Wikipedia:Reference desk/Archives/Science/2009 December 11

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December 11

What kind of machine is used to sew patches over holes in the middle of the roof of a M.A.S.H. tent?

I can't find anything on how to use a sewing machine to sew a patch in the middle of a large sheet of fabric (50' x 50').

The problem I'm running into is that there isn't enough room between the needle and the base of the sewing machine for that much fabric.

But I've seen patches done in large tents, tarps, sails, blankets, etc.

What's the secret?

(Is there another type of machine that is used for this?)

Simple Simon Ate the Pieman (talk) 03:52, 11 December 2009 (UTC)

There are special "long arm" machines for that kind of thing. Like this one, for example. SteveBaker (talk) 05:56, 11 December 2009 (UTC)

The example machine is said ominously to have "All fear driven hook mechanism". Cuddlyable3 (talk) 21:02, 11 December 2009 (UTC)

The thing I need to sew is a GP Medium (M.A.S.H. tent). See http://www.youtube.com/watch?v=lAlYK3W7FAA The thing is huge, and the rips are in the middle of the roof. I would need a sewing machine with an arm 6 feet or more long just to pull enough of the fabric past the needle to be able to reach the holes!

Is there a machine that can sew a lockstich from one side of the fabric? Or without an arm? If so, what is it called?

Simple Simon Ate the Pieman (talk) 00:49, 16 December 2009 (UTC)

On the number of exoplanets that transit their stars in the milky way

I'm a pretty avid observer of the amateur transit watch community and I was just wondering something. Obviously, astronomers looking for transits are hoping that the rotational plane of the observing planetary system will coincide with our line of sight, thus having planets crossing the light of the star every once in a while giving us much more direct evidence for their existence. To measure the radial velocity of stars we watch how quickly the star moves towards and away from us, and thus, a similar transiting orbital orientation works best for radial velocity measurements as well.

Now my question is, is there a tendency for planetary ecliptic planes to orient themselves in a certain way relative to the orbital plane of the galaxy? Obviously this comes down to the rotation of the star in its early stages, so I could ask then do the rotations of stars in any way reflect their orbit around the galaxy? I couldn't find out the planar angle of the solar system relative to the milky way, but from visual memory it doesn't seem to be very close, which means that this will be much help to astronomers looking for transiting stars, but I guess it's also possible that local planetary systems will have similar planes, due to similar environment, age, etc.

Does anyone have any insight on this? Thanks! 219.102.221.182 (talk) 05:02, 11 December 2009 (UTC)

I doubt there is any special tendency for the plane of the stellar ecliptic to line up with the galactic ecliptic. So the probability of an exoplanet eclipsing it's parent star basically comes down to it's orbital radius and the diameter of the star (and to a much, much lesser extent, the diameter of the planet). The angle through which the orbit will occlude the star is arctan(starRadius/planetaryOrbitRadius)x2.0 - if the angle of the star's ecliptic is random then you can easily figure the probability that it'll happen to line up with the position of the earth. So to answer your question - we'd need to know the distribution of diameters of the stars in our galaxy - but we'd also need to know the statistics of the planetary orbital radius...and there's the problem. We haven't got that information until we've already found the exo-planets! We could probably make a guesstimate.

So if we said that our sun was typical - and we were looking for planets out as far as (say) Saturn - then that eclipse is only visible over an angle of about 0.06 degrees...out of 180 degrees. That's about a one in 3,200 chance. There are a lot of stars out there - so there are a lot of chances - and lots of planets are a lot closer to their stars than Saturn - so the odds are probably a lot better than that. This is a worst-case - where the plane of the stellar ecliptic is completely random. If there is some tendency for the stellar ecliptic to line up with the galactic ecliptic - then for more distant stars - that greatly increases the probability of eclipses - but for closer ones - it decreases the probability. SteveBaker (talk) 05:36, 11 December 2009 (UTC)

Thanks! 1 in 3200 chance, that would actually be quite a bit better than I had expected (for random planes). Is there any particular reason you feel that the planes should be random? The fact that the solar system, earth, and most satellites occur on the same plane would seem to hint at a trend, and I have no reason to assume in the other direction. Though I have to say I totally didn't realize the consequences this would have for near star systems... I guess I was thinking too 2-dimensionally. Also it's worth noting that if the sol system isn't nearly planar with the galaxy, but there indeed is a general trend for other planetary systems to be, that could also lower the odds of catching a transiting planet considerably. 219.102.221.182 (talk) 06:34, 11 December 2009 (UTC)

