This may be interesting. Feel free to correct any stupid math mistakes and/or deviations from Borges' description of the library.

Assumptions:

1. each shelf is four meters long

2. each hexagon is three meters high

3. the hexagons are arranged with hexagonal air shafts between them such that exactly half the total volume is hexagons and half is air shafts

4. exactly one copy of each possible book exists in the library (the theory is true)

4a. the letters on the spine don't matter

5. the library actually exists in the "surface volume" of a hypersphere. This would mean that is really is endless and cyclical.

There are 410 pages/book * 40 lines/page * 80 characters/line = 1312000 characters/book, and therefore 25^1312000 books. There are 35 books/shelf * 20 shelves/hex = 700 books/hex. If a hex is 4 meters on a side, that gives 12sqrt(5) m^2 as the floor area, and 36sqrt(5) m^3 as the volume of a hex.

1312000 * log(25) = 1834097.29

10^1834097.29 books (3% accuracy in the number of books is more than enough here)

(10^1834097.29 books / 700 books/hex) * 36sqrt(5) m^3/hex = 10^1834096.35 m^3

The formula to get the surface volume of a hypercube from the radius is:

V = 2 pi^2 r^3

(10^1834096.35 m^3 / 2 pi^2) ^ (1/3) = 10^611365.02 m

So, the library exists on the surface of a hypersphere with radius 10^611365.02 m.

For example: 10^611365.02 m is 10^611352.04 light years, or somewhere around 10^611342.04 times the radius of the observable universe. There are estimated to be about 10^80 particles in the universe. Let's say you wanted to put universes end-to-end from the center of the hypersphere to the Library. If you had a universe just like ours for each one of those particles, and also one for each of the particles in each "sub-universe", and so on, you'd need to go through 7642 levels of recursion to have enough universes to accomplish your goal.

This still doesn't help much, but it does instill a healthy respect for combinatorics.

(And each book only contains about 6 MB of data - anyone want to write about the CD-ROM Archive?)

posted by Serf at 12:36 PM on May 15, 2000 [2 favorites]

Assumptions:

1. each shelf is four meters long

2. each hexagon is three meters high

3. the hexagons are arranged with hexagonal air shafts between them such that exactly half the total volume is hexagons and half is air shafts

4. exactly one copy of each possible book exists in the library (the theory is true)

4a. the letters on the spine don't matter

5. the library actually exists in the "surface volume" of a hypersphere. This would mean that is really is endless and cyclical.

There are 410 pages/book * 40 lines/page * 80 characters/line = 1312000 characters/book, and therefore 25^1312000 books. There are 35 books/shelf * 20 shelves/hex = 700 books/hex. If a hex is 4 meters on a side, that gives 12sqrt(5) m^2 as the floor area, and 36sqrt(5) m^3 as the volume of a hex.

1312000 * log(25) = 1834097.29

10^1834097.29 books (3% accuracy in the number of books is more than enough here)

(10^1834097.29 books / 700 books/hex) * 36sqrt(5) m^3/hex = 10^1834096.35 m^3

The formula to get the surface volume of a hypercube from the radius is:

V = 2 pi^2 r^3

(10^1834096.35 m^3 / 2 pi^2) ^ (1/3) = 10^611365.02 m

So, the library exists on the surface of a hypersphere with radius 10^611365.02 m.

For example: 10^611365.02 m is 10^611352.04 light years, or somewhere around 10^611342.04 times the radius of the observable universe. There are estimated to be about 10^80 particles in the universe. Let's say you wanted to put universes end-to-end from the center of the hypersphere to the Library. If you had a universe just like ours for each one of those particles, and also one for each of the particles in each "sub-universe", and so on, you'd need to go through 7642 levels of recursion to have enough universes to accomplish your goal.

This still doesn't help much, but it does instill a healthy respect for combinatorics.

(And each book only contains about 6 MB of data - anyone want to write about the CD-ROM Archive?)

posted by Serf at 12:36 PM on May 15, 2000 [2 favorites]

are we supposed to do all of that math for *fun*? :)

rcb

posted by rebeccablood at 2:25 PM on May 15, 2000

rcb

posted by rebeccablood at 2:25 PM on May 15, 2000

If you read "Death and the Compass" instead, you only have to count to four.

posted by snarkout at 4:26 PM on May 15, 2000

posted by snarkout at 4:26 PM on May 15, 2000

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This thread has been archived and is closed to new comments

For more on Borges, check out the Borges page at The Libyrinth.

posted by snarkout at 10:54 AM on May 15, 2000