Wikipedia:Reference desk/Archives/Science/2011 August 12

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August 12[edit]

Pogo stick on trampoline[edit]

I was wondering what would happen if you tried to trampoline on a pogo stick. Could it be an Olympic Sport? And what about if you put all four corners of the trampoline on pogo sticks? And put a platform under that and have more pogo sticks under it at the corners? And so on. Could you go into orbit like that? Btw, awaotnmpi, what ever happened to pogo sticks? I've seen push scooters and hula hoops go in and out, and back in, but I've never seen pogo sticks come back. Kangaroos use a natural analog of pogo sticks to bounce around. It's actually very efficient. Do any other animals use pogo stick technology? Myles325a (talk) 01:37, 12 August 2011 (UTC)[reply]

Nowadays they're probably considered unsafe. ←Baseball Bugs What's up, Doc? carrots→ 02:05, 12 August 2011 (UTC)[reply]
I'm also pretty sure you couldn't get into orbit on trampolines and pogo sticks, lol... Both have weight and friction, i suspect once you got to about 3-5 trampolines and pogo sticks, they would stop "adding" to the effect, i.e., the trampolines/pogo sticks on the bottom wouldn't even leave the ground. Vespine (talk) 02:30, 12 August 2011 (UTC)[reply]
Come to think of it, in effect you're creating a big Shock absorber. Vespine (talk) 02:47, 12 August 2011 (UTC)[reply]
The effect would depend entirely on if you could get the trampoline and the pogo stick into resonance. If the two were not in resonance, it would probably do nothing interesting. If they were in resonance, you could likely bounce quite high. Until you came down and smashed your skull open, that is. --Jayron32 02:53, 12 August 2011 (UTC)[reply]

What a fascinating image. μηδείς (talk) 20:47, 12 August 2011 (UTC)[reply]

Pogo sticks are always coming back in, and most children I know in the 7-10 age group have had a go on one. The problem is that, like many toys, they're now perceived as belonging to a younger and younger age group, and pogo sticks are tricky things for the very young. So you can get easier versions designed for younger children, which only increases the perception of them being aimed at younger children. 82.24.248.137 (talk) 09:19, 12 August 2011 (UTC) [reply]
A request for consideration of pogo-on-trampoline as an Olympic sport should be addressed to Jacques Rogge, President of the International Olympic Committee corporation, Lausanne, Switzerland, where I am sure he will be intensely interested. Cuddlyable3 (talk) 10:25, 12 August 2011 (UTC)[reply]
Pogo sticks come in, and then they go out, and then they come in, and then they come out... Plasmic Physics (talk) 14:06, 12 August 2011 (UTC)[reply]
It's been done. Not exciting. Cuddlyable3 (talk) 19:43, 12 August 2011 (UTC)[reply]

Hard sandpaper grinding wheel[edit]

I'm looking for a sandpaper grinding wheel with a grit close to the hardness of diamond dust, but hopefully far cheaper. Any ideas ? 68.79.97.98 (talk) 04:53, 12 August 2011 (UTC)[reply]