It is also worth mentioning that gravitational microlensing techniques such as the proposed The Galactic Exoplanet Survey Telescope (GEST) are expected to be good for small planets some distance from the star. Both observed transits and radial velocity techniques require a large planet close to the star (which microlensing in not very good for), hence microlensing is expected to find many more planets and of a smaller size than previous methods. SpinningSpark 10:46, 11 December 2009 (UTC)

I agree that one in 3200 seems like unexpectedly good odds - I tried to make it come out worse - but the numbers insisted and we must obey! It's not so much that I have a reason to believe that the ecliptic planes of stars should be random with respect to the plane of the galactic ecliptic - so much as that I can't think of any reason why they shouldn't be random. (Maybe that's the same thing?!) The plane of the Milky Way galaxy is at 60 degrees to the Sun's ecliptic - so unless we're rather special - I'm pretty sure there is no correlation. However, if there is a preferred direction then that's a problem. The deal is that the galactic spiral is quite thick - maybe 1000 light-years. So all of the stars within about 1000 light years of us are much more likely to be above or below us as than they are to lie in the same plane. If there is some tendency for planetary disks to lie in the galactic plane (meaning that our sun is weird in that regard) - then we'd be unable to see any eclipses for almost all of the stars within about 1000 ly. That would be bad news for astronomers because the really nice, easy-to-measure stars are going to be the closest ones...and those would be the problematic kind. However - as I've said - there doesn't seem to be a good reason for them all to line up like that (and the Sun certainly doesn't) - so I think the 1:3200 number is about right...at least for stars the size of the sun with planets at the distance of Saturn. Of course, exo-planets that are closer to the parent star will be seen to eclipse their star over a wider angle - and planets further from their star - less so...so the 1:3200 number is just a ballpark figure based on our Solar System. Also - some of those eclipses will be briefer in duration than others if the planet only just eclipses the very edge of the star - so nice long-duration eclipses would be rarer. SteveBaker (talk) 13:48, 11 December 2009 (UTC)

Check out Methods of detecting extrasolar planets#Transit method - it has some of the probabilities. For an Earth-like planet in an Earth-like orbit (which are the most interesting), it's about 1 in 200. Pretty good odds. --Tango (talk) 14:00, 11 December 2009 (UTC)

Binoculars in Games & Movies.

When game makers and movie makers want to depict the idea that we're seeing the world through a pair of binoculars - they often use the trick of masking off the edges of the screen with a pair of overlapping circles kinda like this:

  #################################
  #######      #######       ######
  ####            #            ####
  ###                           ###
  ###                           ###
  ####            #            ####
  ######       #######       ######
  #################################

I'm pretty sure that out here in the real world, binoculars don't look like that when they are properly set up (sadly, I don't own a pair to try) - it's really just a single circle. Is this a true statement? ...and (because I need to convince some people about it today) is there a cogent explanation as to why that is the case.

Finally - is there a name for this kind of mask - maybe some kind of movie jargon?

TIA SteveBaker (talk) 13:32, 11 December 2009 (UTC)

You are correct that only a single circle is seen in the real world, however that would waste a lot of screen real estate (cf a "gun scope / crosshairs" view as seen in any typical James Bond film opening.) As far as terminology, in Final Cut Pro, they just use the term "binoculars filter". --LarryMac | Talk 13:50, 11 December 2009 (UTC)

Yeah - and the 'real-estate' issue is under debate too. I'm building a computer graphics simulation where realism is very important - so loss of screen real-estate is taking second place to "getting it right". What I may do is to use a single circle that's only a little bit narrower than the screen - and let it cut off at the top and bottom. A compromise solution.

So what is a convincing explanation (to my Producer - who also doesn't have a set of binoculars at hand right now - and who wants a double-circle) of why we only see one circle?

