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May 23

Solar-powered vending machine indoors?

Could a low-cost photovoltaic cell mounted to the top or side of an electronic vending machine provide enough power to run it reliably? The machine would be indoors, but in a vestibule with glass double doors and overhead lighting during the evening (at least during the hours when it would need to run). Neon Merlin 02:25, 23 May 2009 (UTC)[reply]

One important question is whether this vending machine needs to have a refrigerator in it, as it would if it's vending cold soft drinks. Tempshill (talk) 03:33, 23 May 2009 (UTC)[reply]

No fridge. This is for trading cards. Neon Merlin 06:05, 23 May 2009 (UTC)[reply]

There are Mechanical vending machines, condom machines are often just a place to put cash, pull a handle and get your prize. Zero electricity - the room that it is in is already lit up so no need for electricity. Similarly there are apparently 'eco' vending machines but not sure how good they are (from a brief read - not entirely free of grid-electricity use - the link from google is 'blacklisted' but search for 'eco vending machine'). ny156uk (talk) 10:02, 23 May 2009 (UTC)[reply]

If the panels are directly driving the machine - then I'd say no. But if they are charging a battery that operates the machine - then perhaps if the machine is used sufficiently infrequently, it'll have time to recharge fully between 'vends' - in which case you'd be OK. But if you had a whole bunch of kids arrive at once - all wanting to buy trading cards - then you'd run the risk that the battery would go dead. I agree with Ny156uk - there have been plenty of purely mechanical vending machines where the weight of the coin against a spring releases the ratchet on of a wheel - allowing it to rotate. The purchaser drops in the coin - then turns a knob to rotate that wheel and thereby to release the product. When the wheel rotates past a certain point, it reveals a slot through which the coin drops - allowing the ratchet to re-engage against the wheel and prevent it from rotating again until another coin is dropped in. This approach is much more practical than messing around with solar panels and rechargeable batteries. SteveBaker (talk) 15:21, 23 May 2009 (UTC)[reply]

The only way I think it's possible is if the electricity is for verifying the payment and perhaps light a small display, and unlocking something, after then the user would use his own energy to pull a lever or crank something. Modern newspaper vending machines outdoors in Canada usualy have a battery inside for electronics inside that verify the payment and unlock the door. I can't imagine the energy use be too much energy for that battery to be trickle-charged by a solar panel. 209.148.195.177 (talk) 10:23, 31 May 2009 (UTC)[reply]

Insect identification

This tree in Bavarian countryside was entirely covered in a kind of web and had caterpillars crawling all over it, yellow to light brown with two dotted black stripes down their backs, thin, about 1.5 - 2 cm long (I think, I didn't measure them). Can anyone tell me what the caterpillars are? N p holmes (talk) 08:25, 23 May 2009 (UTC)[reply]

The caterpillars.
The tree.
Another view.

Try Small Tortoiseshell The German page has a pic with the "web" de:Kleiner Fuchs 71.236.24.129 (talk) 09:38, 23 May 2009 (UTC)[reply]

Thankyou. I looked at some pictures: the Small Tortoiseshell caterpillars look fatter and hairier than these (and the web was only a little thing on a couple of nettle leaves). So I'm slightly doubtful. Any other suggestion? N p holmes (talk) 10:13, 23 May 2009 (UTC)[reply]

I think I've narrowed it down now to some kind of Ermine moth. I don't know how to distinguish the kinds. N p holmes (talk) 10:38, 23 May 2009 (UTC)[reply]

Most likely the Bird-cherry Ermine, yponomeuta evonymella, from the images of the larvae on this website (9th out of 30) [1]. Mikenorton (talk) 14:33, 23 May 2009 (UTC)[reply]

Thank you. That seems likely. N p holmes (talk) 07:47, 24 May 2009 (UTC)[reply]

Air vortex force calculation

If an air vortex has enough force to move a 150lb object 1 foot from 30 foot distance from the launcher. Is there any way to estimate how far it will be able to move the same object from 5 foot? Or is the degredation of force within a vortex unique for each vortex depending on the speed of the spin of the vortex, and the speed which it is traveling?

