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February 16

Webmail

I am currently a student at a Queensland State school using a government hosted Webmail for most of my emailing, and each time I try to send something I get a message from postmaster@eq.edu.au telling me that I'm over quota and it won't send. I've used the same thing for a few years and never encountered this. I'm using 6.7 MB out of a 10MB inbox and there is no limit on the total number of messages. Does anyone know what would be causing this and how to overcome it? Mix Lord 01:20, 16 February 2007 (UTC)[reply]

There's nothing stopping a government-run service from suddenly cutting the quota, or merely crapping out. In fact, the amazing thing is that it has been working for years! You are going to have to go to the root of the problem, and penetrate the bureaucracy! I wish you luck. --Zeizmic 01:24, 16 February 2007 (UTC)[reply]

lol. Thanks

Mix Lord 03:44, 16 February 2007 (UTC)[reply]

Is your total webspace quota different than your inbox size quota? (Sounds weird but I have seen it before.) The problem sounds to me like it can't copy the message to your Sent Mail folder correctly because of a quota problem -- you might look into whether that folder has some sort of quota. --24.147.86.187 01:34, 17 February 2007 (UTC)[reply]

Actually, if that was the case, that it can't copy the message to the Sent Mail folder, then the message would have been already sent, as the postmaster is replying. Must be a server issue. BTW: I hope your QLD government webmail is better than my shitty NSW government webmail! --wj32 ^{talk | contribs} 09:35, 18 February 2007 (UTC)[reply]

Well I think it's good now. Something to do with the inbox of the guy I sent the emails too having a full inbox. Thanks anyway

Mix Lord 23:48, 18 February 2007 (UTC)[reply]

How do I tell if a windows install is 64 bit?

How would I tell if a particular windows install is 64 (or 32) bit? I've tried the msinfo32 tool, but it will just say "Microsoft Windows XP Professional" under OS Name. It doesn't seem to say it anywhere under System either. Is there any tool or location in Windows that will state clearly "Windows XP Pro x64" or x32? The Wikipedia article on Windows XP x64 is no help in this regard. Any ideas appreciated!--inks ^T 03:08, 16 February 2007 (UTC)[reply]

Whoops, figured it out. Under system properties, the x32 version will say "Microsoft Windows XP Professional", the x64 version will say "Microsoft Windows XP Professional x64 Edition".--inks ^T 03:14, 16 February 2007 (UTC)[reply]

It's not x32, but x86. --wj32 ^{talk | contribs} 09:32, 18 February 2007 (UTC)[reply]

Looking for an HTML tag

Let's say I want to write an HTML tutorial on my HTML based website. I want to have the text " example text " on my website to show how to use the HTML language, but if I simply type this into my own HTML file for the website, it will show up as example text which is not what I want. Is there a tag to "turn off" HTML analogous to how <nowiki> turns off wiki markup? Thanks, Dar-Ape 03:38, 16 February 2007 (UTC)[reply]

PS adding something like this </html>bold<html> in the middle of a file still produces bold. Dar-Ape 03:43, 16 February 2007 (UTC)[reply]

You can 'escape' HTML elements by using < and > which produce < and > respectively: not bold. -- zzuuzz^(talk) 03:56, 16 February 2007 (UTC)[reply]

Excellent, thank you very much! Sometimes the quickness of the RD is almost scary... :-) Dar-Ape 04:03, 16 February 2007 (UTC)[reply]

Pen drive/ Flash drive

What is the difference between Pen drive & Flash drive? —The preceding unsigned comment was added by Milindnpatel (talk • contribs).

Nothing significant, they are synonyms. "Flash" denotes a specific technology, where "Pen" denotes a form. Droud 05:32, 16 February 2007 (UTC)[reply]

Yep.. you could say that a pen drive is a type of flash drive. SubSeven 10:38, 16 February 2007 (UTC)[reply]

electrons revolve

electrons revolve why is it that electrons revolve around the nucleus? the force of an atomic nucleus is far greater than that of a single electron. if that is true, the electrons must be attracted and they must fall into the nucleus. why is it not happening?... 59.92.92.173 13:16, 16 February 2007 (UTC)[reply]

You will find better answers at the Wikipedia:Reference desk/Science. --cesarb 14:01, 16 February 2007 (UTC)[reply]

(after edit conflict):You might get a better answer from the Science desk, but if I'm not mistaken, the "orbital model" of the atom is quite (over)simplified. Even within that model, to say that "electrons orbit the nucleus like planets orbit the sun" is not necessarily to mean that the same physical mechanisms apply. –RHolton ≡– 14:08, 16 February 2007 (UTC)[reply]

Your question is framed in a way that's not entirely correct; when you write "the force of the atomic nucleus is far greater than that of a single electron," you are confusing the force exerted by the nucleus with the inertia of the electron. In fact, an object with enough momentum can orbit very close to an object that exerts a very strong force on it, which is (in the simplified model we're using here) exactly what electrons are doing.

