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November 13

my camera for my experiment recorded video with a 0.80x resampling factor. When calculating centripetal acceleration and curvature, how do I scale?

I run the video through a program and it plots the trajectories of tracked objects over time (based on timestamps). The only issue, my video is 25% faster than reality. Video is taken for 5 min and 35 seconds, but only 4 min 30 seconds of video is recorded. The video is being time-scaled to match, i.e. it is not truncation. I have since corrected this problem, but I have two months of data that were taken in this "time-compressed" fashion.

Am I right to assume that curvature will remain unchanged, that is, the scale corrections applied to velocity and x' * y - y' * x will cancel out?

However for centripetal acceleration, since it is v^2 / r or v^2 * curvature, and curvature will not change with time scale, my values that I am calculating for centripetal acceleration will be (1.25)^2 faster, that is I have to apply a correction factor of 1/1.56 to my values. 137.54.63.225 (talk) 00:12, 13 November 2012 (UTC)[reply]

(EC) Why would you assume that that the time scaling would cancel out? Have you tried the calculus yourself? I think to say anything definitively, you'd have to know exactly how the camera samples the frames in time. Can you document that it's uniform? If not, then we cannot say for sure without knowing how frames are sampled over time. Also, is this the fruit fly project that IP questioners have discussed here before? SemanticMantis (talk) 02:13, 13 November 2012 (UTC)[reply]

I know it's uniform because VirtualDub explicitly says the resampling rate is 0.80x. When I disabled a time compensation feature (it was because the camera takes 15 FPS in the dark, rather than 30 FPS), it goes back to 1x. Yeah it's the project, but I just want to confirm my intuition. Using dimensional analysis, s^-1 implies one-fold framerate dependence, and s^-2 implies two. Curvature is specified as a inverse length scale, and has no time component, so it won't be affected by a sped-up framerate. However centripetal acceleration has a two-fold framerate dependence, as seen by c = v^2 * k and also looking at the dimensions of centripetal acceleration. Can somebody confirm my intuitions? 71.207.151.227 (talk) 08:35, 13 November 2012 (UTC)[reply]

Chemistry problem

I was wondering how to do part B of the following problem.

1) A saturated solution of sulfur dioxide was prepared at 27 °C and 760 torr (millimeters of mercury). A 5.00-milliliter sample of this solution was diluted and an excess of potassium iodate solution was then added. The iodine liberated was equivalent to 32.80 milliliters of 0.100-molar sodium thiosulfate solution. The reactions that occur are given in the equations below:
5 SO2 + 2IO3¯ + 4 H2O --> 5 HSO4¯ + 3H+ + I2
2S2O32¯ + I2 --> 2I¯ + S4O62¯
(a) Calculate the moles of I2 liberated.
(b) Calculate the volume of sulfur dioxide needed to prepare 1.00 liter of the saturated sulfur dioxide solution.

I got .00164 moles of I2 for part a (I am almost positive this is correct, if someone would like to double check feel free to do so and let me know).

NOTE:

The answer for part B is 40.3 L, but I am not sure how to do this.
Part B is not a limiting reaction, it is a gas law problem, and may or may not involve stoichiometry.
You don't need to write a detailed explanation for your steps, if you just write the equations as you progressed and the answers for them, I should be able to figure it out on my own. If I need more help I'll contact you. Or you can just write the steps you'd take.

Alternatively, if someone can't solve it but can just explain what Part B is asking, that'd be appreciated. SO2 and sulfur dioxide are the same, so isn't 1.00 Liter of satured sulfur dioxide the same as the volueme of SO2? I don't understand what they're looking for.

