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> A fully-populated 128-bit storage pool would contain 2^128 blocks = 2^137 bytes = 2^140 bits; therefore the minimum mass required to hold the bits would be (2
by jsjohnst 7y ago
> A fully-populated 128-bit storage pool would contain
2^128 blocks = 2^137 bytes = 2^140 bits; therefore the minimum mass required to hold the bits would be (2^140 bits) / (10^31 bits/kg) = 136 billion kg.
> Thus, fully populating a 128-bit storage pool would, literally, require more energy than boiling the oceans.
Very definition of excess as well, so thanks for proving my point!
- throw0101a 7y ago> Very definition of excess as well, so thanks for proving my point! Except you're skipping over the part about running out at 2^64: > Some customers already have datasets on the order of a petabyte, or 250 bytes. Thus the 64-bit capacity limit of 264 bytes is only 14 doublings away. Moore's Law for storage predicts that capacity will continue to double every 9-12 months, which means we'll start to hit the 64-bit limit in about a decade. 2^64 bits ought to be enough for anybody. -- jsjohnst
- jsjohnst 7y agoNever once did I say 2^64 ought to be enough for anybody, so how about stop being a jerk? Try reading what you quoted again, nobody is running out yet. There are a lot of intermediate steps between 2^64 and 2^128! ;)