I believe Corundum sandpapers are widely available: though not cheap, they are surely cheaper than diamond sandpaper would be. {The poster formerly known as 87.81.230.195} 90.197.66.65 (talk) 05:18, 12 August 2011 (UTC)[reply]
Thanks. Anything between that and diamond ? 68.79.97.98 (talk) 05:22, 12 August 2011 (UTC)[reply]
Yes, the article Mohs scale of mineral hardness gives some examples, including rhenium diboride, tantalum carbide, titanium diboride, boron nitride etc but I've never seen these made into "sandpaper". What are you trying to sand down? The main problem is usually not in the hardness of the abrasive, but in the glue that holds it to the "sandpaper". An alternative to corundum is Silicon carbide (carborundum), but I'm not sure whether it is any harder than corundum. Dbfirs 06:40, 12 August 2011 (UTC)[reply]
I'm trying to sand down rubies (corundum), so need something harder than 9.0 on the Mho's scale. Cubic boron nitride seems to be hard enough, and does appear to be used in grinding wheels. I've not been able to find a place to buy it yet or a price, though. Any help would be much appreciated. 68.79.97.98 (talk) 07:04, 12 August 2011 (UTC)[reply]
Drat, I just found the price of cubic boron nitride powder, and it seems to run about 2.5x that of synthetic diamond. 68.79.97.98 (talk) 07:18, 12 August 2011 (UTC)[reply]
Surely if you're trying to abrade rubies, it would be far easier to use an appropriately composed grinding wheel or similar solid tool rather than sandpaper? Have you thought of finding a local engineering shop and having an informal chat with someone about what you're trying to achieve? They might be able to suggest a better approach. {The poster formerly known as 87.81.230.195} 90.197.66.65 (talk) 07:30, 12 August 2011 (UTC)[reply]
Yes, grinding wheels are more appropriate. I e-mailed 3 people who are contacts for local lapidary clubs (South-Eastern Michigan), but got no responses. Ideally I'd like to find a place where I can rent or borrow the necessary equipment, but I've had no luck. 68.79.97.98 (talk) 07:43, 12 August 2011 (UTC)[reply]
This search found a plethora of sellers of diamond grinding wheels. Cuddlyable3 (talk) 09:51, 12 August 2011 (UTC)[reply]
This editor has been on the RefDesks a number of times: he is currently trying to find a way to turn the ruby rod from a laser into a set of jewel-cut rubies so he can sell them. Looie496 (talk) 15:08, 12 August 2011 (UTC)[reply]
Not quite. I'm trying to grind them into hemispherical cabochons, not cut them into faceted stones, and I intend to give them away, not sell them. (I've concluded I can't make a profit, but these might make a nice "homemade" gift.) 68.79.97.98 (talk) 18:37, 12 August 2011 (UTC)[reply]
A difficult endeavor without proper training. Googlemeister (talk) 15:21, 12 August 2011 (UTC)[reply]
Yes I noticed that. Since as GM said, it already sounds like a difficult endeavor, I wonder whether the OP has considered using the gemstones which may or may not be diamonds to help cut the rubies? Perhaps manufacturing a home made grinding wheel would be a useful additional task? Nil Einne (talk) 17:36, 12 August 2011 (UTC)[reply]
No, that wouldn't work, as diamond dust is needed for grinding, and it wouldn't make much sense to pulverize diamonds to make dust for grinding. Instead of a grinding wheel, I plan to do the reverse, and put the ruby rod in a drill chuck. This should allow me to avoid the flats which are almost inevitable when using a grinding wheel. While this should assure it is circularly symmetrical when viewed from the end, there's still the issue of how to ensure a circular 5 mm radius profile, when viewed from the side. Ideally I'd have a 5 mm radius circular "bowl" coated with fine diamond dust for this finishing work. Can anyone find anything like that ? 68.79.97.98 (talk) 18:41, 12 August 2011 (UTC)[reply]
Cabochons are said to be prettier elipsoidal than spherical. Cuddlyable3 (talk) 19:37, 12 August 2011 (UTC)[reply]
Perhaps, but they wouldn't fit in the ring where I intend to mount it. 68.79.97.98 (talk) 20:15, 12 August 2011 (UTC)[reply]
Are you sure about that? [1] & [2] suggests diamond dust does in fact normally come from pulverised diamonds. Some may arise from the bits and pieces left over that are too small for anything else but I'm guessing traditionally low quality diamonds were indeed pulverised and used to make diamond dust because there wasn't otherwise enough of the left over bits and something had to be done with the lower quality diamonds. Nowadays in the era of cheap synthetics I suspect natural diamonds aren't used much but since the things may or may not even be diamonds I don't know if that applies. Nil Einne (talk) 07:30, 13 August 2011 (UTC)[reply]
The "diamonds" are large, faceted stones. Even if simulants, they would still be worth more than the dust which could be generated from them. Furthermore, I have no method available to pulverize them, and, if not real, they might well be softer than ruby, making their dust useless for grinding them. 68.79.97.98 (talk) 12:10, 13 August 2011 (UTC)[reply]