However, the classical model of electrons orbiting doesn't hold so well, because we expect them to lose energy through bremsstrahlung and fall into the nucleus. But that doesn't happen, and it used to be a big mystery a century ago why not. It turns out to be because, according to quantum mechanics, the angular momentum of the electrons is quantized; that is, they are only allowed to "orbit" at certain "speeds" and at certain distances from the nucleus. Electrons at the lowest such distance simply aren't allowed to fall further. -- SCZenz 14:22, 16 February 2007 (UTC)[reply]

Expanding on what SCZenz said, the Rutherford model is of an atom is with orbiting electrons. In the same way that the moon or satellites don't crash into the Earth, electrons don't crash into the nucleus. This is because to crash, the electron would have to lose energy (slow down). There's no friction, so the electron can happily sit in the orbit indefinitely.

In fact, the Rutherford model is wrong. The electron would lose energy by radiating it away (in the form of light). This is because it has an electric charge, and charges which change speed or direction (like in an orbit) will radiate in this way.

Rutherford got around this by simply stating that the radiation doesn't happen in this case (he bodged it!), and this was accepted because the model otherwise agreed with measurements of the atom.

The current model for electron behavior is using quantum mechanics to say the electron is spread out in a hollow sphere roughly where the atom should orbit (this works because quantum mechanics does not deal with actual positions, but only with the probability that it would be at a given position. It is the probability which is spread out). This model also explains the way electrons are organised in shells around the atom. See electron for amazing detail. --h2g2bob 08:49, 17 February 2007 (UTC)[reply]

ADSL Modem Connection Speed

I have an ADSL2+ modem which has both ethernet and USB connectivity.But on the ethernet connection, it connects at 100Mbps, but on USB it connects only at 10Mbps.But the maximum supported bandwidth of the ADSL is 24Mbps.So what would happen if you would connect the modem through USB at maximum connection bandwidth of 24Mbps?Since maximum speed of USB 2.0 is 480Mbps, is it possible to tweak the driver or net connection to get 100Mbps through USB???

Sign your posts with ~~~~. If you cannot get 100 Mbps from the underlying medium, how can you get it by changing to USB? Splintercellguy 16:16, 16 February 2007 (UTC)[reply]

It is possible that you have USB1.1, which maxes out at 12Mbps. Either way, if ethernet is faster, use it! Freedomlinux 17:35, 16 February 2007 (UTC)[reply]

Well what I really wanted to know is if we can form a lan connection between the modem and comp at 100Mbps rather than 10Mbps as I have USB2.0 and not USB1.101:54, 17 February 2007 (UTC)

Lan/ethernet connection is different from what USB version you have and have no effect on the ethernet speed. --antilived^{T | C | G} 09:02, 17 February 2007 (UTC)[reply]

Guys, he's said that his modem has USB connectivity to his computer... --⁪froth^T 02:29, 18 February 2007 (UTC)[reply]

Help with Gentoo linux

I am a linux user but now a newbie in linux administration. I am using gentoo linux.I have 2 gentoo servers (one mail server and 1 gateway server).If i want add a new user, in which of the servers do I add the user.And again how do you tell which group the already existing users belong to.

I also want to set webmail in Gentoo using exim MTA.Lastly I want to set up an auto-response/reply service so that our senders of emails to our users can receive this message when they (users) aren't at their desks.

My client machines are windows. I also forgot to say that I use ADSL connection from telkom.So I have telkom router connected to my gateway server.

The problem is I'm new to Gentoo.Can someone kindly help.

thanks in advance

Derick

I know enough Linux to get me into trouble, and when I do, then Google is my best friend. You aren't going to get a short answer here. I would go back to the manuals and howto's and beef up on mailservers, and Windows connectivity. You've got about 5-10 hours of homework here. --Zeizmic 14:55, 16 February 2007 (UTC)[reply]

Why would a user need to use the gateway server? Use group user to determine what groups a user named user is in.

