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September 30

Photosynthetic Animal

Hey. I'm trying to remember a certain animal that I think I recall seeing on a nature documentary, it was like, some kind of newt or something. But its most remarkable feature was that it shared a symbiotic relationship with algae that lived just underneath its skin, the algae would absorb the sunlight and effectively create food for the newt to live off of. Does anyone have an idea of what animal I'm thinking of? 62.49.131.66 (talk) 00:56, 30 September 2008 (UTC)[reply]

Elysia viridis ? It's a sea slug and I don't think it assimilates the intact algae, only chloroplasts. I've never heard of a "solar-powered newt" so to speak, but I'd sure love to know if there exists one :) --Dr Dima (talk) 01:48, 30 September 2008 (UTC)[reply]

Or possibly the acoelan flatworm Convoluta roscoffensis. Our article doesn't explain it very well, but one of those other sites mentions symbiotic algae. 'Pedians to the barricades - we need references dammit! Franamax (talk) 06:31, 30 September 2008 (UTC)[reply]

Coral, specifically tropical hard corals used to make tropical reefs. I don't know of other animals, but I guess there could be. Sabine's Sunbird talk 19:03, 30 September 2008 (UTC)[reply]

Power consumption vs. hour of day

I want to know, in a typical home, what hours of the day tend to show peak electrical power usage, what hours of the day show minimal electrical power usage, and the relative differences throughout the day. Basically, I'd like to see a graph of power consumption vs. hour of day in a typical home in a city. —Lowellian (reply) 02:44, 30 September 2008 (UTC)[reply]

It is difficult to have a typical home by hour of day. Most power consumption occurs when people are at home. So, you have to start by generalizing to the point of being incorrect by deciding which hours of the day people are at home (and awake) in a "typical" home. -- kainaw ™ 04:43, 30 September 2008 (UTC)[reply]

Here's some articles I found through a google search: [1]. About 1/4 of the way down the page is a graph of the power consumption over a single winter day in Humbolt County, California. This one:[2] shows the last 24 hours power consumption in Ontario, Canada. This one: [3] (see figure 2) shows an average graph by season, though I don't see that one's methodology. There's 3 for ya. Good luck! --Jayron32.talk.contribs 04:55, 30 September 2008 (UTC)[reply]

The power consumption through the day will vary by season and by region. See Load profile. A Google Book search shows lots of books which discuss such load profiles or load curves. Your own utility doubtless has data on residential load curves which they might provide you. Edison (talk) 05:28, 30 September 2008 (UTC)[reply]

(ec)I wrote a paper on the energy saving benefits of daylight savings time (there isn't one, by the way), and the timing of people's power usage was huge part of it. Suffice to say that it varies with climate, season, and culture. Some people need to heat their houses all night and some need to air condition it all day. I don't know what you are using it for, but you might be interested to know that industrial power consumption is more than 3 times larger than domestic consumption. Plasticup ^T/C 05:29, 30 September 2008 (UTC)[reply]

Off-topic, but since you mention daylight-saving time - now that the powers-that-be have decided (solely on energy-saving grounds mind you) to extend DST to cover more than half the year, shouldn't we just call it "Time"? The thing we do in winter should now be called "Daylight Deficit Time". Franamax (talk) 06:19, 30 September 2008 (UTC)[reply]

While we're discussing it, in 1973 Daylight Savings Time was observed all year! Plasticup ^T/C 14:31, 1 October 2008 (UTC)[reply]

Tyvek house wrap

I live in central Ontario, Canada. My house is about 30 years old. It is a two story with brick on the first level and siding on the second. My question is this: is it worth the effort and expense to remove the siding, wrap and seal the second level with Tyvek and then put the siding back on? Modern houses are wraped in Tyvek but I don't believe that they were 30 years ago. Thoughts? —Preceding unsigned comment added by 209.161.212.209 (talk) 04:27, 30 September 2008 (UTC)[reply]

If it were my house, I would not do it. I question what the payback period would be for such a retrofit. Edison (talk) 05:23, 30 September 2008 (UTC)[reply]