SteveBaker (talk) 14:09, 11 December 2009 (UTC)

Steve, if this is for work, this is the perfect occasion to buy a pair of binoculars on company expense ;-). --Stephan Schulz (talk) 14:13, 11 December 2009 (UTC)

It's all about frame of reference... The point of looking through binoculars is so that both of your eyes can experience the visual of the down-field magnification. Your brain, doing what it always does with information from your eyes, stitches the two circles together to form one stereoscopic view. A looking glass, spotting scope, monocular, etc. are all single optic versions of the same thing, if you aren't interested in the stereoscopic view. The answer that I am getting to is that there is no good computer screen approximation for the function of binoculars, unless you have a stereoscopic display of some sort, so "getting it right" with binoculars is kind of out of the question. If you want my opinion, either go for the movie cheeze double bubble view, change the device in the game to a monocular and make it a circle, or make it a computerized spotting device of some sort that doesn't use discrete optics but still has the wide form factor of binoculars, so that the onscreen representation can be 4:3 or 16:9 or whatever it is that you are going for. --66.195.232.121 (talk) 14:29, 11 December 2009 (UTC)

Actually, I can't change the device. What we do at is "Serious Games" - using games technology and game development techniques to produce simulations for training real people for the jobs they actually do. They could be firefighters, county sheriffs, black-ops guys, campus security people...you name it. But if what they carry is binoculars - we have to simulate binoculars - the actual binoculars they'd really have with the right field of view, depth of field and magnification. We don't have the luxury of being able to make it into some kind of futuristic gadget. Whatever that person would be issued with in reality is what we'll give them...to the degree of fidelity possible with the computer we're running on. SteveBaker (talk) 22:44, 11 December 2009 (UTC)

...and therefore no good computer screen approximation for the function of eyes. 81.131.32.17 (talk) 17:21, 11 December 2009 (UTC)

(ec) Try this one on for size — even without special equipment, you already have binocular vision. You're looking through two eyes, but you only see one image. The brain is very good at fusing two images into one, and the same merging process happens when you use a pair of binoculars.

If you don't have a pair of binoculars handy, then you can do a quick and dirty demo with a matched pair of hollow tubes. (Note - this is original research. I just tested this with a toilet paper tube cut in half, as it was the only tubular object I had handy.) Hold the tubes side-by-side, directly in front of your eyes. You want the center of the tubes to be as close as possible to being in line with the pupils of your eyes.

Now, focus your eyes on an object in the distance. Notice how the inner surfaces of the tubes (as seen from each eye) get merged together, and you have an apparently circular field of view? Presto! You may have to wiggle the tubes a bit to get the positioning right, but without a pair of binoculars it's probably the most convincing demo you're going to get. That said, if your boss wants the double-bubble silhouette, just give it to him/her. Or try to persuade him to equip the in-game character with a spotting scope, telescope, or other single-eyepiece device. TenOfAllTrades(talk) 14:37, 11 December 2009 (UTC)

I think the important point is not so much that your eyes can fuse the two images, since that doesn't preclude the sort of "double-bubble" effect, but that the fields of vision provided by the binoculars to each eye is nearly identical; they almost completely overlap. In contrast, your normal vision without binoculars is much closer to this double-bubble thing, since the left side of your field of vision is only seen by the left eye and the same with the right. Only the center area is in the overlap seen by both. With binoculars you adjust the position of the two telescopes specifically so that they provide each eye the same view in order to get binocular vision of what you're looking at. Rckrone (talk) 15:28, 11 December 2009 (UTC)

Surely the why-for is so that it is instantly clear to (most) people that the vision we are seeing on screen is (as if) 'through binoculars'. Binoculars defintely show just a single-circle when used - though if you adjust the 'width' you can make it look more like a side-ways laid number 8 (infinity sign?) too. I'd imagine it's a simple short-cut by film-crews to make it clear what we're supposed to be seeing, and as others not it takes away less of the view on screen than a circle would. 194.221.133.226 (talk) 15:32, 11 December 2009 (UTC)

To answer the question of what this is called in cinematography, if it is done with "soft" edges, as it invariably is, it is called a "vignette" (and the technique is called "vignetting"). If the edges are hard it is simply called mask/masking. SpinningSpark 16:06, 11 December 2009 (UTC)

The problem with the game/movie thing is that with real binoculars you can actually see depth, and know the difference between a binocular image and a spyglass image. In a game/movie, you cannot, and cannot know the difference otherwise. So you're already going to have to sacrifice the biggest "reality" aspect of binoculars for your game (seeing in 3-D) just by the nature of it (unless you are working on something a bit more cool than I am guessing). At some level, the amount of "reality" is somewhat arbitrary, given how much you are already abstracting. --Mr.98 (talk) 16:15, 11 December 2009 (UTC)

