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could an exothermic reaction underwater create, as a byproduct, ice: however, doing so by decreasing the density of the affected area greatly.

I am thinking of a cubic centimeter of water that we mark so we can trace it. Then an explosion of that cubic centimeter into a cubic meter. The greatly reduced density could cause it to freeze, could it not, yet the explosion from cubic centimeter from cubic meter is, well, explosive. Which leads to the question posed above. Please let me know if I seem to indicate that I am laboring under false assumptions of either physics, chemistry, or thermodynamics. Thanks. (Note that I don't make any assumptions about the actual process involved! It could be a special chemical, a physical process, whatever...just that it's exothermic. I also don't care if, as a result, the water outside our new 'cubic meter' gets hotter/undergoes an increase in entropy: indeed I expect it to.). --80.99.254.208 (talk) 06:54, 28 February 2012 (UTC)[reply]

This is quite possible if pressure is sufficient. All you need is for the heat removed by the expansion to be greater than the heat needed to be removed as heat of sublimation. If the water is super critical, ie the pressure is above the critical pressure 22.064 MPa for water, the heat of sublimation is essentially mimimum (about 6 MJ/kmol) when the triple point temperature is reached. However, if the pressure is well above critical, the specific heat of the super critical phase is mimimised, making it harder to make ice. This is quite easy to see if you look at a plot of temperature versus internal energy. Ratbone124.178.178.1 (talk) 07:08, 28 February 2012 (UTC)[reply]

I'm confused, 1 milliliter into 1000 liters, wouldn't that just vaporize the water not freeze it? Maybe I'm not understanding what you mean by "explosion". Vespine (talk) 21:52, 28 February 2012 (UTC)[reply]

Vespine, obtain a graph of temperature versus internal energy, and compare it with the phase diagram for water, and it will become clear. At standard pressure (~1 atmosphere), water transistions from ice/solid, to water/liquid, thru to gas/steam. But at sufficiently high pressures the liquid will be a supercritcial liquid, which is a liquid that is someting like a gas (you can expand it and reduce its temperature) but does not comply with the ideal gas laws - that it the normal straitforward relationship between volume and temperature doesnot hold. So, under high pressure, the liquid H2O becomes supercritical - and temperature drops without vaporisation. Ratbone120.145.22.14 (talk) 02:00, 29 February 2012 (UTC)[reply]

"The greatly reduced density could cause it to freeze, could it not(?)". No, it could not. I don't know why you think it would. The only thing that comes to mind is the negative thermal expansion of water, but you have to keep in mind that this only occurs between 0°C and 4°C and is not a reduction in density in the order of magnitude that your talking about. I agree with Vespine that increasing the volume will just vaporize the water. To increase the volume (explode) the 1mL of water you can either heat it (which in and of itself is by definition an endothermic process) or decrease the pressure, in which case it will vaporize and expand to fill the void (also endothermic). 203.27.72.5 (talk) 03:45, 29 February 2012 (UTC)[reply]

Vespine, did you look at the phase and internal energy diagrams as I suggested? If the water is above critical pressure, it can be expanded without vaporisation, by definition. Liquids are only "incompressible" if outside the critcal region. Ratbone124.182.60.34 (talk) 04:49, 29 February 2012 (UTC)[reply]

That IP is not me.. but I still don't get it..I found a phase diagram for water, I can't find "temperature versus internal energy" for water. I'm not sure what it's going to show.. What I don't understand is how you are keeping the water "above critical pressure" while expanding its volume by six orders of magnitude? Vespine (talk) 06:10, 29 February 2012 (UTC)[reply]

It would be difficult indeed to keep it above ~22MPa while expanding 6 orders of magnitude. But the 6 orders of magnitude came from the OP, not me. I assumed that wasn't the core of what he was asking - he was just clarifying his idea. For water, at triple point temperature, you only need to extract 6MJ/kmol to freeze it. So if its only just above that temperature, you only have to expand it enough to remove 6MJ/kmol. That won't require 6 orders of magnitude expansion, it won't need even 2:1 expansion. Now, if I knew how to insert a picture, I would give you a temperature - internal energy diagram - but I don't. If someone explains how its done, I will upload it. Ratbone124.182.159.25 (talk) 07:57, 29 February 2012 (UTC)[reply]