I wonder if there are tax deductions for improving the insulation in your house. That might be something to consider. Plasticup ^T/C 05:30, 30 September 2008 (UTC)[reply]

Tyvek (TM) forms an air-tight/water-proof barrier, so the question is whether your existing siding is inadequate to form the air-seal. Call Ontario Hydro (Hydro One now, or your local electric utility) and ask if they still have the cut-rate efficiency assessment programs. Whichever heating supplier you have (such as Enbridge) is a good starting point too. If you're going to pull off the siding, you should also be thinking about putting on 40mm or 50mm of that blue styrofoam insulation. (You'll need 2" corner-caps if you do that though) Franamax (talk) 05:42, 30 September 2008 (UTC)[reply]

And rather than strike the above, I'll note instead that if it's an Ontario house, 30 years old, it will already have an air/moisture-seal - that black wax-coated paper you used to always see on houses as they were being built, before Tyvek figured out how to print their brand name in big letters across the whole sheet. If you have no moisture seal at all, you may have a problem - pry up a bit of the siding and look for the black paper. Franamax (talk) 05:51, 30 September 2008 (UTC)[reply]

My neighour just added another layer of insulated cladding on the outside of their wall. However they complained about the heat of the sun in summer, so they just altered the west facing wall. Once it was coated and painted it looked just the same as before. Graeme Bartlett (talk) 21:12, 30 September 2008 (UTC)[reply]

It can be worth doing the Tyvek if your already planning on replacing the sideing. Otherwise... unless your having water intrusion problems (mold, etc) the cost is not worth it. ---J.S (T/C/WRE) 07:36, 2 October 2008 (UTC)[reply]

Charging device: sometimes it works

I have a charging device (for my laptop) that when it's plugged, sometimes works and sometimes not. Chances are 1/10 that it will work. I ruled out a bad contact, since I am not shaking or moving the charging device itself, only plugging the cable to it, and I used two different cables with the same charging device.

Why does it sometimes work?

80.58.205.37 (talk) 10:15, 30 September 2008 (UTC)[reply]

It could be that heat, in the cable or the transformer, at some point causes the internal connections to move apart. I had a similar problem with a poorly-connected cable running into the circuit breaker box in my house. Only after a given appliance had been running for a while did heat build up, causing the circuit breaker to trip. The fix involved having an electrician redo the connection. Not that you asked, but with such unreliable performance, I'd be inclined to buy a new charging device. But then, I have two power cord for my laptop -- one stays at my desk, the other stays in the laptop's travel bag, so I don't have to remove and reconnect the desk cord all the time. --- OtherDave (talk) 10:43, 30 September 2008 (UTC)[reply]

Oh, that brings me to the idea that perhaps it is the heat in my case. When I connect the power cord a couple of times, the charging device is cold. After it warmed up, it starts to work. I going to test it more thoroughly though. 80.58.205.37 (talk) 10:59, 30 September 2008 (UTC)[reply]

And to be thorough, you need to try charging a different laptop with your charger and cables. How do you know yet that it's not a bad connection within the laptop? Franamax (talk) 11:04, 30 September 2008 (UTC)[reply]

Because the charger has a led indicating if it is working.80.58.205.37 (talk) 11:21, 30 September 2008 (UTC)[reply]

OK, now exactly what do you mean by "it is working"? Do you mean that if the LED is on, it will definitely always charge the laptop? (Sorry, troubleshooting usually has to go one step at a time) Franamax (talk) 12:11, 30 September 2008 (UTC)[reply]

Yes, if the LED is on, the charger is charging the battery. Mr.K. (talk) 15:20, 30 September 2008 (UTC)[reply]

It could still be in the laptop though. Some batteries have built-in temperature sensors that detect when the battery is overheating and turn off the charging voltage until they cool off a bit. If your battery temperature sensor was intermittant - then that could account for the behaviour. You've really got to do this by a process of elimination. If you know someone with a similar type of laptop the trying the "known good" charger on the "iffy" laptop will either eliminate or implicate the charger. Then, if you establish that the charger is OK (eg, if it successfully charges a different laptop) then you can try exchanging batteries and see whether the fault "follows" the battery or "stays" with the laptop. That'll narrow things down still further. If you can't find someone else with a similar machine then you have to do it the expensive way - buy a new charger - if that doesn't help, buy a new battery - if that doesn't help - junk the laptop or live with the problem! SteveBaker (talk) 19:21, 30 September 2008 (UTC)[reply]

