I wrote this text about two years ago for my german blog; now I'll more or less just translate it to english. ;-)
Update: Someone posted a comment on stumbleupon saying "The internet is not the web. Not all web pages are the same size. His figures are just stupid." Well... he's somewhat right. While writing this post, I didn't think of the fact that there's more on the internet than just "the web". So in order to be absolutely correct, you should replace all occurences of "internet" in this text by "the web" (or "the websites" or whatever you think is the best term). ;-)
During the course "Introduction to Computer Science" we had an interesting idea: Let's print out the internet. The whole internet. That way we would'nt need a laptop to surf the net during boring courses. ;-) So I took the time to make a few calculations on this idea.
On August 11, 2005, Google knew
8,168,684,336 (over 8 billion) Webpages. I got this value from wikipedia, since google doesn't show their statistics on the front page anymore. And even this number isn't correct: On the one hand there is the fact, that google doesn't know
all webpages. On the other hand there is also the so called "Deep Web", which consists of that part of the web which isn't searchable by normal search engines. The Deep Web (again according to wikipedia) consists of about 500,000,000,000 (500 billion) pages. But for now I will just stick with the 8 billion pages of google.
My first question: How much space would all these 8 billion pages use?
Since there was a package of 500 pages lying right in front of me, I took a measurement of it: One page of DIN A4 paper (DIN A4 ist the normal paper size in Germany, it is (IIRC) just a tiny bit shorter than the US letter format) has a size of 210x297mm. 500 pages have a height of 5cm, so one page has a height of 0,1mm.
Now we have to remember, that a normal web page won't fit on a single sheet of paper. So I'll just assume that the average webpage needs three sheets of paper when printed out.
So we would need 24.506.053.008 sheets of paper. If we built a single stack of all this paper, it would result in a
height of 2,450,605,300.8mm = 2,450,605.3008m = 2,450km. For comparison: The ISS is orbiting the earth in a height of about 400km, so this stack would reach the ISS six times.
All those printed pages would have a
surface of 1,528,442,526,108,960mm² - that's 1,528,442,526m² or 1,528km². With all this paper you could cover the whole city of New York two times!
About the weight: Normal paper wighs 80g/m².
So the stack would have a weight of 122,530t. That's so much, I can't even think of a suitable comparison.
How long would it take to print all the pages?
At the time of research, I found an article written on Septemer 12, 2005 by macwelt.de about the fastest laserprinter on earth: The IBM Infoprint 4100 is able to print 1354 pages per minute. That's really fast - but not fast enough:
The print job would take 18,099,005 minutes - that's 12,568 days equaling to 35 years. As soon the printer would be done with this job, you'd have to start it all over if you're not interested in antiquated web pages...
If you want to finish the print job a lot more quicker (let's say within a month), you'd need 420 of these printers. Well, one of them costs about 1 million dollars...
And of course you'd have to download all this pages before you can start printing them.
Let's say the size of an average webpage is 20 KByte and you have a reasonable fast internet connection of 100 MBit/s.
Then, you'd have to download about 163 Terabyte (163,000 Gigabyte), which would take you 151 days (5 months).
Conclusion: The internet is huge. Really huge. And you should carefully think about any plans of printing out the whole internet. Very carefully. ;D
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Comments
Tue, 06.01.2009 20:28
Good information and content.. Will use THanks AL
Thu, 23.10.2008 13:34
I think is is bigger than you think... = ~1. [...]
Mon, 24.03.2008 15:41
very interesting numbers. I do n't like the w [...]
Tue, 29.01.2008 16:28
so it was just a bullshit gues s then?
Sat, 26.01.2008 21:46
Well, in order to be able to c alculate the a [...]