Substances whose conductivity lies between that of a conductor and an insulator is called semi-conductor.But why isn't it called semi-insulator. — Preceding unsigned comment added by Aditi keerti (talk • contribs) 08:29, 28 February 2012 (UTC)[reply]

It could have been. Go find the person who chose that convention and shoot him with a semi-manual gun. StuRat (talk) 09:26, 28 February 2012 (UTC) [reply]

The definition of a conductor is that which has charge carriers (typically valence electrons fully mobile within crystals), can be ions in liquids), thereby permitting the carrying of current. You can thus have a material with not very freely available carriers (typically mid-valence electrons & missing electrons partially/semi mobile with crystals), so "semi-conductor" makes some sense in this case.

The definition of an insulator is that which has NO charge carriers, so cannot carry ANY current (in practice, with the right instruments, very minute currents may be detected, due to impurities, leakage thru contaminants on the surface, etc). It's just semantics, I guess, but semi-nothing doesn't make as much sense in English. Semi-something is still something; semi-nothing is an oxymoron at best. Ratbone124.182.55.62 (talk) 09:58, 28 February 2012 (UTC)[reply]

At sufficiently high electric fields, even the best insulators will start conducting (Breakdown_voltage) - a perfect conductor exists in reality (superconductor) but a perfect insulator doesn't. 83.134.160.77 (talk) 06:33, 29 February 2012 (UTC)[reply]

But breakdown hardly affects the explanations given, as the explanations are based more on the quirks of the english language than the properties of materials. Having said that, with the exception of breakdown within a sealed container, breakdown means permanent chemical damage to the insulator - it is no longer an insulator because at least part of it has been changed to a different chemical form. While gaseous breakdown, breakdown is due to a change to plasma; as soon as you take the voltage away, it changes back to the original insulating gas. You can push current thru semiconductors and conductors as many times as you like without the slightest chemical change, subject to not overdoing it and overheating. And you never see in insulators current varying in proportion to voltage, but you always do in conductors, and you also see it approximately so in semiconductors. Ratbone138.217.245.71 (talk) 07:37, 29 February 2012 (UTC)[reply]

When a ball rolls down a incline, how do we know how much of the potential energy gets converted into kinetic energy and how much gets converted into rotational energy? Widener (talk) 11:52, 28 February 2012 (UTC)[reply]

See Moment of inertia and List of moments of inertia. Dolphin (t) 12:08, 28 February 2012 (UTC)[reply]

If you know the radius of the ball, it is easy to calculate (if it isn't gliding, but rolling) what the relation is between the rotation- and movement velocities. Using the moment of inertia calculated for the ball, you can derive the ratio between rotational and kinetic energy. -- Lindert (talk) 12:37, 28 February 2012 (UTC)[reply]

How does one use the moment of inertia to calculate the ratio between kinetic and rotational energy? Widener (talk) 20:39, 28 February 2012 (UTC)[reply]

The rotational energy is (1/2)Iω², where I is the moment of inertia and ω is the angular velocity. The translational kinetic energy is (1/2)Mv². If the object is rolling without sliding then v = rω where r is the radius, so the ratio of the translational kinetic energy to rotational kinetic energy is Mr²/I. Rckrone (talk) 00:42, 29 February 2012 (UTC)[reply]

Oh, I see; it makes sense that v = rω Widener (talk) 06:24, 29 February 2012 (UTC)[reply]

(edit conflict) You also need take friction into account. Is there sufficient friction that the ball is purely rolling or will it also slide a bit? If there is no sliding, then you can easily relate the linear velocity to the rotational velocity (distance travelled in one rotation equals the circumference of the ball). Once you've got the velocities, you can work out the kinetic energy using E=1/2mv² and the rotational energy using the moments of inertia Dolphin links to. If there is sliding, then it gets rather more complicated (friction, generally, is complicated - there are massive simplications usually taught in schools that might get you somewhere, but to get it accurate is really hard). --Tango (talk) 12:40, 28 February 2012 (UTC)[reply]

