Wikipedia:Reference desk/Archives/Computing/2007 November 17

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November 17

Importing DVDs into iTunes

It seems so ironic to me that they make it so easy to illegally download a movie off Limewire and put it on your iPod, but a DVD that you have bought and paid for is so difficult. =o) I've tried googling it, but everything I've found talks about such complicated things like file extensions and converters and stuff. So, in dumbed-down layman's terms, what do I have to do to get a movie from a DVD and onto iTunes? When I go to My Computer, then click on the D drive, then there's a little folder that's labeled VIDEO, but when I click on that, there's a whole bunch of little files instead of one neatly packaged little video file. I'm so confused! 131.162.146.86 (talk) 02:56, 17 November 2007 (UTC)

I assume by your mentioning of My Computer that you are using Windows (you need to tell us that kind of stuff!!). I don't know the specifics for Windows but in general you need to:

1. Get a DVD ripper, which will grab the raw VOB video from the DVD for you (it'll be a couple of gigabytes in size)
2. Get a video file converter/compressor that can resize and compress the VOB file so that it will fit onto your iPod

Some DVD rippers can do both of these functions at the same time (that is, they will let you rip the DVD video into a 320x240 MP4 file, which is what iPods use for video). Perhaps someone can recommend an easy one. --24.147.86.187 (talk) 16:03, 17 November 2007 (UTC)

Yes, I am using Windows. Sorry, didn't realize that was important! And forgive my ignorance and complete stupidity where computers are concerned, but what does VOB mean? 131.162.146.86 (talk) 03:54, 18 November 2007 (UTC)

VOB stands for Video OBject. But mainly, it's the filename extension of the file on DVD that contains the actual movie. We even have an article on it (called VOB of course). SteveBaker (talk) 04:13, 18 November 2007 (UTC)

Okay, thank you. 131.162.146.86 (talk) 18:04, 18 November 2007 (UTC)

Deezer down?

Why might the music service [www.deezer.com] Deezer be down? When I try to access the website, it says that the server cannot be found. Acceptable (talk) 03:49, 17 November 2007 (UTC)

Works for me, maybe it was a temporary server crash? · Dvyjones ^Talk 09:09, 17 November 2007 (UTC)

I'm in Canada and all the Canadian computers I've tried (school, library, friends) will not connect to Deezer. Could it be that they blocked Canadian IP's like Pandora? Thanks. Acceptable (talk) 23:15, 18 November 2007 (UTC)

Splitting 1 audio into instrumental music, human voices and something.

As the title says. Well, I don't know how. (Here's software I have; Sony Vegas 7.0 and GoldWave)--JSH-alive ^(talk)_(cntrbtns)^{(mail me)} 12:46, 17 November 2007 (UTC)

We get this question a lot. It's not possible to get high quality tracks, but you can use GoldWave to get a "Karaoke" track that will probably sound bad but might sound all right depending on the song --ffroth 16:28, 17 November 2007 (UTC)

Yeah, there's software that tries to do the job, but it's inherently impossible to do with a high degree of accuracy. What people do in real life is to retain the source material as separate tracks indefinitely, so they can be mixed as needed whenever you want. Of course, when you're starting with something already mixed, you have to settle for some less good solution. Friday (talk) 16:45, 17 November 2007 (UTC)

The trick for removing vocals for Kareoke is to note that on most music with vocals, the stereo mixing is set up to place the voice in the middle of the stereo image with the instruments off to the sides. If you subtract the left-side image from the right, then anything that's common to both sides will disappear. This works surprisingly well for vocals. We use 'Audacity' (an free/OpenSourced audio processing package) - and it's vocal-removing system is quite amazing. But translating that idea to removing (say) the trumpet from a piece of music when it is in no special place in the stereo image would be much harder. SteveBaker (talk) 17:42, 17 November 2007 (UTC)

Steve, so then if you subtract the resulting Kareoke track from the original will you get a vocal only (well, maybe not "only") track? hydnjo talk 20:06, 17 November 2007 (UTC)