Statistics generator in C

I've got three variables representing statistics: str, dex and con. I want to generate values for each one, from 1 to 20. Assume I already have a RNG function called getrand(). The easiest way to do this would be to assign, for example, str = getrand();. However, this obviously produces completely random weightings: The chances of getting 1, 1, 1 and 20, 20, 20 are equal. How can I "weight" the outcomes so that it balances out? For instance, if I roll str at 20, I'm going to want the other stats to be much lower to compensate.

If anyone could explicitly suggest an algorithm or popular process used to do this, that would be great. I've so far considered a points method in that it has a fixed number of points to spend on each stat but I can't seem to pull it off.

Personally, I would randomize the possible total, and work from there. In example, since you have three statistics with a maximum value of 20, we can say there is an average of 30 points to be split between the three statistics. Randomize a number between 30[+-5], getting a value in the 25-35 range. This will be the total amount of points to split between the statistics. When you roll a value in one of them, substract the total from the calculated value if possible. If it is not possible, use what remains if it is the last statistics, or throw a random between 1 and the amount left to see the value of the statistic. In example, suppose you get 30 points to split, if the user gets 20 str points, it leaves 10 points to split between dex and con. Throw another random, from 1 to 9, and assign it to dex, and the rest (if dex is 9, the rest is 1) and assign it to con. Note that, since 20 is the maximun, it is very likely that if the user gets a 20, one of the other statistics must be fairly low. -- ReyBrujo 15:49, 16 February 2007 (UTC)[reply]

As implemented, this is probably not what you want; in order for, say, con to get a large value, both of the others have to get small values, but for the first (str) to get a large value, it only needs to get a large value (once). So the distribution of excellent statistics will be badly skewed towards the front of the list of variables. What you want to do depends on what exactly you want the "compensation" to be: ReyBrujo is quite correct that a fixed (or nearly fixed) total value is a good choice, but there are others. In particular, is (20,20,20) impossible to get, or should it merely be unlikely (in which case the total can't be fixed near its average value)? For the "impossible" route, start with a total t as suggested but then also generate the statstics normally; finally normalize their sum to t. (Making the choices in the normal fashion and then using t to fix the total preserves the wholly-uncorrelated nature of the statistics as generated independently.) For the "unlikely" route, you can in general define a

P({\vec {x}})

probability function, and then — with some relatively painful math — decompose it into a series of individual probability functions:

P(x_{1}=t)\equiv \sum _{x_{2},\ldots ,x_{n}=\ldots }^{\ldots }P(\left\langle x_{1},\ldots ,x_{n}\right\rangle )

,

P(x_{2}=u|x_{1}=t)\equiv \left.\sum _{x_{3},\ldots ,x_{n}=\ldots }^{\ldots }P(\left\langle x_{1},\ldots ,x_{n}\right\rangle )\right/P(x_{1}=t)

, etc. Then you can invert the probability functions by tabulating them and searching with a random parameter.

As an example, let us consider 3 statistics on [1,4] and suppose that we use

P({\vec {x}}):={\frac {k}{w+(\left\|{\vec {x}}\right\|_{1}-o)^{2}}}

where k is determined by normalization, w controls the strength of the weighting, and o is the "optimum" total; take

w=1,o=7

. Then the sum of all

4^{3}

probabilities implies

k={\frac {1105}{30248}}

. Then the probability sums for

x_{1}=\{1,2,3,4\}

are

\{975,1105,975,726\}/3781

. (Note that these add to 1.) So roll a 3781-sided die and pick a value of 1 for all sides up to 975, etc. Let's suppose we rolled a 3000 and got "3". Then we consider the probabilities for

x_{2}

given that

x_{1}=3

:

\{2431,2431,1989,949\}/30248

. These add to the

{\frac {975}{3781}}

that we've already selected, so divide them by that to make them sum to 1:

\{187,187,153,73\}/600

. So roll your 600-sided die, and suppose we get 200, giving

x_{2}=2

. Finally consider the individual probabilities for

x_{3}

:

\{1105,2210,1105,442\}/60496

, which sum to the

{\frac {2431}{30248}}

that is the product of the probabilities selected so far. Divide them by that to get

\{5,10,5,2\}/22

: roll the 22-sided die and get, say, 10, yielding

x_{3}=2

for a total value of (3,2,2), which sums to 7 (our o) as a plurality of results will. This can be implemented most easily with floating-point math; the dice then have non-integer numbers of sides but the result is the same. Does that help? --Tardis 23:02, 20 February 2007 (UTC)[reply]