Other names for the anesthetic lidocaine are "xylocaine" and "lignocaine". The prefixes xylo- and ligno suggest this has something to do with wood, but the Wikipedia article doesn't say anything about it being made from, or being chemically similar to, some substance extracted from wood. What, if anything, does lidocaine have to do with wood? -- Finlay McWalterჷTalk 13:16, 28 February 2012 (UTC)[reply]

According to the Oxford English Dictionary, 'xylocaine' got it's name because of the chemical relationship with xylene (and cocaine), which in turn is obtained from wood-spirit ('Crude methyl alcohol obtained from wood by destructive distillation'). It is also called 'lignocaine', because ligno- is the Latin equivalent of 'xylo'. -- Lindert (talk) 13:35, 28 February 2012 (UTC)[reply]

If the Earth were a perfect sphere and had no atmosphere, then a pod could accelerate fast enough to reach Orbit right at sea level, couldn't it? (well, a few inches/feet above sea level, let's say). So, at g's that are acceptable to tourists/passengers, how long would this acceleration take, both in terms of time taken and distance travelled in terms of our physical geography/Earth miles or kilometers? What if we increase the acceptable G's from tourist/passenger to whatever is the human maximum, however uncomfortable? Then how long would it take and what distance would be covered? Thanks. 188.6.78.231 (talk) 14:19, 28 February 2012 (UTC)[reply]

also could some confirm my premise itself, that at the right speed, orbit a few cm/feet above sea level in a perfect-sphere (uniform) earth without an atmosphere scenario is possible? 188.6.78.231 (talk) 14:27, 28 February 2012 (UTC)[reply]

Orbital speed for a circular orbit at zero altitude is about 7.9 km/s. If we assume a lateral acceleration of 5 m/s² (half a g, equivalent to the acceleration in a fast car) then it would take about 1600 seconds or about 27 minutes to reach this speed, in which time you would have travelled around 6,400 km, which is a little more than the distance from London to Chicago. Gandalf61 (talk) 15:00, 28 February 2012 (UTC)[reply]

Thank you! Do you think this is reasonable for a tourist/passenger to undergo for 27 minutes? (The acceleration of a fast car?) Would it be possible I don't know to have the pod turn toward the direction of travel, so everyone just feels a bit heavier instead of being pushed BACK (and not down) into their seats? Do you think this is a reasonable tourist experience? Thanks again for the calculations. 188.6.78.231 (talk) 15:17, 28 February 2012 (UTC)[reply]

On the other hand, at a crushing 10g, it would take 80 seconds and 3000 km. SpinningSpark 15:14, 28 February 2012 (UTC)[reply]

wow. Okay, here's one. How long would it take at a g so low that you can't notice that you're moving? Would you already go around the earth / several times before you reached orbital velocity at such a speed? 188.6.78.231 (talk) 15:17, 28 February 2012 (UTC)[reply]

Just as an aside: the slower you accelerate, the more energy you waste to gravity burn. Using the "high school algebra" formulation, as discussed in this excellent NASA page from Glenn Research Center, you need to modify your ideal rocket equation, yielding a loss term proportional to the time of the burn, "-g0*tb" - energy (or, fuel, or, dollars), that is purely wasted and gains absolutely nothing for the final orbit height or velocity. Nimur (talk) 18:00, 28 February 2012 (UTC)[reply]

To reach orbital velocity at sea level you'd likely use something like a maglev train in a vacuum tube, which should deal with gravity more efficiently than a rocket. OP should make plots of time vs horizontal g-forces (sqrt(radius of earth * gravitational acceleration) / time) vs distance-to-orbital-velocity (.5 * sqrt(radius of earth * gravitational acceleration) * time^2) to see what's plausible. --81.175.230.91 (talk) 18:44, 28 February 2012 (UTC)[reply]

How should a maglev train "deal with gravity more efficiently than a rocket"? Do you have a reference for this statement, or did you simply decide that you could ignore the principle of conservation of energy? Perhaps you are confusing non-conservative work against gravity with structural load bearing by a hypothetical "space elevator." Nimur (talk) 19:42, 28 February 2012 (UTC)[reply]

How much energy do ordinary trains spend not falling down? --81.175.230.91 (talk) 22:00, 28 February 2012 (UTC)[reply]