In the meantime, I am pretty convinced that the problem lays in the charger. My laptop doesn't receive any power when the LED of the charger is off. It doesn't matter if the battery is being charged or not. I have tested several times, and the charger doesn't work when it is cold. I always have to wait a couple of minutes until it has warmed up a little and is able to deliver power. I am going to live with the problem for a while and buy a new one at the end of the month. Mr.K. (talk) 09:16, 1 October 2008 (UTC)[reply]

Methane Clathrates are melting...NOW!

Oh-oh...it looks like the end of the world just started (something to take your mind off of the global economic meltdown at least):

We've talked about the hypothetical possibility of deep-ocean Methane Clathrates melting being the "tipping point" beyond which it might be impossible to prevent 'runaway' global warming. The two reports above indicate pretty clearly that we just hit that point.

I'm puzzled though - why there? You'd think the arctic waters would be coldest for longest - how come we aren't seeing this happening in more southerly latitudes first? If methane is a 20x worse greenhouse gas than CO2 - should we not set light to these methane plumes? (Is that even possible?)

SteveBaker (talk) 11:02, 30 September 2008 (UTC)[reply]

Bah! Peer review? Correlation of atmospheric temperature rise with sea-bottom temperature rise with clathrate release? Causative conclusion or always-been-doing-that? And they're only talking megatonnes of CH4, even at Ce20 - call me when you get a gigatonne. :(

I'll watch my favourite mag closely to see their interpretation. As you say, seems strange to see this. I believe there's an increased flux of freshwater into the Arctic basin, but I would think that would stratify the waters and keep a cold layer below. Regardless of this result, the concerns remain, including potential initiating methane production from soil. Sticking in a big straw and striking a match at the top might not be such a bad idea though... Franamax (talk) 11:27, 30 September 2008 (UTC)[reply]

I say we burn it all as it comes out of the ground. 96.242.34.226 (talk) 17:56, 30 September 2008 (UTC)[reply]

Well, it's coming out of the deepest parts of the ocean - not the ground. Burning it would convert it to CO2 and water - and the CO2 would still cause an increase in the greenhouse effect - just not as bad as methane. But I'm not sure whether the methane that's appearing in this case is sufficiently concentrated in one place to burn. It could be argued that we should mine the clathrate deposits and use them as fuel as a way to stop the raw methane from making it into the atmosphere - but that's really an admission of defeat, the goal is supposed to be to keep the planet cool enough that they don't melt in the first place. I'm horrified to see that this doomesday scenario already starting to happen. I had imagined that this would be 100 years from now. SteveBaker (talk) 13:08, 1 October 2008 (UTC)[reply]

Actually, burning Methane (CH₄) instead of coal (dirty C) or even oil is producing much less carbon dioxide per unit of energy released. So using the clathrates before they melt would be a double gain. Of course, knowing us, we would probably burn both the methane and the coal... --Stephan Schulz (talk) 15:20, 1 October 2008 (UTC)[reply]

There is also the teeny-tiny catch in that we have no clue how to mine stuff that's that cold - that deep under the deepest oceans. Learning out how to mine the stuff before it's all gone may be an impossibility anyway. SteveBaker (talk) 01:19, 2 October 2008 (UTC)[reply]

Methan Clathrates form under particular conditions, which include "cold enough". But that just means they form deeper in lower latitudes. They will be released not if surface water reaches a certain temperature, but if the water at the reservoir reaches a certain temperature. Arctic waters have warmed most, so the reservoirs in the Arctic melt first. This is, of course, only a first and preliminary popular press report, so it's hard to assess the actual impact. But I'm getting the feeling that we might be on our way to experimentally solve the Fermi paradox. --Stephan Schulz (talk) 18:24, 30 September 2008 (UTC)[reply]