Please, if anyone can solve Part B and explain their steps here, it would be much appreciated! Or offer even some steps or help. I've tried this multiple times, but have never been able to get 40.3 (and I know this is the answer). --Jethro B 02:45, 13 November 2012 (UTC)[reply]

Part b is ambiguous. Does it assume that all the gas you have is dissolved?--Jasper Deng (talk) 03:46, 13 November 2012 (UTC)[reply]

Part b isn't ambiguous. I'm a little sleepy right now, and just got done a bunch of chemistry tutoring, so I'm a little fatigued, but you can calculate the moles of SO2 in solution using simple stoichiometry, and you can use Henry's law to calculate the partial pressure of SO2 needed to form a solution of SO2 at that concentration. I don't feel much like doing the math, but Henry's law is the key to solving this. You would need to know the Henry's law constant for SO2 dissolving in water, but there are tables of values for that, availible on the internet via google; type "Henry's law constant" into google to find a bunch. --Jayron32 03:58, 13 November 2012 (UTC)[reply]

Unfortunately, I want to do this without Henry's Law, as it was done in 1969 (when the problem was published) without Henry's Law. Henry's Law could work, but it's possible without. Some equations that may be of use here that I can think of are PV=nRT (or v=nRT/PV), and MM=gRT/PV where MM=Molar Mass. --Jethro B 04:22, 13 November 2012 (UTC)[reply]

I'm really not sure how to do it without Henry's law, it is the only way I know to calculate the relationship between the concentration of a gas in solution based on the pressure of the gas over the solution. Also, I'm not sure what 1969 has to do with it, Henry's law had been established for 166 years at that point. It was hardly unknown. Most general first-year chemistry classes teach it today, and though I was not yet alive in 1969, I can't imagine it wouldn't have been covered if the above problem had been given in class. --Jayron32 04:29, 13 November 2012 (UTC)[reply]

Addendum: I thought perhaps that Raoult's law would be applicable here, but that only works at ranges where the solute is below its boiling point. In this case, it isn't, as SO2 is a gas at room temperature. --Jayron32 04:34, 13 November 2012 (UTC)[reply]

Addendum #2: The key here is saturated is the word in the problem that leads me to Henry's law. But OK. Let me play around with the numbers to see what I can get. --Jayron32 04:37, 13 November 2012 (UTC)[reply]

.00164 x 5 = 0.0082 moles SO2. At the above conditions, PV=nRT gives (1 atm)(V) = (0.0082)(0.08205)(300) so V = 0.201843 liters for a 5 ml sample. To prepare a 1-liter sample of the same concentration, you'd need .201843 * 1/.005 = .201843 * 200 = 40.36 L. QED. So you were correct. Henry's law wasn't needed. Does this all make sense? --Jayron32 04:55, 13 November 2012 (UTC)[reply]
- Thank you, it looks good. I'm just going to write out what you wrote here on paper and make sure I see each step, and then I'll let you know. But thank you so much for the help. --Jethro B 04:59, 13 November 2012 (UTC)[reply]
- Yes, I got it. Thank you so much for your help! --Jethro B 05:02, 13 November 2012 (UTC)[reply]

Momentum

Why is p the symbol for momentum? --168.7.228.22 (talk) 06:18, 13 November 2012 (UTC)[reply]

Nobody knows. Seriously. —Tamfang (talk) 06:29, 13 November 2012 (UTC)[reply]

What he said. The other letters were already taken up, and p was still left over. It was basically like being the last guy picked for the kickball team. If it were "m" or "M" it would have been confused with mass, which would have been bad in equations. Since we couldn't have 2 Ms, it got hung on p. Otherwise, there is no other reason. --Jayron32 06:48, 13 November 2012 (UTC)[reply]

If you have as much trouble remembering silly things like that as me, think of it as "pomentum". StuRat (talk) 07:14, 13 November 2012 (UTC)[reply]

If you have less time, think at it as p for 'pull'. OsmanRF34 (talk) 17:38, 13 November 2012 (UTC)[reply]

We are a reference desk, not a plead our own ignorance desk. So here are several references [1][2][3] each claiming that "p" was inspired by the Latin "petere", which is the root of the English word "impulse" (Latin: "impetere"). It is claimed that the concept we refer to as "momentum" was called "impulse" at the time of Newton, and that this led to the choice of "p" to represent this concept (noting that "i" would have been a poor choice since it was already in use for other purposes). Now, I don't know for sure that the "petere" story is accurate, but it seems plausible enough and clearly has some support on the interwebs. Dragons flight (talk) 12:01, 13 November 2012 (UTC)[reply]

"Do You Feel Like An Insider Or Outsider?"