Thank you —Preceding unsigned comment added by 79.68.229.253 (talk) 11:12, 23 May 2009 (UTC)[reply]

Some information that may help is that the starting pressure within the vortex is 14,7 psi and the speed is mach 1 and the size of the vortex is 2ft (24 inches) in diametre —Preceding unsigned comment added by 79.68.229.253 (talk) 11:43, 23 May 2009 (UTC)[reply]

Well - the air vortex is going to dissipate sideways - so the pressure will decrease as you get further away at a rate that's proportional to the increase in cross-sectional area. Probably a good estimate would be an inverse-square-law kind of thing - as you double the distance from the source, the pressure decreases by a factor of four. However, it's going to depend critically on the size of the object being moved. For an object who's cross-sectional area is small compared to the diameter of the air vortex, the area over which that pressure is exerted doesn't change as the air flow diameter increases - so the inverse-square drop in pressure equates to an inverse-square decrease in force applied. But for an object that's much larger than the airflow, as the pressure drops off, the area it's exerted onto increases at the same rate - so the force is about the same no matter how far away you are. But in truth, this is a very rough estimate. The details matter a lot. SteveBaker (talk) 15:11, 23 May 2009 (UTC)[reply]

Hmmm that helps a little but wont the slowing down of the air due to the distance decrease the impact as well? If the vortex grows at a very small rate the area effected would not change much, but the air speed would slow due to drag. how much does drag decrease air flow?

User: Robin (talk) 16:24, 23 May 2009 (UTC)[reply]

The air can really only slow down by spreading out. Think about it...if the air leaving the nozzle is moving at speed X meters per second with a cross-section of Y square meters - then the volume of air coming out of the nozzle is X.Y cubic meters per second. If it then slows down to X/2 - then the air is only moving out of the way at X.Y/2 cubic meters per second...so where does all the extra air go? As the air slows down, it's cross-sectional area has to increase at the same rate - so if the speed drops to X/2 then the cross-sectional area has to go up to 2.Y...unless the pressure is steadily increasing somehow. But if the pressure is bigger than ambient - then the air in the jet can push the ambient air out of the way - so it'll spread out. There is really no avoiding it. SteveBaker (talk) 03:50, 24 May 2009 (UTC)[reply]

Fancy swordcraft and damaged clothing...

I'm guessing not, BUT...

is it actually possible to damage clothing but not the wearer in a controlled manner ala Zorro (or many other films & shows)?

61.189.63.185 (talk) 11:49, 23 May 2009 (UTC)[reply]

It is possible, but unlikely - especially at the speeds Zorro slashes at. Fine motor skill is handled by small muscles, while quick slashing motions is gross motor skill. The two processes are handled by different parts of the brain and executed by anatomically different muscle and tendon groups. Nimur (talk) 12:13, 23 May 2009 (UTC)[reply]

Ala zorro, very unlikely. Zorro carved complex patterns into another persons clothing; a Z. As stated above, this would take a combination of fine motor and gross motor skills. However, the kind of swordplay often seen in humorous disney films, i.e. peter pan, where peter slashs the belt off of captain hook, and his pants drop, would be a very simple maneuver. The only real difference is that in the animation it shows a slash across, while in real life it would have to be a downward slash, to cut the belt. It is unlikely this would make the pants drop though, unless the target was wearing extremely loose pants.

Calculating oversteer/drift in a car

Hi all. I'm trying to calculate the amount of oversteer of a car, based on one main variable, the centre of mass, and forward velocity. The main result I want to achieve is to be able to calculate, for a set numbers of degrees of drifting, how the center of mass affects the forward speed and stability of the drift, so the car doesn't slide out.

I've tried to solve it in terms of opposite torques. For the front tires, I've assumed they aren't slipping, so have used the force into the drift as (coeffient of static friction)(normal force of tires)(center of rotation), and taking the centre of rotation as the intersection of the normals of the inside tires (think that's right, haven't derived it.)

The rear tires are where I'm having the most trouble. Assuming they are slipping, the only force I can think to use is mv^2/R, but I don't think that's right because there should be a force straight on, since the car is angled? And I'm not sure what figures to use for R, the center of rotation, and r, the length of the lever arm for the torque equation.

The end result I think I want it (coeffient of static friction)(normal force of tires)(center of rotation) > (mv^2/R(?))(r) so the car doesn't slide out.