Indeed. Maintaining a constant altitude doesn't use energy. The whole reason gravity drag is a bad thing is because the energy is wasted - if it were a conservation of energy thing then it wouldn't matter how long it took you to get into orbit, the energy requirements would be the same. --Tango (talk) 00:19, 29 February 2012 (UTC)[reply]

This debate confuses me. "Gravity burn" isn't defined at that web page, but I assume it simply refers to the fuel burned by a rocket to provide acceleration equal to that of gravity pulling it down, as it goes up. Clearly any solution that offers a solid support with no fuel burned, e.g. maglev or space elevator, should avoid that cost. Wnt (talk) 18:09, 29 February 2012 (UTC)[reply]

If you travel around the earth's axis of rotation (in the same direction as the earth's rotation) at the earth's widest point east along the equator, then you are already travelling at about 0.5km/s. That means you'd only need to accelerate to ~7.4km/s which takes only 1480 seconds (24 minutes 40 seconds) and during that time you'd only travel about 5500kms which would be from Dakar in Senegal to Manaus in Brazil. 203.27.72.5 (talk) 04:22, 29 February 2012 (UTC)[reply]

So I was cooking some ground beef the other day, and I drained the fat, and I looked at the slimy mess and I thought to myself, "Is it possible to fuel a car with this stuff?". So is it possible? ScienceApe (talk) 15:34, 28 February 2012 (UTC)[reply]

Animal fats can be turned into fatty acid esters (fatty acid methyl esters, in particular) that make suitable biodiesel. Using the fat directly would present practical difficulties. 148.177.1.210 (talk) 15:43, 28 February 2012 (UTC)[reply]

Animal fats used to be used for all sorts of things, including as a cooking and lighting fuel. See Tallow, Lard, Schmaltz, Suet, for some animals fats which have both culinary and non-culinary uses. Also see Soapmaking#Soap_making_processes. Making soap from animal fat is an ancient, and relatively simple, process. Soak woodashes in a bunch of water. Filter off the ashes; the water now contains a bunch of alkali. Put a bunch of the alkali water together with a bunch of tallow. Heat for a few hours, seperate the glycerin layer; and let it cool. What you're left with is soap. --Jayron32 21:11, 28 February 2012 (UTC)[reply]

What speaks against slaughtering animals with a guillotine? It seems more reliable, faster and less painful than stunning and cutting the throat. XPPaul (talk) 18:34, 28 February 2012 (UTC)[reply]

Probably not faster because it takes time to position the head correctly in the guillotine, adding to the distress the animal feels and increasing the risk of injury by slaughterhouse personnel. Also, the guillotine would have to be checked, cleaned and reset after each operation. Dominus Vobisdu (talk) 18:40, 28 February 2012 (UTC)[reply]

That don't make sense. Cattle in slaughter houses for years have been lead in to a squeeze to have their throats cut. A guillotine at this stage could likewise be automated. The problem with a guillotine would be is now one has a carcass in two parts. This requires an extra operations to recover the head and that complicates the following processes of carcass inspection, because the are two bits which need to be kept together for veterinary inspection (check for diseases extra, which is legal requirement in many countries). So, this imposes extra costs and chances of getting parts mixed up. It is reckoned that cattle loose consciousness before that realise what has happened with the current process. This requires the knives to be kept sharp, just like a guillotine blade. If guillotines were better they would already be in use. Gosh, all this talk of food is making me feel hungry again and I'm supposed to be on a strict diet. --Aspro (talk) 23:54, 28 February 2012 (UTC)[reply]

They'd almost certainly be conscious for far longer after being guillotined than they are after being stunned. Dmcq (talk) 00:29, 29 February 2012 (UTC)[reply]

Decapitation would require slicing through the spinal cord or even brain stem which may release disease causing prions that are known to exist in bovine and other ruminants. 203.27.72.5 (talk) 04:27, 29 February 2012 (UTC)[reply]

So does a captive bolt ploughing through the brain. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 19:36, 29 February 2012 (UTC)[reply]

From the article you just linked to; "(The non-penetrating stunner) has undergone a resurgence of popularity due to concerns about mad cow disease." 203.27.72.5 (talk) 23:22, 29 February 2012 (UTC)[reply]