Road-safety for trucks

You know the reverse parking sensors you get on some cars? The ones that beep and help you park. I guess they work based on radar on infra-red technology. My question is...Is there anything to stop placing these along the side of trucks trailers and having a light-panel on the dashboard/somewhere that lights-up while vehicles are along-side trucks? This would reduce the risk of side-swiping/trucks pushing vehicles off the road overtaking (or whatever) when not able to see there is a vehicle along-side them? It was just an idea I had when I watched a show about how little visibility there is on trucks and thought...why not employ the same technology but instead of beeping it could light up to show when a vehicle is within X-metres of the side of the truck and thus help prevent accidents. Or is the technology totally unworkable when moving at say 50mph with vehicles alongside that are doing the same? 194.221.133.226 (talk) 11:09, 30 September 2008 (UTC)[reply]

Well, in the auto industry, "safety gadgets" have extremely high litigation costs. The first time that one allegedly fails, its a contingency fee lawsuit waiting to happen. Many times, the cost of defending a lawsuit is more expensive than just settling the lawsuit. There are so many car crashes each year, that its a bad business plan thus the entrepreneur would never get venture capital. Sentriclecub (talk) 11:21, 30 September 2008 (UTC)[reply]

They'd go off everytime the truck drove close to the concrete barriers at the sides of the road. But I suppose they could be turned on only when the driver switches on his turn signal or something. SteveBaker (talk) 13:18, 30 September 2008 (UTC)[reply]

well in my head they would just light-up so if you were driving close to a wall/barrier it would be lit up on the appropriate side. Is there any reason that the technology wouldn't work at the speeds mentioned? —Preceding unsigned comment added by 194.221.133.226 (talk) 13:49, 30 September 2008 (UTC)[reply]

So long as it's not using an acoustic (sonar) type of technology, speed ought not to matter - but the idea of having a whole bunch of lights flashing in the cab all the time there is a wall to the side of the truck would drive most people crazy! Also, once a warning light has flashed a few bazillion times, you'd start to ignore it so it wouldn't help when it actually mattered. But as I said - if it only turned on the hazard lights when you hit your turn signal in that direction, then it should be OK. You wouldn't often be indicating (say) a left turn when you're in the left lane on the freeway with concrete lane dividers triggering the sensors. SteveBaker (talk) 19:11, 30 September 2008 (UTC)[reply]

This what 194.221.133.226 (talk · contribs) asks about is undergoing very active development, particularly in Europe, for passenger and heavy vehicles. Front, rear, and lateral proximity-alert systems are being tested and implemented, as well as more integrated setups that can actually intervene on the driver's behalf to avoid a collision, lessen the severity of a collision, and/or lessen the consequences of a collision by means of steering assistance, braking assistance, and even acceleration assistance. These systems are variously based on cameras (increasingly using tiny camera-on-chip systems) or on ultrasonic, radar, or lidar sensors. All of these systems fall under the category of driver assistance systems. I haven't (yet) looked to see what articles we might have on related topics, or what condition they're in. —Scheinwerfermann (talk) 13:31, 1 October 2008 (UTC)[reply]

This exact sort of product is available on Volvo motorcars (and likely, more brands in the near future).

Atlant (talk) 21:28, 1 October 2008 (UTC)[reply]

Is Thrust a usual vector? (i.e. same properties of all vectors I have learned so far?)

I checked on that page, and it is a force. All forces are vectors, thus all white horses are horses right? Also, is this ^{${\overrightarrow {T}}$} the symbol for thrust(in n>1 dimensions)? I'll make some minor improvement to the article asap. Or is ^{${\overrightarrow {F}}_{thr}$}

Secondly, the symbol for impulse. My book says to use J because 'I' stands for electric current, not impulse. (ever heard p=iv? that means power=current times voltage, not impulse times voltage). (source = impulse discussion page) however MIT OCW professor Walter Lewin uses I so I'm curious if the rate at which college students are being taught J is increasing and likely someday to be more popular? Afterall, change is mathematical conventions and symbols are very slow. Sentriclecub (talk) 11:14, 30 September 2008 (UTC)[reply]

The choice of symbol depends on what field you're in also. There just aren't enough letters in enough alphabets to cover all of the things we might need to represent mathematically. In aerodynamics, 'I' for electric current is unlikely to be a concern since we're only rarely considering electric current at the same time as the reaction force from a bunch of jet engines - so aerodynamicists may well choose 'I' for impulse. If you design electric cars - maybe it could get confusing - so you use 'J'. <shrug> Not all forces are vectors - some of them are fields...although you may be able to treat them as vectors locally.