Sadly (and surprisingly to me), no - I don't think you can. My first reaction to your question was "well, of course you can!"...but when I started to figure out how, I couldn't. To understand why not, you need to do this with a bit of basic algebra: if A is the sound coming from instruments in the left side of the stage and B is the sound coming from the instruments on the right with V being the vocals - then the sound on the left channel of the original stereo recording ('L') is L=A+V, the sound on the right ('R') is R=B+V. L and R are our 'givens'. We have two equations and three unknows (A, B and V) - which first year algebra says is not something you can solve for all three variables. Fortunately we CAN say L-R=(A+V)-(B+V)=A-B - so we can calculate A-B without V. We'd really like a stereo signal with A and B separately - but we can't do that because if we could, we'd have performed the magic of solving three unknowns with only two equations. But A-B is a mono signal - just one value - but with no 'V' in it! In audio terms, a proper mono signal would be A+B - but negating an audio waveform just switches the phase of the signal 180 degrees - and that doesn't sound too bad...certainly good enough for Kareoke! So if (for example) we subtracted our kareoke track from the original L and R signals to try to get the vocals by themselves, we'd get L-(A-B) which is (A+V)-(A-B) which is V+B, similarly R=B+V and subtracting the kareoke track gets us (B+V)-(A-B) which is V-A. In other words, instead of getting rid of the instruments A and B - we just swapped them over. We could try making a mono track first (L+R) and subtracting our mono kareoke track from that - but then we'd have (A+V)+(B+V)-(A-B) = 2B+2V - still we have B mixed up in it. There simply isn't any way to do this...which surprises me! SteveBaker (talk) 04:05, 18 November 2007 (UTC)

Yeah, I actually came to that realization myself while falling asleep last night. Amazing how clarity arrives when our everyday environmental noise departs! Also, the A-B kareoke track must have some weird stuff going on as the original A+V track probably had varying amounts of B in it an so on. hydnjo talk 14:03, 18 November 2007 (UTC)

Ehh maybe you addressed this, but why not mix the 2 channels of song together and 2 channels of karaoke together, invert the karaoke, and superimpose the mono waveforms? It's no stereo signal but as far as I can tell it would work. Also, audacity is the biggest piece of crap software I've ever layed eyes upon- even the geek who occasionally has to work with audio will tell you that GoldWave blows it out of the water. --ffroth 06:46, 18 November 2007 (UTC)

Do the algebra:

mix the 2 channels of song together - OK M=L+R=A+B+2V

and 2 channels of karaoke together, - the Kareoke track is already mono - K=A-B A-B

invert the karaoke, - OK: K=-(A+B) K=-(A-B)

and superimpose the mono waveforms - Result = (M+K)/2 = (A+B+2V + -(A-B))/2 = B+V.

Nope - that's the same as the original right channel.

You can't do it because you have two equations and three unknowns. SteveBaker (talk) 16:42, 18 November 2007 (UTC)

Well despite your step 3 coming out of nowhere (K should = -(A-B)) that sounds right.. but what about the other K.. K=B-A? There's your third equation. It seems distinct from K=A-B.. it is, right? --ffroth 19:57, 18 November 2007 (UTC)

Yeah - sorry - that was a typo. It's fixed (above) now. I got the final line right though. The bottom line is the same - you can't get three unknowns from two equations no matter what. SteveBaker (talk) 21:44, 18 November 2007 (UTC)

Geesh Steve, I wuz jist askin', ya know? This is amazing! hydnjo talk 22:28, 18 November 2007 (UTC)

Lower screen resolution for speed?

I may have come across this question before. Would lowering the screen resolution on a Windows Vista machine allow the computer to run faster when graphic editing, gaming and everyday surfing? Acceptable (talk) 17:52, 17 November 2007 (UTC)

Well, high resolutions do take more processing power and require more work by your graphics card. So a lower res should run faster, the question is whether it would be appreciably faster or not, and that would probably depend on your processing speed, your graphics card, and what sort of programs you are intending to use. Photoshop is going to be a processor hog no matter what resolution you run at, for example. --24.147.86.187 (talk) 18:58, 17 November 2007 (UTC)

It depends whether the applications you are using are CPU-bound, vertex-processing-bound or fill-rate-bound. Only the fill rate is affected by the screen resolution. Games are most likely to benefit - but even then, only if they run full-screen. I doubt you'll notice much benefit. SteveBaker (talk) 20:52, 17 November 2007 (UTC)

Games can be made to run faster anyway by reducing the quality settings (e.g. Anti-Aliasing) in-game. --Dave the Rave (DTR)talk 21:32, 17 November 2007 (UTC)

I didn't think about games, but in that case resolution can matter a LOT, depending on the game. My computers are usually a bit slow for games (or are running them through virtualizers) so I usually end up running very low resolutions (640x480 or 800x600) with them (my native resolution is 1400xwhatever, so that's a pretty serious cut!), but it speeds them up a huge amount. My machine generally can't even do them at anything close to a native resolution; it requires way too much out of its puny graphics card. --24.147.86.187 (talk) 22:02, 17 November 2007 (UTC)

If your graphics card can handle it, and it's not trying to do other things like render a 3D scene, then keep your resolution high. It will make no difference to lower it. --ffroth 06:16, 18 November 2007 (UTC)