You might try showing a movie clip of an accurate depiction, the only one i can recall is from The Missouri Breaks.—eric 17:08, 11 December 2009 (UTC)

This problem is equally as challenging as asking somebody to describe the "shape" of their field of view without binoculars. Nimur (talk) 17:21, 11 December 2009 (UTC)

Well, I learned from this pseudo-educational host segment from MST3k, they're called "Gobos" or "Camera Masks". WP's articles don't seem to fully back this up. But google shows me that that "gobo" is at least sometimes used in this context. Google also seems to suggest "Binocular masks" APL (talk) 17:54, 11 December 2009 (UTC)

As a compromise, when you switch to the binocular view, you could start with two separate circular images (arranged like a MasterCard logo, with the two images identical) that converge to a single circular field. The action should be sort of irregular, with some jerkiness and overshoot. This would mimic someone picking up binoculars and adjusting the interpupillary distance to their eyes, and I think it would clearly depict "binoculars" to the user. -- Coneslayer (talk) 18:57, 11 December 2009 (UTC)

What a touching lament "sadly, I don't own a pair to try" so close to Christmas.... Cuddlyable3 (talk) 20:58, 11 December 2009 (UTC)

(No,no,no! Do NOT confuse Santa. I carefully wrote my Xmas list (in my best handwriting) and shoved it up the chimney already - what I want is a Kindle - I don't own binoculars because I neither want nor need binoculars! I have not been naughty this year - at all - ever.  :-) SteveBaker (talk) 22:44, 11 December 2009 (UTC)

What never? If you ask for a pair of binoculars Santa won't think of an unauthorised verse sung to the melody of the Eton Boating Song

It was Christmas eve at the harem

All of the eunuchs were there

watching the beautiful houris

combing their pubic hair

Just then from the regal apartment

the sexy old sultan calls

What do you want for Christmas boys?

The eunuchs as one shouted Balls.

Cuddlyable3 (talk) 00:02, 13 December 2009 (UTC)

You shoved a piece of paper up the chimney? You won't get a Kindle, you're getting kindling! -- 128.104.112.87 (talk) 15:59, 14 December 2009 (UTC)

Coneslayer's idea sounds good. I could also suggest that at the edges of the circles you blur it a bit and darken. You could even have the whole picture go out and in focus a couple of times, increase the chance of missing something in the view. And don't forget to ray trace the internal reflections off the lenses if you are looking near something very bright! And there would be a little bit of unsteady shaking. Graeme Bartlett (talk) 21:54, 11 December 2009 (UTC)

I like the idea of doing a quick bit of faked "adjustment" - but maybe only the first time you use the binoculars in the game...it wouldn't be cool to do that every time they picked them up though! There is a fine line between "cool effect" and "bloody annoying"! SteveBaker (talk) 22:44, 11 December 2009 (UTC)

Although if you are going for realism "bloody annoying" may be more realistic than the "cool effect". Ideally (i.e. for maximum realism) anytime the binoculars are used straight out of their case (i.e. from folded) the faked adjustment should happen. Also there should be a little bit of focusing delay when shifting views to objects at different distances - similar to what Graeme suggests. Reading about what you do (cool job by the way) simulating this usage delay could be fairly critical for some stuff. For example, if a police officer gets in the habit of keeping multiple people in view just by swinging their binoculars around, they are going to be unprepared for the need to constantly refocus when doing this in the real world. I speak from birdwatching experience, keeping multiple subjects at different distances "under surveillance" is quite challenging. Putting focus delay into the simulation would train people to consider carefully how many subjects they can watch at once, an annoying but important real world limitation. Apologies if you are way ahead of me on this one or I am getting too pedantic. 131.111.8.99 (talk) 01:12, 12 December 2009 (UTC)