When asked the above question, is this a psychologically sound question? Is it a diagnostic question, or criterion for a DSM illness?Curb Chain (talk) 06:27, 13 November 2012 (UTC)[reply]

Can you define "Insider" and "Outsider"? You may be interested in our articles on solitude and emotional isolation. Someguy1221 (talk) 07:24, 13 November 2012 (UTC)[reply]

Innie and outie? 124.150.45.187 (talk) 09:00, 13 November 2012 (UTC)[reply]

The Outsider (Colin Wilson). μηδείς (talk) 17:37, 13 November 2012 (UTC)[reply]

electrons and time-varying electric fields

There is a free electron placed in an electric field that varies as E = r-> * cos(w*t) where r-> is some constant vector. How do I come up with a complex function for its velocity? If the time-varying electric field is external, won't that set up a time-varying magnetic field that will radiate EM waves away from the source? But now the electron will be moving too, and will be radiating away its own energy. 71.207.151.227 (talk) 09:40, 13 November 2012 (UTC)[reply]

You seem to be overthinking this. Work out the force f=Ee (vectors here). Then work out acceleration of the electron based on force and its mass. Velocity is time integral of acceleration. Assuming this is homework, you can calculate the resultant formula. You may have to consider initial velocity as well. Graeme Bartlett (talk) 11:31, 13 November 2012 (UTC)[reply]

But if the electron accelerates, won't it produce its own changing electric field, that it will respond to? 71.207.151.227 (talk) 12:07, 13 November 2012 (UTC)[reply]

In an introductory physics class, which I assume this is, second-order concerns such as how the electron's field will influence the experiment are almost always assumed to be negligible. Dragons flight (talk) 12:42, 13 November 2012 (UTC)[reply]

I'm studying graduate-level optics, where EM waves are very important. I last took E&M two years ago. I need to know these secondary effects. Eventually, I have to estimate very small effects like light pressure. However, I missed two lectures and now the homework is ver confusing for me. 137.54.43.148 (talk) 19:27, 13 November 2012 (UTC)[reply]

For a lot of electrons as in a plasma you can treat it as a differing dielectric constant k=1-(Ne²)/(ω²mε₀) m-mass of electron ε₀ is permittivity of free space, ω angular frequency N number of electrons per unit volume, e is electron charge. The dielectric constant is real unless you start to worry about collisions between the particles. This is from my textbook on radiowave propagation, where electromagnetic waves meet ionised gas. It says the positive ions can be ignored because they are so heavy. The book does not mention magnetic effects. These will become comparable if the electron is moving at close to the speed of light, which it might do if the wave has extreme intensity. But will not be big normally. To workout magnetic effect you can consider the magnetic field from that electromagnetic field and velocity of electron. However since the electric field is location independent, it is not travelling anywhere, so is not actually an electromagnetic wave. Graeme Bartlett (talk) 11:23, 14 November 2012 (UTC)[reply]

My professor asked me to "employ complex form" to compute the velocity of the electron. But the velocity of the electron doesn't have two components. How is this possible? 76.123.35.31 (talk) 23:04, 14 November 2012 (UTC)[reply]