With 3D applications, there is always a bottleneck that limits the frame rate - but unless you know where it is, you can't know how best to optimise it. The CPU could be overloaded by the AI, the pathfinding, collision detection, etc - if it can't feed triangles to the graphics card fast enough then no amount of messing around with the graphics card or the display resolution will make any difference to your frame rate. Similarly, it might be that the game draws an enormous number of very small triangles - in this case the vertex processing stage of the graphics card will be overwhelmed and speeding up the CPU or messing with the display resolution won't help. Only if the CPU has time to spare and the vertex processor is blocked waiting for pixels to be pushed to the frame buffer memory will reducing the display resolution (or reducing the antialiasing quality which is almost exactly the same thing) help.

The trouble is that different games have different bottlenecks - and unless you are equipped with the source code and a boatload of specialised tools, it's hard to know which games have which bottlenecks. Worse still, this information isn't published anywhere because it depends too much on your precise computer setup - something that's CPU-bound on a 2GHz CPU may be vertex limited on a 3GHz processor. If it is the case that the final stage is the bottleneck then you'll find that reducing the display resolution a bit improves the frame rate a bit. However, dropping the resolution down further may not help because you've already removed the pixel fill rate bottleneck and now the system is blocked somewhere else. So even if reducing the frame rate helps, you might want to try a range of different screen resolutions to see which one gives you the best frame rate - but retains the most pixels on the screen.

If you're still with me at this point - I guess I should complicate the story still more by pointing out that many games are CPU-limited in some areas of the game map, vertex-limited in others and fill-rate limited in yet others. In producing the optimum 'game experience', designers have to trade these various limitations on the grades of hardware they expect their users to have (and, probably, on a couple of console platforms too). Some parts of the game might not need blazingly fast frame rates because you are on some kind of a slow-moving stealth mission - but other areas that involve fast movement through the world (driving a vehicle for example) may demand higher frame rates.

SteveBaker (talk) 16:37, 18 November 2007 (UTC)

I was talking about just for desktop use, but all that is true! When playing F.E.A.R. my Yonah can push out all Maximum settings for the "CPU" graphics options, but my low-end-mobile 3D card can barely handle the Minimum settings for the "GPU" graphics options. (they appear in separate columns in the options menu) --ffroth 19:50, 18 November 2007 (UTC)

I have found it to be true that lowering the screen resolution AND the color depth speed up CPU intensive programs on computers that share video memory with main memory, i.e. use main system memory for video memory (which most of the lower priced ones seem to do these days). Must be because of bandwith to the main memory. Bubba73 (talk), 01:54, 19 November 2007 (UTC)

Yep - that's to be expected. In 'unified memory' systems like that texture mapping and even simply writing to the screen all use main memory bandwidth - which will clobber the CPU (and heavier CPU activity will clobber your fill rate). The efficiency of these systems depends heavily on how well their texture caches work - which is a REALLY complicated thing for developers to deal with. If the game gives you the option, make sure you have MIPmapping turned ON and Anisotropic texture filtering turned OFF. You should probably prefer disabling antialiasing to reducing screen resolution - but you may need to do both. SteveBaker (talk) 03:55, 20 November 2007 (UTC)

Converting Matlab code to PHP/Actionscript/Javascript/whatever

Somebody helpfully gave me some Matlab code to get X,Y points for a given set of latitude and longitude coordinates with a Robinson projection. But I can't make heads or tails of how it deals with arrays. Could someone convert it into PHP, Actionscript, Javascript, and/or just pseudocode for me? It's totally opaque to me in its current form, but someone who has used Matlab could probably convert it pretty easily. (Note that you can just substitute a fake interpolation function if you want—I have interpolation functions I can use, you don't have to write me one.)

Matlab code snippet

robval = [
00 1.0000 0.0000 
05 0.9986 0.0620 
10 0.9954 0.1240 
15 0.9900 0.1860 
20 0.9822 0.2480 
25 0.9730 0.3100 
30 0.9600 0.3720 
35 0.9427 0.4340 
40 0.9216 0.4958 
45 0.8962 0.5571 
50 0.8679 0.6176 
55 0.8350 0.6769 
60 0.7986 0.7346 
65 0.7597 0.7903 
70 0.7186 0.8435 
75 0.6732 0.8936 
80 0.6213 0.9394 
85 0.5722 0.9761 
90 0.5322 1.0000 
];

robval(:,3) = robval(:,3) * 0.5072;
robval = [robval(end:-1:2,:);robval(1:end,:)];
robval(1:90/5,[1,3]) = -robval(1:90/5,[1,3]);

rvals2 = interp1(robval(:,1),robval(:,2),latitude,'cubic');
rvals3 = interp1(robval(:,1),robval(:,3),latitude,'cubic');
y = -rvals3;
x = rvals2/2.*longitude/180*2;

