User talk:Jorgenev/MFD'd/Wikipedia:Recent Changes coefficient

something for meta? Martin


 * It is a little addendum to Are You a Wikipediholic Test, so I have put in the same namespace. - Patrick 01:34 19 Jul 2003 (UTC)

I don't really get this. Shouldn't it be the other way around? If you have a low C value you're adept at following up on recent changes. If you have a score(rate of addiction) of almost 1 then you are very slow at getting up to speed! Correct me if am wrong... You could reverse the c value by making the formula something like this n/(c-1)=time needed. Eruantalon
 * The larger c (below 1) the longer it takes, even my wrong version, which I have now corrected, had that right.--Patrick 23:40, 15 Apr 2005 (UTC)


 * If c<1, after being away for n hours it takes n/(c-1) hours to catch up.

This is impossible - for c<1, this results in a negative number. Daniel


 * You are right, my formula was wrong, but the way it was changed was not an improvement. I have corrected it now. To process the changes which occurred while you were away takes nc hours, but in the meantime there are more changes, taking nc2, etc. The sum of the geometric series is nc/(1-c).--Patrick 23:34, 15 Apr 2005 (UTC)

c=1 or more? So, essentially, c>1 would imply you're such a wikipedaholic that you can anticipate and react to changes before they're made. Yes, that is indeed scary. --Saforrest 23:30, 10 March 2006 (UTC)

Switch to watchlist
Perhaps, since RC is so overwhelming these days, we should be more concerned with one's watchlist? If one is keeping up with one's watchlist a good portion of the day, then that is addicted indeed. --maru  (talk)  contribs 06:55, 29 May 2006 (UTC)
 * Well that would not be too efficient as the number of watchlists users have varies considerably. -- S iva1979 Talk to me  20:34, 9 June 2006 (UTC)

My reversal
I have changed the old c to $$(c)^{-1}$$ as if you take less than 1 hour to catch up on the RC of 1 hour, you're anticipating and reacting to changes before they're made. 4 T C 14:58, 29 December 2009 (UTC)


 * No, you're not. The scenario is that you go away for an hour, then return and start to catch up. --Tango (talk) 09:23, 21 March 2010 (UTC)

Equation is wrong?
The expression given is negative for c>1. Is this a mistake? NotAnonymous0 (talk) 06:17, 21 March 2010 (UTC)


 * Yes. I've reverted the most recent change. There has been a very slow edit war on this page about the maths... people keep making it not make any sense and I keep fixing it... --Tango (talk) 09:22, 21 March 2010 (UTC)