Better still, train the slaughter-men to stare at cattle in the same way the U.S. Army stare at goats. Sorry – couldn't resist. --Aspro (talk) 19:09, 1 March 2012 (UTC)[reply]

I was reading about the Google X Prize team entry Odyssey Moon that submitted its entry on December 6, 2007 to send a probe to the moon. The Wikipedia article has nothing as to the current status or if the team has actually built anything so far (note, the picture in the Wikipedia article is not their thing), and as far as I am able to tell, their official website doesn't plainly say how much they have actually done either (or maybe they have, I'm not fluent in marketerese). What have these people actually made as of the asking of this question? Have any of the people in that fundraising engineering operation released any estimate of a possible launch date in any way, shape, or form? 20.137.18.53 (talk) 19:06, 28 February 2012 (UTC)[reply]

Apparently nothing concrete. Their latest press release says that they teamed up with Paragon to establish a "Lunar Oasis", or greenhouse on the moon. Sounds impressive until you see it: [[1]]. Lots of marketing pie-in-the-sky talk, but still in the fundraising stage. Dominus Vobisdu (talk) 19:25, 28 February 2012 (UTC)[reply]

http://en.wikipedia.org/wiki/Time_dilation In the two mirror experiment when the light returns to the bottom mirror it would take some time for this event to propagate back to the observer. Is time dilation a consequence of this delay? Please clarify — Preceding unsigned comment added by 199.89.103.11 (talk) 19:29, 28 February 2012 (UTC)[reply]

The relationship you're calling "signal lag" is expressly and rigorously defined as the retarded time, which is a crucial intermediate concept that you can use while deriving the lorentz transform from classical electromagnetic wave theory. Time dilation is a consequence of the fact that we observe Maxwell's equations to be valid in all reference frames, including those frames that are moving relative to our own frame. As a result, the relationship between the change in an electromagnetic field with respect to both space and with respect to time follow the same law in both reference frames. (In fact, this applies in all reference frames). To satisfy this constraint, we must define an adjusted time, whose value depends on the relative motion between reference frames. If you work the math, you find that this adjusted time is defined by the lorentz transform. Nimur (talk) 19:48, 28 February 2012 (UTC)[reply]

I guess what I'm asking is whether in time dilation experiment the time is measured using retarded time. If not then what would be the consequences if retarded time is used? 199.89.103.11 (talk) 20:04, 28 February 2012 (UTC)[reply]

In which time dilation experiment? For example, the NIST Physics lab for time and frequency measurement has a great webpage explaining practical details of Two Way Time Transfer. Indeed, in actual experiment, most of the time, the effects of special relativity are dwarfed by practical engineering considerations. In this 2007 paper published in the journal Nature, Test of relativistic time dilation with fast optical atomic clocks at different velocities, time is measured using ion-counters and a frequency-domain technique (which is really the state of the art). So, "no, retarded time is not used." In as few words as possible, accurate modern clocks use differential frequency measurement to experimentally validate and constrain relativistic time dilation. Nimur (talk) 20:30, 28 February 2012 (UTC)[reply]

In the moving mirrors thought experiment: http://en.wikipedia.org/wiki/Time_dilation does the observer compute the speed of light using retarded time? If not what what be the outcome if he measured it using retarded time? 199.89.103.11 (talk) 20:44, 28 February 2012 (UTC)[reply]

Signal lag doesn't enter into the definition of time dilation. The time in time dilation is the coordinate time, which is defined by means of a bunch of comoving synchronized clocks. There's a clock near every event of interest, and the coordinate time of an event is the time reading that you see on the nearby clock at the same time that you see the event. This removes any dependence on how far you are from the event, how fast you're moving, or anything else about you. Only the event and the nearby clock factor into the measurement.

Time dilation is the fact that if you have three clocks moving like this:

       A  B  C
       |  | /    ^
       |  |/     |time
       |  *      |
       | /|      |  
       |/ |       -------->
       *  |       position
      /|  |
     / |  |

where A and B are Einstein-synchronized, then the difference between the readings on C at the two crossing points (the asterisks) is smaller than the reading on B at the second crossing point minus the reading on A at the first crossing point.