One adjustment to my own answer: see simplex for better information about the "fixed sum" approach in particular. --Tardis 03:57, 26 February 2007 (UTC)[reply]

How about using two "dice" instead of one for each stat. That is, generate two random values 1-10 and then add them to get the final tally. This will center all values "around" 10, using a binomial probability, and make 2,2,2 (you couldn't get a 1 with my scheme, all though the second dice could be 0-10 instead) and 20,20,20 exceedingly unlikely, and it would prevent the possible skewing of values that is quite possible in ReyBrujos suggestion. Oskar 15:55, 16 February 2007 (UTC)[reply]

Or hey, why not do both? Generate a total like ReyBrujo suggests, but generate the values to subtract using my way. That way there won't be much skewing, the values will still be pretty random, and all users will have a fair total. I think that's an excellent solution Oskar 16:01, 16 February 2007 (UTC)[reply]

The chances of getting 1,1,1 and 20,20,20 are equal however the odds that non-homogenous choices are chosen are much higher than extremes at any end of the spectrum. Just pointing that out. It is true that the odds of getting 0,0,0,0,0 in a truly random sequence is the same as getting any other sequence, however the odds of getting a more "mixed" sequence is much, much higher than getting one of all the same numbers, to state it another way. --24.147.86.187 14:59, 17 February 2007 (UTC)[reply]

While you are correct, it is also true that there is a high probability that the total will be skewed from user to user (that is, it is unlikely that, say, 10 users will have the same, or almost the same total score) Oskar 19:11, 18 February 2007 (UTC)[reply]

Homepage click-through rates

A well-known voluntary organisation with several hundred pages on its website discovers from traffic logs that only half the people who visit its homepage go on to look at any other pages on the site. Is that a good or bad click-through rate, or have I not given you enough information to give a meaningful answer? --OpenToppedBus - Talk to the driver 17:03, 16 February 2007 (UTC)[reply]

I suppose a lot depends on the content. If all the other pages require payment, then 50% clickthrough is outstanding! If you're wondering whether clickthrough can be improved, perhaps mention the website for comments? —EncMstr 17:51, 16 February 2007 (UTC)[reply]

I can't name the actual site I'm referring to, I'm afraid. None of the other pages require payment, though. As an example of a site with a similar depth of content, have a look at The Children's Society. --OpenToppedBus - Talk to the driver 18:11, 16 February 2007 (UTC)[reply]

I wouldn't say 50% is bad. --⁪froth^T 19:06, 16 February 2007 (UTC)[reply]

I think that's pretty good, infact Oskar 19:10, 16 February 2007 (UTC)[reply]

Is the glass half full. Or is the glass half empty. It depends on the way you look at it as well as the type of content. A news site would have a different follow-through rate than say, Microsoft, because the purpose of the homepage is different. Harryboyles 10:57, 17 February 2007 (UTC)[reply]

Digital Audio Workstation question

Hello,

I've done searches, and I can't find the answer to this question...

When looking at the computer screen in a digital audio workstation such as Pro Tools or Audacity, there is a visual representation of the recorded sound over time. What does this image actually represent - is it volume, intensity, frequency, or some other quality or mix of these qualities? In other words, what does that actual visual graph of the audio have as its axes?

Thanks!

Eliza

Likely amplitude against time (amplitude on y, time on x). If you zoom in close enough you'll see the individual peaks and troughs - the closer these are, the higher the frequency. — Matt Eason ^{(Talk  Contribs)} 20:20, 16 February 2007 (UTC)[reply]

(edit conflict) The vertical (Y) axis is amplitude; the horizontal (X) is time, so it is the somewhat-hard-to-understand "total waveform". Zoom in very, very closely, so the timescale across the screen width is 0.1 second or less. For normal sounds, you'll see the waves-on-an-ocean aspect of sound. The amplitude alternates both positive and negative. Many tools also will perform a frequency analysis: in Audacity's case, it summarizes the selected sample as the maximum amplitude as a function of frequency. —EncMstr 20:28, 16 February 2007 (UTC)[reply]

Thank you for the responses! Eliza

Laughing

Put whoever you want in there, it works. Beat that. --Darkest Hour 21:43, 16 February 2007 (UTC)[reply]

See Here. Thats why I laugh.

As I've told you, that's no different than your example before. You're still not entering text in the box directly as a comment. --⁪froth^T 17:59, 17 February 2007 (UTC)[reply]

Wikipedia:Reference desk/Archives/Computing/2007 February 16

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