Anti-aging[edit]

Is it possible to maintain skin like a young person after the age of 50? Will it is possible in the near future to look like a young person even after 60 with the development of anti-aging medicine and life-extension technologies? --HHH000xyz (talk) 07:51, 12 August 2011 (UTC)[reply]

Not sure how near it is, but genetic engineering could be used to grow replacement skin from your own stem cells. The problem, though, is replacing your own skin with this skin, which would require surgery, leave scars, etc. Repairing your own skin might be less traumatic, but also less effective, due to cumulative genetic damage. If we could find a way to replace not just a segment of the genetic code in each skin cell, but the entire genome (or else locate and repair the damaged locations in each DNA strand), then that approach might work. StuRat (talk) 08:00, 12 August 2011 (UTC)[reply]
Alternatively, is it possible to change community attitudes so that a few natural wrinkles don't bother people? HiLo48 (talk) 08:03, 12 August 2011 (UTC)[reply]
If you're talking about sexual attractiveness, I'd say you'll never get more than a minority to consider wrinkles sexy. This is because they are an age marker which makes it apparent that this person is beyond their prime reproductive years, and millions of years of evolution have shaped us to desire people who are able to help us pass on our genes. On the other hand, if you're talking about people from whom we "seek wisdom", wrinkles might indicate they are more likely to have enough experience to provide good advice, with the possible exception of advice about sunbathing (I wonder if old vampires get wrinkles :-) ). StuRat (talk) 17:02, 14 August 2011 (UTC)[reply]
It is possible now to skin a young person and freeze the skin. The skin will look the same for hundreds of years. Cuddlyable3 (talk) 09:47, 12 August 2011 (UTC)[reply]
Are you male or female? It's harder for females because of the effect of oestrogen on the skin: as production switches off after the menopause, the skin naturally becomes thinner and more translucent. Men don't have such problems.--TammyMoet (talk) 11:32, 12 August 2011 (UTC)[reply]
Biologically speaking, we've always known that there is a way to get fresh new skin out of old people: just combine sperm and eggs and let them do their thing... So it's definitely possible to get fresh new skin again, the only question is how to actually do it. But obviously nobody has licked this problem yet. Wnt (talk) 14:48, 12 August 2011 (UTC)[reply]
As I understand it, the stiffening of the skin results mainly from the formation of advanced glycation end-products due to glycation of collagen -- in other words, due to cross-linking of the collagen fibers that mainly give skin its mechanical strength. These deterioration end-products are very stable, and there is currently no good way of removing them or preventing them from building up over time -- researchers are working on a variety of things, though. Looie496 (talk) 15:22, 12 August 2011 (UTC)[reply]

Circuit training[edit]

What does circuit training do and how effective is it as part of a strength/mass building and endurance improvement programme? Thanks. Clover345 (talk) 12:17, 12 August 2011 (UTC)[reply]

See Circuit training. --Jayron32 12:32, 12 August 2011 (UTC)[reply]

relativity of time[edit]

Velocity of light is medium dependent.Then how can relativity in time be proved on its basis.being medium dependent it will travel at different speeds in different media. Suppose there exists a huge mass of water in between path of light being observed by us. If we are unaware of such mass, the change in wavelength observed can be wrong? — Preceding unsigned comment added by Shubahmdiwe (talkcontribs) 15:27, 12 August 2011 (UTC)[reply]