My guess is that, if you are starting a garage band, and if if the number "7" is taken, then use "8". That's my creative story of why J=impulse. Also, a lowercase j is resemblant of an i. Sentriclecub (talk) 12:23, 30 September 2008 (UTC)[reply]

Or an uppercase J, if your handwriting is as bad as mine. Plasticup ^T/C 14:28, 1 October 2008 (UTC)[reply]

Isobaric peptides

In scientific literature, I sometimes see a reference to "isobaric peptides" (meaning peptides of the same mass). As I understand, it "isobaric" is a reference to pressure, not peptide mass, so is it acceptable to use "isobaric" in this way? 11:11, 30 September 2008 (UTC)

~~I dont know if that is accurate. Is it an online source? Give me the link, i'll take a look.~~ Confirmed, its an accurate way to use the word. As accurate as Pound (mass) guess I'll have to live with another one. Sentriclecub (talk) 11:17, 30 September 2008 (UTC)[reply]

This is on the wikipedia page for the term isobaric. A quick search on google also list various research papers that use the word isobaric in this same sense. This link: [4] also cites (from The American Heritage® Medical Dictionary) that the adjective "isobaric" means "Having equal weights or pressures". So this term can definitely be used thus.Leif edling (talk) 11:51, 30 September 2008 (UTC)[reply]

What's wrong with Pound (mass)? Its usage predates Newtonian mechanics by centuries. AlmostReadytoFly (talk) 08:01, 1 October 2008 (UTC)[reply]

Sunrays and satellites

Why the Sunrays defend th Satelite signals? —Preceding unsigned comment added by 61.2.206.4 (talk) 16:20, 30 September 2008 (UTC)[reply]

Your question is unclear, can you re-state it? You may be discussing the ability of solar storms to disrupt electronics (satellites included), however. If so, the phenomenon is well-covered there. — Lomn 19:56, 30 September 2008 (UTC)[reply]

Solar outage is due to the sun's radiation overwhelming the satellite signal. It occurs at intervals in October in the northern hemisphere. It happens as the dish pointing at the satellite also points at the sun, which is giving off microwaves too. Graeme Bartlett (talk) 21:31, 30 September 2008 (UTC)[reply]

black hole

I appreciate if you could help, i would like to know

what is in the black hole
what happens if you fall in

thank you —Preceding unsigned comment added by 89.100.97.189 (talk) 17:28, 30 September 2008 (UTC)[reply]

See Black hole. Feel free to come back here if you have questions not answered in the article. Algebraist 17:55, 30 September 2008 (UTC)[reply]

1. What is in the black hole: A large number of things that fell in, only extremely small and probably warped beyond recognition. 2. What happens if you fall in: You would be extremely small and warped beyond function.--ChokinBako (talk) 22:37, 30 September 2008 (UTC)[reply]

It's not so much warped as broken up into tiny pieces by tidal forces. And that's only until you hit the singularity. What happens then is anybody's guess. --Tango (talk) 00:04, 1 October 2008 (UTC)[reply]

So what's inside a black hole if atoms and even sub-atomic particles are torn away from each other? Degenerate sub-atomic particles? What kind? --Kjoonlee 03:20, 1 October 2008 (UTC)[reply]

What's inside an evaporating micro black hole? --Kjoonlee 03:21, 1 October 2008 (UTC)[reply]

Sigh...it's time for a L-O-N-G explanation isn't it.