Thanks a ton! (No, this is not homework at all—I'm well out of my homework phase of life! I'm just working on a Flash project which uses a Robinson projection and it's driving me a bit nuts.) --24.147.86.187 (talk) 20:45, 17 November 2007 (UTC)

Uncommented Matlab is largely write-only. But here is an explanation in English/pseudocode:

- Define robval as a three-column, 19-row matrix with given values
- Multiply third column (in all rows) of the matrix by 0.5072
- "Mirror" the matrix.
    The matrix now contains the original matrix backwards (minus first row), followed by
    the original matrix, like so:
        90 0.5322 1.0000 
        85 0.5722 0.9761 
        80 0.6213 0.9394 
        75 0.6732 0.8936 
        ...
        05 0.9986 0.0620 
        00 1.0000 0.0000 
        05 0.9986 0.0620 
        10 0.9954 0.1240 
        ...
        85 0.5722 0.9761 
        90 0.5322 1.0000 
     (Of course the third column would have been multiplied by 0.5072)
- Invert all values in rows 1 to 18 (i.e. the backwards part), in columns 1 and 3.

- rvals2 = interpl(column 1 of robval, column 2 of robval, latitude, 'cubic'); ← for all rows of robval
- rvals3 = interpl(column 1 of robval, column 3 of robval, latitude, 'cubic'); ← for all rows of robval

- y = -rvals3;
- x = rvals2 / 2. * longitude / 180 * 2;

Hope this helps. ›mysid (☎∆) 22:25, 17 November 2007 (UTC)

God, that's totally unobvious from the code as was given! Thank you. (What godawful syntax Matlab uses!) --24.147.86.187 (talk) 23:30, 17 November 2007 (UTC)

But oh-so-powerful... :) And by the way, note that rvals2, rvals3, y and x are all matrices (or arrays if you wish) as well, so the operations are actually performed on every element. ›mysid (☎∆) 00:12, 18 November 2007 (UTC)

Powerful but totally opaque. Question: if all of them are matrices/arrays, how does it know which values to assign to x and y in the end? Or are x and y matrices themselves? I'm confused. --24.147.86.187 (talk) 06:16, 18 November 2007 (UTC)

Yes, y and x will be matrices as well, with same dimensions as rvals3 and rvals2 respectively, but with the said operations applied. I guess they are one-dimensional, assuming that interpl returns one-dimensional matrices when fed with one-dimensional parameters. ›mysid (☎∆) 08:26, 18 November 2007 (UTC)

Jetman Cheats

Does anyone know any cheats for the game Jetman on Facebook? Much obliged
мιІапэџѕ (talk) 21:49, 17 November 2007 (UTC)

Firefox's Wikipedia search function

I just had to reinstall Windows XP on my computer, and afterward I reinstalled Firefox 2.0. I then added the tool that allows Wikipedia to be searched in the little window in the upper right-hand corner, which I had before the reinstall. Previously, when I entered a term in that search box, it took me straight to the article (much like typing an article name in Wikipedia's own search box and hitting "Go"). Now, it takes me to Wikipedia's search page, where I have to click on the article name to get there (much like hitting "Search" in Wikipedia's search box). Does anyone know how to fix it back to the convenient way? Also, previously I could go to a Wikipedia article simply by typing "wp article" in the URL box at the top of the screen; that doesn't work any more. Any ideas how to fix that? —Angr 22:23, 17 November 2007 (UTC)

There are two "English" Wikipedia tools for that thing. You want the one that is called "Wikipedia (EN)". The other one (forgot what it was called) does go to the search results instead of the page. -- kainaw™ 22:52, 17 November 2007 (UTC)

The one I'm using says "Wikipedia (EN)" grayed out when it's empty. Isn't that the right one? —Angr 22:55, 17 November 2007 (UTC)

To go straight to an article by typing "wp article" in the address bar, go to Bookmarks > Organize Bookmarks > New Bookmark. Type "Wikipedia", "http://en.wikipedia.org/wiki/%s" and "wp" in the first three fields, hit OK and close the Bookmarks Manager. — Matt Eason ^{(Talk • Contribs)} 10:27, 18 November 2007 (UTC)

That is a really cool trick! Thank you! It seems to be case sensitive, though, which it didn't use to be. —Angr 12:34, 18 November 2007 (UTC)

WOW that's _exactly_ the keyword and bookmark that I use. wp EXPLOAD!!1. --ffroth 19:47, 18 November 2007 (UTC)