You can do a Fourier transform of the electric field and the electron location/velocity. The complex result will reflect the amplitude and phase of the electron response. And you can express the electric field with e^jωt and if you want the real part is the real part of that magnitude or cos (ωt). The Fourier transform of this would be δ(ω). And one more clue I will give you is that the velocity of electron is 90° or π/2 out of phase with the field. So that would give you a factor of i, making the velocity purely imaginary. This situation happens in the night time ionosphere with radiowaves being reflected back to earth. When you transform the derivatives or integrals with time factors of ω will pop out, but since I can't think of what they are without looking them up, I won't write them here. Graeme Bartlett (talk) 11:06, 15 November 2012 (UTC)[reply]

0.45 vs 0.2 micron filters (for life sciences)

I'm looking for information regarding the use of 0.45 vs 0.2 micron filters? I guess the 0.45s are cheaper? Can bacteriological sterility be assured with a 0.45 micron filter? — Preceding unsigned comment added by 129.215.47.59 (talk) 10:04, 13 November 2012 (UTC)[reply]

0.2 micron filters are usually considered adequate to ensure bacterial sterility, while 0.45 micron filters would generally not be adequate. There are occasional reports of ultramicrobacteria that may have slipped through 0.2 micron filters as well, but this would be pretty unusual. A 0.45 micron filter will still catch many bacteria, and are sometimes used for that purpose, but they are unlikely to stop all of the bacteria present under typical conditions, and hence the filtered fluid is unlikely to be sterile. Dragons flight (talk) 11:01, 13 November 2012 (UTC)[reply]

Thanks. Why then do we have 0.45 um filters? 129.215.47.59 (talk) 12:49, 13 November 2012 (UTC)[reply]

Bigger holes generally filter faster (everything else being equal), which is an advantage. For some experiments the filter residue (i.e. the material remaining on the paper) is of greater interest than the fluid that passes through, and in those cases the difference between catching all cells and catching most cells may not matter much. For example, if you care about larger cells, such as eukaryotes, then you wouldn't necessarily care what happens to the bacteria. Dragons flight (talk) 13:02, 13 November 2012 (UTC)[reply]

It's also worth pointing out that light scattering samples need to be filtered to remove dust (> 1um) and other static scatterers (really big things) from solution. If what you're looking at is bigger than 200nm than you wouldn't really want to use a 0.2 um filter because you'll be filtering out the thing you're trying to analyze. I guess the take-home message is that such filters aren't used solely for sterilization purposes. (+)H₃N-Protein\Chemist-CO₂(-) 15:06, 13 November 2012 (UTC)[reply]

One manufacturer says that their 0.45 um filter reduces Serratia marcescens titer by a million-fold. [4] One good reason to use them is for "pre-filtration applications", to reduce "bioburden" as they say (or, I suppose, to get rid of precipitates/insoluble matter etc.) In other words, as a first rough filter to keep all these unwanted goodies from clogging up the one you use for your actual sterilization. Wnt (talk) 05:05, 14 November 2012 (UTC)[reply]

Dyson sphere - "Gravity"

Hi,

I have a question regarding the "gravity" caused by a dyson sphere (or ring if you prefer) at the equator due to the centrpetal force of the spinning sphere/ring. I realise that such a structure in unfeasable, but view this just as a thought experiment. Lets assume that the spere is 1 AU in diameter and is spinning at such a rate that the centripetal force is equal to that of gravity, and there are Earth like features (cliffs/mountains etc) on the inside of the sphere. My question is such: What happens if you were to run off the edge of one of these cliffs, would you "fall"? Is this indestinguisable (physically) from the notion of "falling" on Earth (assuming effects of curvature inward/outward are negilgable on both at such a small scale)? Eg. distance travelled, speed attained etc. Also, what happens if you adjust the spin speed of the sphere/ the radius of the sphere with this problem (aside from the obvious heat/cold problems). I am just having trouble imagining this, thanks for any help! Matt 80.254.147.164 (talk) 13:28, 13 November 2012 (UTC)[reply]