Ultimately - at the very center of a black hole is a 'singularity' - this is an object with literally zero size. It's not just small - it has no size at all. Within such a thing, all of the mathematics that we might use to analyse it fall apart. Density becomes infinite for example. Gravitation - also, infinite. We really can't say anything about the singularity itself except (oddly) it's mass, it's charge and how it's rotating (I think that's all - someone please correct me if I missed one).

But when we casually talk about black holes, we're usually talking about the region within the so-called "Event horizon" - which is larger. If you think about (say) a planet like the earth, it has an "escape velocity" - if you launch an object into space at a speed lower than the escape velocity then it'll eventually fall back down again. If you launch it at greater than the escape velocity then it'll never quite slow down to a dead stop and fall back - so it has "escaped" the earth's gravity. Well, the escape velocity of a black hole is greater than the speed of light. Since nothing can go faster than light, something that's close enough to the singularity to have an escape velocity that's greater than light is doomed to fall into the singularity and be scrunched up so tight by the infinite gravity at the singularity that it too has zero size. As you get further away from the singularity, the gravity falls to lower values until eventually, the escape velocity is less than the speed of light and fast moving things can (in theory) escape. The distance at which that happens is the "event horizon".

What makes the black hole "black" is that even light can't escape if it starts out inside the event horizon. Hence no light whatever from within the event horizon can ever reach the rest of the universe and the "hole" is black. There are deeper things to consider here. If no light and no matter can escape from within the event horizon, then no information can escape either - since information has to be carried on either light or matter. Hence we can never know what's inside the event horizon - all we can do is predict what happens using math...and since even the math gives up on us at the singluarity - we're going to be somewhat ignorant of the inner workings of these beasts.

The "evaporation" thing is a bit more subtle. All black holes are believed to evaporate - but the big ones do so amazingly slowly and easily make up for that evaporation because their huge gravity is pulling in more and more stuff from the outside universe. The mechanism involved in evaporation is weird. In a pure vacuum, it is believed that fundamental particles (like electrons) spontaneously pop into existance - along with an exactly equal number of anti-particles (positrons, for example). Generally, these "virtual pairs" of particles almost immediately crash back into each other and vanish again...so the vacuum stays empty. But if this happens right next to the event horizon of a black hole, it's possible for one of the pair to accidentally slip across the event horizon and be unable to return to it's partner outside. When that happens, it seems that a particle just suddenly popped into existance AND DIDN'T IMMEDIATELY VANISH AGAIN! So in the region just outside of the event horizon, these particles are popping into existance and (rarely) have enough energy to escape out into the big wide universe without getting 'eaten' by the hole. So if you stand back far enough - it looks like the black hole is slowly emitting particles. Since there has to be an energy balance - this means that the black hole is slowly losing mass. It's "evaporating".

Since all black holes evaporate - there is no special distinction to be made "inside" the ones that are...because they all are.

As for the "tearing up" (spaghettification is the "correct" (if silly) term): Consider the earth and the moon...the moon's gravity causes tides on the earth. That's because on the side of the earth nearest to the moon, the moon's gravity is a bit stronger than on the opposite side. That pulls the ocean towards the moon - which is why we have high tides when the moon is overhead (you also get tides on the opposite side of the earth - but we're not going to get into that right now). When you were falling towards the earth - feet-first let's say - then the gravity of the earth is pulling very slightly more on your feet than it is on your head because your feet are a tiny bit closer. Just as the moon's gravity causes tides on the earth - the earth is causing tides on your body. This "tidal" effect on our bodies is not even slightly noticable because the earth's gravity is pretty feeble and we don't get close enough to the center of the earth to feel their largest effects. But if you were falling feet-first into a black hole - then even if you're outside of the event horizon, the gravity at your feet is a LOT stronger than at your head - to the point that it's going to stretch you out...as you fall closer and closer to the hole, this tidal effect gets worse and worse - you get pulled out into a long, thin streak - like a piece of spaghetti (hence the scientific term "spaghettification"!)...the closer and closer you get, the more this tidal force starts to affect you - until your body is shredded to molecules - then the molecules start to feel different gravity on the atoms at one end of the molecule than at the other - and they too get shredded (into atoms)...then the atoms get shredded and so on all the way down to the most fundamental particles. At the point where the singularity is, this tidal force becomes infinite and we can't even calculate what happens.