We have 'subject matter experts' who we consult about small details like this. It's always interesting to discover what features matter and what don't. For example - play one of the very latest combat games: Call of Duty: Modern Warfare 2 - as you're walking around in "first person" mode - look for your shadow. You'll be surprised to find that while every other character in the game casts a shadow - you don't! It may not be immediately obvious why this is important - but if you are hiding around the corner of a building - hoping that the bad guy doesn't know you're there - in the real world, your shadow can give you away...if you have a shadow! You'd like someone who is trying to do that to pay attention to where their shadow is falling - so CoD isn't a great training aid (for lots of reasons - not just this). It's actually surprisingly difficult to produce a good shadow in a first person game because the animation of your avatar is hard to get right when you're doing joystick/mouse-driven actions while the pre-scripted "run", "walk", "shoot" animations are playing...so they just don't bother! We don't have that luxury to just ignore things when they are difficult. SteveBaker (talk) 04:10, 12 December 2009 (UTC)

He is right though, and I think an animation is the only solution to this. There have been games in the past that go for pseudo-realism using similar tricks. Heck, Counter-Strike tries to make the whole "pull a machine gun out of your pocket" thing a little more realistic by varying the time it takes to get the gun loaded and ready, although the animations can generally be overridden by shooting. If annoyance is something you want to avoid, then you could always just shorten the animation after the first use. I don't see any reason it would be that annoying though, as long as the zoom settings don't reset every time you use it, it shouldn't take that long to get ready. 219.102.221.182 (talk) 23:48, 13 December 2009 (UTC)

Need a reliable value for the sun's illumination under variable Earth conditions

Firstly, I am after approximate ranges for the sun's illumination on earth and in orbit under various conditions (clear day, overcast, etc.) for the purpose of calibrating CG lights in a 3d computer application. The confusion has arisen as a result of trying to research a good value for an overcast day; I tried using Google for that and encountered wildly varying values with some surprising implications. I have already looked at the Wiki article on Sunlight for these values. The opening section of the article gives an (uncited) value for the sun's illumination of 100,000 lux at the Earth's surface. Now, this sounds all well and good adn would match the default ranges for sun objects in the doftware package I am using. However, I then ran into, for example, these values on this page; the value given for 'direct sunlight' is indeed 100,000 lux, but it then describes 'full daylight' as 10,000 lux, and an overcast day as 1000 lux. This page gives the same values without specifying the 10,000 lux value for 'full daylight'. The implications of the first page are that, if direct sunlight is 100,000 lux, and 'full daylight' 10,000 lux, then the Earth's atmosphere absorbs or reflects 9/10ths of the sun's light on a clear day (!), and that an astronaut in orbit would be receiving 10 times the amount of light as someone on the Earth's surface; and that a cloudy overcast day provides around 1/10th or 1/100th the illumination of a sunny day (depending on which figures you look at), and 1/100th the illumination in orbit. . . All this sounds surely wrong to me. Firstly I was under the impression that, as the Wiki article suggests, the 100,000 figure is roughly correct for Earth's surface on a clear day, although I am a bit puzzled by the figure of 1/100th that for an overcast day - that sounds a bit too dark. Similarly, I would have assumed that the amount of visible light absorbed or reflected by the atmosphere is comparatively slight (given, for example, Earth's average albedo of around 0.31, ie. 3/10ths), and that lighting conditions in orbit aren't too far off those at Earth's surface on a clear day - brighter yes, but not that bright. Please help if you can, I seriously need some reliable values for these conditions, and with pages contradicting each other I don't know which values I'm supposed to follow. Thanks in advance for any helpful answers. LSmok3 ^Talk 17:14, 11 December 2009 (UTC)

Insolation is usually measured in Watts per Square Meter. The solar constant is pretty clearly measured at about 1370 watts per square meter at the top of the atmosphere (reliable sources are cited in our article); depending on conditions, anywhere from ~ 100 to 1000 watts per square meter reach the Earth's surface; but since you need a luminosity and not an incident power (irradiance), you might want to look at lux and see the conversion procedure. Nimur (talk) 17:26, 11 December 2009 (UTC)

The albedo of cloud is betwen 0.5 to 0.8 (see albedo). So we'd expect the total amount of sunlight reflected back out on a cloudy day to be 50% to 80% of the total sunlight. All that isn't reflected is either absorbed by the cloud - or scattered around so that it ends up down here on the ground as ambient sky-light - so we'd discounting absorption by the cloud - you'd expect the ratio of light on a clear-day to an overcast day to be between 2:1 and 4:1 - with multiple layers of cloud and some idea of what is being absorbed - I could easily believe the 1/10th of sunny day illumination. The 1/100th value seems more dubious...perhaps they are talking about the light coming fron the direction of the sun itself - neglecting the scattered light...that could easily explain the 1/100th number. For patchy cloud conditions between those two extremes, it all depends on whether the sun happens to be behind a cloud or not from wherever you happen to be standing.