Gravitational attraction of any object appears to act from a point source located at the centre of gravity. In the case of a thin narrow dyson ring this means that the centre of gravity is at the centre of the sphere. So if it is rotating just fast enough for centrepetal force (acting outwards) to be equal to the gravitational attraction (acting inwards) at the ring, folk located at the ring will appear to be weightless. If they climb an inward facing mountain, the gravitation distance is less but the rotational speed is also less. So gravity is stronger so they will fall towards the centre. Clearly, if the mountains face outwards, centrepetal force will be the stronger of the two. Gravity is weakening in proportion to the inverse square of the distance to centre and centrepetal force is increasing in proportion to distance to centre, that is, faster than gravitational attraction. So folk on the outside of the ring fall away.

For a dyson sphere, the rotation velocity falls off toward zero asr the rotational poles are approached. So folk on the sphere at the poles can only fall inwards, regardles. Everywhere else, they must be between being balanced as on the equator, or at the poles where they always fall inwards. So, everwhere else, they must fall inwards readless of whether the moutains poke inwards or outwards, unless the mountains approach a height just touching an imaginary cylinder tangent to the equator and of radius equal to the sphere radius. On any moutain exceeding that cylinder, they must fall outwards. Floda 120.145.145.21 (talk) 14:45, 13 November 2012 (UTC)[reply]

For the full sphere, the shell theorem applies, so the gravity of the sphere cancels out. Only the gravity of the sun (and potential planets) is relevant. That does not change the core statement - rotating a Dyson sphere would be a bad way of generating (pseudo-)gravity, since would change from the equator to the axis of rotation. On a Ringworld, the effect of the centrifugal force is very nearly the same as that of gravity on Earth. The major difference is that the Coriolis force affects things slightly differently. But this effect is rarely noticed directly in daily life. It might cause weather to do interesting things in a ringworld. --Stephan Schulz (talk) 15:33, 13 November 2012 (UTC)[reply]

OK, that makes sense, but what if we assume that the mass of the ring is zero and the ring is spinning fast enough so standing on the inside of the ring the centripetal force is "pulling" you towards the ring (ie the star would appear to be directly above you as if you were on earth) with the same force as gravity on earth. Also the height of the "mountains"/"valleys" is negligable compared to the distance to the central star. I am mostly interested with what happens when you jump off a "cliff" on the ring, really. I would think that since you are on in contact with the ring, the only forces acting is the force of the star, so you would orbit it or fall towards it. 80.254.147.164 (talk) 15:27, 13 November 2012 (UTC)[reply]

No, you would still "fall" towards the ring, because even though you're no longer touching it, you are still moving at the same speed as it, due to inertia, so centrifugal force will still act on you. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:29, 13 November 2012 (UTC)[reply]

Okay, call me Mr Pedantic, but someone has to say it. There is no centrifugal force. When you jump off the cliff you move in a straight line at constant velocity (ignoring air resistance and the effect of the star's gravity, which will be much smaller than 1g) until that straight line intersects the ring. You appear to be falling towards the ring but in reality the ring is rising to meet you. Gandalf61 (talk) 17:26, 13 November 2012 (UTC)[reply]

No, Mister Bond, I expect you to die! (as linked above ;-) --Stephan Schulz (talk) 17:42, 13 November 2012 (UTC)[reply]

Given the equivalence principle, the centrifugal acceleration is just as real as the gravitational acceleration that it's intended to replace. -- BenRG (talk) 20:42, 13 November 2012 (UTC)[reply]

That makes much more sense, the idea of the ring exerting force on a body it wasn't in contact with was confusing me. If someone wants to be extra nice , they could calculate the RPM/ velocity of the disc to require an earth like "gravity" - although when my brain is working i'll get round to it 46.64.43.100 (talk) 19:29, 13 November 2012 (UTC)[reply]

Or you could just read Ringworld, which is worth doing anyway. IIRC Niven got 770 miles/second. Of course his world wasn't exactly 1 AU in radius and the surface gravity wasn't exactly 1 g, but there's an order of magnitude for you. --Trovatore (talk) 19:57, 13 November 2012 (UTC)[reply]

