Comments on: FACT: a CD track is five kilometers long http://lsned.com/facts/compact-disc/ Learn Something New Every Day Thu, 23 Feb 2012 22:01:49 +0000 hourly 1 By: Ryan http://lsned.com/facts/compact-disc/comment-page-1/#comment-1875 Ryan Sat, 01 Jan 2011 05:07:06 +0000 http://lsned.com/?p=687#comment-1875 Wow! Mucho thanks for the clarification. Wow! Mucho thanks for the clarification.

]]> By: tristan http://lsned.com/facts/compact-disc/comment-page-1/#comment-1873 tristan Sat, 01 Jan 2011 02:32:58 +0000 http://lsned.com/?p=687#comment-1873 Actually there are 44,100 samples in a second, not bits. A CD uses 16 bit samples which encode the amplitude of the sound wave at that instant as a number between 0 and 65,535. Plus they're stereo so there's 2 sound waves encoded side by side. So we need 2* 16 * 44100 = 1,411,200 bits per second to represent our sound. However, on the disc a series of ones or zeroes (pits or flat spots) will cause the laser to "fall out of the groove" and so each byte (8 bits) is encoded in what's called 8 by 13 encoding where 13 bits are used to represent 8 in such a way that there are nonmore than 3 flat spots in a row. Therefore one needs 1,411,200 * 13 / 8 = 2,293,200 reflection tests in a second. Actually there are 44,100 samples in a second, not bits. A CD uses 16 bit samples which encode the amplitude of the sound wave at that instant as a number between 0 and 65,535. Plus they’re stereo so there’s 2 sound waves encoded side by side. So we need 2* 16 * 44100 = 1,411,200 bits per second to represent our sound.

However, on the disc a series of ones or zeroes (pits or flat spots) will cause the laser to “fall out of the groove” and so each byte (8 bits) is encoded in what’s called 8 by 13 encoding where 13 bits are used to represent 8 in such a way that there are nonmore than 3 flat spots in a row. Therefore one needs 1,411,200 * 13 / 8 = 2,293,200 reflection tests in a second.

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