If the light passed through a huge mass of water, we would be able to tell by spectrum analysis. thx1138 (talk) 15:33, 12 August 2011 (UTC)[reply]
Um, well, if the light passed through a huge mass of water, we would be able to tell by seeing a huge mass of water where the light is coming from. Looie496 (talk) 16:11, 12 August 2011 (UTC)[reply]
What do you mean by "relativity in/of time"? What change in wavelength are you talking about? --Tango (talk) 18:39, 12 August 2011 (UTC)[reply]
I guess the OP refers to astronomical Redshift as change in wavelength. For a long time the observed speed of light in (what we thought was) a vacuum was explained by an all-pervasive medium called Luminiferous aether, not water. But trying to interpret the Michelson-Morley experiment's negative result persuaded Einstein that the Luminiferous aether idea was unworkable and instead he gave us the Theory of relativity. Cuddlyable3 (talk) 19:08, 12 August 2011 (UTC)[reply]
The frequency of light (or any other wave) doesn't change when it passes from one medium to another. The only effect in known physics that can cause a uniform frequency shift is the Doppler effect. The wavelength does change when entering a medium, but it changes back when leaving it. -- BenRG (talk) 17:04, 13 August 2011 (UTC)[reply]

Instantaneous travel[edit]

Since the Lorentz factor increases without bound as you approach c, could you not theoretically travel anywhere in the observable universe seemingly instantly (from the viewpoint of the traveller) provided you could go fast enough? --Goodbye Galaxy (talk) 18:57, 12 August 2011 (UTC)[reply]

No. Cuddlyable3 (talk) 19:09, 12 August 2011 (UTC)[reply]
The word 'instantly' is going to give you trouble here. As a traveler approaches the speed of light, their percieved travel time decreases substantially. However, only at the speed of light limit would the trip be perceived as instant. The traveler would perceive the universe to be shrunk in their direction of travel (also by the Lorentz factor I seem to remember). All sorts of complications arise when one is coming back, because their velocity must change. This is discussed in the Twin paradox article. Rosilisk (talk) 19:12, 12 August 2011 (UTC)[reply]
Also, Unless you are willing to go past 1g of acceleration to get to c, it is going to take you about 350 days to get up to speed, and the same to get back down to a stop, the majority of that time not seemingly instantaneous. Googlemeister (talk) 19:20, 12 August 2011 (UTC)[reply]
Sorry, when I said "seemingly instantly" I meant to the average human. I'd call anything less than a second "seemingly instant," but yes, that should have been more clearly defined. 1 second to get to the Andromeda. --Goodbye Galaxy (talk) 20:11, 12 August 2011 (UTC)[reply]
As far as I'm concerned, that's theoretically possible. Getting up to that speed, of course, is completely unfeasible with any imaginable technology. However, speed could be calculated that would allow the traveler to perceive such a fast travel. If you're keen to get an exact number, I could crunch it out. Rosilisk (talk) 20:25, 12 August 2011 (UTC)[reply]

To traverse the 2.5 million lightyears to Andromeda in one second, you need to travel at a gamma factor of 7.88*10^13. At that speed, the cosmic background radiation photons that are colliding with the spacecraft will have a typical energy of 100 GeV. If all the energy from these photons is abosorbed by the spacecraft, the total power per unit area is approximately 2*10^22 Watt/m^2. Count Iblis (talk) 00:00, 13 August 2011 (UTC)[reply]

Googlemeister's calculation that you can get up to the speed of light in a time of c / 1g = 350-ish days is incorrect. In order to keep from getting squished like a bug, you need the 1g to be proper acceleration, not acceleration as measured in the Earth's (or sun's) reference frame. As your speed relative to the Earth increases, 1g of proper acceleration becomes equivalent to an increasingly smaller acceleration as measured in the Earth's reference frame, such that you can never actually reach a speed of c.

An object moving with constant proper acceleration like that is said to undergo hyperbolic motion. More details about hyperbolic motion is discussed at Rindler coordinates.