Some of the confusion may arise from the fact that the sky isn't black. If you measure the light coming from the direction of the sun alone - you get a much lower value than if you average it over the entire sky. —Preceding unsigned comment added by SteveBaker (talk • contribs)

Okay, thanks for the quick answers, but I'm afriad I'm still a bit lost. Firstly, the watts to lux conversion is a bit beyond me - I'm an artist, not a mathematician: the lux page directs me to the page for the luminosity function, which is required to make any such calculation, and that specifies variables I don't have, like the actual wavelength. Secondly, all I'm after are guideline ranges and averages for a few conditions - Earth orbit, clear day at surface (noon - sunrise/set), and fully overcast (noon - sunrise/set), from which I can approximate all I need. The point about ignoring cloud absorption seems a bit odd, as that is a determining factor in surface illumination based on cloud cover, ie. exactly what makes an overcast day, (and remember, I was only quoting the average albedo of the Earth). I also seem to have run into the same trouble here as I encountered on other pages, namely a lack of any clear definition of terms and contradictory figures. Helpfully, the lux page gives various values for different conditions, but makes a distinction between 'full daylight' and 'direct sunlight' (10-25k and 32-130k respectively) and gives the value of 1000 lux for overcast: without any actual definition, I would assume that 'full daylight' refers to the brightest Earth-surface conditions, and 'direct sunlight' to unfiltered, non-atmospheric, ie. orbital conditions. This matches the pages I had found which caused me the confusion in the first place, and suggests that the sun lights in the application I'm using default to outer-space conditions (which is a bit silly; and to quote from the documentation: "Intensity: The intensity of the sunlight. The color swatch to the right of the spinner opens the Color Selector to set the color of the light. Typical intensities in a clear sky are around 90,000 lux."). However, the daylight value links to the Wiki page on Daylight, which also provides guideline values; it doesn't use the term 'Direct sunlight', but gives similar values for 'bright sunlight' and 'brightest sunlight', and gives the value of 10,000 - 25,000 lux for an overcast day: 10-25 times higher than the lux article. So, what am I supposed to follow? LSmok3 ^Talk 18:46, 11 December 2009 (UTC)

Among the problems in converting solar radiant flux into an equivalent "candela" is the difference between specular and diffuse lighting. "Candela" really only applies to an approximately point-source light - but sunlight, whether the day is cloudy or clear, is illuminating the ground diffusely - in other words, the entire sky "glows" and lights the object. So, trying to model the lighting as a single point-source (the sun) is obviously flawed - there is no equivalent brightness or luminosity for the sun which would result in equivalent lighting conditions. In computer graphics, this is usually dealt with by applying an ambient lighting condition - a "uniform" illumination from a particular direction. Alternately, you can model the sun as a point-source at a great distance, and then model the atmospheric diffusion - but that will be very computationally intense, and will yield about the same visual effect as an ambient lighting source. Nimur (talk) 19:58, 11 December 2009 (UTC)

Yes, I am well aware of that issue. And no, I'm not looking for the candle power of the Sun. Indeed, in modeling a daylight system (including the IES Sun and Sky system I quoted the documentation for above), two lights are generally used, one to represent the contibution from the sun as a point source (ideally Photometric) light (although in recent years that's tended to be an area light purely to allow for the modelling of realistic shadows due to scale and diffusion), and another to represent diffuse scattered 'fill' from the sky. But that doesn't mean I don't need realistic values, especially given that the systems may mix photometric with non-physical lighting, and with global photometric controls the relationships between all lights in a scene are affected (for example, street lighting at dusk would feature a sun and sky system in combination with artificial lighting, so the relationships must be correct). (And in fact, the reason I was originally looking for realistic lux values for an overcast day - the whole thing - was to calibrate just such a fill to realistic proportions, bearing in mind that it may also be photometric; again, if the relationships aren't accurate, not only is it bad practice, but problems would also occur if, say, I was animating a shift from clear to overcast conditions.) Also, don't assume I am only interested in Earth surface figures; as I already said, I also need realistic values for space scenes. So are there no reliable values that actually agree for sunlight in lux? That seems a bit strange. It should be possible for just about anyone to get such figures by going outside under the given conditions armed with a light meter set to give values in lux and measuring at exposed ground - I'd do it myself but I don't own one, and can't help assuming those values must surely be well-established; for some reason my documentation suggests values that Wikipedia claims are burn-your-retina space lighting and the reason why astronauts where mirror-visors, and no one seems able to agree on overcast values. LSmok3 ^Talk 20:50, 11 December 2009 (UTC)