Fortunately, black holes are 'cloaked' in this information shield - the event horizon - so no matter how weird and impossible to calculate things get, the bizarre things that happen there can't ever affect the rest of the universe. This has been termed "Cosmic censorship" - and it's almost like the universe doesn't want us to know what's going on in there so it wraps the impossible math in an event horizon to keep the rest of our physics nice and clean and sensible!

SteveBaker (talk) 12:58, 1 October 2008 (UTC)[reply]

Infra-Red

On an episode of Mythbusters ages ago, they were testing various movie stories relating to cat burglars getting through various alarm systems, including infra-red lasers. Night-vision goggles did not work on the infra-red lasers, as, apparently, night-vision goggles just intensify remaining light. How would infra-red goggles work in detecting these lasers? Surely they would see them? I assume that the reason they were not used in the episode is because infra-red goggles do not exist....--ChokinBako (talk) 21:39, 30 September 2008 (UTC)[reply]

"Active infrared" night vision technology is widely available, though I can't comment on the particulars of the night vision stuff used by MythBusters. The single frequency used by the laser might have been out of the range of what the NVG could detect. — Lomn 21:56, 30 September 2008 (UTC)[reply]

Lasers are difficult to see even in visible light - you can only see them if they are either pointed directly at your eye or if the heroine of the story blows her make up into the beam to scatter the light. If you had something to scatter the light then night-vision goggles could well allow you to see them, but you have to bare in mind that the infra-red range is very wide (it goes from around 750nm to around 1mm, whereas visible light is just from roughly 380nm to 750nm - put another way, the wavelength of the top end of visible light is just under twice the wavelength of the bottom end whereas with infra-red it's over 1000 times longer), so just because the night-vision goggles see infra-red and the laser is infra-red doesn't mean they are actually anywhere near each other. --Tango (talk) 22:08, 30 September 2008 (UTC)[reply]

So, you would need adjustable goggles so you could set them to the right frequency to see? By the way, the heroine in Mythbusters did blow her make up onto the beams, but this scattered the light so much that it broke the beam and set off the alarm. Wouldn't just blowing cigarette smoke be better? Not very movie-ish and lady-like in these PC times, but, if it works, hey!--ChokinBako (talk) 22:30, 30 September 2008 (UTC)[reply]

But risky... cigarette smoke or foundation powder may lower the intensity of the light striking the detector enough to trip the alarm. Anything that scatters the beam to make it visible is likely to also diminish its intensity. --Jayron32.talk.contribs 23:46, 30 September 2008 (UTC)[reply]

It's guaranteed to diminish its intensity - conservation of energy, and all that. --Tango (talk) 00:02, 1 October 2008 (UTC)[reply]

"If you had something to scatter the light then night-vision goggles could well allow you to see them," Is a key phrase here. You can not ever see the beam of a laser like in star trek unless it's hitting something. (Smoke particles.) Even if you have smoke particles, the laser might be an invisible color of light. It's no use making some smoke light up in infrared if you can't see infrared.

So if you're trying to see an infra-red laser you've got two problems. One is that lasers beams don't show up in air, and the other is that you can't see infrared. The smoke solves the first problem, and the goggles solve the second. With any luck. APL (talk) 00:59, 1 October 2008 (UTC)[reply]

Most night vision goggles DO see into the infrared - the only difference between them and the kinds of thermal sensors the fireman and the military use is the range of frequencies within the infrared spectrum that they see. There is a huge variance between the cheap ex-Russian-army goggles a lot of people buy and modern, state-of-the-art US military NVG's. It seems that the more modern the goggles are, the further into the infrared they see. The problem the Mythbusters had was that air is pretty transparent - so not much of the laser light is scattered sideways as the laser crosses the room - that's pretty much as true for IR light as visible light. With that little light being scattered - even a super-sensitive NVG cannot see it. The trick of putting powder into the path of the laser is a good one - because it causes the light to scatter so you can see it - but that also cuts down the amount of light reaching the detector and (as the Mythbusters found) that sets off the alarm. If NVG's did work, the people protecting buildings with IR lasers could simply leave the room lights turned on. Light that bright simply overwhelms the goggles and they can't see a damned thing. SteveBaker (talk) 00:30, 1 October 2008 (UTC)[reply]

Sound speed / loudness

When studying the ear, I learned that one of the ways sound is amplified between the environment and the hair cells in your cochlea (the other two being the three ear cells, and the change in area between the eardrum and the oval window) is the vibrations being tranferred from air to a fluid in the cochlea, and since sound waves travel faster in water, this makes the sound louder. But how does travelling faster make a sound louder?