A trip which starts out at rest on Earth, accelerates at a constant 1g proper acceleration halfway to the Andromeda galaxy, then decelerates at a constant 1g proper acceleration for the other half of the trip, will take about 28.6 years of proper time. Plug the 2.5 million light-year distance to the Andromeda galaxy into the "Long Relativistic Journeys" calculator here. Red Act (talk) 00:32, 13 August 2011 (UTC)[reply]

The math is simple. If you accelerate from rest at a uniform acceleration a, then after a proper time of τ, you'll be at coordinate position (a (cosh aτ − 1)) and coordinate time (a sinh aτ). Solving the first one for τ, to go a distance d takes a proper time of (1/a) cosh−1 (1 + d/a). With factors of c back in, that's (c/a) cosh−1 (1 + dc²/a). More "realistically" you have to accelerate half way and decelerate the other half, in which case it takes (2c/a) cosh−1 (1 + dc²/2a). Coincidentally, c/g, where g is Earth gravity, is very close to one year, so at one gee acceleration it takes (2 years) cosh−1 (1 + d / 2 light years). For large arguments, cosh−1 is very close to ln (natural log), so for large distances that's τ = (2 years) ln (d / 2 light years). Plug d = 2,500,000 light years into that and you get τ = 28 years. There are two problems that make this impossible in practice: lack of fuel and the fact that space isn't empty (see Count Iblis's post re the second one). -- BenRG (talk) 16:52, 13 August 2011 (UTC)[reply]

libya[edit]

why dont Apache helicopters in libya seem to have the same cameras that they did in iraq that highlight people in white like this http://www.youtube.com/watch?v=G3CnpGwsFL8&feature=related

That's an infrared camera used for night-viewing. The video you saw from Libya was likely taken during the day, when infrared isn't necessary or as useful, since people tend not to be any hotter than the background in Libya during the day. StuRat (talk) 19:13, 12 August 2011 (UTC)[reply]

no it was at night — Preceding unsigned comment added by Von1235 (talkcontribs) 19:48, 12 August 2011 (UTC)[reply]

It would help if you could link to the video in question. One possibility is that the background was the same temperature as the people, which would make them no longer "glow" on an infrared camera. It being summer in Libya, I'd expect this to be the case. If the background was hotter than the people, they would appear darker (although some cameras might have the ability to invert brightness to make them appear brighter again). StuRat (talk) 20:10, 12 August 2011 (UTC)[reply]

heres 1

http://www.telegraph.co.uk/news/worldnews/africaandindianocean/libya/8556413/Night-strikes-by-French-Tigre-helicopters.html — Preceding unsigned comment added by Von1235 (talkcontribs) 13:31, 14 August 2011 (UTC)[reply]

OK, that just doesn't seem to have close-up images where you could make out individual people (I think I may have seen some, but they could also just be poles at that resolution). Interestingly, it does show the operator inverting brightness a couple times. StuRat (talk) 16:55, 14 August 2011 (UTC)[reply]

Rationale against antistars and antigalaxies[edit]

I have read (and yet admittedly not understood) many responses to this question, and yet I don't seem to see a clear reason why we (as in most particle physicists) think that all of the antimatter was completely destroyed shortly after the big bang. The best reasoning I saw is here http://en.wikipedia.org/wiki/Baryon_asymmetry#Regions_of_the_universe_where_antimatter_predominates , however it seems to me easy to speculate that, especially since galaxies may move apart faster than the speed of light, and also that through years (billions of them) of interaction that if there were regions of anti-matter dominated space, or an antimatter galaxy, that the 'average density of intergalactic space' is less here and that matter and antimatter were no longer interacting. This would leave, say, an antimatter galaxy with similar gravitational effect and light production as other galaxies, and perhaps a potential solution to the Baryon Asymmetry problem.

