Business Week states that, "In ten years, the volume of online data accessible either on the Internet or on corporate networks is expected to approach a yottabyte..."

Currently, the largest named measure of data is the yottabyte (YB), which is exponentially greater than the zettabyte, exabyte, petabyte, terabyte, and so on. The yottabyte is approximately equal to 5X * 10^6, where X is equal to all printed matter that exists on our planet. More definitively, 1 YB=2^80 Bytes.

The highest-density digital tape (which is the highest-density medium currently in common use) made by one of the world's leading manufacturers stores data at a density of 124,000 bpi (bits-per-inch). Assume zero overhead for error-correction and -detection.

At a tape thickness of 8.9 microns, what must the diameter of a roll of tape be in order to store one YB of data, assuming that it is wrapped around a spool with a diameter of 0.5 inches and assuming that there is no space between the layers of tape?

Well, 1 byte is 1024 bits, and 1 YB is 2^80 bytes, so 1 YB = 1.23794003928x10^27 bits, divided by 124,000 bpi is 9.98338741355x10^21 inches long of tape. Analysing the growth of the perimeter in every loop, the sum of all the loops must be equal to the total lenght of tape already calculated (9.98338741355x10^21 inches)2*Pi*R1+2*Pi*R2+2*Pi*R3+2*Pi*R4......2*Pi*Rn = 2*Pi*(R1+R2+R3+R4.....Rn) = 9.98338741355x10^21 inches. We can see that R1=.5, R2=.5+8.9x10^-6, R3=.5+2*8.9x10-6, etc. We can observe that .5 is added n times and 8.9x10^-6 is added n*(n-1)/2 times. From 9.98338741355x10^21=2*Pi*(.5*n+(8.9x10^-6)*n*(n-1)/2) we get the equation n ²+((1-8.9x10-6)/8.9x10-6)*n-(9.98338741355x10^21/(Pi*8.9x10-6))=0, we find the roots of the 2nd degree equation and neglect any negative root. This gives n=1.88959626281x10^13, we finally calculate de radius of the roll of tape R=.5+(8.9x10^-6)*n=168174067.89in
The total diameter is 336348135.78in= 5,308.52487027mi

Posted by Antonio on 2003-09-03 15:38:11