So if anyone can make sense of that, and can help, thank you very much :) 203.206.34.183 (talk) 11:50, 23 May 2009 (UTC)[reply]

On second thoughts, mv^2/R may be not a valid equation for this, instead using the kinetic friction force. But then velocity doesn't come into it at all, and it should... I think I need to use both, on thinking that sounds better.203.206.34.183 (talk) 14:07, 23 May 2009 (UTC)[reply]

I think you are dramatically underestimating the complexity of the system you're trying to analyse. In real cars, the weight is not evenly distributed over the wheels - and as the car accelerates or turns, that weight shifts as the suspension takes up the strain and the car leans or pitches. In some cars, one wheel may actually lift off the road altogether. Chassis flex and the effect of anti-sway bars change this weight shift in complicated ways. Front wheel drive versus rear wheel drive makes a difference. Static versus dynamic friction and side-wall flex in the tires complicates the answer too. This is an insanely complex thing to try to analyse mathematically. SteveBaker (talk) 14:52, 23 May 2009 (UTC)[reply]

Yeah, I know it's a heavy bit of analysis. I was planning on simplifying it by assuming the front and rear wheels are both on live axles, so they can be looked at together. It is viable for me to ignore suspension and roll affects for my problem. Rear wheel drive. Sidewall flex would be insignificant as well. I'm not looking to include every possible force, just the most significant ones. 124.169.20.149 (talk) 00:29, 24 May 2009 (UTC)[reply]

But my point is that these other things are FAR from insignificant. There are 'approximations' that are reasonable - and there is ignoring major features of the system...and you are well into realms of the latter!

I've competed in many Autocross events in my bone stock MINI Cooper'S - and videos taken by spectators clearly showed that in hard cornering (long before 'drift' set in), the inside rear wheel would lift completely off the ground. Just think about that - with that wheel off the ground, the frictional forces on the rear end of the car are halved! Later, I added a stiffer rear antisway bar (going from a 20mm torsion bar to 25mm) - that keeps that inside wheel glued to the road and that added close to a third of a g to my lateral cornering force! That's a SPECTACULAR improvement...not by any means a negligable effect and it's an extremely subtle suspension tweak. Ignoring all of these kinds of effects makes your results all but meaningless. You might just as well save yourself the effort of doing the math and just guess! SteveBaker (talk) 01:57, 24 May 2009 (UTC)[reply]

Today's featured pic

Hi, in todays (23) featured pic, it shows a volcano erupting. There's a curious thing; the smoke from the volcano is drifting to one side, while the clouds around the volcano seem to be drifting to another side. What is the explanation for this? What is happening here? 202.129.232.137 (talk) 13:56, 23 May 2009 (UTC)[reply]

My guess is that the clouds are actually flowing in the same direction as the smoke - and are being constrained by the other two islands. But as the cloud reaches the volcano, the air temperature rises steeply. Since the amount of water that the air can hold increases with temperature, the water vapor in the clouds simply evaporates.

Alternatively - if your theory as to the direction the clouds are travelling is true - then perhaps it's merely that the smoke is at much higher altitude than the clouds - and that perhaps the wind at those altitudes are blowing in a different direction from the winds at lower altitudes.

SteveBaker (talk) 14:31, 23 May 2009 (UTC)[reply]

Those "clouds" look pretty brown to me. That couuld be a downhill flow of a heavy portion of the initial ash eruption, similar to a pyroclastic flow, but with much less material. -Arch dude (talk) 18:05, 23 May 2009 (UTC)[reply]

No - that's not possible. Look at Mount_Cleveland_(Alaska) and you'll see that this is a very steep-sloping stratovolcano. Now, look again at the photo - notice that the mountain isn't casting a shadow - the sun is sufficiently high in the sky that there is no shadow being cast. Next, look at the shadow of the brown stuff - you can see it cast clearly onto the mountainside beneath. With the sun being as steep as it is - that means that there much be considerable vertical distance between the smoke plume and the mountain. Look (for comparison) at the shadows of the clouds - notice that the shadow is MUCH shorter - meaning that the clouds are much closer to the ground than the smoke plume is. It's not close to the ground. SteveBaker (talk) 18:37, 23 May 2009 (UTC)[reply]

