Jeff Thompson | Blog

A lot of math today: if all the zeros and ones on my hard drive weighed 1 gram, they would form a pyramid running from the center of the Earth to a base of 6.2cm and 5.5cm respectively.  If tipped on its side, that would mean the pyramid would run from NYC to Torino, Leipzig, Berlin, Senegal, or La Paz.

Map radius calculations via: Free Map Tools

Ratio of square/page in Malevich’s “Black Square”:  ~3/4

Ratio of pupil/eyeball in average adult human: ~3/8

Following yesterday’s experiments with the Random Pi Walk (hat tip Alex Bellos), I’ve upped the ante.  The above image is the decimal expansion of the square root of two, following the first one million digits.  The data is thanks to Stan Kerr via Project Gutenberg.  Each decimal digit 0-9 results in a change of direction of 36 degrees and, in this case, travels 3 pixels in that direction.

The resulting image is MUCH larger than the previous visualizations – click here for the full resolution version.

Can’t say I really understand what they mean, but found these “ideal circular membranes” for tuned drums.  The black number is the ideal value, plus kettle drum (red) and tabla (blue).

Via: +plus Magazine

Doing some Friday afternoon math about the audio files on my computer.

~22 days of music = 80,129,302,838 samples*
Sample values range from -32,768 – 32,768

If each value were to be cut using a standard CNC mill with 0.000125″ accuracy, the resulting object would be:

10,016,162.8″ = 834,680.2′ = 158.08 miles long, and only…
8.192″ tall!

* calculation from Adam Caprez at the Holland Computing Center; thanks Adam!

Via lots of sources today, the “Batman Equation”.  Written out by ipi31415, the equation is:

((x/7)² Sqrt[Abs[Abs[x] – 3]/(Abs[x] – 3)] + (y/3)² Sqrt[Abs[y + (3 Sqrt[33])/7]/(y + (3 Sqrt[33])/7)] – 1)(Abs[x/2] – ((3 Sqrt[33] – 7)/112) x² – 3 + Sqrt[1 - (Abs[Abs[x] – 2] – 1)² ] – y) (9 Sqrt[Abs[(Abs[x] – 1) (Abs[x] – 3/4)]/((1 – Abs[x]) (Abs[x] – 3/4))] – 8 Abs[x] – y) (3 Abs[x] + .75 Sqrt[Abs[(Abs[x] – 3/4) (Abs[x] – 1/2)]/((3/4 – Abs[x]) (Abs[x] – 1/2))] – y) (9/4 Sqrt[Abs[(x - 1/2) (x + 1/2)]/((1/2 – x) (1/2 + x))] – y) ((6 Sqrt[10])/7 + (3/2 – Abs[x]/2) Sqrt[Abs[Abs[x] – 1]/(Abs[x] – 1)] – (6 Sqrt[10])/14 Sqrt[4 - (Abs[x] – 1)² ] – y) == 0

Via: Boing Boing

Yes, a 30-minute video about sorting algorithms… and it’s amazing.

By George Hart (download the STL files and print your own)
Via: Make blog