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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 (2^1
by Keyframe 5y 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.
136 billion kg:
- ≈ 0.64 × mass of trash produced in the United States in one year ( ≈ 2.36×10^8 sh tn )
- ≈ 0.35 × estimated wet biomass of all humans alive ( ≈ 385 Mt )
- ≈ 1.3 × estimated dry biomass of all humans alive ( 105 Mt )
thanks wolframalpha!
- _Microft 5y agoAnother one: 136 billion kg is the mass of a block of water with a footprint of a square kilometer and a height of 136 meters. Put this way, it does not sound that much anymore, in my opinion.