To specifically answer what would happen when you fall off a cliff, you would fall as normal. Everything would seem equivalent. Until you fell far enough to notice the curvature of the ring, nothing would seem strange. Someguy1221 (talk) 22:59, 13 November 2012 (UTC)[reply]

This question is suspiciously coincidental with the release of Halo 4, which features both a set of concentric dyson spheres, and the usual ringworld the series is known for. That game has some crazy stuff in it, like solids that aren't; fake stars, objects that float in place. Apparently, it can all be explained by ideas like spacetime manipulation, quantum phenomena, exotic particles, etc. Plasmic Physics (talk) 09:08, 14 November 2012 (UTC)[reply]

No doubt, a lot of it is pseudoscience and quantum fruitloopery, but it is interesting to think about. Plasmic Physics (talk) 09:10, 14 November 2012 (UTC)[reply]

Certain electric weapons, use micro wormholes to teleport electrical current instead of using wiring, resulting in strange concepts such as the weapon fireing a few atoseconds before you actually pull what counts as a trigger. Plasmic Physics (talk) 09:53, 14 November 2012 (UTC)[reply]

Not sure what the grandfather scenario says about that one. If I pull the trigger the same instant as he does, will we both get shot before we pull the trigger? And more importantly, does that count towards a kill streak??? :D - ¡Ouch! (^{hurt me} / _{more pain}) 07:49, 16 November 2012 (UTC)[reply]

Atmosphere and oceans on Dyson ring

In the scenario above, with the ring rotating quickly enough to produce 1 g for anyone standing on the interior edge, I wonder about oceans and atmosphere on such a ring. Clearly, you would need edges on the ring to prevent them from flowing over the edge. But, would you need an inner ring or would centrifugal/centripetal force be sufficient ? StuRat (talk) 18:31, 13 November 2012 (UTC)[reply]

It depends on the time scale. The classical Ringworld had no cover, being open to the sun (with a secondary ring of shades providing day and night). If you have not read the book, do it now! Niven's ring had rims 1000 miles high - more than two times higher than the limits of the exosphere, and about four times the altitude of the ISS. Of course some atmosphere spills over, but little enough to be only noticeable over geologic time periods. And then you probably have atmosphere plants to fix it... --Stephan Schulz (talk) 18:58, 13 November 2012 (UTC)[reply]

Sufficiently high walls will do the trick (I think Niven's Ringworld called for 1000 km walls). You'll still get some atmospheric loss over the edges (disproportionately weighted towards non-critical stuff like hydrogen and helium), but if you can postulate "create a Dyson ring", "replace lost air" is comparatively trivial. — Lomn 19:00, 13 November 2012 (UTC)[reply]

Earth's atmosphere is gradually losing air too - as long as your ringworld doesn't lose air faster than Earth, I don't think anyone can really complain. --Tango (talk) 23:14, 13 November 2012 (UTC)[reply]

Well, at least Niven's Ringworld does not have tectonic processes that replenish the atmosphere with new volatiles. But on the other hand, if it's only good for 10 million years or so, that should be good for most civilisations ;-). --Stephan Schulz (talk) 23:21, 13 November 2012 (UTC)[reply]

You could catch comets ... —Tamfang (talk) 08:15, 17 November 2012 (UTC)[reply]

Do they even let people graduate high school without having read Ringworld? μηδείς (talk) 03:12, 14 November 2012 (UTC)[reply]

Yes. Smaller rings are all the rage nowadays. Someguy1221 (talk) 03:24, 14 November 2012 (UTC)[reply]

OP here, I ironically graduated with a maths degree as Niven did, and haven't read ringworld Oops! 80.254.147.164 (talk) 09:27, 14 November 2012 (UTC)[reply]

Fatigue=drunkenness?