Okay, how about Reference luminous solar constant and solar luminance for illuminance calculations, (J. Solar Energy, 2005), or Determination of vertical daylight illuminance under non-overcast sky conditions, (Building & Environment 2009)? Table 2 in the first article specifies a lot of parameters, including luminance of about 1.9x10⁹ cd/m². For the very detailed cases you are considering, you may want to read the entire articles, since you are particularly clear that their assumptions differ from yours. There is an entire section devoted to the assumptions made and the impact these have. Nimur (talk) 22:20, 11 December 2009 (UTC)

Well, I'm afraid the problem with that is the principle of having to pay around $60 for the information, which as I say I would have thought would be pretty well-established. The whole point of Wiki is that information is free, and as it is, I don't even own a credit card or have a paypal account to buy it, especially without being sure it actually contains what I need. Having said that, I did find this link through Google to the same site, which suggests that the values quoted in my software are correct and the ones on Wiki are wrong. So never mind. LSmok3 ^Talk 07:43, 12 December 2009 (UTC)

If you don't have access to the online journals (which are not free), then you can try requesting hard copies of those journals through a library. Unfortunately, not all information is published under a free license, so immediate access to download over the internet is not always possible. Alternately, you can ask your school or library to purchase a web subscription to those journals so you can have immediate access in the future. Nimur (talk) 20:32, 12 December 2009 (UTC)

I actually wouldn't be that surprised if direct sunlight really was 100 times more "bright" then on an overcast day. The sun is literally blindingly bright. And anyway, isn't it one of those logarithmic scales? Something with 10 times the energy only looks twice as bright? So 100 times it appears 4 times as bright? If you are really interested in getting good results, I would be borrowing a light meter and taking measurements myself. Vespine (talk) 03:10, 14 December 2009 (UTC)

trialkyloxonium salts and beta elimination

Most of these are unstable, right? My question is, what happens if you have no beta proton on the oxonium salt? It can't possibly eliminate via the Hoffman mechanism? Will it decompose onto ether + carbocation anyway?

Also, can you undergo transalkylation with ethers and an alkyl halide via SN2?

Are carbamates prone to beta elimination?John Riemann Soong (talk) 22:52, 11 December 2009 (UTC)

Triethyloxonium tetrafluoroborate is pretty stable. Graeme Bartlett (talk) 23:47, 11 December 2009 (UTC)

How does it decompose in water? Does water act like a base? John Riemann Soong (talk) 02:01, 12 December 2009 (UTC)

It would appear that water is acting as a Lewis base. If you do the electron pushing for this reaction:

• [(CH₃CH₂)₃O]⁺BF₄⁻ + H₂O → (CH₃CH₂)₂O + CH₃CH₂OH + HBF₄

You can see pretty easily how an electron pair from water "grabs" an ethyl group from the trialkyloxonium and then loses a proton to the tetraflouroborate anion. --Jayron32 05:28, 12 December 2009 (UTC)

Can tetraalkyl ammonium salts undergo SN2 substitution with a nucleophile instead of beta-elimination? An alkene product seems to be favoured for ammonium salts, but an alkane-alcohol product for oxonium salts? John Riemann Soong (talk) 06:04, 12 December 2009 (UTC)

Both types of reactions are possible. Depends (as usual) on sterics and electronics of the tetraalkylammonium structure and sterics and electronics of the "other" reactant, solvent effects, etc. All the usual concerns for deciding between these two competing reaction modes. DMacks (talk) 07:50, 13 December 2009 (UTC)

I know ammonia is a pretty bad leaving group, but is it a worse leaving group than hydroxide? John Riemann Soong (talk) 18:27, 13 December 2009 (UTC)

Ammonia (NH₃) is stable, so it's a plausible LG. Hydroxide (OH^–) is not stable, so it would only be a viable LG if there were some other large gain in stability in the reaction. DMacks (talk) 04:59, 15 December 2009 (UTC)