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Here is a small list I gathered (each name followed by a .com) -
Real (not sure - the page jumps)
I thought no one would really get around to asking this.
The reason I ask is this -
We know that google has indexed 3,307,998,701 pages on the net.
We also know that the sum total of the PR on the web is 3,307,998,701
We also know that the Google Toolbar is exponential logrithmic scale. But no one really know the scale of WHAT?
If we do a reverse calculation of the PR10 sites down to 3,307,998,701, we can actually figure out the true base.
So far, my calculations say, the base is a factor close to '8' if we have about 24 PR10 sites.
Get the idea?
Google has page rank 11 [webmasterworld.com] Especially check Chris_R's msg 36 and the link to his PR 11 analysis.
There was at least one other pretty good thread on this but I can't seem to find it right now.
Wouldn't this change as they add pages to their index, and links are created?
The base of the log wouldn't be very sensitive to this at all.
Pages on frames sites
PDF files with no putbound links
Pages that only link out with a form
We also do not "know" that toolbar PR is exponential and automatically generated.The PageRank papers hint at this, and it does seem to be that way, but it could also contain other adjustments to give it a more useful distribution. If Google is an 11, and they had 80 times the pagerank of the next site, it would be in their interest to tweak the numbers so that there are 9s and 10s.
I'll admit that playing around with this stuff is fun, and it can lead to some additional understanding, it really is of little practical use. Just be aware of all the places where errors can sneak into your calculations, and don't run around declaring your results as fact.
Nice though perhaps, but you might be better off applying your math skills to a jelly bean counting contest :o)
PR means absolutely nothing these days so why bother?
If Yahoo! was PR3, do you think that they would have 21 million pages in the index?
And in areas where there are some real high PR sites, they still tend to dominate the searches, without applying specific SEO techniques.
PR2 pages simply do not hold positions ahead of PR8 pages that are on topic. It isn't all due to the PR of the page, a good part of it is due to the reason that page achieved PR8. But the PR is still part of the algo, and it still counts for something.
Just because it is no longer everything does not mean that it is "absolutely nothing".
through invisible penalty
One of my distant partners has PR9 and links written in footer to 9 affiliate sites, some of which my company has developed. The site is a huge resource that amounts to nearly 1K pages with footer links almost EVERYWHERE across!
None of the footer sites get any links from this domain at link: query i.e. seemeingly no actual PR is transfered.
Does this look like a clearcut case of invisible google penalty of some sort?
Otherwise they might have NOFOLLOW in the robots tag or file, telling Google to index the page but not follow the outgoing links. Or the links might not be spiderable. Or it might be the site uses java or frames to hide the true page which has the link. I've seen all sorts of tricks!
Seems like a pretty obvious 'fix' to me.
How many of the sites out there have "Search powered by yahoo" links?
They have a link off of all of macromedia's pages (and how many of those are PR10)
How about w3c.org?
Hey, look at that, python.org has a quote with a link on their home page, and a powered by google on their search page.
Look they run the search for the World Heath Organization.
Apple has a link from the PR9 description of the Safari browser.
I would not be surprised if Yahoo had more total PR for all their pages, but google gets more big links directly to their home page.
I think it was pointed out on this forum, that if the link is part of the site's template, it doesn't transfer PR. (?)
Couldn't find the exact thread but in this thread
Brett said that "those guys paying for pr and having their link stuck in a template aren't getting much bang for their buck"