My question is, IF this were the case, and there was an antimatter galaxy with no matter-antimatter cancellation front, would we be able to detect it? Would its emission spectrum be different? Would it cause perhaps some other noticeable difference, or would it be completely indistinguishable from a normal galaxy without a cancellation front? — Preceding unsigned comment added by Ehryk (talkcontribs) 19:19, 12 August 2011 (UTC)[reply]

An anti-galaxy would be identical to a normal galaxy. But I don't see how the cancellation front could be avoided. It's also not easy to explain how the matter would have gotten separated from the anti matter since they are normally produced in pairs. The Baryogenesis article talks about the three conditions necessary for the observed asymmetry to form. These conditions were first stated by Andrei Sakharov and are called Sakharov conditions.
  1. Baryon number B violation.
  2. C-symmetry and CP-symmetry violation.
  3. Interactions out of thermal equilibrium.
Dauto (talk) 05:26, 13 August 2011 (UTC)[reply]

Not seeking for answer, but explanation[edit]

prolate spheroid for explanation by medeis

Why do atoms have a cone (or paraboloid) shape under electron microscope(Do you have any answers but"it's just the way it is"?)?--Irrational number (talk) 19:42, 12 August 2011 (UTC)[reply]

I believe it is not the electron microscope, but the Scanning tunneling microscope which can resolve atoms. Perhaps our article about them or some of the sources cited in that article will provide an explanation. Jc3s5h (talk) 19:50, 12 August 2011 (UTC)[reply]

that's right, scanning tunneling microscope, that's what I meant...--Irrational number (talk) 19:54, 12 August 2011 (UTC)[reply]

I looked in that article but didn't see anything cone shaped. Dauto (talk) 05:12, 13 August 2011 (UTC)[reply]
Neighboring atoms form chemical bonds with each other; this can cause the electron clouds to develop a distinctly non-spherical appearence. An STM just probes the overall electronic structure of the entire material; the bonds cause the electron cloud to "spread out" a bit, which could kinda, sorta, maybe be interpreted as a roughly "conical" shape. You don't see a bunch of isolated spheres; but then again you shouldn't. --Jayron32 05:18, 13 August 2011 (UTC)[reply]

You mean it could be the effect of the microscope itself?"observer effect"?(and Dauto,there are no cone-shaped in the article, but I suggest a google search,too like [[3]])--Irrational number (talk) 14:40, 13 August 2011 (UTC)[reply]

If you read that article you just cited, it itself contains citations of earlier articles; you could likely find the information you seek by following those trails. Or you could read the very same article you cited, since it provides a nice explanation for the cone shapes. --Jayron32 17:17, 13 August 2011 (UTC)[reply]

Sorry, but there wasn't any information about what I wanted in the very same article, but I'll read the references... — Preceding unsigned comment added by Irrational number (talkcontribs) 18:30, 13 August 2011 (UTC)[reply]

The image is generated by the reflection of particles which are presumed to travel in straight lines. If you were looking down from above on a beachball upon which you were dropping tennisballs, you would only detect tennisballs which bounced back up at you from the upper surface of the beachball. Tennisballs that hit the lower surface would not be able to come back around the "equator" of the beachball, so their lower surfaces would remain obscure. Basically the conical appearance reflects the fact that there is no way, given the size of the observed particles and the probes being used to see them, to view the underside of the atoms. μηδείς (talk) 19:21, 14 August 2011 (UTC)[reply]

