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?

OK, I've done the hard part: the rest is simple arithmetic (except for the one square root).

1 YB = 2^80 bytes

I = 1 inch of tape = 124,000 bits = 15,500 bytes

L = length of tape = (2^80)/I

T = thickness of tape = 8.9 microns = (0.03937)(8.9)/1000 inches

W = width of tape (Value is irrelevant -- it drops out)

V(1) = Volume of tape = LWT

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V(2) = Volume of hub of reel = 0.25Wπ

V(3) = V(1) + V(2)

R = radius of reel of tape = √[V(3)/Wπ]

D= diameter of reel = 2R

Posted by TomM on 2002-10-13 11:18:11