But a pyroclastic flow hugs the ground. Look at the picutre at Pyroclastic flow. You see the flow going opposite from the wind plume. Flow is determind by slope and in independent of wind. Also follow the link to pyroclastic surge and read about 'base surge. I do not think the "clouds are an active surge: I think they are the aftermatch of such a surge, basiclaly dust left in the air by teh surge that has not yet fallen out. -Arch dude (talk) 18:48, 23 May 2009 (UTC)[reply]

This image (taken at roughly the same time as the one our OP is asking about) makes it much clearer. Firstly, you can see more clearly how the low altitude clouds are being channeled by the nearby mountains (as I suggested) - which indicates much more clearly how they are moving in the same direction as the dust plume. Secondly, this photo makes it easier to see how high the dust plume is above the terrain. This photo (taken by a satellite one day later) shows that the dust plume is drifting away from the actual volcano (something that would absolutely not be possible with the 'pyroclastic flow' hypothesis...and the photo has that plume clearly labelled "Ash Plume". This video of an eruption about a year later - under very similar-looking cloud conditions makes it even more obvious. QED. SteveBaker (talk) 22:47, 23 May 2009 (UTC)[reply]

Thanks, Steve. I never thought the plume to the SW was a pyroclastic flow or surge. I thought the crescent-shaped billow to the NE might be the aftermath of such a flow or surge. However, thanks to your excellet work we now know that it is merely the leading edge of a very large area of low-lying clouds. -Arch dude (talk) 05:20, 24 May 2009 (UTC)[reply]

Oh! I see what you're saying. Yeah - now I understand where you're coming from, it seems more plausible - but then you'd still have had to explain why it's drifting in the opposite direction to the other plume...but those other photos prove it's just clouds behaving oddly because they are close to the ocean and are being funnelled around these tall mountain/islands. It's a very cool photo though. SteveBaker (talk) 14:24, 24 May 2009 (UTC)[reply]

Looking for a quote by Taiichi Ohno

At some point in the past few weeks I have read a quote by Taiichi Ohno that Lean Production should not be separated and codified as tools, but should be adopted by organisations as a whole system. Now that I'm writing my dissertation I can't find it, and it's driving me bonkers. Can anyone who has read any of Ohno's books identify the quote, and give me a citation for it please? Otherwise I'm going to have to re-write my dissertation in order to not miss the deadline. -- role^player 13:59, 23 May 2009 (UTC)[reply]

Breaking stength of an extension cord.

Someone at my car club spotted a pickup truck towing an Accura using a yellow extension cord!! I was wondering whether this was really as foolhardy as it sounds. Does anyone have any idea of the breaking strain on one of those things? Some way to get a rough estimate? SteveBaker (talk) 14:23, 23 May 2009 (UTC)[reply]

Human perception has weird effects. When on the road things have a tendency to look smaller than they really are or would look in, for instance, a small room. I suspect the cord could have been far more heavy-duty than it looked, and thus able to hold vehicles together... 69.140.12.180 (talk) 15:59, 23 May 2009 (UTC)Nightvid[reply]

Well, clearly it was able to hold vehicles together, since that's what it was doing. I don't think that's in any doubt. The question is whether it can do so reliably, which isn't really a question of perception... -- Captain Disdain (talk) 16:26, 23 May 2009 (UTC)[reply]

If nothing else there's the danger that there will be damage to the cord that you can't see because it's covered by the outer insulator. APL (talk) 16:37, 23 May 2009 (UTC)[reply]

You can estimate it, but you have to make tons of assumptions. So, lets assume that it is a twisted copper wire. Copper has about 1/3 the tensil strength of steel (from memory - please verify). So, if you assume that it is safe for a 1/4" twisted steel cable to pull a car, then it should be safe for a 3/4" twisted copper cable to pull the car. You can make different assumptions and estimate away. -- kainaw ™ 17:25, 23 May 2009 (UTC)[reply]

You mean sqrt(3)×1/4 = less than 1/2 inch. But anyway, if you've ever cut into an an extension cord you know that most of the material is not solid copper -- most of the cross-sectional area is insulation. --Anonymous, 17:55 UTC, May 23, 2009.

