69 (talk) 21:34, 9 May 2009 (UTC)

Real life problems require many years of experience and training to solve and developing the solution can take months or even years. Homework on the other hand needs to be solved in 30 minutes max with no more than a calculator. Of necessity the questions are artificially contrived and it is pointless and unnecessary to complain about it. SpinningSpark 00:00, 10 May 2009 (UTC)

Indeed. Any air-conditioner I know of uses an AC motor, so I'd rather expect the current intensity to be the maximum discharge rate of the battery, and if the battery had 120V output, my answer would be "fire extinguisher". However, that's not the spirit of the question and it doesn't even ask whether the ACU is turned on. :) Franamax (talk) 01:07, 10 May 2009 (UTC)

The question shows (apparently) that the teacher is unaware that light bulbs are far from Ohmic resistances. Their resistance varies dramatically with the voltage across them. The light bulb filament's resistance increases dramatically as voltage increases. A hot tungsten filament has perhaps 13 times the resistance of a room temperature tungsten filament, with more recent estimates being 15 times the resistance. Answer the question in a simple minded way and you will likely get the teacher's answer. Or use the curve from the second ref and get an accurate answer. There is absolutely no reason that light bulbs and air conditioner motors could not be operated from a 120 volt battery. I have worked with 120 volt batteries and with light bulbs operated from them. I have worked with powerful motors operated from 120 volt DC batteries. Edison (talk) 01:29, 10 May 2009 (UTC)

The small-text indent was supposed to show a light-hearted comment, sorry if that didn't come across. The teacher is obviously making an assumption about steady state operation and expecting straight series-parallel math. Do you have a source for how a lamp filament operating at 60V has appreciably less temperature (and thus lowher resistance) than one at 120V? Not that it matters to the OP, but since you brought it up, I'd be happy to learn. Oh yeah, what was the exact model of air conitioner you ran off a 120V DC battery? Franamax (talk) 02:13, 10 May 2009 (UTC)

Note the portions of my answer in blue. If you click on them, they magically take you to books. As for DC air conditioners, ones operating on 24 volts DC are readily available at up to 3000 watts input. No reason the motors could not have been wound for 120 volts DC input. (I would exercise caution in connecting 5 of them in series across 120 volts DC (locked rotor and all that)). Certainly an invertor or motor-generator could be used to run an AC air conditioner from 120 volts DC, which to the battery would look like a 120 volt DC air conditioner. Edison (talk) 02:57, 10 May 2009 (UTC)

How appropriate, to have Edison explaining about light bulbs! Looie496 (talk) 03:15, 10 May 2009 (UTC)

Balanced symbol equation

A chemistry question, above, mentioned in passing a balanced symbol equation, also calling it a symbol balanced equation. What would one of those be? Even google thinks we might not know. Anyone fancy putting an article together? --Tagishsimon (talk) 19:24, 9 May 2009 (UTC)

The "reversible/equilibrium reaction" symbol is the ⇌ character (or the wikipedia template {{eqm}}). DMacks (talk) 19:35, 9 May 2009 (UTC)

I'm not sure that we need a new article. Assuming that "balanced symbol equation" simply means a chemical equation that is balanced, then the concept is already covered by Chemical_equation#Balancing_chemical_equations. --Heron (talk) 20:37, 9 May 2009 (UTC)

Reversible reaction uses "<math> \rightleftharpoons </math>" to get $\rightleftharpoons$. hydnjo (talk) 21:15, 9 May 2009 (UTC)

Unfortunately, unicode &#8652; doesn't have an HTML entity name that I can find. DMacks (talk) 21:39, 9 May 2009 (UTC)

And yet another image. hydnjo (talk) 22:25, 9 May 2009 (UTC)