This is geometrically the same thing as the fact that if you have two vertical lines and a diagonal line, the distance between the crossing points along the diagonal line is larger than the difference between the Y coordinates of the two crossing points. In fact the ratio of the two values is ${\displaystyle {\sqrt {1+(run/rise)^{2}}}}$, while the relativistic time dilation factor is ${\displaystyle {\sqrt {1-(v/c)^{2}}}}$, and the similarity of those formulas is not a coincidence.

Of course, the measurements made in actual experiments depend on the experiment. In the Hafele-Keating experiment, the clocks A and B are the same clock and C crosses that clock's worldline twice. This probably should not technically count as "time dilation". It's more like a version of the twin effect/paradox (which is geometrically the same as the fact that a straight line is the shortest distance between two points) -- BenRG (talk) 22:49, 28 February 2012 (UTC)[reply]

In the moving mirror thought-experiment there are Einstein synchronized stationary clocks at A and C to establish the coordinate time of the stationary frame. This is practically the same setup as the example I gave above. You could eliminate the mirror at B and move C to its mirror-image position in the diagram (above and to the right of B) or you could eliminate C and put the clocks at A and B, which amounts to the same thing. The advantage of the mirror is that you only need one clock (moving with the lower mirror) to measure the round-trip travel time in the moving frame. Otherwise you would need a moving clock at B also, and once you have two clocks you need to synchronize them, which would make this thought-experiment less convincing since the usual synchronization process uses beams of light. -- BenRG (talk) 02:10, 29 February 2012 (UTC)[reply]

Was Fukushima or Chernobyl worse? --108.227.26.244 (talk) 22:00, 28 February 2012 (UTC)[reply]

Define "worse". --Jayron32 22:08, 28 February 2012 (UTC)[reply]

See Comparison of Fukushima and Chernobyl nuclear accidents. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 22:11, 28 February 2012 (UTC)[reply]

Independent of definition: Chernobyl was worse, with more direct and indirect deaths, and more radiation released. XPPaul (talk) 22:27, 28 February 2012 (UTC)[reply]

Chernobyl also has a larger affected area, but more of Fukushima's reactors were involved. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 23:35, 28 February 2012 (UTC)[reply]

I can't think of a reasonable definition of worse where the number of affected reactors is significant in itself... --Tango (talk) 00:22, 29 February 2012 (UTC)[reply]

There were more alarms up on the control room operator's DCS screen at Fukushima. 203.27.72.5 (talk) 04:36, 29 February 2012 (UTC)[reply]

I would argue that Fukushima was worse in at least one measure, namely the sigma level, i.e. the "unlikelyhood" of the event (and thus the induced shock in the general public). A major disaster in the Soviet Union was more or less expected, at least it did not come as a too big surprise. Fukushima on the other hand was pretty much a perfect storm. This does not only refer to the sequence of events leading to the scenario, but also that it should happen to such a high-tech nation as Japan. 109.149.46.78 (talk) 01:28, 29 February 2012 (UTC)[reply]

Surprising to whom? That's always the question. I think it was surprising to most people that Soviet engineers would almost purposefully blow up their own reactor. I think Fukushima was not as surprising as it ought to have been to the people who had already run flood simulations on the reactor setup and found that it was severely lacking in adequate protections. Fukushima wasn't a "perfect storm" — it was a reactor that was known to be in a seismically active zone, yet was not equipped to deal with heavy flooding after a tsunami, and the engineers were inadequately trained for emergency situations. --Mr.98 (talk) 03:34, 29 February 2012 (UTC)[reply]

http://www.mylovetechnology.com/others/airpod-car-runs-on-compressed-air/

A car that can go long ranges on compressed air - something Big Oil would hate.

Why isn't that car happening in mass-numbers already? Perhaps it's because Big Oil Lobbyists went to congress to prohibit these cars on a pretext of something inconsequential and different. (If you try to argue to the contrary, are you working for Big Oil?)

And lobbyists could also gag the media from saying a thing about it, now could they?