Is it true that fatigue produces most of the same effects as drunkenness? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:22, 13 November 2012 (UTC)[reply]

No. OsmanRF34 (talk) 16:15, 13 November 2012 (UTC)[reply]

Fatigue, however, can be one of the sequelae of drunkenness. μηδείς (talk) 17:35, 13 November 2012 (UTC)[reply]

... and some of the symptoms of ordinary fatigue could be mistaken for those of drunkenness (e.g. impaired driving ability). Dbfirs 18:00, 13 November 2012 (UTC)[reply]

No, not fatigue. But Oscar Wilde made an important discovery: that alcohol taken in sufficient quantities produces all the effects of intoxication. :) -- Jack of Oz ^[Talk] 21:51, 13 November 2012 (UTC)[reply]

Also, lack of sleep due to to staying awake for a long time (say 40 hours). Count Iblis (talk) 03:39, 14 November 2012 (UTC)[reply]

Ah, yes, that reminds me of summer camp when I was twelve. Several of us stayed up all night twice in a row playing cards and were giddy on the third day so that it would indeed have looked like drunkenness. μηδείς (talk) 17:10, 14 November 2012 (UTC)[reply]

Two of the assistant MythBusters (Kari and Tory) drank almost enough beer to get busted for DUI, tested their driving skilz, then stayed awake for thirty hours and tried again. Sleepy was much worse than buzzed. (MythBusters (2010_season)#Episode 152 - Arrow Machine Gun) —Tamfang (talk) 08:14, 17 November 2012 (UTC)[reply]

Why are flat screens not flexible?

Are they too thick? Conductive materials can be made flexible, so why aren't they flexible? OsmanRF34 (talk) 18:49, 13 November 2012 (UTC)[reply]

"Can be" and "must be" are not synonyms. Most flat screens are constructed with one or more layers of glass which needs to be thick enough to maintain structural integrity and thus inflexible. Roger (talk) 18:55, 13 November 2012 (UTC)[reply]

Well, I see a couple of advantages for a product with a flexible screen, or a foldable screen. If some screens are an array of LEDs, why didn't we get yet the iPhone with a huge foldable screen, but in pocket size? OsmanRF34 (talk) 19:03, 13 November 2012 (UTC)[reply]

LCD screens are made on a silicon substrate, which is not flexible. In the not-too-distant future there will probably be OLED displays, which can be built on a flexible substrate -- see also our rollable display article. Looie496 (talk) 18:57, 13 November 2012 (UTC)[reply]

Nice link. OsmanRF34 (talk) 19:03, 13 November 2012 (UTC)[reply]

The problem with such displays will be durability. After you fold a map a few times, it tends to want to tear at the folds. The same would be true of a screen. So, folding is out of the question. On the other hand, rolling up a screen inside a tube could protect it, much as it does with engineering drawings. However, if you want a screen 2 meters by 3 meters in size, you still end up with a roll over 2 meters long, so it's not all that portable. Another problem is that after having been rolled up for a while, it won't want to lie flat, but a fold-out frame, possible stored inside the tube, could hold it flat. Another option is sufficient weight so that gravity will hold it flat, but, in the above example, that would mean a 3 meter long roll, if you want a "landscape" display. Also, lugging around that weight might be unpleasant. StuRat (talk) 21:06, 13 November 2012 (UTC)[reply]

The United States Army funds the Flexible Display Center, where the goal is to advance the technology for reliable, full-color video displays. Like most Army research, this technology would be incredibly useful if it worked. There are plenty of neat videos on their website, and you can even visit in person. If you're flush with cash, you can even buy in to the technology. Nimur (talk) 22:11, 13 November 2012 (UTC)[reply]

And note that the calculus is a bit different for the military, where they are willing to pay quite a bit more for portability than the average consumer. For example, a front-line infantry unit could use one to view the video feed from a surveillance drone. StuRat (talk) 22:21, 13 November 2012 (UTC)[reply]