That's very very far indeed from my reading of Scanning tunneling microscope. Do you know what you're talking about, or are you making it up? --Tagishsimon (talk) 19:25, 14 August 2011 (UTC)[reply]
What particularly do you take exception to, my use of analogy? No, the atoms aren't beachball-like spheres with discreet surfaces. But if you had to guess what a beachball in a dark room looked like by dropping tennisballs on it and recording the scatter pattern you would get something that looked like the results you get from this type of observation with the what I would call egg-crate like appearance caused by the relative indelicacy of the probe compared to the observed object. That's not objectionable to you, is it? (You could also imagine trying to guess the shape of a rack of billiard balls in the dark by running a whiffle-ball bat over them.) μηδείς (talk) 21:30, 14 August 2011 (UTC)[reply]
In particular: the reflection of particles analogy, and the suggestion that what is measured is only that which is directly beneath the tip of the probe. I read the article to say that changes in current are being measured (which admittedly is a function of electronics) and I imagine that the current changes in relation to the substance happening to be close to the tip rather than merely beneath it. --Tagishsimon (talk) 22:35, 14 August 2011 (UTC)[reply]
Okay, I see what you mean. I was trying to leave it simple and visual by saying that you can't distinguish the underside of the atoms with the type of probe being used. To emphasize what you are specifying, you could imagine the image as a graph of the vertical area of the atom under the probe. Imagine an atom as a spherical tomato on a cutting board. You will cut it into a large number of thin slices while holding it upright on the board. Each slice will slide down to the board, so that, viewed from the side, the middle slice will still have the same height, while the slices on each end will drop down to the cutting board. The mass of slices will go from being a sphere to a shape with a flat bottom and a rounded peak of a top. You will now end up with something that looks from the side like the long half of a plastic easter egg shell with its flat surface on the cutting board, half of a prolate spheroid. This will approximate the long upper half of a plastic easter egg. Hence the "cone" shape you see. μηδείς (talk) 00:43, 15 August 2011 (UTC)[reply]

small desalinization plant cost[edit]

I would like an example of the construction cost of a desalinization plant for a small town (around 1,000 people using ocean water as the input). Location and power consumption of that plant would also be informative. Googlemeister (talk) 20:47, 12 August 2011 (UTC)[reply]

I imagine that would greatly depend on the technology used, which in turn would depend on the location. If you are on a vacant coast in a desert area, I would think a rather inexpensive solar-powered desalination system could be used, where water flows in at high tide, is heated by sunlight and evaporates, then cools at night and condenses for collection. If, on the other hand, you are on the coast of Antarctica and need it to work in winter when both sunlight and fuel might be scarce, and you need to pump it from a depth below where ice forms, then you might need to build a nuclear reactor. StuRat (talk) 00:25, 13 August 2011 (UTC)[reply]
You might be interested to read about Reverse Osmosis Water Purification Unit and Reverse osmosis plant. Dbfirs 16:49, 13 August 2011 (UTC)[reply]

The hourly cost of second-hand smoke?[edit]

Is it possible to do a back-of-envelope calculation of the hourly cost of secondhand smoke? It'd be much appreciated. 66.108.223.179 (talk) 23:05, 12 August 2011 (UTC)[reply]

Hourly cost to who? Looie496 (talk) 23:29, 12 August 2011 (UTC)[reply]
According to a site that offers medical advice, to which you will not see a link here, in Scotland 850 to 950 deaths were the result of the exposure to second hand smoking among non-smokers in 2000. Since the site also sells a homeopathy product it must be very reliable. Cuddlyable3 (talk) 23:55, 12 August 2011 (UTC)[reply]
Another joke ? You really need to add a smiley so people won't take you seriously by mistake. StuRat (talk) 00:20, 13 August 2011 (UTC) [reply]

Deaths of people older than about 50 lead to savings for society. Count Iblis (talk) 00:35, 13 August 2011 (UTC)[reply]

That rather depends upon the manner of death, no? I was under the impression that ridiculous measures taken to extend things by a couple of weeks, days, or hours was a huge burden on the health service. Am I mistaken? Egg Centric 00:44, 13 August 2011 (UTC)[reply]
Smokers have had the cost worked out (typically they take more from the health service – which is a different situation in countries with a free-at-the-point-of-use system). This says that people in the US have to spend $10 billion in medical bills and lost work. Some of that money stays in the system, depends what sort of 'cost' you want. That's £6.1 billion/year; 5 to 6 pence (8 to 10 cents) per person per day. Grandiose (me, talk, contribs) 11:09, 13 August 2011 (UTC)[reply]
One also has to consider the lower life expectancy, leading to less pension payments. Count Iblis (talk) 16:32, 13 August 2011 (UTC)[reply]
I've seen that $10 billion figure, but it didn't occur to me to do such a calculation. Thank you. 66.108.223.179 (talk) 20:12, 13 August 2011 (UTC)[reply]