There is a lot of explaining to do here, you all. Thanks. --129.130.208.81 (talk) 22:50, 28 February 2012 (UTC)[reply]

I see no evidence of a conspiracy - just a lack of evidence for practicality. Can you start by explaining why a system that "uses a small motor to compress outside air to keep the tank full" should necessarily be any more efficient than one that used the 'small motor' in a more conventional way - to turn the wheels? AndyTheGrump (talk) 22:57, 28 February 2012 (UTC)[reply]

This IEEE Spectrum article written 3 years ago seems like a much more realistic evaluation of the AirPod. Vespine (talk) 23:08, 28 February 2012 (UTC)[reply]

I doubt that Big Oil can indeed prohibit, or lobby for the prohibition of, a harmless invention. No matter if it's the air engine (which is doable, but seems impractical) or if it's the water powered motor (which is the invention of some crackpot). XPPaul (talk) 23:22, 28 February 2012 (UTC)[reply]

And America doesn't dominate car production like it once did. The conspiracy would need to reach deep into every other car company and nation where cars are produced. StuRat (talk) 02:54, 29 February 2012 (UTC)[reply]

Actually, Compressed-air vehicles seem to pre-date 'big oil'-powered ones: "In 1903, the Liquid Air Company located in London England manufactured a number of compressed-air and liquified-air cars. The major problem with these cars and all compressed-air cars is the lack of torque produced by the 'engines' and the cost of compressing the air". AndyTheGrump (talk) 04:25, 29 February 2012 (UTC)[reply]

Tata Motors is supposedly releasing a compressed air car soon. We'll see how it works. There are extreme tradeoffs in terms of speed, distance, and safety (weight), to say nothing about the infrastructure you need to be able to refill one of these things. I could see them for city use, though, as something similar to a Smart car. I wouldn't take an air car on the freeway (at least not the freeways where I live!). --Mr.98 (talk) 13:16, 29 February 2012 (UTC)[reply]

I don't know about you, but I would be alarmed at the amount of noise involved, both from the engine and from recharging, if the video I saw a few years back was any indication. Wnt (talk) 18:03, 29 February 2012 (UTC)[reply]

For recharging, an enclosure that absorbs sound would be easy enough. While running, a good tuned muffler and/or active noise reduction might work, although white noise may make both a challenge. StuRat (talk) 21:08, 29 February 2012 (UTC)[reply]

Easy enough, but have you ever confronted someone with a machine that makes noise? Unless it's a person exercising 'free speech' on a political topic, it's treated like a God-given right. Wnt (talk) 22:47, 29 February 2012 (UTC)[reply]

Hi all, what's the simplest formula for adding or subtracting 2 sine waves of the same frequency? I know that you can add or subtract two arbitrary sine waves and get a fairly complex formula like on this page (under Constructive and Destructive Interference), but if I understand correctly, two waves of the same frequency should result in a simple sine wave expressible in the basic form ${\displaystyle y(t)=A\cdot \sin(\omega t+\phi )}$. Is there a way to get that simple function for the resulting added (or subtracted) wave? Thanks! — Sam 66.31.201.89 (talk) 22:53, 28 February 2012 (UTC) (Yes, I know I first posted this in WP:RD/MA, but it's really a physics question, and that forum has only had one person post in over four hours and not even had a single actual answer all day...)[reply]

As the instructions at the top of the page say, it can take several days to get a complete answer. We're all volunteers and answer questions as and when it is convenient for us, so please be patient. That said, I don't understand your question - the first formula on the page you link to in the section you name is a formula for sine waves of the same frequency... --Tango (talk) 00:26, 29 February 2012 (UTC)[reply]

This is very easy to do if you understand vectors. But if not its still easy. If each sine wave is specified by an amplitude and a phase angle, convert each to a vector, then add the two vectors, then if required convert back to amplitude and pahse representation. For any sinewave of y volts at angle a, the vector is R = y sin(a) and I = y cos (a). You then add the two R's and I's. The ammplitude of the combined wave is the square root of [(R1 + R2)^2 + (I1 + I2)^2]. Keit121.221.233.252 (talk) 00:55, 29 February 2012 (UTC)[reply]

If we ingest milk, or some other food which contains tryptophan, will this tryptophan reach our brains and have an effect?XPPaul (talk) 23:12, 28 February 2012 (UTC)[reply]

