# Data storage and transmission

Data storage and transmission

Q: How much data storage is needed for various large amounts of data, eg

• A telephone directory for the world;
• A video of a person's life;
• The DNA required to clone everyone in the world;
• A black-and-white photocopy of the earth's surface?

A: We work to 1 significant figure, but even this often isn't justified. Define some units for measuring large amounts of data:

• 1 fibresecond: amount transmitted per second by the London universities' new (mid-1990s) fibre-optic ring = 2e7 bytes.
Hence 1 fibreyear = 3e7 fibreseconds = 6e14 bytes.
• 1 genome: amount stored in a human cell's DNA = 2e8 bytes.
Hence 1 population = 5e9 genomes = 1e18 bytes.
• 1 lorry: amount stored in a lorryfull of Exabyte tapes. 8Gb per tape, 6 x 8 x 50 = 2400 tapes in a cubic metre, 8 x 2 x 2 = 32 cubic metres in a lorry so 6e14 bytes.
Hence 1 ship = 500 lorries = 3e17 bytes.
• 1 kg: amount stored in a kilogram of nano-memory. 1000 atoms per bit, 2e26 nucleons/kg or 2e24 atoms/kg so 2e20 bytes.

Note 1 lorry = 1 fibreyear. To send 1 lorry of data up to about 5e8 m, say to the moon, the lorry is quicker than the fibre. To send 1 ship of data up to about 1e11 m, say to the sun, the ship is quicker than the fibre.

So we have

• A telephone directory for the world: 100 bytes/person so 5e11 bytes.
7 fibrehours, 2500 genomes, ~0 lorries, ~0 kg.
• A video of a person's life: 1000x1000 pixels, 3 bytes/pixel, 30 frames/second, compressed by factor of 100 so 1e6 bytes/s so 2e15 bytes.
3 fibreyears, 1e7 genomes, 3 lorries, 0.01 grams.
• The DNA required to clone everyone in the world: 1e18 bytes.
2000 fibreyears, 1 population, 4 ships, 5 grams.
• A black-and-white photocopy of the earth's surface: 300dpi, 4e14 m2, compressed by a factor of 10 so 8e20 bytes.
1e6 fibreyears, 800 populations, 3000 ships, 4 kg.

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