There are yellow ropes. What made the viewer think it was an extension cord? Cuddlyable3 (talk) 17:42, 23 May 2009 (UTC)[reply]

Well, the plug on the end of it dangling off the tow-bar of the truck was a clue. SteveBaker (talk) 18:19, 23 May 2009 (UTC)[reply]

ALthough it is a really stupid thing to do except in a dire emervency, it is possible, using extreme care, to tow a vehicle without placing a lot of strain on the tow rope. Don't go up hill, use extremely low acceleration, and the guy in the towed vehicle must not use the brakes unless the rope is slack. Therefore, your friend may have seen someone actually do this, but it's really dumb. -Arch dude (talk) 19:00, 23 May 2009 (UTC)[reply]

Maybe it was an electric car and, suffering from a spot of MS paranoia , they were driving "unplugged" :) --Cookatoo.ergo.ZooM (talk) 19:39, 23 May 2009 (UTC)[reply]

Okay, this is where I went: Extension cord, links to power cable, links to cross-linked polyethylene used in insulation. Googling its shortened name came up with something all about exactly that. Nifty. Does it help? Vimescarrot (talk) 19:46, 23 May 2009 (UTC)[reply]

Yes. It gives a tensile strength of 10.9MPa and a breaking strain of 224%. I'm not sure how to turn those into the answer we need, but I imagine someone here can. --Tango (talk) 22:31, 23 May 2009 (UTC)[reply]

Seems like it's "safe", then (this is not hauling advice). A one-ton car accelerated at no more than, say, g/4 (0-60 in 10.9 s), would be 2.45 kN, so a cylinder of radius 8.5 mm of the plastic should be able to support it. --Tardis (talk) 22:44, 26 May 2009 (UTC)[reply]

Hmm — I suppose the real issues are whether the stress is evenly distributed (it isn't, especially where the cord is bent!) and whether several lengths of cord were being used in parallel (as by looping the cord around both attachment points a few times). --Tardis (talk) 22:46, 26 May 2009 (UTC)[reply]

Mystery sea creature

What is the sea creature that appears at about 0:52 in this video, which the speaker calls a "flying turkey"? 69.224.113.202 (talk) 15:25, 23 May 2009 (UTC)[reply]

I've looked hard to try to find this - but without luck. The only references to it online are in other places hosting that same video. It's possible that "flying turkey" isn't the name of the creature - just a description this guy gave it in his narration - and it's also quite possible that this is the first and only time it's ever been seen. People who do these kinds of crazy deep dives often report that they see species that are new to science every time they go down. Perhaps someone else will have more luck. SteveBaker (talk) 01:43, 24 May 2009 (UTC)[reply]

I've looked at pictures in the book The Deep and found a plausible match: a sea cucumber called the "deep-sea Spanish dancer"(scientific name Enypniastes eximia). Here's a video of it in action: http://www.youtube.com/watch?v=5PdRt31FqDc 15:43, 24 May 2009 (UTC) —Preceding unsigned comment added by 69.224.113.202 (talk)

Here's another video, which I think makes this very plausible. [2] Mikenorton (talk) 17:05, 24 May 2009 (UTC)[reply]

5d optical recording

I've just read this article about a 'new' DVD technology that could allow up to 300 DVDs to be recorded on a single disk. I've had a look at the article on 5D, but as a layman I cannot understand head nor tail of it. How does this work. What is the 5th dimension and more to the point, why don't we just invent a 700D DVD that could store every movie ever made? russ (talk) 21:18, 23 May 2009 (UTC)[reply]

OK - they call it "five dimensional" but that's not what's really going on. They are saying that they are recording data using five parameters:

How far from the center of the disk are you?
How far around the circumference of this 'track' are you?
What layer are you on (they claim 10 layers)?
What color of light is reflected at this point?
How does this point polarize light if it is reflected by it?

These five numbers are what they are calling dimensions - but that's just marketing-speak. They haven't invented five-dimensional space-time (which is what Fifth dimension is talking about).

Let's examine their claims carefully:

Existing DVD's can be 'multilayer' (two layers on each of two sides - four layers in total) - so they have (1),(2) and (3) above. At each point in that three-dimensional space, there can be something that absorbs laser light or reflects it - representing either a 'one' or a 'zero' - a binary code. So all these guys are proposing is to use polarization plus color to allow each point in that 3D space to represent more than just a 1 or a 0. Contrary to what the article says, a standard 4 layer DVD can already store 17Gbytes (they said 5Gbyte?!?) - the new technology claims 1.6Tbytes - which is actually less than 100 times more than a "standard" DVD - and only 30 times more than a BluRay disk (which can only have 2 layers). So they are getting 5x more than the BluRay disk by having 10 layers instead of 2 - and that leaves only a factor of 6 gain from their fancy color and polarization tricks. So if they could distinguish 2 different colors and 3 different polarization planes (or vice-versa) - then they can reach their claimed capacity. There is no magic going on here - it's just a logical extension of what we already do.