Why do you think warm milk is one of the things recommended for trouble falling asleep? Or why the most popular day for taking a nap in US is the fourth Thursday in November? (Turkey is rich in tryptophan). The injested tryptophan ends up being converted into the neurotransmitter serotonin, which has a calming effect on the mind. Dominus Vobisdu (talk) 23:20, 28 February 2012 (UTC)[reply]

I was asking because it could be a common misconception. Being recommended doesn't mean a thing. We get every kind of crazy recommendations. XPPaul (talk) 23:27, 28 February 2012 (UTC)[reply]

Read the serotonin article for details. There is a section of the effects of food content. Dominus Vobisdu (talk) 23:33, 28 February 2012 (UTC)[reply]

It looks like you are wrong. — Preceding unsigned comment added by XPPaul (talk • contribs)

(edit conflict) Tryptophan is an amino acid and quite commonly present in many proteins, including meat and dairy proteins. It is an Essential amino acid, which means that a) you need it to live and b) your body cannot synthesize it; which means that you must get it from your diet. Contrary to the urban legend that dietary tryptophan can make you "sleepy" or have other psychoactive effects, tryptophan's main use in your body is to basically do what all amino acids do; which is to act as primary building blocks for making proteins and enzymes. Now, it is true that tryptophan is a precursor to certain neurotransmitters like seratonin and melatonin, which are involved in maintaining circadian rhythms (i.e. day-night cycle). Here's the important bit: Without tryptophan, you probably couldn't make melatonin; however an excess of tryptophan does not directly lead to an excess of melatonin. Your body makes melatonin as needed, not merely because there is an excess of tryptophan around. There is always an excess of tryptophan around, and your body uses it for lots of things. Because tryptophan is converted to melatonin, and melatonin is associated with sleep cycles, people make the erroneous conclusion that dietary tryptophan leads to sleepiness or other effects on the brain. In reality, you either get enough tryptophan or you don't, and under normal circumstances, having more than the minimum requirements doesn't lead to any extra effects on the brain. Lots of this information, as well as the "dietary tryptophan causes drowsiness" myth, are covered in the tryptophan articles. In summation: yes, tryptophan is used in processes which go on in the brain, yes you need to get it from diet, but these "effects" are a part of the process known as "being alive" and there isn't anything "extra" that tryptophan does when you eat it. It has an important role, but eating lots of it doesn't make it do those jobs "more". There's also nothing special about tryptophan from milk. It's the same molecule regardless of where it comes from. --Jayron32 23:30, 28 February 2012 (UTC)[reply]

Thanks. XPPaul (talk) 23:50, 28 February 2012 (UTC)[reply]

The above answer is very good, but I already found this reference so I'll still post it: From Serotonin#Biosynthesis: Tryptophan and its metabolite 5-hydroxytryptophan (5-HTP), from which serotonin is synthesized, can and does cross the blood-brain barrier. Vespine (talk) 23:39, 28 February 2012 (UTC)[reply]

yes, but it hasn't any direct effect. So, no glass of warm milk --> getting sleepy association.— Preceding unsigned comment added by XPPaul (talk • contribs)

Yeah, the issue is that tryptophan itself doesn't have any direct effects on the brain; it needs to be converted into something else, and the rate of conversion into those other things isn't directly affected by the concentration of tryptophan. It is to a point, in the sense that too little tryptophan will hinder the conversion into melatonin and seratonin; however having excess tryptophan, above what is being demanded at any given time, does not directly lead to drowsiness or other effects. --Jayron32 23:58, 28 February 2012 (UTC)[reply]

There appear to be hints of use for tryptophan in PMID 20184811, PMID 19925721, etc.; a recent review PMID 19715722 speaks of sedative effects. It should be clear that most often tryptophan is not making up for a deficiency, nor curing a disease; rather, it is a component of food which helps induce sleepiness in people who normally eat and tend to be sleepier afterward. Note that some vagueness in the mechanism is inevitable - it is a component of thousands of proteins and a precursor to many important small molecules used in the nervous system, and there's no particular reason why any of them should be completely unaffected by changes in the level. Wnt (talk) 17:58, 29 February 2012 (UTC)[reply]