My guess is, if a vibration travels faster, then less energy has time to be lost due to friction. Is that accurate?

Thanks Jonathan ^talk 22:16, 30 September 2008 (UTC)[reply]

The sound is non-existent. Loudness is not about amplitude or pressure or energy. Its about neurons. Loudness is just your brain's way of interpreting simple harmonic oscillations. Sentriclecub (talk) 23:01, 30 September 2008 (UTC)[reply]

What he is implying is that the amplitude of the sound waves increase in the ear. While loudness is about perception (after all, one can be exposed to high-frequency sounds which one cannot hear, and thus would not notice the loudness of them, however these sounds can have an amplitude high enough to damage the ear). Its not very helpful to the question asker to play games with semantics. If we replace the word sound with vibration, and the word loudness with amplitude, perhaps someone could answer his question? I'm curious myself... --Jayron32.talk.contribs 23:43, 30 September 2008 (UTC)[reply]

I don't think the amplification is due to the liquid. It comes about because a relatively large entrance to the ear is progressively reduced in size through the cochlea. If the amplitude of the sound waves at the entrance to the ear were (say) 1mm and suppose the entrance to the ear was 100mm². The volume of air being moved is therefore 100mm³. Towards the narrow end of the cochlea, it's probably reduced to a cross-sectional area of 10mm² - but more or less the same volume is moving back and forth - so instead of 1mm of amplitude, you have 10mm of amplitude. This increase bends the little hairs more strongly - resulting in a louder sound. It's just like making a cone of paper and sticking the pointy end in your ear - it collects air movement over a large area (at the big end of the cone) and concentrates it to the narrow area at the pointy end. So everything sounds a lot louder. SteveBaker (talk) 00:22, 1 October 2008 (UTC)[reply]

When a wave's speed increases its amplitude decreases. When the acoustic energy passes from air to liquid, the energy rate doesn't change (though some energy is lost in the transition). Since the energy/time is constant, a higher speed of sound will cause a lower amplitude. (This parts a guess, but I expect that the liquid will couple with the stereocilia better than air would.) Saintrain (talk) 00:49, 1 October 2008 (UTC)[reply]

See Underwater acoustics. Water transmits sound waves faster than air and with lower loss per metre. Therefore sounds carry further than in air (whale song carries hundreds of miles). For scuba divers these two factors also mean that it's a lot harder to locate the source of a given sound. TrulyBlue (talk) 11:35, 1 October 2008 (UTC)[reply]

That's true but the losses over the couple of centimeters the sound travels within the ear are truly negligable compared to the losses over many meters before the sound reaches the ear. That can't be the reason the cochlea is filled with liquid. It's much more likely to be that the liquid is incompressible and therefore somehow transmits the vibrations onto the sensing hairs more efficiently...something like that. SteveBaker (talk) 12:22, 1 October 2008 (UTC)[reply]

Maybe amplitude is not really the important thing about the fluid in the cochlea... maybe it's more about the clarity of the sound, or maybe it helps keep the cochlea's shape (the same reason the eye is filled with fluid). I bet the stereocilia would vibrate better in a fluid where they're less likely to get randomly squished by outside forces. Jonathan ^talk 15:37, 1 October 2008 (UTC)[reply]

The fluid functions as a potassium resevoir for the hair cells to create impulses. Donek (talk) 16:30, 1 October 2008 (UTC)[reply]

I believe the shape of the cochlea has evolved to aid the discrimination of different frequencies of sound - as it gets smaller, progressively higher frequencies are detected.