These kinds of fancy high-capacity storage systems are claimed by researchers all the time - but 99% of them fail because they are often unable to actually deliver that in a form that's cheap to manufacture (DVD's cost about 50 cents each to make - and that includes the case, the printed inlays and shrink-wrapping!) - and which withstand the rigor of the mishandling they'll get in daily life. There are also issues of the cost of the players and whether such players will be able to play existing CD's, DVD's and BluRays as well as their new format.

So don't hold your breath waiting for these things!

SteveBaker (talk) 22:29, 23 May 2009 (UTC)[reply]

By "standard DVD" I guess they mean single-layer. That also fits with them saying DVDs are 2D. The article says there are 2 polarisation angles, so I guess there must be 3 colours. While calling it 5D is, in a sense, correct, it is a little misleading since 3 of the dimensions are discrete (have only a finite number of positions). In topology discrete spaces aren't usually counted as dimensions. The article says the "material cost" would be less than 5 cents, I don't know what it would actually cost to make. I think the main flaw in this new design isn't cost, though, it's speed. There is no point having lots of data if you can't access it. Lots of storage space would be great for really high definition movies, but if you can't read 1 second's worth of data per second, it doesn't work and the article suggests data access is pretty slow. I can't think of a real purpose for this technology. I know better than to say no-one will ever find a purpose, but I'm drawing a blank. --Tango (talk) 22:41, 23 May 2009 (UTC)[reply]

Yes - I suppose you could stick 300 movies on it so you could store your entire movie collection on a handful of disks - then when you want to watch a particular movie - you tell the machine to copy the movie (S-L-O-W-L-Y) onto a conventional hard drive so you can watch it later. If the replay mechanism was (say) 10x slower than required - then you'd have to request the movie you wanted the day before you wanted to watch it...which is kinda inconvenient.

Personally, I think we're more likely to be downloading movies in the future - and the storage issue goes away...you need enough space locally to store whatever new movies you want to watch over the next few weeks - the software slowly downloads new ones on demand so you always have enough material to keep you happy. I simply don't see the value of a 1.6Tb non-recordable media. What could you possibly want to buy in one chunk that consumes that much storage space?

But bear in mind that storage capacity isn't really the limiting factor here - the cost of buying a movie is what kills you. They could already sell movies on DVD at $1 each and make a decent profit if the cost of making the movie was low enough. But we pay $20 for a DVD - that's $19.50 for the movie and $0.50 for the disk/box. Even if these things were $0 to make - the cost of the 300 movies that it would hold would still be something like $6,000. NOBODY is going to spend $6,000 on a movie disk - and the movie companies can't afford to drop the price of the actual content to the $0.10 per movie it would take to pack 300 movies onto a $30 disk. You can't even argue form the point of view of convenience. You can already buy 1Tbyte hard drives - 1.7Tbyte are presumably only about a year away. So your hard drive is by far the best place to keep movies...now it's just a matter of how to distribute them - and the Apple iTunes mechanism is by far the more likely way for this to happen. That's possible with today's tech. Take a 1Tbyte drive - with a $50 computer - pack it into a shiney white $200 box with an Internet socket and a TV video output and a little remote controller and you're done. Apple could make a product like that tomorrow if they wanted to - the only issue is pursuading the movie companies to sell their movies that way.

If such a disk format were to have a use, it would be in computer games - in the games business, we're always bumping into the limits of what we can pack onto a DVD. But lack of speed would kill us in that application. SteveBaker (talk) 23:12, 23 May 2009 (UTC)[reply]

You can already get 2TB hard drives. CD style storage media are good for transferring data, not for storing it. I'm not sure if these things are intended to be writeable to on a desktop, but if they aren't then they'll only be useful for when you want to buy a terabyte of data all at once, which no-one is likely to want to do. Games is a possibility, I suppose - you would have to fully install it to a hard drive to play, though, and at those sizes it might be easier just to sell it on a hard drive! --Tango (talk) 23:34, 23 May 2009 (UTC)[reply]

Ah - I wasn't aware that we'd hit the 2Tbyte level with hard drives - but it doesn't surprise me. Anyway - from the way I read that article, it doesn't sound like a writable version is likely...quite how you'd change the orientation of these nanoparticles in order to change the plane of polarisation seems tricky - and changing the chemistry of the particles on the fly to change the color just using a laser seems impossible. Installing a couple of Terabytes onto your hard drive would take HOURS - possibly days if this drive is as slow as we suspect. SteveBaker (talk) 01:40, 24 May 2009 (UTC)[reply]

I've had a quick read through of the actual letter in Nature ([3] subscription probably needed, I'm on a uni network). The only number they offer for speed is the recording speed of up to 1 Gbit/s (and they have an idea for dramatically increasing that), so it seems production is reasonable. They don't say how quickly it can be read, but it sounds like it could actually be quite fast (if I'm reading it right, they have a way of reading all the polarisations and colours at the same time, although I didn't fully understand that). It's a little difficult to understand - it's a scientific paper, not a technical specification, and concentrates more on how they do it rather than what it can actually do. --Tango (talk) 14:36, 24 May 2009 (UTC)[reply]

Oh! So it can be written to? Wow. SteveBaker (talk) 16:18, 24 May 2009 (UTC)[reply]

Well, they've got to be writeable to at some point! I don't know if they could be written to by a desktop drive or if you would need an expensive thing only big manufacturers could afford. It seems they are written to using a single laser, though, so they probably are writeable to at home (how reasonable the cost would be, I don't know). --Tango (talk) 16:22, 24 May 2009 (UTC)[reply]

Not in that sense. Pre-recorded CD's and DVD's are pressed out of plastic just like the old vinyl records were - I suppose that technically that is "writing to them" - but if so, then the entire content of the disk is written in a tiny fraction of a second as the plastic is pressed...it would be meaningless to talk about the "recording speed" in that case. SteveBaker (talk) 22:02, 24 May 2009 (UTC)[reply]

I believe the ability to write them is one of the advantages they are pushing "Holographic methods take all of the information to be recorded and encode it in the form of a graph showing how often certain frequencies arise in it. That means that the recording process is a complex, all-at-once, all-or-nothing approach that would be difficult to implement on an industrial scale. By contrast, 5-D recording is "bit-by-bit", like current CD and DVD writing processes in that each piece of information is read sequentially." They do acknowledege the cost/development issues: "For the moment, Dr Milster says, the equipment needed to write the data would make a commercial system expensive. However, that has not stopped the development of optical storage solutions in the past. "For example, a Blu-ray player is not an easy system to realise; they've got some wonderful optics in there," Dr Milster said. "People thought that would be pretty difficult to do, but others managed to do it." " Nil Einne (talk) 21:55, 27 May 2009 (UTC)[reply]

Hurrah! Another query solved by the RD. Thanks guys -russ (talk) 20:59, 25 May 2009 (UTC)[reply]

Irresistible force paradox

I have read and re-read this article. Assuming (for sake of arguement) that both an irreststible force and an immovable object did exist (perfectly, of course)and they were to meet or collide, wouldn't they just cancel each other out and the end result would be nil? —Preceding unsigned comment added by 216.154.17.116 (talk) 21:27, 23 May 2009 (UTC)[reply]

Well, then the force wouldn't really be irresistable. Recury (talk) 21:42, 23 May 2009 (UTC)[reply]

Just because you say that such things can exist doesn't mean that they can. Neither irresistable forces nor immovable objects can actually exist - so it's truly meaningless to ask what would happen. Saying "for the sake of argument" doesn't allow you to logically debate illogical impossibilities. There simply is no argument, no paradox, just some meaningless words on a page. SteveBaker (talk) 22:06, 23 May 2009 (UTC)[reply]

The paradox is proof that such concepts are nonsense. That's what paradoxes always are - proof that your assumptions are flawed. --Tango (talk) 22:28, 23 May 2009 (UTC)[reply]

Indeed - "God is omnipotent - so can he create a rock so heavy that he can't lift it?"...poof! No God. QED. SteveBaker (talk) 23:36, 23 May 2009 (UTC)[reply]

Our article Omnipotence paradox is fortunately more lengthy than this. (It's a featured article! Nice.) Tempshill (talk) 02:24, 24 May 2009 (UTC)[reply]

Well, that's just proof that a naive definition of omnipotence doesn't work. You need a 5 point scale! --Tango (talk) 23:46, 23 May 2009 (UTC)[reply]

If Wikipedia has an article about everything, can Wikipedia have an article about the things not on Wikipedia? SpinningSpark 00:11, 24 May 2009 (UTC)[reply]

Maybe you don't understand the full implications of having an article on everything... we have several articles about things that are not on Wikipedia... Nimur (talk) 00:23, 24 May 2009 (UTC)[reply]

The real point is that, even if both these things could exist (which they can't), they could not exist simultaneously. The existence of an irresistible force means that there's no such thing as an immovable object, because if there were, no force would be irrestistible. That's a logical contradiction. And vice-versa. So they could never exist simultaneously, not even theoretically. -- JackofOz (talk) 02:34, 24 May 2009 (UTC)[reply]

Hmmm - I suppose they could both exist if they were the same thing. An immovable object would require infinite mass (F=ma) - and an irresistable force would require infinite energy - but since E=mc², our infinite force provider will conveniently have infinite mass...which means it's also an immovable object! Since it can't be moved - there is no paradox. SteveBaker (talk) 03:41, 24 May 2009 (UTC)[reply]

What if there were two of them? --Tango (talk) 14:38, 24 May 2009 (UTC)[reply]

Since they are both immovable - they'll never meet! :-P SteveBaker (talk) 16:15, 24 May 2009 (UTC)[reply]

I'm not going to let you get away that easily! What if they came into existence already touching? --Tango (talk) 16:19, 24 May 2009 (UTC)[reply]

Mathematically speaking, the answer to your question is everything. For example, if an irresistible force is exerted on an immoveable object, Elvis is alive. See the Paradox of entailment for details. — DanielLC 05:50, 24 May 2009 (UTC)[reply]

If such a thing existed, wouldnt the force be reflected back at 180 degrees? 89.243.84.208 (talk) 10:17, 24 May 2009 (UTC)[reply]

You can't reason about it - it's just impossible - beyond logic. SteveBaker (talk) 14:20, 24 May 2009 (UTC)[reply]

Thanks everyone. I think I'll just stick with "they would cancel each other out leaving nil" cause that is easier for my wee brain to handle! —Preceding unsigned comment added by 216.154.23.87 (talk) 17:56, 24 May 2009 (UTC)[reply]

I answered this question at school when I was five years old. If there were such things as an irresistible force and an immovable object, the source of the irresistible force would end up moving toward the immovable object, attracted by its own force. Totally logical answer to a seemingly impossible question. --KageTora - (영호 (影虎)) (talk) 18:06, 24 May 2009 (UTC)[reply]

A similar paradox is this: a man in Ancient China advertises the most superior weponary; his sword, which will cut down all walls, and his shield, which will defend against attacks of the strongest arrows. A customer then asks him, "what if you put your own sword to your own shield, what will happen?" The answer is that the man advertising this items has to walk away, because if the sword breaks the shield, then the shield is inferior, and if the sword does not penetrate the shield, then the sword is inferior. ~A H 1^(TCU) 20:36, 24 May 2009 (UTC)[reply]

I am hit by 50 trillion neutrinos every second and survive. I conclude that neutrinos are irresistible and I am immovable (at any relative velocity). QED. --62.47.156.70 (talk) 20:40, 24 May 2009 (UTC). Ooops, not logged in, --Cookatoo.ergo.ZooM (talk) 20:47, 24 May 2009 (UTC)[reply]

You are not "hit" by them . They pass through the mostly empty space that you inhabit. Sorry. That's a fact, not a personal attack.Julzes (talk) 15:18, 26 May 2009 (UTC)[reply]

The anecdotal "counterexample" (not really the right word) I was given many years ago was: You can no more imagine the I.F. meeting the I.O. than you can imagine two men, each of whom is taller than the other. Hope that helps! --DaHorsesMouth (talk) 02:35, 25 May 2009 (UTC)[reply]

In Exalted, defender wins.

"Defense has primacy. If an unstoppable force meets an immovable object, the object stays still."
— Exalted, second edition, page 179

:-) Axl ¤ [Talk] 11:10, 25 May 2009 (UTC)[reply]